# Freshman Organic Chemistry II

## CHEM 125b - Lecture 18 - Linear and Cyclic Conjugation Theory; 4n+2 Aromaticity

**Chapter 1. Why is Diene Stabilization Small? Orbital Mixing and the Semicircle Mnemonic [00:00:00]**

**Professor Michael McBride:** OK, so last time we started talking about conjugation, the special properties that functional groups have when they’re adjacent to one another, in particular dienes. We talked last time about the allyl system.

The question is, when does conjugation make a difference? That is, when is it simple enough and accurate enough to think in terms of isolated functional groups with pairwise atom interactions?

So we started last time with giving experimental evidence that allyl cation, anion, and radical are all stabilized by something in the order 13 kilocalories per mole. And in fact the transition states, when they’re allylic, are also stabilized.

And then right at the end last time you guessed for me, and did the right thing, to show that the diene at the bottom is special because it’s conjugated, as opposed to the other dienes. So the heat that’s involved in removing the double bonds is less, because it started out more stable.

But here conjugation is worth only about four kilocalories per mole evidently, compared to 13 kilocalories in allylic situations. Another way of measuring it is to look at how difficult it is to rotate about the single bond between the two double bonds in butadiene.

So for confirmation we can have syn, sort of like eclipsed, and anti. And in between those there’s a twisted form where the two ends are perpendicular to one another, and of course there’s then no overlap across the middle. The two π systems on the ends are orthogonal. So you might expect that to the extent that conjugation is worth something, the interaction of these π things, you’d lose it in this twisted form, because they don’t overlap.

So you might expect that the energy curve would look like that as we twist it. In fact it’s more or less like that, but there’s another factor, which is up at the top left there you can see there’s going to be repulsion between the groups when they’re syn, so that’s going to raise the left end of this plot. So that’s what it actually looks like, as determined by Raman spectroscopy.

So the important thing is to look at the how big the barrier is. You can see the barrier is about six kilocalories per mole in this case, versus about 60 kilocalories per mole for twisting the interaction between two *p* orbitals across a normal double bond, an isolated double bond. So the one between the two double bonds is 10 times easier to rotate than the one that’s involved in a genuine double bond.

Even though, when you look at it, there are just four *p* orbitals in a row. It’s not immediately obvious that there’s anything special about the one in the middle that would make it easier to rotate, than the ones at the ends, which we know would be to be hard to rotate.

So why is conjugation worth more in allylic intermediates, on the order of 13 kilocalories per mole, than it is in , where it’s only a handful, three to six kilocalories per mole? Well, one would be tempted to say it’s because we can draw reasonable resonance structures in the case of an allylic intermediate. Those two at the top there are good. They’re equivalent to one another. The one at the bottom is bad, because we have to separate charge.

But that’s not really a physical theory, as we know; the resonance theory. It’s a very handy thing to remind us of what really happens, but it’s not what really happens. We have to look at how the orbitals interact with one another. So we’re going to now, having looked at some experimental evidence for the extent of stabilization due to conjugation, we’re going to look at the theory behind it.

And we’re going to work in terms of simple Hueckel MOs. The PowerPoint chooses that particular color to give links. If you link there, you can get a simple-minded program to run on your computer that’ll let you make simple Hueckel MOs.

So Erich Hueckel was a physicist, a theorist, and not long after Schroedinger developed quantum mechanics, he began to apply it to some organic-type molecules. In particular, the interaction of *p* orbitals with one another. His brother was an organic chemist, wrote a very influential text book of advanced organic chemistry, and no doubt influenced him some in this interest.

But it’s a very simple-minded theory, as you might have expected from being so soon after Schroedinger developed wave mechanics. But anyhow, these results are due to that, and I don’t think it’s appropriate to take the time to go into the limitations of the theory. But let’s just take it as qualitative guide to the way things work in quantum mechanics.

OK, so suppose we have four *p* orbitals in a row. There’s the energy of an isolated *p* orbital. Now there are two ways to look at butadiene. You can look at it as a whole, or you can look at it as two double bonds, and see how they interact with one another. Let’s compare those two ways of doing it.

So first let’s look at it as a whole. You take those four *p orbital*s in a row, and you combine them to make the best possible molecular orbital, ψ_{1}, then the next best, then the next best, them the next best. So that’s what they look like. And they have what you would anticipate in terms of nodal structure. The lowest one has no nodes, except the node of the *p* orbital. The next one has one node, two nodes, three nodes.

But there’s a little bit more to it than that. You notice that the best one is big in the middle. Why are the *p* orbitals big in the middle rather than all of them being the same size to get the best overall overlap? Elisa, what do you say?

**Student:** The middle ones have two overlaps.

**Professor Michael McBride:** Right, the ones in the middle have two overlaps. And in fact, you can write a simple equation that will maximize the product of adjacent orbitals, that is the overlap, subject to the sum of their squares being one, and you get that shape. Right?

So to maximize the bonding-orbital overlap, the central atomic orbitals are large in ψ_{1} and for the same reason they’re small in ψ_{2}. Why should they be small?

**Student:** You don’t have much opposing overlap.

**Professor Michael McBride:** Right, because the overlap in the middle is hurting you now, so make it small. Get the overlap on the ends. OK. Notice incidentally, that three and four turn out; they’re not as good as they could possibly be, but as bad as they could possibly be. Here the one that’s in the middle where it’s favorable is small. Here the one where it’s unfavorable is large.

That turns out to be the flip side of – Do people know what flip side is? It occurs to me that that’s something that people might not know anymore. When you had vinyl records, you’d buy a song that would be, and there would be some other song on the back that wasn’t very good, right? So you’d play the side of the record that was the real one, then you have the flip side, which wasn’t so great.

So the flip side of having these be as good as possible, is those have to be as bad as possible, the bonding and the antibonding ones. Ok anyhow, there’s a little cultural history for you.

So we have four *p* electrons, and they go into those two orbitals and get stabilized. And that’s the energy then, the π energy of the butadiene, four *p* orbitals in a row.

Now let’s look at the other way. Let’s see if we just look at localized π bonds on each end, how different it is. So we get two π orbitals, one on each end, two π*s, one on each end, and the four electrons go down into the two π orbitals.

And what we want to do is see how that compares with the delocalized orbitals on the left. How different are they in overall stability?

Well, let’s analyze it this way. We know that in fact we could mix these two with one another. This is π+π and π-π. So we’re getting things that look a little bit like this, but they’re not distorted to optimize the overlap. And if we– this goes down, and this goes up the same amount.

You know in truth the one that goes up goes up a little more than the one that goes down goes down. But that’s part of the Hueckel approximation, is that they go up and down the same amount. So the average is the same as where it started, but it’s a little worse than it was on the left when we optimized things.

Now you can optimize things on the right too, and see why it’s not very important to do this change in size. So here are the π* orbitals that we got just by adding the two ends. But now we can mix among these.

So look what happens if we mix this one, the two nodes, with the no nodes. We can make it so it reinforces in the middle and cancels on the end, and get a thing that looks like that. And that will make it a little bit lower in energy to do that, to mix this one with this one.

We started with just isolated p orbitals, and now we’re letting them interact with one another, to see how much difference it makes. But this secondary mixing – the first mixing was making these from here – the secondary mixing is minor, because the energy match is so bad. You don’t mix very much of this into this one to get that, or of this into this one to get that.

At the same time, this one can mix with this one, and that one will go down just a little bit. But notice that the amount by which they went down is very small, because of the bad energy match.

But then the ones on the far right are the same as the ones on the left, if we let these interact with one another in two stages. But that second stage, the one that made it big in the middle and small on the ends here, didn’t account for much energy. Because the energy match was so bad, of the things we were mixing. And the same here.

Now you might say, that it’s only about three kilocalories per mole, max. That’s only the barrier to rotation of ethane; it’s not very much. Now why did we ignore this mixing? Because there the energy match is pretty good. Can anybody see why we didn’t consider that mixing? There it might have been more important.

What’s the overlap of this with this? Notice that despite the better energy match, it doesn’t lower the energy, because these are orthogonal to one another. They won’t mix. What is gained at two of the positions, is lost at the other positions. That is, if you adjusted on the left to be big in the middle and small on the outside, on the right it’ll be just the opposite. So you won’t gain anything.

So you consider only this one mixing with that one, this one mixing with that one, and it doesn’t count for much. That’s why you don’t get much out of a diene.

So, although the total energies are nearly the same, with and without conjugation. That is, the total energy that we calculate here is almost the same as the total energy we calculate there. There’s just that few kilocalories difference.

But the energy of the HOMO and the energy of the LUMO are quite different here from what they are here. So if you’re interested in reactivity, then the conjugation makes a difference. If you’re interested actually, not the total energy, but the energy of the HOMO and the energy of the LUMO, and also if you’re interested in color. Because if you want to take one of these electrons with light and put it up into the antibonding orbital, it’s a much bigger jump here, when it’s not conjugated, than here, when it’s conjugated. So you get a very different color. And we’ll talk about that later. For example carotene, the stuff that makes carrots orange, is a bunch of double bonds in a row.

So we see that when we mix these, they go up and down. Is there a limit? If we put a whole bunch of these π orbitals in a row, as in carotene or something even longer than that? Do we just keep getting lower, and lower, and lower, and lower energies for the lowest energy?

So let’s consider a few of these and just think about it sort of approximately quantitatively. So there the chain length is two for ethylene. And the normalized atomic orbital size is 1 over the square root of 2. So when we square and sum, we get 1. So the overlap, when you have this one [gr[pi bond, is 1/2. It’s 1 over the square root of 2 here, 1 over the square root of 2 here. Multiply them together to see how much they overlap. And you get 1/2 in some units.

Notice we’re ignoring the smallish influence of overlap on normalization. Because remember when we take overlap into account, we have to make them a little bit smaller– less than one, greater than one, kind of thing. We’re ignoring that, just to get an idea of what’s going on.

OK, so the overlap is 1/2. There’s only one π bond in that case, so we get 1/2 times one. The total overlap stabilization is 1/2 in these sort of weird units. And notice that if you compare this with a textbook, you’ll see that they use a unit called beta which turns out to be half as big as this unit is. But anyhow, I think it’s easy to go through it this way.

Now suppose we have four in a row. Now the chain length is 4. The normalized size is 1 over the square root of 4. So that you could square and sum them up and get 1. And now the overlap between each pair of neighbors is 1 over the square root of 4 times 1 over the square root of 4, so it’s 1/4. And there are three such bond interactions, here, here, here. So there are 3 times 1/4, or 3/4.

So indeed it’s more stable, that lowest-energy orbital, the one that’s approximated like this. Now suppose we have eight in a row. Now the normalized atomic orbital size is 1 over the square root of 8. 1/8 is the overlap per bond, and there are seven bonds, so we get 7/8 as the amount of stabilization.

So it’s going down, down, down, down. And obviously each time we add a pair, it’s going to go down more and more. Is there a limit? Does it just keep going? Obviously it’ll get a little bit better, a little bit better, but does it approach a limit? Can you see what it would be if N were infinite? If the number of the chain lengths were infinite? Yeah.

**Student:** It approaches one.

**Professor Michael McBride:** It approaches one, right? Because if you have N, it’s 1 over the square root of N. And there, 1/N is the overlap, and there are N-1 of these pairs, right? So it’s (*N*-1)/N, which approaches one as N gets very large.

So there’s a limit to how good it can get. And incidentally, that limit is twice what it started out here, here, right? It was 1/2, and it approaches 1.

And incidentally we also get bad combinations, the anti-bond. But it’s exactly the opposite of this. So the worst antibonding combination is, the destabilization is twice that of π star. So these approach a limit. And that suggests, although it doesn’t prove, this construction that allows you to view what the energies of the orbital will be in this simple approximation for any length of chain.

So you draw a semicircle. The vertical axis is the energy. Zero is a p orbital. Here’s plus 1, minus 1. Those are the limits if you have a very long chain.

But it turns out there’s this geometrical analogy. That if what you do is draw this circle, whose radius is twice the stabilization of the two p orbitals in the π bond, and has that limit. And then you put points along this circle, along the circumference, evenly spaced along the circumference, it turns out that the height of the intersections, of those points, is the energy of the various orbitals you can have.

So if you have just one p orbital, you put it here, equidistant between top and bottom. And it has the energy 0. So that’s the energy of an isolated p orbital. That was definition, actually.

OK now we take ethylene, so we put two evenly spaced points along here, to divide it into three along the circumference. And now we can put into electrons in ethylene. And that’s the energy they’ll have in a π-orbital. It’s 1/2, halfway down here.

And now we take N equals 3. That’s the allyl system. Put three equally spaced dots. And there are the energies in the allyl radical. Two of the three p orbitals, of the electrons in the p orbitals, are in the lowest-energy molecular orbital, and one is up here.

Now, what’s the stabilization? Compared to what, right? Compared to what you had if you had just the π bond and a p orbital, a radical. How much better is it to have them next to one another and overlapping in an allyl system? Well you can see that if you had just the p orbital and a π bond, you’d have these energies. In the allyl, you have these energies.

So that’s the allylic stabilization. Notice the orbitals, the electrons that got stabilized were the ones in the lowest orbital. In the difference between having them separate, and having them together.

And what that means is, that it doesn’t make any difference whether you have the electron there or not. If that electron is not there, you have a cation. If the electron is there, you have a radical. If there are two electrons there, you have the allylic anion. But in each of those cases, the amount of stabilization you get is the same. So you get the same two-electron stabilization whether you have cation, radical, or anion.

And incidentally, when we look at this one, remember that’s just an isolated p orbital. But in the allyl case it’ll look like this. It’s the one with one node. Plus on one side, minus on the other. But that’s going to be generally true, however long the chain is.

If you have an odd number, odd chain, here we had N equals 3, but if you had N equals 5, you’d have blue, nothing, red, nothing, blue.

And notice that in this orbital, there’s no overlap between occupied atomic orbitals. They’re far apart. So there’s no interaction between them. Therefore their energy will be the same as an isolated p orbital, in this crude approximation. So every odd chain is going to have an orbital that has the same energy as an isolated p orbital, that’s going to have a node at every second carbon.

Now suppose you have N equals 4. Now you have these energies for the four electrons, compared to this that you’d have, if you had two π-orbitals. So the favorable orbital is favored, compared to the electrons in the π-orbitals when they were isolated. But this one is less stable. It’s the one that’s antibonding in the middle, remember?

But this one, remember, was big in the middle, and this one we made small in the middle. So this one went down more than this one went up, and the difference is the resonance stabilization. But it’s not as big. The difference between these two is not as big as how much those lowest two got stabilized in allyl.

So that’s why conjugation is worth more in these allylic intermediates, like 13 kilocalories per mole, than it is in a diene. Because in a diene some go up and some go down.

So as the conjugated chain lengthens, more and more levels get crowded between minus one –here’s the limit of the lowest one. There’s the highest one. And we’re putting dots along here. But as we put more and more, longer and longer chain, we’re going to crowd more and more energy levels between those two limits. So that means the HOMO-LUMO gap is going to decrease. So as we make it longer and longer, you’ll have this gap, and then this gap, this gap, this gap, this gap, this gap.

Why did I draw it like this? Why did I use those colors? Because as the energy gets smaller, it absorbs longer and longer wavelengths. So you can tune the color of a conjugated polyene– by how long you make the chain. And we’ll see examples of that too. It’s a good thing to talk about now, because very soon the spring migrant birds will be coming back to East Rock Park in these beautiful tropical colors they’ve acquired, and most of them have to do with that, as you’ll see.

**Chapter 2. Benzene, Hückel’s 4n+2 Rule, and the Circle Mnemonic [00:22:36]**

Now let’s look at heats of hydrogenation and conjugation. So here’s cyclohexane, and here’s cyclohexene. The heat of hydrogenation, to change this to that, is 28 kilocalories per mole. It gives off heat.

Now let’s take two double bonds, and they’re conjugated. That turns out to remove both of those and go down here is 54. That is, the second double bond was worth 26. Is everybody with me on that? So 26 isn’t as big as 28. That means this was a little bit lower than we would have expected. Why? Because it’s conjugated. It’s better than just two isolated double bonds, because they’re conjugated. The lowest orbital went down a little more than the next one went up. That’s what we just talked about. But that’s not very much to write home about. It says that conjugation is worth about two kilocalories per mole by this measure.

Now let’s put another double bond in the ring, and tell me what you think that’s going to be? What do you think the gap is going to be here? Any ideas? Let’s get a couple hypotheses out there. Cassie, what do you say?

**Student:** Maybe 24.

**Professor Michael McBride:** Why do you say 24? Why not 26?

**Student:** Because 26 was less than 28.

**Professor Michael McBride:** Yeah, it was a little bit less, maybe a little less again. Another reason it could be less again is this last double bond interacts on two ends, not just on one end. So it might get two of those interactions. So maybe 24, maybe 22, if it depended on how we’d calculate that. That would predict that we’d get about 78. The heat of hydrogenation of benzene to give cyclohexane should be about 78 altogether. You know what it is experimentally? 49.

So this conjugation in a ring, in benzene, is really worth a lot. Not just a little bit more because you get the same thing on two ends, but a lot more. It’s worth about 30 kilocalories per mole. Here we were talking about two kilocalories per mole. This is 30. So this is a big deal. And this is such a big deal, that it has a special name associated with it. It’s called aromaticity.

The name originally had to do with the odor. Benzene has a characteristic odor. And other things that have benzene in them, that are volatile, have a characteristic odor. They were called aromatic. But now the idea of aromaticity is much more associated with this special stability, when you have a bunch of double bonds in a ring.

So bringing the ends of a conjugated chain together to form a ring gives the lowest occupied orbital with another bond. We’ve got all these p orbitals in a row, remember, and we made it a little bit bigger in the middle, a little smaller on the end, to optimize the overlap. But if we bring them around, and touch the ends, for no additional charge we get an extra bond here. A whole new π bond. So that’s much more effective than just these little changes that are made by adjusting the sizes.

So the lowest molecular orbital now will have N of these overlaps, rather than N-1. Are you with me on that? When we had a chain, the number of pairs down the chain was one less than the number of atoms. But if we bring them together and let the ends touch, then we get another one. So it’s N/N. So it goes to minus 1, the lowest possible energy.

So in a conjugated ring, the nodes as you go around the ring must come in an even number. Let’s look for example at a three-membered ring. So here are three p orbitals viewed from the top. So there are no nodes, obviously. And the energy of that is going to be minus one, on the scale we were talking about. Each of these is 1 over the square root of 3. The overlap here is 1/3. 1/3, 1/3, 1/3, totals to 1. Minus 1 in energy. And there are no nodes.

But now let’s suppose we want to make a node. But now here we have an energy that’s 1/2. It’s the same as ethylene. But notice we can’t have just one node as you go around. Because if we go red to blue, we have to go back. And someplace we’re going to have to have another node as we go around there to get back to the pink again.

You can’t just have one node. Unless, just to look ahead a little bit, you twist it and make a Moebius strip. But that’s quite a job. We’ll see that it has been done, but it’s quite a job.

OK, so anyhow, the nodes come in pairs. So there are two like that. But the other aspect of that – I was going to say the flip side – is that you get another way of doing the same thing. So you can get two nodes here. You can also get two nodes that way. So there are two different ways to get two nodes.

So you have a lowest one, and then you have a pair that have… There’s one with no nodes, then there’s a pair with two nodes. And they have the same energy.

So now we can look at the energy shift that happens when you form one of these rings. That is, how much is the aromaticity worth? So there was butadiene, which we did already. Now suppose we bring the ends together. There was the lowest one, and now we have an end-to-end interaction. So those interact with one another, and obviously that’s favorable, and it’ll go down in energy Also they’ll become of the same size, once you do that, because it’s no longer advantageous to have it big in the middle and small on the ends, because there isn’t an end anymore.

How about that one, when you bring the two ends together and let them touch. What will happen to its energy? Ellen, what do you say?

**Student:** Decrease.

**Professor Michael McBride:** Pardon me?

**Student:** Decrease

**Professor Michael McBride:** What do you mean decrease? Will the energy go up or down when you touch the two ends?

**Student:** Down

**Professor Michael McBride:** Why do you say so?

**Student:** Because there’ll be another bond.

**Professor Michael McBride:** There’s another interaction. But look at the interaction, look at the colors. When you touch the two ends together, when you bend it around so this touches that. That’s antibonding. That’s cute, huh?

OK, so that one’s going to go up in energy. Now how about the next one? That one’s going to be favorable. It’ll go down in energy, and end up the same place this was. Remember there was a lowest one, and then two pairs. Two which had a pair of nodes. And the next one, that’s going to be unfavorable again. So the structure when you bring the ends together is going to be different.

And fortunately there’s a mnemonic device, a geometric one just like this, that works in that case as well. And notice that the shifts alternate, down, up, down, up. And they must do that because of the number of nodes. If you have an even number of nodes, it’ll be favorable because you’ll have the same sign on the two ends. If you have an odd number of nodes, it’ll be unfavorable. So it has to be that down, up, down, up, as you go up the scale of MOs and curl them to touch their ends.

So when you bring the ends of a chain together, the odd numbered MOs, 1, 3, 5, decrease in energy. Those are the ones with odd numbers, therefore an even number of nodes. Because the very lowest one, 1, has no nodes. Sorry for the confusion.

While the even numbered ones increase in energy. Thus, if you have an odd number of occupied MOs, there’ll be more going down than going up. Is that clear? If you have an even number of electron pairs, they’ll be down, then up. That will cancel. Down, then up. That will cancel. Down, then up. That will cancel.

But if you have an odd number, if you have down, then up, then down, there’ll be more downs than up, if you have an odd number of pairs of electrons. So they’re going to get overall stabilization, special stabilization, due to this aromaticity, if you have an odd number of electron pairs.

Of course, you also have to take into account the sigma bonds. Because when you bend things into a ring, you may introduce ring strain, like in the three-membered ring. There’s obviously that.

But as far as the π electrons go, there’s going to be special stability if you have an odd number of electron pairs. And this gives rise to Hueckel’s rule, as it’s called. Which is, if you have 4n+2 π electrons, that’s going to be unusually favorable in a conjugated ring. Compared to what? Compared to a chain that’s not touching the ends.

Linear versus ring is going to be especially stable if you have 4n+2 π electrons. That means an odd number of electron pairs. So, more of these going down than up, when you bring the ends together. And as I said, there’s a circle mnemonic. Now remember for a chain, we had a semi-circle mnemonic. For a ring you have a circle. It’s the same radius; minus 1 to 1 is the limit.

And now you inscribe a regular polygon with a point down. Because remember that lowest one is always going to have minus 1 as its energy, so the point has got to be at the bottom, like that. Now we put dots on there for the energies we’re going to have. And notice after the first one, they’re going to come in pairs.

So if you have cyclopropenyl, those are the energies you’re going to have, if you have two electrons. So the cyclopropenyl cation, just two electrons that’s the energy you’re going to have. Had you had the allyl, the three p orbitals in a chain rather than in a ring, then we already saw you’d have that.

So now we’ve got strong stabilization of two electrons coming down from here to here. So there’s a lot of aromatic stabilization of the cyclopropenyl cation.

Now suppose you have the radical, the cyclopropenyl radical. Here’s the allyl radical. In the cyclopropenyl radical, two of the electrons will be there, but one has to be up there, in one with two nodes. So this one is going up more than those are going down. But there are two going down here, and one going up. If you have the anion they’re going to be two going up and that’s bad. You lose more here, than you gain here.

Now remember, there’s always an MO at minus 1. So what happens if we have cyclobutadiene? Four p orbitals, and four electrons. Now there are our energies. There are the energy levels. We’re going to put two electrons down here and one in each of those, like that. And we’re going to compare it with normal butadiene which we saw before has a pair here, and a pair here. Notice that these go up more than these go down, so it’s bad. That’s antiaromatic. This stabilization that we saw in benzene, aromatic stabilization, is the opposite here. It’s bad, overall, to put the ends together, although it’s good for the best orbital.

And notice also that these will be singly-occupied orbitals. So they’ll be a little bit like free radicals. So this is going to be a very reactive species, cyclobutadiene.

Now if you have benzene, you have six, and they look like that. And now we have six electrons to put in. And we compare it with hexatriene, as a chain, which has that. And now you notice these went down a little. These went down a lot. These went up, but not very much. So this is where we got that 30 kilocalories of special stabilization in benzene, a really good example of aromatic.

So 4n+2 is the number of electrons in a ring that will give special stability, because there’ll be more going down than going up.

**Chapter 3. Generalization: Nonbenzenoid Aromaticity [00:35:59]**

Now that’s interesting, and it prompted organic chemists when this became–although Hueckel did it in the early 1930s, it didn’t reach the consciousness of most organic chemists, even though his brother was an organic chemist, until the late 1950s or 1960s. Then people started trying to make all kinds of rings with p orbitals with different numbers of electrons, and see which ones– could they confirm this rule. And indeed they did confirm it.

But much more interesting was how the same concept of cyclic conjugation applies to reactions, and to transition states to make reactions faster or slower. And in that context we’re going to talk about cyclo-additions and about electrocyclic reactions.

But first I want to mention also heteroaromatic compounds. So you can have aromatic compounds that aren’t just carbons in the ring, like for example nitrogen here. Now think of the electrons on nitrogen. It’s forming sigma bonds to carbon on either side. It has an unshared pair of electrons. But that’s not part of the π system. That’s in the plane, right? It’s not part of the p orbitals, which is what we’re talking about. It’s an *sp*^{2} orbital.

But there is an electron in the *p* orbital. Incidentally, this is redundant. I drew a double bond here. I’m also counting that electron, but it’s actually participating in that double bond. But I’m just showing it there so that we can keep track of what heteroatoms are doing.

So there’s a *p*-electron from each of the five carbons, a *p*-electron from nitrogen, 6 π electrons. 4n+2 where n equals 1. That’s stable; it’s aromatic.

Now furan, a five-membered ring with oxygen. Oxygen has two unshared pairs. So it has two that are in the sigma system that are irrelevant for our purposes. But it also has two in the π system. So now you have one from each of the carbons in the π system, two from the oxygen, six altogether. Great. Aromatic.

Pyrrole. Now the nitrogen has an unshared pair in that one. So it’s got one, two, three, four, five, six. Good, that’s aromatic too. Unusually stable.

Imidazole. This is a tougher one to count. One of the nitrogens has an unshared pair that’s in the plane. Don’t count that one. The other one has an unshared pair that’s coming in and out of the plane. That’s part of the π system. And there’s a single electron on this nitrogen, as there was over here. So now we have one on each of three carbons, four, five, six. That’s aromatic too, imidazole.

Incidentally that structure, it occurs in the amino acid histidine. And that’s really interesting, in that it is involved in long-range proton transfer in enzymes. Suppose you have a proton here, but where you need it is way over here. What you can do is this. So this hydrogen goes up, this hydrogen goes onto the ring, and the ring sort of looks like it’s turned over. But a hydrogen has gotten up where you want it, without actually having to move the proton all that way.

So these are all what are called heteroaromatic compounds, because they have atoms other than carbon in the ring, but they’re stable for the same reason.

We could look at the orbitals of furan, the one with an oxygen in it. There’s the lowest-energy orbital. It’s what you expect, *p* orbitals all overlapping favorably all around the periphery. No antibonding nodes.

But then there are two that have antibonding nodes. There’s one that has an antibonding node at the back and in the front here. So it’s plus on one side, minus on the other. And there’s another one that has an antibonding node that goes across. So it intersects the ring both here and here. So it’s plus in back, minus in front.

And remember there are six electrons in this. Two here, two here, two here. These are the antibonding orbitals. And they, as you can see, have four. Each of them has four nodes as you go around the periphery. So for example, here there’s a node on the oxygen, a node between the carbons, a node between the carbons, a node between the carbons. Four nodes as you go around the ring. So these are a pair of the same energy.

Now if you want you can go to that program you can get on the web, and draw crude calculations to see this kind of thing. So here’s benzene, and here are the energies we’re talking about. A lowest energy orbital with a pair of electrons, and then the next two, “degenerate,” as they’re called, pair. And then antibonding, worst.

Now what happens if you put a nitrogen in the ring? Now the lowest energy orbital is bigger on nitrogen, than on the other atoms. Why? Anybody got an idea why it should be bigger on nitrogen? Amy, you got an idea?

**Student:** It’s more electronegative.

**Professor Michael McBride:** Yeah, it has a bigger nuclear charge; it’s a better place for the electrons to be. It’s more electronegative, as you say. OK. And so it’s lower in energy, see, than it was here. It’s larger on the N, and it’s lower in energy.

Now if we look at this, the next orbital here, it has two nodes, here and here. Here it was in benzene. Here it is it in the nitrogen analog. And notice it’s the same energy, here and here, because it’s the same orbital.

In fact, what it is, this is a π bond, here’s a π bond. They don’t interact with one another. So the energy is minus one. It’s the energy of a π bond. And it’s the same here. And nitrogen doesn’t have anything to do with it. But the other one, there was another one here. So those were identical in shape and in energy.

But there’s another one that has the two nodes across the middle here, and across here. And now it does involve nitrogen. So this one is high electron density on nitrogen, and therefore a little bit lower energy. So these two orbitals that we just looked at that had two nodes, were the same energy here, but different energy in pyridine. One with nitrogen.

Now suppose you have 10 atoms in a ring. Cyclodecapentaene. Here are the structures, here are the energies. And now here’s the lowest one, same as you go all around. The next ones have a node here, that is, two nodes as you go around the periphery. But notice these, right and left, don’t interact. There’s no orbital here.

So this is just like butadiene, and it’s the same energy as butadiene. And same for this one, which has the horizontal node. Same energy, here, as they would have been in butadiene. Now the next one has four nodes as you go around, and four nodes as you go around.

Now, naphthalene is two benzene rings next to one another. How can you go from this to naphthalene, to two six-membered rings? What you can do is take the atom that’s in the middle on the top and in the middle on the bottom, and pull them together so they overlap, like this. So then you have two benzene rings together. That’s naphthalene.

Now let’s look at the energies of that and see why they come out to be the way they do. So the very lowest one, we brought this one to touch that one, so we introduced a new bond. So it got lower in energy. And it’s big in the middle, as you expect, because it has more interactions.

Now the next one. If we take this one and draw the top and bottom together like that, it makes no difference because are no orbitals there. It’s on a node. So that has the same energy, here, here, and here. But if we bring the top to bottom together here, what’s going to happen? Noah, what do you say? If I touch this to this, is that favorable or unfavorable?

**Student:** Unfavorable.

**Professor Michael McBride:** It’s unfavorable. So the other one will go up in energy to there. And now, how about this one up at the top? If we bring this to touch that, makes no difference. There’s no orbital there when you have a node. So that comes straight across.

But how about if we touch this to this? Noah?

**Student: **Favorable.

**Professor Michael McBride:** That’s going to be favorable. So although that one goes straight across, the next one goes down in energy. So we can understand the special stability of naphthalene compared to cyclodecapentaene.

And notice, not only that, but this one which has nodes here and here, has just isolated pairs of overlapping p orbitals. So it’s the same energy as ethylene would be, because it’s the same interaction.

OK now can you make that compound, cyclodecapentaene? I’m going to quit with this one, OK? And in fact it was made by Heinz Dieter Roth, who was a post-doc here, later. But he did this in Germany with Emanuel Vogel.

You can’t make a ring that’s that big, without having enormous carbon-carbon-carbon angles, and have it be flat. But you could bend it in, like this, except for the fact that those two hydrogens touch one another. That’s not good. So one thing you could do is just make that a bond. And we’ve already talked about that, that’s naphthalene. That was the previous slide.

But another way is to put a bridge across, like CH_{2}, from this carbon to this carbon, so that those two atoms don’t run into one another. That’ll distort the periphery a little bit, but it will leave all the *p orbital*s around the edge.

So the way they made it, was to first add dichlorocarbene to a double bond in the middle. Then remove the chlorines. And now this thing could open that central bond to give this, and it does that. Here’s a picture of what it looks like. So here’s the CH_{2} bridge across the middle, but all around the periphery you have 10 atoms with p orbitals.

And there’s one of the orbitals. And you see that’s the one that has two nodes, here and here. So you have five p orbitals here, and five p orbitals here.

Now there’s another criterion for aromaticity, which is the chemical shift in proton magnetic resonance, and we’ll see that next week.

[end of transcript]

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