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Dr. Laurie Starkey brings in her love of organic chemistry and helps students learn through visual models and real world examples. Focusing on Chemical Structures, she covers topics from Alkanes to Aldehydes and Amines. This course is crucial for students who wish to excel in Organic Chemistry in order to satisfy their degree or pre-medical requirements. Lessons go in depth and are followed with numerous examples similar to those found on organic chemistry exams and qualifying tests. Additional topics include everything from Stereochemistry to Nomenclature and Spectroscopy. Dr. Laurie Starkey is also the author of “Introduction to Strategies for Organic Synthesis” (Wiley) and earned her Ph.D. in Chemistry from UCLA. She has been teaching Organic Chemistry at the university level for over 15 years and most recently won the 2013 Provost's Award for Excellence in Teaching, Cal Poly Pomona's highest teaching award.

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Hi welcome to educator .com.0000

This is our first lesson in organic chemistry and let us start by defining what it means to be organic.0002

These terms were starting to be developed hundreds of years ago.0007

Things were being classified either organic or inorganic.0011

Organic, was described as something that came from living things plants and animals and the matter form those species and inorganic was everything else so things like rocks and minerals.0016

Rocks and minerals never came from living matter, metal, glass for example.0031

These are some materials that are described as inorganic and other than these sorts of things they are known as organic.0041

Now we have expanded that definition to not just being things that come from a natural source but we have expanded it and defined as the chemistry of carbon.0048

This is because our life is carbon based and that it encompasses the original definition.0058

So it is true that things that living things and things that came from living sources are going to be organic.0066

But we can expand our definition to include a lot of other interesting things.0071

Now, what I love about organic chemistry is that it surrounds us in our everyday lives, and I was really fascinated by the topic when I took this as an undergraduate.0075

Let us go through some examples of things we will encounter throughout the course.0083

The products that we used that improve our lives, we’ve heard lots of examples of organic chemistry in medicines, drugs, about the entire pharmaceutical industry is that majority of those products are organic compounds.0088

Carbon containing compounds, including pesticides, herbicides, fungicides and things that protect our crops, and help feed people; dyes, inks, things that are of colored, pigments, things at non colored to things colored to be organic.0100

Paints, in the paint industry it’s very important to study organic chemistry and understand how we can make the paint be longer lasting or more environmentally friendly.0116

Gasoline, fuels, the petroleum industry is using organic molecule.0127

We distill crude oil into its various components.0135

Each of those components are organic compounds and we find it that they burn very quite nicely, it provides energy.0139

Cosmetics, shampoos and perfumes and those sorts of things -- so the personal care industry is also something that definitely involves organic chemistry.0146

A lot of materials that we use in our world, natural materials as well as manmade materials...0156

So if we take a look at the structure of paper or cotton you’ll see that those are carbon chains, carbon containing compounds as well as rubber and tires and things that are made up of rubber.0162

But then there are a lot of manmade examples of these compounds which are polymers so they have these long chains of molecules like we do for cotton or rubber.0174

So some manmade ones or things like nylon and polyester those are organic compounds.0184

All plastics that you can imagine vinyl that’s organic as well as adhesives.0190

So anything that is sticky or gluey is also an organic molecule.0195

Liquid crystals if we think about our LCD displays that we have on our phones and our televisions, those are organic molecules, a lot of interesting research going on there, as well as nanotubes.0199

Nanotubes are all carbon containing structures that are extremely strong and this is one we are looking into, nanotechnology.0209

Nanoorganic chemistry is really leading the way and I’m sure we’ll have some very exciting breakthroughs in the years to come.0220

And In nature remember that organic chemistry is initially defined as having this relationship to living systems and it’s still quite true.0228

So if you look at the structures of proteins, fats, sugars those are all organic molecules so anything we eat or we digest is an organic molecule.0235

The vitamins that are also essential to our existence, things that we can taste or smell those are organic molecules that we are interacting with flavors and fragrances,0244

hormones, steroids, structures of DNA.0254

When you look at the field of biochemistry, these are all organic reactions in living systems.0258

And so that’s why it’s critical to have a strong foundation in organic chemistry before we move on to biochemistry because you are going to be studying organic reactions that are taking place in living system.0274

As you could see there’s a lot of interesting things that are in store for us and as an organic chemist there is a wide range of industry in which you could be involved, the pharmaceutical industry, petroleum, food, health and beauty, polymer the list goes on forever.0288

As a matter of fact, I was a Biology major when I took organic chemistry as an undergraduate and I was so excited about the topic and found it so interesting that I ended up changing my major to chemistry and studying organic chemistry in graduate school.0304

Let’s see what we have in store before we can get into some with some of the really fun stuff.0318

Let’s make sure we review some chemistry basics because anyone taking organic chemistry should have a year or so of general chemistry.0323

While you’ve learned a lot in general chemistry, there are certain things that are going to be really critical to succeed in organic chemistry so I want to make sure we review those so that you’re ready to hit the ground running.0330

So for example, let’s think about electrons.0341

Electrons are the things that are going to be doing all the bonding, doing the reactions and so it’s really important we have an understanding of how the electrons look like, how they behave.0344

Remember these are held in atomic orbitals around the nucleus so we have s, p, d, f those sorts of orbitals.0353

For organic chemistry we’re mostly just dealing with this s orbitals and p orbitals that’s what carbon has available to us and we are studying the chemistry of carbon.0360

If we think of the s orbitals and p orbiatals, let’s consider their energy and their relative energy.0371

If you remember the very first shell of electrons has just a single orbital, the s orbital that’s available to it.0380

That is the 1s orbital we call it.0390

Then we move it to carbon it now also has a second shell of electrons and in that shell we have two types of orbitals that are available.0393

We have an s orbital but because it is now farther from the nucleus it’s going to be higher in energy so what I’m showing here is relative order of energy.0400

The 2s orbital is higher in energy than the 1s.0411

Then the p orbital, it has a node, it is dumbbell shaped; these are further still from the nucleus.0414

The p orbitals are higher in energy still and if you recall there are going to be three different types of p orbitals in that second shell so we can draw them at the same level and this is the 2p.0423

What we want to remember this general description of relative energies and the s orbital is going to be lower in energy meaning more stable than the p orbital.0436

It’s going to be important to us down the road so I just want make sure we have that mind.0450

Let’s take a look at the periodic table and again we’ll see here just a section of the periodic table because carbon is the atom we’re going to be most concerned with, we going to be involved throughout this whole year of organic chemistry.0457

Really very few other atoms are going to be involved in our organic molecules, definitely hydrogen.0472

A lot of hydrogen in our molecule is attached to carbon, there may be nitrogen, oxygen some of these halogens.0479

Possibly we’ll see some of these down here, the silicon, the phosphorous even the sulfur.0485

But really it’s just very small region of the periodic table that we’re going to spend the most of our time with.0492

But some of our reagents and reactions do involve some other atoms as well.0497

You’re going to remember what’s very important about the periodic table is that these atoms are the most special here.0502

These are the noble gases and these elements are extremely stable.0507

Every other atom in the periodic table every other element is trying to walk like the noble gases.0513

What that means is they want to have the same electronic configuration.0520

That imparts a special stability because an atom is going to be stable if it has a filled valence shell.0525

That’s going to be the ultimate goal that all atoms are seeking and we know this as the Octet Rule.0533

Hydrogen can never have an octet because it only has a 1s orbital so it can only hold 2 electrons.0540

So for hydrogen, it will have a filled s orbital and that would be very stable.0549

But for carbon and nitrogen and all the others we want to have in our outer shell two electrons in our s orbital.0554

We want to have six electrons in our p orbitals, this would be a filled shell.0561

Now it moves all the way over to give a configuration that will be like Neon, Argon or Krypton depending on what atom you are looking at.0566

For example if we take a look at sodium, sodium metal this is not a stable atom on its own because it has just one valence electron but if sodium was to give up a valence electron to be Na+ that would bring it back to have the electronic configuration of Ne making it extremely stable.0576

So anytime we see Na as part of the formula, we know that that sodium has to be Na+ if it’s going to be stable.0599

How about something like Ca or Mg something in the second column here.0607

Well, these have two valence electrons that’s not a filled shell so the way that they can end up with the filled shell is to get rid of those two electrons so we have thing like Mg2+ it’s very stable Ca2+.0613

Those atoms like you have +2 charge.0628

Again anytime see them in the formula we know that we can think of that charge, Li, K will have a +1 charge.0631

How about the halides, F, Cl Br what would make them stable.0640

Well, to get all the way to Cl, we have to go 1,2,3,4,5,6,7 Cl has 7 valence electrons so it will have just one more then it will have the electronic configuration of Ar that would be very stable because it will have a filled octet.0645

So how can I do that, well, it can gain an electron and have a negative charge that would give it instead of 7 valence electrons it will have 8 valence electrons and will have a filled octet.0661

The periodic table is a critical resource for us and just a quick glance of the periodic table is going to remind what charges certain atoms like to have of course the atoms on the far left would like to have positive charges, the atoms in the far right will have negative charges.0672

We also learned something about electronegativity and that’s going to be a concept that is important in organic chemistry allows us to predict certain things.0693

The one rule that we know here is that F is the most electronegative element so the periodic trends are such that as you move towards F moving either to the right or moving up that is going to increase your electronegativity.0703

What does it mean to be electronegative?0727

It means you pull electron density towards yourself.0733

Fluorine is very very good at pulling electron density towards itself.0736

Now I’m not going to share with you the table of the numbers because we don’t need to memorize the numbers, we need to build just kind to have a feel for who’s more electronegative and who is not.0740

A few things that we should point out - F is the most electronegative atom.0752

You should also know that O is the second most electronegative atom.0756

So anytime you see an O you know right away that that O is pulling electron density towards itself.0766

That’s a good fact to know.0775

It’s also difficult to compare things that are not within the same column or the same row so it’s also good to know that carbon has very similar electronegativity to H now they are not the same number but there is not a significant difference between those two.0777

So it’s good to remember that those have had similar electronegativities.0796

When you look at the electronegativities of N and Cl those are equal so that kind of help to balance it out.0801

Typically you are not having to memorize this numbers if you know that F is extremely electronegative and we know that O is also extremely electronegative that’s going to get us through a lot.0813

If you’re comparing any two atoms that happen to be in the same column or the same row then for sure you would be able to compare their electronegativities.0823

Comparing C to N, which one will be more electronegative?0832

Here’s C, here’s N. Since N is closer to F then N is going to be more electronegative.0838

Now let’s talk about bonding, how do these elements come together to form bonds?0845

We have two types of bonds, the ionic bonds and covalent bonds.0853

Ionic bonds is what we’re going to have if we combine atoms which have large differences in electronegativities so if we take something like Na which is on the far left of the periodic table and combine it with Cl which is on the far right of the periodic table.0857

I have shown them here with just their valence electrons, Na has one Cl has seven, neither of these species are happy these are very unstable they are very reactive because neither has a filled octet.0872

The way that they can end up satisfying this problem is we can have a transfer of an electron; the Na can give up its electron and give it to Cl.0887

What happens now is that Na loss an electron so it’s positively charged, the Cl gains the electron so its negatively charged.0901

As a result of these transfer of electrons you have stable filled octets.0912

So we are going back to that octet rule and we are going back to achieving that noble gas configuration.0920

In this situation it would be very useful to transfer that single electron and both would be satisfied.0928

This is described as a salt and ionic bond is a representation of a salt if you have a + and – charge and it’s a network structure.0937

So it’s not correct to describe NaCl as a molecule.0948

There is no such thing as a molecule of NaCl.0956

There’s 1:1 ratio of those two.0958

But every Na+ is surrounded by Cl- in the salt crystal structure.0960

Try not to call it a molecule although that does happen sometimes accidentally but instead it is a salt and it is a network structure.0966

Let’s take a look at a different example where a transfer of electron will not be so good and when we are going to have that is when we are going to have covalent bonds rather than ionic is when we have atoms with similar electronegativities.0977

With Na and Cl, Cl really really wanted that electron because it is so electronegative and Na did not want that electron because it is so electropositive and so the transfer makes sense.0992

In a case where like this where we have a C, C has 4 valence electrons and if we combine that it would be four H atom each with a single valence electrons.1004

The idea of each of those H donating electron to C to fill its octet is not a good idea because C then would have 4 extra electrons and would have a -4 charge that would be highly unstable.1012

So instead, what we do is we share the electrons.1025

This is still unstable just like before because no one has a filled octet but if we were to share the electrons between the nuclei so that this H now that sees 2 electrons (remember that’s the filled octet for H and that it gives it the noble gas configuration).1028

The C now sees 8 electrons, everyone gets a filled octet but we don’t have any charges.1050

This is a stable situation and that these three lines kind of mean that the equivalent of this is instead of drawing this two shared electrons as 2 dots what we do is we draw a line to represent what is known as a covalent bond.1057

So every time we draw a line connecting two atoms this represents two shared electrons.1082

Let me call that a covalent bond.1091

C is very good at forming covalent bonds that’s why so many of its structures are so stable and drew out so many different types of interesting products and compounds.1097

By doing this bonding and sharing these electrons we now have again stable filled valence orbitals so this is a good arrangement.1109

And what we have now is this methane (we will going to be learning these names shortly) and this is a molecule.1124

This is a discrete molecule.1133

This is a discrete stable group of atoms that is independent of other molecules, so this is a methane molecule and it has a covalent bonding.1136

So the way we decide between ionic and covalent is we compare the atoms that are coming together and decide if they want to be shared or if they want to transfer the electrons.1146

If they are coming from the same area of the periodic table like C and all the neighboring atoms around C those are going to be preferred to be covalently bonded.1155

If we see anything off on the far left of the periodic table the metals those are going to be charged species and those will always be ionic species.1163

Is there anything between strictly covalent or strictly ionic?1176

In fact this is kind of a continuum it is not always clear quite whether something is completely ionic or completely covalent.1183

Especially when you are looking at a covalent situation, when you have two atoms that are sharing electrons those two electrons does not necessarily means being equally shared between the two atoms because those atoms may not have identical or very similar electronegativities.1190

In that situation we have what is known as a polar bond or polar covalent bond and that if so we have a significant difference in electronegativity.1205

So let’s take a look at the C - O bond do these have the same electronegativity or similar electronegativity?1215

What do we know about O?1222

We know O is the second most electronegative atom in periodic table so we know that this O is very electronegative and what is happening is that it pulls electron density , this two electrons being shared by the C and O are being pulled toward the O and so they are not equally shared.1224

And what we have is a polar situation or a dipole situation here where the O is δ -; we use this δ symbol to represent a partial charge and the C is δ+.1242

So there’s a couple we can show this polar bond we can draw this arrow line points in the direction of the electrons being drawn or we can have a δ+ or δ- to represent to represent the electron deficiency on the C or the electron richness of this O.1258

How about the C-H bond?1278

Do those have a significant difference in electronegativity?1281

Again we need to know this by looking at the table, in fact they do not have a significant difference so this is described as a non-polar bond because there is no significant difference in the electronegativity.1284

This C-O bond we describe is a polar bond and this is a non-polar bond.1300

We need to be able to describe bond this way.1316

How about if we have a C attached to a halogen, the letter X is going to represent a halogen?1320

Whenever we see that in organic chemistry the X will represent something like F, Cl, Br, I.1327

Remember F is the most electronegative so for sure that’s the most electronegative and in fact it does decline throughout but except for iodine it’s a little less extreme.1333

But for the most part when you attached an X to a C you do expect electron density to be pulled toward the halogen so we have a δ- and a δ-+ and these are also polar bonds.1349

This is going to be important to us because this is going to help describe the relationship of these carbons when they are in this structure, this C is electron deficient and it’s going to have certain reactivities as a result of that.1360

Now when we are looking at the molecule as a whole we could decide whether that molecule is polar or non polar but in doing so what we would have to consider is the geometry of the molecule or the shape of the molecule.1374

Here’s an example, this is a CO2 and the geometry of this molecule is shown, we’ll know later how to predict the geometry of the molecule but this is only a molecule.1387

Although this C-O bond is polar, when we see those two vectors, we see that they are pulling equally in opposite direction which means they cancel out and they have no net dipole moment.1399

So this is actually a non polar molecule.1417

So in determining the polarity of the molecule, not only do we need to look the polarity of the bond but then we have to determine the relationship of those polar bonds with another see what the overall dipole moment is.1419

Here we have a molecule which is a bent molecule we have the same C-O bonds that are pulling in the direction of the O and while they both have the right and the left components cancel they both have an upward component and there is nothing to cancel that in the downward direction.1432

So this does have a net dipole moment and so we describe it as a polar molecule.1452

How about this next one, here we have a bent shaped as well but what polar bonds do we have?1458

We know that C-C bonds are non polar, we also know that C-H bonds are non polar.1469

So because there are no polar bonds there can’t exist a net dipole moment so in this case it does not matter what the geometry is because we have no polar bonds.1474

So this is a non polar molecule.1482

Any molecule with just Cs and Hs will have to be non polar.1486

Let’s look at this last one NaOH.1492

What we know about the bonding of this species, NaOH?1494

I just said, if we have a formula containing something like Na, what do you know about Na?1500

How does Na exist in order to be stable?1508

It must be positively charged.1510

So this is actually an Na+ which means this OH must be an OH Na-.1513

This must be anionic compound.1518

You are going to be able to make that evaluation if you are looking at formula.1520

So what does it really look like?1523

It looks like we have Na + then for an ionic bond we don’t draw a line between the two we just show the two ions next to each other and we have an OH-.1526

This is the actual structure of sodium hydroxide NaOH.1535

This is ionic. We don’t describe this as a polar molecule we describe it as an ionic molecule because it has a full charges.1541

It doesn’t have a δ+ and δ-1550

We could also show up here that this O is a δ- and this C has some δ+ character because of that polarity.1553

This has full charges not partial charges.1564

We describe this as an ionic rather than polar.1569

An OH-, this here is described as a covalent ion because it is an ion it is negatively charged but it has covalent bond as well, the bond between the O and H shared covalent bonds.1571

We call it a covalent ion.1587

So NaOH has both ionic and covalent bonds in the structure.1589

Finally, as a little bit of an intro and a catch-up let’s talk about the use of line drawings for organic chemistry because this is a shorthand notation that organic chemists use constantly as a way to save some time.1595

What we do is we draw a line to represent carbon chains and the definition is that any endpoint of a line represents a carbon and any intersections or bends represents a carbon atom.1612

For example this carbon chain has 1,2,3,4,5,6,7,8 Cs; so the way that this can be represented is the 1,2,3,4,5,6,7.1630

So this zigzag line or line drawing notation represents the carbons.1646

This end point is a C and any bend here or intersects to this two lines 2,3,4,5,6,7,8.1652

What we do is we draw the lines to represent the Cs and then we omit those Hs that are attached to those Cs.1663

We can do that because we know that C likes to have four bonds to be stable.1673

So we know that this end C has just one bond shown to a C we know the other three bonds must be the Hs.1678

So this must be a CH3. Any line that ends at a point is a CH3.1692

And this C, what do you know this C looks like?1701

It has two bonds to Cs and so the other two bonds must be the Hs. This is what a -CH2 looks like.1705

Very quickly we can draw very complex organic molecule by using this notation and we need to have some experience with that to recognize that there is a lot of Hs on these structures that are not really shown.1712

So this eight C organic molecule known as octane we are going into some nomenclature down the road and we’ve heard of octane that is one of the components of gasoline.1727

You have the octane reading when you fill your car up.1740

This is one example of an organic molecule being good fuel with these some other compounds that we use in our car.1744

What do we do if we have an atom other than C?1753

So we just draw that atom so we have two C chain 1, and then we have an O attached to that.1757

Now this H you have to include.1764

We only omit the Hs attached to Cs so all other Hs are shown.1767

So this represents this structure, this is called ethanol or ethyl alcohol.1772

There are many alcohols we will learn as a class of compound but this is the alcohol that is grain alcohol.1780

It’s the kind that is drinkable.1788

This molecule we can draw this in line drawing what we will be doing is replace those Cs.1790

So if we have a double bond or a multiple bond we just connect two points with a double line instead of a single line.1800

This molecule is called acetic acid and this is the compound that is in vinegar that gives its characteristic smell and taste.1807

So this carboxylic acid groups looks like easy to find, have acidity to them that’s because of that bite and kind of burns a little bit when you taste vinegars.1820

So there is again organic chemistry in flavors and fragrances.1830

How would we represent this one?1833

Well we have a C=O and then we have a C and then we have a double bond to another C so you could draw it this way.1837

Or you could draw…it doesn’t matter what angle we do we’re going to be showing actually how this molecules kind of rotate around and you’ll see some different shapes and conformations.1846

In a lot of times in this particular case, we like to show these Hs.1856

This is kind of a one exemption to the rule where we leave off the Hs usually shown.1862

This H when it’s attached to a C=O we use to draw it that way.1868

It’s acceptable to leave it off but don’t be surprised when you see it on there still.1877

Carbons not only can make very long chains but they can also for rings so have cyclic compounds.1881

So when we have a six-member ring like this we simply draw a hexagon.1888

So cyclic compounds are very easy to draw with line drawings by just drawing the polygon, a pentagon or a square.1893

This also has three double bonds so we would just show that every other C is connected by two bonds and we could leave off all these Hs we don’t need to do that1900

we know because there’s 1, 2, 3 bonds shown so there must be a fourth bond and that’s the H.1913

So this is the representation of benzene, this molecule is called benzene.1919

That is an interesting molecule.1924

This is actually known to be carcinogenic that means it causes cancer.1928

Benzene is one of the many dozens of carcinogenic compounds that are in cigarette smoke.1932

We will be seeing chemistry of benzene related compounds down the road.1938

Some of the exercises that you want to do is being able to convert from condensed formulas to line drawings back and forth to get some experiences because we really want to become fluent with this line drawings.1943

It’s going to save you so much time down the road.1955

You’re also going to have to be able to work with them so you need to be able to interpret that.1957

For example, this is pretty complex, we have a C with three CH3 groups1961

so here is that C and we just draw three lines coming off to represent those CH3's and then it is attached to a C and see all this C have no Hs on here.1967

That tells me that there’s a triple bond.1980

There’s a triple bond between those carbons because there are no Hs, and then we have another C with two chlorines on it.1984

So very funny with the line drawing for a complex molecule but this condensed formula takes a lot of deconvoluting as you work it out as well.1995

We are going to be spending a lot of time talking about how to draw Lewis structures and come up with these various line drawings and formula so this is just a brief introduction into the process and rules.2008

But what’s nice is that now that you have an understanding of what this line drawing represent now we can take look at some examples of just how diverse organic molecules can be.2024

For example this very first structure is called vanillin, why do you think it’s called vanillin?2036

This is the major component of vanilla extract, the vanilla bean and so this is what tastes and smells like vanilla.2043

You know if you are to buy artificial vanilla, it would be this compound, just the vanillin or the natural extract will have mostly vanillin but then also dozens or maybe hundreds of other constituents in there because it is very rich, deep flavor compared to the artificial which has just this one component.2052

Look at this structure, this is it.2072

It is just a benzene ring, we can see a benzene ring in here and just carbons, oxygens and hydrogens, that’s it, but depending how those atoms are put together they can have completely different reactivities and behaviors.2075

This next molecule is interesting again just carbons, hydrogen, oxygen.2087

This molecule is called carvone.2092

I built a model of carvone here, we’re going to be looking at some of these three dimensional shapes called stereochemistry when you consider the three dimensional shape of a molecule.2094

We have a six-membered ring in here and we have some double bonds.2104

And what’s so interesting about this molecule is on the six-membered ring we have this group, if this group is in this position, compared to this position it will have a very different flavor and fragrance.2108

In one position it will taste and smell like spearmint, like you might have a gum.2123

In the other position, it smells and tastes like a kerwin.2127

We’ll see some really fascinating changes that happen by considering the stereochemistry, the three dimensional shape of certain molecules.2131

Now I said everything that we eat and digest is an organic molecule, let’s take a look at some of those structures.2140

This is the structure of a protein.2146

So when we heard of proteins these are amino acids that are linked together.2148

This R group here, whenever we use the letter R it just means that we have some kind of carbon chain.2153

So there’s a wide variety, I just gave you a basic structure here depending on what that R group is and depending on the order in which these are put together and how long it is it’s what defines what protein you have and how does that behave as an enzyme, what functions does it have in the body and so on.2159

But look, it’s a carbon and nitrogen chain but definitely organic molecule.2175

This is a structure of sucrose it’s called table sugar or cane sugar has this.2184

All sugars, starches, saccharides are organic molecules, carbon, hydrogen, oxygen, that’s all we put them together this way.2191

Here we have a six-membered ring with an oxygen, here we have a five-membered ring with an oxygen and they are linked by an oxygen bridge.2199

Sometimes we draw bonds with some shading, we draw this as a wedge and that’s another way to represent three-dimensionality, stereochemistry.2206

If the bond is wedged that means it’s projecting out toward you, if it’s dashed it means it’s behind the page so it’s kind of hidden.2217

That’s something that maybe you can understand what you’re seeing when you’re looking at these line drawings.2224

Down the road we’ll explain in much more details so you will really have a great feel for these structures.2231

This is the structure of triglycerides so when we’re looking at lipids, that’s oils, greasy things that has this general structure where it has this three-carbon chain with this three oxygen groups on it.2236

I just put an n here because these are really really long carbon chains.2251

Sometimes they have double bonds in them, sometimes they are all single bond and that makes the difference between saturated or polyunsaturated or monounsaturated fat so we know how important those are.2255

That is all explained by looking at the structure of the organic molecules involved.2267

So fats, sugars, proteins everything that we are digesting is organic.2272

All of the compounds undergoing reactions involved in our body are organic.2276

This is the structure of cholesterol, it’s a steroid.2284

Any steroid in the body has a similar structure.2287

This group of four rings kind of defines the steroids structure then various groups attached to it make it one or the other.2292

We know how important cholesterol is to our health.2300

Here’s the structure of vitamin C, again carbons, hydrogens, oxygens that’s it but if we put together this way, this now is an essential vitamin and keeps us very healthy, really important part of our diet.2303

And you can see here that that structure have been elucidated with some stereochemistry was shown.2317

Here is a one component of a DNA; we know DNA is that double helix that goes on and so these are the groups that link out to make each of those long chains.2322

You could see again kind of a sugar backbone here, kind of looks like the sugar part is only a five-membered ring and then some groups that look a little like benzene ring with some nitrogens in there and some phosphate group here.2335

DNA are blueprint for our reproduction is also an organic-based molecule.2348

Then of course the pharmaceutical industry and drugs of any kind, things that affect our central nervous system, on our body in any way, almost all of these are organic molecules.2357

This is the structure of aspirin, very simple molecule of carbon, oxygen and hydrogen with the benzene ring in there.2372

That’s the structure of aspirin.2380

Remember I said benzene itself is carcinogenic, but just having the benzene unit as part of the structure does not mean it will also be carcinogenic.2383

Obviously aspirin is actually good for your health in many cases and it has pain-relieving and fever-reducing effects.2394

As soon as you attached groups to the benzene ring it’s no longer benzene itself and has very different physical and chemical properties.2409

Here’s a structure of nicotine, so that’s the component in cigarette that makes it so addictive.2418

Here’s caffeine, you could see a lot of drugs have nitrogens in them they have a lot of reactivity that we will be studying throughout organic chemistry.2424

Here’s a compound that is colored, so this called Orange-2 which is used as a dye and it turns out that having these double bonds one after another like this, having a large network of double bonds like that.2433

It’s called the conjugated pi system, having those interact with light in such a way that they we can see them as colored,2446

They reflect certain colors of the spectrum and so that’s an example of a pigment that is organic based.2454

This is the structure of Roundup a commercial herbicide that is used as weed killer and again if you have weeds growing in your yard or termites invade your house or cockroaches we are expecting that there’s a product out there that will kill those or take care of the weeds.2463

So there’s a research to make them safe for humans as possible, safe for crops as possible but get rid of the pests, and the fungus and all that and the weeds so that we can protect our crops and that they’ll have to grow in healthy and safe way.2487

This molecule always fascinates me, this is an example of pheromones.2504

A pheromone is an example of the chemicals that insects used to communicate with one another.2509

This is a sex pheromone and so this is what the housefly uses to attract the mate.2516

It lets off this chemical and them a fly of the opposite sex is going to be attracted with that and so they can be together to mate.2523

And look at this structure, it’s just a giant carbon chain, has one double bond here with this particular stereochemistry with both pointing to the same direction and that’s it, if you take that double bond no effect on the housefly.2533

If you have another groups somewhere else no effect.2546

I really find it fascinating and interesting.2548

And you might wonder why do we know with the sex hormone of the housefly looks like, what application could we gain from that.2552

Well, let’s think about maybe a roach motel type thing where you have bait that you have some poison in for ant trap or roach trap or something like that or a yellow jacket trap those are things you might have seen.2560

Well, we put them under the sink we can’t just hope that they randomly walk in there and eat on the poison, we need to find some way to attract them.2576

Wouldn’t it be a great idea to take the sex hormone for that insect and put it in a box so that when a roach is walking by kind of turns around and says --2583

“hey, there’s a party in there and let’s check that out”2593

And they go in and of course there’s just poison for them to feed on and then they take that back to their colonies and then they can wiped out the whole colonies.2596

That’s not by accident that they are being drawn to this box and so pheromones could have an important role in that.2603

Here’s a brief structure of a carbon nanotubes; so if you look at it very closely you could see that it’s a benzene ring fused with another and another and another.2610

When you fuse benzene rings together to get a sheet, to get a flat structure. in fact that is the substance called graphite, that’s the structure that we have in pencils and that allows you to trace a line across a piece of paper.2620

That is a graphite.2635

Well if you take that sheet and roll it up you’re going to make a tube, and this is a tube that you could extend for a very long distances because C-C bond is so strong.2636

It’s a flexible yet extremely durable strong lightweight material and so nanotubes are really revolutionizing a lot of materials.2647

You’re seeing it being used for creating a really strong durable bicycle that’s super lightweight that you can carry easily, those sorts of applications.2659

Do keep an eye up in the future for nanotechnology and you most often see organic chemistry at the forefront of that.2670

Then here’s an example of a polymer and a polymer is described as a long chain of carbon atoms with repeating units.2679

This is called polyvinylchloride so you could see we have a chlorine here periodically these goes on and on in each direction and it is called PVC for short.2691

So if you have heard of PVC piping that the piping that we use for water pipes and if you have sprinkler system in your yard it called PVC pipes.2701

So this is an example of what an organic polymer looks like and by varying the group that’s hanging off and the spacing of those groups and maybe the interaction between the chains you can have a tremendously different physical properties.2710

Let’s take a look at some of those.2725

I have a slide here which only have a very small set of examples of polymers, we just talked about PVC, that is something that you have in your pipes, your shower curtains, how great it is to have something that is water resistant and mold resistant in order that we can use it in our showers.2727

The car seats we have in our car that are vinyl.2746

Polyurethane is something that we can make kind of foamy so things like mattress or soles of your tennis shoes or they can be used as a fiber for spandex clothing very stretchy so some polymers might have that stretchy physical property.2750

Polystyrene is also something that you can foam up so it is very lightweight.2769

Styrofoam is a polystyrene or depending on how it is processed and how it is cast you can also make it hard and see through like for plastic cups or not see through and the plastic toys are made up of varieties of different polymer.2774

Polyethylene is the plastic that you used for some plastic wrappers, sandwich, there’s a few different polymers that you could use for that.2795

Polyacrylamide is the polymer which is super absorbent of water so that they really can hold a lot of water and last a long time.2804

Adhesives are also polymer based so things that make tapes stick, post it notes, all those sorts of things, anything sticky or gluey is going to be as an example of polymer .2817

Teflon is an example of polymer, that nonstick coating that we have in our pots and pans is a polymer.2829

Polymethylmethacrylate is what is used as a plexiglass so the huge walls that you have in aquarium that are really thick and really durable, those are examples of plexiglass as is the plastic in eyeglasses and contact lenses those are also polymers.2837

The fibers and fabrics, obviously some clothes are 100% cotton but those do not stretch very well, they kind of sag over time. They don’t hold the color very well.2856

And so by using manmade fibers we can not only dramatically vary our clothing but also make it a lot more durable and stain resistant and all those interesting properties so things like nylon, ryon, polyester additives to your clothes.2869

Polyacrylonitrile is what we have for carpets so carpet fibers also are synthetic to make them a lot more durable and longer lasting.2888

Then you can even have polymers that are incredibly durable even stronger than steel in some case like kevlar which a polymer that is used for lightweight bulletproof vests, on tires they use them for bolt holes to make them very strong.2899

If you think back to thousands of years ago on how did someone protect themselves in battle.2920

Well they put on a metal helmet and a coat of armor that is all metal and they had to walk around with this and probably needing help getting dressed and its very heavy and you get very fatigued very quickly.2926

You have no visibility because they did not have any plastic shields they could go over.2941

Now you’ll think of today’s military and the lightweight, movable fabrics that they have that are bullet resistant and really protect you but also allow you to function, the goggles that are plastic and see though and are durable and protect us.2950

Really as you can see in just with this very brief introduction that what we’re going to learn in organic chemistry.2969

It really does affect in our everyday lives and we’ll find examples of it all around and I look forward to exploring more with you and we’ll do that in the future.2976

Thanks very much for visiting educator.com.2989

Welcome back to Educator.0000

Next we are going to talk about a class of reactions called elimination reactions.0002

The first reaction we will study is called an E2 reaction for an E2 mechanism; and the name of that reaction comes from the fact that it is an elimination reaction; and it is bimolecular.0008

It is a one-step mechanism very similar to the Sn2; the other mechanism we saw that was a single step was also bimolecular.0020

That has the same 2 that the E2, as the Sn2 backside attack substitution reaction.0031

Something that the E2 mechanism needs is a strong base; so what are things that we know of as strong bases?0038

Things like hydroxide or alkoxide; maybe an N- would also be a very strong base; so pretty short list of species that we would describe as strong bases.0044

Those are the kinds of reagents we will need to do an E2 elimination; and here is an example of one.0059

Let's react... we will start with an alkyl halide; and we will react it with sodium hydroxide; that is a nice strong base.0065

What do bases do?--bases go after protons; the proton we are going to attack is located right here.0072

It is not on the same carbon as the leaving group; remember the halide is going to behave as a leaving group just like it did in the substitution reaction.0080

The hydrogen that gets attacked is not on the same carbon as the leaving group; it is the next carbon over.0089

We call this not the α position but the β; this is called a β hydrogen.0095

The hydroxide or whatever strong base we are using is going to come in; and in a single step, it is going to grab that proton, form a π bond, and kick off the leaving group.0101

The product we are going to get will be an alkene; we are going to get an alkene product; we are going to be forming a carbon-carbon double bond as a result of an elimination.0115

The other products that are being formed... if I used hydroxide, I am also forming water, the conjugate acid of hydroxide; and of course I always lose my leaving group, so I also form bromide as well.0131

This is described as an elimination reaction because you have eliminated both the β hydrogen and the leaving group.0144

We lost HBr so sometimes this reaction is described as a dehydrohalogenation; because you have lost the hydrogen and the halogen.0152

What do you think an energy versus POR diagram would look like?--well, it would be a lot like our Sn2 mechanism because that was also a single step reaction.0162

Our energy versus progress of reaction, we would start at some combined energies for our starting materials; the hydroxide, the alkyl halide.0174

We are going to end at some final product energy; assuming it is a favorable reaction, that it is going to be an exothermic reaction.0184

In order to get from the starting material to the product, as usual for our one-step reaction, we are going to go through just a single transition state to go here.0192

So really, the energy diagram looks very similar to that for the Sn2.0202

What does the transition state look like?--the transition state has a lot of bonding changes taking place all in a single step.0206

First of all, we are forming a new bond between the hydroxide and the hydrogen; so forming bonds, we show as partial bonds in the transition state.0215

We are also breaking this carbon-hydrogen bond; so breaking bonds, we are going to draw as a partial bond in the transition state; let me draw the rest of my carbon chain here.0227

We are also forming a π bond between these two carbons so we will show a partial double bond being formed between those two.0239

Finally our leaving group is leaving so we are breaking the carbon-bromine bond; and we will draw that as a partial bond as well; so four bonds involved in this mechanism, being formed or broken.0248

How about any partial charges that we need to account for in our transition state?0260

In our starting materials, our oxygen is negatively charged, but it is neutral in the product so that charge is disappearing; we have a partial minus on the oxygen.0267

Our bromine starts out neutral but ends up as bromide; so we have a negative charge developing on the bromide; so that is what our transition state looks like for the E2 elimination mechanism.0277

What about the stereochemistry for the E2?--if we are dealing with a chiral molecule and we have some asymmetric centers, what is the ramification of that stereochemistry when we do an E2?0292

It turns out that there is a special relationship we have to have between the β hydrogen and the leaving group that got eliminated; they need to be anti-coplanar.0305

That is the relationship we described when we looked at Newman projections; we said two groups were anti if one was straight up and one was straight down; anytime they have 180 degree dihedral angle.0314

That is the required relationship of the leaving group and the β hydrogen; and for that reason, the E2 is described as anti-elimination.0326

Let's see an example of this; here we have... I had to hand draw some of this... we have an alkyl halide--here is our leaving group that is going to be eliminated.0336

We are not just eliminating a leaving group; we also have to eliminate a hydrogen on the next carbon over.0349

We go this carbon and we see there are no hydrogens; so we go to the next carbon, and here we find there is only one possible β hydrogen.0355

But when we look at this conformation as drawn, the H and the Br are not anti to one another; they are both dashed bonds, dashed lines.0365

If we were to look at this from the side, we see that they would be gauche to one another; they are close to one another.0377

But that is okay; this is a molecule that can be rotated; we could rotate around that carbon-carbon single bond.0382

That is what we will do; we are going to rotate first and rearrange it such that the hydrogen and the bromine are 180 degrees.0391

One way to draw that very easily is to show the hydrogen pointing straight up, the bromine pointing straight down, both in the plane; that must mean they are anti-coplanar.0400

Now the next two drawings we see, the H and Br are anti; and that is the necessary conformation; we have to be able to do this before we can do the E2 elimination.0411

But if we are going to rotate this, that is going to move these other groups; where are they going to go?0429

Let's start with, let's look at this bromine; and I built a model of this; so this bromine right now is pointing away from us; it is a dashed line.0436

What we need to do is we need to rotate it into the plane; then we need to bring it forward.0449

What does that do to this phenyl group?--it was in the plane, but when we rotate it, it is like we grab onto this group and turn all three groups clockwise.0455

That pushes the phenyl group backwards; and so on this carbon now, the phenyl is going be a dashed bond.0463

This hydrogen was a wedge; it is still a wedge; it was pointing down a little bit; and now it is pointing up a little bit; but it still pointing towards you.0472

Let's see if we can do this second carbon; we want the hydrogen to be pointing straight up and in the plane; so what did we have to do?0481

We had to grab onto that hydrogen and bring that forward as well; so we grab onto that molecule and twist that carbon.0490

What is going to happen to this ethyl group?--is it now going to be a wedged bond or a dashed bond?--well let's take a look.0498

Here is our drawing; our ethyl group is in the plane; our methyl group is a wedge; hydrogen is a dash.0507

We want to bring this forward so that it is in the plane; and when doing so, that is going to push our ethyl group back.0515

So the ethyl group will also be pointing backwards, CH3CH2; and our methyl group is still a wedge.0524

So wWe are going to need to manipulate our molecule a little bit to get the leaving group and the β hydrogen anti to one another.0533

Another perspective of this drawing is if we stand here and look at it this way.0541

On this front carbon, we see the hydrogen is pointing straight up; on the back carbon, you see the bromine pointing straight down.0548

This is that relationship of anti that we described back when we first learned about these Newman projections.0554

Once it is in the proper conformation, now my base, whatever base I am using, can come in and do the elimination.0562

One, two, three arrows; grab the hydrogen, form the π bond, kick out the leaving group.0568

What happens is because it is a concerted mechanism, all the other groups that are on the carbon chain end up getting frozen in place the relative positions they were in this anti conformation.0574

Notice that the phenyl and the ethyl are on the same side here; after I do the elimination, the phenyl and the ethyl are required to still be on the same side.0587

I could see it in this drawing too here; the ethyl is a dash, the phenyl is a dash; they are both behind the page.0598

So when we flatten it out... remember when we do the elimination, we are going to a planar product.0605

Everything is going to be pushed up or down a little bit; but the phenyl and the ethyl are still going to end up being on the same side.0611

This going to be the only stereochemistry that is observed; in other words, we are not getting this possibility where the ethyl and the hydrogen are on the same side.0618

This product would arrive from a different conformation, one that does not have the hydrogen and the bromine anti originally.0632

Let's try another example; here we have two phenyl groups; let's identify our leaving group; that is always easy to find; it is going to be a halide, maybe a tosylate, but a halide would do great.0639

Where do we have our β hydrogen?--this is another example where there is only one possible β hydrogen right here.0654

Are they anti right now?--they are not anti right now, they are both dashes again; so let's rotate this.0663

The easiest way to rotate this compound would be to bring that bromine slightly forward, bring that hydrogen slightly forward, so that they are both in the plane with one up and one down.0670

Let's fill in our dash and wedge that we know are here and see where those phenyl groups go.0683

What do you think?--this bromine, when you bring that bromine forward, what does that do to the phenyl group?0690

If you bring the bromine forward, that is going to push the phenyl group back; and when you bring this hydrogen forward, that is also going to push this phenyl group back.0697

This is the proper conformation that we have where they are now anti to one another; so what we did here was we just rotated so that they are anti.0710

Now we could have our base come in and do our elimination; three arrows--base attacks the hydrogen, forms a π bond, kicks out the leaving group0722

Since it is just a one-step mechanism, it is a great idea to go ahead and draw the arrows so you have a better chance of drawing the product accurately.0734

We are going to get an alkene product which is planar; and what is the relationship of those two phenyl groups going to be?0741

Because they are both behind the board, one is a little up and one is a little down, they are both going to get flattened out, but they are both still going to be behind the board.0750

So whether you draw them on the top or the bottom, it doesn't matter; but you need to show them on the same side of the double bond.0757

We have two phenyls on the same side; and what are the other two groups?--we have the methyl and a hydrogen; those were both wedges pointing out towards you; and they still will be.0763

Again it doesn't matter if you draw the phenyls up, or you flip it over and draw the arrows down, that is the same exact product.0773

Anti elimination is the way we describe the stereochemistry of the E2 mechanism.0778

What about the regiochemistry of the E2?--regiochemistry is when we ask about what region or site reacts.0786

If we have a choice of more than one site on a molecule to undergo a reaction, how do we make that decision?--and here is such an example.0799

Here we have an alkyl halide; we see our leaving group; we treat it with a strong base; let's say we want to do an E2 elimination.0807

Actually I think there is more than one possibility here so let's consider both possibilities.0816

If we wanted to do an E2, we have our leaving group is on this carbon, where do we have β hydrogens?0823

Well, we have a hydrogen on this carbon; and on this carbon, we have two hydrogens; we have one that is a wedge and one that is a dash.0829

The hydrogen on the far left, we will call that HA; and the hydrogen that is a dash, we will call that HB.0840

And let's call this D; I will refer to that hydrogen in just a moment.0848

What is another thing that can happen?--let's think about this reagent that we are using, hydroxide; what do you know about hydroxide's reactivity?0857

What can it do?--is it an acid or a base or an electrophile or a nucleophile?0866

It certainly can be a strong base; then that is what we are considering right now, doing the E2 elimination as a strong base and attacking a proton.0871

But isn't it also a strong nucleophile?--it definitely is a strong nucleophile; and what mechanism do we associate with strong nucleophiles?--backside attack, Sn2 mechanism.0879

So another thing that hydroxide can do is it can attack in this position; let's call that path C; and that would be an Sn2 path; that would be a backside attack path.0892

Let's look at what those three products would look like A, B, and C; A, if I took this hydrogen, would end up with a double bond between these first two carbons.0905

B, if I took the hydrogen from this middle carbon, would give me a double bond between these middle two carbons; now I went ahead and took that hydrogen and left the carbon chains where they are.0916

Because notice this HB is a dash and the bromine is a wedge so they are already in the anti-conformation; so I don't have to do any rotations.0931

If I wanted to take hydrogen D and do an elimination there, what I would have to do is I would have to twist the molecule to get those two, the hydrogen and the bromine, anti.0941

That would give product D like this; D isn't even formed, but we will draw that up here so you can see what that looks like.0951

What does product C look like if you do the Sn2 mechanism?--backside attack; that means you replace your leaving group, substitute your leaving group for your nucleophile.0960

Tell me about the stereochemistry of that reaction; since the bromine is a wedge, the hydroxide must have come from behind; and you would get a dash here.0968

These are in fact the three major products; 16% of A is isolated; 75% of B and 9% of C; so Sn2, we get a very small amount here.0980

We are going to talk a little later about how we would make that prediction of the E2 versus the Sn2 mechanism.0995

But of these, these are both E2 eliminations, but there is a huge difference on which is the major product.1001

B is favored by far; why is that?--it is the major product because it is the more stable alkene.1012

This is known as Zaitsev's rule; and Zaitsev's rule states that when you have a choice between different alkenes that you can form, you are always going to want to form the more stable alkene.1019

Can you tell between B and D, why B would be favored over D?--why is that the major product?--let me fill these in; looks a little better.1032

Why is B better than D?--well, this is cis versus trans; and the cis is going to have some steric hindrance so that is definitely going to be less stable than the trans.1042

That explains why no D was formed even though it is a possible product.1055

But other than cis and trans, what are some other things we should know about alkene stability?1060

If we want to predict which is the major product, we have to understand which alkene is more stable; let's take a look at some factors that will affect alkene stability.1066

One thing is that if you are ever given conjugate, if it is ever possible to be conjugated, that is going to be a more stable alkene, a more stable diene.1078

If you have multiple π bonds, then it is better to have those double bonds alternating double bond, single bond, double bond, than isolated.1086

Here they have no relationship to each other; here they have one right after another; we have a p orbital here and a p orbital here, and right away we have more p orbitals.1095

This relationship is more stable because we could have resonance delocalization of these π electrons.1104

If you happen to already have a double bond in the structure, when you go to form a second double bond, if it is possible to put that new double bond in conjugation with the original one, that would be a good thing.1110

Another rule is that as you increase the number of alkyl substituents on the double bond, you increase stability.1123

What do I mean by alkyl substituents?--these are the number of carbon groups on the carbon-carbon double bond.1131

What is the maximum number of carbon groups that you can put on a carbon-carbon double bond?1143

Remember each of these carbons wants four bonds; so there is two bonds over here and two bonds over here; that would be a very stable carbon-carbon double bond.1148

Let's take a look at some examples; here is one such example of that; this alkene has four alkyl groups attached; we describe this as being tetra-substituted.1158

This is called a tetra-substituted alkene; that is the most alkyl groups you could possibly have; and this is the most stable alkene that there is; it really likes having those alkyl groups on there.1175

That is going to be more stable than an alkene with just three alkyl groups; we call those tri-substituted.1186

That is more stable than any di-substituted pattern; and look at all these possibilities; these are all di-substituted.1197

But within the di-substituted patterns, we want to know that the trans arrangement where the two alkyl groups are as far apart from each other as possible, that is going to be the best arrangement.1205

That is more stable than cis; cis is not so good because it has some sterics here; remember there is some steric hindrance in cis that makes it less stable, higher energy.1219

That is not such a great arrangement; and this one isn't so great either; this is called geminal or gem for short; having two groups on just one end of the double bond is not a very stable arrangement.1232

I usually don't care too much on trying to compare these two, but for sure you should know that neither of these is as good as the trans relationship; but typically cis is better than geminal.1246

All of these are better than having just a single alkyl group; we would describe that as a mono-substituted alkene.1257

Or you could also describe it as terminal; it is called a terminal alkene because it is at the terminus; it is at the end of a carbon chain if it has just one alkyl group on it.1269

That is not a very good arrangement for alkyl groups as well; this is the least stable.1279

It turns out that this is the same general trend in stability we saw for carbocations; carbocations like to have alkyl groups attached.1286

That one is a little easier to understand or maybe predict because it is an electron deficient species.1293

Why is it that alkyl groups also like to have... I'm sorry, why is it that alkenes also like to have alkyl groups attached?1300

That discussion of that stabilization is a bit beyond our scope here; so we simply need to know that the more alkyl groups we have, the more stable it is.1306

Let's take a look at a few examples and see if we can apply Zaitsev's rule; here is an example--draw both E2 products and select the major.1318

We have our leaving group here; how is it possible to get two different E2 products?--well, let's look for our β hydrogens.1328

There is no hydrogens on this carbon; it is the next carbon over; we could grab one of these hydrogens or one of these hydrogens.1339

Would those give two different products?--actually those would both give the exact same product; so β hydrogen type A is going to be right here or right here.1345

Where is the second type of β hydrogen?--how about going in this direction right here?--going to this methyl group; β hydrogen type B.1359

Product A would form a double bond within the ring; and like I said, whether you go to the right or left, it is the same product.1370

Tell me about this methyl group; is that still a dash?--is that still pointing behind the board?1378

Remember what an sp2 hybridized carbon looks like, an alkene carbon?--this is planar; and so that methyl group now after the elimination is completely flat; and it should be drawn as a straight line.1385

Now you can see that if you just flip it over, taking the hydrogen from either side would give the exact same product.1396

But what is the other possible product we can form?--instead of going to one of the side ones, we can go up to this top carbon.1403

What does he look like now?--he used to be a CH3; it is now going to be a CH2 because one of those hydrogens was taken away.1411

These are the two possible products we can get; and now which is the major product?1419

We are going to look to see which one is more stable; and for that, we are going to see which one is more highly substituted.1426

How would you describe the number of alkyl groups attached to this first carbon, to this first alkene?1431

Here is our alkene; we have one, two, three carbons attached to it; in other words, there is only one hydrogen.1438

This is a tri-substituted alkene; and how about this second one?--we have two hydrogens up here and two carbons.1448

We are not counting the carbons in the double bond; we are asking how many carbons are attached to those double bonds; this is a di-substituted alkene.1459

Which would be the major product?--the tri-substituted would be the major; why?--it is usually not such a great idea to just say because Professor Zaitsev told me so or just invoke the rule.1471

The reason it is the major product is because it is the more stable alkene; that is how we determine the regiochemistry of E2 elimination.1486

Let's take a look at another example--if our leaving group is attached to a six-membered ring.1502

Now it is the chair conformation of that compound that needs to be taken into consideration when we look at the stereochemistry and the regiochemistry.1510

Here again we have a strong base; we forgot to mention that up here just noting we were using hydroxide; now here we are using methoxide, also a strong base.1522

When we look at it in this conformation, if you think about where you are going to take your β hydrogen, we have hydrogens over here and we have hydrogens over here.1532

Which one is more tempting to take based on Zaitsev's rule?--for regiochemistry, where do you think you want to go?1542

I am guessing we might want to go to this side because that would give us a more highly substituted double bond; but there is a problem with that.1549

What did we learn about the stereochemistry of the E2?--let's draw in that hydrogen and see if you can find a problem with that hydrogen.1557

That hydrogen is a wedge; that bromine is a wedge; is it possible to rotate that to get them anti to one another?--it is not; so because this is not anti, I can't eliminate in that direction.1567

Instead I have to look over to this side; this is a CH2; one is a wedge and one is a dash; this is the only anti β hydrogen; so there is only one possible product we can get.1580

When you look at the chair conformation, the most stable chair here has a bromine up; we could put that in an equatorial position.1595

And it has a methyl down; we can also put that in an equatorial position; so the most stable chair has the leaving group in an equatorial position.1603

Unfortunately the hydrogen we want to take next door is not anti; those are not anti to one another; again those are going to be gauche to one another when you take a look at that Newman projection.1614

What we are going to have to do is we are going to have to flip our chair into the other conformation.1628

Let's see; that means we are taking this carbon and bringing it down; so that is down here; this hydrogen is now pointing straight down.1642

This carbon got flipped up; so my bromine up is now axial up instead of equatorial up; and this carbon has my methyl group down; it was equatorial and now my methyl group down is now axial.1650

In this chair conformation, now my H and Br are now anti; we call that anti-diaxial as the necessary conformation.1668

In a chair conformation, the leaving group must be in an axial position in order to be 180 degrees from the β hydrogen.1683

Our base can come in, grab that proton, kick off the leaving group; and what does our final product look like?--we can go back to our line drawing here.1692

We are going to form a double bond on this side; and our original methyl group is still there; and that is the only possible product we can draw.1703

This product was really just a question of regiochemistry only in order to draw the product; but I wanted to point out that the required stereochemistry requires a specific conformation of our chair.1713

That is going to be relevant because depending on the substitution pattern throughout the chair, this chair flip might not be such an easy thing.1727

You will find that some cyclohexyl halides do very fast E2 elimination; some our very very slow; this is an important thing to take a look at when you are considering such cases.1734

Now that we know what an E2 mechanism looks like, let's consider its competition with the Sn2 mechanism.1747

The Sn2 mechanism was described as backside attack; and what is required for the Sn2 is that it needs a strong nucleophile.1754

In an Sn2 substitution, a strong nucleophile attacked the carbon, kicked off the leaving group; that backside attack; we needed a strong nucleophile.1768

Another thing that was important with that backside attack was something about steric hindrance; remember that backside of the carbon had to be very accessible.1778

Let's just write that steric hindrance... so it needs a strong nucleophile; and steric hindrance is bad; it is not a good thing; that slows down the reaction.1787

What do we know about the E2 mechanism?--what does it require?--E2 is when we have something attack a β hydrogen and form a π bond and kick out a leaving group.1802

We are attacking a hydrogen; the E2 needs a strong base... it needs a strong base.1812

It turns out that this deprotonation reaction is not affected by sterics; so the sterics is going to be a major factor in making the decision between an Sn2 and E2, we will see.1821

When do they compete?--when do we have to make this decision?--well, most definitely we are going to see it for reactions that involve hydroxide, that is HO-, or RO-, that is known as alkoxide.1835

In other words, we can have methoxide, ethoxide, propoxide, so an alkyl group with an O-.1853

What is special about these species is they are both strong nucleophiles, meaning they love to do backside attack; they are great for an Sn2 mechanism.1858

And they are strong bases, meaning they can also go after a proton and do an E2 elimination.1868

We really need to be on the lookout when we are using hydroxide or alkoxide; and we will have to make a decision between the E2 and the Sn2.1875

Let's compare some rates; if we took a variety of alkyl halides and treated them with sodium methoxide and ethanol.1885

We could have either a substitution reaction take place; what would the product look like if methoxide decided to act as a nucleophile and do a substitution?1895

That means instead of RBr, we have our ethoxy; so our nucleophile has replaced the leaving group; that is our Sn2.1908

Why don't we draw out the ethoxy here; so I'll do some more abbreviations on this page; it is real good to start getting familiarized and seeing back and forth between these abbreviations.1922

This would be our Sn2 product; what would it look like if we did elimination instead?--well, that of course is going to depend on which alkyl group we started with.1934

But let's just say plus some kind of alkene or mixture of alkenes; in other words, we could do an elimination and form a double bond instead.1942

Let's take a look at these series of alkyl halides and try this reaction and see what product composition we get.1953

The first one we have here is methyl bromide; we could just abbreviate this MeBr for methyl bromide.1961

What kind of carbon is bearing the leaving group in this case?--it is simply just a methyl group; we just describe it as a methyl group.1970

This next one has one, two, three, four carbons; you could call that a butyl bromide or n-butyl bromide since it is the normal butyl group.1977

What kind of carbon is bearing the leaving group here?--well, because we have just one carbon group attached to it, we describe that as a primary carbon.1987

This next one is another common arrangement of four carbons; when we have something that looks like the isopropyl group here, but you have an extra carbon, this is called the isobutyl group.1997

If we had isobutyl bromide, this is still primary; but right next door to that on the β carbon, the next carbon over, we have some branching; so let's call this primary with β branching.2014

The second one, this three-carbon arrangement with the attachment at the middle carbon, that is the isopropyl group.2031

So this is isopropyl bromide; it is an example of the secondary carbon bearing the leaving group.2037

Then once again we come back to good old t-butyl bromide; tert-butyl bromide is just our classic example of a tertiary carbon bearing the leaving group.2043

What have we done?--why have we chosen this series of compounds?--because as we move down this list, we see that we are increasing in our sterics.2054

We have increasing sterics about the carbon bearing the leaving group.2065

What proportion of products do we see as a result?--well, when we have methyl bromide, it turns out that we see 100% Sn2 reaction; no E2 at all.2072

Let's think about what that E2 product would look like; what does E2 mean?--it means you remove the leaving group and a β hydrogen and you form a carbon-carbon double bond.2083

How does that look for methyl bromide?--it is looking pretty impossible because not only does it not have any β hydrogens, it has no β carbons.2095

This is let's just put NA here because there is no β hydrogens; you can't form a carbon-carbon double bond if you only start with one carbon; so of course we expect 100% for Sn2.2103

But when we move to primary, now the E2 is possible; and we do get a mixture of the two; but it is heavily in favor of the Sn2; we get about a 90:10 mixture.2116

Almost all Sn2, but we start to see a little bit of the E2 side product.2125

Adding just the slightest bit of β branching here, slightest bit of sterics, now pushes it in the other direction where the elimination is favored.2132

That is just how sensitive the Sn2 substitution, that backside attack, is to steric hindrance.2140

In fact if we go to a secondary carbon, then it is favored in the opposite direction even more so; where E2 is the major product by far, 4:1.2147

Now we are really shifting; and how about tertiary?--what do you think about the tertiary Sn2 reaction?--does that look like a good reaction?2159

No, we know that there is way too much steric hindrance here; so let's say there is essentially 0 and 100 now in this switch; remember tertiary alkyl halide is no reaction with the Sn2.2166

It is no reaction with the Sn2; but a reaction in fact does take place; instead it is the E2 elimination reaction.2183

Let's continue looking at this comparison and come back to tert-butyl bromide.2196

What we are asking it to do when we react it with ethoxide, or some other strong base that can also be a nucleophile, is it has two choices.2201

It can either attack the β carbon... I'm sorry, attack the carbon bearing the leaving group, do a backside attack; that is the Sn2 path.2210

Or what can it do?--it can attack the β hydrogen; that is the E2 path; attack the β hydrogen, form the π bond, kick out the leaving group.2220

Which of these is possible for the tertiary center like t-butyl bromide?--the Sn2 is impossible; it is impossible to get in here with all this steric hindrance and do backside attack.2231

But look how accessible that β hydrogen is; you can see that sterics is not going to be a problem; anytime we are going after a tiny little hydrogen, eliminations are no problem.2243

Let's just make a note here that the backside attack is too hindered and that the β hydrogen is more accessible.2254

We get 100% E2 with a tertiary leaving group; that is kind of the ultimate in the examples of Sn2 versus E2.2272

Another possibility is to vary the base that you are using; what if you had a primary alkyl halide?2285

Here we have fifteen methylene units; fifteen of these carbons; so it is just a really long carbon chain with a bromine at one end.2292

What if we reacted this with a species that could be either a base or a nucleophile?--what kind of mixture do we get, distribution between Sn2 and E2?2300

If we use a small base like methoxide, what we see, the same thing we saw on the previous slide, is that Sn2 is favored because it is very little steric hindrance for a primary carbon.2311

We get something like 96% of the Sn2 and maybe just a little 1% of the E2.2326

However if we use this guy, what is this structure?--that is a strange looking structure.2331

Remember potassium is just a K+; so that means we have O-; and this is the t-butyl group... t-butyl group.2336

This could be abbreviated as tBuOK; tBuOK is potassium tert-butoxide.2344

When we have a very very bulky base now, that is going to hinder that backside attack; and we see a push in the other direction; now elimination can be favored.2352

This guy is called t-butoxide; he is a very bulky base because he brings along with him that t-butyl group.2363

I brought some models here again just as a reminder about the sterics like we are seeing here.2371

The methyl is so tiny; but the tert-butyl, when you have three methyl groups attached to a carbon, that really inhibits this backside attack.2377

If we are thinking about this coming in and acting as a nucleophile, this is going to have the same problem.2388

This is going to have a hard time getting in to do a backside attack on a carbon bearing a leaving group because it is bringing steric hindrance with it.2395

T-butoxide is a very bulky base so it is never good for an Sn2 unless maybe you are attacking a methyl group that can't do an E2.2402

Really these are classic E2 reaction conditions; if you really want to favor an E2 elimination, then t-butoxide is a great base to choose.2415

By the way, addition of heat also favors eliminations so we should get used to seeing heat as part of the reaction conditions.2424

That is because we are breaking the molecule up into different pieces so that is going to favor the entropy of the reaction; so elimination reactions are also usually done in the present of heat.2430

Let's summarize; what would we say the preference is for E2 over Sn2?--looking at a reaction type, which type of alkyl halide would really like to do the E2 elimination?2443

The tertiary is going to be the best; this is pretty much all E2 and no Sn2 at all; secondary still prefers E2, but you are going to have a little substitution.2455

Here is where we see that big jump when we are looking at primary; this now is essentially all Sn2; so only if it is a primary alkyl halide can you still expect to do an Sn2 with a strong base.2472

As soon as you get to secondary and of course if you are at tertiary, if you take a look at that data from the previous slide, you will see that E2 is now the major product.2487

So only when there is absolutely no steric hindrance, because you have a primary leaving group.2496

Of course, we are leaving off the methyl here intentionally because methyl has no β hydrogens; that would be impossible to do the E2 in that case.2500

Let's take a look a second elimination mechanism; we saw the E2 elimination which is the one-step mechanism; base attacks a β hydrogen, forms a π bond, kicks off the leaving group.2513

Another mechanism that we can have for doing a dehydrohalogenation, losing a hydrogen and a halogen, is an E1 mechanism; that stands for elimination unimolecular.2525

Let's take a look at an example of that; if we take this alkyl halide and we treat it with water, how would we describe water as a reagent?2538

Water is a weak nucleophile; it is weak everything; it is a very stable molecule so it is a weak nucleophile; it is a weak base.2549

But we can still undergo substitution-elimination reactions here; in fact, we do get the substitution product.2559

Would you expect that substitution to occur by Sn2 mechanism, backside attack?--no, because this is such a weak nucleophile and because this is a tertiary leaving group; so this can happen by Sn1.2570

It is also possible to get some elimination product; could this elimination be the one-step E2 mechanism?--no, because we don't have a strong base.2583

Instead a different mechanism must be in operation and it is the E1; so the Sn1 and E1 we will find are always in competition.2598

Just like we saw the E2 and the Sn2 are often in competition, can be in competition, the same is true for Sn1 and E1, these unimolecular reactions; and the Sn1 is usually the major product.2608

In most of the reactions we are going to see, there will be a nucleophile around; it is going to prefer to add; and so we will get this substitution.2627

E1 is maybe a side reaction; so you might see a little bit of elimination and this is how you get it.2635

We will see one example where the E1 is the primary mechanism in action when we do the dehydration of alcohols; but with that exception, Sn1 will be major.2646

The mechanism for an E1 is going to be two steps; and it is going to be via a carbocation; again very similar to Sn1; that is why they compete.2655

Our first step is going to be loss of a leaving group; our chloride is going to leave which results in a carbocation intermediate; notice this is the same slow rate determining step as Sn1.2667

I lose my leaving group and I form a carbocation; but now instead of having that carbocation get attacked by a nucleophile to do a substitution reaction, instead we can do an elimination.2683

What did it mean to eliminate?--you eliminate a leaving group plus a β hydrogen; so what have we done so far?--we've lost our leaving group.2699

What's next?--we lose our β hydrogen; and we have a word for loss of a proton; we call it deprotonate.2707

We are going to take our carbocation, we are going to take a look at one of these β hydrogens, and we are going to eliminate that β hydrogen.2718

Some base comes in; again you can use chloride if you want; I think water is a better choice because that is going to be present in this reaction as well; and it is your solvent so you have more of that around.2731

We are going to do a deprotonation; we grab the proton; why can carbocations get deprotonated so easily?2744

Because when you fill in those electrons as a π bond, you are going to get rid of the carbocation; you are going to add a new bond and get to a stable alkene product.2751

Let's consider the regiochemistry; why did I select this β hydrogen?--we did see that that is the major product; but is that the only β hydrogen that is available?2766

Here is our carbocation; no, this is a β hydrogen; this is a β hydrogen; I could have also made this product.2775

But why was that not listed as one of the products formed here?--this is actually not formed; how would we make that decision?2782

It turns out that this elimination also follows Zaitsev's rule just like the E2 did; and we are going to get the most stable alkene possible.2793

This would be a tri-substituted alkene; that is going to be more stable than a di-substituted alkene.2802

And so as usual we are going to go for the more substituted β hydrogen to give the more substituted alkene.2810

Let's take a look at the kinetics for the E1 mechanism.2820

Because it had the same slow rate-determining step as the Sn1 mechanism, it has the same rate expression, the rate is proportional to simply the concentration of the alkyl halide.2823

That is all you need to lose your leaving group; so that is why it is described as unimolecular like Sn1.2836

The E1 and Sn1 are both referring to the fact that just a single species is present during the rate determining step.2844

It is directly proportional to the rate of the reaction; its concentration is directly proportional.2851

The rate is not dependent on the concentration of water; it doesn't matter how much of that base we have in there; it won't affect the rate.2857

That also reinforces the fact that water is not involved in the rate determining step or whatever base you have; tt is simply the loss of the leaving group from the alkyl halide.2865

How could we speed up this reaction?--well, once again, the more stable the carbocation you have, the faster it is going to be formed.2874

If you bring down the energy of your carbocation intermediate, that is going to have a lower energy transition state as well; and you have a lower energy of activation, faster reaction.2885

Again the faster the carbocation is formed, once you are at that carbocation, you can now do substitution or elimination.2897

The same thing that speeds up the Sn1 reaction is also going to speed up the E1 reaction; and we come back to that competition.2905

So the E1 rate is exactly the same as the Sn1 rate; looking at the type of carbons bearing the leaving group, these are all great carbocations; so they are going to be fast E1 eliminations.2912

Allylic means you have an allylic carbocation; so in other words, an allylic leaving group... let's put a chlorine here.2928

That would be a great substrate to do an E1 elimination just like an Sn1 elimination.2940

Benzyl means it is next to a benzene ring; a leaving group next to a benzene would be great because that would be a resonance stabilized carbocation.2945

And then of course tertiary is what we just looked at because that is also a good carbocation; those are better than secondary.2954

This is where we have our huge jump for carbocations; primary and methyl are so bad; these are poor carbocations so it is going to be a very very slow E1 elimination.2961

Actually let's cross out that methyl again because you can't have an E1 elimination with just a methyl halide because you don't have a second carbon to form the carbon-carbon double bond.2979

How about the stereochemistry of the E1; is there any kind of relationship that we need to have between the leaving group and the β hydrogen?2992

Remember E2 elimination, what did we need?--our leaving group and our β hydrogen had to be anti to one another; one up and one down; they had to be anti-coplanar, 180 degrees.2999

Remember the E1 mechanism is a stepwise mechanism that goes through a carbocation; that carbocation is achiral; it is planar; so this is another case where we are going to see loss of stereochemistry.3011

This is the same story for the Sn1; remember we described it as racemization because we were dealing with chiral carbons in that case; here we just describe it as loss of stereochemistry.3021

Let's see an example; if we take this substrate and we treat it with a strong base, hydroxide; strong base, what does that do?--that is going to come out and attack; this likes to do E2.3031

Let's just define it; hydroxide is a strong base and it is a strong nucleophile; as a strong base, what mechanism can you do?--E2.3045

As a strong nucleophile what mechanism can you do?--backside attack, Sn2; which one is going to be favored in this case?3059

We take a look at the carbon bearing the leaving group; it is a secondary carbon; secondary--E2 is going to be favored.3069

We add in a little heat; that doesn't hurt; but even without that, with secondary, we are going to go with E2; there is just too much steric hindrance to do the backside attack instead.3077

We've conveniently drawn it already so that the hydrogen and the leaving group are anti-coplanar; so we could come right out and we don't have to rotate at all.3088

We could just take our hydroxide, grab the proton, form the π bond, kick off our leaving group; that is going to give us an alkene product.3097

What groups are going to be on the same side?--you have a phenyl that is a dash; and you have a methyl that is a dash; so those are both on the same side of the alkyl halide.3108

They will be trapped on the same side of the alkene; it doesn't matter whether you put them at the top or bottom; the important thing is they just need to be on the same side.3119

What other two groups do we have?--we have a methyl and a hydrogen; those are still on the structure.3128

Remember, because this is anti-elimination, this is our only product; it requires those two groups to be anti; and so this would be the only possible stereochemistry.3136

Now let's compare this same starting material and let's just react it with water; no hydroxide; so now we have a weak base and a weak nucleophile.3152

What kind of reactions are possible there?--is a weak base going to go and attack a β hydrogen?--no.3167

Is a weak nucleophile going to go and do a backside attack and kick off a leaving group?--no way.; so this water does reactions like Sn1 and E1; and of course the E1 is probably going to be major.3174

The Sn1 is probably going to be major, but let's take a look at the E1 to see what its stereochemistry would be.3190

What is going to happen here is because we have no strong base or strong nucleophile, the only other thing that can happen is the leaving group can just leave on its own; so that is exactly what happens.3197

When that chlorine leaves, the resulting carbocation is now planar; it is flat.3212

When we go to deprotonate our β hydrogen, the β hydrogen can be removed now or we can have some rotation; it can rotate here.3223

There is no longer anything trapping these two methyl groups into the trans position; it can rotate around.3237

Because it is a multistep mechanism, we lose that stereochemical relationship; once that leaving group leaves, it is gone.3245

When water comes in and it deprotonates, we now have a mix of relationships between these methyl groups.3254

We are going to get two products; we are going to the one where the phenyl is on the same side as the methyl; and we can also form the one where the methyls are on the same side.3264

You are going to get a mixture; and of course you are also going to get Sn1; and Sn1 is actually going to be your major product probably; but we just want to look at our E1 product composition here.3282

These are both possible; does that mean they are going to be formed in equal amounts?--remember Zaitsev's rule is still holding true; we still want to form the most stable alkene possible.3294

Is there a difference in this case in their stability?--the phenyl remember is a benzene ring; that is going to be larger than a methyl group.3306

This product has more steric hindrance than the other one; so this is going to be our minor E1 because of sterics; and this is going to be our major E1.3315

For anti-elimination, we only have so much say on which alkene we can get; it has to be from anti-elimination.3332

With na E1 elimination coming from a carbocation, we could rotate around and get to the most favorable carbocation conformation.3342

Which leads to the most favorable transition state, leading to the most stable alkene product.3349

Let's talk just a little bit about carbocation rearrangements; because like the Sn1 mechanism, we are going through a carbocation.3362

Carbocations can rearrange; so let's just briefly review what we saw for this Sn1 mechanism.3376

This was a rearrangement when our leaving group left; it gave a secondary carbocation; and that secondary carbocation could have a hydride shift to give a tertiary carbocation.3386

That would be a favorable rearrangement; so that is how we can get this substitution product where the nucleophile has come in to a different position.3405

What is important to remember is that this new carbocation can undergo E1 elimination products.3425

Sometimes our double bond is going to end up in a different place from where our leaving group used to be.3434

Because both of these mechanisms deal with carbocations, and carbocations can rearrange, a lot of times we can get some interesting product mixtures.3441

So these end up becoming excellent practice problems for mechanisms; in other words, how can this combination of reaction conditions lead to all one, two, three, four, five of these products?3450

This is definitely a reaction that looks like it is coming from the carbocation because the carbon chains have rearranged; the positions of the substituents have rearranged.3467

This is another example where our initial secondary carbocation can rearrange, reorganize itself to become a tertiary carbocation.3482

In this case, it is by rearrangement of one of these methyl groups, a shift of a methyl group.3495

If this carbon group moved over here, that carbon has lost its carbocation; it has got four bonds again; but this carbon is where the carbocation is now.3504

Hopefully you can see how that can lead to this substitution product; and it could lead to these two elimination products.3515

This original carbocation can lead to this substitution product and this elimination product down here.3522

Just considering the various carbocations that are possible, and the various Sn1 and E1 products that can result from those, is how we can get to these various products.3530

Of the elimination products, which would you say would be the major elimination product... be the major elimination product?3541

Again the substitution products are going to be the favorable one; and probably this is going to be the major overall because it comes from the most stable carbocation.3550

Because we do have a nucleophile present, that is why we do get a substitution; that nucleophile will attack the carbocation more often than not.3563

This will be our major overall; and of the elimination products, which do you think would be the major elimination product?3570

I am thinking the one where we can have a one, two, three, four, a tetra-substituted alkene.3578

And that comes from the more stable carbocation; so both kinetics and thermodynamics are favoring this as the major E1 product.3588

Let's try a few examples where we are looking at substitution and elimination competition; the ones shown here... let's just focus on the Sn2 and the E2 and see if we can decide which will be major.3600

First of all let's take a look at our starting material, our starting carbon compound, and see what we have; OTs is called a tosylate; that is a good leaving group.3618

What kind of carbon is it on?--is it on a primary carbon?--it's not even on a primary carbon; this carbon has no carbons attached; so we just call him a methyl; so we have a methyl tosylate.3635

What are we reacting it with?--the sodium means that I have a CH3O-; and what do you know about methoxide?3648

This methanol, what is this methanol doing here?--well, it is very common that when you are using methoxide as a reagent, you use methanol as your solvent.3657

But because he is neutral and stable and less reactive, we are not going to consider him for our reaction.3664

It is going to be the less stable, more reactive species that is going to dictate the path of this reaction.3670

What do we know about methoxide?--acid, base, electrophile, nucleophile?--well, it is a strong base; and it is a strong nucleophile.3676

So what mechanisms does it have?--these are perfect for E2 or Sn2.3689

We said that alkoxides like this; like methoxide is a perfect substrate to have this competition between E2 and Sn2.3695

Who is going to win in this case?--if you have a leaving group on a methyl carbon, then it is going to be impossible to do the E2 because there is no β hydrogen.3704

I am expecting to do the Sn2; that means backside attack; attack the carbon, kick off the leaving group.3713

My methoxy group is going to now be attached to my methyl group; it is going to be a substitution, backside attack.3722

We have another alkyl halide; this time it is on a secondary carbon; so this is a secondary leaving group.3731

What do you know about ammonia?--what kind of reactivity does ammonia have?--is it a strong acid, base, electrophile, nucleophile?3737

He is a very good nucleophile; which means he would love to do the Sn2; but is he a strong base?3744

He definitely is basic, but he is not one of the strong bases we identified for the E2; he doesn't have a negative charge on it.3751

Not a really really strong base... so in the case of a neutral amine, our favored reaction is going to be the Sn2 which means backside attack.3760

Tell me about the stereochemistry of that nitrogen; since our leaving group is pointing out towards us, the nucleophile has to come in from behind.3772

Let's keep our carbon chain fixed; and so now instead of the wedge pointing down, we have a dash pointing down.3780

What happens to that nitrogen when it gets a fourth bond?--one, two, three, four; nitrogen wants five; it is going to be an N+.3790

Which means... a lot of times, we draw that leaving group in there because the bromide is not going to get very far; it is going to have an ionic bond with the ammonium.3799

So a lot of times we draw it as the salt here as this product.3808

Who is DMF?--it is another commonly used solvent in organic chemistry; it is an example of an aprotic solvent.3811

A lot of these abbreviations or names, you should be familiar with and get used to seeing them there.3820

Even if we don't draw a solvent, all of our reaction are run in a solvent; so it is good to get some practice seeing those and recognizing those.3826

In most cases, the solvent is not going to be dictating the mechanism we choose.3836

But of course it is possible, if this solvent is the only variation, that you are going to see a difference in the mechanism that you pick.3840

How about this next one?--we still have our same secondary leaving group; but instead of ammonia, we are using this guy, tBuONa.3848

This is Na+ so this is tBuO-; what do you know about tBuO-?--it is t-butoxide; strong bulky base.3856

He loves to do the E2; not so good at the Sn2 because he is so bulky; so even if he was primary, he wouldn't be able to do the Sn2; this looks like perfect E2 conditions.3869

We look for our β hydrogens; we have some over here and we have some over here.3880

First we should consider regiochemistry; where would we prefer to deprotonate?--well, we want to get the most substituted alkene possible; so we are not going to go to the end carbon.3890

Then how about stereochemistry?--well, we know stereochemistry needs to be anti; but remember this is an acyclic structure; so we can rotate; it is not locked into any one hydrogen.3902

In fact, if it took this hydrogen, it would keep the carbon chain the same; and if it took this hydrogen, meaning it had to flip first and then do the elimination, it would give this product.3915

Which of those two products do you think is the better product?--how about stability of the alkene as a guide?3932

Sure, the trans alkene is more stable; so that is going to be the one that we get; so don't be fooled by the conformation that happens to be presented.3938

Remember that you can rotate the molecules and manipulate them around; these are in solution; they are mixing around; they can be any way they want.3952

So we are going to do whatever we can to get the most stable alkene possible, assuming that it is still anti-elimination.3959

Here are a few more; this one is sodium methoxide in ethanol so you could just abbreviate it as EtONa and EtOH.3968

What do we know about ethoxide?--acid, base, electrophile, nucleophile?--strong base, meaning it can do the E2; strong nucleophile, meaning it can do the Sn2.3977

How do we decide which is going to happen?--we take a look at our carbon bearing our leaving group; what do you see?--where is our leaving group?3993

Well, there is a problem; what do you expect to find as a leaving group?--chloride, bromide, iodide, tosylate?--is hydroxide a leaving group?--no, not a leaving group.4003

If you don't have a leaving group, you can't do a substitution; you can't do elimination.4018

It doesn't matter which; you can't do any of the four mechanisms that we've learned for Sn2, E2, or Sn1, E1; no reaction.4022

We learned a couple strategies where we could make this reaction happen; if we had a made a tosylate first, then we could do a substitution-elimination.4031

If we had a strong acid present, then it is possible to do a substitution; but that is not the case here.4039

We have strongly basic conditions; so nothing is going to happen with that alcohol.4045

Same thing is going to be true in this next step; we have sodium azide and acetone; azide is a very good nucleophile.4050

It is not a strong base; it is a good nucleophile; another nucleophile we should get used to seeing for the Sn2.4057

But because this is not a leaving group, again no reaction; nothing it can do.4063

How about our next one?--we have hydroxide; hydroxide is a strong base; it is a strong nucleophile; so this is a great example where we have our E2, Sn2 competition.4072

Do we have a leaving group this time?--yes, we have an iodide; we are back to our alkyl halide so we have a leaving group that we can get rid of.4089

What kind of carbon is that leaving group on?--it is a one, two, three; it is a tertiary carbon; a tertiary leaving group.4096

What do you think?--is that backside attack?--is tertiary good for a backside attack?--no way, impossible because of sterics; so this is going to be all E2.4105

Now we have to think about regiochemistry; we have to think about stereochemistry; we've already identified our leaving group; where is our β hydrogen?4117

We have this β hydrogen; we have either one of these β hydrogens; this is very similar to a substrate we worked with earlier.4127

We are going to want to eliminate one of the inside, one of the ones within the ring; so then we end up with this tri-substituted product, more substituted.4136

Just a little reminder, we don't want to eliminate in this direction because that is going to be di-substituted; this is less stable; we always want to get the more stable alkene possible.4145

Finally here we have our same tertiary leaving group; and we have sodium cyanide; what do we know about sodium cyanide?--great nucleophile; great nucleophile; it loves to do the Sn2.4160

Is it a strong base?--we have our list of strong bases; this is not one of them; this is a weak base so it cannot do the E2.4175

Let's take a look at our Sn2, our backside attack; does it look like a good Sn2?--tertiary leaving group?--no way.4190

So guess what?--we have a great nucleophile; but our carbon bearing the leaving group is too hindered; and it is not a strong base; so it can't do E2 elimination.4198

This is another example where no reaction is going to happen.4207

You might ask, well can't we have the leaving group just leave and make a tertiary carbocation in this case and maybe do an Sn1?--and do the substitution that way?4210

Well, here is one case where the solvent is going to come into play; the cases where we did see an Sn1 reaction like in the solvolysis reactions, we always had a protic solvent.4222

That is going to be true for a carbocation formation; it actually needs a protic solvent; like water or an alcohol.4238

Protic solvents have an acidic proton; they are extremely polar; and they are needed to stabilize if it is an extremely unstable polar carbocation.4249

If you have an aprotic solvent, it is not going to be possible to form a carbocation.4259

Most textbooks aren't testing that much detail; but this is an example where really having an understanding of the details of the solvent capabilities is going to help us determine something about a mechanism.4266

In this case, I was specifically asking you to compare E2 versus Sn2; so with those restrictions, for sure we could conclude that it is no reaction.4279

Sn1 would be a good guess in this case; but it turns out that would not be possible either.4288

That wraps it up for looking at elimination reactions and considering their competition with substitution reactions.4293

Hope to see you again soon; thank you.4299

Welcome back to Educator.0000

We've already had a brief introduction on how to name alkanes; some basic IUPAC rules; methane, ethane, propane, and so on.0003

Let's take a look at some more complex alkanes and cycloalkanes and a variety of other functional groups we will encounter.0010

If we have a ring in a compound, we call those cycloalkanes; we simply take the name of that alkane and we put the prefix cyclo- before it to indicate that those carbons are forming a ring.0018

If there is a single group that is attached, if it is a mono-substituted ring, then we assume that the point of attachment is carbon number 1, defined as carbon number 1.0031

Because when we have a carbon chain... when we have a carbon chain, we have to decide which end to number from.0040

But when the carbons are in a ring, any carbon within the ring can be carbon number 1; we have to decide which one to assign as number 1.0046

For this first example, we have a five-membered ring; a five carbon alkane is called pentane; so this is called cyclopentane.0054

Then we take... that is our parent; then we alphabetize and number our substituents as usual and put them out front.0066

We have a two carbon chain here; that is an ethyl group; this is simply called ethylcyclopentane, all one word.0073

There is no number that needs to be drawn because any place you put the ethyl on that ring, you are going to get the same compound; so this is called ethylcyclopentane.0082

If we have two more groups, now we have to decide which carbon is defined as carbon number 1.0092

If there is one carbon with two groups on there, then that would be the best place to assign as number 1 because then we would have two groups at position number 1.0097

Remember we always want to get the lowest possible numbers that we can; so we will define that as carbon number 1.0107

Now we have to decide, are we going to number in a clockwise fashion or counterclockwise?0112

We want to number in the direction closest to the next substituent so that that substituent gets as low a number as possible.0117

Let's number here in a counterclockwise, this direction; two, three, four, five, six; this is a cyclohexane parent; cyclohexane.0123

We have a fluoro and a methyl and then this group is a little more complex than a simple methyl, ethyl, propyl, that sort of thing; how would we name that?0139

We have some instruction here; to name a complex substituent, we need to define the point of attachment to the parent as carbon number 1.0151

Then we perform a second, like a little mini IUPAC, to describe the name of that complex group.0159

For example, this is the group that... this is the point of attachment; so this has to be described as carbon number 1 for this substituent.0165

Then we ask what is the longest carbon chain from that point on?--we see that we can have a three carbon chain here; this is actually a propyl group.0174

But it is more than just a propyl; what does it have on carbon 1?--it has a methyl group; we are going to describe this as a... this whole group is described as a 1-methylpropyl.0184

That is how we will handle more complex substituents; now let's alphabetize these and list them in the proper order; the fluoro comes first so we have a 1-fluoro.0200

Then methyl comes before methylpropyl; just like if you look that up in the dictionary, the shorter word would come next; we have a 1-methyl.0211

Then at carbon 3, we have this big whole group; what we do is we write a 3, and then in parentheses, we write this big name of the complex group.0220

Let's see if I can not completely run out of room; we have a 1-methylpropylcyclohexane.0229

How about this last case?--we have a four-membered ring; we call that cyclobutane.0241

Here we have two groups; we have to decide now which one should be carbon 1 because we could either have 1,2 or 1,2; we would get the same set of numbers either way that we numbered it.0250

In that case what we do is the carbon bearing the group that comes first alphabetically is assigned number 1.0263

That way the substituents are listed alphabetically and the numbers are listed in increasing order.0271

What we have here is a methyl; then this is going to be a longer word; we will have that next.0277

You know what?--let's make this an ethyl... sorry... so this comes first alphabetically; we will just change that.0285

This is... we will do one and two and three and four; this is another complex substituent that we are going to have to name.0293

It looks like a t-butyl group; that would be nice if we were allowed to use a common name; we could just describe it as a t-butyl.0302

But an IUPAC name can't use that term; so instead we go through our rules again; now this is carbon number 1; what is the longest carbon chain from that point on?0308

No matter which direction you go, the most you can have is two carbons; so this is actually an ethyl group.0320

But it not just an ethyl; it has a methyl here on carbon 1 and another methyl on carbon 1; what we are going to call this is a 1,1-dimethylethyl.0327

This is the IUPAC name for the t-butyl group; it is a 1,1-dimethylethyl; and where was this whole group?--this whole group was on position 2.0345

On position 1, we had an ethyl; we could double check, see how we alphabetized e- comes before m-; when we go to alphabetize, we ignore the prefixes like di- and tri-.0357

So we have a 1-ethyl-2-(1,1-dimethylethyl)cyclobutane; we have the ring for the cyclo; we have the four carbons.0372

Bit by bit, step by step, in a systematic way, we are going to build these complex names up.0383

If we have a carbon-carbon double bond in a structure, those compounds are called alkenes; the way we are going to name them is we are going to look for the longest carbon chain.0390

But that longest carbon chain must include both carbons of the carbon-carbon double bond; it is no longer just where is the longest carbon chain throughout the whole structure?0399

Now that we have a functional group in the molecule, we need to incorporate that functional group into the parent; so the longest carbon chain must contain both carbons of the carbon-carbon double bond.0407

Then we are going to number from the end that is closer to the carbon-carbon double bond so that the double bond will get the lowest number possible.0418

After doing that, what we are going to do is we are going to take the ?a in the word alkane, methane, ethane, propane, butane, and we are going to change it to an -e.0427

We are going to call it an alkene instead; and we are going to put a number out front that tells the reader where the double bond starts--what is the first carbon of that carbon-carbon double bond.0436

We are going to call it a #-alkene or you can insert that number inside the name; this is actually the preferred way for IUPAC.0447

But I think it is a little more difficult for beginning students to look at that convention.0456

For example, our simplest alkene is a two-carbon alkene; he has a common name; this is called ethylene; ethylene is a gas that is generated by fruits in the ripening process.0463

Ethylene is actually sprayed on fruit to help them ripen when they are picked earlier for packaging and for handling.0473

This is commonly known as ethylene gas; but the IUPAC name, it would be an ethane derivative because it is two carbons; instead of calling it ethane, we call it ethene.0481

Not a big difference between the common name and the IUPAC name; but the IUPAC name is ethene; this tells the reader... that E tells the reader that there is a double bond in this molecule.0496

In this case, I don't need to indicate where the double bond is because there is only one place for it to be--connecting those two carbons.0506

This compound is commonly known as allyl chloride because you have a chlorine that is allylic--next to a double bond.0512

It is known as allyl chloride; but let's do its IUPAC name; we have our three carbon chain is right here; that is our only possibility for our longest carbon chain.0519

When we go to number it, we are going to number it from the end closest to the double bond; we will start from this end; one, two, three.0528

Three carbons means it is a propane derivative; because of the double bond, we call it propene; this is 1-propene.0538

We are going to indicate the first carbon of the carbon-carbon double bond; this implies that there is double bond starting at 1 and going to 2.0547

What else do we have on here?--we have a chloro group at carbon 3; so this is 3-chloro-1-propene; remember there is dashes between numbers and letters.0555

Sometimes we leave off the 1; in this case, that might be implied; but it is a really good habit to leave that 1 in there.0569

We don't want to make that assumption then make a mistake; so this would be the best name for that.0576

How about this case?--where is our longest carbon chain that contains both carbons of the double bond?0582

It looks like we will come across here but then we will move down in this direction because that would be a longer chain than over here.0588

How are we going to number it?--if we numbered it from here, our double bond would start at carbon 2.0596

If we numbered it from here--one, two, three four, it wouldn't start till carbon number 4.0601

We would rather have it be at carbon 2; one, two, three, four, five, six; a six carbon alkene is going to be called hexene; it is a 2-hexene.0605

By circling off and blocking off the parent, it is very easy for me to see what other groups are attached.0622

We still have this two-carbon group; that is called an ethyl group; at carbon 3, we have an ethyl.0628

How about this next case?--our longest carbon chain would be this one; but the longest carbon chain containing the carbon-carbon double bond would be down here.0638

We have to make sure to include that; and how long is that?--one, two, three, four, five, six, seven, eight carbons; that is an octane derivative; we are going to call this 1-octene.0647

At carbon 2, we have a one, two, three carbon chain; that is a 3-propyl; 3-propyl-1-octene.0663

We can incorporate a double bond into a cyclic structure as well; remember, with a ring, we have to decide where to start numbering our carbon chain.0673

One of the double bond carbons is going to be carbon number 1; we are going to have to number through the carbon chain and then continue on to also give the substituent the lowest possible number.0685

If I numbered from here, one, two, three, four, five, six, the substituent wouldn't be until position 6.0695

If I numbered this way, one, two, three, I get the lowest number for the double bond; that has to be at carbon 1; then I also get the substituent at the lowest number.0702

Four, five, six; this is a six-membered ring; we would call that cyclohexane because it is a ring; because it has a double bond, we call it cyclohexene.0712

Where is our group?--what is this group called?--the benzene ring is called a phenyl group; we have a 3-phenylcyclohexene.0725

Here is a case where we really don't need the number; the double bond has to be between carbons 1 and 2; that is how you define a ring.0734

There is no other place to put the double bond since that is our highest priority; if you want to put that in there just to be safe, that probably won't hurt if you want.0741

What if we have two double bond?--if we have two double bonds, we are going to indicate both numbers of where the double bonds start.0751

We are going to call it a diene or a triene or a tetraene and so on; so we are going to include that, we are going to put that in there.0758

Again, we want to give our double bonds the lowest possible number; and I need to give both double bonds the lowest possible number.0768

The only way to number this to give the double bonds the lowest possible number would be as so; one, two, three, four, five.0778

What do we have on carbon 5?--we have a bromo; we have another 5-bromo; we are going to call that a 5,5-dibromo... a 5,5-dibromo.0787

Then at carbon 1 and carbon 3 is where we have our double bond starting; this is a 1,3-cyclopentadiene.0797

Notice we stick that -e back in; it would be called cyclopentene, but when we have the diene or the triene, we put the -a back in just because it is easier to read that way.0812

So it is called cyclopentadiene or hexadiene or some such thing.0822

Another thing that we need to be concerned with with double bonds is the fact that we can have stereochemistry associated with the double bond.0830

Where if we have two groups... remember we have cis or trans as a relationship that those double bonds can have; all the examples here don't have that shown.0837

This molecule can have cis-trans isomerism, but I haven't shown one particular stereoisomer; I haven't shown the geometry of this molecule.0848

I have drawn this as a linear molecule with these bonds hanging straight down; we wouldn't be able to tell which stereoisomer I was dealing with here.0857

When you have a double bond in a ring, in a five or six-membered ring, that double bond has to be cis.0866

Those two alkyl groups have to be pointing in the same direction in order for them to reach each other and form a small ring.0872

So this must be cis if you are looking at a seven or fewer... for small rings, let's just say... for small rings.0879

So we don't say that it is cis-cyclohexane because there is no such thing as trans-cyclohexane; it would be impossible to build that.0892

Let's take a look at some examples that do have stereochemistry involved because we are going to want to include that as part of the name.0901

If you have an alkene simply with two groups and two hydrogens, then we could describe those two groups as being cis or trans.0909

Cis means they are on the same side; trans means they are on opposite sides.0917

For example, we have an alkene here; let's number from the end closest to the alkene; one, two, three, four, five, six; this is hexene; this is 2-hexene.0926

But one particular stereoisomer is shown of 2-hexene, where the two alkyl groups are pointing on the same side; we are going to describe this as cis-2-hexene.0937

We are going to put that stereochemistry all the way out in front of the name to describe the stereochemistry of the double bond.0948

Here is another example; we have this in a ring; one, two, three, four, five, six, seven, eight, nine, ten; this is a ten-membered ring.0955

We would call this a cyclodecane if it was an alkane; because it has a double bond, we call it cyclodecene.0966

In this case now, we have one carbon group pointing this way and one pointing this way... let's write the name first, sorry... cyclodecene.0975

But in this case, the two carbon groups are not pointing in the same direction; they are pointing in opposite directions.0989

If you want to look at the hydrogens to confirm that, you drawn in those hydrogens for those trigonal planar carbons.0994

You see that in fact they are pointing in opposite directions; so this is an example of trans-cyclodecene.0999

If we had looked at this molecule instead, if the double bond were here, this would be an example of cis-cyclodecene because those two carbon groups are pointing in the same direction.1006

When stereoisomers can exist, we need to look carefully at the three-dimensional picture that is shown to decide which stereoisomer it is that we are trying to name.1020

Here is another example; now we have three alkenes; let's number from here so we get the lowest numbers possible--one, two, three, four, five, six, seven.1031

Seven carbons is a heptane; when we have three double bonds, we call this heptatriene; where are those three double bonds?--they start at 1 and 3 and 6.1041

If it is a triene, we need to have three numbers out front; the triene, one, two, three, tells us where those three double bonds are.1057

Is there any stereochemistry in this problem?--there can't be any stereochemistry on this first or this last one because there is only one group attached.1068

But look at this middle one; there are two groups attached; what is the arrangement of those two groups?--they are on opposite sides; this is trans.1076

Sometimes we might miss that; if it is a line drawing and it is obviously drawn in the same side, then we might notice that it is cis.1084

But when we leave it in its regular zigzag notation, sometimes we might miss that that must actually be the trans isomer that is shown; this is trans-1,3,6-heptatriene.1090

I don't need to say which double bond is trans because it is the only double bond that can have stereochemistry.1101

If we only have groups, we can use the word cis and trans; if we have more than two groups, the term cis and trans doesn't mean anything more.1108

Cis means the two groups are on the same side; trans means the two groups are on the opposite side; if I have three groups on the alkene, the term cis and trans is now ambiguous.1116

We need to learn new terms to describe those stereochemistries; the new terms we have are going to be E or Z.1125

Z is what means the same side; so if we are similar to cis, we are going to call it Z; and E means opposite sides.1134

Where do these letters come from?--E and Z, they come from German words; if you speak German, you will have no trouble remembering these rules.1149

If you are like me and these words are not familiar to you, then you need some kind of way to remember which is E and which is Z and what do they mean.1156

I have a little mnemonic that I came up with that I will share with you; see if that helps.1165

Here is how we assign it; what we do is we look at one carbon at a time and we assign the priorities to those two carbons.1172

We are going to use the same... we are going to prioritize them based on simply their atomic number.1180

We are comparing a chlorine to a carbon; chlorine has a higher atomic number so chlorine is a higher priority than carbon.1186

Then we go to the second carbon and we do the same ranking; carbon versus hydrogen; carbons wins; it is a higher priority, higher atomic number; that is also number 1.1194

We ranked them 1,2 and 1,2; now what we do is we compare the number 1 groups and we ask: are those number 1 groups on Zis same side?1205

If we say it with a little crazy accent, then we remember that Zis same side means that it is the Z stereochemistry.1215

In other words, since these number 1 priority groups are cisoid to each other, we are going to describe this molecule as being the Z stereochemistry.1221

We put that all the way out in front in parentheses; this is italicized as is the cis and trans are italicized, you see when they are in print; this is going to be called (Z)-2-chloro-2-butene.1230

The other stereoisomer, if I swap the chlorine and the methyl, the chlorine and the CH3, now you see that my number 1 priority group is here and my number 1 priority group is here.1241

Now they are transoid to each other; we ask: are the higher priority groups on Zis same side?--they are not; now it is the opposite of Z; it is E.1250

We need to remember what both of those names are; of course, the more you work with these, the more you will become familiar with which is which; this would be called the E isomer.1258

Let's just take a look at a couple quick examples here; how would we name this alkene?--the longest chain that contains the double bonds is right here.1267

What side do we number from?--one, two, three; one, two, three; this is a case where we would we have a tie for the double bond.1279

How do we break that tie?--now we want to consider the substituents and try and give them also the lowest possible number.1288

We will number from the end closest to the first substituent; one, two, three; this would be the better direction to go, the correct direction to go.1294

This is a six carbon chain; we call that hexene; the double bond starts at carbon 3 so we call this 3-hexene.1304

What else does it have attached?--we have a methyl and a methyl and a methyl; we group those all together; we call this a 2,3,4-trimethyl.1315

Sorry, I ran out of room here... 2,3,4-trimethyl-3-hexene; which isomer have we shown?1330

Again, I can't... what does the word cis mean here?--or trans?--it has no meaning; we have to define it as E or Z.1336

What we do is we break the double bond in half; we look at one carbon at a time.1343

This carbon... we are comparing a carbon and a carbon; that is a tie; how do we break the tie?--we are going to move the next atom and so on until we find a difference.1348

This has three hydrogens attached; this has another carbon attached; this group wins; this is number 1; this is number 2; in other words, an ethyl beats a methyl.1357

Over here, we have again a tie--carbon versus carbon; but this carbon just has three hydrogens on it while this carbon has two other carbons attached; this is number 1 and this is number 2.1368

Which isomer do we have here?--do we have E or Z?--we compare the number 1 groups; are they on Zis same side?--they are not; they are on opposite sides.1382

We call that the E stereochemistry; that goes all the way out front; we describe this as E.1391

How about last one?--what is our longest carbon chain?--we have to come down in this direction to find the longest carbon chain containing the two carbons.1398

We will number from this end to give the double bond on carbon 2; this is a pentene derivative; it is pentene; it is 2-pentene.1408

We also a chloro and a methyl; who will come first?--we will do alphabetically; 1-chloro-3-methyl.1425

Finally, do we have stereochemistry?--we need to check every time; is there stereochemistry in this case?1436

There is because we have three different groups on here; if I flip these two groups around, it would be a different isomer; it would be a different compound.1441

How do we describe this isomer in its name?--we rank the two groups on carbon 3; we have a methyl and an ethyl; we have done that ranking before; the ethyl is the higher priority.1449

What are the two groups on carbon 2?--we have a carbon and a hydrogen; there is just a hydrogen here; so this chloro substitute group is the higher priority.1461

Here we see our two higher priority groups are on Zis same side; they are on the same side; they are cisoid to each other; we describe it as the Z stereochemistry.1471

When we have a carbon-carbon triple bond in a molecule, we call it an alkyne; we are going to very much follow the same rules we did for alkenes.1488

Except instead of replacing the -a in the alkane, we are going to replace it with a -y instead of with an -e; we are going to name it as an alkyne.1495

Once again, we have to put the number out in front or insert it in the middle to say where the triple bond is on the carbon chain.1505

We are going to try and give the triple bond the lowest number possible; if we had two or three or four alkynes, we would call it a diyne or a triyne or a tetrayne.1514

Let's again look at some simple examples; this is the simplest alkyne; it is called acetylene; that is another fuel; acetylene is what is used in torches, very hot burning fuel.1527

This is called acetylene; its IUPAC name, since it is two carbons, means it is an ethane derivative; we are going to call this ethyne.1537

We are going to call it ethyne or acetylene--is another name that you should be familiar with.1547

The word propargyl means next to a triple bond; just like we use the word allylic to mean next to a double bond; propargyl means it is next to a triple bond.1553

This could commonly be known as propargyl bromide; but let's do its IUPAC; we have a three carbon chain in this case; numbering to give the triple bond the lowest number.1566

This is a propyne derivative; it is 1-propyne; at carbon 3, we have a bromine; we are going to call it 3-bromo; 3-bromo-1-propyne.1576

How about this case?--what is our longest carbon chain?--longest carbon chain will go here and here and don't be fooled by this CH3 group.1593

I am not going to stop at the end of this carbon chain because my carbon chain is not done; it extends up here.1602

There is actually a one, two, three, four, five, six carbon chain here, with a triple bond at carbon 1; we are going to call that a 1-hexyne... a 1-hexyne.1608

What else do we have?--we have a methyl and we have a phenyl... we have a phenyl; which comes first?--L-M-N-O-P, methyl comes first; so we have a 4-methyl and a 5-phenyl.1621

That is our hexyne; if we have multiple double bonds and triple bonds, our names get pretty ugly pretty quickly.1639

Here we can number one, two, three, four, five, six, seven, eight; it is an octane derivative; but we have a double bond at 1 and 6; it is an octadiene; plus we have a triple bond at carbon 3.1646

Here we have to... because we have two different functional groups in the same molecule, we have to break the name up and insert the number right before the suffix it is modifying.1663

This means I have a triple bond at carbon 3 and my diene is at carbons 1 and 6, starting at 1 and 6.1674

Or you could break up the name more completely and insert the numbers in both cases immediately before the functional group that they are representing, that they are defining.1684

Either one of these is fine; but the one where the numbers are inserted are becoming increasingly more common.1700

If you have to make a choice between an alkene and an alkyne, who gets the lower number?--it is the alkene that wins.1707

In this case, when we look at our longest carbon chain right here, we want to number from this end so that the double bond gets the lowest number possible.1713

Then the triple bond just ends up where it ends up; this is a seven carbon chain; what do we call that?--that is a heptane derivative.1726

But now we have an ?ene and an ?yne; let's break up the name; we have a hept-2-ene and then a 6-yne; hept-2-ene-6-yne.1734

We also have a phenyl group here on 7; we can throw in a 7-phenyl; then that continues all one word hept-2-ene-6-yne.1754

If you an OH on a carbon chain, we are going to describe that molecule as an alcohol; that is going to define the parent chain.1766

It is going to be the carbon containing... the chain which contains the carbon bearing the OH group; and it is going to be named as an alcohol.1776

We will find the longest carbon chain that has the alcohol attached to it; and we are going to number from the end closest to that alcohol so that that alcohol gets the lowest possible number.1785

What we are going to do is we are going to drop the ?e at the end of the name and we are going to add a new suffix; we are going to put the ?ol at the end of it.1795

So it is going to be called methanol, ethanol, propanol, and so on.1805

We need to also put a number out in front to tell the reader where along the carbon chain that OH group resides.1809

Once again, you can sometimes see the number inserted immediately preceding the suffix so that it is no question to what the number refers.1817

For example, this looks like the isopropyl group, when we have three groups and attached at the middle; this is called isopropyl alcohol.1828

You may have heard of that compound; that is used as rubbing alcohol; it is an astringent; you can use that as a cleaning agent.1835

You might find that in your medicine cabinet for facial cleaners and such.1842

Isopropyl alcohol, that is the common name for this; its IUPAC name would follow the rules described above--longest carbon chain containing the OH group; the OH defines our parent.1847

It doesn't matter in this case which way we number it because both of them would put the carbon bearing the OH as the carbon 2.1862

We are going to call this 2... instead of propane, it is propanol... propanol.1870

We dropped the -e and we replaced it with an ?ol; the number 2 tell us where that alcohol is, where that hydroxyl group, that OH group is.1876

Here is another example; we have a four carbon chain; now we would have to number it in this direction to give the OH the lowest possible number.1887

This is called a... it is four carbons so it is called 2-butanol; but this also has a common name.1895

What does this arrangement look like?--when we have a four carbon chain and we've attached something at the second carbon, the secondary carbon?--we call this sec-butyl.1902

This could also be called sec-butyl alcohol; those common names just give us a picture so quickly that it is really useful to be familiar with them.1911

How about this case?--we have a couple OHs; this is where we are going to now call it a diol; our longest carbon chain must contain both OHs.1926

Once we find an OH on our structure, we have to... that defines our parent carbon chain; it is no longer just about finding the longest carbon chain; that is if all we have is an alkane.1937

Let's go up here; it looks like that is the longest carbon chain that contains our two OHs.1948

We will number it to the end closest to the first OH; we will number down here one, two, three, four, five so that we have an OH at carbon 1.1955

Let's see, a five carbon alkane is called pentane; with one OH, we would call it pentanol; with two OHs, we call it pentane diol.1965

Notice we add the -e back in; we actually at the ?e back in because it is easier to read that way than just saying pentandiol; that would be difficult; so it is called pentane diol.1975

We have to tell everyone where these OHs are; they are at 1 and 4; these numbers tell us where to find the OHs.1989

What else do I have on this chain?--I have a 3-methyl; and I have a one, two, three, four, five, six; I have a hexyl; that comes first; we have a 2-hexyl and a 3-methyl... 1,4-pentane diol.2002

Once I see an OH on our chain, that is going to have the highest priority number; I no longer care where the double bond or triple bond is; I always want to make sure my OH gets the lowest possible number.2023

For example, in this ring, since any carbon can be carbon 1, we have to decide which is carbon 1; it is the carbon bearing the OH; that must be carbon 1.2034

Then we are going to number again either clockwise or counterclockwise to take care of the rest of the functional groups.2044

Of course, now we will number in a counterclockwise direction so that the double bond gets the lowest possible number.2049

This molecule is called 2-cyclohexene-1-ol; the 2 refers to the double bond and the 1 refers to the OH group.2056

Again, when you have two functional groups with numbers on them, it is sometimes a little clearer to break up the name and insert both numbers directly into the name.2066

But you clearly have to break it up a little bit; otherwise, you can't just put the 2 and the 1 out in the front, because then you wouldn't know which number describes which functional group.2077

We have to break up the name at least this much; and you insert a number immediately before the suffix.2085

How about this next one?--longest carbon chain is right here; how do we number it?--which end do we number it from?2094

I don't care what number the triple bond gets; I care instead what number the OH gets so I am going to number from this direction; one, two, three, four, five.2104

Let's see, five carbons would be a pentane; the triple bond makes it a pentyne... pentyne; this is where the triple bond changes this middle letter.2116

Because of the OH, we call it a pentyne-ol; let's put the 2 in here so we know where that alcohol group is; and the triple bond was at carbon 4; this is 4-pentyne-2-ol.2130

Or you could say pent-4-yne; the triple bond starts at carbon 4; -2-ol; the suffix always goes at the end; the ?ol is always the very last thing that you have.2145

We take care of all this parent stuff; then we list out front everything else that is attached.2159

We have already handled the OH; we've already handled the triple bond; the only thing left is this methyl group at carbon 2; we alphabetize all those, put them out in front.2164

Notice how we keep building on these systematic IUPAC rules; so it is really important to get that foundation down of your alkanes.2175

Then gradually build on them with each new functional group so that you can see how they work together when you have multiple functional groups in the same structure.2183

If you have an ether in your compound, an ether means that we have an oxygen attached to two carbons.2194

An ether is not something that changes the parent name like an alcohol does; we are not going to learn a new suffix for it.2201

Instead, it is simply a substituent that you might find attached to the parent; it is just a group; how do we name that group?2208

If we have an OR group hanging off of a parent, normally we would call that R group, have a -yl ending like a methyl, ethyl, propyl.2215

What we are going to do is we are going to change that -yl and we are going to make it ?oxy instead; instead of ethyl, we have ethoxy; so we have ethoxy, propoxy, butoxy, and so on.2224

If we have very simple ethers, sometimes we just use common names by listing the two alkyl groups on either side; again it is nice to be familiar with the common names; but that is not IUPAC.2237

For example, this could be named isobutyl ethyl ether as a common name because it has an isobutyl on one side, has an ethyl on the other side.2250

That is one name for this ether; but let's see how we would do it by IUPAC.2260

With an ether, what we have to decide is of the two alkyl groups on either side, or the two carbon chains on either side of the oxygen, which one is going to be the parent.2264

We are going to look for either one that has a functional group on it or one that is a longer carbon chain; here they are both just alkanes so we will go on this side with the longest carbon chain.2272

Here is our longest carbon chain; now it is an alkane; so we treat it like any alkane; we are going to number it so that the substituents get the lowest possible number.2283

We will number from this end; one, two, three; this is a propane derivative; what does that propane have on it?--it has a methyl.2291

Then this group, what do we call this group?--it has one, two carbons, plus an oxygen; we are going to call that an ethoxy group; this is an ethoxy group.2301

Then we alphabetize it just as usual; ethoxy comes before methyl; so we have a 1-ethyoxy-2-methyl; pretty complicated name for this molecule.2312

You can see why we might prefer to just call it isobutyl ethyl ether because that gives a much quicker picture to those who are familiar with those common names.2324

How about this guy... I'm sorry, I skipped some information on that previous slide; it pointed out that these aromatic groups, the benzene with an OH is called phenol.2336

This parent could be named as phenol; the OH is defined as carbon number 1; in the 3 position, we have a methoxy group; this is called 3-methoxyphenol.2351

This compound is called diethyl ether as its common name; again just two ethyl groups attached to an oxygen; that is usually how we describe this molecule.2370

But what would its IUPAC name be?--you have to pick one of those ethyl groups to be your parent; it is no longer an ethyl group; it is now ethane as your parent.2379

What does that ethane have attached to it?--it has an ethoxy group; ethoxyethane.2391

What if we have an OH and an ether again?--how do we decide who has priority and where do we go for the parent chain?2405

Again, as soon as you see that OH, that means your molecule is an alcohol and that carbon attached to that OH has to be your parent.2413

Let's define that as our parent; here is our carbon chain containing the OH group; we will number it from this direction to give the OH the lowest possible number.2421

This is a 3-hexanol... 3-hexanol because it has an OH group; what are we going to call this group?--again, this is not a simple methyl, ethyl, propyl.2436

What we need to do is we need to figure out how to name this; what we are going to do is our little mini IUPAC here.2452

Remember, by definition, the point of attachment to the parent must be defined as carbon number 1; we number from there; one, two, three, four; this would be a butyl group.2459

But it is just not a butyl group; it has a 2-methyl; it would be called 2-methylbutyl.2472

But it not a 2-methylbutyl because we also have this oxygen; it is going to be called 2-methylbutoxy; it is a 2-methylbutoxy group.2479

Without this, it would just be a butoxy, but now we have a 2-methylbutoxy.2492

We put that all in parentheses and where is that entire group attached to the parent?--to carbon 5; step by step, we will build up these names.2497

Last case here, we have a triple bond; that is going to define our parent--any functional group that modifies the parent name; so instead of having ethane, now you have ethanol or ethene or ethyne.2509

Anything that modifies the name of the parent--double bond, triple bond, or an OH group; that is going to be your higher priority compared to any other ordinary substituent that you have.2524

This is an alkyne; we are going to name it as such; we will number it from here to give the lowest number possible; a four carbon alkyne is called butyne; this is 1-butyne.2534

What do we have attached here at carbon 3?--if this oxygen weren't here, we would call this cyclopentane group, we would call it a cyclopentyl group, right?2546

If it was just five straight chain carbon, we would call it pentyl; now that is called cyclopentyl.2557

With the oxygen, we call it cyclopentoxy... cyclopentoxy; where is this cyclopentoxy?--carbon 3; 3-cyclopentoxy-1-butyne.2563

If we have a nitrogen containing compound, we call those amines; as usual, we are going to find our parent chain that contains the nitrogen.2582

We are going to drop the ?e at the end of the name; we are going to add a new suffix; we are going to add the word -amine as our suffix.2592

It is going to be named as an alkanamine, ethanamine, propanamine, butanamine, so on.2598

It is possible for a nitrogen to have more than one alkyl group; one of those alkyl groups is going to be the parent; the other alkyl groups would be described as substituents.2605

Because they are attached to the nitrogen rather than a carbon chain, we don't give the number of the carbon; we just list a capital N to indicate where it is being substituted.2615

Again, we will find some common names; if it is a simple alkyl group, we will just call it an alkyl amine; we will see some examples of that.2626

We will also see some examples where maybe it is not the highest priority functional group.2634

For example, this amine has four carbons and a straight chain; it would just be called butyl amine or N-butylamine; that would be a common name for that.2639

But its IUPAC would be... longest carbon chain; one, two, three, four; it is a butane derivative; at carbon 1, we have an amino group, we have a nitrogen group; this is called 1-butanamine.2648

We've dropped the -e of butane and we've replaced it with the word ?amine; 1-butanamine.2665

This next one, we have two alkyl groups; one of them is going to be the parent; we will pick the longer one; this is our parent.2674

One, two, three to give the nitrogen the lowest possible number; this is propanamine; this is 1-propanamine.2681

But it is not just propanamine because it has on the nitrogen, it has an ethyl group attached; the way we are going to describe that is with a letter N; N-ethyl-1-propanamine.2691

Just like a number refers to one of the carbons in the carbon chain, a capital letter N, it is italicized, refers to the nitrogen in the compound as the location of the substituent.2703

When an NH2 is on a benzene ring, we call that molecule analine; we could use that as a parent name; you might come across that name.2716

This NH3 all by itself was called ammonia; that is not really an organic molecule; it has no carbons in there; but we will see ammonia a lot; you should be familiar with that name.2726

Let's see a few more examples; this is another one that has a very simple common name; how would you describe the alkyl group attached to the nitrogen in this case?2736

It is four carbons arranged, attached to the middle carbon; this is the t-butyl group; we could just call this t-butyl amine as a common name; t-butyl amine.2745

But its IUPAC name, we would have to follow the rules--find our longest carbon chain bearing the nitrogen.2758

That is right there; it is a three carbon chain; one, two, three; this would actually be propanamine.2766

What else does it have attached?--on carbon 2, it has a methyl group; we could call this 2-methyl; 2-methylpropanamine.2775

If we have an OH and an NH2, we have to decide which one is the higher priority because we can only have one suffix; we can't have an ?ol and an ?amine suffix.2787

The rule is that OH is the higher priority; we would have the same carbon chain in both cases; but in this case, now we are going to name it as an alcohol.2798

We want to give the OH the lowest possible number; so we will number from this direction; one, two, three, four, five, six.2815

This is a six carbon alcohol; we call hexanol; it is a 3-hexanol; remember the ?ol ending tells us there is an OH in the molecule.2822

Now we just simply name the NH2 group as a substituent, as a group hanging off; just like we have a methyl or a phenyl or a chloro.2835

What we call an NH2 group is an amino group; we have a 5-amino-3-hexanol; amino just like methyl or ethyl or propyl or bromo or chloro.2842

How about this one?--it looks like we can use that analine as the parent; analine; how is this different from analine itself?2859

It has a methyl group on the nitrogen and a methyl group on the nitrogen; as usual, we combine the common substituents all together.2872

Where are they?--it is a dimethyl, but they are both on nitrogen; we are going to call this N,N-dimethyl; N,N-dimethylanaline; that means each methyl group is on the nitrogen.2882

Here is an example; now we have an ether and an alcohol and a nitrogen; how do we handle this?--thus far, our highest priority group is the oxygen, the OH alcohol.2898

The longest... the highest priority group is the one containing... the parent chain is the one containing the OH; we give the OH the lowest possible number.2910

Everyone else just gets whatever number they end up with; this is actually a propanol; 1-propanol derivative; what do we have attached to it?2920

This is a one, two, three carbon chain; we would call that a propyl; but with the oxygen, we call it a propoxy.2933

Here, if it was just a nitrogen, we would call that an amino group; but it has something else attached to it; what is attached to it?2942

It has a phenyl attached to it; just like IUPAC, this is not... we just do a little IUPAC here; instead of just amino, it is a phenylamino.2950

We don't have to locate the position of it because it is just one atom; it has to be attached to the nitrogen.2960

That would come first alphabetically; we will list that first; 2-phenylamino and then 3-propoxy-1-propanol.2969

A little bit more about amines; we could describe an amine as alkyl or aryl depending on the type of carbon group that is attached.2987

If you have a benzene ring attached, we would call it an arylamine; otherwise, it is an alkylamine.2995

We can also describe the amine as being primary, secondary, tertiary, or quaternary; we define it by describing how many carbon groups are attached to the amines.2999

A primary amine, which we designate with this 1° sign, that means primary, means we have just one R group; then it is an NH2.3011

Remember nitrogen likes to have three bonds and a lone pair to be neutral.3019

Because this has just one alkyl group, it is described as a primary amine; this would be methylamine or methanamine would be the IUPAC.3023

But if we had two alkyl groups attached, like two methyl groups here, we describe it as a secondary; it is just an NH group.3032

Here we see our two alkyl groups; we would call this diethylamine as the simplest name; here is the IUPAC; you can check that; N-methylmethanamine would be its IUPAC name.3041

Tertiary is what we have when we have three alkyl groups or aryl groups, three groups attached to the nitrogen.3052

This is called triethylamine--would be a common name for this; that is usually how we call it because look how big and awful the IUPAC name is; that is called a tertiary amine.3060

It is even possible to have four bonds to nitrogen; but any nitrogen with four bonds is now going to be a charged nitrogen; this is called a salt.3070

It is called a quaternary ammonium salt; this is called tetramethylammonium iodide; ammonium means we have an N+.3082

So it is possible to have even four groups on a nitrogen; but that is going to be an ionic compound.3090

Let's move into compounds that have carbonyls, C-O double bond; the carbonyl is what we call a C-O double bond; there are many functional groups that contain carbonyls.3099

The first one we will talk about is an aldehyde; an aldehyde has the formula here, RCHO; we have a carbon chain on one side and a hydrogen on the other; that is described as an aldehyde.3111

When we have an aldehyde, we are going to find the longest carbon chain that starts at the carbonyl; because there is a hydrogen here, the carbonyl is necessarily at the end of a carbon chain.3127

We start from here and we count our longest carbon chain; the carbonyl carbon is always carbon number 1; we don't say that it is... we don't list it as carbon 1; that is the definition of an aldehyde.3138

What we are going to do is we are going to drop the -e of the alkane name and we are going to add the suffix ?al; it is going to be called an alkanal.3151

The ?AL ending means we have an aldehyde; notice how close that is to an -ol ending we saw for the alcohol.3161

It is very important to have neat penmanship when you are doing these nomenclature problems because the difference of a single letter can mean a right or wrong answer or a totally different structure.3167

As I mentioned, no number is given to indicate where the carbonyl is.3178

Our simplest carbonyl is right here, simplest aldehyde actually; it has a hydrogen on both sides; that is the simplest aldehyde; he is called formaldehyde.3183

If you have ever worked in a biology lab, you are familiar with formaldehyde because this is used as a preservative for dead tissues if you ever dissected a frog or something.3192

That smell that is associated with that is the formaldehyde coming from that preservative solution.3201

This is the simplest case, just a single carbon; that means it is a methane derivative; what do we call this?--we drop the ?e from methane and we add the letters ?al; this is called methanal.3208

Methanal means I have an aldehyde; but almost always this is called formaldehyde instead; it is such a common name.3221

This one where we have the next simplest aldehyde--has just a methyl group on one side; he is known as acetaldehyde; we are going to see this acet common name pop up again and again and again.3228

This acet group or the Ac group represents a carbonyl and a methyl group attached; a COCH3, a carbonyl with a methyl group attached.3240

Because this is the aldehyde derived from this acetyl group, we call this acetaldehyde; we are going to see this name pop up again and again.3252

But that is its common name; what would its IUPAC name be?--longest carbon chain; we have two carbons; one, two; instead of ethane, we have ethanal.3260

Notice there is no number; we don't say 1-ethanal; that is impossible; it is just called ethanal because the carbonyl has to be a carbon number 1.3271

Here is a more complicated molecule; here is our parent; we must number this way; one, two, three, four; this is a butanol.3281

I have already shown that the carbonyl is the higher priority because that is the fact; the prioritization of functional groups is based on oxidation state.3292

A carbon with two bonds to oxygen is going to be a higher than one with one bond to oxygen; any carbonyl is going to beat out an oxygen, be a higher priority.3301

So we are no longer going to name this molecule as an alcohol; we are going to name this as an aldehyde; now we have to deal with this OH group simply as a substituent.3312

What are we going to call it?--we are going to call it a hydroxy group; we are going to call it a hydroxy group.3322

At carbon 4, we have a hydroxy; this is called 4-hydroxybutanol; notice it is not hydroxyl; there is no -l there; it is just hydroxy.3329

This is benzaldehyde; the aldehyde that is derived from benzene; it is called benzaldehyde; what would its IUPAC be?--let's see if we can come up with that.3341

What is our longest carbon chain containing the carbonyl?--there it is, just one carbon; this is actually a methanal derivative.3351

What does it have attached?--it has a phenyl group; you could call this phenylmethanal.3359

We wouldn't find it named that way very often; but it is important to know both the common names and the IUPAC names.3368

A lot of times, if you go to a chemical supplier catalog and you are looking it up, sometimes it is just as likely to be listed in its IUPAC name as its common name.3372

So it is really important to go back and forth; a lot of times, you pick up a reagent bottle in the lab, and it might have the common name rather than the IUPAC name.3383

That is why they are stressed so often; it is a short list of common names; you do want to be familiar with them so that you can communicate properly.3391

How about this last one?--this is carbon 1; it has to be part of our parent chain; this is where our parent chain starts.3401

We will go like this for our longest carbon chain; we also want to include that double bond; one, two, three, four, five.3409

Let's see, it started out as pentane; the carbonyl at carbon 1, the aldehyde makes it pentanal; the double bond makes it pentenal.3418

We have at carbon 2, we have pentene; then ?al at the ending; remember there is no number for the aldehyde.3432

This number 2 must be referring to the location of that double bond; there is nothing else for it to mean; so 2-pentenal.3441

What else do we have?--we have a methyl group at 3 so 3-methyl; 3-methyl-2-pentenal; anything else for this problem?3448

How about stereochemistry?--let's take a close look at that double bond; is there more than one way to draw this?3460

If I had this carbonyl down in this position, would that be the same molecule?--that would be a unique molecule.3467

We have to show only this stereoisomer was drawn; we have to indicate that stereochemistry as part of our name.3474

Can we use the word cis or trans here to describe this alkene's stereochemistry?--no, because we have more than two groups on the alkene; we have to use the E and Z description.3480

Remember what we do is we separate the two carbons of the alkene; our higher priority group on carbon 3, we have a methyl versus an ethyl; ethyl wins.3492

On carbon 2, we have a carbonyl versus a hydrogen; the carbonyl wins; so our higher priority groups in this case, are they on Zis same side?--nope.3504

They are on opposite sides; opposite sides means E; this is the E stereochemistry.3515

A ketone is another carbonyl containing compound; but in this case, now we have an alkyl group on both sides, a carbon group on both sides of the carbonyl.3525

That distinguishes an aldehyde which is at the end of a carbon chain because it has a hydrogen attached compared to a ketone.3535

What complicates things for the nomenclature of a ketone is you have a carbon chain and that carbonyl can be anywhere along the carbon chain.3543

We are back to the case where we find the longest carbon chain bearing the carbonyl carbon.3551

Then we number it from whichever end is closest to the carbonyl so that the ketone gets the lowest possible number.3556

What is the suffix for a ketone?--we drop the ?e ending and we add ?one; it is called an alkanone; we have to put a number out in front or in the middle to indicate where that carbonyl is.3562

Let's see some examples; the simplest ketone we can have is acetone; the common name is acetone; where does the name acetone come from?3577

Remember that acet, acetyl group we had?--the carbonyl with the CH3?--if we made that a ketone, the simplest way to make it a ketone is to add another methyl.3587

That is the ketone of the acetyl; we call it acetone; acetone, of course, very commonly used solvent in organic chemistry.3597

We use that in the lab a lot; sometimes for washing glassware, dissolving organic compounds; awesome solvent.3604

Nail polish remover, most nail polish removers are acetone based to dissolve the organic nail polish.3610

It is also very good for dissolving organic inks and permanent inks; you can use acetone; you can use your nail polish remover to dissolve paints and varnishes and all sorts of things like that.3619

So it is a great organic solvent for dissolving organic things; and it is an example of a ketone because we have a carbon chain on either side of the carbonyl.3629

What would its IUPAC name be?--we have a three carbon chain; one, two, three; it is a propane derivative; we are going to call this propanone.3638

Actually, you don't even need to put a number because there is only one propanone you can have; it must be on carbon 2.3648

But we can put that in there just to be safe; it is 2-propanone; ?o-n-e tells the world we have a carbonyl at the number given.3653

How about this next one?--where is our longest carbon chain containing the carbonyl?--let's go this way, number it from the end closest to the carbonyl.3664

We will number this way; one, two, three, four, five; this is a pentane derivative; this is called pentanone; it is going to be 2-pentanone... 2-pentanone.3676

What else does it have attached?--it has a 3-methyl and a 3-ethyl; sometimes students get confused; when they see them at the same position, they try and group them together somehow in the name.3691

But it doesn't matter if they are at the same carbon or different carbons; we list all our substituents as usual; we alphabetize them and we put them out in front.3702

Ethyl comes first; we have a 3-ethyl; methyl comes next; we have a 3-methyl; very simple; no need to panic if they are attached to the same carbon.3709

What if we have a ring here in a double bond?--how do we incorporate all this?--again, anytime we have a carbonyl in a ring, that carbon must be carbon number 1.3723

Because we are going to name it as a ketone; now we will number in this case in a clockwise direction so that the double bond also gets the lowest possible number.3731

Let's see, six carbons in a ring is called cyclohexane; with the double bond, we call it cyclohexene; with the carbonyl ketone, we call it cyclohexenone; this is cyclohexenone.3742

We have a 2-cyclohexene-1-one; 2-cyclohexene, the 2 refers to the double bond; the 1 refers to the carbonyl.3757

Of course, we can break it up completely and put the 2 right before the ?ene and the 1 right before the ?one; you will see that pretty regularly so you should be able to work with either of those.3772

Remember, we don't have to say that this is cis double bond; it has to be cis in a six-membered ring or a five-membered ring.3785

It is when we get to eight or higher that we have the possibility of having cis or trans.3792

This is another common ketone; it is nice to know the name for it; it is called benzophenone when we have a benzene ring and a phenyl group; benzophenone is the common name.3800

How about the IUPAC name?--how about the IUPAC?--we have a one carbon ketone; that is tough to do; that would be the IUPAC; this is a methanone derivative.3808

What do we have?--we have two phenyl groups; we would call this diphenylmethanone; very strange name; IUPAC names end up that way sometimes.3819

How about this last one?--what if we have two ketones?--just like we can have a diol or a diene, we are going to call this a dione.3834

Here is our longest carbon chain; we number it to give the carbonyl the lowest possible number; run into the first carbonyl; we want the lowest possible number.3845

Seven carbons is a heptane; one carbonyl would make it heptanone; this is going to be 2,4-heptanedione.3859

Just like we stuck that ?e back in there when we did a diol just so it is easier to read, we do the same thing anytime there is a di- or tri- or tetra- prefix we have to put in there.3872

So 2,4-heptanedione; what is this group attached to carbon 6 here; what are we going to call that?3882

Without the oxygen, we would just call this a phenyl group; with the oxygen, we will call it a phenoxy; at carbon 6, we have a phenoxy.3889

If we have a benzene ring, we would name the compound with benzene as the parent; let's take a look at some other aromatic compounds that we might encounter in our nomenclature problems.3904

Benzene is our... again, as I said, would be described as a parent as benzene; but there are several heteroaromatics; these are called heteroaromatic compounds.3918

That means we have aromatic rings with something other than carbon, a heteroatom--nitrogen, oxygen, sulfur, something like that.3932

These are also aromatic compounds; and these are very common names that you should be familiar with.3940

Benzene with a nitrogen replacing one of the carbons is called pyridine; this is called pyrrole, furan, and thiophene; those are real common ones that are good to know.3946

When we go to number a benzene, any of these carbons can be defined as number 1; you make that choice; but in a heteroaromatic, the heteroatom is defined as position number 1.3955

This would be 1; then you number in whatever direction you want; one, two, three, four, five, six.3967

A 2-methylpyridine means that the methyl group is attached on the carbon right next to the nitrogen; same with pyrrole oxygen; the heteroatom is always position number 1.3971

We can have some fused aromatics; it is called fused when some aromatic rings share sides; this guy is called quinoline, for example; two benzene rings attached to another is called naphthalene.3984

Naphthalene is used for moth repellants; moth balls have the same kind of smell that naphthalene does; very aromatic actually; which is where the name came from.3998

Anthracene, when you have three together, and so on; there is lots of them; just showing a few examples here; but when you have a fused aromatic, let me show you the numbering.4010

We start with one of the carbons next to the point of fusion; then we go around one ring--one, two, three, four; then we jump to the next ring--five, six, seven, eight.4019

We do not number... we skip the numbers for these because those carbons already have four bonds to carbon; there is no way you can add a substituent to those positions.4028

That is interesting; a 1,5-dimethylnaphthalene means that we have them in this position and this position.4039

When we have two benzene rings directly attached to each other, that is called a biphenyl; that is a common name--is biphenyl.4048

You may have heard of PCBs; that stands for polychlorinated biphenyls; these were used as insecticides and such.4056

They have all sorts of long lasting environmental impact so those have all been phased out; so PCBs are an important chemical that we are trying to mitigate in the environment.4068

How about the IUPAC for a biphenyl?--what would you call that?--one of these benzene rings would have to be the parent; you just pick one; that is our parent so this is a benzene derivative.4080

That is our parent; what does that benzene have attached to it?--it has a phenyl group; phenylbenzene would be the IUPAC name; phenylbenzene is another way to say biphenyl.4092

It is also nice to be familiar with these names of aromatic compounds because a lot of times the common names for other compounds are derived from the aromatic name.4108

For example, this molecule looks a lot like furan; it has the same skeleton as furan except we've gotten rid of both of these double bonds.4120

So this molecule is described... it has a common name; it is called tetrahydrofuran; because it is the same thing as furan, but we have added four hydrogens to it.4129

This is called tetrahydrofuran; it is abbreviated THF for short; that is a very common solvent that is used in organic chemistry.4140

If you have substituents on the benzene ring, we can number the ring just as usual--one, two, three, four, five, six, and describe it that way.4150

But if you have just two groups on a benzene ring, there is only three possible arrangements for those two groups; so we have common names to describe those possible relationships.4157

If we have two groups that are 1,2 to each other, meaning they are on adjacent positions, we describe that as being ortho di-substituted.4167

We either can put the word ortho in italics out in front or we could just use the letter o- will represent ortho; that describes where the two groups are attached.4176

When they are 1,3, meaning they skip over one carbon, we call that the meta arrangement; when they are 1,4, they are opposite to each other, they are called para.4186

Those are the only three possibilities; if I move the substituent over here now to the other side, this is meta again; and if I move if up here, it is ortho again.4195

So ortho and meta and para are the only three possibilities.4203

When we start getting into aromatic chemistry and learning about the reactions of benzene and benzene derivatives, we are going to be using the terms ortho, meta, para again and again.4205

You are going to eventually become very familiar with these; they can be used to very easily name aromatic compounds if we could do some common names.4217

For example, we have a bromo and iodo here; they are 1,3 to each other; they are meta; we could just describe this as meta-bromoiodobenzene.4226

Benzene is the parent; I have a bromine and an iodine; they are meta to one another; it is a flat molecule; it doesn't matter where you put the bromine, where you put the iodine.4239

As long as you put them 1,3 to each other, you are going to end up with the same compound every time; so it is really easy nomenclature.4248

If you want to number it and do it that way, now you have to be concerned with who gets carbon 1 and so on; for this, we are going to do it just like we did for cycloalkane.4253

We are going to pick the one, the substituent with the lowest... that comes first alphabetically, and we will give that to be carbon 1.4264

The bromo comes first; he would be carbon 1; then we number to give the iodine also the lowest number; we would call this 1-bromo-3-iodobenzene; it is a lot more complicated than just calling it meta.4271

We call it benzoic acid; we will see this... we will talk about carboxylic acids in a moment, on how to name those; but he is called benzoic acid; so this could be called ortho-hydroxybenzoic acid.4290

Ortho describe the relationship; so ortho, meta, and para are good to do that; or you could call it 2-hydroxy as another possibility instead of ortho.4309

But then when you have more than two groups, you can't use the words ortho, meta, and para anymore because those words have no meaning.4324

Here you could say that the two methyl groups are meta to one another; you could still use it as an adjective to describe the relationship of two groups.4332

But this molecule can no longer be described as an ortho, meta, or para molecule because there is more than two groups.4340

In this case, we are going to go back to our nomenclature rules where we find whoever comes first alphabetically, gets the lowest number.4346

We have an ethyl versus a methyl; this is number 1; then we number either clockwise or counterclockwise depending on which will give the lowest number to the next substituent.4353

We will number this way; one, two, three, four, five, six; our parent is benzene; our parent is benzene and what substituents do we have?4364

We have a 1-ethyl; how do we handle our 2-methyl and our 4-methyl?--we group them together and we call it 2,4-dimethyl; so naming a benzene is pretty straightforward.4374

There is never any stereochemistry to worry about because it is a planar molecule; every carbon can only have one substituent on there so that simplifies things.4389

But when we get into the heteroaromatic compounds, then that nomenclature can expand our vocabulary a little bit.4399

There is also a lot of common names that exist for simple compounds and simple mono-substituted benzenes; these are very useful to be familiar with.4410

In fact, you may even be required to learn them; that is quite likely.4422

For example, methylbenzene, when we just have a methyl group on a benzene ring, that is known as toluene.4426

Toluene is a very nice solvent that we use for organic chemistry; it is preferred over benzene because it is not carcinogenic; so that has some uses as well in the lab; that is called toluene.4432

It is called phenol when you have an OH on a benzene ring; it is a very special type of alcohol when you have an OH attached to a benzene ring; it is called phenol.4447

Notice the ?ol ending is just like we have for an alcohol; but it is not called benzol; it is called phenol.4457

I see I have a little typo here, sorry... this is called anisole; we will copy and paste here... it is called anisole when we have a methoxy group here instead of a methyl; it is called anisole.4465

When we have a nitrogen, this is called aniline; when we have two methyl groups, those compounds are known as xylenes.4477

Notice that this bond is pointing to the middle of the ring; that is because there is three possible places to put that second methyl; we could put it ortho or meta or para to the first methyl.4486

So there is three isomers of xylene that exist; ortho-xylene, meta-xylene, and para-xylene; together we call these the xylenes if maybe you had a mixture of those dimethyl benzenes.4498

Finally, it is called a cresol when you have an OH and a methyl; so some of these very simply substituted benzene rings have common names.4511

You are guaranteed to encounter them so being familiar with them is really going to pay off.4521

For example, this one has a methyl and an OH; they are para to each other; so this molecule can be called para-cresol.4527

Or it could be called para-methylphenol; para- or maybe 4-methylphenol; or it could even be called hydroxymethylbenzene; all sort of names you can come up with for aromatic compounds.4536

A lot of times, we get to a situation where there are more than one acceptable names; you definitely want to check with your instructor to see what their expectations are.4555

This is an interesting aromatic compound; it is a toluene derivative; see the word toluene in here?--it is called trinitrotoluene; here is the toluene; that means methylbenzene.4565

It is a derivative of that because it has these four nitro groups on it; guess what this is called?--trinitrotoluene; this is called TNT.4576

So this is the structure of dynamite; that name, when you explore that name, it comes from the common name of this aromatic derivative; very explosive.4585

Let's take a look at carboxylic acid; this is another carbonyl containing compound; a carboxylic acid has this formula here.4597

We have a carbonyl and an OH group attached; this is no longer now an alcohol; it is no longer a ketone or an aldehyde.4607

This arrangement of atoms is combined together, taken as one, as one functional group, called a carboxylic acid; it will be named as a carboxylic acid.4618

It must be at the end of a carbon chain because it has a... the oxygen stops the carbon chain; once again, like the aldehyde, the carboxylic acid defines the beginning of the carbon chain.4629

We number from that point onward to decide how long our parent chain is; but we never have to designate that the carboxylic acid group is a carbon 1 because that is the definition; it must be.4642

What we are going to do is we are going to drop the ?e ending of our alkane, and we are going to add two words; we are going to make the first word ?oic as our suffix.4655

Then we are going to do a space and then the word acid; so we are going to have an alkanoic acid; that tells us we have a carboxylic acid.4663

Again, a few common names here; our simplest acid with just one carbon is known as formic acid; this has a real interesting history for this name.4672

This is the acid that can be extracted and isolated from ants; if you take some ants and smash them up and extract the organic or acidic component, you will be able to isolate formic acid.4682

The Latin word for ant is formica; this was called formic acid since it is the acid derived from ants.4694

Again, very commonly called, typically called formic acid; but its IUPAC would be derived from the fact that it is just one carbon; it is a methane derivative.4704

But instead, we call it... we drop the ?e ending of methane and we add the suffix ?oic; then we make a space; then we put the word acid; so this is called methanoic acid.4714

Methanoic acid means I have a one carbon carboxylic acid; I have one carbon and, on that carbon, I put a carbonyl and an OH; that is what makes it a carboxylic acid.4728

This guy is called acetic acid; why do we call it acetic acid?--there is our acetyl group, our carbonyl with a CH3; when we make it a carboxylic acid, we will call that acetic acid.4739

Acetic acid is actually the acidic component of vinegar; that is what makes vinegar smell and taste that characteristic smell; it has that bite, that acidic bite.4749

Most often, we call it acetic acid; we abbreviate it AcOH by the way; a lot of times you will see that abbreviation, the Ac group, representing the carbonyl with the CH3.4762

What would its IUPAC name be?--what is its IUPAC name?--here we have a one, two carbon chain; it is an ethane derivative; we are going to call this ethanoic acid; two words, ethanoic acid.4773

How about a more complex carboxylic acid; here is our parent; we have to start at the carbon bearing the carbonyl.4788

Both of these functional groups... both of those atoms are going to be incorporated into our parent name; one, two, three, four, five; five carbon carboxylic acid is called pentanoic acid.4797

I do not need to say 1-pentanoic acid; there is no such thing; it has to be at carbon 1; what else do I have on carbon 4?4812

Now we take our substituents, we alphabetize them, we put them out in front as usual; we have a 4-methylpentanoic acid.4819

Benzoic acid is what we call the carboxylic acid with a benzene ring attached to it; benzoic acid; that is good to know.4829

We also need to know, now that we have learned about carboxylic acids... remember the priority of functional groups depends on oxidation state.4837

A carboxylic acid is the most oxidized we can have for a chain; so that is going to be our highest priority.4845

That means we are going to have to find a name for any other carbonyl that is in the structure; an alcohol group is also going to be named as a hydroxy.4855

In this example, we have three functional groups; we have an alcohol and we have the ketone and we have a carboxylic acid; all three of those functional groups exist.4865

Which is the one that is going to define the IUPAC name?--the highest priority is the carboxylic acid.4873

This is our parent chain; we are going to number to give the carboxylic acid carbon 1; so this actually incorporates the carboxylic acid; if you want to circle that, that is even more clear.4881

This is butanoic acid; no numbering there to put in our name; this is butanoic acid; we know that the alcohol group is the... the OH group is going to be called hydroxy.4894

What is a carbonyl going to be called?--it is not a keto group anymore... because it is not a ketone anymore; it is going to be called an oxo group; oxo means you have a carbonyl; not oxy; oxo.4907

Who comes first?--hydroxy comes first; we will list that first; 4-hydroxy-3-oxo-; and this is all one word, 3-oxo-butanoic acid.4919

Oxo is what we will call a carbonyl; hydroxy is what we call an OH; both of those are lower priority than the carboxylic acid.4932

Remember you can only pick one suffix; every other group then becomes named as a substituent listed out in front.4940

Now that we know how to name a carboxylic acid, let's take a look at some related functional groups.4950

These are called carboxylic acid derivatives because they can all be derived from a carboxylic acid; they are all related to a carboxylic acid.4954

We can have a... what we are doing is we are varying the group attached to the carbonyl; a carboxylic acid had an OH; and we are varying the group.4963

What all these have in common is that each of these groups has lone pairs; a halogen, remember X represents a halide.4975

A halogen has three lone pairs; oxygen has two; oxygen has two; nitrogen has one; you can see that the general structure for these carboxylic acid derivatives looks like this.4984

It has something attached to the carbonyl with a lone pair; we are going to describe this as a leaving group; this is actually going to define the reactivity of carboxylic acid derivatives.4996

We are going to see that being involved in reaction mechanisms as a leaving group; we will get to that down the road.5009

The nitrile doesn't look a lot like a carboxylic acid; but it is related just like this.5016

All of these have a carbon with three bonds to heteroatoms--an oxygen, and an oxygen, and something else with a lone pair.5023

This also has three bonds to a heteroatom, three bonds to nitrogen; so even though he looks a little different, he is also related; this is called a nitrile.5030

It is called an amide when you have a nitrogen attached to the carbonyl; it is called an ester when you have an OR group; so it is no longer an ether.5040

An ether means I have R... alkyl, and then an oxygen, and then an alkyl; that makes it an ether; when one of those groups is a carbonyl, we call it an ester.5048

When both groups on either side of the oxygen are carbonyls, we call that an anhydride; if we have a halogen attached to the carbonyl, we call it an acid halide.5058

One by one, we will go through the nomenclature for these; but I just wanted to put it in context, showing you that each of these is related to the parent carboxylic acid.5067

Because if you know how to name the parent carboxylic acid to which these are all related, then it is going to be a little easier to name the particular derivative we are looking at.5078

We will start with an acid halide; what we will do is we will find the longest carbon chain that starts at the carbonyl carbon.5090

We have to start at the carbonyl carbon; it is always carbon number 1; so we don't even mention that, just like a carboxylic acid.5097

What we do is we drop the ?e and we add the letters ?oyl; we are going to get an alkanoyl halide; then we are going to list the halide, making it an acid halide.5104

For example, again coming back to our common names, a common name for this molecule would just be called acetyl chloride; this is the acid chloride derived from acetic acid; we call that acetyl chloride.5118

What would its IUAPC name be?--we have a two carbon chain; one, two; it is an ethane derivative; but we are going to drop the ?e and add ?oyl; ethanoyl.5131

We make a space; this is another two word name; and we list whatever halogen happens to be here; it is most often chlorides that we deal with.5144

You will see that nine times out of ten; but it is possible to have other halides as well; this is called ethanoyl chloride.5151

Again, pay very close attention to these details; if you drop the -o, that name doesn't exist.5157

If you drop the -y, then you are talking about an alcohol and you've made a mistake; so be very careful; -o-y-l as our ending for an acid chloride; ethanoyl chloride.5163

How about this one?--what is our longest carbon chain?--we can include the acid chloride if you want; there is our longest carbon chain.5174

Be very careful, when you have a cyclic substituent, that you don't include the carbon of the ring as part of your carbon chain; you can't count it for both.5183

Here it is the group that is attached to the carbon chain; this is a one, two, three, four carbon chain.5192

What is a four carbon acid chloride called?--it is called butanoyl; butanoyl, space, chloride; butanoyl chloride.5201

What is attached to carbon 4?--it would be called cyclopropane if that was the parent; but as a substituent, we call it a cyclopropyl group; cyclopropyl; 4-cyclopropyl... 4-cyclopropylbutanoyl chloride.5213

How about this one?--what is our parent carboxylic acid we have?--if this were a carboxylic acid, if this were an OH, we would call that benzoic acid.5235

We are going to call this benzoyl bromide; that is our parent; kind of related to benzoic acid; we call it benzoyl bromide.5246

What do we have attached?--we have another bromo substituent; it is in the meta relationship; we could just call this meta... I'm sorry, 3-bromobenzoyl.5262

Or we could call this meta-bromobenzoyl bromide; those would both be good names for this compound.5275

Benzoyl bromide means I have something that looks like benzoic acid--a benzene with a carboxylic acid; but it is a bromine there instead of an OH.5281

An anhydride name is also very closely related to the parent acid to which it is related; let's just take a look at the structure of the anhydride.5293

If you think of the anhydride structure, it kind of looks like we have two acid pieces that have come together.5302

If you think of these two acids, what would they have to lose to result in the structure shown?--it looks like we've lost the OH on one and the H on the other; so we lost H2O.5306

If we lose H2O, these two carboxylic acid groups can come together and form the structure, the product, shown; that is why it is called an anhydride--you've lost water.5323

In the simplest case, where these two alkyl groups are the same, then we simply name it as the parent acid; but instead of calling it an alkanoic acid, we call it an alkanoic anhydride.5336

For example, this came from acetic acid; this is related to acetic acid; we could call it just acetic anhydride as its common name; this is acetic anhydride.5347

The abbreviation for that is Ac2O; it has two Ac groups, acetyl groups, attached to the middle oxygen; but what is its IUPAC name?5362

For the IUPAC name, let's take a look at just one half and name that acid; that acid would be a two carbon acid; it is an ethane derivative; this would be ethanoic acid.5372

Ethanoic acid, so now we call this ethanoic anhydride; we just replace the word acid with anhydride.5385

What if it is not a symmetrical anhydride?--we call those mixed anhydrides, meaning it came from two different acids if you will; then all we do is list both acid groups before the word.5395

What carboxylic acid did this come from?--one, two, three; this was a propanoic acid; and over here, this came from one, two, three, four; this came from butanoic acid.5408

When we brought those two acids together, if we think about that as a way to come up with the IUPAC name, we just list them alphabetically; this is butanoic propanoic anhydride.5424

That is so simple; it seems like that would be a common name but that is actually how you do the IUPAC name; you just list the two halves to the anhydride.5442

If you had a diacid like this... again, there are common names associated with most of these diacids; oxalic, molaic, succinic, glutamic, and so on.5451

These are... you might come across these common names; it is going to be useful in this case; this is called succinic acid.5465

Let's try and do the IUPAC for this one just to see if we could do that; it is a four carbon chain; one, two, three, four; one, two, three, four; it is a butane derivative.5472

If we had just one carboxylic acid, we would call it butanoic acid; because we have two, we call this butanedioic acid.5484

Remember we add that -e back in so we can fit the d-; butanedioic acid; remember ?oic acid is our suffix; just like we would have diol or dione, we could have dioic acid.5498

If we take succinic acid and lose water from this structure, what can happen is the two carboxylic acid groups can come together to form a cyclic anhydride.5512

In fact, we are going to be seeing the mechanism of this eventually; what we will get is a five atom ring that looks like this.5522

That is an anhydride structure; you have the oxygen with a carbonyl on both sides; this is definitely anhydride.5532

The way we describe this anhydride is by referring to the parent acid it came from; this is called succinic anhydride; succinic anhydride.5539

Believe me, you don't want to know the IUPAC name for this; this is really how most of us refer to it--is just by its common name.5554

Some of these cyclic anhydrides you will see are named... their name comes from the common name of the parent diacid.5561

I think esters are a little tricky to name because an ester has two R groups; it has the R group that is attached to the carbonyl side; and it has the R group that is attached to the oxygen side.5572

It is possible to mix those up when you are first learning; but let's see how we can keep them straight.5587

I think the way to do that is to immediately take a look at an ester structure and immediately break it down to--here is the parent acid to which this ester is related.5594

This looks like the carboxylic acid group; this is the parent chain; this is the parent right here; and this is just an additional alkyl group that added on to make it an ester rather than a carboxylic acid.5605

What we are going to do is we are going to drop the suffix ?oic acid and instead we are going to use the suffix ?oate to tell everyone that it is an ester instead.5621

What we are going to do then is we are going to list in front, we are going to list the alkyl group that makes it an ester.5632

Ultimately, our name is going to be an alkyl, space, alkanoate; the alkyl group is this part that is attached to the oxygen; the alkanoate is our parent just like the alkanoic acid was.5640

A lot of words; let's just look at some examples and I think it will make a lot more sense.5653

For example, this is acetic acid; we call that acetic acid; its IUPAC is ethanoic acid; ethanoic acid is related to this ester because we still have the components of ethanoic acid.5658

We still have the two carbons and the carbonyl with the oxygen attached; but it is now an ester because there is an R group here; there is an alkyl group instead of a hydrogen.5675

Instead of calling it ethanoic acid, we are going to call it ethanoate; we are going to use the suffix ?oate to indicate that it is an ester.5686

There are a lot of ethanoate esters you can have; all sorts of different esters you can have because this alkyl group can vary.5697

Which ester do we have in this case?--we have the methyl ester; we just list the word methyl out in front; this is methyl ethanoate.5704

Its common name for an ester would be acetate, for this acetyl group as an ester; we would call this methyl acetate; that is a very good common name to know as well.5715

Let's try another example; this is an ester because we have a carbonyl with an OR group attached.5726

How do we name it?--first we really want to identify the parent acid; here is the carboxylic acid that is related to the ester of which it is a derivative.5735

How would you name that acid?-it is a three carbon acid; one, two, three; it is a propane derivative; you would call it propanoic acid.5747

We are going to name this propane ester, we are going to call it propanoate; we are going to call it propanoate; drop the -e and add ?oate; so it is propanoate.5757

Which ester is it?--what do we have over here?--this is an ethyl group making it an ester; we call this ethyl, space, propanoate; esters are named as two words; ethyl propanoate.5770

Let's try another one; this one is a little more complex; here is our parent ester, a parent chain because that came from the carboxylic acid.5786

Remember the carboxylic acid is the highest priority functional group; that must be our parent; we are going to number from there; one, two, three, four, five.5800

It is pentane that we started with originally; but we have a double bond here; so this is actually going to be called 4-pentene... 4-pentene.5811

But it is not pentene; it is an ester; what suffix do we use for an ester?--we drop the ?e and we add ?oate; this is called 4-pentenoate.5823

Remember double bonds and triple bonds, we can always sneak into our name because they affect a different part of the parent name.5836

They just replace this -a with a -e or a ?y; so we can always incorporate those in there.5844

4-pentenoate; the 4 must be referring to the double bond because the ester needs no number; the ester must be on carbon 1; this is good so far.5849

Now we ask what is the alkyl group that is making it an ester?--we come over here; this is a little complicated now.5859

This is a complicated group; we just have to recall that the carbon point of attachment is defined as carbon 1.5868

What is our longest carbon chain?--that is it; it stops right there; this is a methyl group... this is a methyl group.5878

But it is not just methyl; what does it have attached to it?--it has a phenyl group attached to it; we are going to call this phenylmethyl; phenylmethyl, space, 4-pentenoate.5884

Phenyl methyl, I just realized was on the previous slide; I forgot to mention that with the aromatic nomenclature.5900

This group, there is a common name for a benzene ring with an extra carbon; it is called a benzyl group.5908

The carbon next to a benzene ring is described as a benzylic position; this is called a benzyl group.5917

A common name for this might be benzyl 4-pentenoate; that would be the common name; but the IUPAC name is phenylmethyl, space, 4-penteneoate.5922

Once we figure out how to name esters, it turns out we can use that same nomenclature to describe carboxylate salt; it is called a carboxylate when you have an RCO2-.5932

RCO2-; it is like a carboxylic acid that has a lost a proton; it has been deprotonated; we call that a carboxylate; we use the exact same rules to name this.5945

Again, we look at the parent acid from which it is derived; this would benzoic acid; now we call it a benzoate with the negative charge there; this is called benzoate.5958

In this case, it is not an alkyl group that is making it the benzoate; it is a sodium; we just list the counter ion in front; sodium benzoate.5974

So you could have sodium acetate; you could have... you could have lithium pentenoate here if this was the lithium salt.5983

You will come across those names as well when you are looking at carboxylate salts; so that is good to be familiar with that nomenclature.5995

Amide nomenclature goes as follows; an amide is when you have a carbonly with a nitrogen attached.6005

It is no longer an amine; that carbonyl now makes it an amide; again all the difference that a single letter makes.6013

Once again, this must be carbon 1; we count the carbons from that point forward, find the longest carbon starting from there.6020

What we for our name is we drop the ?e and we add the entire word -amide; so we get an alkanamide as the parent; the suffix is the entire word ?amide.6028

If we have groups attached to the nitrogen, as usual just like we did for the amines, we will identify those as N-substituent.6040

Again let's look at the very simplest example; acetamide is what we call it when we have the acetyl group that is turned into an amide.6049

What would its IUPAC be?--we just have a two carbon chain; that is an ethane derivative; this is an ethane derivative.6058

We drop the ?e of ethane and we add the word ?amide; ethanamide; ethanamide would be the name of this compound.6066

This guy is called N,N-dimethylformamide; let's see if that name makes sense to us; where does formamide come from?--it comes from formic acid.6077

Formic acid was our simplest carboxylic acid; that is the one that came from ants; like formaldehyde.6087

That is where the name formaldehyde comes from--is related to formic acid; that is the aldehyde with this one carbon structure.6096

This is the one carbon amide; we call that formamide; it is N,N-dimethyl because there is two methyls on the nitrogen; that is N,N-dimethyl.6103

Formamide is its common name; that is a good one to know because it is abbreviated DMF; this is a solvent that is used in organic chemistry; so you might see the abbreviation DMF occasionally.6112

How would this name change if we wanted to do IUPAC?--because formamide is not an IUPAC name; what would the IUPAC name be?6125

This is just a one carbon chain; that means it is a methane derivative; we would just call it methanamide; we drop the ?e of methane and add -amide.6134

It would still be the same beginning; it would still be N,N-dimethyl; but it would be N,N-dimethylmethanamide rather than N,N-dimethylformamide.6145

Let's try an example here; we have a lot going on; let's identify the highest priority functional group and let that set which one will be our carbon chain, our parent chain.6156

Right here, this is the carbon that is most important because that has the carbonyl and the nitrogen; that is going to be our parent chain.6170

This is the longest carbon chain from that point on... from that point on; this is one carbon, two, three, four; this is a four carbon chain.6177

What is that going to be called?--it looks like a butanamide; butane derivative; now butanamide; butanamide means that I have an amide at carbon 1; carbonyl plus a nitrogen.6188

What else do we have here?--we have an N-ethyl; a lot of times the N substituent is listed all the way out in front of the molecule before the carbon substituents are sometimes mixed in.6203

It maybe depends on the publisher; this is called an N-ethyl because it is an ethyl group on the nitrogen.6216

What is this group called?--what is that group called?--if you could use common names, there would be a simple name for this; it looks like the isopropyl group with an oxygen.6224

We could call this an isopropoxy; we could call it an isopropoxy but that is not IUPAC; let's figure out an IUPAC name for this substituent.6235

Remember the point of attachment has to be carbon 1; this is a two carbon chain; this is actually an ethyl group.6248

But it has a methyl at carbon 1; it is a 1-methylethyl is how we would call an isopropyl group; but it also has the oxygen here; we are going to call this 1-methylethoxy.6257

It is actually an ethoxy group; here is the ethoxy; but it is a derivative of that because, at the 1 carbon, it has an extra methyl; this is called a 1-methylethoxy.6274

Like I said, usually the end substituent come first; it turns out that is also alphabetically correct; so we are in luck here.6286

This is N... I'm sorry, N-ethyl is listed first; then at position 2, we have this big whole group; -2- and then parentheses to keep it all together; a 1-methylethoxy.6295

Our nomenclature rules are building on one another bit by bit so that we can eventually name some pretty complex molecules.6313

Finally let's take a look at a nitrile; a nitrile is RCN; so we have that C-N triple bond group; that is also a carboxylic acid derivative.6323

It also indicates that the first carbon, the carbon of the nitrile, defines the beginning of the carbon chain; just like the carboxylic acid derivatives.6336

I typically don't bother trying to distinguish who has a higher priority among the carboxylic acid derivatives.6346

Assuming you have just one of those, that is going to be higher priority than any other functional group we saw.6353

But if you want to get into the details of how do they rank amongst each other, you can look that up and have those challenging problems.6359

We are going to find our longest carbon chain; then we are going to not even drop the ?e of our alkane name; we are just going to add the word nitrile.6367

That must be the simplest suffix to remember because you don't even have to drop the ?e; you just stick the word nitrile on the end; that magically turns your carbon chain into a nitrile instead.6377

This is a tricky one to recognize; this molecule's common name is acetonitrile; but you can see that it is the two... just like... remember this was the acetyl group.6391

It is the two carbon derivative that we had all these names for--acetone, acetic anhydride, acetic acid, acetamide, etc; the two carbon nitrile is called acetonitrile.6402

Again, even though it looks quite different from the others, this is a common name; another very common solvent to use in organic chemistry; so you might see acetonitrile around.6415

But what would the IUPAC name be for acetonitrile?--our longest carbon chain is right here; it is just a two carbon chain; that is an ethane derivative; he is called ethanenitrile.6423

That is it; you just add the word nitrile to the end; ethanenitrile; notice that the cyano group is part of the carbon chain; it is part of the carbon chain; that is important here.6435

You might at first glance, at the way this is drawn, you might think that your parent is down here and then you have some kind of cyano group attached.6445

It would be called a cyano group if we had some higher priority functional group here; but we don't; the cyano group is the highest priority functional group.6453

So that is actually part of our parent; our carbon chain has to start right here; from that point, our longest carbon chain goes in either direction because it is symmetrical.6460

But this must be our parent; one, two, three, four, five; this is actually a pentane derivative; we are going to call that pentanenitrile... we are going to call that pentanenitrile.6470

What do we have attached?--we have some methyl groups; and then we have another one of these complex substituents; let's see if we can tackle that; how do we start?6488

We look at the point of attachment to the parent; we define that as carbon number 1; then we number from there; what is the longest carbon chain starting from that point?6496

The longest carbon chain would be one, two, three; it doesn't matter if you number up here or down here; one, two, three; this is actually a propyl group.6506

It is a propyl group; but that does not... it is not just a propyl group; it has a methyl at carbon 1 and a methyl at carbon 2.6514

What are we going to call this?--we are going to call this a 1,2-dimethylpropyl; it is a 1,2-dimethylpropyl.6524

We are going to put that whole big thing in parentheses because it is all one group that is attached to the parent; it is attached to the parent at carbon 2; it is called a 1,2-dimethylpropyl.6537

I listed that last... I listed that second because the other groups we have are methyl groups and methyl comes before methylpropyl.6551

Remember we ignore the di- for alphabetizing; but methyl comes before methylpropyl because methylpropyl is all one word.6559

At 3, we have a methyl; at 4, we have a methyl; we are going to call that a 3,4-dimethyl.6565

Make sure you give yourself plenty of room when you are doing these nomenclature problems because the names can get really big and long.6572

Make sure you draw your letters nice and large and clearly so that there is no mistaking an ?e for an ?a or ?o because, as you can see, that makes all the difference in getting these problems correct.6578

Let's do this last one; this looks like a doozy doesn't it?--we have an aldehyde; we have a ketone; we have an alkene; we have a nitrile; which of these groups wins?6591

Remember it is based on the oxidation number; any carbon with three bonds, two oxygen, or two heteroatoms is going to win; so this is our highest priority functional group.6604

He will be named as a nitrile; all the other groups are just substituents attached to the parent because we can only have one suffix.6615

Let's include that nitrogen in our parent so we make sure that we have already accounted for it; one, two, three, four, five, six, seven; seven carbon chain is heptane.6624

Double bond makes it heptene; we have a 2-heptene... 2-heptene; how do we make this a nitrile?--we just add the word ?nitrile to the end.6643

That means the first carbon of that seven carbon chain is part of a carbon-nitrogen triple bond; we have 2-heptenenitrile.6659

What do we have at carbon 4?--remember we would name it as a ketone if that was the highest priority functional group; then we would use the ?one suffix.6667

But now it is a substituent; he is not called a keto group; that is a really common mistake; instead he is called an oxo group; any carbonyl is called an oxo substituent.6679

We have an oxo at 4 and we have an oxo at 7; again now an aldehyde or a ketone, they are both the same thing, they are just carbonyl substituents.6689

We are going to call this a 4,7-dioxo; just like we would have a dimethyl or a dichloro or a diphenyl; this is called a dioxo; excellent.6700

That accounts for all our functional groups; but there is one last thing that is missing here; is there any stereochemistry that might be relevant?6717

Look at this double bond; how would you describe that stereochemistry?--both of these groups are pointing in the same direction, on the same side.6724

So yes, actually the cis isomer has been shown here; we need to show that; it is okay to just call it cis here because you have only two groups.6731

It is also okay to call it Z; it is never wrong to use the E and E nomenclature; if you want to get used to doing that, then that is a good idea.6739

Z means these two groups are on Zis same side; in fact, in some cases, sometimes you will learn that unless you have two identical groups, they prefer not to use cis but to use Z instead.6750

But if you have two hydrogens, it is quite unambiguous if you call it cis; that is why they accept it as an okay name as well.6763

That wraps it up for all the different functional groups that we are going to learn the nomenclature for; I realize nomenclature is not the most fascinating topic in the world.6774

But just like you have to memorize the alphabet before you can learn how to spell and learn how to put words together, you do need to memorize these nomenclature rules.6782

It is one of the few things in organic chemistry that really you do have to memorize; you do have to know the names; because we need to be able to communicate.6791

If every time you see one of these IUPAC names, it is a huge roadblock for you when you are going to have to spend ten minutes figuring out what it means; that is going to be very costly.6799

So it is worth to maybe put together some flashcards and start to get familiar with these; of course, the more problems you work on, the better you are going to get.6808

And the more it is going to be second nature to you just like the alphabet is now that you have had plenty of time to practice that.6817

Hope to see you again soon; thanks for coming.6824

Welcome to Educator.0000

Next we are going to talk about alkenes; there is actually a couple lectures we are going to have on alkenes.0002

We will start by looking at the structure of alkenes and the synthesis of alkenes; an alkene is defined as a molecule containing a carbon-carbon double bond.0006

When we look at the structure of a carbon-carbon double bond, we remember that it is a planar structure; each of these carbons has sp2 hybridization; that requires a trigonal planar geometry.0018

Connecting the two carbons are two bonds; we would describe these bonds as one σ bond and a π bond.0041

The way a σ bond is made is we join the hybrid orbitals on each carbon; this is an sp2 overlapping with an sp2 hybrid orbital.0055

For the π bond, the way we form a π bond is have a p orbital on each of these carbons and that is overlapping to form the π bond.0070

This model has a very nice picture of what an alkene looks like; here in the gray, we see the σ bonds; they are all planar.0077

On each carbon, each sp2 hybridized carbon, we have a p orbital; remember that is the dumbbell shape type orbital.0089

It is overlap of the top half of that orbital and the bottom half of that orbital that comprises a π bond; a π bond is a cloud of electrons above and below the plane of the molecule.0095

We could describe the π bond as being the overlap of a p orbital and another p orbital.0107

If we want to do a 3D sketch of this molecule, one way to draw it is to draw all the σ bonds in the plane; that is certainly an easy way to draw all the σ bonds.0114

Trigonal planar means these bond angles are about 120 degrees; but then our 3D sketch should also show the π bond and where the π bond is.0127

If the molecule is in the plane, then the π bond is perpendicular to that plane, sticking straight out and straight back.0135

We can sketch that as maybe a wedged and a dashed lobe, a wedged lobe and a dashed lobe, one on each carbon; then we can show some overlap at the front and back.0142

What we need to show is that the p orbitals are perpendicular to the sp2 plane; or orthogonal; we need to show that they are perpendicular.0153

Actually another way that we can draw this molecule is to draw it so that the p orbitals are in the plane, the π bond is in the plane.0164

We could draw our p orbitals nicely shaped and our π bond, again, top and bottom shows overlap.0174

But when we do that, what happens to the hydrogens?--if there are hydrogens here or the σ bonds, whatever they are attached to?0181

The σ bonds are now--these two are projecting out toward you and the others are pointing back; the rest of the molecule is now perpendicular to the plane.0186

We need to show them as a dash and a wedge; when we go to draw this, what we do is we tilt it just a little bit so you can see the wedged bonds and the dashed bonds as well.0197

You want to make sure you use the same angle though; either tilt it up a little or tilt it down a little; you don't want to try and twist the molecule and draw it like that.0206

Here I drew both wedged bonds pointing down toward the bottom of the page; that would be an acceptable drawing.0215

Either of these 3D sketches look really good as a way to draw an alkene, a carbon-carbon double bond.0222

What else do we know about π bonds?--we've seen a little picture of them; we should also know that π bonds are higher in energy than σ bonds.0229

The p orbitals are higher energy, that are coming together to form those π bonds; that is going to make them more reactive.0238

We are going to see lots and lots of reactions that break π bonds and all sorts of reactions of alkenes that we will talk about in our next lecture.0245

Also, we want to note that π bonds are electron rich; they are a good source of electrons; those two electrons are fairly loosely held; they are fairly available.0255

What are some things we associate with electron rich species?--they could be a nucleophile; what does it mean to be a nucleophile?0263

That means you can react with an electrophile; we will see a lot of reactions of alkenes with electrophiles.0269

It could also be a base; what does it mean to be a base?--an acid is something that donates a proton; a base is something that accepts a proton; in other words, it can be protonated.0278

Many of our mechanisms for alkenes, we are going to start with reaction with an acid and our very first step is protonation of that π bond.0290

We should also review some things that make alkenes stable; some of the things we've already seen before; but some other things that we want to know.0299

One feature is that an alkene is more stable if it has more alkyl groups attached to it; more carbon groups, the better.0309

If we take a look at this carbon-carbon double bond, it has a possibility of having four alkyl groups attached.0318

In this case, we have one, two, three of those positions filled with alkyl groups; we describe this as being tri-substituted.0328

We know that is going to be more stable than... for this alkene, our carbon-carbon double bond is here; how many alkyl groups do we have attached?0340

We have just one and two carbon groups attached here; this we could describe as di-substituted; it is less stable; a di-substituted alkene is less stable than a tri-substituted alkene.0348

Another way you could describe these two particular alkenes is you could say that this one is internal to the carbon chain while this one is terminal.0360

The second one is terminal; it is at the end of a carbon chain; that is also not a good place for a double bond to be.0369

We can look for features like that to determine whether or not our alkene is more or less stable; we can also look for the relationship of the groups that are attached on a double bond.0374

For example, if we have a di-substituted double bond, we know that the... we could have a cis arrangement or a trans relationship.0388

This cis one is when they are both facing on the same side of the double bond; trans is what we call it when they are on opposite sides of the double bond.0397

The trans is more stable; that is because by forcing these two alkyl groups in the same direction, they have some steric strain, some steric hindrance there.0404

There is some more crowding; cis is going to be less stable than trans.0417

The exception is when we have a double bond within a ring; if we take a look at something like this--cyclopentene or cyclohexene.0421

In order for these two groups on the double bond to connect in a ring, they actually must be pointing in the same direction on the double bond; it has to be the cis conformation.0432

For small rings like these, it is cis only; it is possible to have trans if you get to a larger ring.0442

It is possible if it is seven; that is still pretty unstable; but as soon as you get to an eight-membered ring, you can have...0450

For example, cyclooctene can either be cis-cyclooctene where we draw our eight-membered ring and we put a double bond in there; now we see that those two alkyl groups are cis to each other.0456

Or it could be trans-cyclooctene; when we draw this, it gets a little more complicated because the molecule ends up being twisted because we put one alkyl group on the opposite side of the other.0475

But if the ring is large enough, then those two alkyl groups can still come around and be connected; so it is possible to have cis or trans.0488

Let's see, we have one, two, three, four, five, six, seven, eight carbons there; this is kind of passing behind the double bond; this is the 3D sketch.0495

Try building these models and you can get a feel for what cis and trans looks like.0505

But if it is a smaller ring, if we just have cyclopenetene or cyclohexene, we do not put the word cis as part of that name because it is impossible to have trans there.0509

Cis is assumed; and cis is going to be much more stable because it is a more stable ring that way.0517

If we happen to have more than one π bond, something we can look to for stabilizing the π bond is conjugation.0524

Conjugation is when we have a π bond; then a σ bond; and then another π bond; if our π bonds are separated by just one single bond, that is going to be a good relationship.0531

Because what happens... this is an example where we have a double bond, single bond, double bond; this is conjugated.0544

And this is not conjugated because separating these two π bonds, we have an sp3 hybridized carbon.0556

We take a look at the p orbitals here; we know that this π bond is a p orbital and a p orbital; and this π bond is a p orbital and a p orbital.0565

Look what happens when they are conjugated; we have p orbital, p orbital, p orbital, p orbital.0573

We end up getting delocalization of those π electrons over all four of these atoms; what we get is resonance stabilization.0578

We will be looking more at these conjugated systems down the road; but for now, we want to see it.0588

If we ever see it, we want to recognize that that is something is being stable; and that is more stable than having this π bond totally isolated from this π bond.0594

And there is no relationship between the p orbitals from one to the other; this is going to be more stable and the non-conjugated is going to be less stable.0603

That is something else we can look to stabilize a double bond if we can.0618

Finally, if we have something like this; this is an example of a bicyclic compound; it is a bridged compound; we have two rings in this molecule.0624

When you have a double bond in a bicyclic compound like this, we are going to look for the bridgehead carbons.0636

The bridgehead carbons are defined as those that are shared between multiple rings, where the two rings come together and there is a juncture; this is bridgehead carbon; this is a bridgehead carbon.0642

When you look at this molecule from the perspective of the bridgehead carbons, you see that these carbons are connected by three bridges.0652

Here is a one carbon bridge; here is a two carbon bridge; here is another two carbon bridge; we call these bridgehead carbons.0660

It turns out that having a double bond attached to one of those bridgehead carbons is very unstable; this molecule is not going to want to exist; it is going to be highly unstable.0666

It is okay to put a double bond somewhere else in the bicyclic compound; it is okay to have a bicyclic compound; over here would be okay because the double bond avoids the bridgehead position.0678

This is known as Bredt's rule; you might come across some examples where you are looking for where to put a double bond as a result of a reaction; we want to avoid bridgehead carbons.0689

This alkene stability can be measured; this is something that we can find some evidence for, these various rules that we are looking at.0701

By doing a reaction called a hydrogenation reaction and measuring the heat of that hydrogenation.0709

We are going to be talking about that in our next lecture when we look at some of the... that is one of the many reactions that we will be studying that alkenes can undergo.0713

How would we make an alkene?--if we want to put a carbon-carbon into a structure, there is really two major approaches that we would want to take.0723

One of them is going to be an E2 reaction on an alkyl halide; let's see an example of that.0731

How about if we wanted to transform this given starting material into the desired product; what a transform problem looks like is it means provide the necessary reagents to convert one to the other.0737

More than one step might be possible; it might not just be one reagent that you are putting in there; it might be multiple steps to convert one to the other.0750

If we compare our starting material to our product, we see that an elimination has taken place.0759

It must be an elimination because we've formed a double bond; we've lost the bromine--the leaving group; but we've also lost a hydrogen.0766

What we would need to consider is what mechanism do we want to employ in doing this elimination?--we have two choices; it could be either an E1 elimination or an E2 elimination.0780

There is some benefits to some; some might be more suitable; let's think about an E1 elimination; that was the mechanism that was the multistep elimination mechanism.0790

It started by loss of a leaving group to form a carbocation; carbocations are involved in the E1 mechanism; would this be a good substrate to form our carbocation?0803

We have our leaving group on a primary carbon; that would give us a primary carbocation; that is not a very good carbocation; that would be highly unstable.0814

I don't think an E1 is a good idea here because we have a primary leaving group; that would give a primary carbocation; that most certainly would rearrange.0826

I think we should avoid trying to do some kind of unimolecular elimination and count on a carbocation; instead we are going to do an E2.0841

E2 elimination was where we had some strong base attack in a single step mechanism; attack the β hydrogen, form the π bond, kick off the leaving group; that is our E2.0850

What we need here is we need a strong base; we need a strong base; if you had to think about a strong base, maybe sodium hydroxide comes to mind; that would definitely be a strong base.0861

But let's take a look at this starting material, this n-bromobutane or 1-bromobutane, and think about its reaction with sodium hydroxide.0882

Yes, it could do the E2; but is there another reaction that hydroxide can do?--remember that hydroxide can be both a base and a nucleophile; we have our competition here between Sn2 and E2.0892

Because we have a primary alkyl halide, what would be favored?--there is no steric hindrance; Sn2 is going to be favored.0906

Sodium hydroxide wouldn't work because that would give us the substitution product; how can we force the elimination then?--how can we suppress the substitution, the backside attack of the Sn2?0915

If there is some way we can increase the sterics of that backside attack, then that would help us; that would force the E2 be favored over the Sn2.0926

How about if we just used a different base?--what would be a bulkier base than sodium hydroxide?--how about t-butoxide as a base?0936

T-butoxide has a tert-butyl group; now it is very bulky; this is bulky; so E2 is major; even with a primary halide, the tert-butoxide favors the elimination.0945

All we need to do, in this case, our reagents... it can be done just in a single step... is we use t-butoxide.0960

Maybe we can throw in some heat if you want; that usually favors the elimination as well; but all we need in this case is a strong bulky base.0970

The E2 is a very good method for forming alkenes; very reliable single step; let's review some of the features of the E2.0978

Remember the stereochemistry of the E2; there was a relationship between our leaving group and our β hydrogen.0986

They had to be anti to one another; we call that anti elimination... anti elimination; we need to be able to achieve that stereochemistry in order to do the E2.0991

Our regiochemistry, this was governed by Zaitsev's rule; we wanted to get the most stable alkene ;in this case, we didn't form a very stable alkene; it is terminal; it is mono-substituted.1011

But because there was only one β hydrogen that was... one type of β hydrogen that is possible, this is the only E2 elimination product that is possible; this would be in fact our major product.1031

Zaitsev's rule comes into play when we have more than one β hydrogen from which to choose; the one we select is based on the one that would lead to the most stable alkene product.1043

Finally, how about the Sn2 versus E2?--because we are going to have that competition with our strong bases; they can also be nucleophiles.1056

As usual, as we increase our sterics, we know that is bad news for the backside attack for the Sn2; that is going to increase the proportion of the E2 elimination product.1065

We need to... if we want to increase our sterics, we need to have a nice repertoire of bulky bases from which to choose.1078

Of course, the t-butoxide is the one we are most familiar with; the t-butoxide; but there is some of other ones you can do as well.1087

There is some nice amine bases; amines have a nitrogen in them and they usually have some groups attached on that nitrogen.1096

For example, triethylamine, a nitrogen with three ethyl groups attached to it, like the tert-butyl group, very big and bulky; that is very good for doing eliminations.1105

Or diisopropylamine has a nitrogen with two isopropyl groups on it; look at all that steric hindrance, all that bulk; diisopropylamine is another example of a base that is good.1114

If you put that in there with an alkyl halide and warm it up, then you can have a good bet that E2 is going to be your major product there.1129

A second approach, besides doing the E2 elimination, is to start with an alcohol and do a dehydration reaction; here is an example of an alcohol.1140

In order to do a dehydration, we are going to react this either with H2SO4 and heat or maybe H3PO4; these are the ones we will probably see most often.1149

Some strong concentrated acid and heat--what happens is we get an elimination reaction to take place; we form an alkene product; we start with an alcohol and we form an alkene.1160

What did we just lose?--what has been eliminated in this elimination reaction?--we lost the OH, of course; but remember we also lost a β hydrogen; we lost an OH and an H.1175

It means we have a loss of H2O; that is why we call this reaction the dehydration reaction.1188

Because just like when you are out in the desert or you are running and you are dehydrated, you are low on water, dehydration means you are losing water; the other product in this reaction is water.1196

Let's think about the mechanism for this; how can this alcohol get converted to an alkene?--how can it lose water?1211

Clearly, we have very strongly acidic reaction conditions; what do you think our first step should be?--in a strong acid, we need to protonate something.1217

Let's look at our alcohol and think about where to protonate; we really only have one choice--on the oxygen; that in fact is going to be our first step of the reaction.1225

Let's just use HA to represent our strong acid; two arrows to do a proton transfer; and we can protonate the alcohol OH group.1235

Great idea for our first step because we are in strongly acidic reaction conditions; so protonate; what does that do for us though?--why would that be something that might move us toward our product?1251

Remember, once we protonate an alcohol, we turn that OH into a very good leaving group... very good leaving group; it is going to be... it would be water once it leaves, very stable molecule.1265

We now have a leaving group to do our elimination; now we should think, do you think it is going to be an E1 elimination mechanism?--an E2 elimination mechanism?1281

E2 means we have a strong base come in and attack the β hydrogen to kick out this leaving group; is that what happens?--do we have any strong bases in these reaction conditions?1288

No, of course not; we have very strong acid conditions; it can't be an E2 elimination; it can't be an E2 because there is no strong base.1299

Our only other choice is to be an E1; what does it mean to be an E1?--it means our leaving group just leaves on its own.1312

In acidic conditions, that is what we would get; our leaving group leaves on its own; this next step is loss of leaving group to give a carbocation intermediate.1320

This carbocation can go on to be a carbon-carbon double bond, to be an alkene; this how the E1 continues; how do we do that final step?1337

Remember we are losing water; we are losing the OH leaving group; but we are also losing a β hydrogen; what we do is we look to one of the hydrogens on the neighboring carbon.1349

I'm going to choose over here because that is going to give me a double bond that is more highly substituted, that is more stable; you could see that is the major product that was shown here.1359

I could just use A- as a good base; that was formed in that first step; but I could use A- to come back in here and deprotonate; carbocations can be deprotonated to form double bonds to form alkenes.1367

This last step we could describe as deprotonate or loss of the β hydrogen; we know that has to be part of an elimination; we want to put in that context.1385

We are going to protonate, lose our leaving group, and deprotonate; that is our E1 mechanism for dehydration.1398

What are some other things that we want to know about this dehydration mechanism?--because it is the E1 mechanism, it is going to give the most stable alkene; this will also follow Zaitsev's rule.1409

It involves a carbocation; we know carbocations can rearrange; if there is a possibility that we can have a hydride shift or an alkyl shift to go to a more stable position.1422

A more substituted carbocation, that can happen; that will happen; taking a look at our reaction conditions, remember that we are dealing with a very very strong acid.1433

We need that to make OH a good leaving group; we really have to have something like sulfuric acid as our catalyst in this reaction.1442

In my mechanism, I just used HA to represent sulfuric acid; that is a safe species to use in a mechanism to represent a strong acid.1451

But let's just remind ourselves what H2SO4 looks like; this is H2SO4.1460

After it protonates something, after it is used as an acid, we are left with A-; that is this species right here.1468

Why is sulfuric acid such a great acid?--why is it such a strong acid and so willing to donate its proton?--we could take a look at its conjugate base.1478

This is the conjugate base of sulfuric acid; do you think this is a pretty stable molecule or is this pretty reactive?--what do you think about it?1490

It has an O-; sometimes we consider that to be a strong base, but this is not a strong base; that is because this O- is resonance stabilized.1499

We can draw several resonance forms for this; I will just draw one here; this negative charge can also be delocalized on the third oxygen, the other oxygen down here.1510

That negative charge is equally delocalized over all three of these oxygens; which means it is very very stable and is resonance stabilized.1524

We have a very weak conjugate base; this is very stable, very unreactive, very weak; we would expect that for something that is the conjugate of a strong acid.1537

Because it is so stable, that makes it non-nucleophilic; in other words, there is going to be no Sn1 competition.1552

Normally up till now, every time we've seen a carbocation, we've used that carbocation; it had a nucleophile add to it and we've done a substitution reaction.1565

The only time we've seen elimination, we said that was just usually a side product for our substitutions; substitution is usually favored.1573

The dehydration reaction is the one example that deviates from that norm; it is simply because, in these reaction conditions, we have no nucleophile to add to the carbonyl, to the carbocation.1580

All we have is a strong acid and its weak conjugate base; once you form the carbocation, your only choice is to dehydrate, to eliminate and form the alkene double bond.1591

You can think of the A- like a spectator ion; you can use it as a mild base to deprotonate something that needs deprotonating.1603

But it is not going to be something that is going to be nucleophilic; that is not going to be participating in the reaction.1614

What I wanted to do is I wanted to compare a strong acid like H2SO4 with another strong acid like HCl.1621

What we just said about H2SO4 and heat, this looks like what kind of reaction conditions?--this looks like you are going to lose water; it is going to be dehydration.1629

Strong acid and heat on an alcohol, what would that product look like?--we have four carbons; we are going to lose water.1639

We would form the double bond between the middle two carbons here--that is more stable than being at the end; we can keep this trans relationship; that would be more stable.1649

Remember we could also form the cis-butene product; but that is going to be less stable; that would not be our major product.1658

Our major product is the most stable alkene we can have after rearrangements if they are necessary; in other words, we would expect E1 elimination to take place with H2SO4.1667

How does that differ when we react it with HCl?--HCl is unique because, not only is it a strong acid and a source of proton, it is also a source of Cl-.1679

It is a strong acid, plus it is a nucleophile, it is a source of a nucleophile; what happens when we react an alcohol with a haloacid like HCl?1691

We protonate to form a good leaving group; then the halogen replaces that leaving group; we get a substitution mechanism taking place either by the Sn1 mechanism or the Sn2 mechanism.1705

Both of those can happen with the halide; we can form a carbocation or the halide can do a backside attack.1717

But with an acid like HCl, we get substitution because, if there is a nucleophile present, that nucleophile will want to attack the carbocation.1724

With H2SO4 and heat, there is no nucleophile present; once we form the carbocation, our only choice is to eliminate and get the alkene product.1732

Let's try another example here; here we are given a reaction and we are asked to provide a mechanism; it looks like we have these two methyl groups here.1742

Remember our line drawing means these are methyl groups; it looks like those methyl groups used to be on the same carbon; now they are on different carbons.1757

I think we are going to have to do some kind of rearrangement in our mechanism to account for that.1766

How do you think we should get started?--I see that strong acid; I am going to protonate as my first step; that gives you HA to protonate.1771

Every step in this mechanism, the dehydration mechanism, is reversible; this oxygen now has just one lone pair; it has a positive charge; I protonate it.1783

After protonating, now I have a great leaving group that can leave; I show this bond breaking and leaving with water; this is where I just kicked out my water molecule; now I have a carbocation.1796

Let's put those CH3s back in again so we can see them a little more clearly; we are at the carbocation stage; now we can rearrange.1811

Remember a carbocation rearrangement is where we look over to one side or another and we try and find and steal some electrons, steal a bond that would result in a more stable carbocation.1825

What this carbocation sees is, if one of these methyl groups picked up and moved over to the next carbon over, we call that a methyl shift... a 1,2-methyl shift.1836

We now have one methyl up here and one methyl down here; where does that place our carbocation?--this carbon now has four bonds; it had only three before.1846

But this carbon just lost one of its bonds; the carbocation is now down here; why is that rearrangement favorable?1855

It is favorable because we have gone from a secondary carbocation to a tertiary carbocation which is more stable.1865

We always want to be on the lookout for any mechanism involving a carbocation; we want to be on the lookout for rearrangements; because if it can find a way to stabilize, it will.1876

How do we get to our alkene?--how do we do our last step here?--we need to form the double bond; now we need to deprotonate.1885

Where is the proton that we want to go for, we want to deprotonate?--it is not going to be at the carbon of the carbocation; it is going to be one of the adjacent carbons, one of the β positions.1893

Why did we not deprotonate in this direction?--because that would give us a double bond that is less highly substituted.1907

If we go for this hydrogen, that is going to give us the double bond between the two methyl groups; it is going to give us a tetra-substituted, which is the most stable we can have.1915

In this case, because it gave us the major product, we didn't have to make this decision; I just want to keep that in mind.1930

If you had to predict the product here, this is the product; this is why you would get the product you did.1936

We need to deprotonate; what can we use to deprotonate?--A- is the safest bet here; we just formed A- in the first step; you can also see that this is catalytic in acid.1941

For every protonation step, there is a deprotonation step; we grab that proton and use those electrons to forms a double bond.1950

We protonate, lose water, and then, in this case, we rearranged; then we deprotonated to form the double bond.1959

One more example, let's think about a transform; if I had an alcohol here; I am starting with an alcohol; I want to go to an alkene.1971

It is important to consider the functional groups in our starting material and our product to decide how we are going to convert one to the other.1983

We are back to our original problem; we want to form an alkene; the way we form an alkene is by eliminating to form the double bond.1991

You know our choices are E1, E2; those are the only elimination mechanisms we know so far; those are the only ways we know how to make alkenes so far; we will learn about some other down the road.2000

We have an alcohol; what if we took this and we treated it with H2SO4 and heat; we just learned that alcohols can be dehydrated; certainly this alcohol could be dehydrated.2010

Would that give this product that is shown as the major product?--let's think about that mechanism; we know this is going to be an E1 mechanism; it is carbocation conditions.2023

We are talking about a primary carbocation, highly unstable; but with immediate rearrangement, we can get to a secondary carbocation, maybe even a tertiary carbocation.2040

But most definitely, we would rearrange this carbocation; what do you think that major product would look like if we had to predict this?2050

I think we would end up forming the tri-substituted double bond, the internal double bond, as the major product.2060

Because we don't want that as our target molecule, as the thing we are trying to synthesize, then this would not be a good synthesis.2069

What is another way that we can put in that double bond?--we need to do an E2 elimination to form the double bond at the end here.2077

Why don't we treat this with t-butoxide?--we talked about t-butoxide as our bulky base... throw in a little heat... as our bulky base.2088

What would happen if you take an alcohol and you react it with t-butoxide?--would you expect the E2 elimination?--what is wrong with trying to do an E2 here?2098

You lose a β hydrogen and what?--a leaving group; we do not have a leaving group here with hydroxide.2107

This is no reaction because there is no leaving group; we wanted to try and do the E2; but that is not going to happen here.2113

How can we possibly do the E2 elimination?--what do we have to do?--we have to make the OH a good leaving group; then we can do the E2.2122

An ideal way to do that is to convert the alcohol to a tosylate; we can use some tosyl chloride in pyridine; that keeps the alcohol right where it is.2130

But turns the OH into an OTs which is now a great leaving group; now that you have this great leaving group, now you can treat it with t-butoxide and heat and do the E2 elimination.2145

If you don't want the tosylate, you can maybe use something like thionyl chloride, SoCl2, or PBr3 to convert it to a bromide or chloride.2163

Then we have a good leaving group and we can do an E2 elimination.2173

We want to keep in mind, in synthesizing alkenes, we need to know the mechanisms very well for the E1 and the E2 and decide which is appropriate in each condition, in each situation.2176

Next we will be talking about the reactions that we can do for alkenes once we have a carbon-carbon double bond in our structure.2188

Thanks for visiting Educator.com.2196

Hi; welcome to Educator.0000

Next we are going to talk about reactions that alkenes can undergo; many of the reactions we are going to be seeing are described as electrophilic addition reactions.0002

One such reaction is the addition of HX across the double bond; we call this a hydrohalogenation reaction since we are adding a hydrogen and a halogen.0011

For example, if we react an alkene with HBr, what is going to happen in these electrophilic addition reactions is we are going to break the π bond.0021

Then we are going to be able to add a group to each carbon; we are going to see the same pattern again and again throughout this lecture.0032

When we are reacting with HBr, those are the two groups we are going to be adding--a hydrogen and a bromine; you can see we have a choice on which carbon gets the hydrogen and which gets the bromine.0039

It turns out that the hydrogen prefers to add to this end carbon and the bromine goes to the middle carbon; this is the only product that is formed.0048

Just so we are clear on what is not happening, the other choice we could have made is that the hydrogen goes to the middle carbon and the bromine goes to the end carbon; that product is not formed.0057

We describe this reaction as being a regioselective reaction since it reacts with just one region, one site, preferentially.0072

This is a question of regiochemistry--when we see that there is more than one place that we can react and we need to make that decision.0080

The regiochemistry that is shown is described as Markovnikov's addition; or we can call it Markovnikov's rule.0086

What Markovnikov's rule states is that the carbon with more hydrogens gets the hydrogen; the carbon with more hydrogens originally, initially, gets the hydrogen in the electrophilic addition.0093

We go back to our alkene; we see this carbon has just one hydrogen; the end carbon has two hydrogens; the hydrogen goes to the carbon with more hydrogens; that is exactly what is observed.0109

If we want to understand why we follow Markovnikov's rule, we need to look at the mechanism; it is a two-step mechanism; it is going to start with our alkene reacting with HBr.0124

What do we know about the reactivity of HBr?--we know it is a very strong acid; in fact, that is going to be our first step of our reactiom--is our alkene is going to act as a base.0137

HBr is going to act as an acid; we are going to protonate that alkene as the very first step; usually when we have done a protonation, it has been a lone pair on a base.0147

A π bond can act as a base as well; we just start with that π bond, those two electrons attack the proton, kick off the Br-.0157

This is a proton transfer as usual--two arrows; but we are starting from the π bond in this case.0166

Right now in this very first step... let's call this step one... in this very first step, we have to decide which carbon gets the hydrogen.0173

We know based on the product observed that the hydrogen goes to the carbon with more hydrogens, the end carbon in this case.0181

What happens to this carbon?--it just lost one of its bonds; this carbon now has one, two, three bonds; if we do its electron count, we get one, two, three; we know carbon wants four.0192

We end up with a positive charge on this carbon; we are going to get a carbocation intermediate; that is what happens when you protonate an alkene with acid--you get a carbocation.0203

Let's think about that other possible regiochemistry; if I instead added the proton to the middle carbon, that would give me a carbocation on this end carbon.0218

Take a look at those two possible carbocation intermediates and think about why it is that this first carbocation is favored and leads to product and the second carbocation is disfavored.0231

Do we know anything about carbocation stability?--how would you describe this one?--it has one, two carbon groups attached to the carbocation carbon; we call that a secondary carbocation.0243

This is a primary carbocation; we know the more alkyl groups donating into that carbocation, the better; so this is more stable.0254

This primary is less stable; primary carbocation is really awful looking; that is why this again is not formed; we get this carbocation; let's redraw it down here so we can do something else with it.0263

What is going to happen to that carbocation?--it is now going to react with the Br- that was just formed in that first step.0278

The carbocation acts as an electrophile; it looks like a great electrophile; it has a full positive charge; it is electron deficient.0292

The Br- acts as a nucleophile; we simply have the bromine bonding with the carbocation; our reaction is done; we now have our addition product as we were expecting.0300

We added H and Br across the π bond; it is a two-step mechanism--first we protonate and then our nucleophile attacks, adds to the carbocation.0321

Considering that mechanism now, let's see if we can get a good understanding of why that Markovnikov's rule is followed.0332

It turns out Markovnikov addition is favored because that is the one that favors the more stable carbocation intermediate; it is all about the stability of the intermediate.0338

If you have a more stable intermediate, that means it is lower in energy; if the intermediate is lower in energy, that means the transition state leading to that intermediate is lower in energy.0348

That means the energy of activation is lower leading to that transition state; there is a smaller hill to climb; simply the reaction is faster.0361

What we are looking at here in the observed regiochemistry is kinetics in play; it is just simply which reaction is faster; in other words, it is not about product stability.0369

We don't look at the two possible regiochemistries and think: where would the bromine like to be?--is that interesting to us?--no, it is simply about the stability of the carbocation intermediate.0384

If we think about our reaction here, our energy diagram, our starting materials are at some combined starting energy; our products are typically going to be at a lower energy.0395

We have a pathway to get here; this is a two-step mechanism going through an intermediate; our carbocation is going to be a high energy intermediate; we could put that here.0412

If this represents our secondary carbocation, the other path leading to the primary carbocation would be higher in energy; these are the two possible paths that we can take.0423

Our starting material can go to the secondary carbocation and then onto product; or the starting material can go up all the way to the primary carbocation and then down to the product.0435

If we look at these two paths and we ask which one is faster, it is going to be going from the starting material energy up to that transition state energy... my curve missed that a little bit.0450

Here is our transition state; this is our energy of activation; trying to go up to this transition state is going to be a larger hill to climb; this is the faster reaction.0462

In this part, formation of that carbocation is going to be our difficult step; this is our rate determining step; it is going to be very difficult to protonate and form the carbocation.0477

That is an uphill battle; because this is a faster step, that means overall, it is a faster reaction since it is our rate determining step.0494

Let's see an example--if you want to predict the product here; I also pointed out that, rather than just predicting the product, let's also think about the stereochemistry of that product.0511

We have an alkene; we are reacting with HCl; we are going to be breaking the π bond; we are going to be adding a group to each carbon; we have a hydrogen and a chlorine.0521

Where do you think the hydrogen should go?--it wants to go to the carbon with more hydrogens; we could think about our line drawing; we have one hydrogen here; we have two down here.0533

Once again, the end carbon is going to be the place that we would want to put that hydrogen; so this is our addition product; that is our regiochemistry.0544

What about the stereochemistry?--what do you think... we have a chiral center here on this carbon; what do you think the orientation of that chlorine is?--is it a wedge or a dash?0559

In fact it is going to be a mixture of those; we could just draw it flat; but what we really mean is we are going to get some with the chlorine as a wedge and some with the chlorine as a dash.0571

In fact we are going to get exactly 1:1 of this mixture; what is the relationship of these two products?--they are enantiomers of each other; in other words, we are going to get the racemate.0584

Stereochemistry is going to be a big issue for us in looking at reactions of alkenes; because for alkenes, we are starting our planar; we are starting out flat; these are achiral starting materials.0598

But when we break the π bond and add something to it, we have the potential for creating new chiral centers in the new tetrahedral centers, sp3 hybridized centers that we are making.0611

Stereochemistry is going to be an issue constantly; if we think about that protonation, we can protonate to form this carbocation intermediate; there is a hydrogen here.0620

This carbocation intermediate is sp2;--it is still planar just like the alkene starting material.0637

But now when the chlorine attacks this planar species, it can attack from the top face... attack from the top; or it can come in and Cl- from the bottom.0646

This is how we can get to either enantiomer; you can imagine your planar intermediate or your planar starting material.0664

Attacking from one from one face or the other face is going to lead to your two possible stereocenters.0673

What we need to keep in mind here, what this is demonstrating, is that any new chiral center or chiral carbon is generated with both R and S configurations.0679

We have an achiral starting material; so if we come up with a chiral product, we must get a racemic mixture; if we come up with a product that is chiral, it must be formed as a racemic mixture.0704

How do we normally draw this product?--we normally draw... it is assumed we know it is a racemic mixture because that is always the case.0720

So we usually just draw it flat; we don't need to draw any stereochemistry here; but technically what this is representing is an equal mixture of both enantiomers.0727

In some other examples where we create two new chiral centers in the addition to an alkene, we will find that we are going to have to talk more about the relative stereochemistries of those two groups.0736

We will see those in some of our next examples; what else can we add across a double bond?--we can also add water across the double bond; we call this the hydration of alkenes.0746

We are going to break the π bond; what are the two groups that we are going to add to that π bond if we are adding the components of water?--we are going to be adding an H and an OH.0762

We are going to break this π bond; we are going to add these two groups; if you had to take a guess on where the hydrogen goes and where the OH goes.0775

What if this followed Markovnikov's rule just like addition of HBr did or HCl?--then we would expect the hydrogen to go to this end carbon; guess what?--that is exactly what happens.0786

We add H and OH; we get Markovnikov addition of the H and OH; we started with an alkene; we end up forming an alcohol product; this is one way we can make alcohols--is by hydrating an alkene.0802

Our reaction conditions for this is we need water of course if we want to add water; we also need acid; this is acid catalyzed mechanism.0824

I just want to reflect for a moment on what happens when you take a strong acid like H2SO4 and you put it in water.0832

What you end up with is H3O+; sometimes your reaction conditions for this hydration might just show H3O+.0840

Sometimes it will show H2O and a strong acid; it can have a few different ways that it will be presented.0846

But if you ever see H3O+, it means that you have both HA, some strong acid, and H2O present; it implies that you have both of these.0852

The reason we can assume we have H3O+ is by definition a strong acid is something which fully dissociates in water.0862

Dissociates in water... in other words, it is going to protonate water and give A-.0873

A strong acid, when we represent HA in aqueous conditions, if we ever we use HA, the strongest acid we have is going to be H3O+, the hydronium ion.0888

You no longer have H2SO4 molecular in an aqueous solution; you have H3O+ and you have the conjugate base of that strong acid.0897

When you see these reaction conditions, you can just go ahead and use H3O+ in your mechanisms knowing that this is the mechanism by which that H3O+ was formed.0909

Let's take a look at that mechanism; what do you think our first step should be in these clearly acidic reaction conditions?--I think we need to protonate; we can use our π bond for that.0922

I am going to go ahead and draw out H3O+ since I know that is the acid in these reaction conditions; I am going to have my alkene act as a base.0936

I am going to protonate my π bond; just like our mechanism for addition of HBr, it starts the exact same way--it starts by protonating.0947

Remember this protonation step is the point at which we determine our regiochemistry--which carbon is going to get the H+.0963

Just as a quick reminder then again, the other possibility is that we had the proton to the middle carbon; if we do that, our carbocation ends up on the end carbon.0971

Exactly like we saw for addition of HBr, we want to form the more stable, in this case, secondary carbocation.0984

Always going to form the more stable carbocation; every time we go to protonate an alkene, we will keep that in mind.0995

We have a carbocation; where do we go from here?--we want to form this alcohol; we need an oxygen to attack.1002

How about using hydroxide?--how about if we wanted to use hydroxide to add to the carbocation?--our mechanism would be finished very quickly there.1010

But there is a problem with that; what is the problem with using hydroxide?--let's take a look at our reaction conditions.1021

Do you expect to have a lot of hydroxide present in these strongly acidic conditions?--no way; there is no hydroxide around.1027

Who is our nucleophile?--who is the strongest nucleophile we have in these conditions?--it is going to be water.1036

Water is going to be the strongest nucleophile that we have and that we can use; that is what is going to attack the carbocation.1043

Let's think about this oxygen now that has just attacked; it still has two hydrogens on it; what else does it have?--it still has one lone pair and a positive charge.1055

My water can attack; we can call this second step--attack of our nucleophile onto the carbocation.1068

We have added a group to each carbon; but we are not done yet because this oxygen has a positive charge; it is very unstable; we need to get rid of that positive charge.1077

How can we do that?--we know oxygen just wants to have two bonds and two lone pairs; it is one of these protons that can be removed.1085

Our last step is going to be deprotonate--that is what we call it when we remove a proton; what could we use to deprotonate?1093

We can come back and use this water that we just formed in our first step; and we are done; a couple things we can note--that it is catalytic in acid; we need that acid there.1103

But for every step where we protonate, there is step where we deprotonate and get that H3O+ back; we just need a tiny bit of acid to go along with our water.1129

We could take a look at this pattern; it is going to be something we are going to see again and again with acid catalyzed reactions where we protonate and then we attack and then we deprotonate.1141

We are going to see that pattern again and again; something else to keep in mind is that if this mechanism seems at all familiar or these species that we have.1152

It is because we have actually seen this mechanism before; but we have seen it in the reverse direction; we have seen it in the reverse direction.1163

What do we call it when we take an alcohol and we lose water to go to an alkene?--that is the dehydration of an alcohol; how did we do that?1170

We protonated to make it a good leaving group; then it left; then we deprotonated the carbocation; same exact intermediates.1181

When we compare hydration for the dehydration, we will see that the hydration mechanism is the exact reverse of the dehydration; same number of steps.1193

We just saw this part when we want to hydrate a π bond, we use water and H2SO4 to go from an alkene to an alcohol.1202

How do we dehydrate?--what conditions do we do to get rid of water?--we use H2SO4 and heat; very similar conditions in that they both have a strong acid; they both need acid.1214

But when we have, in the presence of water, we do hydration; we add water; these reaction conditions are aqueous with some added acid.1230

Where these conditions--look we have H2SO4 and heat; these are strong concentrated acid... concentrated acid.1246

We have an absence of water; there is very little water; that is how we do a dehydration; of course the heat also helps in doing the elimination.1255

But take a look at your reaction conditions; they are very similar; but the key here is that we are adding water, we are removing water.1266

But this is truly an equilibrium; you can force a reaction in either way by your reaction conditions; also by having an excess of one or the other or removing your product as it is formed.1276

Let's try an example of this hydration reaction; we have an alkene; we have H2O, H2SO4; this looks like we are going to add water.1289

We are going to add an H and an OH across the double bond; this looks like hydration conditions; here is where we have our alkene.1304

Notice that these are not going to be reactions of benzene π bonds; benzene π bonds are not alkene π bonds.1314

All the reactions we are going to be studying in this lecture are not relevant to benzene rings; just isolated double bonds, separated double bonds.1320

What we have to decide is which carbon gets the hydrogen, which gets the OH; we could think about Markovnikov's rule.1333

Markovnikov's rule tells us that the more stable... I'm sorry, the carbon with more hydrogens, gets the hydrogen; that doesn't help us here; each of these carbons has just a single hydrogen.1341

You can see very quickly that Markovnikov's rule is only going to get us so far; it won't help us predict every case.1354

We need to have a better understanding of the mechanism and the foundation underneath Markovnikov's rule.1360

That foundation is the notion that in our mechanism we want to have the most stable carbocation intermediate.1367

In our first step, we are going to protonate this π bond; that will give us a carbocation either here or here; where is the better carbocation?1374

How would you describe this carbocation in this position?--let's imagine the carbocation here; this is a secondary carbon; it would be a secondary carbocation; that is not a bad one.1384

How about if we had the carbocation in this position?--it is also secondary; again we are seeing that we have a very similar condition, very similar situation.1395

But this not just secondary; we have this benzene ring here; what do we call the position that is next to a benzene ring?--we describe that as benzylic.1405

A carbocation in this position would be secondary and benzylic; that is better than just being secondary; we want to think about the stable carbocation.1419

We want to know the mechanism and think about the mechanism, what is important to it; you might want to think: are sterics important in this case?1436

Sterics have nothing to do with carbocation formation; that is if we are trying to do a backside attack maybe.1445

But we really want to think about what is the governing thing in this situation; it is really about the carbocation stability.1452

What we want to form is the benzylic carbocation; how do we get to the benzylic carbocation?--what does that mean?1460

That means we protonate in the other position; the hydrogen goes here so that the carbocation goes in the position we want.1471

This is resonance stabilized; try it; that is a good exercise; can you show the resonance stabilization of this benzylic carbocation?1479

Once we know the carbocation that is going to be formed, now we can go to the product hopefully without doing a complete mechanism because where the carbocation was is where our OH group will be added.1494

Remember we are adding an H and an OH; in this case, we would want to put the OH on the benzylic carbon because that would be the more stable carbocation.1509

Notice I am not showing any stereochemistry here; we did form a new chiral center; but we just know that is going to be racemic; we could just draw them all as straight lines.1517

Addition of H3O+ is one way to hydrate a π bond and add the components of water across the π bond.1529

But there is two other alternative hydration methods that we are going to consider; we will see some benefits of that down the road.1535

This first one is called oxymercuration-demercuration; as the name implies, it is a two-step process.1548

It is good to pay attention to the name of this because that is going to help us understand more about the mechanism and remember some of the details of this.1555

Oxymercuration-demercuration looks something like this; we start with an alkene; the first thing we do is we treat it with Hg(OAc)2; that is called mercuric acetate.1563

OAc represents the acetate group; this is called mercuric acetate; and water; we also add water in this step; this adds a mercury group and an OH group.1580

A little reminder--the Ac group, every time we see Ac, it means we have COCH3; we have a carbonyl and CH3; this is like an ester group here.1596

It breaks the π bond and it adds a mercury and an OH; that is our first step; the second step, we treat this with NaBH4; that is called sodium borohydride.1608

It replaces the mercury part; it replaces the mercury with a hydrogen; we call this a reduction reaction because we are introducing hydrogen; we are increasing the number of C-H bonds.1626

Sometimes actually this process is called oxymercuration reduction; some textbooks might have it described as that process because the second step is both.1640

It is a demercuration; we are getting rid of the mercury; because we are replacing it with a hydrogen, it is a reduction.1648

What happens here?--we look at our structure; this mercuric acetate group gets replaced and it is just a hydrogen.1654

What happens overall?--let's look at the big picture comparing our starting material to our product; overall we broke the π bond and we added two groups; we added an H and an OH.1662

This is a way of hydrating an alkene; how would you describe the regiochemistry of that addition?--the hydrogen added to the carbon; it already had a hydrogen.1675

That is going to be described as Markovnikov addition; this is an alternate way of doing Markovnikov addition; why do we need an alternate way because we already have H3O+?1686

It turns out that this is going to be higher yielding; in the laboratory, this just works better than H3O+.1700

There are a lot of other functional groups that react with H3O+ or might not be stable to those reaction conditions; so these are alternative reaction conditions.1705

The mechanism does not involve a carbocation; we saw that H3O+ does; that is going to be a benefit; it is not going to have any rearrangements in the reaction.1715

Experimentally this is a superior way to do the same thing--add water across the π bond with Markovnikov regiochemistry.1724

Let me just briefly show you this mechanism here for this first step, the oxymercuration step; we have our mercuric acetate.1732

What happens is the π bond attacks the mercury and kicks off one of the acetate groups; but then the mercury comes back and adds back into the π bond to form a three-membered ring.1743

That gives a positive charge on the mercury; I am showing you this because we are going to see some intermediates later in this chapter and in future chapters that look a lot like this.1767

After we get a more thorough explanation of those species, you can come back and maybe have a better understanding of this species.1779

But it turns out that this now gets attacked by the water; it chooses to attack on the carbon that is more substituted; again we are going to look at that in detail more later.1787

That opens up this three-membered ring to form the OH here and the HgOAc group here; the other reason I want to point this out is because instead of doing this reaction in water.1800

If we did this reaction in the presence of an alcohol as our solvent, then it could be the alcohol oxygen that opens up the mercurium ion.1816

We would end up forming an OR group; what would happen if we did step two here, NaBH4, and replaced our mercury?1826

We could get a product where instead of adding the components of water across the π bond, we add the components of an alcohol across the π bond; this could be a way that we make some ethers.1838

Another possibility for adding water across the π bond, our second one, is called a hydroboration-oxidation; again with that big name, that tells us that it is another two-step process.1853

We start with an alkene; the first thing we do is we do a hydroboration; we add BH3; BH3 is in quotes because BH3 is a very unstable molecule.1865

What we are going to use are various reagents that are the equivalent of BH3; we could use B2H6; it is called diborane; this dimerizes; that is a little more stable.1876

Or we could have BH3-THF where the BH3 is complexed with a solvent molecule, the THF molecule.1890

Pretty much any boron reagent that you might see is all going to be doing the same reaction--that is a hydroboration; we are going to be adding a hydrogen and a boron across the π bond.1900

Our second step is an oxidation step; that means we are adding some oxidizing agent; in this case, what we typically use is hydrogen peroxide.1915

In our oxidation, we are going to replace the boron group with an OH; our boron used to be here; now there is an OH in that place; hydroboration-oxidation.1924

What did we do overall?--we started with a π bond; now we have an alcohol; what did we add across the π bond?--we added water across the π bond; we added the H and the OH.1948

How would you describe the regiochemistry of the H and the OH?--the hydrogen went to this carbon; the hydrogen went to the carbon with less hydrogens; would you describe that as Markovnikov addition?1960

No, that would expect the hydrogen to add to the end carbon; we describe this as anti-Markovnikov addition, anti-Markovnikov regiochemistry.1973

This is interesting because now it gives us a method we can use to get the opposite regiochemistry of the other methods.1994

Let's take a look at the mechanism to see again a little bit of this mechanism so we understand where it goes and some details of the product that unfolds.2003

This first step, the hydroboration, is our first step; we add a hydrogen and a boron; this is a one-step mechanism; we describe it as a concerted reaction.2016

We are going to form a bond from the hydrogen to one end of the double bond and the boron to the other end of the double bond; both of these bonds are going to get at it simultaneously.2025

Our mechanism--our alkene acts as a nucleophile; that is electron rich; this boron acts as an electrophile; boron, again we talked about why it is so unstable.2037

It is because a BH3 with just three σ bonds has no lone pairs; it is missing an octet; it is highly electrophilic, highly reactive.2049

What happens is the π bond attacks the boron and forms a bond here; but at the same time, the boron gives back one of its hydrogens and adds that back into the double bond.2059

We have formed these two... these two arrows are moving these four electrons all at one time; our product then is we have a boron attached and we have a hydrogen attached.2072

This is going to explain two things; this is going to explain the regiochemistry; this is a regioselective reaction where the boron goes to the end carbon and the hydrogen goes to the middle carbon.2086

Remember we call this anti-Markovnikov; there is a few explanations for this; one of them is just simply about sterics.2097

Because the boron is bigger than the hydrogen, it prefers to go to the less hindered carbon.2106

There is also an electronic argument saying that again, because this is the electrophile, as this bond breaks it is better to have some positive charge building up on this carbon.2114

There is more than just a sterics argument here; but we can keep it simple by pointing to that; it turns out this reaction is also stereoselective.2125

We describe it as being syn addition; that is because, since the boron and the hydrogen both added at the same time, they had to add to the same face of the alkene.2136

We call that syn addition when they come from the same side; they add at the same time to the same face; this is called syn addition.2148

Let's see an example then; if we wanted to do a hydroboration-oxidation on this alkene; the first thing we are going to do is BH3-THF; how do I interpret that reagent?2160

I know I am going to add a boron and a hydrogen; that is what all of our boron reagents are going to do; they are going to deliver a boron and a hydrogen.2174

We are going to break the π bond; we are going to add a group to each carbon; now we have to decide where to add what.2183

Is it going to follow Markovnikov or anti-Markovnikov?--it is going to be anti-Markovnikov; the hydrogen is going to go the carbon with less hydrogens.2191

That is because this big boron, remember that is going to be larger; sterics plays a role here; the boron is going to prefer to go here; that is the regiochemistry.2201

Remember we are thinking about that; how about the stereochemistry?--because this is a concerted mechanism, they have to add to the same face.2213

When we draw our product here, the way we can show them adding to the same face is, for example, we can draw them as both wedges.2223

If I add the hydrogen to the top face, what happens to this methyl group?--this CH3 can no longer be in the plane anymore; this is a tetrahedral carbon.2235

That methyl group gets pushed back, gets pushed down; it can add to the top face or it can add to the bottom face; it is your choice whether you want to show them both as wedges or both as dashes.2242

Let's take a look at that other product where they are both dashes and see if they are the same thing.2254

Sometimes they are going to be the same thing; sometimes they are not; we will have to evaluate each case.2260

If they added to the bottom face, would this be the same intermediate product or are they different?--it looks like they are mirror images of each other; it looks like they are enantiomers.2265

We could either draw both of them or we could just say +enantiomer when we see that we have made a chiral product.2279

This is because this is chiral and we came from an achiral starting material, we know we can't just form this single chiral molecule; we always have to form its enantiomer as well.2290

We can either just say +enantiomer or we have to draw them both; drawing both usually takes a little too much time; there is our hydroboration reaction--stereoselective, regioselective.2300

Our next step is oxidation; what we need to know is that this oxidation step retains the stereochemistry.2312

I am not going to go through the mechanism for this oxidation on how it is that the boron gets replaced by the oxygen; but you could think of it as just inserting itself into the carbon-boron bond.2324

If it was a wedge, the oxygen will end up as a wedge; if it was a dash, it will end up as a dash; we will draw that for our product here.2334

What used to be a BH2 as a wedge is now an OH as a wedge; the other carbon stays the same; and +enentiomer of course.2344

If you want to see what that enantiomer looks like, that is where both of our groups have been added as dashes; an OH here and a hydrogen here.2358

The methyl group must be pushed into the wedged position, must be pushed up, as the hydrogen added from the bottom face.2374

A couple things to point out--a racemate is formed; as always now we have to be very careful that is a right +enantiomer.2381

Because we have added dashes and wedges to our product and we have drawn one specific enantiomer, now we have to point out that the other enantiomer is also present.2389

We are going to keep that in mind and see lots of those examples in this unit; it is going to be anti-Markovnikov regiochemistry.2399

The hydrogen went to the more substituted carbon, the one with less hydrogens; and it needs to reflect syn addition of H and OH.2409

With the hydroboration-oxidation, it is one of the most complicated reactions we will be studying for alkenes.2421

Because not only does it have complicated reagents that we have to take a look at--this BH3-THF and this H2O2.2428

But we also need to know what does it add, what is the stereochemistry, and what is the regiochemistry?2435

Here is an example; it is coming back to just what I pointed out; these are the questions we need to ask when it comes time to predicting a product.2446

We have to interpret the reaction conditions and decide what the groups are adding; then we have to decide where are they going and what is the stereochemistry.2453

Where are we adding it is another way of asking what is the regiochemistry; then what is the stereochemistry--dashes and wedges.2461

Let's take a look at this two-step process; we are almost always going to be seeing those two steps back to back.2472

We are not so concerned about that intermediate product, what it looks like; we want to get in the habit of being able to draw a final product.2478

I see a boron here; this looks to me like hydroboration-oxidation; hydroboration-oxidation; we are going to be adding an H and an OH; this is one of our hydration reactions.2486

Here is our two... we are going to break the π bond; we are going to add a group to each carbon, going to add an H and an OH.2501

Where should the hydrogen go?--because hydroboration-oxidation is anti-Markovnikov, we want the hydrogen to go to the carbon with less hydrogens.2508

This carbon has one hydrogen; this carbon has none; that is a good place to do it; again if you draw that exaggerated boron here, then that helps you remember that you want to avoid some sterics with it.2518

What is the stereochemistry?--we need to do syn addition; that means we can add them from the same face, either from the top or the bottom; that is your choice.2530

But we need to show that in our product; we can do that over here... where the boron adds, in our final product, we are not going to have a boron there; what are we going to have?2543

We are going to have the OH; where the boron goes is where the OH ends up; we can draw that as a wedge right here.2559

The tricky part I think is right here--the hydrogen; where do we draw that hydrogen as a wedge?--it is not correct to draw it down here because this is not what a tetrahedral carbon looks like.2566

Let's try that again; this is our OH as a wedge; where do we put a wedge on this carbon?--the dash and wedge are pointing up in this direction.2578

Remember the H and the OH both came from the top face; but because we have a zigzag structure, they are coming in from slightly different directions; that is what our product should look like.2590

What happens to this methyl group?--he is no longer in the plane; he is being pushed back because we have a dash; so we must have a dash in that position.2599

Being able to predict these products, we are going to draw on all that experience we have in drawing 3D structures and understanding tetrahedral carbons.2608

And being able to draw something that represents the three-dimensional shape; let's double check; we added an H and an OH.2616

We added it with anti-Markovnikov regiochemistry; we added them syn; they are both wedges; very nice; is that my only product that I formed?2625

That is going to be hard to do because this is a chiral product and I started with an achiral starting material; what is missing?2635

We have to write +enantiomer; or we could say that it is racemic; we could say racemic if we want; we could either say +enantiomer or we could say racemic.2641

Some instructors do something like they say +/- indicating that we are getting a mixture of both isomers, both enantiomers; so we will have no optical rotation.2651

Either one of these is fine; but you have to do something; it is improper to draw a single chiral carbon as a product.2664

A lot of textbooks get a little sloppy with this; try not to pick up on those bad habits.2671

Again why do we need to learn about all these different hydration methods?--it really comes to being synthetically useful; there is a very good reason that we need to be able to use these.2679

One thing is that we can add water with either Markovnikov or anti-Markovnikov regiochemistry.2693

These are wonderful techniques that are complementary to each other and let us go in different directions synthetically.2698

These alternative methods have no carbocation in the mechanism; therefore we have no rearrangements; carbocations can do all sorts of weird things.2705

Here is a perfect example; if we wanted to hydrate this alkene and we just used plain old H3O+, you would not get the alcohol that you would normally expect.2712

Because it will rearrange; try this; try and do this; try protonating, looking at the carbocation, and see how we could rearrange.2728

What we have is all these methyl groups over here right next to where you would have the carbocation; one of those methyl groups is going to shift over to give a more stable carbocation.2737

This is the case where if you wanted to hydrate this π bond with Markovnikov regiochemistry.2746

Instead of using H3O+, you would do oxymercuration-demercuration or oxymercuration-reduction.2751

In doing that, because there is no carbocation, there is no possibility for rearrangement, you would get exactly the product you expect.2757

Hydrogen goes to the end carbon; OH goes to the middle carbon; perfect.2766

Or if you took the same alkene and you did hydroboration-oxidation, then you would break the π bond and you would add an H and an OH.2772

But in this case, you would now do it with anti-Markovnikov regiochemistry, where the hydrogen goes to the inside carbon and the OH goes to the outside carbon.2782

Why did I not... we know this is syn addition of the H and the OH; why did I not show any dashes and wedges in this particular case?2794

I didn't show any dashes and wedges because neither of these is a chiral center; so there is only one stereoisomer that is possible here.2803

It is not a chiral molecule; there are no chiral centers; in that case, there is no stereochemistry that needs to be shown.2812

It is only when both carbons involved are going to be forming its new chiral centers that we need to show the dashes and wedges and the configuration of those chiral centers.2820

If we are not dealing with chiral centers, then stereochemistry is not so relevant; how do we keep all of these reagents straight in our minds?2831

Because this is just the start and there is a lot more reagents to come even in the reactions of alkenes.2841

Let me give you just a little suggestion on how to use flashcards; I think flashcards are very useful for studying and learning organic chemistry.2846

Because it is a great way not only to organize your information, but to store it somewhere that is handy that you can refer to very quickly.2855

You could test yourself on the various information we are going to have for these chemical reactions; we are going to be seeing many many many reactions of this nature.2861

Where we start with some kind of starting material and we react it with some sort of reagent or set of reagents and it gives one or more products; we have these three possibilities.2871

One way that you can test the reaction is you can have on one side your starting material and your reagent; in other words, set up the reaction and make it like a predict-the-product.2885

Then of course, on the back, you have the answer; that is a way that you can test yourself to see whether or not you can do that problem; but that is not the only way you can or should test yourself.2900

Another way that you can test yourself is to say, if I had this starting material and I wanted to go to this product, how could I do that?--that is a transform-type problem.2911

Those are a little more challenging because now you have to plan what reagents and what reaction conditions are going to be necessary to do the mechanism that you want to employ.2922

But that is still not the only possibility; you can also say, given the following reagents and given this is the product, what starting material did we start with in order to go there?2932

This kind of problem is going to be very useful as a way to test yourself and to organize, to remember, as these number of reagents and reactions pile up and accumulate.2947

You need to remember which functional group is associated with which reagent; we just learned about hydroboration; we saw B2H6.2960

You need to be able to associate B2H6 as a reaction of alkenes going to alcohols; seeing it from all those directions is going to be very useful.2970

I think managing information and developing those good study habits is one of the most important things that you get out of organic chemistry.2980

This lecture is really good place to start thinking about those things; what else can we add to double bonds?--we saw addition of HBr; we saw addition of water.2987

How about the addition of a halogen like Br2?--this is called the bromination reaction; we could also do chlorination.2999

It is described as anti-addition; that is because when we react an alkene with Br2, it is common to see carbon tetrachloride as a reaction condition.3006

That is simply a solvent that is used in this reaction; try not to let that distract you too much; it is the bromine that is doing the chemistry here.3019

What happens as you might expect is you break the π bond; you add a group to each carbon; for Br2, the two groups you are going to be adding is a bromine and a bromine.3026

What happens is the bromines end up adding to give the trans product only; we get one up and one down in the case of a cyclic compound; you can see that very nicely--trans product.3035

We call the mechanism anti-addition; we call the product trans; the opposite, let's see what is happening; we are not getting syn addition; we just saw an example of syn addition.3049

Syn addition is what we called it when the groups add to the same face, if they were both wedges for example; that would give cis bromines; that does not happen.3064

Just getting to know some of this terminology, some of this vocabulary--syn and anti describe the mechanism; cis and trans describe the products that result from those mechanisms.3081

I showed it trans; this bromine can be the wedge and this could be the dash or the top bromine can be the dash and this can be the wedge; aren't these the same product?3092

We are really going to have keep making these decisions about stereochemistry again and again; what do you think?3102

Is there any way you can flip one over or rotate it, do anything to get them to superimpose?--no, indeed; because this is a chiral compound, it has no plane of symmetry.3107

Then these in fact are enantiomers; we are going to get, as usual, we are going to get the racemate formed here.3122

You don't have to draw both; you could just draw one and say racemic or +enantiomer; just for argument sake, let's think about this syn addition product.3129

If I added these bromines to the same face, added them to the top let's say, would I also have to add them to the bottom?--would that be a unique product?3137

Is it right to say +enantiomer here?--actually not in this case because if I added the bromines to the bottom face, it would be the exact same product.3147

The difference here is because this is a meso compound, it is symmetrical; it is achiral; it does not have an enantiomer; we are really going to have to keep this in mind.3158

It is not enough to just go through as you are doing predict-the-product and say +enantiomer, +enantiomer, +enantiomer; you need to be thoughtful about that and think about it.3170

Back to bromination, what we observe is anti addition; we will end up with these trans bromines; we describe this as a stereospecific reaction; we get one specific stereochemistry.3178

If we want to understand why we get this anti addition, we are as usual going to have to look at the mechanism; let's do that.3190

The bromination mechanism starts with our two ingredients; we have an alkene and we have bromine.3201

What we can imagine happening is the alkene acting as a nucleophile, the bromine acting as an electrophile.3211

You could have the double bond attack one bromine and kick the other bromine off; let's see what that what that would give us.3221

That would attach a bromine to one carbon; that would form a carbocation on the other carbon.3232

Like protonating a double bond gives a carbocation, adding a Br+ to an alkene also can give a carbocation.3238

However this is not formed because this is less stable than another arrangement this can have.3249

Because this bromine has these lone pairs, those lone pairs can come back in and can instead form a three-membered ring.3260

This bromine has two bonds and still has two lone pairs; it is an unusual looking bromine; we are used to it just having one bond.3274

Let's think about the formal charge for this bromine; we have one, two, three, four, five, six; bromine wants seven; it is missing an electron; this is a Br+.3282

This is called a bromonium ion; that is actually the intermediate that we get in the bromination reaction; we don't form this carbocation.3295

Instead we add the bromine, the Br+, to both carbons of the double bond to form this bromonium ion.3309

It turns out that the reactivity for the bromonium ion involves each of these carbons; they are partially positive; that makes them electrophilic.3317

Those are electrophiles; those are electrophilic carbons; we will see what happens to those next.3328

This is an introduction to the bromonium ion; let's step back and look at the complete... when we want to start the bromination mechanism from the beginning again.3339

We have our alkene; we have our bromine; we know that every time we have bromine in the presence of an alkene, we are going to go to this bromonium ion.3348

I am actually drawing this as a wedge; they are both bonds pointing in the same direction; we know we are going to get this bromonium ion--how do we get there?3357

The mechanism involves both the double bond coming out and attacking the bromine and kicking off the bromine; like we thought was going to happen in the first place.3371

But at the same time, one of the lone pairs of this bromine adds back in to the second carbon of the double bond.3381

We end up forming two bonds to the double bond in one step of the mechanism; that is why they end up adding to the same face; that is why we are showing them both in this case as wedges.3388

This is an electrophile; do we have any nucleophiles around in the reaction?--sure, we just formed in this very first step, we formed Br-.3399

This Br- is going to attack the bromonium ion; it will attack either one of these carbons; we can show it attack this one.3414

If it attacks that carbon, what happens is it treats this carbon-bromine bond as a very good leaving group; it opens up that bromonium ion ring.3427

What does this mechanism look like?--have we ever seen a mechanism where a nucleophile attacks a carbon and kicks off a leaving group?3440

In a single step, attacks a carbon and kicks off a leaving group?--absolutely, this is the Sn2; we describe that as backside attack.3448

If my bromonium ion is a wedge, if it is sticking out at you, where does that second bromine have to come in?--it has to come in from the opposite side; it has to come in from the back.3459

If it attacks this carbon on the right, then that new bromine is going to be a wedge and that original bromine is still here now with its three lone pairs; it is back to being neutral.3468

We can call that path A if you want, product A; or we could have it attack the other carbon and do an Sn2 on that carbon, path B; that is how you can see we can get to our other enantiomer.3481

If we attack that carbon, then this bromine would be the one that is back and this one would be out; we get our 1:1 mixture of enantiomers.3500

It explains the backside attack is why we get the anti addition; because whatever face the first bromine added to to form the bromonium ion, the second bromine has to come in from the opposite face.3513

We always get anti addition, one up and one down; it is possible to add more than Br2 in this fashion.3529

We can have other nucleophiles add to the bromonium ion in something that is known as forming a halohydrin.3540

The example here would be if we had an alkene reacting not just with bromine but bromine in water or maybe bromine in an alcohol; what happens in that case?3547

We form the bromonium ion; but instead of the Br- attacking it, it is the water that is going to attack it.3561

The two groups we end up adding will be a Br and an OH instead of two Br's; this is called a bromohydrin; when you have a Br and an OH, it is called a bromohydrin or a halohydrin as a general term.3568

We are going to get trans stereochemistry again; it will still be anti addition.3593

Of course, we have to say +enantiomer because it was totally random but I chose to make the bromine a wedge and the OH a dash; the other one will also be formed.3599

Let's think about a mechanism; how do we combine these three ingredients to make this bromohydrin?3609

The first thing that is going to be happening is this bromine is going to be reacting with the alkene; like I said, anytime you see bromine and alkene, we know we are going to form the bromonium ion.3616

I think it really helps to draw the bromonium ion that we are expecting; a lot of times we draw it as a wedge; you could draw it as a dash if you want; but this is often how we see it.3628

I know this is the bromonium ion I am going to get; I am going to be adding a bromine, Br+, to one face of the alkene; that helps me draw my mechanism.3638

This π bond is going to attack one bromine, kick off the other bromine; then that first bromine takes its lone pair and adds back in to form the second bond.3647

You can see there is still two lone pairs left; one lone pair has gone to form this bond; the π bond has gone to form this bond; here is our bromonium ion.3658

We have an electrophile with a bromonium ion; we look around for a nucleophile; what nucleophiles do we see?--we have water as a nucleophile; and we have Br- as a nucleophile.3669

Who do you think would be a better nucleophile?--I see a negative charge with this bromine; I think he would be a very good nucleophile.3684

However what we are looking at here is a matter of statistics because water is our solvent.3693

Water is surrounding all these molecules including the bromine atom; it is solvated; it is totally surrounded by water.3698

So statistically, as is this bromonium ion, statistically the nucleophile that is going to be most likely to attack and therefore leading to our major product is going to be the water instead.3711

Water does in fact, when you have water present, water will be the most significant nucleophile to attack; again Sn2 mechanism; we have our backside attack; it will come in as a dash then.3722

The bromine on this left-hand carbon is still here; he is still a wedge with three lone pairs now; he is neutral.3738

What do we have on the other carbon?--we have an oxygen bonded in the back position, in the dash position.3744

What else is it on this oxygen?--we still have our two hydrogens; and we still have one lone pair; and we have a + charge.3753

We formed a bromonium ion and we attacked the bromonium ion; is our mechanism done yet?3768

Once again, anytime we use water as a nucleophile, our mechanism isn't done until we get rid of that positive charge; how do we get rid of that positive charge?3774

We can deprotonate; you can use anything you would like; some books use Br- to be nice and tidy.3782

But really again, because water is our solvent, water is surrounding everything, that is going to be the most prevalent base that we have; we can deprotonate; then we are done.3792

In our mechanism, we usually show the mechanism just to get one of the enantiomers, knowing that we would have the same mechanism with the opposite stereochemistry to get the other enantiomer.3807

We can alter our reaction conditions a little bit, put in a solvent that could be nucleophilic; this is a new nucleophile that we are adding in here.3819

Therefore when we form a bromonium ion, it is going to be that solvent that attacks to give a halohydrin.3828

Because we are adding two different groups across the double bond, we need to think about regiochemistry again.3837

In other words, which carbon gets the bromine?--which ones gets the OH?--the regiochemistry that is observed is shown here.3843

The major product formed is the one on the top here where the OH goes on the middle carbon and the bromine goes on the end carbon; the other product is not formed.3852

We need to explore this a little more to understand why this would be the major product that is formed.3868

In order to do that, we need to look at our bromonium ion intermediate; both of these products would come from the same bromonium ion.3874

Here is our bromonium ion; our only question is: when water attacks this bromonium ion, where does it prefer to go?--in order to answer this question, let's take a look at the transition state.3882

Let's imagine that water attacking here and starting to form a bond, starting to open up the bromonium ion ring; let's show those as partial bonds.3896

As that carbon that carbon-bromine bond stretches, the positive charge on the bromine is now being shared with the carbon.3906

We have a partial plus on the bromine and a partial plus on the carbon that is being attacked by the nucleophile.3912

Or we could have the nucleophile attack this carbon; our partial plus will be on the bromine and this carbon.3920

If you take a look at these two transition states and we try and decide which one is more favorable, which one is going to be a faster reaction again?3927

We take a look at this location of this partial positive charge; how would you describe the carbon bearing this partial positive charge?3937

It is attached to two other carbons; this is a secondary partial positive; in this case, on the end carbon, it is a primary partial positive.3946

Just like we know for carbocation stability, we are going to extend that same logic to our partial positives here.3957

We will acknowledge that the secondary partial positive is better; it is more stable and therefore it is a lower energy transition state.3964

When the water goes to attack this bromonium ion and has to decide between these two positions, it is actually going to attack the carbon that is more highly substituted.3984

Because this carbon, let me write this down here, it is more partial positive; it is the better electrophile; that means the nucleophile adds here.3995

Water does not add to this end carbon; it is drawn to the more substituted carbon because that has more of a partial positive.4012

What is interesting about this is our mechanism is an Sn2; it is attacking this carbon and breaking open the ring.4022

But it is an Sn2; but it is going to the more substituted, more sterically crowded site; it is not just a simple Sn2; we describe this as an Sn2 with some Sn1 character.4031

The reason it has some Sn1 character is because of the positive charge; because we do not have just a neutral electrophile being attacked, it is not a plain old Sn2.4047

Because there is a positive charge on the leaving group, electronics beat the sterics; electronics win over the sterics.4061

Even though it is more sterically crowded, the partial positive character, the electron deficient character, is significant enough that that is where the nucleophile goes in spite of increased the sterics.4073

That explains our regiochemistry then; the nucleophile goes to the more substituted carbon; we can't really describe this as Markovnikov or anti-Markovnikov because we are not adding a hydrogen.4086

But this is similar to Markovnikov's rule that we have seen; although we are not adding an H+ in our mechanism, we are effectively adding a Br+.4099

Just like the H+ prefers to go to the carbon with more hydrogens, the Br+ prefers to ultimately end up there as well.4117

Br+ goes to carbon with more hydrogens just like H+ did.4125

Hopefully you can see, you can tie it in a little bit with the addition of HBr or the addition of water; those mechanisms had carbocation intermediates; there we were looking at competing carbocations.4136

Here, because both of these have the same bromonium ion intermediate, there is only one intermediate; what we are looking at is the path from that intermediate to the product.4148

Who is going through the lower energy transition state--therefore that is the faster reaction; slightly different explanations; but still they have some similarities.4157

Let's try and predict the product here; how about if we reacted this alkene with bromine and methanol as our reaction conditions, Br2 and CH3OH?4169

Again what are the questions we need to ask?--what are we adding?--if all we had was Br2, then we add a bromine and a bromine.4182

But because we have a solvent that can be a nucleophile, that can be a nucleophile, it will be a nucleophile.4194

The two groups you are adding are going to be Br and... what do we get if CH3OH was our nucleophile?--would we add the OH group?--no, we get an OH if we use water.4202

We get the OCH3 group when we have an alcohol; we add the OR group; in this case, with methanol, we would get the methoxy group.4215

Where are we adding it?--that is our regiochemistry; that is our regiochemistry; we are breaking the π bond; we are adding a group to each carbon.4226

Which one gets the bromine?--which gets the OCH3?--we need to consider that; we need to think about the stereochemistry.4237

What is the relative stereochemistry of those two things?--for that, we need to understand the mechanism.4243

We need to understand the mechanism; that is going to tell us the relationship of those two; let's look at some choices for our products and see if we can decide.4250

We said we are going to add a Br and an OCH3 to these two positions; let's take a look at those choices; all these are chiral so they all should say +enantiomer.4263

Assuming these are all racemic, which one has the proper combination of the regiochemistry and the stereochemistry?--first of all, the stereochemistry; let's think about our mechanism.4276

We go through a bromonium ion; the bromine adds to one face; then our nucleophile, our methanol, comes in from the opposite face; remember it is anti addition.4290

Anytime we have bromonium ion, we have to get anti addition; which are the ones that have anti addition?--here our OCH3 and our bromine added to the same face; that is syn addition.4303

Here the bromine and the OCH3 added to the same face; that is also syn addition; I know those answers are wrong.4316

But here the bromine is a dash and the OCH3 is a wedge; that is good; here same thing--bromine is a wedge and the OCH3 is a dash; perfect.4324

These are both anti addition; they have the stereochemistry right; how about the regiochemistry?4334

When we make this bromine ion... let's imagine that; let's imagine that bromonium ion; let's do a little work here to think about that.4340

The methanol is coming in and attacking; where is it going to go?--is it going to be governed by sterics?--no, it is going to be governed by the positive charge in this case.4350

Because this carbon is tertiary, it has a higher partial positive; so the nucleophile goes to the more substituted carbon of the bromonium ion.4362

Which product is that?--it looks like D is our best product because it has the proper regiochemistry as well; the methoxy went to the carbon that was tertiary.4373

Interesting again, an Sn2 happening on a tertiary center; this is not an ordinary Sn2.4385

It is definitely because of that charge that we are seeing that Sn1 character and we are allowing this to happen; very good.4390

Another reaction we can have for alkenes is one called catalytic hydrogenation; as the name implies, hydrogenation, we are adding hydrogen across the double bond.4402

We will break the π bond and we add two hydrogens; what you can see here from this example is that the hydrogens have both added from the top face.4413

We describe that as syn addition; we describe the hydrogens in the product as being cis to one another.4424

The reaction is a catalyst; that is why we usually call it catalytic hydrogenation; we will talk about catalysts in just a second.4432

It is considered a reduction reaction; that is because we are increasing the number of C-H bonds; we are increasing the number of C-H bonds; that is going to be a reduction.4438

Overall it is an exothermic reaction; like most of the addition reactions we are seeing for alkenes, this is exothermic; it is releasing energy.4453

That is because overall we are breaking a weak π bond and forming a stronger σ bond.4461

All these addition reactions are going to be favorable because overall the net change is breaking of a π bond and forming stronger σ bonds; that is usually good news.4475

Let's talk about the catalyst a little bit; what is going on here?--we have some kind of metal catalyst.4490

It is going to provide a surface upon which this reaction can happen; it is going to speed up the reaction.4495

What happens is the metal adsorbs the hydrogen gas; the metal also grabs onto the alkene; then the metal delivers the hydrogens to that alkene; in doing so, it delivers them to the same face.4502

It delivers them to just a single face; that is why we get syn addition; the hydrogens end up on the same face of whichever face of the alkene it goes to.4519

Our catalyst can be described as either heterogenous catalyst or homogenous catalyst.4530

The heterogenous catalyst, the ones we see most often--you will see palladium nine times out of ten as the complete reaction conditions for catalytic hydrogenation.4535

These are metals that are added into the mixture; they don't dissolve in the solvent; so we just stir them very rapidly or we shake the bottle maybe.4546

Because they don't dissolve, we need some way to mix them in with the gas and with the alkenes that is dissolved in the solution.4556

At the end of the reaction, you just filter the catalyst off and you can isolate your product.4564

The other class of compounds are called homogenous catalysts; they are called homogenous because the reaction mixture in these cases are homogenous.4569

They are a solution; they don't have any solids in there because they do dissolve in organic solvents.4577

That is because these are organometallic species; they have both organic components attached to the metal.4583

We have a variety of transition metals like rhodium or ruthenium or iridium; they have various organic ligands attached to them; this guy is called rhodium tris(triphenylphosphine) chloride.4590

Each of these phosphorus has three phenyl groups, has three benzene rings attached to it; then there is three of those coordinated around the rhodium.4607

A lot of benzene rings in there; that makes it organic, soluble; so we can use organic solvents; they are called homogenous that way.4614

By varying the ligands that we have here, we can have a huge variety of catalysts; you can do some cool things with that; you can have some increased selectivity.4626

For example, if you have a diene, if you have two alkenes in a structure, you could use this catalyst to just react with the less hindered alkene; it would be sensitive to sterics in that case.4634

You can't really do that with catalytic hydrogenation, with just ordinary catalysts; that just reduces every π bond it sees.4651

But we can do some more interesting things with the organometallic catalyst; we might see some of those too.4658

The heat of hydrogenation of this reaction is going to be useful to us; it can tell us something about the stability of the alkene starting material that we had.4667

For example, if we take a look at this series of alkenes, 1-butene, cis-2-butene, and trans-2-butene, what is interesting about this series of alkenes?4679

If we take any of these alkenes and you react it with hydrogen and palladium, you will break the π bond; you will add hydrogen to each carbon.4690

The product you are going to get is the same; in every case, you are going to get butane.4701

This is an interesting series of compounds to study because they all give the same product with the same energy at the end of the reaction.4706

We can measure the heat of that reaction; we see it here given at kcals per mole: -30, -28.6, -27.6; we see a difference in the energy given off.4718

This is going to be related to the stability of the alkenes; we already know something about the stability of alkenes; which of these do you think is going to be the least stable?4733

We know that the less substituted an alkene is the less stable it is; so we know that the terminal mono-substituted is going to be the least stable.4742

These are both di-substituted; of the di-substituted, who is the most stable?--the trans is the most stable; the fact that these energies are going in this trend is a reflection of that.4753

We can show a diagram here to show how we could relate the heat of hydrogenation data.4768

If we take a look at our alkene A, we know it is higher in energy than alkene B which is higher in energy than alkene C.4777

But when they get hydrogenated, they all go to the same place; they all go to butane as the final product; they end up with the same final energy.4788

To go from A to butene is a longer path than to go to B or to go to C.4797

Because A starts out as the least stable, the highest energy, it gives off the most heat when it is hydrogenated; that is why this number is the biggest number.4806

Where C, because it is the most stable, it starts out with the lowest energy, it does not have as much heat to give off when the hydrogen takes place.4819

If given some heat of hydrogenation data, we should be able to convert that, interpret that into something about the stability of our starting alkenes, of the alkenes that we are comparing.4830

This is also useful when we are looking at dienes; if we had two double bonds, what are the relationships we can have for those two double bonds?4841

There is three possible relationships they can have; they can have one where they share a carbon; they are right on top of each other, right next to each other; we call these cumulated systems.4849

The p orbitals on that middle carbon, because it is sp hybridized, the p orbitals on that carbon are orthogonal.4863

The p orbitals for the first double bond are perpendicular or orthogonal to the p orbitals for the second double bond.4870

We can have double bonds that are unrelated to each other; we call those isolated π bonds because we have these π bonds here and these p orbitals and these p orbitals.4877

Because of this sp3 center in between, there is no relationship to them; or we could have conjugated π bonds.4887

That means that the p orbitals on the first double bond nicely align with the p orbitals on the second double bond.4894

When we have conjugated, it means we have a π and then a σ and then a π like that; that allows for overlap; it allows for resonance.4901

Can we see that in our heat of hydrogenation data?--absolutely; when we have these isolated π bonds, we get about 60kcals of energy taken off.4916

That is approximately the heat of hydrogenation for an alkene, for an ordinary alkene times two.4928

An alkene was somewhere around 30kcals; an isolated π bond, we get about 60kcals because we have twice as many π bonds to hydrogenate, to react.4938

But when we force those double bonds to be cumulated to each other, it actually increases the amount of energy given off for hydrogenation.4949

That tell us that that arrangement now is less stable; this is the least stable of all three of these.4957

It destabilizes; putting those two π bonds perpendicular to each other destabilizes the π bonds.4963

Where putting them in a conjugated relationship actually stabilizes it; this is the most stable; that is because of that resonance.4970

We can use the heat of hydrogenation data to get some evidence for which arrangement of double bonds is more or less stable; what I forgot to point out is these are all pentadienes.4979

Again if we have excess hydrogen and palladium, if you have as much hydrogen as you need, you will reduce every π bond there is and convert it to an alkane.4992

We will get just pentane as our final product; it is useful to study this series because they all go to the same product.5004

We can do the same cartoon again for our energy diagram; A--the cumulated π bonds are unstable, are less stable than an ordinary diene; having them conjugated is more stable than an ordinary diene.5012

They all go to pentane as their final product; A has the highest distance to go; B and C has the lowest distance to go.5027

This -54.1, the smaller the number, the smaller the magnitude of it... they are all going to be negative; remember this is always going to be an exothermic reaction.5037

But the smaller that magnitude tells us the lower in energy our alkene or diene started before we did the hydrogenation.5045

Let's see if we can predict a product here; if we reacted this alkene with H2 and catalyst.5056

One thing I wanted to point out with this example is here we have a carbonyl; a carbonyl has a π bond but it is not a carbon-carbon double bond; so there is no reaction here.5066

Catalytic hydrogenation is a reaction for alkenes, carbon-carbon π bonds; we will see it for alkynes also when there is a triple bond.5078

We will leave that carbonyl there; we can draw our backbone; this is our carbon chain; we are adding hydrogens; do we know anything about the stereochemistry of that addition?5085

Regiochemistry is not an issue of course because we are adding a hydrogen and a hydrogen; there is no decision to make there; but how about the stereochemistry?5099

Let's think about that mechanism of the catalytic hydrogenation; the hydrogen is bound to the metal; it is going to be delivered to the same face of the alkene; we get syn addition.5106

How do we draw those hydrogens coming from the same face?--we could draw them both as wedges or we could draw them both as dashes.5117

If they are wedges, that means for the zigzag down, our wedge is pointing down; but for the zigzag up, our wedge is pointing up; here they are coming from the same face, all from the top.5125

What does that do to these two methyl groups?--that pushes them back into the plane; it is your choice on which carbon part of the chain you want to keep in the plane.5139

But you want to make sure that the other one gets pushed from being in the plane; it is no longer in the plane.5152

This is a chiral product; we want to make sure we say +enantiomer or we say racemic so that we know that having the hydrogens come from the bottom face would give a different product.5161

That product is also formed--the enantiomer; another way we can draw this... there is a lot of different ways to draw this.5171

It is important to keep that in mind when you compare your answer that you have worked on to the answer in the textbook or the solution manual.5179

That if theirs is slightly different, you need to be able to evaluate whether your answer is still right or not or if you have done something wrong.5185

Another possibility is we can say I want them to both come from the top face and pointing down here for example; we could do that; that would keep this first carbon the same.5191

But the second carbon, if the wedge is down here, then the bond that is in the plane is this methyl group is still in the plane.5209

It is the ethyl group that got pushed down to behind the board or behind the screen; a lot of different ways that you can draw this; but get into something that you are comfortable with.5217

And that again, with a lot of practice, you are used to drawing these chiral products and these new stereocenters that are going to be formed when we form new chiral centers in addition reactions.5230

The next group of reactions we are going to look at are called oxidation reactions; there is several ways that we can oxidize an alkene.5243

First though let's take a look at the word--oxidation; we know it comes hand in hand with reduction; we have learned about redox reactions in the past.5251

But in organic chemistry, we view it a little differently; it is a little easier to see than some of the redox reactions that you may have seen in general chemistry.5259

Let's just take a look at organic examples of oxidation and reduction reactions; let's just look at a one carbon species.5268

Every carbon has four bonds; we have this example with methane; all four bonds are to hydrogen; then we replace one of those bonds with an oxygen--this is methanol.5278

Then we replace two of those bonds with oxygen--this is formaldehyde; then we could have three bonds to oxygen--this is formic acid.5288

Then finally all four bonds can be to oxygen--this is carbon dioxide; we have gone from four bonds to hydrogen and one by one we are replacing a hydrogen bond with an oxygen bond.5299

This process in going from left to right is described as an oxidation; all these are oxidations.5313

For example, you could say that methanol could be oxidized to formaldehyde or even it could be oxidized to formic acid.5325

The way that we can identify it as an oxidation is we are increasing the number of C-O bonds while decreasing the number of C-H bonds.5336

Any process that trades a C-H bond for a C-O bond, we know is an oxidation; as a result, the oxidation number will increase.5348

But we don't have to calculate the oxidation state of each carbon in order to make that determination; it really can be done visually by looking at the oxygens.5362

This reverse process in going from right to left, these are all described as reductions; what happens in a reduction is we have an increase in the number of C-H bonds.5372

At the same time, there is a few different things that can happen; in this case, we are seeing a decrease in the number of C-O bonds; other bonds can be lost instead.5387

In general, as soon as we are seeing we are increasing in the number of C-H bonds, then we are going to see it as a reduction reaction; like catalytic hydrogenation, we call this reduction reaction.5396

As a result, by increasing those CHs, our oxidation number of the carbon will go down, will be reduced just like in a reduction reaction.5406

It is not changing the definition of a redox reaction; we are still having a change in our oxidation state; but we are just going to look at it more simply in this case.5414

There is going to be three... as you can see, there is going to be three types of oxidation reactions we are going to be studying for alkenes.5427

This first one will give as a product--we break the carbon-carbon double bond and we put in its place a three-membered ring with an oxygen in it; this is known as an epoxide.5433

It is called an epoxide when you have a three-membered ring with an oxygen; that is one possible oxidation.5448

If you react a double bond with KMnO4, what happens is you break the π bond and you add an OH to both carbons; that forms a diol or a glycol.5455

It could be called a glycol if you have two OHs right next to each other or a vicinal diol.5467

Or you can react it with ozone, a reaction called ozonolysis; in that reaction, we actually break the carbon-carbon double bond and we replace it with carbon-oxygen double bonds.5474

We get these carbonyl products; we are going to get either ketones or aldehydes or some combinations of those depending on what your substitution patterns were on the original alkene.5489

Handful of new reagents; very different looking products; but do you see why these are all grouped together and all described as oxidation reactions?5505

In each case, you have lost carbon-carbon bonds; what have you replaced them with?--carbon-oxygen bonds, carbon-oxygen bonds, carbon-oxygen bonds.5514

You are increasing the content of oxygen in your structure; we call that an oxidation; let's look at our first one--the epoxidation reaction.5521

If we take an alkene like we said and we react it with this guy... we will define that in minute; it is called a peroxy acid.5532

The product we get is an epoxide; we are going to add both of those C-O bonds from the same face; we call that syn addition; I have drawn both of those bonds as wedges.5539

I added the oxygen from the top face; what if I added it from the bottom face?--is that another product?--do I say +enantiomer here?5556

Would it be a unique product if I drew both of those bonds as dashes?--or would it be just the same thing flipped over?--same thing flipped over.5564

Because this is a meso compound, it has no enantiomer; this is an achiral product; this would be a case where we just draw the stereochemistry as either wedges or dashes.5573

Then we are done; we don't have to say +enantiomer; it would be wrong to say +enantiomer.5588

Let's take a look at this formula RCO3H; it looks very similar to a formula we may have seen before, RCO2H.5593

RCO2H, when you expand that, you have a carbonyl and an OH; that is a carboxylic acid.5600

That is an ordinary compound; we will study its reactions down the road; but as the name implies, it is just an acid; it is not an oxidizing agent.5610

But now the formula we are looking at is RCO3H; just by looking at the formula and recognizing that it deviates from the carboxylic acid, it has extra oxygen in it.5619

That should tip you off to the fact that it must be an oxidizing agent; it must be something that causes oxidations.5632

Let's look at that structure; it still has a carbonyl; but now it has two oxygens; this is described as a peroxy acid.5641

This oxygen-oxygen bond is called a peroxide; this is very unstable; this is dying to get rid of an oxygen--very easy to cleave that bond.5657

That is what leads to its reactivity and the fact that it is possible to oxidize things.5672

Another example of a peroxide is this guy, H2O2; what is that called?--that is called hydrogen peroxide.5680

This is another oxidizing agent; this in fact would also work to do this epoxidation reaction; that would work just fine because you still have the peroxy group.5694

Look at the formula, H2O2; it has an extra oxygen compared to water which is nice and stable; this is now instead a strong oxidizing agent.5702

We might just draw a peroxide; we might draw a generic formula like RCO3H to know that we are dealing with a peroxide or we could pick a specific peroxide reagent.5715

One that is very commonly used is this guy; he is called meta-chloro; meta describes the relationship between these two groups--that they are 1,3.5727

We have a chloro and we have a peroxybenzoic acid; without this double oxygen, this would be benzoic acid; but because it has the extra oxygen, we call it peroxybenzoic acid.5736

Luckily we can abbreviate this guy as mCPBA; mCPBA stands for meta-chloroperoxybenzoic acid; if we saw the formula, what part of the formula would clue you in that this is an oxidizing agent?5746

Right there--that extra oxygen; if we saw the name, we need to just recognize the name, very common to just use the acronym here; again another great flashcard to get to know mCPBA.5764

Just a general mechanism; for the most part, we are not going to be studying detailed mechanisms for these oxidation reactions because they are outside the scope of our organic chemistry mechanisms.5776

But I want to give you a little bit of an idea of what the mechanism is so it is not just this total magical black box thing.5786

What happens, by having these two oxygens attached to one another and having this acetate group here, is we have a leaving group attached to an oxygen.5793

Very unusual; that is again what makes it such a great oxidizing agent; what happens is the π bond of the alkene attacks the carbon and kicks out the leaving group.5804

But at the same time, this lone pair adds back in and forms the three-membered ring; we saw this similar mechanism for the oxymercuration, forming the mercurium ion.5817

We saw this with Br2 to form the bromonium ion; in this case, we are seeing that this would be a way to form a three-membered ring with oxygen, not as an intermediate but as a final product.5828

This mechanism happens again; because it is concerted, because both of those bonds happen in a single step, that is why they must come from the same face; we get syn addition.5839

It is concerted so we have syn addition; let's look at another way that we can make an epoxide; what would happen if we took an alkene and we react it with bromine and water?5850

I know that bromine always makes that bromonium ion; would this add a bromine and a bromine?--or do we have to pay attention to this water here as our solvent?5870

We do in fact because that it going to win over the bromine, the Br-, as what gets to attack the bromonium ion.5883

This is going to add a Br and an OH across the π bond; what about the stereochemistry of those two groups?5894

Again because we know the bromonium ion is the intermediate, that means if the bromine comes from the top face.5902

Then the water must come from the bottom face due to that backside attack for the Sn2 to open up the bromonium ion.5908

We can show... anti addition is what we call it; we are going to get the OH and the Br trans to each other; +enantiomer; it doesn't matter who you draw as a wedge and who you draw as a dash.5916

We saw this reaction as a way to make bromohydrins; what I want to show now is there is a nice application for bromohydrins.5934

If I took this structure and I treat it with sodium hydroxide, what can happen?--sodium hydroxide is a base; it is a nucleophile; you might think maybe we can do some substitution, elimination.5942

Anytime you see something that is a base, we want to look first to see is there any place that it can act as a base?--is there any proton that is unusually acidic?5958

Sure enough, we have an OH group here; in the presence of base, we are going to deprotonate some of that alcohol group to make an O-.5967

What can happen here is now we have a great nucleophile, very strong nucleophile, in a structure where we also have a leaving group.5979

What can happen is that nucleophile can attack the carbon and kick off the leaving group; this is an intramolecular Sn2 if you want to call it that; it is a backside attack.5989

Those are possible; we have usually seen it just for five or six-membered rings; but three-membered rings are possible too.6015

Now we are seeing that if we did that here, we would get a product we call an epoxide.6021

What is nice about this synthetics scheme is it relies on some newer reactions we have seen with addition reactions to alkenes.6025

And some older reactions we have seen where we are doing a substitution of a good nucleophile on a carbon bearing a leaving group.6035

Why might I use this instead of just using mCPBA of some other oxidizing agent?--maybe I have some other functional groups on this molecule that are sensitive to oxidizing agents.6045

It might not be suitable to do an oxidation reaction; but I could still convert the alkene to an epoxide by doing electrophilic addition and then a base-promoted substitution reaction.6055

The next oxidation reaction we will study is dihydroxylation; dihydroxylation means that we are adding two hydroxyl groups to the alkene.6074

The way we do that is with either KMnO4 as a reagent or OsO4 as our reagent, either of these.6086

You can see the similarity in their structures; we have a metal with four oxygens on it.6092

In fact, our reaction conditions are usually more complicated; they have some other additives in there; some other oxidizing agents.6097

But these are the key ingredients we need to see that will tell us we are doing a dihydroxylation.6103

What we do is we break the π bond; we add an OH to each group; this turns out to also be a syn addition; we get syn addition of two OHs.6109

We call this a dihydroxlyation, a syn dihydroxylation, meaning the OHs end up cis to each other.6130

Any enantiomer in this case?--I drew the OHs both as wedges; how about if I drew them both as dashes, would that be the same compound or would that be a different one?6138

I don't see any plane of symmetry in this molecule; this is a chiral molecule; that means it must have an enantiomer; this is a case where we would want to say +enantiomer.6148

How do those two OHs get added in a syn fashion?--again just a little bit of a mechanism to give you some background on that, not a detailed mechanism.6160

But if you think of this potassium permanganate or the osmium tetraoxide, if you think of either of these reagents as some metal with four oxygens.6170

What is going to happen is that metal is going to take two of those oxygens and add them at the same time to the double bond.6179

It is going to be another concerted mechanism; we are going to form this bond and this bond all in one step.6185

Our mechanism looks something like this--we form a bond here; this π bond moves; this π bond moves; something like that.6192

Because they are adding in the same step, the oxygens have to add to the same face.6204

It is concerted; it goes to be a cyclic intermediate; when we follow our electrons around, we get something like this.6213

Again the exact details of the metal species is not important because it is going to vary whether it is manganese or osmium.6220

But the idea is we have added them in one step; that is why we must get syn addition; like the epoxidation, any concerted mechanism is going to give you that syn addition.6228

How does this get to this?--how do we end up with the two hydroxyl groups?--again we do some kind of workup; we cleave off the metal and we free those oxygens to be OHs.6245

But again carbon-oxygen bond is now set and the stereochemistry is fixed.6255

Finally let's talk about ozonolysis; this is our third oxidation reaction; as the name implies, it uses ozone; let's take a look at this ozonolysis.6264

We have two words here; the ozone tells us that we are using O3 as our reagent; what does lysis mean?--what does it mean to lyse something or to cleave it?6277

It means we are breaking apart; we are going to be breaking or cleaving the molecule with ozone.6289

Just knowing the word ozonolysis, knowing the name of this reaction is going to help you predict the product properly.6299

But first let's talk about ozone just for a second because we have all heard of ozone before; it is interesting that we are using it as a reagent in the lab.6306

The question I want to ask is do you think that ozone is something that is good or something that is bad?--it kind of has a dual nature.6313

We know that there is an ozone layer and we are very concerned that there is a hole in the ozone layer; that let's us know that ozone is a good thing and the destruction of ozone is a bad thing.6323

That is true that ozone can be good; that is when it is in the stratosphere--that is the layer of the atmosphere, the outermost layer that is protecting us from outerspace.6336

That is because what the ozone does is it absorbs UV radiation; the sun is coming down with a ton of this ultraviolet radiation.6356

The ozone layer is doing a good job of absorbing a lot of that so only a fraction of that makes it down to earth.6370

That is a very good thing because this radiation is... meaning it is coming from sunlight I should say... that is a good thing because this UV radiation is damaging.6376

It can do damage to your skin; that is what gives you a sunburn; that is what causes the simplest cases; it is a sunburn.6391

But you can also damage DNA; if it penetrates, you can have damage there, cause skin cancer, and so on... put that here too.6402

The more UV we have coming down to earth, the worse shape we are; so the ozone is very good for that; it is protecting the earth; we don't want to put anything in there that might destroy the ozone.6413

But it is bad, ozone is something that is bad when it is in the atmosphere; in other words, if that is something that is surrounding the earth.6427

If it is something that I am interacting with, then ozone is not a good thing; it acts as an irritant to your mucus membranes and so on.6436

It is a smog component; it is one of the parts of smog that is tracked for health alerts and that sort of thing because it can cause irritation in your nose and your eyes and that sort of thing.6448

It also... we are seeing it is an oxidizing agent; it is a very strong oxidizing agent; if we have a lot of ozone around.6463

Then that can do a lot of damage to just materials that we have that might have double bonds in them; we are seeing that it reacts with double bonds; obviously ozone will do that rapidly.6469

Just the oxygen in the air is also an oxidizing agent; it could cause some of these damaging reactions; but ozone is even more reactive and will do it more rapidly.6481

For example, if you have ever taken a rubber band and tied up some bills or something and thrown it in the attic or thrown it in a box and stored it for a while in your closet.6490

What happens, when you go back to that box and you are down the road and you take that thing that was strapped with the rubber band, what happens to the rubber band sometimes?6499

Have you ever seen it just crumble or crack or fall apart or become stiff, something like that?6507

That is an example of oxidation that is happening to it where it is breaking down the structure of the rubber.6513

We can have oxidation reactions like rubber bands for example; or your tires on your car, they will crack; they will wear because of oxidation reactions; and so on.6521

What we have is we have a little ozonizer; we have a little instrument that we can bring in the lab and it generates a stream of ozone gas.6535

You stick that into your reaction; you can bubble some ozone through it and we can cause this oxidation reaction to occur.6542

Let's see what happens at the molecular level when it reacts to the double bond.6548

What is going to happen when it reacts with ozone... again not doing a detailed mechanism here but just giving you some idea... is we are going to completely cleave all bonds between the two carbons.6553

The two carbons, you still have two bonds connecting them; the ozone inserts three oxygens in between the carbons so the two carbons are no longer connected.6564

This is known as an ozonide intermediate; that is always formed as your first step in an ozonolysis; that is why there is always a second step in an ozonolysis reaction.6574

It is always followed up by some other set of reagents; these are typical ones we might see like zinc or dimethyl sulfide; these are reducing agents; this is known as a reductive workup.6583

What happens is we simply take these carbons; this carbon has two separate bonds to oxygen; after our workup, we are just going to put both those bonds to a single oxygen.6596

This carbon has two bonds to oxygen; after workup, we are going to have two bonds to a single oxygen; that is all.6610

With this reductive workup, there was a hydrogen here; there was a hydrogen that started out all along; that hydrogen is going to remain in our product.6618

Depending on the type of workup you have that might change; sometimes we can do an oxidative workup where we get rid of that hydrogen.6627

But this is a typical procedure for ozonolysis where we react step one with ozone and step two with some kind of reducing agent.6634

Our product, how do we predict our products?--we completely break the carbon-carbon double bond; we replace the carbon-carbon double bond with a carbonyl at each position.6644

We insert a C-O double bond called a carbonyl; in this case, we got one ketone and one aldehyde; we describe it as an aldehyde if it has a hydrogen attached to it.6654

But we will get a mixture of ketones and aldehydes depending on what alkene we started with.6665

Let's look at a few examples of predicting the products of ozonolysis; again looking at our reaction conditions and trying to decide what to expect here.6671

I see ozone; there is exactly one reaction we know that uses ozone; that is ozonolysis; right there in the name, we find out what we are doing to this molecule.6681

We are finding our double bond and we are completely breaking that double bond; when we go to draw the products then, we will now have two separate products.6695

We have broken our molecule into two pieces; we are going to have a carbonyl on the five-membered ring and we are going to have a carbonyl...6704

You can leave it upside down; you don't have to flip it over; it is fine to leave it in the orientation it had; we are going to have a carbonyl with this three carbon chain.6712

One great check that you can do when you are doing an ozonolysis is to take a look at your starting material; I had eight carbons in my starting material.6719

When I am done, I should still have eight carbons; I do--I broke it up into a five carbon ketone and a three carbon ketone.6729

Because ozonolysis will take every double bond in a structure and break it; if you have multiple double bonds you will end up with lots of fragments for your product mixture.6739

In fact ozonolysis is a nice way to study the structure of a compound--is by breaking all the double bonds and studying the fragments that you get.6752

You can help determine where the double bonds were located in your structure; those are some good exercises too--is looking at ozonolysis products and putting them back together.6761

How would we do that in this case?--that is a good problem; if you said that these are the ozonolysis products and you were asked what is the starting material that gave these products?6771

How would you approach that reaction?--you know that ozonolysis forms carbonyls from carbon-carbon double bonds; we are going to take those two carbonyls and we are going to reattach them.6786

We are going to reattach them as double bonds to each other; take a look for some of those problems; those are good exercises as well.6797

How about this one where we have a double bond within a ring?--how is that going to affect our product?--again ozonolysis and zinc reductive workup; looks pretty typical.6806

Here is the double bond that we are going to be breaking; but now because we are cleaving a ring, we are going to cause the ring to open up.6818

This is a really good idea to number our carbons so we don't get lost.6826

What I recommend actually as a good way to handle some of these problems is to redraw your starting material with your carbons numbered--one, two, three, four, five, six.6831

Then literally erasing the bond that you are breaking; that way I know I haven't lost any carbons or gained any carbons.6843

I am breaking that double bond; what am I replacing it with?--a carbonyl at carbon 1 and a carbonyl at carbon 6.6850

You don't have to straighten it out and draw it as a nice zigzag line; you can leave it in whatever conformation you want.6859

The only thing I will point out here though is because I made an aldehyde, it is okay to leave your line drawing like this.6866

But by convention, we usually draw in the hydrogen of an aldehyde; that is the only hydrogen attached to a carbon that we normally leave in a line drawing.6873

I am just going to feel a little more comfortable adding those hydrogens in because that looks a little better to me rather than having a carbonyl dangling at the end of the chain.6885

But that is optional; even though they weren't shown on carbons 1 and 6, of course there is a hydrogen here and a hydrogen here; those are still preserved with our reductive workup for the ozonolysis.6892

Finally let's look at some additions to alkenes that follow a completely different mechanism; that is a radical mechanism; let's think back to what we know about radical reactions.6908

The one reaction that we have seen before is the free radical halogenation of alkanes; we take an alkane and we react it with bromine and light for example; we would form an alkyl halide.6920

What we are doing is we are taking our alkane, we are plucking off a hydrogen, and we are replacing it with a bromine; we call that free radical halogenation.6933

Let's review the mechanism for that so we can look at a new radical mechanism; the very first step in any radical mechanism is some kind of initiation--something that forms radicals.6941

In this case, the reaction, the exposure of Br2 to light, to hν--we represent it that way; it causes homolytic cleavage of that bromine-bromine bond where we split it.6952

One electron goes to each bromine; we are going to form a radical; we are going to form two of these; we are going to form two bromine atoms; that is our initiation.6964

Then what that bromine does is a hydrogen atom abstraction; it is going to look for a hydrogen on the alkane to remove.6975

It is going to remove the one that results in the most stable radical possible; I am going to go for one of these internal ones rather than one of these end ones.6985

What an atom abstraction looks like is you break the carbon-hydrogen bond; one electron pairs up with the radical; the other electron stays behind.6995

We just formed a molecule of HBr; in fact I forgot to write up here--that is the other product you get in free radical halogenation.7006

I usually am more concerned with the organic product; but you are also forming HBr here.7014

The other product we get is this secondary radical; what does this radical do?--same thing--atom abstraction; that is what radicals do.7020

In this case, it is going to go after bromine; it is going to take a molecule of bromine and it is going to do a bromine atom abstraction.7031

We are going to, just like we did before, break this bromine-bromine bond; one electron pairs up with the radical; the other electron stays behind.7041

That is how we get our product; that is our bromination product; the other thing we formed here was Br·.7052

What was nice about this radical reaction is it also reformed the Br· which can come back and do another hydrogen atom abstraction and so on and so on; we could say et cetera here.7061

We get this chain reaction mechanism that started with the initiation; that is because the majority of the steps that are taken in a radical mechanism are described as propagations steps.7072

We start with one radical; it reacts with some stable species; then as a result, we get a new radical; here we had a Br· reacting with an alkane; that gave an alkyl radical.7087

Then we had an alkyl radical reacting with Br2 to give a bromine radical; it is propagating because one radical begets a new radical.7101

There is two general types of propagation steps we can have; in free radical halogenation, the only one we saw was this first one--for example, atom abstractions.7112

The radical comes after an atom from a stable molecule such as a hydrogen atom and it removes that hydrogen; we saw it can pluck off a hydrogen; it can pluck off a bromine from Br2.7121

We are going to make a molecule here of a hydrogen attached to the radical; we are going to form a carbon radical in this case.7136

Free radical halogenation relies on only atom abstraction as our propagation steps.7145

But if you happen to have an alkene around, a carbon-carbon double bond, there is another propagation step that can happen; that is called addition to an alkene.7152

What does it look like?--a radical comes in and it is the π bond that breaks; one electron pairs up with the radical; the other electron stays behind.7160

Let's take a look at that product; what has happened now is this radical has bonded to the alkene and formed a radical where the π bond used to be.7173

We call this an addition to an alkene; since we are studying reactions of alkenes, let's take a look at some examples of radical additions to alkenes and see what those products look like.7183

The most common reaction to study in the first place is the addition of HBr; we have already seen... let's review the reaction we have already seen.7195

That is addition of HBr; just normally what happens is we break the π bond, we add an H and a Br, and we do it with Markovnikov regiochemistry.7209

This is our electrophilic addition; we saw that mechanism; that mechanism was protonation with our acid and then addition of Br-.7218

But if we do this reaction in the presence of ROOR, which is the formula for some kind of peroxide, a different mechanism occurs which is a radical mechanism.7228

As a result of that mechanism, we get this product which has the bromine on the end carbon and the hydrogen on the middle carbon; we describe that as anti-Markovnikov.7242

It is possible to add HBr with Markovnikov regiochemistry or anti-Markovnikov regiochemistry; we control that by whether or not we have added in some peroxides.7252

Let's see how this mechanism works, the radical mechanism works, and why is it that we get the anti-Markovnikov product.7264

As usual, our reaction has to start somewhere; we have to initiate a radical; that is the job of the peroxide; this is a radical initiator.7272

It is this oxygen-oxygen bond that is so weak and just loves to cleave homolytically and form two radicals, alkoxy radicals in this case.7280

We have a source of radicals now; our reaction conditions are--we have an alkene and we have HBr.7295

What this radical is going to do is it is going to do a hydrogen atom abstraction on HBr and pluck off a hydrogen to give Br·.7302

Ultimately the result of this reaction of HBr with peroxides is we are forming bromine radical; we are forming some Br·.7312

How is that Br· going to interact with the alkene?--if we have an alkene around, it can add to that alkene; we have a four carbon alkene here.7325

We are going to break the π bond; one electron joins up with the radical; the other electron stays behind; the bromine adds and forms a radical.7338

This is the step at which the regiochemistry has been determined; the first group you add has to go to one carbon or the other; that is when you decide the regiochemistry.7356

Let's look at the choice we have to make; the bromine can add to the end carbon or the bromine can add to the middle carbon; that would give the radical over here.7364

Let's look at our two possible intermediates, our competing intermediates, and decide which one is better and wide.7375

Radical stability turns out has the same general order as carbocation stability; if we take a look at the secondary one and this primary one, we will see that the secondary is going to be favored.7385

Just like we have seen before, we are going to follow the path that gives the more stable radical intermediate; that is important.7401

It turns out because radicals are not quite as unstable as carbocations... this is not the only issue that is going on here.7421

It turns out that it is also nice that the Br· adds to the less hindered carbon.7428

For this radical reaction, actually steric hindrance does play a little bit more than it would for just the addition of HBr with our other electrophilic addition mechanism.7439

But either way, we can see that the bromine wants to add to the end carbon; we now have this radical; how do we get to our product?--we wanted to add HBr.7448

Let's come back and redraw our radical; what can that radical do to get to the final product?--we only have two choices; it can either do an atom abstraction or it can add to an alkene.7456

Because we have plenty of HBr around, it is going to choose to do an atom abstraction; it is going to pluck a hydrogen atom off of HBr; we have just formed our product.7471

What else do we generate in this step?--we just generated some more Br·; he can continue to come back and add to a new radical and so on.7489

Once again, these propagation steps go on and on and on; this is the way that we make our product--is we start with a radical and then we make a new radical and we keep going and keep going and keep going.7500

We saw that we have some techniques to add water across a π bond with Markovnikov or anti-Markovnikov regiochemistry; that is very important.7515

Now we see we can do the same thing with the addition of HBr; we can add it with Markovnikov or anti-Markovnikov by adding some peroxides to our reaction mixture.7523

Let's take a look at one more interesting reaction that has a radical mechanism involving alkenes; that is the idea of doing a polymerization starting with an alkene.7536

This molecule is called vinyl chloride because it has a chlorine attached to a vinyl group; it is attached directly to a double bond; this is described as a monomer because it is just a single unit.7550

If you added some kind of radical initiator, then we would expect that initiator to add into the alkene and form a radical.7562

Let's suppose that in these reaction conditions all we have is this monomer; all we have is a ton of vinyl chloride, just a tiny bit of radical initiator to get a radical mechanism going.7573

What product do we expect to have happen here, to observe here?--we have this radical; if this has plenty of vinyl chloride around, then this new radical is going to do another addition to an alkene.7586

Now we have two units of vinyl chloride here; we call this a dimer because we have two of them; if we still had vinyl chloride around, then this could continue to add, break the π bond.7600

We could get a trimer and a tetramer and so on; it can continue going until we get a polymer--many many many repeating units.7615

This is one of the ways that polymers can be built--is by counting on this radical chain reaction; we can see here that we have a repeating unit.7624

This is what we see; we see our vinyl chloride components at two carbons with a chlorine on there.7638

Depending on the monomer that we start with, we can get a different polymer each time; depending on what groups are attached, then we can get very different properties.7645

This is PVC--this is polyvinylchloride; it is used for water piping, for example, for sprinkler lines, that kind of thing is PVC piping.7655

But also a wide variety of fabrics and plastics and materials that we can get from this polymerization reaction.7665

It is also possible to do the same sort of polymerization reaction with an acid catalyst.7671

Because if we form a carbocation in the presence of lots of other alkenes, then that alkene can act as a nucleophile.7682

And so on, giving a new carbocation which then can add to another carbocation and so on.7702

That would be another way of doing a polymerization; but radical mechanisms are very well suited for that as well.7709

That sums it up for the many reactions we can have for alkenes; we will talk to you later; thanks for coming to Educator.com.7716

Hello and welcome back to Educator.com.0000

Today we are going to talk about alkynes; an alkyne is a molecule that has a carbon-carbon triple bond in it.0002

We could describe these three bonds connecting the two carbon atoms here as one σ bond and two π bonds.0009

The hybridization of each carbon in this alkyne because it has just two regions of electron density around it--the triple bond and the single bond.0022

We describe it as being sp hybridized; both carbons are sp hybridized.0032

The way that the σ bond is formed is the sp hybrid orbital from one overlaps with the sp hybrid orbital on the other; we describe that as an sp σ bond.0038

Of course, π bonds as usual come about from an overlap of a p orbital with a p orbital and another p orbital with a p orbital.0049

When we try to do a 3D sketch of an alkyne, it tends to get a little messy because we have six electrons here we have to find homes for in a small area; we will get a little practice in that.0058

Let's take a look at a model so maybe you can visualize this before we go to draw it; what we have here is a model of an alkyne; of just acetylene as our simplest alkyne.0071

The two black atoms are the carbons; we just have a single bond out to hydrogen here; notice the geometry of the alkyne with that sp hybridization is linear.0082

We can mention that with our sp--is that it is linear, meaning these bonds are going to be 180 degrees.0094

If you take a look at the π bonds, as usual the p orbitals are going to be perpendicular or orthogonal to each other.0101

With the purple, we have one set of p orbitals; that will be one π bond; that is going to be drawn in the plane of the page, perpendicular to the sp linear.0112

The second set of p orbitals are coming straight out and straight back.0123

If you imagine that the p orbital that was used to hybridize is here along the x-axis, then the π bonds are from the py axes and the pz axes.0127

We can add that little detail if we would like to here--is that we have two aligned p orbitals in the y-axis and two aligned orbitals in the z-axis; we could do that too.0140

When we go to do a sketch, we want to draw the carbon-carbon σ bond as well as the σ bonds going either direction; that is going to be 180 degrees.0151

Then 2hen we draw our p orbitals for our π bonds, we can draw one in the plane very nicely, two lobes of the p orbital.0163

Then we show some interaction with the top and the bottom half there; that represents one π bond; where is the second π bond?0172

The second π bond, that p orbital is going to be coming out and going back; what we could do is just maybe twist the molecule a little bit to the side so we can see both halves.0180

Maybe we can draw one as a wedge and one as a dash, one as a wedge and one as a dash; then try and show some connectivity between these front halves and these back halves.0189

Again, the larger you draw your 3D sketch, the better chance you have of having enough room to show everything.0200

Maybe we could show the bond angle over here since it is a little less messy; we can say that it is 180 degrees.0208

We want to note that our p orbitals are perpendicular to the sp linear line; that is what we are showing in our 3D sketch.0214

So really, an alkyne is a linear molecule; you can see that you have as a cylinder electron density around the carbon-carbon σ bond; that is what a triple bond looks like.0223

A triple bond has room for bonding on either end; it has two things that are going to be attached to it.0237

One other bit of terminology we are going to use to describe alkynes, we are going to describe them as being internal if there is a carbon group on either side of the triple bond.0244

In other words, the triple bond is within a carbon chain; we are going to describe it as terminal if it is at the end of the carbon chain.0253

What makes a terminal alkyne unique is that it has a hydrogen attached to the triple bond carbon; we are going to see some unique reactivity for that.0261

We will see some reactions that are good for all alkynes; then we are going to see some reactions that are specific just to terminal alkynes.0269

One of the reactions we can do for an alkyne is we could do a reduction; if we do catalytic hydrogenation much like we did for alkenes.0279

In other words, we react with hydrogen gas and some kind of catalyst like palladium or platinum, something like that typically.0286

If we have an excess amount of the hydrogen gas available, then what we expect to happen is much like we did for the alkene.0293

We break one π bond and we add a hydrogen to each carbon; then we are going to break the second π bond and do the exact same thing.0303

If we do an exhaustive hydrogenation like this, we can convert an alkyne to an alkane; you would end up adding two equivalents of your hydrogen gas.0313

However if we used a different kind of catalyst, if we still had plenty of hydrogen around.0326

But instead of just using palladium, we use this combination of palladium with some kind of salt, like barium sulfate or calcium carbonate, and quinoline.0333

The molecule of quinoline is here; it has two aromatic rings with a nitrogen replacing one of the carbons.0341

This combination is known as Lindlar catalyst or Lindlar's catalyst sometimes; sometimes you might see all these ingredients listed out or sometimes you might just see it referred to as Lindlar catalyst.0347

This is described as a poison catalyst; what happens with a poison catalyst is that there is no reaction with alkenes.0361

We are going to only partially reduce the carbon-carbon triple bond; we are going to add one equivalent of hydrogen; but then we are going to stop at that point.0376

Remember the mechanism for the catalytic hydrogenation; the metal is going to absorb the hydrogen; the metal is going to grab onto the alkene or alkyne in this case.0387

It is going to deliver both of those hydrogens to the same face of the π bond.0397

We are going to be able to observe that in this case because the two hydrogen that are added are going to be added to the same face.0402

We are going to get... I forgot there was a prime here... we are going to get a cis alkene as our product when we do a poison catalytic hydrogenation; a cis alkene.0411

Up here when we added two equivalents, yes those hydrogens were added in a syn fashion.0423

But because each carbon received two hydrogens, we didn't have any stereocenter there; so there is no way to observe that stereochemistry.0429

There is no stereochemistry to show here; but here we want to be sure to very carefully draw the cis stereochemistry when we add a single equivalent of H2.0435

There is one additional reduction reaction we can do for alkynes; that is called dissolving metal reduction--that is when we use either sodium metal or lithium metal in ammonia.0446

Ammonia is being used here as our solvent; it is also part of the mechanism; it is a proton source in the mechanism.0457

What happens in this case is we reduce the alkyne partially and we get the trans alkene as our product.0466

I am not going to go through a mechanism for this; that is something that you can find in a textbook or some other resource.0480

But just to give you an understanding of what is going on in this mechanism--is we are using sodium metal; this is sodium metal, lithium metal; it is not the cation.0489

Each of those metals has a single electron in its valence shell, very readily loses that single electron, makes a great reducing agent because it donates that electron to someone else.0498

The alkyne is accepting that electron, getting reduced; what is happening is our Na0 is going to Na+ after it donates the electrons.0508

The sodium is getting oxidized as the alkyne is getting reduced; this is a redox reaction like we have seen for electron transfer reactions.0521

What is important to note here is that, as opposed to the Lindlar catalyst, we are going to get the opposite stereochemistry.0532

We are going to get the trans alkene; that is due to the stability of the intermediate that is formed in this reaction.0539

A lot of different options for reducing an alkyne; you can do it completely to an alkane; or you can do it either to the cis or trans alkene.0549

When it comes to synthesis problems, we have a lot of variability, a lot of versatility, with an alkyne starting material.0557

We just talked about reductions of alkynes because we were increasing the number of C-H bonds; oxidations of alkynes means we are increasing the number of C-O bonds.0567

There is a few different ones; the one that you will see most often is this one--it is called ozonolysis.0577

When we react with ozone, just like we did for an alkene, we are going to completely cleave this carbon-carbon triple bond.0585

Because we have three carbon-carbon bonds that we are breaking, this is an oxidation; we are going to be replacing those three with carbon-oxygen bonds.0595

How do we a draw a carbon with three bonds to oxygen?--we are going to make one a carbonyl just like we did for ozonolysis of an alkene.0607

We replace the double bond with a carbonyl, C-O double bond; then we are going to add a third bond to oxygen which will result as an OH group.0616

We are going to convert all three carbon-carbon bonds to three oxygen bonds; that is going to happen on both sides of the triple bond.0627

The functional group we make in this reaction is a carboxylic acid; so oxidation of an alkyne would be a way to make a carboxylic acid functional group.0637

You are also cleaving your carbon chain; you are going to be losing some of your carbons and getting a carboxylic acid as a result.0649

Most of the reactions that we are going to be studying in this unit involve addition reactions; again similar to alkenes where we break the π bond and we add groups to either carbon.0659

One example of such a reaction is the bromination of an alkyne; if we used Br2 and we had an excess amount of Br2, what do you think might happen?0669

I am thinking we are going to break the carbon-carbon triple bond, break one of the π bonds, and add a bromine and a bromine.0682

Then we are going to break the second π bond; we are going to add a bromine and a bromine as well.0690

That is exactly what happens; it adds two equivalents of the Br2; both π bonds react just like the single π bond in an alkene.0695

What is nice about this reaction is that there is no regiochemistry to consider; in other words, since we are adding a bromine and a bromine, we don't have to decide who goes where.0705

There is really only one place to put them; there is also no stereochemistry to be concerned with in this reaction.0714

We know that bromination occurs with anti addition because of that bromonium ion intermediate.0720

But because we end up adding two bromines to each carbon, again there is no chiral centers; so there is no way to observe that anti addition.0725

This ends up being one of the simplest products to predict because it is almost impossible to get it wrong; all we do is break the π bonds and add four bromines.0733

It is a little more interesting when we add HBr across a triple bond because now we are adding two different groups and we have to decide where does the hydrogen go, where does the bromine go.0746

Remember we learned about Markovnikov's rule for alkenes; it turns out Markovnikov's rule holds for alkynes as well.0756

Just as you might predict from what you know for alkenes, we are going to break the π bond; we are going to add a group to either carbon.0767

We are going to add the hydrogen to the carbon with more hydrogens; that is what Markonikov's rule tells us; we are going to the hydrogen to this end carbon and the bromine to the middle carbon.0776

It turns out that the second equivalent of HBr that adds follows the exact same regiochemistry.0788

We are going to break the second π bond, add a group to each carbon; hydrogen again goes to the end carbon and the bromine goes to the middle carbon.0793

It is essentially the same mechanism as the alkene; but let's go through that because it is a little interesting.0801

After we add our first HBr, we want to think about the influence that might have when we go to add the second equivalent of HBr; let's take a look at that mechanism.0807

It is going to start with the reaction of the alkyne with HBr as a strong acid; we are going to start by protonating the π bond.0819

This is an acid; the alkyne, any alkene or alkyne around, any double bond is going to act as a base--is going to get protonated.0830

This the point at which we decide our regiochemistry; we are going to want to put the hydrogen on the end carbon so that our carbocation goes on the more substituted carbon, the more stable carbocation.0841

This is as usual the foundation for Markovnikov's rule and the foundation for our regiochemistry determination--is we want to go through the lowest energy intermediate possible.0857

This is interesting; it is called a vinyl carbocation; we have never seen one of these before--having a positive charge on a triple bond; but it can happen; it will happen.0866

What happens from this point forward?--we just made Br- in this first step; that is going to act as a nucleophile; it is going to add to the carbocation.0879

We are going to have... after this, we will have added our first equivalent of HBr following Markovnikov's rule as we expect; but let's think about this second equivalent.0892

If we assume that the second protonation occurs with the same regiochemistry, that is going to yes put the positive charge on the more substituted carbon.0904

But that carbon... it is secondary versus primary; but that carbon also has a bromine attached to it; we can't just assume that that bromine is not going to have an influence.0921

We need to consider that to see if this still is the better carbocation or maybe the carbocation wants to be on the other carbon away from the bromine.0932

We need to ask: is the bromine good or bad for the carbocation, for the positive charge?--is that a stabilizing thing to have the bromine or is it destabilizing?0939

You might think bromine is more electronegative than carbon so bromine is pulling some electron density away; that is not a good thing for a carbocation.0953

It is already electron deficient; pulling electron density would make it less stable; that would not be good; but take a look at what else bromine has.0960

Bromine has lone pairs; that is a source of electron density; what can happen is these lone pairs can actually be shared, can donate to the carbocation and offer some resonance stabilization.0968

It turns out that yes the carbocation chooses to go here; in fact it is going to be stabilized by that bromine by resonance; it is resonance stabilized.0985

I just wanted to point that out and have you think about that; if you ever had a mechanism with a carbocation and you have a heteroatom like this--a bromine with a lone pair.0999

It is something that could be an electron donating group; it could stabilize the positive charge; that would be a good thing.1009

That is just a little note here for our resonance; let's redraw our first form so we are not making too much of a mess here.1021

We have this carbocation now; after our second protonation, what do we do to finish up our mechanism?1028

We simply attack with our second equivalent of bromide, Br-; and our mechanism is done.1034

Ultimately it is the same mechanism as the addition of HBr to an alkene; we just do it twice; protonate and attack.1043

Then protonate again with the same regiochemistry, the same Markovnikov regiochemistry, to give the more stable carbocation; and then attack.1051

Let's take a break for a second and see if we can do a transform problem beginning with an alkyne starting material and some of the reactions we have seen; we know about alkynes.1061

Let's say we wanted to get to this product; we have gotten rid of the triple bond; the two bromines are new; but there is something else that is new in this product; what do we have?1070

One π bond is gone; the other π bond is also gone though; so there is also two new hydrogens in this molecule that need to get added in throughout the course of our synthesis.1085

How about if I just used... I need to add an H and a Br, an H and a Br; so how about if I added HBr?--an excess of HBr?--would that give me my product that I'm expecting?1101

That would break both π bonds and it would add HBr twice; but where would the location of those bromines be?--they would not be on opposite carbons; they would be on the same carbon.1113

Remember we get Markovnikov addition of both equivalents; the hydrogens both go to one carbon; the bromines both go to the other carbon; that would give the wrong regiochemistry.1123

What reaction have we seen that puts a bromine on one carbon and another bromine on the carbon next to it?--what did you need to add to... what reaction have we seen that does that?1132

How about the bromination reaction of an alkene?--if we think about a retrosynthesis of this, in other words, what starting materials do I need to plan this synthesis?1143

If I want two bromines, then I need to have a double bond; if I had a double bond in this position, then I would be able to add Br2 to that; I would now have two bromines.1159

That is good; I know I need to get to the alkene; but the second question we have is how about the stereochemistry?1177

If I were to have this alkene and add Br2, what do we know about the stereochemistry of that bromination reaction?1184

Remember the first bromine comes in to make the bromonium ion; then the second bromine comes in to open that up; we get anti addition because of that backside attack, that Sn2.1193

If these methyl groups were on the same side like they are here, would you get the two bromines adding from the same face?--that would not give us the right product.1206

Something is wrong here; the stereochemistry of our alkene is wrong; maybe we can... another way to look at this is we can rotate this.1216

And say: I know that bromination occurs anti so let me look at the product in a slightly different way where the bromines are drawn anti--they are drawn one up and one down.1228

In other words, if I just rotate this... I just did that backwards, excuse me... when I rotate this bond, if I just turn around here and bring that bromine down.1243

Then what happens to my methyl group?--it was a wedge; when I rotate it 180 degrees, now it is going to be a dash.1256

If we look at it from this point of view, now when we think about what starting material do I need to do the bromination, what is the actual relationship that we need between those methyl groups?1269

One is back and one is forward; they need to be trans to each other in order to get the proper stereochemistry in the product.1281

We can confirm that; if I had this alkene with the trans methyls and I brominated, add a bromine, one to the top and one to the bottom, yes I would get this product out.1293

This is now... now we have solved our problem as to how we are going to get there; what we need to do to our alkyne is first convert it to the trans alkene.1304

Then we can take that trans alkene and add the bromine to it; the second part is just simply Br2; we will add the bromines trans.1319

How do we go from an alkyne to a trans alkene?--it looks like we have done some kind of reduction; we have added hydrogens to the carbons.1332

Catalytic hydrogenation would totally get rid of the triple bond; poison catalytic hydrogenation adds the hydrogens to the same face; it would give the cis double bond.1344

How do we get the trans?--that is the dissolving metal reduction; sodium metal and ammonia solvent, NaNH3, is dissolving metal; that reduces it to give the trans alkene.1355

Then bromination gives the final product; so we need to think about regiochemistry; we need to think about stereochemistry when we propose a synthesis.1368

We have to check every step of the way, is this actually the product, the product we are expecting?--is that actually the product that would be formed?1377

One quick thing I want to mention about these reagents--Na and NH3 is not the same as NaNH2; that is another reagent we are going to be seeing shortly.1383

NaNH2 is a salt; it means we have Na+NH2-; that is a completely different reagent than having sodium metal and NH3 solvent.1394

You could list them separately or make sure you put a comma between them; you don't want to squeeze them together and confuse it with NaNH2.1408

What else can we add to a triple bond?--we have seen adding hydrogens; we have seen adding bromines; we have seen adding a hydrogen and a bromine.1417

How about if we wanted to do a hydration of an alkyne?--that means we are going to be adding an H and an OH; we are going to add the components of water.1426

What do you think might happen based on what we know about the addition of HBr?--I think I am going to break the π bond; let's just do this one at a time.1436

I am going to break the π bond; I am going to add a group to each carbon; where do you think the hydrogen should go of H2O?1445

If it has a mechanism similar to all the other addition reactions, then I expect the hydrogen to go to the carbon with more hydrogens so that it follows Markovnikov's rule.1452

This in fact is what happens; but this is not a final product; you might think it is going to continue and we will do a second addition.1463

If it were exactly analogous to addition of HBr, we would add one equivalent; then we would add a second equivalent; and we would be done.1474

The problem is that this functional group, having two OHs attached to the same carbon, is extremely unstable; that is a very very rare arrangement of functional groups.1482

This is not the product that is observed; what happens instead is we add a single equivalent of water; this product is described as an enol.1497

It is called an enol because it has both an alkene, a carbon-carbon double bond, and an alcohol on the same carbon; one carbon has both the double bond and the OH.1509

It is no longer an alkene; it is no longer a simple alcohol; it is called an enol.1521

Enols do something very special; they will undergo a process called tautomerization; this will tautomerize... tautomerize... I thought I spelled it wrong, sorry.1526

What is going to happen is it is going to rearrange; instead of having an enol, we have a carbonyl; we are going to get a ketone as our product.1548

This is actually called a keto-enol tautomerization; we will take a look at that mechanism; it is a two-step mechanism; let's look at the mechanism for the complete process.1560

You will notice in our reaction conditions, we have water as you might expect; we have acid as we also typically have for hydration; it is an acid-catalyzed mechanism.1578

We have this last guy--mercuric sulfate is usually thrown in here; this is simply a catalyst... and this is a catalyst.1588

If you... you might be able to do a mechanism using that catalyst; but let's just take a look at a simple mechanism where we don't use that catalyst.1598

What do you think our first step will be for a mechanism in these reaction conditions?--clearly we have acidic conditions.1610

Anytime we have an acid and we have a π bond, a good first step is to protonate the π bond; let's just use a J to represent an acid.1620

We will protonate the π bond; we are going to do that with Markovnikov regiochemistry to give this more stable secondary vinyl carbocation rather than a primary carbocation.1631

What could we do next?--we look around for a nucleophile to add in; we have water as our solvent most likely; but that is the best nucleophile we have around.1647

So water is going to attack the carbocation... I will just condense this to be a CH2 here.1661

What do we still have on this oxygen?--it still has the two hydrogens and still has one lone pair of electrons; the other lone pair is right here now being shared.1672

This looks like it is charged; one, two, three, four, five; one, two, three, four, five; oxygen wants six; it has five; this is an O+.1683

We do the same mechanism we had for HBr; we protonate and then we attack; but because water is a neutral nucleophile, after we attack, we have a charged species.1692

We need to get rid of that charge; the way we do that is the usual; we deprotonate the oxygen; we remove one of these protons.1701

Let's bring in A- again, grab the proton, leave the electrons behind; and we are done with our addition, our first addition.1711

We are here; this is where we are--at the enol; now we need to go to the... we are going to want to go to the ketone.1725

This is going to... I think it helps to think about the rest of this mechanism by drawing the product that we know that we get; we don't get the enol; we get a carbonyl where the enol carbon was.1739

Let's draw that product out; there is our product; now by seeing where we are going, we can think more carefully about how we get there.1751

If you compare the structure of this enol to the structure of the ketone, we know we have to do this tautomerization mechanism; what changes do we have to accomplish?--what differences are there?1763

The π bond moves; but that is part of our mechanism; that is part of the resonance in our mechanism; but here we have an OH; now we have just an oxygen; we have to remove that H+.1774

The way we could describe that is we have to deprotonate here; that is one of the things we need to accomplish; we need to deprotonate that oxygen.1788

What other change is taking place?--this double bond carbon used to be just a CH2; now he is a single bond; he is a CH3.1796

How would you describe what has to happen here on that carbon?--I have to protonate here; that is it; those are our two steps for our tautomerization.1805

It is a two-step mechanism; it is simply protonations and deprotonations--are what describe any tautomerization process.1819

The only thing we have to decide is what should we do first?--what should be our first step, a deprotonation or a protonation?1828

How about we consider our reaction conditions; we are still in our acidic reaction conditions; we are here at a neutral molecule; where do we go from this point?1834

Let's protonate something; let's use our acid; let's protonate; this is going to be step one is protonate; and step two is deprotonate.1843

I will bring in HA again... let's just think about this for a second; how is it that I can protonate this carbon; it doesn't have a lone pair of electrons; how can I protonate it?1852

I can use the π bond attached to it; what we are going to do is we are going to protonate this π bond as a way to put the hydrogen on the carbon.1864

That is step one--is we are going to protonate that CH2; now let's think about going from here to here; this is step two; we need to deprotonate.1881

Can you think about what our deprotonation mechanism is going to look like?--I am going to grab this proton; where am I going to put these two electrons?1892

Rather than put them on this oxygen and have an O- and a C+, all I need to do is bring those two electrons down to be a π bond.1902

There we go; we have our carbonyl; we have our ketone product; we have our carbonyl product; this is called a tautomerization.1911

We are going to be seeing lots more tautomerizations down the road when we are studying carbonyl compounds and the mechanisms they undergo; but this is our first example of it.1918

Where we are going to see it is in the case of hydrating an alkyne; what we do is we add just a single equivalent of H and OH.1928

Then that resulting enol, we have to get used to converting an enol structure to a ketone structure, a carbonyl structure.1939

The carbon that used to have both the OH and the double bond is now a carbonyl; the carbon that used to be part of the double bond will have a new extra proton.1947

That is how we get our ketone structure out; if we use water and acid, we are going to get our Markovnikov addition as usual.1957

Remember when we did hydration of alkenes, we had a few different mechanisms that we could do, a few different reaction conditions we could do; for alkynes, we have those options as well.1967

If you recall the process of hydroboration-oxidation, that was another way that we could add water across a π bond.1979

But this is the one that did it with anti-Markovnikov regiochemistry; this is the one that was complementary to the others and did the opposite.1987

We are going to do the same... we could do the same thing for an alkyne; if we have a terminal alkyne, we can do this reaction and get a specific regiochemistry.1996

This guy is called disiamylborane; it has two bulky siamyl groups attached to the boron; but like every boron reagent, the most important thing is that it has a boron and it has a hydrogen.2009

The reaction that it does is called hydroboration; in a single step it adds the hydrogen and the boron; that helps explain the regiochemistry.2025

The regiochemistry that we see is that we add the hydrogen to the internal carbon and we add the boron to the terminal carbon.2035

I say this has to be a terminal alkyne because if the triple bond has a carbon group on either side, then you are not going to get a fixed regiochemistry because there is no significant difference between the two carbons.2045

But if there is a hydrogen here and an alkyl group here, then the boron clearly goes to the less hindered carbon; we will get this anti-Markovnikov regiochemistry.2056

The second step of this process is typically the oxidation; we treat it with an oxidizing agent like hydrogen peroxide; what happens is this boron gets converted to an OH group.2066

We are going to break the π bond and add just a single equivalent of water; we are going to do it in anti-Markovnikov regiochemistry.2081

The hydrogen goes to the carbon on the inside; the OH goes to the carbon on the outside; what do you think happens next?2090

Are we going to add a second equivalent of water?--or can this structure do something else that is going to be more favorable?2099

Once again, every time we add H2O across a π bond or the components of water across a π bond, we are going to end up with an enol intermediate.2107

That is going to not be our final product and not react further to add the π bond.2117

Instead it is going to undergo a rearrangement which is an equilibrium but highly favored in the forward direction where we have, instead of an enol, we have a carbonyl.2125

This carbon becomes a carbonyl; this carbon gets an extra hydrogen.2136

When we do hydroboration-oxidation on a terminal alkyne, the oxygen goes to the end carbon; the product we get here is an aldehyde.2147

When you have a carbonyl at the end of a carbon chain so there is a hydrogen attached to that carbonyl, we call that an aldehyde rather than a ketone.2156

We still call this a tautomerization; but sometimes the enol might go to a ketone; sometimes it might go to an aldehyde depending on what other groups are attached to the enol carbons.2163

Let's think about some syntheses we can do; if we wanted to make an alkyne, how could we do that?2180

We have already talked about the reactions that alkynes can undergo; but if we wanted to create a triple bond, where could it come from?2187

We have two good ways to make an alkyne target molecule; one option is to do a dehydrohalogenation; what does that mean?--it means we lose H and a Br; we do a dehydrohalogenation.2193

If we start with a carbon chain with two bromines here and two hydrogens here and we treat it with some really strong base like KOH or NaNH2.2211

Very strong base, we have NH2-; we have hydroxide; you should recognize both of those as very strong bases.2224

We add heat here; these are perfect conditions to do an elimination reaction; do you think this is going to be an E1 elimination mechanism or an E2 elimination mechanism?2231

We have a really strong base so I am thinking it is going to be an E2; it is going to end up doing this twice; we can show our first equivalent of base; let's just use the hydroxide.2243

We could show the mechanism; he E2 is a one-step mechanism where our strong base attacks a β hydrogen, forms a π bond, kicks off the leaving group.2257

We do anti elimination; we haven't really shown stereochemistry here; but you could maybe see the stereochemistry up to this point.2271

But again typically we don't stop here; we have heat; we do more vigorous conditions; we need a really strong base to do this double elimination; we are going to a second elimination.2278

Even though this is harder, but we can have the hydroxide or the NH2- come in and grab and even eliminate a vinyl bromide; we get a vinyl bromide intermediate here.2297

Up to now, we have never seen an E2 elimination occur when a leaving group is already on a double bond; but in fact you can force those conditions and you will be able to form a triple bond.2312

We are losing HBr times 2; we call this a dehydrohalogenation reaction; just like we can use an E2 to form an alkene, if we have two leaving groups, we can use an E2 elimination to form an alkyne.2328

Let's see if we can apply this strategy to the following alkyne synthesis; if I had this alkene, how could I convert it to this alkyne?2349

Once again we really should be thinking about these retrosynthetically and plan our synthesis first; let's look at our product and say what starting materials do I need?2362

What structure could I have started with that I know I can convert into this target molecule, into this alkyne?--what did we just learn?2373

We said if there were two leaving groups, if we had a bromine on each carbon like this, then that would work as an elimination; if I had this, I would be able to do a double E2 to get the triple bond.2383

I look at where I want to be; then I look at where I am starting and say how can I get from the double bond to the intermediate structure that has the two leaving groups?2408

All I have done is I have removed the π bond and replaced it with a bromine and a bromine; what reagent will do that?--all we need to do is add Br2.2421

It doesn't really matter what our stereochemistry is at this point; because once we do our double elimination, all the stereochemistry is gone; we are left with a linear molecule.2435

In this kind of synthesis, it doesn't matter whether you get one stereochemistry or another; that is not a concern in this case.2443

But we could add the two bromines; then we could add some strong base like NaNH2 and heat.2451

It is a real good idea to put that heat in there because we definitely need vigorous condition; that always favors the elimination; but in this case, it definitely favors the double elimination.2458

Then we can do our alkyne synthesis; so one route to creating a triple bond is by having two leaving groups in a position that can do a double elimination.2468

A more common way to synthesize an alkyne is to start with a triple bond already in place but then to functionalize it--to add groups on either side of the triple bond.2483

What we are going to be using is if we had this kind of an intermediate with a negative charge on that carbon, then we could use that as a substrate to make different alkyne products.2494

How do we get a negative charge on a carbon?--what we are going to do is we need to deprotonate that carbon; let's talk a little bit about the potential acidity of an alkyne.2511

How willing would an alkyne be to lose its proton to make this anion?--what I would like to do is I would like to compare an alkyne to an alkene to an alkane.2522

Let's look at these three different types of CHs and think about how easy or how difficult it is to deprotonate them.2533

I have shown you their pKa's; alkyne has a pKa somewhere around 26, an alkene about 36, an alkane about 49; again huge differences in their numbers here because these are orders of magnitude.2540

Who is the strongest acid here with these three numbers--26, 36, 49?--what is the relationship between pKa and acidity?--the lower the pKa, the higher the acidity.2555

This number of 26, this is the most acidic; it is ten to the ten times more acidic than an alkene; the alkane is by far the least acidic.2567

How can we justify that?--how can we explain that?--as usual what we want to do is we want to look at the conjugate bases.2588

In other words, let each of these acids be an acid; let them donate a proton; let's see what we end up with after that happens.2594

In other words, let's have some base come in and remove this proton; where I used to have a hydrogen, now I have a lone pair and a negative charge.2604

I can do that for the alkyne and the alkene and the alkane; this is a C- versus a C- versus a C-.2614

There is no periodic trends we are looking at here; there is no difference in electronegativity of these atoms; but the difference we do have is the hybridization of those carbons.2620

We can see that the alkyne is an sp hybridized, the alkene is sp2, and the alkane is sp3; we have another way to describe these hybridized orbitals.2632

We could say that the sp orbital... because in order to make an sp hybrid orbital, we took an s orbital and a p orbital and mixed them together.2651

The new hybrid orbital has--50% of its character is like an s orbital and 50% of its character is like a p orbital; we describe an sp hybridized carbon as having 50% s character; it is 50% s-like.2659

How about an sp2 hybridized orbital; how did we get an sp2?--we took one s orbital and two p orbitals.2676

Of the three total orbitals only one of them is s; so one-third, we describe an sp2 as having 33% s character.2683

An sp3, we have one, two, three, four hybrid orbitals, only one of which is an s; this is described as having 25% s character.2695

That is this hybridization difference in this s character difference is going to be the difference here; let's remind ourselves the difference between an s orbital and a p orbital.2706

If we take a look at the energy of these various orbitals, an s orbital is lower in energy than a p orbital.2721

Remember a p orbital is further away from the nucleus; it has a node in it so it is a higher energy orbital.2729

Where do you think the energy in sp hybrid orbital is?--it is going to be right in between; because it has 50% character of an s and 50% of a p.2734

How about an sp2?--that is going to be maybe two-thirds of the way up; it is going to be a little closer to the p orbital; and sp3 even closer still.2743

These different hybrid orbitals are of different energies; when you in this case have this negative charge in an sp hybrid orbital, that is a lower energy orbital.2754

That means it is going to be more stable; let's state the facts here; let's describe this; the lone pair and the negative charge is in a lower energy sp orbital.2767

That means that it is a more stable anion; let's stabilize the negative charge by putting that electron density in a lower energy orbital.2794

If you are more stable, what does that mean for your reactivity?--more stable means you are less reactive; this anion is the least reactive; therefore it is the weakest conjugate base.2806

Because he is more stable, he is less reactive; that makes him a weaker conjugate base than the others.2828

If something has a very weak conjugate base, what does that say about the parent?--we have that inverse relationship; this has the strongest parent acid.2835

Is that what the pKa data tell us?--is that confirmed by the pKa?--sure, that had the lowest pKa; so we knew that the alkyne going into this was the most acidic.2848

This explains why; the negative charge doesn't mind being in sp hybridized orbital because it is relatively low energy.2857

We compare that to this sp3 hybridized carbon with a negative charge; this has no stabilization; it has nothing good about it.2864

It is on a carbon; carbon doesn't like having a negative charge; it is on a sp3 hybridized carbon, the highest energy orbital we can have.2877

That makes him, because he is very unstable, that makes him the most reactive and therefore the strongest conjugate base.2886

What do we know about something that has a very strong reactive conjugate base?--it must be a weak parent acid; this has the weakest parent acid.2901

When we look at an alkyne versus an alkene and an alkane, which one is it reasonable to deprotonate?--only the alkyne.2917

What we are seeing here is that alkynes can be deprotonated... alkynes can be deprotonated.2926

They are acidic; they are acidic terminal alkynes, assuming you have a hydrogen attached to the triple bond.2936

Alkenes and alkanes, what do we makes of these numbers like 36 and 49?--that means never; we are not ever going to deprotonate an alkene or alkyne.2945

We are never going to deprotonate an alkene or an alkyne; we are just not going to ever see any cases.2958

There is no base strong enough to remove that proton; that is never going to be a favorable reaction.2962

Again this is something we are going to be seeing a lot in the upcoming chapters--is when can we deprotonate?--when can we not?--who is the stronger acid and why?2970

An alkyne, what is unique about a terminal alkyne is it is one of those few carbon groups that can be deprotonated; normally carbon doesn't like having a negative charge; so this is pretty special.2979

I mentioned that we need some base here to come in and deprotonate it; let's think about what base we can use.2990

We are going to need some very strong base; very strong base meaning not something like NaOH; hydroxide is a strong base; but it is all relative.3000

It is not a strong enough base to deprotonate an alkyne; that would form water in this reaction; that is not going to be as stable as having just the neutral alkyne and hydroxide.3013

Instead we are going to use a stronger base like NaNH2; this guy is called sodium amide; it is NH2- is what we have here; this is ionic.3027

A nitrogen doesn't handle a negative charge as well as an oxygen; this is a more reactive anion; this is a stronger base than hydroxide would be.3042

This could and will deprotonate the alkyne--a lone pair and negative charge; we would have the sodium salt here; if we use sodium amide, we have the sodium salt.3052

Anytime there is an anion, there is always a cation; sometimes we draw it; sometimes we don't.3068

What would the other product be here?--if NH2- was our base, when we protonated that, we would get out ammonia, NH3.3072

I want to point this out because remember any proton transfer is always a fight between two acids trying to give up their protons, two bases trying to take the protons.3081

We need to consider this reverse reaction if we are going to determine whether or not this is going to be favorable.3090

In the forward reaction, we have alkyne as our acid; in the reverse reaction, it would be ammonia as our acid; ammonia has a pKa of about 36.3097

Compared to 26 and 36, who is the stronger acid?--the alkyne is a much stronger acid; the ammonia is a weaker acid.3109

When we take a look at this equilibrium, what direction does it lie?--it always lies in the direction of the weaker acid; in other words, the strong acid and the strong base; this is a very strong base.3121

The strong acid and the strong base are going to react; they are going to transfer their proton and give the products which are more stable.3135

This equilibrium is really a one-way street; the reverse reaction is negligible; so this would be an excellent base if we ever wanted to fully deprotonate an alkyne.3143

What are we going to do with this?--before we move to the next slide, let's take a look at this.3160

What we are saying then is that this is pretty stable; this is relatively stable; this is reasonable to make; this is an anion that we can make; we are going to want to use it.3166

What use do you think you might have for a carbon with a lone pair and a negative charge?--what kind of behavior might that have?3175

Do you think maybe a nucleophile or an electrophile?--electron rich lone pair, it is going to be a very good nucleophile.3183

It is a carbanion; it is an example of one of the few carbanions we can make by deprotonation of a CH; that is how we are going to use it in our synthesis--is as a nucleophile.3192

This is described as an acetylide anion; because if this were acetylene just with the two hydrogens, this would be called acetylide.3206

These are acetylide-type anions; we call it that when we have a C- on a triple bond, an sp hybridized carbon.3213

Let's take a look at a possible synthesis then; how could we use an acetylide anion in synthesizing an alkyne of some kind?3221

The first thing we are going to do is we will start with an alkyne, a terminal alkyne; once again point that out; NaNH2,NH3.3230

Where did the NH3 come from?--again just to get used to seeing that; that is the typical solvent we are going to use when NH2- is our reagent.3243

NH3 is a reasonable solvent; just like if we were using methoxide as our base, we usually use methanol as the solvent; we very often use the conjugate acid as the solvent.3252

We should get used to seeing these conditions; again just a reminder here, this is not the same thing as Na,NH3;.3265

These are so similar; you really want to make sure that you are not confusing those and you make some flashcards, separating them out and delineating them.3275

Na,NH3, this is something that reduces the triple bond to an alkene; NaNH2 is a strong base that will deprotonate an alkyne.3283

I'm sorry... it would reduce the alkyne to an alkene; and this will deprotonate the alkyne; what is this going to do in this first step?3293

We are going to have our strong base; it is going grab that proton and it is going to convert this into the acetylide anion; the acetylide anion is a nucleophile.3301

If we treat it in a second step here with an electrophile like ethyl bromide or bromoethane, where is this electrophilic?--how would I recognize an alkyl halide as an electrophile?3316

Remember the bromine acts as a leaving group, pulls electron density away from the carbon; so this is partially positive; that is where we see that it is electrophilic.3331

What is going to happen?--what do you think?--with this nucleophile and an alkyl halide, what mechanism can happen?--how about the lone pair attacks the carbon and kicks out the leaving group.3341

What do we call that mechanism?--it is the Sn2; it is just backside attack; it is exactly what can happen and what will happen; what we do here is we form a new carbon-carbon bond.3350

This is something that organic chemists get very excited about when you form a new carbon-carbon bond because this is how we can build up new carbon chains and new carbon molecules.3366

It is happening because we have a carbon nucleophile reacting with a carbon electrophile; so we form a new carbon-carbon bond; we just synthesized a new alkene from an original alkene.3375

We started out with a new alkyne from an original alkyne; we started out with a terminal alkyne after this two-step synthesis--deprotonation followed by alkylation.3387

We could describe this as deprotonation followed by alkylation to describe each of the two steps that we did.3401

We now have an internal alkyne; so this is a way that we can add carbon groups to one or both sides of an alkyne starting material.3411

Let's look at a few examples of this; here is one example; we are starting with this alkyne; we are making this new alkyne.3427

We can pretty readily identify the three carbons in our starting material are still these three carbons in our product.3438

Which means this is a new carbon-carbon bond that needs to be formed in the reaction; what we do is we make a disconnection here as part of our planning, as part of our retrosynthesis.3446

We know that is the bond that we are going to cleave; then we need to think about the two carbons involved in that reaction.3463

If we want to form a carbon-carbon bond between these two carbons, that means one of those carbons must have started out as a nucleophile; one of them started out as an electrophile.3469

Which was which?--the carbon that is part of the triple bond, the sp hybridized carbon, what kind of behavior have we seen recently on that kind of carbon?3480

We have seen a lone pair and a negative charge; that means that would be a very good nucleophile; we can identify this one as--this was my nucleophile; which means this guy was my electrophile.3490

When I do my retrosynthesis--asking what starting materials do I need, I need to come up with a reasonable nucleophile.3502

My reasonable nucleophile would simply be a carbon with a negative charge at that position; we have seen that anion; that is stable anion; that is a good nucleophile.3512

How about this carbon though?--that is my electrophile; how do I make this carbon electrophilic?3521

It would be great to just put a positive charge here and say, if I had that carbocation, if I had this benzylic carbocation, and I hit it with this alkyne anion, I would definitely make my target molecule.3528

The problem is that this is not... you can't go to the stockroom and ask for a benzylic carbocation; it is a fleeting intermediate; it is not a stable reagent or starting material that we can use.3540

What we have to think of is what is an actual reagent that is stable that reacts as if it had a positive charge here that will still be electrophilic?3551

The strategy that we do is we simply add a leaving group; if we attach a leaving group here; your choice--bromine, chlorine, iodine.3563

Then we no longer have a carbocation; but we do have a partial positive; that is what we typically see for electrophiles; this would be a great electrophile.3572

What the beauty of doing a retrosynthesis like this is you can test yourself and ask if I had this nucleophile and this electrophile and brought them together, would it make this product?3581

Absolutely, because we know we would do an Sn2--attacks the carbon, kicks off the leaving group; that would be successful.3591

Our planning tells us what we need; now we come back to where we are; we are at propyne; we need to have the anion of propyne; we need the anion of propyne; where does that anion come from?3598

How do we go from a CH to a C-?--we removed an H+; we need to deprotonate; what we need is a very very strong base.3612

NaOH is not going to cut it; that is going to give just a small equilibrium; it is going to form a little bit of this; but it is still mostly going to stay as the undeprotonated species.3621

The strong base that we are going to use instead most often is NaNH2; there are other options as well; but NaNH2.3632

We should get used to seeing NH3 thrown in there as the solvent; that would be effective at doing a deprotonation.3640

Now that I have this, all I need to add is benzyl bromide, phenyl CH2Br, or benzyl chloride or benzyl iodide; your choice, anything like that.3647

Then we have an Sn2; and we have synthesized our target molecule; two-steps again--deprotonation, alkylation.3657

How about the next one?--this is a little more complicated because we no longer have a triple bond in our molecule.3670

Again a good first step is to identify the carbons in your starting material and find those carbons in the product.3677

Here they are; we started with three carbons; now we have four carbons; that means this methyl group is new.3686

That is going to be a disconnection we need to make because at some point in this synthesis I have to bring that methyl group in and form a bond between that and carbon number 3.3692

The other change that has taken place is I went from an alkyne to a trans alkene; both of these transformations we have seen in this unit; so we are capable of both of those.3702

The question we need to ask is: which one should we do first?--let's imagine doing our alkylation first; let's say we wanted to add the methyl group first; then convert the alkyne to the alkene.3716

Let's see if we can fill in those reagents; how do we go from a CH to a CCH3?--how do we replace this proton on the terminal alkene with a methyl group?3739

That is going to be an alkylation of the alkyne; that is our two-step procedure; step one is NaNH2,NH3, a strong base to deprotonate.3753

Step two, after we do that... in a multistep synthesis, you can either draw an arrow and then your product and then an arrow and your next product.3763

Or if it is really long, you can save some time and condense them over a single arrow; but you want to be very clear that the first step is separate from the second step.3771

The second from the third step; you are assuming, after each step, you do a reaction work up; you isolate your product; then you carry it on to the next step.3780

That is clear to do; but make sure you include these numbers here because that is critical to make your synthesis work.3788

After we do that deprotonation, then we are going to add in the methyl group; how do we make methyl an electrophile?--we need an alkyl halide; we will use methyl iodide.3795

When it is methyl, usually the bromine, iodine, and chloride are not a good choice because methyl iodide is a liquid; the other ones are gases; usually for methyl, methyl iodide is a better choice.3805

On paper, it is not as critical because it doesn't float away from paper; but in the laboratory, you would pretty much be using methyl iodide as your best choice.3816

I know how to do this transformation to go from the terminal alkyne to the internal alkyne; how do we go from an alkyne to a trans alkene?3826

This is one of the reduction reactions we saw; which one is it?--we want to do partial reduction; so is it the Lindlar's catalyst with hydrogen gas or is it dissolving metal reduction?3835

Lindlar's adds the syn hydrogens; we get the cis alkene; the dissolving metal, which is sodium and ammonia or lithium metal and ammonia, would do a partial reduction and give the trans alkene.3847

This looks like a reasonable synthesis; I think this will work pretty well; but let's consider what would happen if we tried the other way.3864

What if we started here and we said I want to first reduce the π bond and then I want to add on the methyl group; we know how to do this reduction.3871

We could do dissolving metal reduction here first to get to the alkene; but here is the question now: how do we go from an alkene with a hydrogen here and replace that with a methyl group?3885

If we try to add in a strong base here, can we deprotonate an alkene hydrogen?--no way; no reaction; this transformation can't happen.3905

Remember alkynes are very special that way; only alkyne CHs can be deprotonated; it is alkyne nucleophiles, acetylide type nucleophiles, that we are going to be using in our synthesis, not alkenes.3920

Another problem, even if we could have this happen somehow, how would you control the stereochemistry?3934

How would you be able to deprotonate just this trans one, and not the other one, so that you get the trans alkene?3939

Because there is stereochemistry in this product, we need to think to ourselves when we are doing our retrosynthesis and say: how do I get a trans alkene?3945

What reaction have I seen that gives specifically a trans alkene product?--I know that if I had this alkyne, if I had this alkyne, I could stereospecifically get the trans alkene.3955

That is part of the problem too--is this is a stereochemistry issue; but we also can't invoke reactions.3969

We don't want to mix our functional groups and just start deprotonating any carbon we want now; this is a reaction that is specific for terminal alkynes.3974

One last example, let's imagine going from this alkene to this alkyne; we have one, two, three, four carbons here; and one, two, three, four carbons here.3985

It looks like this is a new carbon-carbon bond; this ethyl group, this CH3CH2 is still a CH3CH2 so I know I should number it that way.4003

But my π bond is gone; that is interesting; I need to form this bond; let's think about how to form that bond; let's do our retrosynthesis asking what starting materials I could have.4014

What I need to consider is if I want these two carbons to come together and form a bond, that means one of them started out as a nucleophile, one of them started out as an electrophile.4028

Who is who?--which is which?--who would be the good nucleophile in this case?--it has to be the carbon that is part of the triple bond; this guy was my nucleophile.4038

Which means this guy must have been my electrophile; it must have been an electrophile for it to be attracted to the nucleophile.4049

What does this two carbon nucleophile look like?--it is a C- with a CH; it is the acetylide anion; that is a good nucleophile.4057

Who is my electrophile?--we have four carbons with a leaving group attached; remember we still have one, two, three, four; be careful with your line drawings.4069

It is easy to lose the carbon at this point; so don't hesitate to number your carbons; number them throughout so that you don't miss any carbons.4079

If I had this alkyl halide--bromide, chloride, iodide, and I reacted it with this nucleophile, would I get this product?--sure, I would expect an Sn2; that looks great.4085

The question is then how do I go from my alkene to this bromide?--I need this bromide in order to do my synthesis; but I am starting with an alkene.4100

It looks like the double bond is gone; I have added a bromine; what else?--there is also a hydrogen here that was new; how about if I just add HBr?4117

I can take an alkene and I can add HBr across the π bond--break the π bond, add the H and the Br; would that give this product?4126

What is the problem with just adding HBr?--what is the regiochemistry you would expect when you add an HBr?--hydrogen goes to the carbon with more hydrogens.4135

That is the end carbon in this case; that would give the wrong product; how do we get to anti-Markovnikov regiochemistry?--there is a few options we have had.4144

We have seen one case where we did hydroboration-oxidation; we could first make the alcohol and then convert it to the Br.4156

But we have actually seen a reaction with Br; it is a good thing we picked Br as our leaving group because we have seen a reaction with HBr that added anti-Markovnikov.4163

That was, instead of using a mechanism involving HBr as an acid, we threw in some peroxides; when we add in peroxides, we get a radical mechanism; and we get anti-Markovnikov regiochemistry.4171

That would be a way of adding the hydrogen to the middle carbon and the bromine to the end carbon; then once we have this, we can add in our sodium acetylide.4196

Which you can assume that you can just get commercially or you could show how you make that; it depends on the instructions for the particular assignment; you can do that as well.4208

One thing that I wanted to point out is, that I forgot to mention, is that this nucleophile is a very strong nucleophile; it is also a strong base.4221

FYI (for your information), if we tried to react this with a secondary leaving group or definitely something with a tertiary leaving group, then we are going to get E2 elimination instead.4235

For example, if I took cyclopentyl bromide and I tried to do a reaction with this to do an Sn2, remember Sn2 is very sensitive to sterics.4249

As soon as we have any kind of steric hindrance like this, rather than do the Sn2, it is going to give E2 as the major product.4267

For this alkylation process that we are talking about, we need to have an unhindered electrophile.4280

Something like a methyl halide, like methyl iodide we just saw, or a primary alkyl halide, primary RX, is a good Sn2.4286

If you don't want to use this peroxide, this radical mechanism, or you don't recall that, the other anti-Markovnikov mechanism that we should know is the hydroboration-oxidation.4299

In other words, we can add water across the π bond in anti-Markovnikov regiochemistry; that was BH3-THF, hydroboration-oxidation, H2O2 and base.4312

A few more reagents to remember if you choose to go this route; but that will work just as well.4329

We also have an extra step because we can't just add in the acetylide; we don't have a leaving group; what we would need to do is we need to convert this to a leaving group.4335

How do we make an OH a good leaving group?--several options; we can make the halide by using something like PBr3; or we could make it the tosylate by using tosyl chloride.4346

Either of those would be good; then we could react that with the sodium acetylide; we can get a Sn2 with our nice primary leaving group; that would be another route to it.4365

This step is a little longer because it requires conversion of the OH into something bearing a good leaving group.4380

This first method would be a little better because it is more succinct and a shorter synthesis.4386

That finishes it up for alkynes; I hope to see you again soon at Educator.com; thank you.4393

Hi and welcome back to Educator.0000

Next we are going to be talking about alcohols; we are going to do two different parts.0002

The first part is going to be discussing the structure of alcohols and the synthesis of alcohols; in other words, how do you prepare them.0006

An alcohol is an organic molecule that contains an OH functional group.0013

Because oxygen is so electronegative, it pulls a lot of electron density toward itself, especially away from this tiny little hydrogen.0018

What we end up with is a huge partial minus and partial plus on these two atoms; so the OH group is a very, very polar functional group.0025

That is going to define a lot of its physical properties; they can undergo hydrogen bonding with each other.0034

In other words, one molecule of an alcohol, partial minus, partial plus, can interact with another molecule of alcohol, partial minus, partial plus.0041

There is such a strong attraction between the oxygen on one and the hydrogen on the other that we actually draw a dashed line connecting them.0056

We call that a hydrogen bond; it is easy for the protons to be transferred from one structure to the other.0064

It results in a very strong association, a very strong association, a very strong affinity for one molecule of an alcohol to another molecule of an alcohol.0073

The physical property effects we see on that is we see an increased boiling point.0080

Because if these are very strongly attracted to one another, it is going to difficult to tear them apart from one another and put them in the vapor phase.0084

We also see an increased water solubility because very much like water, like dissolves like; water is polar; water can hydrogen bond; so a very strong affinity between alcohols and water.0091

Another feature that we are going to be seeing is that the OH group is acidic; in other words, it can be deprotonated.0102

It can act as an acid; it can donate an H+ very easily; we will see reactions of alcohol with a wide variety of bases as well.0111

Let's take a look at some sample boiling points; these boiling points are listed in degrees C so that we can observe the effects on boiling point by various functional groups.0122

If we take a look at a molecule like pentane, completely nonpolar molecule; all it has are carbon-carbon bonds and carbon-hydrogen bonds; all very nonpolar.0137

We compare that to an ether, let's say, where we have an oxygen in here; which now makes it--we have some polar bonds; so there is a small dipole moment here; this is polar molecule.0149

But you don't see a huge difference in boiling points; these are about the same molecular weight; not a huge difference, they are essentially the same.0165

The polarity isn't having... this small amount of polarity is not having a big difference in the boiling point.0174

But instead, if you have an OH group for that oxygen, rather than just an oxygen with carbons on either sides, now look at the huge jump we get in boiling point difference.0179

Even ethanol is now a smaller molecule, yet it has over twice the boiling point; this effect here is because of hydrogen bonding.0190

Those ethanol molecules are very strongly attracted to one another; it is difficult to take them apart, put them in the vapor phase; we have to put more energy into it; that increases the boiling point.0202

When we compare ethanol to butanol, how do we explain that difference in boiling points?--they both have the OH; so they have the same polarity, same hydrogen bonding capability.0213

But now we see the trend where as you increase the molecular weight, you increase the boiling point; so all other things being equal, a larger molecule is going to have a higher boiling point.0225

Then finally I put this molecule on here; this is called ethylene glycol; he is an example of a diol with two OH groups.0235

In fact, I just noticed a typo with my structure here, excuse me; ethylene glycol has just two carbons and two OHs.0242

Notice this boiling point--197; so lots of hydrogen bonding in the case of ethylene glycol because it has two OH groups; a very big network of hydrogen bonding there.0251

That high boiling point is very useful to us; we use ethylene glycol as a component of antifreeze; it is good in that capacity because it is very hard to boil.0267

Antifreeze is th coolant that goes through your car and cools down the engine; it needs to be high boiling because it is going to get very hot.0279

We don't want it to just boil away; you can't just fill your radiator with water because the boiling point is too low.0285

We can see here a clear trend that the presence of an OH group or multiple OH groups is going to have an increase in the boiling point of those molecules.0290

When we take a look at water solubility, we describe water solubility in terms of grams per 100 milliliters of water.0302

If I had a 100mL sample of water, how much of this material could I dissolve in there?0313

If we take again a look at something like pentane, this liquid is completely immiscible with water; they form two separate layers; there is no solubility.0318

Once again, it is because this is a completely nonpolar molecule; it has no affinity for water; in fact, you can describe this as being hydrophobic.0327

Something that is nonpolar is hydrophobic--it fears water; there is nothing about it that is similar to water.0339

Remember, with solubility, we are always talking about like dissolves like; the more similar a molecule is to water, the more it has in common with it, the better the solubility.0345

For example, if you put in an ether, now you have a little bit of polarity; we have a big jump in water solubility.0360

This does have appreciable water solubility; that is because it can accept hydrogen bonds from water.0369

In other words, this oxygen can be attracted to a water molecule and form some hydrogen bonding there; it can be a hydrogen bond acceptor from water.0382

Even though this ether can't hydrogen bond with itself, if you just had a sample of ether, it can undergo hydrogen bonding with water; that increases its water solubility.0392

This is diethyl ether; because it has some water solubility, that is why whenever we use ether in a reaction work up and we are doing an extractive work up.0403

The ether layer is wet with dissolved water and it must be dried.0414

Any extractive work up that you do, where you partition your reaction components between an organic layer and an aqueous layer in a sep funnel for example.0427

If you are using ether, some water will always dissolve into the ether layer and the ether going into the water layer.0437

Part of any work up procedure is going to be taking that ether layer and drying it somehow.0446

Removing that water by using a drying agent like calcium chloride or magnesium sulfate and getting rid of that water.0450

By looking at the structure, we can see the polarity and the hydrogen bonding capability; that would explain why it does have some water solubility.0457

It is also going to be something that is very relevant to us in the laboratory.0466

But then when we move to a molecule with an OH, you might think now there must be a huge jump in water solubility because now we can now both donate and accept hydrogen bonding, a lot more interaction.0470

But when we look at these two molecules, they both have four carbons; but this has a really long carbon chain that is nonpolar.0480

We end up getting this balance, this tradeoff, between this nonpolar part and this polar part, this hydrophilic part and this hydrophobic part.0488

You can see that actually these have very similar water solubilities, not very different at all.0496

However, if you shorten up that nonpolar chain and you look at something like this--this is ethanol; it is completely miscible with water.0502

Now it is a polar enough water molecule; it has this hydrogen bonding capability; it is soluble in all proportions.0513

You can never have too much ethanol so that it separates out as a separate layer from water.0521

This is the alcohol; we are studying alcohols in this unit; we have heard of alcohols before in our day to day lives.0527

One of the alcohols is one that humans can drink without getting too sick; that is grain alcohol; this is the actual alcohol structure.0534

As an organic chemist, our definition of alcohols extend to anything containing an OH group.0544

But in our everyday lives, this is the alcohol that is referred to as alcohol that is drinkable; it is the two carbon chain called ethanol.0550

Of course, the alcoholic beverages that are out there are aqueous solutions with some alcohol in there; the proof tells you what percentage of alcohol is in there; the rest of it is just water.0560

This is something that we might already have seen some evidence for; that ethanol is in fact miscible with water.0572

Let's talk about the other physical property, the other reactivity or feature of an alcohol; that is the fact that it is acidic; it can be deprotonated.0582

We have... let's compare an alcohol OH to an amine NH and see if maybe we can explain why an OH is pretty acidic.0591

If fact, if you look at their pKa numbers, this has a pKa of about 16; that is a very low number, really easy to deprotonate an alcohol; where an amine has a pKa of about 40.0599

Let's just do this comparison as a review of things that affect acidity and basicity and see if we can explain that and understand why alcohols are so easily deprotonated.0610

As usual, we are going to want to consider the conjugate base of each of these to decide why they have such a big difference in pKa.0621

In other words, let's let methanol be an acid and get deprotonated; instead of having an OH, we will have an O-; that is the conjugate base of methanol.0629

The conjugate base of methyl amine has an NH with a minus; we remove an H+ from each; now we compare the two; we have an O- versus an N-.0643

Which of those is more stable?--we look for a difference in the stability of the conjugate bases to explain differences in acidity.0658

We know that oxygen is more electronegative than nitrogen; let's start by stating our facts; oxygen is more electronegative than nitrogen.0666

If you have a negative charge, would that prefer to be on the more electronegative atom or the less electronegative atom?0682

It prefers on the more electronegative; oxygen better handles the negative charge because it is more electronegative and O- is better than N-.0688

That means CH3O- is the more stable and therefore less reactive; if you are more stable, you are less reactive and therefore weaker conjugate base.0703

Whichever conjugate base is more stable, that makes it less reactive; that makes it the weaker conjugate base; and the weaker conjugate base has the stronger parent acid.0721

That can be our last comment; CH3O- has the stronger parent acid; is that what our pKa data tells us?0730

Sure, again pKa of 16; that is a difference of twenty-four units; that is kajillions times more acidic having a hydrogen on an oxygen compared to a hydrogen on something like a nitrogen.0741

It is because oxygen is so electronegative; if you think about the periodic table, we all know that fluorine is the most electronegative atom.0753

Oxygen right next door is the second most electronegative atom in the periodic table.0761

Oxygen, that means it is really good at pulling electron density toward itself; it is very, very happy having a negative charge on itself.0767

That means it is pretty reasonable to deprotonate an OH and convert it to an O- because oxygen doesn't mind having a negative charge.0775

If we wanted to deprotonate an OH to give an RO---these are called alkoxides because we have an alkyl group and an O-.0785

Just like hydroxide is HO-, alkoxide is RO-; how would we prepare an alkoxide if we wanted to?0797

What we would have to do is we would start with the alcohol and we would completely deprotonate it; we want to remove 100% of the H+ from all the OH containing molecules.0804

There is going to be two strategies we can have for this; one option is we can use a very strong base like sodium hydride.0816

Again when I say strong base, what do you think as sometimes a strong base?--you think hydroxide, NaOH; NaOH is not going to do it.0822

Think about it; you can't use an O- to create another O- and be able to do so in an equilibrium that is going to be heavily favored in one direction or the other.0830

This would simply set up an equilibrium with water and the alkoxide; we don't want an equilibrium; we want a one way street.0841

Rather than using NaOH, the base we are going to be using is sodium hydride; that is NaH; that is Na+ and H-; H- is called hydride.0850

This is a very effective base; we just said the alcohol can act as an acid; what happens when you react these two?0863

The base is going to grab that proton, leave the two electrons behind on the oxygen; it is going to give us an O-; it is going to form methoxide here.0871

If we use sodium hydride, we are going to get the sodium salt; this forms sodium methoxide; just a little vocabulary in there on how we are going to name our alkoxides if we have them.0883

What is the other product in this case?--we use sodium hydride as a base; we have an H-; it is combining with an H+.0896

Our other product is hydrogen; what do you know about hydrogen?--it is a gas; it is going to bubble away.0903

Tell me how does this reverse reaction look?--if you wanted to consider whether or not it is going to be going forwards or backwards.0910

Because our hydrogen gas, one of our products, is leaving, the reverse reaction becomes impossible; there is no reverse reaction.0918

That is what makes sodium hydride something that is ideal for making alkoxide; 100% of the alcohol will be converted to an alkoxide.0926

Another possibility is to use a redox reaction to do this, pretty much the same transformation.0938

But instead of using sodium hydride, we are going to use sodium metal; this is sodium zero (Na0) or sodium dot(Na·).0944

This is a metal; it has a single electron in its valence shell that makes it a very good reducing agent.0950

We have seen sodium metal as a reducing agent for alkynes, doing the dissolving metal reduction to a trans alkene.0956

If we were to react with an alcohol instead, what happens is we end up forming two equivalents of sodium methoxide and hydrogen gas.0963

We actually get the exact same products out that we did in this acid-base reaction; but this mechanism is more complicated.0974

It is not a simple proton transfer; it is an electron transfer; we call those redox reactions.0980

We look at this reaction carefully; we see that we are starting with sodium metal and we end up with Na+; that is the oxidation that is occurring.0986

In other words, sodium is releasing that electron, donating it to someone else, getting oxidized itself and causing a reduction in the hydrogen of the methanol.0994

That started out with a +1 oxidation state; but when it is in its elemental form, it is hydrogen gas; it has a 0 oxidation state... 0 oxidation number.1005

We went from a +1 to a 0; that is a reduction; in oxidation... where there is reduction, the gain electrons.1018

This mechanism is not one we are going to study; but it is more of a predict-the-product or providing the reagents.1025

The other thing, like all redox reactions, this reversible reaction is impossible; redox reactions always go to transferring the electron to give the more stable species.1033

What is great about both of these is that they are both irreversible; if we need to form an alkoxide, you have two good choices, sodium hydride or sodium metal.1045

Those are two great reagents that we can count on; sodium metal works; but sodium, lithium, potassium, those are all group 1A; those would all work to do this oxidation.1058

As you move down in the periodic table, your atoms are getting bigger and bigger and bigger; that electron is being held even further and further from the nucleus.1073

That makes it a lot more reactive; so something like potassium, which is bigger, is going to be more reactive.1081

That is useful for certain alcohols like this t-butyl alcohol; this is a weaker acid than something like ethanol or methanol; why is that?1087

Because these alkyl groups, this tertiary alcohol, this is a tertiary alcohol; those are in general weaker acids and less willing to give up a proton than a primary or secondary.1104

Because these alkyl groups are all electron donating; that is not a good thing if you want to be forming and O- here; that would be destabilized by those alkyl groups donating electron density.1117

This would be a very very very slow reaction if we use sodium metal or even lithium metal; but if you use potassium metal, that is going to be strong enough to do this redox reaction.1128

Same idea, our products are going to be an O-K+; we are going to get the alkoxide; the other product here is this H+ that we have--is going to come off as hydrogen gas.1141

When we balance our redox reaction, you see we are using two equivalents of the alcohol and two equivalents of the potassium; so we will get two equivalents of this guy.1158

This is called potassium t-butoxide... t-butoxide or tBuOK; we have seen this before as a nice strong bulky base that we have used.1165

The reason that we usually see this as the potassium salt is because that is the one that is commercially available; that is the one that is typically made.1179

That is how we typically make it--is by using potassium metal to do this redox reaction.1186

One other example of an alcohol, when we are talking about acidity, let's talk a second about phenols; a phenol is a very special alcohol; it is when we have the OH group on a benzene ring.1193

These are significantly more acidic than other alcohols; an ordinary alcohol has a pKa of something like 16 or 18 or somewhere in that range; phenol has a pKa of about 10.1203

Again eight zeros; that is tens of millions of times more acidic; in that case, this NaOH is okay; this is a weaker base--is okay.1214

You don't need to use sodium hydride; you don't need to use sodium metal; you can but you don't have to.1230

Typically we try and use the least reactive base we can because that is going to be easier to handle and maybe cheaper or a simpler reaction to run; so weaker base is okay.1236

When we do the deprotonation here, now we are using hydroxide to do our deprotonation; the products then are going to be sodium phenoxide--is what we call it when we have a pheynl O-.1253

Sodium phenoxide and water as our other product; water has a pKa of about 16 which would not be good if we were trying to deprotonate something with a very similar pKa.1265

But because this is so much more acidic, it is a million times more acidic, we know the equilibrium lies in the direction of the weaker acid-base pair.1275

In this case, the equilibrium would be very much favored in the forward direction.1285

In fact hydroxide is very effective and is suitable for deprotonating a phenoxide and essentially completely converting it to the anion, to the O-, the alkoxide.1290

How would we make an alcohol group?--where does an OH come from in a structure?--there is a few different strategies we can have for this.1306

One way of installing that OH group is by doing some kind substitution reaction.1314

In other words, if I had a leaving group on the carbon chain, I could replace that leaving group with an OH; then my resulting product would be an alcohol.1319

We have a few different substitution mechanisms; we have Sn1 mechanism; we have Sn2 mechanism; let's take a look at both of those.1327

If we wanted to do an Sn2, remember Sn2 is backside attack; we need a strong nucleophile to come in and attack and kick off the leaving group in a single step.1335

The nucleophile I would use here would be hydroxide, very strong nucleophile; it is great at doing the Sn2; the problem is that it is also a very strong base.1344

We have a competition going on between the Sn2 that it would do if it was a nucleophile or an E2 is another thing it could do as a base--it could go after a proton.1358

The way we decided between Sn2 and E2, this competition, is we considered steric hindrance knowing that the backside attack can't have any steric hindrance.1371

When would the Sn2 mechanism be reasonable for alcohol synthesis?--only when our carbon chain has very little if any steric hindrance.1381

If we have a methyl halide or if we have a primary halide, those would be best; maybe it is allylic or benzylic; that would also help support the Sn2 mechanism.1392

For example, if you had ethyl bromide and you treat him with sodium hydroxide, because this is a primary RX, a primary alkyl halide, this is our electrophile.1403

The hydroxide is our nucleophile; no problem attacking the carbon, kicking off the leaving group; this would be a great Sn2.1416

This would be a great Sn2; this would be a reasonable way to make an alcohol; we just made an alcohol by using hydroxide as our nucleophile.1426

This one is a little trickier because we have... if you look at this carbon, it is now secondary; it has some steric hindrance.1438

But because it is allylic, this is something that makes the Sn2 not so bad; that is something that would make this possible.1445

The reason Sn2 is okay here is because the π bond stabilizes the transition state of the Sn2; we have p orbitals here.1459

We form a p orbital as the nucleophile is kicking out the leaving group; that helps make the Sn2 a faster reaction; it competes better.1473

This is another case where, even though it is secondary, you would probably get as your major product the substitution product.1483

But that is a rare case--if it is benzylic meaning it is next to a benzene ring or allylic meaning it is next to a double bond.1493

Otherwise if you are a secondary alkyl halide, you are a secondary leaving group, there is enough steric hindrance where Sn2 is not going to be favorable.1500

For example, if we just have plain old cyclopentyl bromide and we try to do hydroxide, there is enough sterics here that, instead of doing the Sn2, the E2 is faster.1510

Meaning it acts as a base; it attacks the β hydrogen--that is what bases do; forms a π bond, kicks out the leaving group.1522

So the E2 elimination competes with the Sn2 anytime our nucleophile can be a strong base; because we are using hydroxide in this case, for sure that competition is going to be there.1532

E2 is favored in that case; of course, if it is tertiary, there is no chance of having the Sn2; so that is favored as well.1546

Remember if you have a phenyl or a vinyl alkyl halide, we can't do our Sn2 mechanism there because he is on an sp2 hybridized carbon.1553

A leaving group on an sp2 hybridized carbon is not what we have for backside attack; for backside attack, we need a tetrahedral carbon; we need an sp3 hybridized carbon.1564

It is a reaction of alkyl halides, not vinyl halides, not aryl halides; this would be no reaction.1573

Sn2 is one option for doing a substitution synthesizing an alcohol; we could also do an Sn1 reaction with water as our nucleophile.1581

This is a weaker nucleophile; water is not going to come and attack and do a one-step mechanism; but we can get a substitution via an Sn1 mechanism where we have a carbocation involved.1593

We call this reaction solvolysis because it is reacting with the solvent or more specifically hydrolysis since it is reacting with water.1605

What we need in this mechanism is... because we are forming a carbocation, this is going to be a reasonable option only if a favorable carbocation could be formed.1614

For example, if you have an allylic leaving group or benzylic or tertiary, those are all excellent carbocations; those would all be decent Sn1 mechanisms.1626

For example, we have this benzylic carbon with a leaving group; water is a weak nucleophile; that is why we are deciding that it cannot be the Sn2; it has to be the Sn1.1636

What mechanism is that?--what does that mean?--that means because there is no one here to attack, the leaving group just leaves on its own to give a carbocation to this electrophile.1650

We have water react as a nucleophile to an OH2; then we can deprotonate to get rid of that O+.1664

You could use the Br- here to tidy up; you might see that is sometimes; but really water, because we are doing hydrolysis, water is our solvent.1682

Water we have a lot of--that is probably going to be our best base here; we are going to get the alcohol.1689

Tell me about the stereochemistry here; what is the stereochemistry of this C-O bond?1697

Because the carbocation is planar, when water attacks, it can attack from the top face, it can attack from the bottom face.1705

Both of those are going to be equally likely; what we are going to get is a mixture of having the OH as a wedge and an OH as a dash.1711

So Sn1 is not a good situation with a chiral center unless it is okay to have a racemic mixture; then Sn1 is good.1720

But if you want stereocontrol then you need to find something like an Sn2 where you can do that backside attack; here we would get racemic mixture.1729

The other thing to keep in mind with water and trying to do an Sn1 mechanism is that, because you have a carbocation in your mechanism, it can rearrange.1741

If it is not one of these very very stable carbocations like in this case, we would get a positive charge here.1749

We might think that we could, when we do hydrolysis, we might get an OH in this position.1760

But in fact we don't because we are right next to a more substituted carbon; if the carbocation could somehow get over there, then it would be a more stable carbocation.1767

That in fact is the major product we get with rearrangement; we get this substituted product; we always have to keep that in mind.1780

Anytime we want to invoke a mechanism involving the carbocation, we have to make sure that, if there is a possibility of rearrangement, we account for that and that needs to be the product that we are expecting.1790

This is a good mechanism for you to try; see if you can do that complete Sn1 mechanism involving the carbocation rearrangement.1801

Another reaction we have seen in the past that gives an alcohol product is starting with an alkene starting material.1810

If you add water across an alkene double bond, you will end up with an alcohol product.1819

For hydration, remember we saw three different methods for adding water across a π bond.1825

If we used either H3O+ or this one, oxymercuration-reduction or oxymercuration-demercuration; this was the case where we broke the π bond; we add an H and an OH.1831

Where did hydrogen go?--both of these follow Markovnikov's rule; meaning the carbon with the hydrogen over here gets the hydrogen.1844

This last method, hydroboration-oxidation, this two-step method does the opposite regiochemistry.1858

We break the π bond; we add an H to the more substituted carbon, an OH to the less substituted carbon; we add the hydrogen to the carbon without the hydrogens; we have our CH3 there.1864

Our regiochemistry is different here; how about the stereochemistry?--is there anything we knew about the stereochemistry here with the hydroboration?1880

Do you remember that mechanism--hydroboration?--both the hydrogen and boron are added at the same time which means we have to add to the same face.1889

So another result of this is also we get syn addition; that means we need to show the H and the OH coming from the same face.1896

For example, they could be both wedges which forces this methyl group to be a dash; of course, it can come from the opposite face; it can come underneath.1903

Then we would get the enantiomer in which the H and the OH are the dashes and the methyl group is the wedge.1912

When we take a look at these products, we see that this is an alcohol product that we get; so another way that we can maybe put in an OH group on a carbon chain is to start with a double bond.1920

Then we could put the OH maybe on one carbon or the other depending on the reaction conditions we use, the reagents we use.1934

We could also do an oxidation reaction of an alkene; that would give a diol; for example, KMnO4 or if we use OsO4.1942

This was one of the several oxidations we studied for alkenes; this is the one that broke a π bond and did a dihydroxylation and added two OH groups.1951

It added them to the same face; again we would show some stereochemistry here; OH up, OH up; which then forces this methyl group down.1962

We call that syn dihydroxylation; if the it came from the bottom face, that would be the enantiomer.1973

This makes a special kind of alcohol; this makes a diol where we have two OH groups, one right next to each other.1980

If we have that special kind of target molecule, we can consider doing a dihydroxylation of an alkene.1988

Another great way to make an alcohol is to start from a ketone or an aldehyde; in other words, if I had a carbonyl on my carbon chain, that could get converted to an alcohol group.1996

The way we do this is we react it with a nucleophile of some kind because an aldehyde or a ketone... we are going to be studying these later.2009

We will see that these are electrophiles; they are electrophiles because this carbon-oxygen bond of course is polar because oxygen is electronegative.2018

It is also very polar because this has resonance; that has a resonance form with an O- and a C+.2026

As a result, the carbonyl carbon is very... remember this is called a carbonyl; I am going to be using that word a lot; that means the C-O double bond; it is a good word to know.2032

A carbonyl has the following polarity--a partial positive on the carbonyl carbon, partial negative on the carbonyl oxygen; that makes this carbonyl carbon a very good electrophile.2044

If you were to react it with a nucleophile, what happens is it attacks the carbon and it breaks this π bond and puts those electrons up onto the oxygen; we get some kind of intermediate like this.2056

If we were to then treat this with H3O+, some kind of workup procedure, at the end of the reaction, we are going to provide it with some kind of acid, some kind of proton source.2074

What will happen is we can protonate that O- with our work up and convert it to an OH group.2086

When we have a nucleophile adding into a carbonyl, the product we get has a nucleophile now connected to the carbonyl carbon and an OH group where the carbonyl oxygen used to be, the C-O double used to be.2097

Our product here is an alcohol; so this is another very good way to synthesize an alcohol and to come up with an OH as part of our final structure.2111

What nucleophile do we have available to us?--there is several we can look at; one nucleophile we have already seen is this acetylide ion.2126

He would be a very good nucleophile; we saw him doing Sn2s with alkyl halides; but he would also like to add into a carbonyl; he would be good at that.2133

But there is some other nucleophiles that we will be studying this unit that we have not yet seen.2142

This one is called a Grignard; it has an R group; that means some kind of carbon chain with a magnesium and a halide.2148

Anhydride, an example of that is something like LiAlH4; we are going to look at these new reagents and learn something more about their reactivities and what they do.2155

Here we see a C-; I can see how that is going to be a good nucleophile; but what about RMgX?--how do we get a nucleophile out of that?2167

This magnesium has a +2; the chloride, if you think about their oxidation state, the chloride has a -1.2176

That gives the alkyl group, this carbon group, the character of being a carbanion, being an R-.2185

We are going to be looking at things like this Grignard reagent as a source of R-; that would be a very good nucleophile.2193

Anhydride, as given as an example with this reagent, anhydride means we have an H with a lone pair and a negative charge; that would also be a very good nucleophile.2201

We are going to be adding maybe an alkyne group here or an R group here or a hydrogen here to be this final group in the position of where the carbonyl used to be.2212

Let's talk about this RMgX; it is an example of what is known as an organometallic reagent; it is called that because it has both an organic part and a metal.2226

M just represents some sort of metal; an R group is usually some organic component, some carbon chain; these are called organometallic reagents.2236

We are going to two examples of them; one is called RMgX; it has the formula of RMgX; one is RLi.2247

The first thing that we will do is we will talk about how to prepare them; where do they come from?--they are going to be prepared from an alkyl halide, a carbon chain with a halogen on it.2253

For example, if we were to take methyl bromide or bromomethane and react it with magnesium metal.2264

Here you are literally taking shavings of magnesium metal, magnesium turnings, tossing them into your reactions, stirring that reaction, heating it.2270

You will see your magnesium metal dissolve as it reacts; what it does is it inserts itself into the carbon-bromine bond; this is not a mechanism that we are going to worry about.2279

But we want to know what the product looks like; that is where we insert a magnesium into the carbon-bromine bond.2290

The other product here is going to be some salts; we are going to lose the bromide; we are going to make magnesium; that is not too important.2297

What we just said about this guy... he is called a Grignard reagent by the way, named after the chemist who developed the reagent; it has the general formula of RMgX.2308

This R in this case is just a methyl group; we will we see it could be just about anything.2322

The X is very often a bromide; but it also can be a chloride or a iodide; you can have another halogen there as well.2326

As we just mentioned, since the magnesium has a +2 and the bromine has a -1 as their oxidation state and the Grignard is a neutral molecule, that means the carbon has a -1 oxidation state.2334

These are not ionic compounds; you can think of this as having a combined partial +1 and this has a partial -1.2350

If you want to think of it as a partial charge rather than a full ionic charge, either one of those ways to consider it would be good.2359

Because it is a carbon with a negative charge, that means he would be a very good nucleophile; what is interesting is we just came from methyl bromide.2371

What kind of reactivity does methyl bromide have?--here we have a -1 for the bromine; we had a partial positive for this carbon.2378

We went from an electrophilic carbon, and then by attaching a metal to it, we turn it into a nucleophilic carbon.2388

That is what organometallic reagents looks like and that is how they behave--is that any carbon bearing a metal has negative charge character, anionic character.2395

The Grignard is one option; we get that when we react an alkyl halide with a magnesium metal.2404

If we did lithium metal instead, what happens is we do a halogen metal exchange; where we used to have the halogen, we now have a lithium.2409

If we consider the nature of this methyl group, this carbon, what kind of charge do you associate with a lithium?--what is the oxidation state of a lithium?2420

+1; we could say partial +1 if you would like; which again makes this carbon partial -1; this is another example of a nucleophile; all organometallic reagents are going to be nucleophilic.2429

Unfortunately he doesn't have a cool name like the Grignard; this is just called an organolithium... an organolithium reagent.2442

It has this general formula of some kind of carbon group with a lithium attached; we call those organolithiums.2451

What is great about these Grignard reagents, organolithium reagents, and organometallics in general is this R group that you are having the metal attached.2459

The R can be almost anything you can imagine; it can be an alkyl group just like we saw here; here we have a methyl; here we have a carbon chain.2468

You could attach a metal to a random sp3 hybridized carbon group; but you can also have aryl Grignards or aryl lithiums, phenyl lithiums, for example.2475

You can have a metal attached to an sp2 hybridized carbon; or vinyl, you could have it part of a double bond.2484

So huge variety in the types of carbon groups that can be attached to a metal and therefore be nucleophilic.2491

When it comes to synthesis, these organometallic reagents are really going to open up the door for us for making all sorts of interesting and new molecules.2498

Let's think about the reactions that an organometallic might undergo; we just learned how to make them; how do they behave?--what can they do?2506

One reaction that they have is that they are extremely strong bases; in other words, they can be protonated.2515

If we have this magnesium chloride on this carbon chain and we react it with an alcohol like this; we just learned how alcohols can be acidic.2522

This is an acid; this is a very strong base; we are going to get a proton transfer reaction to take place.2534

One way we can show this reacting, sometimes we just see the arrow coming from this carbon metal bond.2542

But I think an easier way to do the mechanism for these is wherever you had the metal attached, you treat that carbon as if it has a lone pair and a negative charge.2549

It doesn't really; it is not an ionic species; but that is a very helpful way to view it; it is a very helpful way to show mechanisms.2560

What we do is we will just put quotes around this species saying we don't really have this but it acts just like it; the quotes let us get away with that.2568

Now we can see this base; clearly it has a lone pair and a negative charge like other bases do; we can clearly see how we can do that proton transfer.2577

The base grabs the proton, leaves the electrons behind; the product we get is my cyclopentyl group--has been protonated; we have a proton in that position.2586

We also made some methoxide here; we deprotonated the alcohol and we protonated the Grignard in this case; what is the salt here?--this is going to be like an MgBr salt of the alkoxide.2598

Typically this is a side reaction; this is not a valuable reaction; this is something we are going to try to avoid.2617

What is very important to know is that Grignards will react with water; just like it will react with alcohol--has an acidic proton; water certainly has an acidic proton.2624

We must use very dry glassware, dry solvents; we can flame dry our glassware to get rid of any residual moisture that might be attached to it.2633

We will store our glassware in an oven to keep it super dry; then set up the reaction under argon or nitrogen, some inert atmosphere, so that it is nice and dry.2642

Our solvents, we will put some drying agents or we will freshly distill it so it is super dry.2654

Any bit of moisture in our reaction mixture or in our atmosphere has the potential of quenching our Grignard reaction, of reacting with a Grignard reagent and destroying it.2660

We are going to be very careful to do this kind of reaction in the appropriate conditions.2671

Of course when it comes to a solvent that we are using, we can't use a protic solvent like water or an alcohol because the Grignard cannot exist in those conditions.2677

Instead we need to use aprotic solvents; things like ethers, diethyl ether, THF stands for tetrahydrofuran--has this formula; some kind of ether is going to be used.2686

Again we should get used to these names of solvents because it is going to be very commonly part of your reaction conditions--is listing the solvent.2699

You don't want to worry too much about that when you see it or try to use it somehow as a reagent; it is just a solvent; all of our reactions have solvents associated with them.2707

There is one way we can make this a useful reaction; that is if we wanted to introduce a hydrogen into our structure or maybe even a deuterium.2718

Deuterium, which is what D stands for, is an isotope of hydrogen that has a neutron in here; it is H2.2733

It is so commonly used, instead of call it H2, we usually just use the letter D to represent deuterium.2743

Sometimes we want to include isotopic labels in a structure; this would be a way that we can do something like that.2750

For example, if we had a bromine here on our benzene ring and we reacted this with magnesium, that would insert a magnesium and turn this Br group into an MgBr group.2757

Then what would happen if we reacted this with D2O, deuterated water, instead of H2O?2769

We would expect that phenyl minus to protonate, deuterate in this case, add a D+; therefore you can get a deuterium installed in your structure; it can be a useful reaction.2774

But by in large, the reaction with alcohols and water just help us, guide us in our experimental conditions to make sure we avoid those so that the side reactions don't happen.2788

Where we are typically going to be using our organometallic reagents like a Grignard is as a strong nucleophile; this is where it is most useful; which means it reacts with an electrophile.2802

An electrophile like a carbonyl, not alkyl halides; we have seen those as electrophiles before; but these do not react with Grignards; we will see an example of that.2812

For example, let's say we had methyl lithium and cyclohexanone with some THF; what if we mixed all those together?2825

The methyl lithium, remember this is like having a CH3-; the lithium gives us a + charge; that means it is like we have a CH3---definitely a nucleophile.2833

I am going to put that in quotes to remind myself it is not really ionic; but it reacts kind of like that.2845

Is the carbonyl in this ketone, is the carbonyl an electrophile?--absolutely, this carbonyl carbon remember is always partially positive.2853

We will be seeing dozens of cases of that when we study carbonyls; what happens when you mix these two?2862

This is going to attack the carbonyl and break the π bond... I moved that up a little higher... going to attack the carbonyl carbon and break the π bond.2870

It doesn't matter if you show it coming from the right or left; but it is going to give me an O- where the carbonyl used to be.2882

Then after work up, step two is work up, I am going to protonate that O-; and I just synthesized what kind of functional group?--I just made an alcohol.2891

This would be a nice way to synthesize an alcohol; furthermore, notice what we just did here--because we had a carbon nucleophile and a carbon electrophile, we just made a new carbon-carbon bond.2909

Again that is unique in organic chemistry; that is a pretty special reaction to study; that is going to be extremely useful.2921

Grignards are extremely useful organolithiums for making alcohols and creating new carbon chains in the process.2928

Let's take a look at this example; we have methyl bromide; methyl bromide is definitely an electrophile because we have a carbon bearing a leaving group.2938

Phenyl magnesium bromide, what does that mean?--we have phenyl minus, put that in quotes.2947

It is just like having a phenyl ring with a lone pair and negative charge on one of the carbons, a strange looking species but that is how it behaves.2955

What would be very tempting here, what makes sense, is we can have this attack the carbon and kick off the leaving group; very tempting to do; however it does not happen.2965

That is just an exception that we really need to get used to seeing is that Grignards do not react with an alkyl halide.2980

Grignards... when I say Grignard, I mean Grignard organolithium; they are pretty much going to be used interchangeably for most of our reactions.2989

Grignards don't do Sn2s; they don't do Sn2 mechanisms; they are not there to react with an alkyl halide.2998

Remember how did you make this Grignard?--you started with the halide; we use alkyl halides or any kind of halide to prepare a Grignard.3008

That would not work at all if the Grignard, as it was being formed, started to react with the starting material.3018

They are compatible with each other; Grignards and organolithiums will not react with alkyl halides.3023

We are just going to have to try to keep that straight because it is really a reaction that is a logical one and one in fact that a lot of students make mistakes for in doing so.3028

But we just need to remember that this is the one reaction Grignards don't do; they are going to react with carbonyls but not with alkyl halides.3040

The other nucleophile we will take a look at for synthesizing alcohols is the hydride nucleophile; hydride again means we have an H with a lone pair and a negative charge.3054

Typically for the reagents we are going to be seeing, especially when it is being nucleophilic... again we are going to put that in quotes.3067

Because these are not ionic species; the hydrogen minus is not floating around on its own; it is always going to be coordinated to a metal.3073

We are going to either use lithium aluminum hydride or sodium borohydride; you could see how similar these species are.3081

That aluminum is going to take one of the hydrogens and deliver it or the boron is going to deliver a hydrogen; but it is going to react as if it was an H-.3090

Let's see an example; if we start with a ketone or an aldehyde, we start with a carbonyl and we react it with lithium aluminum hydride.3100

This actually is often abbreviated LAH for lithium aluminum hydride; in step one, if we react this with LAH, that gives us a source of H-.3107

This is one of those cases where really saying the name of the reagent and thinking about the name of the reagent is going to help us predict the product.3119

The name of the reactive species is part of the name; it is lithium aluminum hydride--right there we see that we have H-; that means we have a nucleophile.3126

We need to look around for an electrophile; this carbonyl will be our electrophile; we attack the carbon and break the π bond; carbonyls do the same thing all the time.3135

We now have a hydrogen attached and an O-; just like with the Grignard, we are going to follow this up with a reaction to get rid of the O-.3151

We are going to do step two as some kind of aqueous work up; step two, we protonate; and our product is an alcohol.3163

This is described as a reduction reaction; we could describe this as a hydride reduction because we increase the number of C-H bonds while decreasing the number of C-O bonds.3182

Ketones and aldehydes can be reduced with a hydride reagent, either lithium aluminum hydride or sodium borohydride, to give as a result an alcohol product.3203

Let's see a few examples of this; we just saw one with lithium aluminum hydride; with LAH, we need a two-step process.3217

First we react with LAH; then we react with H3O+; we do a work up.3224

NaBH4 is less reactive; this is less reactive than LAH; so we don't need to do a stepwise approach; we can actually use as our solvent a protic solvent.3229

The NaBH4 is going to be donating our H-; the methanol here is going to be donating the H+ at the end.3244

It is going to available for, instead of a separate workup step; it is going to be our final protonation step.3253

What happens, same thing as LAH; we attack the carbonyl; let's see if we could just draw a product; we have our nucleophile added to the carbonyl carbon; what used to be the carbonyl is now an OH.3260

We have that same pattern each time whether it is a Grignard or a hydride; we add the hydrogen, we add the nucleophile to the carbonyl and our carbonyl is now an OH.3275

Let's try one more; how about if we took this ketone and we react it with hydride and alcohol; we expect to reduce the carbonyl from a ketone to an alcohol; we could do the mechanism for that.3286

What happens with this alcohol then if we treat it with H2SO4 and heat?--we saw that as a dehydration reaction.3306

This is a way to remove water from an alcohol and form an alkene, form a carbon-carbon double bond.3316

Let's think about the regiochemistry; is there more than one alkene that is possible?--where would the best one be?3324

We have our leaving group here; but our carbocation remember can go anywhere it wants because those can rearrange; so what would be the best location of the double bond in our final product?3331

I think we are going to want to go here so that the double bond is now conjugated with the double bonds that are already here in the benzene ring.3347

This is the major product; it is more stable than having the double bond in this position, in the vertical position, more stable because it is conjugated with the existing π bonds.3356

That means it has resonance; we can have delocalization of these π electrons; we have p orbital, p orbital, p orbital, p orbital; we have all these p orbitals are linked.3375

We can have resonance and delocalization of that alkene; that is going to be a more stable case because it is conjugated.3384

Remember Zaitsev's rule; Zaitsev's rule tells us that the more stable alkene is always going to be our major alkene.3393

Let's look at a few more examples; how about if we wanted to do this transformation?--here we have a one carbon chain; my carbonyl is now an OH and we have a new carbon here.3405

Let's draw this in as a methyl group, as a CH3, expand our line drawing; it looks like this is a new group that is here, that has been added.3422

If we wanted to react with this carbonyl carbon, we know that this is electrophilic; this is partially positive so it is electrophilic.3431

What we need to react with that carbon is we need a methyl that is a nucleophile; how can we make a methyl be nucleophilic?3441

I think all we need to do, if we had a lone pair and a negative charge, that would make it nucleophilic; how do we get to that reactivity and that species?3454

That is exactly what a Grignard reagent does; if we have CH3MgBr, that is nucleophilic; that would react with the carbonyl; we would do that as step one.3465

As step two, after the Grignard, we would need some kind of aqueous work up, H3O+ for example; that would protonate the O- to give an alcohol product.3478

More than doing just a simple reduction of the carbonyl, we also added in a new carbon group; this is how we could use our Grignard reagent.3490

Here is another example with an organometallic; where does this alkyl lithium come from?--how do we make an organometallic?3499

What will we make this, if we ask what starting materials do we need?--we are seeing the reaction conditions here; we have added in lithium metal; what did that lithium metal react with?3509

What did we need in that position?--we needed a halogen of some kind; we need an RX to go to an alkyl lithium or Grignard reagent too.3524

You can pick any halide you want--bromide, chloride, iodide; they would all do a good example.3537

The Grignard and the hydride reagents are new reactions, reaction with a carbonyl; those are new reagents that are very useful in forming the alcohol functional group.3545

That is useful; that is a new strategy for synthesizing alcohols; we can add that to our list of other reagents where we are doing more of a functional group in our conversion.3557

Going maybe from an alkyl halide to an alcohol or from an alkene to an alcohol by doing those sort of transformations as well.3568

The next part of the alcohol unit is going to look at reactions of alcohols; once we have an OH on a structure, what manipulations can we do to it?--what transformations can it undergo?3577

Thank for coming to Educator.com; I will see you again soon.3590

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