CHEM 125a: Freshman Organic Chemistry I

Lecture 32

 - Stereotopicity and Baeyer Strain Theory


Why ethane has a rotational barrier is still debatable. Analyzing conformational and configurational stereotopicity relationships among constitutionally equivalent groups reveals a subtle discrimination in enzyme reactions. When Baeyer suggested strain-induced reactivity due to distorting bond angles away from those in an ideal tetrahedron, he assumed that the cyclohexane ring is flat. He was soon corrected by clever Sachse, but Sachse’s weakness in rhetoric led to a quarter-century of confusion.

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Freshman Organic Chemistry I

CHEM 125a - Lecture 32 - Stereotopicity and Baeyer Strain Theory

Chapter 1. What Is the Source of the Rotational Barrier in Ethane? [00:00:00]

Professor Michael McBride: Okay, so we were talking last time about rotation in ethane, and the fact that there was a barrier, and that you can measure the barrier. And one way to measure the barrier, the first way that the barrier was measured, was remarkably enough, by the heat capacity of ethane; how much heat it takes to warm it to a certain temperature. So that showed that the barrier was about three kcal/mole, but it didn’t say what the geometry of minimum energy was. Was it eclipsed, as shown by the red line here, so that the eclipsed form is the lowest in energy? Or was the favored form staggered? And people differed in their interpretation of that. Ultimately there was some diffraction study, electron diffraction, that showed that it was staggered.

But now the best word on this nowadays is probably from really high-quality calculations. And you can see from this paper, published in the Journal of Chemical Physics in 2003, that there’s pretty good agreement. All these various abbreviations are various really high-quality computational results. You can see at the top, our experimental results, from heat capacity. It’s about 1000 wavenumbers. A wavenumber is 2.86 small calories, not kilocalories, per mole. So there’s heat capacity from infrared spectroscopy, from microwave, from Raman spectroscopy. All these experiments give very close to 2.9 kcal/mole, or about 1000 wave numbers — right? — within a couple of percent; even less than that probably. And then all these calculations give about 2.7 kcal/mole. And it doesn’t make very much difference. The first one of these — some of these aren’t such very high-quality calculations, but some of them are very, very high quality, the best that can be done. And they all agree pretty much. So it’s clear that the staggered form is favored over the eclipsed; despite what everyone wrote in the nineteenth century, when they always wrote them eclipsed, and what whoever it was that drew the picture for the American Chemical Society of their Molecule of the Week, last week, drew too. He drew it eclipsed; or she. Okay?

So anyhow, but then there’s the question why? Why is it that the staggered form is lower in energy than the eclipsed? Why is there a three kilocalorie barrier? Well there’s different points of view on that. One is that the eclipsed form is unstable; that for some reason having it eclipsed makes it unstable. The other one is that the staggered form is unusually stable. And what does this remind you of, in terms of a question?

Students: Compared to what?

Professor Michael McBride: Compared to what? What’s unusual, right? Is what’s unusual the eclipsed form or the staggered form? Well from the point of view of the eclipsed form, why would it be unstable? Well it could be that the hydrogens repel one another, because just in space; their van der Waals radii. Okay, but if you think about that, then if you had a methyl in place of one of those hydrogens, it’s ever so much bigger. Therefore you should have a much bigger barrier to rotation, of a methyl group in propane. But, in fact, the barrier is almost exactly the same; 3.4, not ten or something like that. So the size of hydrogen doesn’t seem to be so very important. Of course, protons repel one another, and the electrons that are in the bonds repel one another; there’s electron repulsion. So that might be why you don’t want to have it eclipsed; that you want to rotate it a little bit to make it staggered. But that’s not easy to do in your head, because there are also attractions between a proton on one side, and the electron on the other, and it’s not clear which of these is going to dominate. So anyhow, there’s the possibility that it has to do with the eclipsed form being unusually destabilized — right? — because of repulsions; that could be.

The other point of view is that the staggered form is unusually stable. Now why could that be? It’s that you have a σ on one side and a σ* on the other side of these C-H bonds; and vice-versa as well. And you can get overlap between this HOMO and LUMO in this way. And it turns out, surprisingly enough, that the overlap is bigger when they’re anti than when they’re eclipsed. You might think, intuitively, that there’d be more overlap when they’re on the same side, than when they’re on the opposite. But it doesn’t take too much work to convince yourself that the other is in fact true, and that this is the better overlap. So you’ll get better HOMO/LUMO interaction from one side to the other; and not only for these two hydrogens, but for the other two pairs of hydrogens as well, if the thing is staggered, so that the opposite ones are parallel to one another. So you can get this kind of mixing, HOMO/LUMO mixing, among σ bonds, which is called hyperconjugation. Conjugation is when normal double bonds, p orbitals overlap, from one to the other, and you get resonance structure. The name “hyperconjugation” was created to talk about the same phenomenon when it’s σ bonds, rather than π bonds that are doing the trick; and it’s usually much, much less important. We’ll talk about more examples of this later on.

But at any rate, that’s a very different point of view from the first one. Which is it, that the eclipsed is destabilized or that the staggered is unusually stable? Or maybe both. And, in fact, this is a little bit a scholastic argument analogous to how many angels could dance on the head of a pin; and you get people that debate about this. Right? So the point is that fundamentally there’s quantum mechanics that controls this, but it doesn’t say which one of these things is what. It doesn’t divide it that way. So some people say one, some people say the other, and we’ll just say maybe a little of both.

Chapter 2. Topicity, Reactivity Difference, and Enzyme Specificity [00:06:34]

Okay, now here’s a digression on the question of what’s called “topicity”; which turns out to be relevant in a lot of biochemical applications. So it’s worth mentioning at this point, this aspect of stereochemistry. Topicity relates to, for example, the question: are two protons equivalent? Now at first glance, that’s sort of a stupid question. Obviously a proton is a proton, so they’re equivalent. But suppose one is the H of an OH group in ethanol, and the other is an H of the methyl group. Now are they equivalent? Well protons clearly still are equivalent. Protons are protons. What’s different about them? What’s different is their environment. So it could be that the place they are is different, even though the protons intrinsically are the same, and therefore they could have different properties. Now, that happens to be the case in ethanol. An OH group has a pKa of about 16. It’s not very acidic but it’s possible to dissociate H+ and leave O-. Right? But an H that’s on a methyl group doesn’t do that. It has a pKa of about 50. So it’s 1034th weaker as an acid, than the OH group. So there’s clearly a difference between these two hydrogens because of the place they’re in. Right? So that the OH on the alcohol group exchanges readily with acidic water; so you can put deuterium in and out of that, wash it in, wash it out; and you can’t do that in the case of methane. So D+ can come in. Low LUMO, attacked by the high HOMO. You get that. And then unzip it the other way. The electrons go back on, H+ comes off. So you can do this exchange. That doesn’t happen over on the other case because it’s so hard to get the proton off, and the carbon has no unshared pair to pick up the new proton.

So these two protons, the red one and the blue one, the OH and the methyl-H, are called “heterotopic”; hetero means different, and the Greek topos means place. So they’re in a different place. They have different properties because they’re in a different place. So ‘topicity’ comes from the same root obviously, and it means the placeness of a group. It could be a hydrogen, it could be another group, two groups, that appear to be the same but are in different places. So this is a pretty obvious case. But there are cases that are less obvious. So protons within the blue group are homotopic, and the green protons are homotopic with one another. Right? They’re in the same kind of place. At least they’re in the same kind of place if we’re discussing constitution, what they’re bonded to, and what those things are bonded to and so on. Right? The nature and sequence of bonds are obviously the same between the two green or among the three blue hydrogens. And both of those are different from one another, and from the red hydrogen. Okay, so from the point of constitution, the blue ones are homotopic, the green ones are homotopic. The red one is its own — right? — but the green is heterotopic with respect to the red or the blue, and so on. So that isn’t telling you much that you wouldn’t already have realized easily.

But if you get into stereochemical topicity, it gets more subtle. So let’s look at stereotopic relationships among these protons. First, we said that the blue ones were homotopic. But are they really homotopic? Let’s consider those two hydrogens. Do they have the same environment? They’re clearly the same with respect to constitution, what they’re linked to — right? — but are they the same stereochemically? Are their environments superimposable, exactly on one another? Yes or no? So are they stereochemically homotopic? Corey, what do you say?

Student: I’d say yes.

Professor Michael McBride: You say they are?

Student: Yes.

Professor Michael McBride: If you took the top hydrogen away, and looked at everything else, and then made another model where you took the bottom one away, and looked at everything else, are those everything elses, the places, superimposable? No, they’re not, because one has a hydrogen down here missing, and the other one has a hydrogen up there missing. You can’t superimpose those unless you do a rotation. Right? So that’s a question of conformation, of rotation, right? Okay, so those are diastereotopic from the point of — that is, different places — from the point of view of conformation, but not from the point of view of [laughs] constitution. Right? Sorry, I’m getting my c’s confused. It’s a long run, from the beginning of September to Thanksgiving. You’ll find that this will have been the longest period, unrelieved period, of your time at Yale. It gets easier after this. Okay, so anyhow they’re diastereotopic from the point of view of conformation, but homotopic from the point of view of constitution. How about those two? Do they have the same environment, those two hydrogens? Cathy, what do you say?

Student: Yes.

Professor Michael McBride: They’re superimposable; that is, if you took the H in front away, you’d have a certain bunch of stuff left; that’s its environment. And if you took the H in back away, and looked at the rest of the stuff, are those two remainders absolutely identical; superimposable?

Student: No.

Professor Michael McBride: What is the relationship between them? Obviously that was sort of a leading question.

Student: They’re mirror images.

Professor Michael McBride: They’re mirror images. So what would you call those two protons?

Student: Enantiomers.

Professor Michael McBride: Not enantiomers. It’s their places that are different. Enantio-what?

Student: Enantiotopic.

Professor Michael McBride: Enantiotopic. So the ones at the top are enantiotopic, and the first ones we looked at were diastereotopic. Right? Now, do you care? Is that going to make a difference in chemistry? I say no, and the reason I say no is that it rotates so fast around a single bond — we talked about that last time; 1010th per second. And when you rotate, you exchange one — one takes the place of another, right? So they exchange places among themselves faster than almost anything could happen; 1010thper second. So who cares? Right? So in truth they’re conformationally diastereotopic, but you don’t care about conformation, in this case, because they would rotate so quickly. So they’re not going to have experimentally distinguishable properties, probably, unless you can look pretty quick. Okay? Now, so these distinctions are only conformational and erased by rotation; in 10-12 seconds I said here. Okay? Yeah, that’s more like it. Okay, now how about the two green ones? What’s the relationship between them? Pat, what do you say? What’s the relationship?

Student: They’re enantiotopic.

Professor Michael McBride: They’re enantiotopic. Now, are they enantiotopic with respect to conformation? Can you rotate one and give it exactly the same place as the other one?

Student: No.

Professor Michael McBride: Why not? What would happen if you tried to rotate, to put the back green one in the place of the front green one?

Student: It’d change the location of the OH.

Professor Michael McBride: It would change the location of the OH. Right? So how could you exchange their environments? The only way, we’d have to break a bond and put the other hydrogen someplace else. So what kind of enantiotopic are these? Is it configuration, conformation, or constitution, that you have to change? Can you do it just by rotation? Is it just conformation? No. Do you have to change what’s bonded to what, to make one? No. So it’s not — so it’s configuration, right? But the important thing is you have to break a bond to change them. Right? So that’s not going to happen easily. So the difference between those two — those are also enantiotopic, but configurationally enantiotopic. So that will last, that distinction, in principle, as long as the bonds endure. So there’s quite a difference. Those two could be different. Now let’s see if we can see a case where they’re different. Now that carbon is not a stereogenic center. Why is it not a — how do you recognized a stereogenic center? Nate?

Student: By what it’s attached to — it’s bonded to four different things.

Professor Michael McBride: You guys shouldn’t sit next to each other. I was talking to the other Nate.

Student: It’s got two H’s on it.

Professor Michael McBride: Can’t hear.

Student: It’s got two hydrogens on it.

Professor Michael McBride: It’s got two hydrogens. It has four different things, right? And this has two of the same. But it’s called, for this purpose of topicity stuff, “prochiral”; that is, it will become chiral if two of the things that are the same become different. For example, if one of them, instead of an H, were a deuterium, then it would be a chiral center. So that thing is called prochiral. It would be chiral if the enantiotopic atoms, or groups — it could be chlorines or it could be methyl groups or whatever; if they were made different somehow, then it would be chiral. Now, so in ethanol, those two green groups are enantiotopic and configurationally. So now “toponymy” — that’s names of places; how do we name them? So we have things — we have groups that are constitutionally homotopic. They have the same constitution, connected to the same things.

Okay, so consider those, the green hydrogen there, the light green one in front; and I’m drawing a Newman projection also to show it. We have to be able to give it a name so people will know what we’re talking about, when we’re — if this turns out to be important for some reason. Okay, so the way we do it is to assign priority, if the hydrogens were different. Right? Then we’ll be able to call it R or S, right? Because we already have a scheme for that. But the two hydrogens would have to be different. So let’s look at the priority. OH at that carbon — the prochiral carbon — is obviously top, and the methyl, the carbon group attached to it, is second; but the two hydrogens come third and fourth. But which one should be third and which one should be fourth? Now here’s the rule. You give higher priority to the one that you’re naming. Right? So if we want to name this one, we’d give it higher priority than this one. So this one will be three and this one will be four. So here’s the lowest priority. You can run one to four, in either direction, high to low or low to high. You get the same sense of handedness. So let’s think about this. Here we have — so if I put my left thumb coming out where this hydrogen is, the number four, and curl my fingers, I go one, two, three,. Okay? So this one would be left-handed, if it were higher priority. Everybody with me on that? So we’ll call that, not S, we’ll call that hydrogen pro-S, because it will become S — okay? — if we make this — for purposes of naming, when we give it a higher priority. Okay? Now, suppose we look at the other hydrogen, that I’ve shown in the darker arrow here. Right? It obviously is going to be pro-R, because the same thing would run in the opposite direction. Okay, so pro-R and pro-S are the name that you can give to distinguish groups that are enantiotopic.

Now, is there a reactivity difference that we would care about between these groups? Suppose you bring up a chlorine atom, and it can attack one or it can attack the other. So suppose it attacks the one on the left. Right? Its SOMO mixes with one of the electrons of the H, and the other electron on the H goes on the carbon, and we break the bond and get HCl. This is the first step, remember, in free radical substitution. Or we could do it with the other one. What’s the relationship between those two pathways? Can you see? The pathway that happens on the left and the pathway that happens on the right, are they the same?

Student: [Inaudible].

Professor Michael McBride: Well they’re not superimposable, okay? So they’re not precisely the same. But what are they? Sophie?

Student: They’re mirror images.

Professor Michael McBride: They’re mirror images of one another. Therefore they have the same energy. So one will be just as — remember, we talked last time about how fast a reaction goes has to do with how high an energy you have to go to, along that path. Right? And these are going to be the same. So they’ll have precisely the same rates. So who cares? Okay? So attacked by a reagent like a chlorine atom, those two versions of the attack, are mirror images, and thus they’re identical in their rate. So who cares? But suppose that you use something that’s handed, like a right hand, to pull it off? Okay? Now suppose you use the same right hand to pull the other one off? Those aren’t mirror images of one another anymore. Right? What do you call — what’s the relationship between these two reactions? What would you call the relationship between them? Angela?

Student: They come out completely different.

Professor Michael McBride: Can’t hear.

Student: They come out completely different. So they’re diastereomers.

Professor Michael McBride: Right, they’re diastereomeric paths. So they’re different in energy. So one will be faster than the other. What does that tell you about the metabolism of ethanol? If you’ve got ethanol inside you, there are things that pull the hydrogen off and make aldehyde out of it. In fact, that’s where the name “aldehyde” comes from. Aldehyde is an alcohol that’s been dehydrogenated; al-de-hyde. It’s interesting, isn’t it? But anyhow, this can be — would your body, if your body was trying to metabolize ethanol, could it distinguish between those two hydrogens?

Student: Yes.

Professor Michael McBride: Why? Who said yes? Sam?

Student: Because whatever enzyme is doing that is going to be a hand.

Professor Michael McBride: Because the enzyme that does it is a hand, a single hand. Right? So it will be able to distinguish between those. Now, can we prove that? Okay? So attacks by a resolved chiral reagent, like an enzyme, are diastereomeric and should have different rates. So horse liver alcohol dehydrogenase removes only the pro-R hydrogen from ethanol. Pardon me, if it removes only the pro-R hydrogen in this deuteroethanol, it should remove the hydrogen and never deuterium, in the case that it’s deuterated like this. Right? Messed that up saying it, but you can read it better than I can. Okay? Now, so that would be a great experiment. If you had this compound, (S)-1-deuteroethanol, and you reacted it with this enzyme, if you always pulled off the H and left deuteroacetaldehyde, that would show it was specific. But if it weren’t specific, then you’d sometimes get one and sometimes the other coming off, and you’d get different acetaldehydes. Some would have deuterium and some wouldn’t, left behind. Okay? Now what’s the problem in doing this experiment? Zack?

Student: How do you know that deuterium’s right there?

Professor Michael McBride: How do you get the compound with the deuterium on only one side in the first place, so you can do the experiment? And this is where there’s a clever way of doing it. Right? That would be the reaction, and you’d get only the deuterium left and never hydrogen left, if it were specific. So that’s a good test, but where do you get the starting material? Now actually, the alcohol dehydrogenase is a catalyst, right? And a catalyst lowers the energy of the barrier you have to go through to go across; lowers the barrier. But that means it lowers the barrier in either direction; because you can go either direction across this. Right? So that means that you can — this is an oxidation; removing hydrogen is an oxidation. But you could also do the reverse reaction; a reduction. That is, you could start with this. So what really does the oxidation — LAD is a catalyst — what does it is this NAD+ becoming NADH, taking H- away. Right? So if you ran this reaction backwards, and started with a deuterium there, then if the thing were specific, you’d put hydrogen only there, when you run it backwards; if it’s specific taking it off, it’ll be specific putting it on. Right? Now how would you know if it’s specific, putting it on? How would you know if it did that? How would you know whether, when it does this reaction, it scrambles and puts it in both positions, and then when you run it backwards it scrambles and takes it off both positions, sometimes one, sometimes the other; or whether it’s specific going on and specific coming off? Lucas, you got an idea?

Student: You can pull it off and check optical activity.

Professor Michael McBride: Ah, if you could measure the optical activity of this stuff, that would be a good way of doing it. But it doesn’t have very much optical activity, if the only difference is between deuterium and hydrogen; not very much.

Student: This is just kind of guess, but if they did it to both things, wouldn’t you get one with both and then one with only one? And then if it only went to one side, you’d only get one with one.

Professor Michael McBride: I think you said it.

Student: Probably not right.

Professor Michael McBride: But I don’t know that everyone understood what you said. The idea is do both. Put on, and then take that product and take it off again. If it’s specific going on, and specific going off, when you go through the whole cycle you’ll get back where you started from. But if it mixes up going on and coming off, then sometimes it will put it on the wrong place, sometimes it will take it off the wrong place. And when you come back, sometimes you’ll have H and sometimes you’ll have D, in that starting material. Right? So you go both ways and see. And that works, right? By starting with the same catalyst and excess deuteride, you do that, to go that way. Okay? So if you do a full cycle, like that, then you come back — pardon me — you come back exactly where you started. But, this proves the specificity; but it doesn’t say which one, it doesn’t say whether it was pro-R or pro-S. For that you need to do other kinds of experiments that we don’t have time to talk about. Okay? So that’s really a neat experiment, that shows that topicity makes a difference, and that enzymes discriminate between stereotopic, enantiotopic groups, or atoms.

Chapter 3. Baeyer’s Strain-Induced Reactivity Theory: Assumptions and Weaknesses [00:27:49]

Okay, next subject is Baeyer Strain Theory, which was proposed in 1885. So that’s ten years after van’t Hoff. Okay? So here’s the group in Munich, in 1893, and here, front and center, is the boss of the group, Adolf von Baeyer. And this picture hangs in the hallway out there, the original picture. You can look at it. And the reason it does is that this guy here is Henry Lord Wheeler. Did you ever see his name?

Student: [inaudible].

Professor Michael McBride: See how observant you are.

Student: [inaudible].

Professor Michael McBride: Pardon? Lexy, did you — Pat?

Student: Is it down in the foyer to this building?

Professor Michael McBride: Yeah, when you come in, there are these names carved on the walls, which were the people who were, in the nineteenth, early-twentieth century, professors. So he was the first organic chemist at Yale, and he went to Munich, where a lot of people went to learn organic chemistry from Baeyer. Baeyer, remember, was the guy who fiddled with the bread rolls, together with Fischer, to make these models. He was also the guy that did experiments on arsenic in Kekulé’s kitchen when he was a student. So Baeyer was the leading organic chemist of the time. Now, in 1885 he — that’s earlier than this picture; that picture was 1893, when Henry Lord Wheeler was there. Notice that he died at what; at thirty-seven, the age of thirty-seven [correction” forty-seven]. I don’t know what he died of, in 1911 [correction: 1914]. So he wasn’t a professor very long.

Okay, so this paper, eight years earlier, was about polyacetylene compounds; so a bunch of triple bonds arranged in a row. We’ve talked about double bonds in a row; you can have triple bonds in a row too. But you don’t have them for very long because, as Baeyer reported in this paper, they explode. He had a collaborator named Dr. Homolka, who he thanks for doing some of the experiments in this paper. Dr. Homolka does not appear in the 1893 picture. [Laughter] I suspect that he just graduated, got his degree and went to be gainfully employed someplace. But who knows? Anyhow, polyacetylenes are explosive. And this got Baeyer thinking, why should just a regular old hydrocarbon be explosive? So he branched out in this paper, and the very first topic was “The Theory of Ring Closure and the Double Bond.” Now that seems a funny thing to talk about; when you’re talking about triple bonds, to talk about making rings, and talk about double bonds. But what he says is: “Ring closure” which was a popular synthetic goal, at that time, to make new kinds of compounds, was to try to make rings. They could make six-membered rings, they could make five-membered rings. In Bayer’s lab they actually made four-membered rings and three-membered rings, right? Although that was quite a chore. Okay, so “Ring closure is apparently the only phenomenon that can supply information about the arrangement of atoms in space.” This is ten years after van’t Hoff. “Since a chain of five or six members can be closed easily, while one of more or fewer members is difficult or impossible, spatial factors are apparently involved.” And you know, we already talked about a case like this, where you tell something about arrangement in space from the ability to form a ring. Do you remember what that is? Anybody? Yeah?

Student: Synthesis of mesitylene.

Professor Michael McBride: Can’t hear very well.

Student: Synthesis of mesitylene.

Professor Michael McBride: Oh, the synthesis of mesitylene formed a ring. But that doesn’t — so he used that to say where the methyl groups were on the ring. And that’s actually a good example, but it’s not the one I was thinking about.

Student: Cyclopropane.

Student: Cyclopropane?

Professor Michael McBride: Dichloropropane?

Student: No.

Professor Michael McBride: Is that what you said?

Student: No, I said cyclopropane with the [inaudible].

Professor Michael McBride: It’s not cyclopropane that I had in mind; although that was one of the compounds that was synthesized in Baeyer’s lab for the first time, and found to be reactive. But there was another occasion. Remember maleic and fumaric acid; one was cis and one was trans? And the one that was cis could lose water to form a ring. Right? So that was what he’s referring to here. Okay, so then he goes on to say: “The previously — ” (So there’s a website you can click on to see this.) “The previously proposed general rules on the nature of carbon atoms are the following: I. Carbon is tetravalent.” Who did that?

Student: Couper and Kekulé.

Professor Michael McBride: Couper and Kekulé, right? “The four valences are equivalent, shown by the fact there’s only one monosubstitution product of methane.” (We’ve talked about that.) “III. The valences are equivalently arranged in space to the corners of a regular tetrahedron.” Who says that carbon is tetrahedral?

Student: van’t Hoff.

Professor Michael McBride: van’t Hoff. Also, incidentally, there was a Frenchman who did it simultaneously, called LeBel. Okay? “IV. The atoms or groups attached to the four valences cannot exchange places.” (So you can’t — you have to break bonds to make new isomers, right? So if you get them one way, they’ll stay.) “The evidence is that there are two tetrasubstitution products, abcd of methane.” (So van’t Hoff and LeBel’s rule, that you can have configurational enantiomers, we would say.) “V. Carbon atoms can bond to one another with one, two, or three valences.” (That is, you have single, double, triple bonds.) “VI. The compounds can form either open chains or rings.” (That’s what he’s talking about here.) “I should like to add the following, to these generally accepted rules. So he’s adding a new property, to the models, the same way van’t Hoff did; van’t Hoff added arrangement in space, to the models that people already drew. What Baeyer is adding is VII.) “The four valences of the carbon atom point to the directions connecting the center of the sphere to the corners of a tetrahedron.” (That’s what van’t Hoff also said — already said.) “But forming an angle of 109°28’ with one another.” (That’s very quantitative and precise. That’s the angle between the center and two vertices in a geometric regular tetrahedron. Okay?) But then this is what’s interesting: “The direction of attachment can undergo alteration” (It doesn’t have to be exactly that angle) “but a strain is generated, increasing with the size of the deflection.” So if you have an angle other than 109°28’, then there’s going to be higher energy associated with strain; or, what he’s actually talking about is not higher energy, but reactivity.

Okay, so he drew this picture, showing how much the angles are distorted from 109.5°. So if you consider ethylene to be a two-membered ring, with two bent bonds, the angle the bonds are going at is where it should be tetrahedral; and in fact they’re collinear. That distortion of each bond is 54°44’, and so on, for three, four; five is perfect, right? It only deviates by 44’. But the six-membered ring is stretched a little bit the other way. It has to go out, rather than sharper. Okay. So he says: “Dimethylene…” — that’s ethylene, the first one — “…is indeed the weakest ring, which can be opened by HBr, bromine, or even iodine.” (We talked about bromine, or chlorine attacking the ethylene already.) “Trimethylene…” (cyclopropane) “…is broken only by hydrogen bromide but not by bromine.” (So it’s not as reactive as the two-membered ring.) “Finally, tetramethylene and hexamethylene are difficult or impossible to break.” They’re pretty stable. So this strain causes reactivity, if the bond angles aren’t right. So he’s becoming more quantitative; unusually quantitative for an organic chemist at this period. Remember, the physical chemists — this is just at the period when physical chemistry is coming into its own — and they were the ones who did heat and energy and so on. All Baeyer is talking about is whether things will react or not. But he’s using geometric precision. Okay?

Chapter 4. Sachse’s Muddled Insights on Cyclohexane [00:36:45]

So the question is, are six-membered rings, like this, cyclohexane, they should be, according to Baeyer, a little bit strained. Right? Because the angles should be — you’d think the angles would be 120° instead of 109. So they’re opened up a little bit too much. But Sachse, a young privatdocent — that is, essentially, between a graduate student and a professor — in 1890 published a paper that had this very funny picture, A,B,C,D,E,F, up on the top there. And people didn’t understand what it was. But what it is, is a thing that if you trace it on cardboard and fold along the diagonals, you get this blue thing — right?; if you fold it along the diagonals and paste the ends together. Right? And then if you paste van’t Hoff tetrahedra on it, you get this thing. And you’ll see that this is just like this. Everybody see that? It’s using — but this is a base on which to build such tetrahedral carbons, so as to make this structure. And what’s interesting about it is that these angles are the normal tetrahedral angles. Right? They’re not strained. Right? Why are they not 120°? Because the ring isn’t flat, it’s puckered. Okay? So that’s what Sachse said. And he gave people this thing so they could make their own models and see it. Right? And he gave the directions for building this more complicated base, where you put it on and have a different form of cyclohexane. Okay? And that one, as you see, looks like this. Okay?

So in 1890 he knew exactly the score that we would know — that we would get with our own models nowadays. Now, when this was abstracted in the journal that published abstracts of all chemical literature so people could go there and find out what was going on in chemistry, Julius Wagner, who abstracted it, said: “It is not possible to write an abstract of this paper, especially since the author’s explanations are hardly understandable, without models.” Now, one conclusion from that should be therefore go and build his models and look at them, and you’ll see what he’s talking about. But that’s not what Wagner said, and it’s not what people did. He said, “Forget that, it’s nonsense” — right? — “you can’t understand it.” Okay. So Baeyer wasn’t happy with this, disputing his theory. So he wrote, in the same year, 1890 — Sachse wasn’t too smart about this because he published that paper just as they were having a big celebration of Baeyer’s contributions to aromatic chemistry in Berlin; there was this big, fabulous dinner with eighteen courses, or something like that, and all — everybody, chemists from all over Germany gathered. And Sachse, this nobody, publishes a paper and says Baeyer’s theory is wrong. Bad timing.

So Baeyer published a response quickly. He said: “A further proposal is that the atoms in hexamethylene…” (six-CH2 groups, cyclohexane) “…are arranged as in Kekulé’s model.” (Now Kekulé’s model, remember, by this time was in fact a tetrahedron; which is what Sachse showed how you could do it. Right?) But he says, “…as in Kekulé’s model, and that the arrangement of the atoms in space is one with minimum distortion of the valence directions.” (That sounds like Sachse.) But what he says is: “Thus, the six carbon atoms must lie on one plane.” (It’s got to be planar.) “and six hydrogen atoms lie in equidistant parallel planes.” (Six above, six below. All the CH2s are just like this around the ring. Right?) “Further, each of the twelve hydrogen atoms must have the same position relative to the other seventeen.” (That is, they’re homotopic; you can rotate it, if the plane is flat. Now how can he say this? It’s not really like Kekulé’s model. Okay?) “The experimental test of the correctness of this assumption is relatively easy; for example sufficient evidence is that there is a single isomer of hexahydrobenzoic acid.”

That this, only one cyclohexane carboxylic acid. That is, if you had this, and put a COOH group here, or a COOH group here, what would be the relationship between those? This place and this place, what would you call — are they homotopic? Are they enantiotopic, this place and this place? Mirror images? No, they’re just different. They’re diastereotopic, right? Now, but you could do a neat trick. So that’s what he says. If it were like Sachse, then you’d have two isomers of a monosubstituted cyclohexane. But what he didn’t take into account was this. This one is sticking up like that. Watch this. Notice the black one is sticking out, down at an angle, and the white one is sticking straight up. But watch this. Now this one’s down at an angle and the black one’s straight up. Did I break any bonds? What did I do?

Student: Rotated.

Professor Michael McBride: I just rotated about bonds. They’re conformational; conformationally diastereotopic. Not configurationally. So they’re different but they can interconvert easily. Right? So Baeyer says he’s right because there’s only one isomer, so you have to regard it as flat, even if it isn’t; although he didn’t say that. “Meanwhile, as long as our knowledge in the field is so incomplete, we must be satisfied that the above assumption is the most likely, and no known fact contradicts it.” Okay?

Now, Sachse didn’t take this lying down either. So he publishes a forty-one page paper in Zeitschrift für Physikalische Chemie, which is published by Ostwald, who hates Baeyer, and vice-versa; the physical chemists and the organic chemists didn’t get along. So he publishes a paper in this other one where he gives all this trigonometry to explain what he means by this structure. Is this something calculated to appeal to organic chemists? Not very likely. Okay?

So this is edited by Ostwald, who didn’t even believe in atoms, remember, and wrote disparagingly of his successor in Riga, who had been an organic chemist: “Scientifically, he had been brought up in the narrow circle of contemporary organic chemistry, and to him the arrangement in space of the atoms of organic compounds was the foremost of all conceivable problems.” That wasn’t what Ostwald thought was the best problem. So Sachse published again. Nobody responded to that first one, so he publishes again, the next year, thirty-four pages again. And here he gives more formulas; again, things that didn’t appeal at all to anybody who was actually interested in the arrangement of atoms in space. And then he died at the age of thirty-one, in that year, and that was the end of Sachse, and the end of his theory. Right? And Baeyer wrote in 1905 — so twelve years later: “Sachse disagreed with my opinion that larger rings are planar. He is certainly right from a mathematical point of view;” (Although I sort of doubt that Baeyer went through every line of this to check it. Right?) “Yet, in reality, strangely enough, my theory appears to be correct. The reason is not clear.” Okay? So that was — remember, Sachse was 1890. So what important lesson should we take from the tale of poor Sachse? Do you want me to give you that as a problem to think about over Thanksgiving? Fine.

Okay, so this, remember is what Sachse wrote, and he drew models like this. And — I’ll just conclude the punch line here — in 1913 Bragg and Bragg determined the diamond structure, by X-ray diffraction of a crystalline diamond. Right? And five years later, Ernst Mohr published a diagram he drew of — from that diamond structure. Braggs didn’t draw the structure, but Mohr did five years later. And here’s what he drew, and that is exactly what Sachse meant. Right? So the arrangement of the carbons in a six-membered ring, in diamond, is exactly what Sachse said; but he got nothing for it. Okay, have a good Thanksgiving.

[end of transcript]

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