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Lecture Comments (26)

1 answer

Last reply by: Professor Starkey
Wed Jun 21, 2017 12:38 PM

Post by Sevada Minasian on April 22, 2017

In the first "practice question" for this lecture  that asks for the hydrogenation of the alkyne with lindlars catalyst, the overall product ends up being trans instead of cis when you check the answer. is this a mistake?

1 answer

Last reply by: Professor Starkey
Sun Sep 27, 2015 11:02 AM

Post by Jinhai Zhang on September 26, 2015

Dear Prof. Starkey:
for the addition of HX of alkyne, when we formed the vinyl carbocation, the Br- attacked from the top or from the bottom?  Is there a stereoselective for the vinyl carbocation? Can I think is a syn addition or trans addition of Br from alkyne to alkene

1 answer

Last reply by: Professor Starkey
Sun Sep 27, 2015 11:01 AM

Post by Jinhai Zhang on September 26, 2015

Prof. Starkey:
For the bromination of alkyne, if we add 1 mole of Br2, does the mechanism have the same as alkene that you talked in previous lecture. I can think it is trans addition when we brominate the alkyne to alkene? Thank you for answering.

1 answer

Last reply by: Professor Starkey
Sun Dec 21, 2014 11:22 PM

Post by Parth Shorey on December 21, 2014

At 38:13, Can I use NH2 for the final bond adding to the alkyne? I got it right when you used OH but then for the 2nd Br to break off with an additional bond I used NH2, is that possible?

1 answer

Last reply by: Professor Starkey
Sun Dec 21, 2014 10:09 PM

Post by Parth Shorey on December 21, 2014

I don't understand what is it about peroxide that replaces B ? In hydroboration-oxidation?

1 answer

Last reply by: Professor Starkey
Sat Dec 20, 2014 10:32 PM

Post by Parth Shorey on December 20, 2014

I don't understand when reacting with NaNh3 in the beginning. How you keep giving out H, so it hoes from NH3 to NH2 but then the next H on the trans, where did that come from? I am doing the mechanism and the it doesn't make sense?

1 answer

Last reply by: Professor Starkey
Sun Nov 9, 2014 10:05 PM

Post by Foaad Zaid on November 9, 2014

Hello, for the hydration of an alkyne reaction. Is it safe to say that the acid is H2SO4, that it does all the deprotonating?

2 answers

Last reply by: somia abdelgawad
Wed May 7, 2014 6:24 PM

Post by somia abdelgawad on May 5, 2014

thank you so much. I think to are inspiration to me to be a chemistry teacher one day. you are a great teacher and great help

1 answer

Last reply by: Professor Starkey
Sun Nov 25, 2012 12:13 AM

Post by cheyenne bolanos on November 20, 2012

Hello Dr. Starkey,

I have been doing my homework and in some examples NaNH2 is used as a reagent and in other instances HC(triple bond)C- is used. Could you verify what the difference is?

Thank you,


2 answers

Last reply by: Jessica Martinez
Sat Nov 3, 2012 10:58 PM

Post by Jessica Martinez on November 2, 2012

Hi Dr. Starkey,

31:25 after you protonated CH2, i think you are missing an "H" it should be CH3COHCH3. I may be wrong

1 answer

Last reply by: Professor Starkey
Wed Dec 28, 2011 11:37 PM

Post by Robert Shaw on December 26, 2011

Could we also use KOH- For the Rx.

1 answer

Last reply by: Professor Starkey
Sun Nov 20, 2011 9:25 AM

Post by Aneseh Ardeshir on November 20, 2011

Dear prof
When we reacting with NaNH2 we suppose to have terminal alkyl, however here we have internal alkyl, I think the second reagent should be koH rather than NaNH2.


Draw the major product(s) formed from this reaction:
Draw the major product(s) formed from this reaction:
Draw the major product(s) formed for each of the reaction:
  • Reduction of Alkynes using Lindlar's Catalyst produces cis alkene:
  • Reduction of Alkynes using Na and NH3 produces trans alkene:
Draw the major product(s) formed from this reaction:
  • This is a dehydrohalogenation reaction.
  • The starting molecule is a geminal dihalide (Two X atoms on the same carbon), which can also undergo dehydrohalogenation similar to vicinal dihalide (Two X atoms on adjacent carbon).
  • The starting molecule will undergo two E2 reactions to form an alkyne.
Draw the major product(s) formed from this reaction:
  • This is a hydration via hydroboration-oxdiation problem
Provide a synthesis for the following reaction:
  • Step 1:
  • Step 2:

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.



Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Structure of Alkynes 0:04
    • Structure of Alkynes
    • 3D Sketch
    • Internal and Terminal
  • Reductions of Alkynes 4:36
    • Catalytic Hydrogenation
    • Lindlar Catalyst
  • Reductions of Alkynes 7:24
    • Dissolving Metal Reduction
  • Oxidation of Alkynes 9:24
    • Ozonolysis
  • Reactions of Alkynes 10:56
    • Addition Reactions: Bromination
  • Addition of HX 12:24
    • Addition of HX
  • Addition of HX 13:36
    • Addition of HX: Mechanism
  • Example 17:38
    • Example: Transform
  • Hydration of Alkynes 23:35
    • Hydration of Alkynes
  • Hydration of Alkynes 26:47
    • Hydration of Alkynes: Mechanism
  • 'Hydration' via Hydroboration-Oxidation 32:57
    • 'Hydration' via Hydroboration-Oxidation
    • Disiamylborane
    • Hydroboration-Oxidation Cont.
  • Alkyne Synthesis 36:17
    • Method 1: Alkyne Synthesis By Dehydrohalogenation
  • Alkyne Synthesis 39:06
    • Example: Transform
  • Alkyne Synthesis 41:21
    • Method 2 & Acidity of Alkynes
    • Conjugate Bases
  • Preparation of Acetylide Anions 49:55
    • Preparation of Acetylide Anions
  • Alkyne Synthesis 53:40
    • Synthesis Using Acetylide Anions
  • Example 1: Transform 57:04
  • Example 2: Transform 1:01:07
  • Example 3: Transform 1:06:22

Transcription: Alkynes

Hello and welcome back to

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; thank you.4393