Enter your Sign on user name and password.

Forgot password?
Sign In | Subscribe
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of Organic Chemistry
  • Discussion

  • Study Guides

  • Practice Questions

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books & Services

Lecture Comments (15)

1 answer

Last reply by: Professor Starkey
Sat Mar 12, 2016 10:00 AM

Post by Kristy Osborne on March 12, 2016

Just wondering is there any way to increase the speed of the lectures?

1 answer

Last reply by: Professor Starkey
Sun Feb 21, 2016 11:58 PM

Post by Hajer Rawag on February 21, 2016

Hi Professor,

Is the cyclobutadiene non-aromatic or anti-aromatic? Thank you!

1 answer

Last reply by: Professor Starkey
Wed Mar 4, 2015 11:45 PM

Post by Krystal Osei on March 4, 2015

Why are we counting the lone pair of electrons for pyrrole in the 4n+2 rule to be 6 electrons instead of 4 electrons? I thought if lone pairs are next to pi bonds they can be in the p orbital for resonance like Furan, so the lone pair should not be counted.

2 answers

Last reply by: Krystal Osei
Tue Mar 3, 2015 12:37 AM

Post by Krystal Osei on March 1, 2015

Do you have a video explaining the difference between aromatic, antiaromatic, and nonaromatic. Your videos are very helpful, thank you

1 answer

Last reply by: Professor Starkey
Fri Feb 21, 2014 10:42 AM

Post by Lalit Shorey on February 19, 2014

Why can't benzene react as an alkene?

1 answer

Last reply by: Professor Starkey
Sun Feb 5, 2012 10:01 PM

Post by jason cowan on February 5, 2012

I'm confused by the last example as well. Doesn't Pyridine have 3 double bonds for 6 Pi electrons + 2 Pi electrons from the Nitrogen - wouldn't this make it anti-aromatic with a count of 8? What is the hybridization on the pyridine nitrogen and how do you know? Isn't it sp2 hybridized because of the double bond with carbon, leaving 2 pi electrons?

1 answer

Last reply by: Jamie Spritzer
Tue Aug 9, 2011 6:31 PM

Post by Costa Sakellariou on August 7, 2011

I would like to ask on your last example how did you count 10 electrons. I see 6 from the double bonds and 2 minus charges which should have been 8 right?

Thank you..

Aromatic Compounds: Structure

Name the requirements for aromaticity.
1. The molecule must be cyclic
2. A molecule must be planar
3. Contiguous p orbitals or completely conjugated
4. Satisfy Huckek's rule: an aromatic compound must contain 2, 6, 10, 14, 18, etc. of π electrons.
How many π electrons are contained in the following molecule:
  • Each π bond contains 2 electrons
14 π electrons
Which compound is more stable?
  • Compound A has a benzene ring
  • Compound A has a smaller ∆Ho than compound B
  • The less stable compound has a larger ∆Ho
Compound A is more stable
Determine if the following compound is aromatic:
  • Circled Carbons are not sp2
Not Aromatic
Determine if the following compound is aromatic:
  • The compound has 12 π electrons
  • Does not satisfy Huckel's rule
Not Aromatic
Determine if the following compound is aromatic:
  • The compound is cyclic and planar
  • The compound has 6 π electrons

*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.


Aromatic Compounds: Structure

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
  • Aromatic Compounds 0:05
    • Benzene
    • 3D Sketch
  • Features of Benzene 4:41
    • Features of Benzene
  • Aromatic Stability 6:41
    • Resonance Stabilization of Benzene
    • Cyclohexatriene
    • Benzene (Actual, Experimental)
  • Aromatic Stability 9:03
    • Energy Graph
  • Aromaticity Requirements 9:55
    • 1) Cyclic and Planar
    • 2) Contiguous p Orbitals
    • 3) Satisfy Huckel's Rule
    • Example: Benzene
  • Common Aromatic Compounds 13:28
    • Example: Pyridine
  • Common Aromatic Compounds 16:25
    • Example: Furan
  • Common Aromatic Compounds 19:42
    • Example: Thiophene
    • Example: Pyrrole
  • Common Aromatic Compounds 21:09
    • Cyclopentadienyl Anion
    • Cycloheptatrienyl Cation
    • Naphthalene
  • Determining Aromaticity 27:28
    • Example: Which of the Following are Aromatic?
  • Molecular Orbital (MO) Theory 32:26
    • What's So Special About '4n + 2' Electrons?
    • π bond & Overlapping p Orbitals
  • Molecular Orbital (MO) Diagrams 36:56
    • MO Diagram: Benzene
  • Drawing MO Diagrams 44:26
    • Example: 3-Membered Ring
    • Example: 4-Membered Ring
  • Drawing MO Diagrams 47:51
    • Example: 5-Membered Ring
    • Example: 8-Membered Ring
  • Aromaticity and Reactivity 51:03
    • Example: Which is More Acidic?
  • Aromaticity and Reactivity 56:03
    • Example: Which has More Basic Nitrogen, Pyrrole or Pyridine?

Transcription: Aromatic Compounds: Structure

Hi, and welcome back to Educator.com.0000

Today, we are going to talk about aromatic compounds.0002

Benzene is an example of an aromatic compound, so if we take a look at benzene and some of the characteristics that it has, we are going to have a better understanding of what all aromatic compounds have in common.0006

What makes benzene special: benzene is a 6-membered ring with alternating π bonds; it has 3 π bonds in there.0016

What makes benzene special is: it has some resonance here, where we can shift those π bonds around...2, 4, 6 electrons, all moving around, delocalizing; and we can draw benzene with the double bonds as shown here, where the vertical one is on the left; or like we have done here, where the vertical bond is on the right.0022

These two resonance forms for benzene--remember, resonance means that neither of these structures really accurately represents what benzene looks like.0044

And, in fact, the true structure is some kind of blend of those discrete Lewis structures.0054

So sometimes, we draw benzene like this: it is represented just with a circle inside of the hexagon.0061

And any time we see this...you might see this notation a lot when we are looking at other aromatic compounds, as well...any time you see this circle, it means you have alternating π bonds.0069

Here we have a 6-membered ring: you are going to put as many alternating π bonds, conjugated π bonds, as possible; so, in this case, we could fit 1, 2, 3 π bonds.0085

If you look at a 3-D sketch of benzene, you know that each π bond comes from the overlap of p orbitals; so we can show one Lewis structure like this, where we have one π bond here and another π bond here and another π bond here.0095

But then, if we take a look at the second Lewis structure (I've kind of tipped it over on its side, so it's a little easier to see), we still have a p orbital on every carbon; but now, the π bonds have shifted to be between these two p orbitals and these two p orbitals and these two p orbitals.0113

And again, recognizing that the actual structure is a blend of these--what that really means is: we have these 6 π electrons that are actually completely delocalized throughout all of the p orbitals.0130

What you can draw--and that is what is nice about this structure that kind of looks like a hex nut--what is nice about that ring representation is: it illustrates that, in an aromatic compound like benzene, we have a cloud of electron density above and below the plane.0144

This is a cloud of electron density above and below the plane.0164

So, benzene is a planar molecule, and we have a cloud of electrons above and below; so there are a few nice models that I would like to share with you, that illustrate this very nicely.0178

This is a model that shows benzene: the 6 black atoms here are the carbon atoms; the σ bonds are in grey; and so, you can see that there is one hydrogen on every single benzene, and all 6 hydrogens and 6 carbons are all in one plane.0187

And what this is showing is: there is a p orbital on every single carbon (these are all sp2 hybridized), and what is nice about this model is that it shows that those π electrons are really delocalized above and below the plane.0203

Rather than having discrete double bonds and single bonds and double bonds, really, benzene has a cloud of electron density above and below, like this.0218

OK, another model that is a lot of fun (but kind of huge) is this one, and this shows--again, same thing: each blue and green pair represents a p orbital (remember, a lot of times we color them differently, because they have different mathematical signs).0226

And so, on every single carbon, we have a p orbital, and we also have a hydrogen pointing straight out, that is in the plane of the molecule; and when we have this kind of a space-filling model of what a p orbital looks like, you can really see how the p orbital on each adjacent carbon is touching an overlapping and interacting and sharing its electrons.0244

So, even though we can maybe think of this as 3 double bonds, really there is no distinction between a double bond and a single bond here, because those electrons are equally distributed throughout the plane.0267

Let's list some of the features that we have seen so far.0282

It is a planar molecule; all of the carbons are sp2--they are all sp2 hybridized--that means trigonal planar with that p orbital on each atom.0287

It is completely symmetrical; in other words, all carbon-carbon bonds are equal in length--so again, that is why this drawing is nice, because this shows the symmetry of the benzene, where, when we draw benzene, it kind of looks like we have some double bonds and some single bonds, but we don't really.0299

All of these carbon-carbon bonds are equal in length, and the length is about the same as 1 and 1/2 bonds.0314

It is shorter than a single bond, but it is a little longer than a double bond, because that hybrid character has some double-bond character, but also some single-bond character.0325

OK, because of this resonance--because we have two resonance forms that equally contribute--that is the best stabilization we can have, so it's very stable because of that resonance.0337

And, as a result, we find that, even though benzene has π bonds, it does not react as an alkene.0349

We have seen reactions of alkenes and things that carbon-carbon double bonds will do; benzene and its fellow aromatic compounds do not undergo the same sorts of reactions, so we will be studying those in the next lesson.0356

In fact, benzene, then, describes a different class of compounds--something called "aromatic."0370

We can abbreviate "aromatic" with an Ar: any time you see Ar, it means you have something like benzene.0376

Another word that you should be familiar with is called aliphatic: aliphatic is kind of the opposite of aromatic.0386

Aliphatic is what we have when we have just an R group, like an alkyl group: aromatic means we have a ring like benzene.0391

One question we can ask is: Just how stable is benzene?0402

We know it has some resonance stabilization, but how much does it have?0405

Well, we can use some empirical data; we can measure the heat of hydrogenation, the energy that is released when we hydrogenate the π bonds on a benzene ring, and see how that compares to the amount of energy we are expecting.0409

So, for example, if you had just a single double bond in a ring, and you had cyclohexene, and you did a catalytic hydrogenation, that breaks the π bond and adds a hydrogen to each carbon; so you get cyclohexane as the product.0423

The heat of hydrogenation for that reaction in kilocalories per mole is -28.6.0439

Let's imagine if we had a theoretical molecule: the theoretical molecule cyclohexatriene.0444

If we put three double bonds in a 6-membered ring, but we didn't allow them to interact--so we had a double bond and then a single bond and then a double bond--if somehow we got rid of that interaction and ignored the fact that they were conjugated, and we did a catalytic hydrogenation of this, this would give the same product: it would still give cyclohexane.0453

We can predict how much energy we would expect to get in that reaction.0473

Each π bond is worth -28.6, and we have 3 π bonds, so we would expect to get out 85.8 kilocalories per mole when we hydrogenate this molecule.0478

So then, when we go to do the experiment, of course (the molecule that has three double bonds in a ring is called benzene)--if we take benzene, and we do a catalytic hydrogenation, suddenly we find there is no reaction.0492

Remember, we just said it doesn't react like an ordinary alkene, so this is a difficult reaction to do.0503

But, if you add some heat, and you add pressure--OK, if you really pound on it--you can in fact hydrogenate benzene to give cyclohexane as a product.0509

When we measure the heat of hydrogenation in this reaction, we get only 49.8 kilocalories per mole.0522

We are not getting as much energy out as we expected: this is what indicates just how stable benzene is.0530

Let's take a look at an energy diagram to try and make some sense of these numbers.0539

OK, what we are imagining is that we have cyclohexatriene; it theoretically would be producing 85.8 kilocalories of energy when it gets hydrogenated down to cyclohexane.0544

But benzene, the molecule--the experimental data actually only released 49.8, so we have a difference: we have 36 kilocalories per mole--benzene is 36 kilocalories per mole more stable than we anticipated.0557

We call this the resonance energy.0571

So, because it gave less energy off when it was hydrogenated, that means it must have started out at a lower energy initially, and so we can't actually quantify it, but it's a huge amount of energy, so there is significant resonance going on in this molecule.0578

Let's talk, then, about the idea of being aromatic and having aromaticity in general.0597

Benzene is an example of an aromatic molecule, but it is not the only one: OK, and there are going to be certain rules that we have to satisfy in order for a species to be aromatic.0603

Now remember: this is the picture of benzene, where we have these clouds of electrons above and below the plane.0613

It turns out that this picture is characteristic of aromatic compounds.0619

Any compound that has this situation, or any aromatic compound, is going to look like this.0629

How do we get there?--well, first of all, it must be a cyclic molecule like benzene: it has to be a ring; and it has to be planar.0638

OK, so we are going to be looking at flat molecules like benzene.0645

We must have contiguous p orbitals: just like benzene, we need to have a p orbital on every single atom.0650

OK, now how do we get a planar molecule that has a p orbital?--typically, we are looking at an sp2 hybridized atom.0658

They can't be sp3, because sp3 hybridized atoms don't have any p orbitals; you have to have a p orbital--contiguous means contiguous: every single atom has to have a p orbital.0670

OK, and finally, we have to satisfy something known as Hückel's Rule: in Hückel's Rule, it says that the number of electrons that are in those p orbitals must be a certain number.0680

They could be either...you could have 2 electrons, or 6, or 10, or 14, or 18, and so on.0690

You have to have a specific number of electrons: now, rather than memorize this never-ending list of numbers, what was determined by Hückel is that these all fit the 4n+2 formula.0696

Sometimes, Hückel's rule is known as the 4n+2 rule: and this is where n equals any integer (0, 1, 2, 3, etc.--any integer).0716

So, rather than memorize this list of numbers, all we have to remember is 4n+2, and then we can see that, if n is 0, then that would come out 2: 2 is one of these magic numbers.0729

If n was 1, we would get 4+2: 6 is a magic number; 8+2, 10; and so on.0740

This is a way that we can look at a very large number and quickly determine whether or not it fits this formula.0745

Let's take a look at benzene: we said benzene is an example of an aromatic compound--let's make sure it satisfies all the rules.0753

OK, it is definitely cyclic and planar--we saw that with our model.0759

They are all sp2 hybridized, so we have a p orbital on every single atom: that is good.0765

And then, let's check the number of electrons: how many electrons does benzene have in those p orbitals?0770

We are trying to see how many electrons are delocalized in that π system.0776

Well, we have 3 π bonds; each has 2 electrons; 2, 4, 6: we have 6 electrons.0781

And is 6 one of our magic numbers?--yes; that fits 4n+2; if you know benzene is aromatic, then that number 6 will always be something that you automatically know is a good number that fits Hückel's Rule, that fits the 4n+2 rule; yes, benzene is an aromatic compound.0787

Let's take a look at some other compounds and see whether or not they are aromatic.0804

OK, pyridine is an example of an aromatic compound, and it looks just like benzene, except we have a nitrogen in the place of one of the carbon atoms.0809

OK, so yes, I see that it's cyclic and planar; OK, every atom is sp2 hybridized--that is good; it has a p orbital on every atom, so that is good; but when I look at the electrons, I see 2, 4, 6, 8; so I have 8 electrons.0820

Is that one of our magic numbers--does 8 ever fit 4n+2?--no, 8 fits the formula 4n: anything that is a multiple of 4 can't be something that would be aromatic.0843

OK, yet pyridine is an aromatic compound: so we are making a mistake somewhere.0860

The mistake that we are making is that we are not counting up all of the electrons that we see: we are counting up only the electrons that are in the p orbitals, that are part of this system.0864

This lone pair is not in a p orbital, because this nitrogen already has a π bond occupying that p orbital.0877

This lone pair is in the sp2 plane: it is in an sp2 hybrid orbital.0885

So, if we imagine this as a model of pyridine instead of benzene, this would be a nitrogen; and instead of having a hydrogen here in the plane, it would be a lone pair of electrons.0892

And so, we wouldn't include these two electrons as part of these electrons, because they are in different regions of space.0903

We ignore these in our electron count; OK, and let's make a sketch of pyridine, so we can have a record of that.0912

OK, what I am saying here is that there is a hydrogen on every carbon that is in the plane, and there is a lone pair on this nitrogen that is in the plane.0926

In addition, we have a p orbital on every atom, as we need to, and we have 2, 4, 6 electrons in those p orbitals: this is just one of the Lewis structures we could have for pyridine, because we know we can have resonance showing another one.0940

OK, so these two electrons are not in the p orbitals, so we don't count it--we don't include that.0958

We don't count them; we don't include that in our electron count.0974

Hückel's Rule is going to involve making sure we know which electrons we are including to see if we fit our 4n+2 rule.0976

How about furan?--now, again, we have 2, 4, 6, 8 electrons that are possible that we see, but we need to figure out whether or not we should count all of those electrons.0987

The key to understanding furan is considering "what is the hybridization of that oxygen?"1000

It could be sp3: normally, we would say that an oxygen...2 bonds, 2 lone pairs, 4 regions of electron density...we would say it's going to be tetrahedral; it's going to be sp3 hybridized.1012

But remember, any time a lone pair is allylic--any time it is next to a π bond...and it can have resonance, like--we can show this...that means that lone pair must be in a p orbital.1024

It is not going to be sp3 hybridized: it is going to be sp2 hybridized: one of these lone pairs is in the p orbital, and the other lone pair is in the sp2 hybrid orbital.1037

So, how many electrons...how many lone pairs do we count for our Hückel's rule?--we only count 2 of the 4, so we have 2, 4, 6: we end up with our magic 6 electrons for Hückel's Rule.1055

We also needed to be sp2 hybridized for this oxygen, because we had to have a p orbital--that is the only way we could have a p orbital on every single one of these atoms.1067

So again, let's take a look at our sketch: if it was sp3, let's compare these two.1075

If it was sp3, that means we have a π bond here and a π bond here, and then this oxygen would be tetrahedral, right?1083

So, we have two bonds in the plane, and then two other regions of electron density are a wedge and dash above and below; we could just kind of put our electrons out there to show, spatially, where they are.1091

OK, this is unfavorable, because there is no interaction of the lone pairs with the π bonds--there is no delocalization.1103

And remember, electrons like to spread out as much as they can--delocalizing is a good thing.1119

So instead of being sp3 hybridized, it is going to prefer to be sp2 hybridized; and if it is sp2 hybridized, then that puts a p orbital on the oxygen; so now, we can become aromatic, because we are cyclic and planar and we have contiguous p orbitals.1124

We are going to have a π bond here and a π bond here, and we are going to put our lone pair of electrons in that fifth p orbital.1145

So now, we have 2, 4, 6 electrons delocalized in the π system: perfect--aromatic.1154

Where is the second set of electrons?--just like in pyridine, it is going to be part of the plane of the furan molecule, and so we don't count these, we don't count these, and we do count these.1160

That is what furan looks like--an example of an aromatic compound.1178

Related to furan: if we had a sulfur in this 5-membered ring, that molecule is called thiophene; this is also an aromatic molecule.1183

In order to do that, again, we would have to make this sp2 hybridized, so this is in the sp2 plane (or one of them--it doesn't matter which: one is in the p orbital, and one is in the sp2)--so analogous to furan...the sketch would look identical.1193

This has 6 π electrons now: 2, 4, 6--6 electrons in the p orbitals.1212

Pyrrole is another 5-membered ring that is an aromatic compound: do you think we should include those two electrons in our electron count?1220

I think, yes, we are going to make this sp2 hybridized once again; so this hydrogen is in the plane with all of the other hydrogens.1230

And then, what do we have in the p orbitals?--we have this lone pair and this π bond and this π bond--2, 4, 6 electrons: pyrrole is also an aromatic compound.1240

These ones with heteroatoms in there (something other than carbon)--we call these heteroaromatic compounds.1252

Thiophene, pyrrole, pyridine, furan--those are all examples of heteroaromatic compounds, and those are all names that you should be familiar with.1259

An aromatic compound can be a charged species: let's take a look at this one.1271

We have a carbon with a negative charge: now, what does it mean to have a carbon with a negative charge?1276

Let's complete this Lewis structure, so that we have a good understanding--if we are going to be counting up electrons, we need to know how many electrons we have.1280

So, in order for a carbon to have a negative charge, that means it has 5 electrons for its electron count.1288

It is going to have 2 bonds; it is going to have a third bond, and a lone pair of electrons; that is what a carbanion looks like.1294

It needs 3 bonds and a lone pair; so we have 1, 2, 3, 4, 5: carbon only wants 4; so that is how we end up with a carbanion.1303

OK, so when I draw out that structure, this now looks an awful lot like the pyrrole we just saw, right?--with the three bonds and a lone pair here.1313

We would expect this carbon, just like pyrrole, to have an sp2 hybridization here; it has to be sp2 hybridized if we want to delocalize that charge and move the negative charge around (which we can do here).1325

And, actually, we can continue doing resonance, and we can move it around to another position.1343

And we can keep going, etc.: in fact, as a matter of fact, this negative charge can be delocalized over all five atoms of the 5-membered ring.1351

So, is this an aromatic compound?1366

It's cyclic; it's planar; we have a p orbital on every atom; and we have 2, 4, 6 electrons--it is an aromatic compound.1368

An abbreviation for this molecule that is very common to see is to draw that circle in the middle and a negative charge, to say that, over these 5 rings, we have alternating π bonds--as many alternating π bonds as we can have.1378

How many can you fit in a 5-membered ring?--you can only fit 2 alternating π bonds, conjugated π bonds; and overall, the molecule has a negative charge.1394

This is a very stable anion, because it's aromatic; and it doesn't just have resonance--it has aromatic resonance.1403

This is called the cyclopentadienyl anion.1411

The cyclopentadienyl anion is a fairly easy anion to make; even though it is a carbanion, this is a very, very stable negative charge.1419

Let's take a look at a carbocation example: again, what does carbon look like--what are the details of that carbon?1429

If it has a positive charge...if it's a carbocation...how many bonds...how many lone pairs...1435

Now we need to have an electron count of 3: we need to be missing an electron; so all we have here is a single bond, so it must be a hydrogen there; so now we have 1, 2, 3 bonds.1440

Carbon wants 4, so it is missing 1; that is what a carbocation looks like.1452

A carbocation, remember, is sp2 hybridized: it only has 3 regions of electron density, so we have seen that before as being sp2 hybridized.1456

It has that p orbital--what is in the p orbital?1465

There is nothing in the p orbital: it has an empty p orbital, but that is fine, because all we need is to have a p orbital: so we now have a ring; it's planar; it has a p orbital on every single atom; and how many electrons does it have in that π system?1469

Let's count them up: 2, 4, 6--6 electrons: that looks like it is going to be extremely stable; it is, in fact, aromatic.1488

As you can imagine, because this is an allylic carbocation, we can move this positive charge around (sorry, I put my carbon on the wrong spot).1497

7-membered rings always look a little strange: now, it's still allylic; I'll just say "etc."--in fact, in this case as well, you can relocate that positive charge onto all 7 carbons.1513

The way that we draw this charged species is: we draw a 7-membered ring with a circle in the middle and a + charge.1524

Again, a pretty famous carbocation--pretty commonly known, because it is aromatic: and what is this one called?1533

It is a seven-membered ring, so that is cycloheptane, but it has 1, 2, 3 π bonds, so this is called the cycloheptatrienal cation.1541

The cycloheptatrienal cation is also an extremely stable carbocation.1558

We can have charged aromatics, as well; and this is an example--this is called naphthalene; this molecule has two benzene rings fused together.1565

This is described as a fused aromatic system; and if benzene is aromatic, then when you stick two benzene rings together, you will still have an aromatic system.1574

Let's check to make sure it follows all of our aromaticity rules.1589

We have a cyclic system: here is our ring on the outside here.1593

We have a p orbital on every atom, π bond to every atom, so there is a p orbital there.1597

They are all sp2 hybridized; and how many electrons do we have total?--2, 4, 6, 8, 10--we have 10 electrons, so this is our first one where it is not a 6-electron system.1601

But is this still one of our magic numbers?--well, this is where it comes in handy to remember 4n+2.1613

It's the 4n+2 rule; so is there any number for n that you can come up with (it doesn't matter what n is, but)--is there any way to make this formula come out to 10?1620

Sure: if you make n 2, then this is 8+2, and this would be 10: so yes, it is one of the magic numbers that satisfy Hückel's Rule; so naphthaline and any fused aromatic system (fused benzene ring system) is going to be aromatic, as well.1629

Let's test it out on a few examples here: let's see if we can figure out whether these are aromatic or not.1650

Here we have a cyclic system: we have 2 electrons; is 2 electrons one of our magic numbers?1658

Well, let's come back again to 4n+2; is there any way you can make that formula equal 2?1667

If n is 0 (which is an allowed integer), then that number disappears, and it becomes 2; so yes, 2 is an OK number.1675

But is it aromatic?--it's cyclic; the three carbons are in a plane (they have to be); do we have a p orbital on every atom, though?--p orbital--yes, p orbital--no.1685

What is the hybridization of this carbon?1696

Let's draw out our line drawing: this is actually a CH2 group; this is sp3 hybridized.1700

There is nothing you can do about that: you can't make it sp2; you can't put a σ bond up in a p orbital; a lone pair you can put in a p orbital, but if you have four σ bonds, you have no choice; it must be sp3 hybridized.1707

So, this is not aromatic--because every single one of the rules has to be satisfied in order to be an aromatic compound.1720

How about this one?--we have a positive charge now on that carbon: does that change anything?1732

It absolutely does: now, we just have three bonds, and so this is now sp2 hybridized; that gives us a p orbital on this carbon.1737

Now, we have a p orbital, p orbital, p orbital; and those two electrons can now delocalize.1746

Those two electrons are the right number; and yes, it is aromatic.1754

OK, what you need is: you have to have a p orbital on every single atom to provide the avenue through which your electrons can delocalize.1759

You have to complete that circuit in order to be aromatic.1767

So actually, this is a pretty famous carbocation--we can abbreviate it like this: this is known as the cyclopropenyl cation, and we draw that circle.1773

Remember, that circle means that you have alternating π bonds; but for a 3-membered ring, you can only fit in one π bond.1789

So, there is one π bond, and there is a positive charge on the carbon that doesn't have a π bond.1794

Cyclopropenyl cation is aromatic.1799

How about this next one?1803

It's planar; we have p orbitals; we have 2 electrons; everything is good...but what is the problem?1806

It is not cyclic, so no: it is not aromatic.1813

It is resonance-stabilized: it is the allyl carbocation--it's a very stable carbocation; it is a good carbocation, but it doesn't have that extra-special resonance stabilization that we define as being aromatic.1822

You have to have those atoms all in a ring in order to be an aromatic molecule.1834

OK, let's try this one: this is cyclooctatetraene: it has eight carbons and four double bonds.1841

Cyclooctatetraene is cyclic, and it could be planar (just like a stop sign, you can put that in a plane); it has a p orbital on every atom, certainly; how many electrons are there?1853

2, 4, 6, 8: we have 8 electrons: is that 4n+2--is that one of our magic numbers?--it is not.1868

It is 4n, and actually, if you have all of the ingredients of aromaticity (cyclic and p orbitals), and you put in the wrong number of electrons (you put in 4n instead of 4n+2), you actually get a system described as antiaromatic.1877

That is especially unstable, because you are trying to force the wrong number of electrons into this delocalized system.1897

So, what happens is: rather than be antiaromatic, it is going to bend; so it is not going to want to be planar.1905

This molecule actually bends so that the π bonds cannot interact with each other; each of these π bonds is not in the same plane as the other π bond, so there is no interaction--there is no resonance whatsoever.1914

This is a more stable form than trying to delocalize the electrons but putting them in something that violates Hückel's Rule.1930

This is a kind of a boat-shaped molecule like this, instead.1941

OK, so we have pointed out how important this 4n+2 rule is; but what makes that so special--why is it that a molecule...that system with 6 electrons is fantastic, but 8 electrons...it's horrible?1948

What is the background for that?--and I would like to just go into a little bit of molecular orbital theory (MO theory) to explain where this aromaticity stability comes from.1960

OK, so let's review what a π bond looks like.1974

You form a π bond (a π bond is described as a molecular orbital, an MO) by overlapping p orbitals (and p orbitals are atomic orbitals).1977

OK, you bring those two p orbitals together to form a π bond; but when they combine, when these p orbitals combine, there are two ways that they can combine (we are actually combining them mathematically).1988

They can either combine in this way, or we can flip it over and combine it that way; so it's like you are adding them or subtracting them.1999

If we describe this as Pa and Pb, let's say they have the same phase (and again, we are using some shading here to indicate the mathematical signs--so let's assume the signs are the same on the top lobes and the same on the bottom lobes)...so if you add these two together, Pa+Pb, you are going to have a positive interaction with the top half and a positive interaction with the bottom half.2007

Your new molecular orbital is going to look like this, where you have a large lobe on the bottom and a large lobe on the top; we describe this as a bonding molecular orbital.2029

We call this bonding molecular orbital a π molecular orbital, and that is exactly what a π bond looks like, right?2045

We have a cloud of electron density above and below.2051

OK, but that is not the only possible combination: if you subtract these two so that the phases don't match up (we describe them as being out of phase), we get this as our molecular orbital: this is described as an antibonding.2055

We call it a π*; the asterisk indicates that it is an antibonding type of orbital, and we are drawing it like this--we are drawing this one higher--because this is higher in energy.2072

What we can see in this picture of the molecular orbital is that we have nodes, and we have a node here, too: any time you change the mathematical sign from one to the other, we describe that as passing through a node.2089

A node is where you have a 0 probability of finding an electron: so remember, these molecular orbitals define an area of probability of where the electrons are most likely to be.2103

So, if you have a π electron, it is going to most likely be in a cloud above the atoms or a cloud below the atoms, but has a 0 probability of being right between the two nuclei.2116

OK, and the antibonding has this node, but it has another node; so that has even more, and what we see is: as you increase the number of nodes, you increase your energy: it becomes higher in energy; it becomes less stable.2130

We call this antibonding, because there is no positive interaction between these: there is nothing: if you had electrons in here, that is not something that would bring these nuclei together and hold them together in a bond.2146

This is the difference between bonding and antibonding.2158

And so, now, let's look at the electrons: if we think of the p orbital of one carbon atom, let's say, coming with one electron, and the other p orbital has an electron; we bring those together to form a bond with two electrons.2163

Where are those two electrons going to go?--the new possible homes that they have (the new molecular orbitals) are the π and the π*; well, they are going to pair up, and as usual, they are going to fill in the lowest energy orbital, the most stable orbital, first.2177

So, we put them down here, and that is what our π bond looks like.2193

We have two electrons fill the more stable molecular orbital, and that is what we describe as a π bond.2199

We have two electrons in a π molecular orbital.2207

OK, so that is the simplest system, when we have just a single π bond; now, let's take a look at benzene.2211

Benzene doesn't have just a single π bond; it has three π bonds, involving a total of 6 p orbitals.2219

And, because these are all interacting and involved in resonance and delocalizing the electrons, all 6 of those p orbitals are going to be combined to form new molecular orbitals.2228

And so, if you have 6 p orbitals, you are going to get 6 new molecular orbitals.2240

We are going to observe that same shift: that, once you do your combination, half of them are going to be favorable interactions; the other half are going to be unfavorable.2246

These 6 new molecular orbitals will have three that are bonding orbitals (so we call them π) and three that are antibonding, that will describe as π*.2257

What I have shown here is an energy diagram: this is called an MO diagram for benzene.2276

We have 3 lower-energy orbitals called π, so down here, we have all of our bonding orbitals--low energy, stable.2282

And then, up here, we have 3 antibonding--all high energy; and this line in the middle--we call this the non-bonding line.2293

Sometimes we have orbitals right at this level: this is kind of halfway between the energies, and in these orbitals, we have an equal number of bonding interactions and nonbonding interactions.2306

OK, so what we haven't shown here is what each of these π molecular orbitals looks like.2320

OK, well, let's think about this lowest-energy one--the most stable one.2327

Let's imagine shading each of these p orbitals mathematically; what would be the best possible interaction that you could have between those p orbitals?2333

Well, clearly, if they were all in phase (so in other words, let's draw that right here: if we had our 6 p orbitals, and for every one of them the top half was shaded in), that would be the most stable; it has just the one node of the plane--it has no additional nodes.2344

This is going to be most stable we can have: that is going to be the lowest energy, and that is exactly what it is.2366

All of the interactions between adjacent p orbitals are bonding interactions, favorable interactions.2372

OK, that is one possible combination; another combination would be if you now introduce a node, where half are one sign and half are the other sign.2380

That is going to be higher in energy, and so that is what these look like (I am not going to draw them all).2391

Then we have a couple of nodes, and we are even higher in energy still; what do you think this worst one looks like?2396

What would be the worst possible interaction between these p orbitals, so that there are no bonding interactions whatsoever--it's completely antibonding?2401

Well, again, that extreme is pretty easy to imagine: what if we had alternating signs (we shade in the top half and then the bottom half, and then the top half and the bottom half)?2412

If we did this, then there is not a single bonding interaction between any of them: it is completely antibonding, and we have a node and a node and a node; we have tons of nodes; and so, that is the highest energy.2426

Something that would be on this nonbonding line would have some bonding interactions, and just as many antibonding interactions; so it kind of cancels out.2441

It is neither bonding nor antibonding; it is somewhere in between.2450

OK, now this is benzene: how many electrons does benzene have in those p orbitals?2455

How many electrons do we have to fill into this diagram?2461

We have 2, 4, 6 electrons, so these are the molecular orbitals into which we can fill with the electrons and place the electrons.2465

Where are we going to put them?--well, as usual, we will start with our lowest energy orbital and fill it in: there are 2 electrons; and then we move up to the next higher energy; we can put 2 electrons here, and we can put 2 electrons here.2476

What do we have?--we have used our 2, 4, 6 electrons, and what we end up with is an MO diagram that looks extremely stable--looks really, really good.2490

What we have here with these 6 electrons is: we have completely filled bonding molecular orbitals.2501

All of our π molecular orbitals are completely filled; we have no electrons up here in the antibonding, and we have all paired up electrons down here.2510

This is going to happen every time we have 4n+2 electrons, and that is what is so magical about that number, and so stable to satisfy Hückel's Rule.2518

This is what benzene looks like: let's take a look at some other sized rings and see if we can come up with their MO diagrams.2531

And what I want to point out is: they are going to have some things in common with this.2538

OK, there will always be one MO with the lowest energy.2542

OK, there is always just one instance, one arrangement, where they are all in phase; that is going to be the lowest point in your MO diagram.2563

You will always have one down there; OK, after that point, your MO's are always going to occur in pairs.2573

OK, so then, we have pairs of degenerate (we call it degenerate when they have equal energy) MO's.2582

We start with one lowest, and then we go in pairs, and then we go in pairs; and how many will we have total?2598

Well, however many atomic orbitals went in is how many molecular orbitals come out.2605

OK, so let's take a look at some examples: and actually, we can point this out in benzene: there is something called a polygon rule, or something that helps you see what the MO diagram should look like.2612

If you look at this MO diagram, and you connect your MO's, you end up drawing the shape of the molecule; so benzene is a 6-membered ring: you can draw this polygon on its point; you can see the hexagon in there.2630

OK, and that is because, if it is a 6-membered ring and every atom has a p orbital, then you have that same number of p orbitals, and you will have that same number of MO's.2648

You will always have the same number of points here that you had in your original system.2658

OK, so let's see if we can try some new ones.2664

OK, how about a 3-membered ring?2668

I pointed out in the beginning here: the number of the MO's is going to be equal to the number of atomic orbitals, which is always equal to the number of atoms in the ring; so we are going to look for that polygon to give us the shape.2671

A 3-membered ring (like this--it's an example here) should have 3 MO's.2685

Now, I have drawn the nonbonding line here, because it's kind of tricky to know what to do with three of them: it turns out that we have one that is low energy, that is bonding (so we label that as a π), and then we have two that are antibonding.2692

That is what a 3-membered ring looks like; we are increasing in energy as we go here.2706

We have one π and two π*s; and let's take a look at this example--let's fill in the MO diagram for this molecule.2712

How many electrons does this have in the p orbitals?2720

It has just 2 electrons, so we put those down in the lowest energy; and then, we are done filling in our electrons.2724

Now, looking at this MO diagram, would you say this looks like an aromatic system?2730

It has just one bonding orbital that is full; therefore, it is aromatic (because, just like benzene, we have filled bonding molecular orbitals--so that is why 2 is a magic number in cases like a 3-membered ring, because all you need is 2 to fill in all of the bonding orbitals).2738

OK, how about a 4-membered ring?--this is: we have 4 atoms, so there must be 4 MO's; OK, and we have...these should not be labeled π; these are nonbonding...2765

I'm sorry: we have a π and a π*, and then these are nonbonding.2782

And now, we have: 2 of them are going to be right in the middle in the nonbonding line; one is going to be lowest energy; one is going to be highest energy.2786

I forgot to put it on our polygon here: you can see our 3-membered ring; here you can see our 4-membered ring, right?2793

We start with one, and then we draw a pair of degenerate, and then we go back to just one, because we only had 4 total.2801

So, 1, 2, 3, 4: now let's fill in our electrons--we have 4 electrons to add in, so where would those 4 electrons go?2809

We would put 2 down here in our lowest; and then we have 2 more electrons, so how do we fill those in?--well, we fill one in each degenerate electron.2818

Remember, we don't pair them up until they are spread out, one in each orbital of the same energy.2828

And so, does this look like an aromatic system--does this look like a very stable system?2834

No, it has unpaired electrons: this is not aromatic; this is not stable.2839

OK, so we knew that because it was 4 electrons; that does not fit 4n+2, so we can predict it--just looking at the molecule and going through our aromaticity rules.2845

But what I want to point out is: by inspecting the MO diagram is how we can prove why this is an unstable system, and verify our predictions.2859

OK, let's see if we can draw some others: how about a 5-membered ring?2873

If we have a 5-membered ring, how do we draw that?--we start with one in the lowest energy.2877

There will always be a π molecular orbital at the lowest energy; that is where all of the p orbitals are in the same phase.2883

And then, we go in pairs: we have 2, and then we have another 2, and how many more?--well, it is a 5-membered ring, so we are going to get 5 MO's, so we are done.2891

We have 1, 2, 3, 4, 5; and with a 5-membered ring, it's a little tricky to know where the nonbonding line is, but it turns out it's right above this.2905

These three are π molecular orbitals, and these two are π*; it's not completely symmetrical, so you have to know where to put that line.2914

And so, let's take a look at furan: furan, we said, was an aromatic molecule.2925

Let's see how its MO diagram looks: how many electrons should I put in here?2931

I have drawn my polygon; I have my 5 MO's; so this is a good MO diagram for furan, a 5-membered ring.2936

How many electrons?--2, 4, 6: this has 6 electrons: remember, we are just counting one of these, because they are in the p orbital, and the other pair is in an sp2 orbital, so we don't count those.2944

We have 2, 4, 6 electrons, and where do we put them?--2, 4, 6; so just like benzene, it has 3 bonding molecular orbitals; so 6 is something that imparts aromatic stabilization.2957

OK, and finally, how about an 8-membered ring?2974

An 8-membered ring: we start at the bottom with one, and then we go in pairs; there is 3, 4, 5, 6, 7, 8; so there is our 8 total.2975

Because it is a symmetrical number of things, we can draw a nonbonding line right in the middle; so these are clearly going to be π's; these are going to be our bonding electrons.2994

These are going to be π*, nonbonding electrons.3005

And here is an example of an 8-membered system: we have 2, 4, 6, 8, 10 electrons; is that a magic number?3012

4n+2 is a magic number; so we are expecting this to be a good structure, so let's fill it in.3029

We have 2, 4, 6; and then 8; and then 10; we have filled in all 10 electrons; they are in bonding molecular orbitals; we dump any electrons in the antibonding.3037

We don't have any electrons that are unpaired, and so this is another stable system.3051

OK, so the MO theory is there to back up what we are predicting with our simple aromaticity rules.3055

And finally, let's look at some examples where we can use aromaticity--now that we know that aromatic compounds exist, that can explain some things we have about reactivity.3065

What we are going to do is: we are going to look for aromaticity as kind of the ultimate in resonance stabilization.3076

We have seen so many examples where resonance was key to explaining why something is so stable and why something reacts the way it does.3080

And so, resonance, now, is just another big example of that.3088

Here is a question: we compare these two molecules, and we ask, "Which is more acidic?"3091

How do you ever answer that question--what does it mean to be an acid?3097

It means you donate a proton: so we are asking which of these is more willing to give up a proton.3102

The way you answer that question is: you let each of them donate a proton; you let each of them act as an acid; and you look at the conjugate base.3108

You look at the conjugate base: in other words, let's imagine a base coming in here and grabbing a proton from each of these; what does the base look like--what does the conjugate base look like?3120

You now have a lone pair and a negative charge, where you used to have that proton; so we have this anion; here we have a 7-membered ring (I don't know if I drew a 7-membered ring there); 2, 4, 6 π bonds...so we have a 7-membered ring.3129

We have this anion and this anion; so we compare our conjugate bases, and then we ask which one is more stable.3149

Well, each of them is allylic; this has resonance, so I can delocalize my negative charge down here; and so on.3157

OK, and this has resonance, where I can move this around (I think I moved it a little extra); I can move this negative charge around and continue; etc.3169

So, based on what we have seen in the past for resonance, what would be your prediction?3187

Which one of these do you think has more resonance?3194

Well, it looks...because this has just the two π bonds, it looks like this is going to have less resonance than the one with three π bonds, because we are going to be able to use all of these π bonds and move it around.3197

OK, so at first glance, we are wondering: does this have more resonance?--is this more stable?3207

Well, that would be a possible conclusion before you knew about aromatic resonance; now that we know about aromaticity, as soon as you see this cyclic system with π bonds and p orbitals, you need to consider aromatic requirements.3215

OK, so let's look at this in terms of aromaticity.3233

How would you describe this first conjugate base--is it aromatic?--because it is cyclic and planar; it does have the p orbitals; it has 2, 4, 6 electrons; this is aromatic.3236

How about this one?--it's cyclic; it's planar; it has 2, 4, 6, 8 electrons (is that 4n+2?--that is not); this is antiaromatic.3253

This resonance is not true: it is not possible for those electrons to delocalize throughout all of those π bonds, because in order to do so, it would force it to be planar, and then we would have one of those unfortunate-looking MO diagrams that we had on the previous slide.3270

OK, so what this molecule does, then, is: it bends so that it disrupts that resonance, and so it does not have all the resonance stabilization that we are expecting.3286

This is antiaromatic.3296

OK, if you didn't know about aromaticity, you would have gotten this problem wrong.3298

So, because this is aromatic, this is the more stable, and therefore less reactive, conjugate base.3302

Less reactive..."weaker"--we can add that in, too: more stable, less reactive, weaker conjugate base.3320

And the weaker conjugate base has the stronger parent acid.3328

Which of these two polyenes is the stronger acid?--this is more acidic, OK?--because aromaticity is the best resonance we can have.3338

So, if we can find that, we know that that beats any other kind of resonance we might have.3357

OK, let's try another example: let's look at these two amines (these two nitrogen-containing compounds, I should say), and ask which is more basic.3364

This is pyrrole; this is pyridine; we have seen both of these; they are both aromatic compounds, heteroaromatic compounds.3375

Let's ask which is more basic...what does it mean to be a base?3381

You accept a proton, right?--so each of these nitrogens can be a base, because they have a lone pair; so we can imagine having it interact with an acid and get protonated.3386

And so, what we need to do now is: we need to look at the conjugate acid.3399

We know each of these starting materials--these are both aromatic.3403

We know we are starting with stable starting materials--they are both aromatic--so let's look at the conjugate acids and see what they look like.3410

In other words, we are going to protonate pyrrole; that gives a nitrogen with four bonds, so it's positively charged now: 1, 2, 3, 4.3418

Nitrogen wants 5, so it's missing an electron.3429

There is a protonated pyrrole, and here is a protonated pyridine.3433

OK, we look at these two conjugate acids, and we ask the same question: Which of these is more stable?3444

Well, pyridine looks like I still have my benzene ring, right?--I still...not my benzene ring; I still have my aromatic ring.3454

It looks like benzene, though: it still has the 6-membered ring; it still has the π bonds; this conjugate acid is still aromatic.3460

How about the protonate pyrrole?3470

We now have...this is now sp3 hybridized, so we have lost our p orbital; so that makes it impossible to be aromatic.3473

We only have 4 electrons, so even if it were still there, we wouldn't have the right number of electrons.3481

And so, this has lost aromaticity.3485

Does that sound like a good process--does that sound like a good reaction?--no way, because you are so stable, and if you give up that aromaticity, you lose that stabilization; you go higher in energy.3490

OK, so because this is still aromatic, this is the more stable, weaker conjugate acid, so it has the stronger parent base.3504

Because this has lost aromaticity, this is the less stable, and therefore stronger, conjugate acid, and that means it has the weaker parent base.3525

What you could say is: you could say that pyrrole's lone pair--why is there a difference here?--because pyrrole's lone pair is part of the aromatic delocalized electrons.3546

Those two electrons are needed to have that delocalization, that resonance stabilization.3562

So, you could say pyrrole's lone pair is tied up in aromatic resonance.3567

OK, and by using it--by reacting it with an acid--you are taking it away from an aromatic system, and you are losing that resonance stabilization.3576

OK, how is that different from pyridine?3585

This lone pair is not part of the aromatic system, so that makes it more available.3588

It is happy to react; it can form a bond with no problem; it doesn't disrupt the aromatic system at all.3599

Remember our sketch of pyridine--we looked at pyridine today: imagine one of these being a lone pair.3605

Forming a bond here has no impact on these 6 electrons that were in the π bonds, that are part of the π system; so that makes pyridine a pretty good base.3611

In fact, where have we seen pyridine before?--we have heard of pyridine before.3619

We have used it as a base: this is a good base--it is a good organic base that is used when we need such a thing, and so this explains why it is a good base, even though it's an aromatic and stable compound.3624

This wraps it up for the discussion about what it means to be an aromatic compound like benzene.3640

What we will move into next: the next lesson is going to start looking at the reactions that aromatic compounds can undergo.3648

I look forward to seeing you then. Thanks for visiting Educator.com.3655