Freshman Organic Chemistry II

CHEM 125b - Lecture 19 - Aromatic Transition States: Cycloaddition and Electrocyclic Reactions

Chapter 1. Aromatic Ions [00:00:00]

Professor Michael McBride: So we’re going to finish up talking about aromaticity, talking then about reactions, as well as molecules that are unusually stable. That is, transition states that are stabilized, that’s Diels-Alder reactions which are cycloadditions, and also other pericyclic reactions like electrocyclic reactions. Then we’ll go on to spectroscopy.

OK, so we’re generalizing the concept of aromaticity. We saw last time that Hueckel thought there would be unusual stability when you had 4n+2 π electrons in a ring, that is, when you have an odd number of electron pairs, because every odd numbered orbital goes down in energy when you form a ring. So if there’s an odd number in total, there are more going down that up. And then we’ll go on to see, as I said, about transition-state aromaticity.

So first just one slide that will be on the exam on Monday, it’s fair game anyhow, which is some things that are not benzenoid, but are still aromatic. Like cyclopentadiene, of course, is a ring, but it’s not a ring of conjugated p orbitals because there’s one intervening one, so the terminal p orbitals of the diene don’t interact, don’t overlap with one another. But if you pull off the proton with a base, then they interact. So now you have five electrons– this anion in the π system– and it’s unusually stable. You could pull off that hydrogen from a carbon with hydroxide. The pK_a of this compound is 15. It’s more acidic than water is. It’s better to pull a hydrogen off that carbon then to pull a hydrogen off oxygen. Now what’s the normal pK_a for pulling a hydrogen off a carbon? Do you remember?

Student: About 50?

Professor Michael McBride: Pardon me?

Student: It is about 50.

Professor Michael McBride: It’s about 50, right? Of course, this is allylic. And in fact it’s doubly allylic, so you might say that helps out. No doubt it does help out. But notice that it has 6 π electrons, same as benzene, and we would expect on the basis of our theory for it to be unusually stable. But if it’s just that it’s allylic, think about doing the same thing with the 7-membered ring. If you pulled the hydrogen off the CH₂ here of a 7-membered ring, you get this anion, which would be stabilized by that resonance structure, that one, that one, that one, that one, that one– an enormous number of resonance structures, but the pK_a is 39. It’s 10²⁴ less acidic than the one on the 5-membered ring– even though you can draw many fewer– only five resonance structures in that case. So it’s not a question of resonance structures.

It’s a question of how many electron pairs there are in that ring. And if it’s an odd number it’s stable, and if it’s an even number of pairs, then it’s not stable. So that one has eight π electrons– 4n. It’s antiaromatic instead of being aromatic.

Another interesting analogue is cyclopropene. So here’s a case of cyclopropene reacting with an unusually stable cation, triphenylmethyl cation. Notice that that cation is triply benzylic. There are three benzene rings that help stabilize the positive charge, or„ to speak more accurately, whose electrons are stabilized by the vacant orbital on carbon. Still, it’s exothermic to transfer the hydride– notice it’s a hydride not a proton pulling off H minus this time onto the phenyl-CH₃ [correction: Phenyl₃CH] to generate this cyclic conjugated cation. So that one is even more stable than the unusually stable triply benzylic one. Now what’s the accounting of that one? The accounting of π electrons? Ruoyi, can you tell me how many π electrons in a ring of that cation? Is it going to conform to 4n+2? OK, we had a double bond, that’s two electrons, and what do we get from that center when we pull off H minus in the p orbital?

Student: Zero.

Professor Michael McBride: So, how many do we have?

Student: Two.

Professor Michael McBride: Two. Is that 4n+2, where n is an integer?

Student: n is zero.

Professor Michael McBride: Yes, it’s when n is zero. So that one you expect to be unusually stable. It’s just got the one pair that goes down in energy when you form a ring. Two π electrons is 4n+2. And notice that now if you go to C₇, here we went in the anion to C₇H₈, pardon me, C₇H₇ anion. Here to C₇H₇ cation, if we do the same trick with that, react this hydrocarbon with that triphenylmethyl cation. And that is also the same. That’s also downhill. It’s favorable to make this. Does that conform to 4n+2? What do you say, Rahul? How many π electrons in C₇H₇ plus?

Student: Six.

Professor Michael McBride: Six. 4n+2, same as benzene. So here are some really weird compounds, that you wouldn’t expect to be so unusually stable, that are because it’s a generalization of the idea of aromaticity.

Chapter 2. Pericyclic Reactions: Cycloaddition, the Diels-Alder Reaction, and Photochemistry [00:05:59]

But it applies not only to ground states but also to transition states, when you get a ring that has a conjugation among a bunch of orbitals where the number of electrons is 4n+2. So in this case the transition states are aromatic, even though the starting material and the products are not aromatic. What that means is that there’s special stability only in the transition state, and therefore the reactions will be unusually easy.

So let’s see if we have some examples of that. We’ll look at cycloadditons, in particular, that Diels-Alder reaction and also at electrocyclic reactions. We’ll start with the cycloadditions. So that Diels-Alder reaction was the subject of a Nobel prize, surprisingly enough to Diels and Alder. OK, it involved a diene reacting with an alkene. So an ene and a diene and they react with one another with three pairs of electrons shifting to form a ring. So it has four σ electrons that are involved in this, and 2 π electrons in the product. In the starting material, it’s 4 π electrons and 2 π electrons. It’s six electrons, three pairs, that are being involved. And we can write curbed arrows like that, but we might want to think about it a little more carefully.

Suppose we just draw the two approaching one another, HOMO to LUMO like that. Now there are two problems here. One is that we’ve got the LUMO drawn upside down. In fact, there are three problems. That we can just write the signs the other way, that’s no problem. Another problem is that the hydrogens which we haven’t shown, the σ hydrogens are going to be running into one another when we do that. And there’s also another problem. Somehow, π electrons have to become σ electrons here and here. It has to be an end on, not a π interaction in order to do that. How can we get around these problems? Yes.

Student: Rotate the ene.

Professor Michael McBride: OK, we can rotate the ene 180 degrees. That’ll make the signs correct. How do we get around the other problems, let’s say the hydrogens running into one another?

Student: 90°.

Professor Michael McBride: If we rotated it at 90°, that’s true, but you don’t get the best overlap that way for forming a new bond. Anybody else got an idea? Brandon?

Student: Can we just move it under?

Professor Michael McBride: Put it under. So make the approach where the p is parallel to the direction of the p axis like this. Now we got around all those problems. So we can form this new– but notice the transition state is folded because of this, because they’re not coming together in a plane. So to see the product then we have to unfold it. We have to grab that bottom black line and pull it up into the plane. So the product is then going to look like that, once we flatten it. You can see the new bonds that will form. Now let’s think a little bit about what we would predict in connection with stereochemistry in this.

Suppose the starting alkene were of a Z configuration. Suppose it had two methyl groups on it. Then if you form those two bonds at once, they’ll still be cis here, and they’ll be cis in the product in methyl groups. How about the diene? Suppose, the diene had the E,Z configuration, so methyl that’s cis here and trans here. Then if we look at the transition state, it’ll be still cis here and trans here. And if we flattened that out, where will the methyls be in the product? How about, there’s going to be a methyl here on that carbon, is it going to be pointing toward you or away when you flatten the ring out? If this is the way it is in the transition state, try to think of it mechanically, we break this bond. Is that going to be, when we pull this up and flatten it out– is that methyl going to be coming up toward us or back into the board?

This is a kind of mechanical thinking that maybe– we used to require people to buy models, you’ve seen those, and that probably helped with this mechanical perception. How many think when you flatten, and when you pull this up, the methyl there is going to be pointing towards you? How many think it’s going to be pointing away? That’s reassuring that the vast majority are right in this. It’s coming toward you. So you can look at this in the privacy of your room, maybe use toothpicks and grapes or something to make a model, and see how it twists like that.

Now, how about this one when you fold this up? What’s it going to be at the bottom carbon? How many think it’s going to be pointing toward you? Have the courage of your conviction. How many think it’s going to be pointing away? The aways have it. Democracy wins again, Debby. It’s away in that case. So you have a little practice to think about how such things work. And there will be more of these as we go along here. So those two methyls will be trans preserving the kind of relation they had in the starting material through that particular transition state.

Now is that true? So here we take a methyl– first we’ll look at regiochemistry, and then we’re going to look at stereochemistry.

So regiochemistry first. OK, so now we can imagine this diene, the methyl substituted diene, to be either approaching this way or to be upside down. It could be either way. So we’re going to slide it over the top of the other one like that and make the new bonds. And notice in this case, once we unfold it– to unfold it, we’re going to have to grab this bond. It’s in the form of a U, when it makes those new bonds on the left, see like that. One is on top; one’s on the bottom. They made these new bonds here. And we’re going to have to unfold it like that to draw a normal picture of it. And when we do that this methyl group is going to come over here. That is the new blue double bond is here on the right, but when we grab and pulled it out to the left it’s going to be there. So spacial perception is entering into the– is raising its ugly head now.

And in this case when we fold it out, the methyl is obviously going to be down. Now, in fact, those two products form. 9% of this and 45% of this. Now why is there’s so much more of this transition state than of that transition state? Incidentally, notice that the diene is not only over this double bond, it’s also as close as it can be to the CO double bond. There’s a secondary interaction of the orbitals that makes it go that way, that’s called endo orientation, but don’t worry about that. I struggled here, how to do this, to make it look realistic. That’s the way it really does.

Most people wouldn’t draw that way. They’d draw it approaching just the diene here, and the alkene on the right. And this would come in, and that would come in, and they would draw the new bonds. But really it approaches above, and I think that’s interesting. So that’s why I took the trouble to draw it that way. But when you do that, you notice that the diene could be either such that it’s close to the CO_2, or it could be out here. So that it if you look at it from the top, it can either be a U, or it could be an S. And it prefers to be a U, to have the diene closer to the carbonyl. OK, so given that that’s true, why might you expect to get rather little of this and more of the bottom stuff. Any ideas? Ayesha?

Student: Is it because when they’re approaching the CO group and the CH₃, they’re sterically hindered?

Professor J. Michael McBride: Yeah, even though the diene is attractive to that group on the bottom which is why it’s formed into a U rather than an S– you would also have it close in this case, but in the case on top, you have those two methyls on top of one another. So it looks like steric hindrance could explain this. But there’s another possibility as well. Perhaps, there’s an intermediate. Perhaps, one bond forms first in the rate-determining step, and then a second bond forms, and you have this intermediate which on this side is an anion adjacent to a carbonyl group. So that is an enolate. And on the left it’s allylic, because we can draw a resonance structure of that. Now if that were true, if you had this intermediate, why might you predict getting more of this than of this? More of the stuff on the bottom. Why is the one of the bottom a better allylic intermediate?

Mary, do you have any idea? The only difference between the allyl on the top and the allyl on the bottom is where the methyl is. In the top, it’s on the middle carbon. In the bottom, it’s on a terminal carbon on the double bond. Let’s think about the resonance structure you would draw. Mary, were you about to say something?

Student: Oh, I was going to say, more substituted alkenes–

Professor J. Michael McBride: Aha! In that resonance structure this methyl is stabilizing that cation. Not in this case, when it’s on the middle carbon, because the allylic charge is on one end or the other. So one could say that this is a more stable intermediate and, therefore, you’d get that. Or it could just be that the transition state is a little unsymmetrical the way we showed in other cases, where a bridge thing like mercury wasn’t in the middle but a little toward one side. So it could just be that both bonds form at once, but not precisely in step with one another, one a little more than the other.

Now, if it went through an intermediate, there’s a test for it, to see whether there’s an intermediate. Because if you have an intermediate, then you would imagine that that could rotate here. So this could move over here before it formed a new bond. Or indeed, this one could rotate around this bond and bring that, instead of on the top where it is now, on the bottom as it approaches this one. So we could have these rotations that would change the stereochemistry of the product. So we can test that.

So let’s look at the stereochemistry of the ene. Now we’ve put two of these groups on, and they’re cis to one another. And if we put that on top, and make the new bonds, and then fold it back, grab this new double bond, the one here and fold it back, we see that these two groups will be adjacent to one another– cis. So we’d expect a cis product, when we form those new red bonds. So a cis alkene should give a cis cyclohexene, and that, in fact, forms in 68% yield. If you take the other stereoisomer of the alkene, the trans one, and do the same trick you get the trans cyclohexene, and that one in 84% yield. And this is what you actually isolate and can sell to your neighbor. You don’t form the wrong isomer. So it doesn’t rotate. So there’s not an intermediate. So the diene just sits down on the ene. It forms the two σ bonds simultaneously from the same face from the top here. So there’s no rotatable intermediate with only one new σ bond. So you didn’t have this situation where you could rotate. So it’s a concerted reaction; you form both bonds at once.

Now, how about with respect to the diene stereochemistry? Here we have E and E, the trans arrangements around those two double bonds. So we put those on top, form the new bonds, and you see that now, as drawn here, the product, in 81% yield, has everything cis. This carbon, this carbon, this carbon, and this carbon. Now these two carbons in the ring being cis to one another are not surprising you couldn’t make those trans, it would be to strained. But that they are cis to these says that this had to be in the U form here, not the S form. And that these are both up shows that you formed both bonds at once as shown here. And this is another place for you to sit in your room, and think about how these things will unfold, and what points and what direction.

Now if you take this E-Z isomer, and do the same trick, then in 15 hours at 150 degrees you get this to be trans, instead of cis. But notice the difference in conditions. This one happens in 5 minutes at 120°. This one takes 15 hours at a much higher temperature, 30° Celsius hotter. So there’s something wrong, something slow about this reaction. Now, can we figure out what might be slow about that reaction? Why might it be bad to come in and do this? Notice that that hydrogen and the methyl are near one another, when we have the rotation around this single bond so that these are adjacent to one another. So it would prefer the s-trans conformation. “s” means the single bond in a diene between two double bonds. So it can be s-cis or s-trans. So this prefers to be s-trans, but to do the Diels-Alder reaction it has to be cis so those terminal carbons can be near the same ene. So because it prefers to be this way, it’s much harder. You have to put in extra energy to make it s-cis, as well as to do the reaction.

Now just to show you some of the variety of the Diels-Alder reaction, it’s not just simple compounds like the ones we showed here, you can use acetylene. So we do that trick in this one at 150°,you get a diene– a cyclohexadiene product. Because the acetylene, when it loses one of its bonds, becomes an ene not a single bond. Or you can have a ring involved like this. And notice that that’s the product, and it happens very fast at 1M concentration, the half-life– or the equivalent of a half-life– is a second at room temperature– actually below room temperature.

So have you ever seen a skeleton like that before? Let me draw it a little differently. Yeah, Rahul?

Student: The Ziegler-Natta, or um…

Professor J. Michael McBride: Yes, it’s two names.

Student: The metathesis with…

Professor J. Michael McBride: Right, this is the starting material when it had hydrogens there for the metathesis polymerization– the ring-opening metathesis. That’s actually absolutely true, but I was thinking of an earlier example of it. Yeah.

Student: Bartlett and Knox?

Professor J. Michael McBride: Ah! The Bartlett and Knox skeleton could be made that way. So you can make bicyclic compounds by a Diels-Adler reaction. Or look at this one, this molecule this unsaturated aldehyde is called acrolein. And you can take two molecules of that, and dimerize them. What’s special about this one? How’s it different from the other ones? What’s special about the 6-membered ring? It has an oxygen in it. It’s not just carbons that can do this trick. So what they all have in common is they form a 6-membered ring with a double bond in it. Although, if you use an acetylene it can be two double bonds. But when you want to do a synthesis project and you see that there’s a 6-membered ring with a double bond in it, you think Diels-Alder reaction. So you see, if you can take it apart into an ene and a diene that would do that kind of thing.

So, the Diels-Alder reaction has this cyclic, six-electron transition state, the one we look at here. And we can actually look at what the transition state looks like and the motion at the time the two bonds are being formed in parallel. So we have the HOMO in front and the LUMO behind, or actually on top, and those will interact to give a HOMO– not the HOMO, but the HOMO-1, it turns out, in the transition state. But it has this node right down the middle that both of the starting pieces had. And if you look at the LUMO on the bottom and the HOMO on the top, then they give the HOMO in the transition state which doesn’t have the node down the middle. So that’s how they go together in the transition state. And here we see, I think, the motion that’s involved in going through the transition state. It’s curious that the hydrogens stay more or less in place and the carbons do most of the moving in forming the bonds.

Now can we do the same thing with two alkenes? Can we take two alkenes and have them come together to form a cyclobutane? What do you think? Any thoughts on this one? How many electrons involved in this ring that we’re making in the transition state, when we are bringing these together and make them overlap? How many electrons are changing in this process? Ayesha?

Student: Two π to two σ electrons.

Professor J. Michael McBride: There is π here, and π here– that’s four electrons– and they become these two bonds here, two σ bonds. So, there are four electrons. So it’s 4n, where n is 1, not 4n+2, as in the Diels-Alder reaction. So this one would have an antiaromatic transition state. Of course, it also is strained here, so the reaction might be expected to go the other way, but that doesn’t work either. It’s the same transition state with four π electrons– four electrons involved. So it doesn’t work in either direction. On the other hand, another way of looking at it is the HOMO of one is orthogonal to the LUMO of the other. That’s the same analysis we did in showing why H₂ can’t add to an alkene.

But if we shine light on it then we take an electron out of the HOMO, and put it into the LUMO, and now you can get stabilization because you’ve changed the symmetry of what the orbitals are. So now in the excited state the HOMO can interact with the LUMO of the other one, because you’ve changed it from π to pi-star in putting light in. So that is when you shift an electron from the HOMO to the LUMO, then you can get the reaction to occur. And it’s an important reaction for most people, because this is a DNA double helix. On the right is A-T-T-G and on the left is T-A-A-C. But if we looked right here in the middle, we see two T’s. So if we zoom in on that we’re looking in the plane of that thing. Let’s rotate it out toward you, so you look down on those two T’s. We’ll look at just that bit of it, and rotate it up, and that’s what it looks like, and we could see what’s in front and what’s in back this way. And notice there is a double bond here and a double bond here. Will they react with one another to make a cyclobutane? No, because it’s four π electrons that are involved. It’s not an aromatic transition state. How could you make them react with one another? Pardon me.

Student: Light.

Professor J. Michael McBride: With light. So if you use UVB and don’t put on sunscreen, then you get that reaction happening, and it forms a cyclobutane. And this thymine photodimerization to do that makes the chain kink, of that chain where the two dimerized. And that inhibits DNA replication and transcription and is believed to be the main source of the mutations that give rise to melanomas. So that’s why you put your sunscreen on that takes out UVB. It’s to stop that photodimerization of thymine.

Chapter 3. Electrocyclic Stereochemistry [00:28:15]

Now we’ve been looking at cycloadditions, which is one class of these things where the pericyclic reactions, where you can have aromatic transition states. And now we’re going to look at the other kind, which is electrocyclic reactions, where you start with a conjugated chain and curl it around to make a ring. So you can start with hexatriene and make a new bond at the bottom, shift the π bonds to make a cyclic compound, cyclohexadiene.

Now, of course, you have groups that are substituted on the ends here, maybe just hydrogens that are in the plane in the starting material, but in and out of the plane in the product. So you’re going to have to twist those carbons to make those things that are in-plane out-of-plane. So changing the π to the σ bond requires a twist in one of two ways. You can either twist it this way, where this comes up and that goes down, and that’s called conrotation, because as we look at it here both of them are going counterclockwise in the same direction. Or we could go through this other geometry and rotate this one counterclockwise, but the one on the left clockwise, and that’s called disrotation, they rotate in opposite directions. Either way would take these four groups here and make them in and out of the plane, as they need to be to form a σ bond. Is there a preference between these?

Well notice, if you do it this way, that the p orbitals that are on top in the starting material are bending in toward one another. The top of one end touches the top of the other end, as they rotate in opposite directions. So that among these orbitals, there’s always going to be an even number of nodes. This has no nodes, but if you change the sign here to make this a node here, you’d have to have a node someplace else as well. So you have to have an even number of nodes. We saw that in benzene too.

So that’s like these things that were involved in in the Hueckel, rule where you have a ring and they all touch one other on the top. You have an even number of nodes. But if you twist in the opposite direction then the top on the right touches the bottom on the left, and now as you go around you have an odd number of nodes. There’s a node here between these two as they’re beginning to touch, but no other nodes. So an odd number of nodes. So it’s just the opposite of what Hueckel is.

Now what name should we associate with that? Have you ever seen something that looks like that? What’s funny about it? You notice if you trace the top of it, you get a node where it has to change from red to blue, if red is one side and blue is the other side of a p orbital. It had to twist so at one place there’s a node as you go around this, if you drew a bunch of p orbitals on that. So you have to have an odd number of nodes if you have that connectivity. And I heard you say that that’s called a Moebius strip. So the opposite of Hueckel is Moebius– two different ways of doing this rotation to form the σ bond, Hueckel and Moebius.

Now here’s the transition state that would do the Hueckel, where the top is touching the top to form a σ bond. And notice that as you do that you preserve a mirror of symmetry. On the left there’s not a mirror. Do you see what symmetry there is on the left? What will change the left hand side of this molecule into the right hand side? What does it look like? Things like this, it’s like a propeller. If I could do it, here. So it has an axis of symmetry that happens all the way from the starting material to the product. So this has a mirror, and this has an axis of symmetry.

Now let’s see what that does. Let’s follow each orbital in the starting material, as it becomes an orbital in the product, and see which starting material orbital becomes which product orbital for these two different ways: the Hueckel connectivity where you always have a mirror of symmetry, and the Moebius where you always have an axis of symmetry.

So of these six– you have six atomic orbitals here, a p orbital here, here, here, here, here, here, six. We have six molecular orbitals, no nodes, one node, two nodes, three nodes, four nodes, five nodes of which three are occupied by the three electron pairs. The same starting material over on the right, and in the product, we have four p orbitals in the conjugated systems, so ψ₁, -₂, -₃, -₄. No nodes, one node, two nodes, three nodes but we have a σ and a σ-star in the product in both cases. But we have the question of where do the electrons that start here end up in the product. That’s our question, to track the MOs.

So let’s look first at ψ₁, and we’ll use as a model the aromatic analog, Hueckel connectivity, where we’re not doing this reaction, but just looking at a benzene. This is what we did last time. Remember if we bring these two ends together, they’ll touch here, and that’ll become more stable. So this one here goes down in the product. We looked at that last time.

So that will happen– here is at the transition state, and you see that when you have the mirror that’s happening. The ones that are red on top are overlapping here, and it’s going to build as it goes on toward product. This will get bigger and bigger because that’s an important overlap. The π-bonds back here get smaller and smaller, and this orbital of the starting material, this one, smoothly converts into σ through that transition state. So ψ₁ becomes σ.

Now think about it on the left. As you begin to do it, as you rotate them in opposite directions, the bottom here touches the top here. That’s going toward a σ* antibond, not toward a bond. In fact how it avoids going way, way up in energy– that electron pair– is to decrease the size in front and increase it in back. But when it increases in back, it forms ψ₁. It makes the nodeless set of four here. So this one doesn’t go to σ the way it did here, it goes to ψ₁.

Now let’s look at– there are three pairs of electrons were talking about, that’s the lowest one of the starting material. Let’s look at the next one. As we brought it together to make a ring in the aromatic case, that was unfavorable antibonding. So, in fact, this pair of electrons went up in energy, and if we look here, we see the same thing. This ψ₂– where we’re starting– is ending up in a π orbital. Notice it’s decreasing in front and increasing in back, but with a node because of that mirror. So it’s not ψ₁ the lowest– the one where all the p’s are in phase– it is the one that has a node in the middle. So it goes to ψ₂ of the product. But over on this side it goes to σ, because if it had a node here in the beginning, now the red on the bottom and red on the top are touching one another. So that’s going to make σ.

Now let’s look at the third electron pair. And we saw that in the aromatic system it went down in energy. And in this case, here it is at the transition state, and it’s going down in energy, and in fact, it’s going to form ψ₁. It’s not so easy to see that here, but you can see that it’s decreasing in front at least. I admit that that’s hard to see. On the other side over here it’s going to decrease in front because that’s an antibonding interaction. But the thing it’s getting in back is blue, red, and red. It’s got two nodes. So it’s not going to ψ₁, we’ve already done that, it’s not going to ψ₂, which has only one node, it’s going to ψ₃, which has two nodes. And that is terrible. Those electrons are– these sort of cancel one another, but these are going way upstairs. Whereas over here, these two sort of canceled one another, but this is going downstairs. So it’s much easier to do the Hueckel then it is Moebius if you have three electron pairs.

Notice that if you had only two electron pairs, you’d have this which more or less cancels versus if you had only two electron pairs this one goes up more. So it goes the other way.

We can also look at what would happen with the other orbitals. And notice that if we took an electron from here, and put it into here, then it’s going to go down in energy. So photochemically, we could make it prefer the left path with six electrons. So, again, that’s what actually happens thermally in a normal reaction, it’s the Hueckel with six electrons not Moebius. So disrotation is preferred for a six electron shift, so that if you have methyls like that, the two methyls will end up on the same side. Say if this one rotates up, this one also rotates up. That is if it’s my little fingers, they go like that, they rotate in opposite directions, not in the same direction where one would go up and the other down.

So stereochemistry shows you the form– and notice that the cis is the less stable isomer. It’s not forming that because it’s more stable. It’s forming it despite the fact that it’s less stable, because it prefers to do that. That’s 4n+2.

So how could you study whether the opposite conrotation would occur like this if it were 4n electrons? What system would you look at? How would you get four electrons rather than six? What starting material would you use? Any ideas? Here you start with hexatriene so what do you go to use? Nathan?

Student: A diene?

Professor Michael McBride: The diene, butadiene. So we’ll do this, we will start with a butadiene that has methyls substituted, so we know stereochemistry. And we expect it then to do– if it does– let’s see so we got methyl on the right here, and on the left here, and if they do conrotation, they’ll come out on the same side. So that’s what we’ll do, but there’s a problem. And the problem is this, this reaction is exothermic by 16 kilocalories because you’re converting a π bond into a σ bond. But up above it’s not so great because you’re forming a cyclobutane which is strained. And that is uphill 11 kilocalories per mole. The reaction won’t work. That’s not so great.

So we can’t do that reaction, right? We can’t study it with that system. Right? Wrong. We can run it the other direction. Start with the dimethocyclobutane and see which butadiene it forms. Because it’s the same transition state when you run one way and the other. If you have a certain transition state that’s low, you can access it not only coming this direction, but also that same low one coming back. So we can do the same test starting with the cyclic form and coming back. And that was done by Brauman and Archie in 1972, and it’s interesting to see how they made the compound.

This what they needed that cis-dimethylcyclobutane. How did they make it? They made it starting with a cycloaddition where an acetylene adds to a double bond to give a cyclobutene. And it had these carbons attached to it which then they reduced off. We will talk about that reaction later. And they made the OH as a good leaving group with a tosylate. And then they used H^- to do the displacement to convert that into methyl. But the crucial reaction was this cycloaddition. How does that look to you to add a triple bond to a double bond to make a four-membered ring? Good or bad? Bad. It’s 4n electrons, two π bonds coming together. But what do you notice? how do they make it work? They wanted to do that. What was special?

Student: They used light.

Professor Michael McBride: It’s a four-electron cycloaddition that should be bad, but they used light to do it. So they put the electron into the orbital of opposite symmetry, and now it will work. So we already saw that that forms the less stable isomer, so it’s disrotation with six electrons. But this is the experiment they did, at 280 degrees– it’s not an easy reaction– the ring opened up and they got this isomer as expected. So this rotated in to the right, and this rotated out to the right, it’s conrotation. Like that. So it’s disrotation with six electrons, conrotation with four, and what was special about them was not only did they measure that that was 99.9%, they measured how much of the other one there was. There was about 0.005%.

This was really, really careful work they did, and it’s an interesting paper to read. So that says– notice that this is the less stable isomer, because it has a cis bond where these are both trans. So there’s the bias that favors– that 99.9 to 0.005 is 11 kilocalories per mole– but the bias in favor of this kind of rotation must be even higher, because it’s being hurt by the fact that this one is strained. So it’s something greater than 11 kilocalories per mole favoring the transition state for conrotation in that, which supports what we predicted by theory. And if you have an eight-membered ring with four double bonds in it, that reaction happens at -10°, and again it rotates like that, it’s a conrotation. So it’s conrotation for 4n electrons here, or here, n is 0, n is 1. But it’s disrotation for 4n+2. So it’s what we expected for these electrocyclic reactions.

And we can look at the transition states for the two and see that that’s what would happen if you could run it forwards, and there’s the transition state. This is the same kind of thing we did before. I’m not going to dwell on this now.

Chapter 4. How Bad is “Forbidden”? Opening Dewar Benzene [00:44:27]

Now, but a particular example of that kind of thing is opening Dewar benzene. Remember last semester we talked about Dewar who sent Kekulé these models of all the different possible structures for benzene, and one of them was this which came to be called Dewar benzene, even though he didn’t advocate that as a structure benzene. He thought it was this, the same way other people did, but he said it’s a conceivable structure of benzene. But notice that this one would be strained, and that one would be stable.

So if you could go from this to this, if you could get this, it would be way down hill in energy, because this is unusually stable being aromatic. This is unusually unstable, because it has two four-membered rings. So it would really help to do that. So it should be a really easy reaction. It really wants to open up. Now does it? Well in 2004, there was a paper where they calculated the energy for isomers of benzene, and they had all these isomers of benzene, there’s benzene, and there were 218 isomers. And, in fact, 84 of them are calculated to be within 100 kilocalories, benzene is the lowest one, but there are many within the strength of a bond above it.

And, in fact, six of them have been prepared, but they don’t go to benzene, because breaking a single bond– these are more than 100 kilocalories above. So if you could go to benzene by breaking a bond it would be very favorable, but you can’t, because if you look at these structures breaking a single bond won’t take you to benzene, you have to break more bonds than that.

However, Dewar benzene, which is this one here, is 74 kilocalories above benzene, and you only have to break one bond to make a change into benzene. But it lasts for two days at room temperature. Why doesn’t it go way downhill? How was it prepared? It was prepared by van Tamelen and Pappas in 1963. And they did it, you notice by an electrocyclic reaction where they take a diene and make it into a cyclobutene. And it had disrotation so they used light to do it again, another example. But it has a half-life of two days at room temperature whereas this one, which is only 11 kilocalories exothermic. Remember that was the one that had to be heated to 280° in order to go, so that one has a barrier of 33 kilocalories per mole. But this one only 25, not what you would expect on the simple analysis. So it’s 66 kilocalories more exothermic, but only 8 kilocalories per mole faster. It is faster, but not very much faster.

So this one, notice, opens conrotatory. We talked about that, that’s what Brauman showed, and that’s what you expect for 4n electrons where n is 2 to make the two double bonds. But in this case– and the HOMOs and LUMOs work there, but in this case if you want to open it up to make benzene, this has to go that way, and that has to go that way to become planer. If one of them went the other direction, you’d have really weird bond angles. Benzene would look like this instead of being a hexagon, it would look like this if you rotate it the wrong way.

So you have to rotate it in the wrong way to do that. But wait a second, how many electrons here? There’s going to be one on that p orbital, one there, one here, one here, one here, that’s six. Six π electrons is supposed to be disrotation. Shouldn’t that work? No. It looks like there’s something more fundamental. It’s more fundamental that when you’re breaking this bond here and making σ*, it doesn’t overlap with the HOMOs here and here to stabilize. So even though it’s the right number of electrons the HOMOs and LUMOs aren’t lining up right. So it’s not numerology we’re fundamentally interested in, it’s whether HOMOs and LUMOs match to get stability.

The next topic is going to be spectroscopy, and we’ll get to that next Wednesday after the exam on Monday.

[end of transcript]

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