Freshman Organic Chemistry I

CHEM 125a - Lecture 31 - Preparing Single Enantiomers and Conformational Energy

Chapter 1. Introduction: Legal Implications of Chirality [00:00:00]

Professor Michael McBride: Last time we heard about some medical considerations regarding stereochemistry, and we’re going to just start today with a little bit about legal considerations about stereochemistry. First, the doctor gave a disclosure last time about her connections. So I’ll do the same thing. I’ve served as a scientific consultant or expert witness to a number of pharmaceutical companies, including Eisai; which is how I knew about that 7389. I take Lipitor, and served as an expert witness for a generic competitor in a case involving the validity of a Canadian Lipitor patent by Pfizer. And my only connection to AstraZeneca and omeprazole is as an occasional consumer of Prilosec. Okay?

This book here is very heavy. It’s about the stereochemistry of organic compounds. And interestingly, it’s from 1994; so it’s less than fifteen years old; and it’s still the bible. And only eight pages of it are about biological properties, and just one of those pages had anything to do with legal things; which had to do, incidentally, with cocaine. It turned out that somebody was on trial for cocaine and the defense had the idea that the prosecution had not tested the cocaine to see which enantiomer it was, [laughter] as to whether it was — might have been artificial; because the enantiomer of cocaine, there’s no law against. It also wouldn’t work. But anyhow, they hadn’t tested it. So they thought they could get their client off. I didn’t actually look up the case to see whether they got off or not. But that’s an interesting stereochemical application. But one that made a big difference was the court rejecting a suit about Nexium marketing. We saw last time, and you did a problem for today, about Nexium and whether the advertising seemed convincing to you; or fair, for that matter. There was a big suit about that, and this — in 2005 there was a news report about that.

“A federal court in Delaware dismissed a class-action lawsuit” alleging AstraZeneca has had misleading marketing of Nexium and thus added billions to healthcare costs. And the “district court judge, Sue Robinson, rejected the suit brought by the Pennsylvania Employee Benefit Trust Fund on behalf of entities that foot the bill in healthcare plans. According to the health plan paying organizations, the big difference between the two drugs is not effectiveness, but advertising.” (So you presumably looked at some of this and have your own opinion about that.) “By selling doctors and patients on the idea that patented Nexium is better than Prilosec, which faced generic competition, AstraZeneca was able to preserve billions in sales.”

So, for example, the year Nexium was introduced, their advertising budget was 260 million dollars, and in 2005 it had declined slightly to 226 million dollars. But the sales were 5.8 billion dollars. So the advertising was negligible, compared to the cost of advertising [correction: compared to the income from sales]. “Judge Robinson said the courts should defer to the U.S. FDA in weighing the differences between drugs, and that since the FDA cleared Nexium’s label, the lawsuit could not stand.”

So that’s interesting, it shows the importance of the FDA.

Chapter 2. Methods for Isolating Esomeprazole into a Single Enantiomer [00:03:37]

Okay, back to chemistry now. In order to market Nexium, the single enantiomer, AstraZeneca had to, of course, prepare it as a single enantiomer, which meant either resolution — so separation of the right and the left-handed forms — or else preparation of only the one form, the (S)-omeprazole. I noticed that the doctor said eeso-meprazole. Maybe they all pronounce it that way, but the reason it got the name is because it’s S; so I say esomeprazole. So how are you going to do it? Resolution. You could do a Pasteur conglomerate kind of thing. You could make temporary diastereomers and separate them, perhaps by crystallization. Or you could destroy one enantiomer, react the racemate with a resolved chiral reagent that would react with only one, or maybe a catalyst that would catalyze the destruction of only one, like an enzyme. Or you could do it nature’s way and prepare only one enantiomer, either from a resolved chiral starting material, or using a resolved reagent or catalyst. And they tried all of these. So let’s look at it.

First, resolution of omeprazole by chromatography. So they had silica — which is what you use probably in your chromatography — but they coated it with something that was a single enantiomer. Now how did they get something solid that was a single enantiomer that they could coat it with? They got it from nature. It was cellulose; trisphenylcarbamoylcellulose. And they did this in 1990, when they were looking forward to the time that they’d lose patent protection on the racemate. So there’s cellulose and phenyl isocyanate; the thing with two double bonds in a row there. Now, what would make, do you think, phenyl isocyanate reactive? Do you see any functional groups there? Any ideas?

Student: A double bond of carbonyl.

Professor Michael McBride: Okay, carbonyl double bond. Good. And how about the cellulose? What’s going to make it reactive?

Student: Hydroxyls.

Professor Michael McBride: Rick? What makes cellulose reactive, any functional groups?

Student: It looks like mostly hydroxy groups.

Professor Michael McBride: So unshared pairs on oxygen. Okay, so an unshared pair on oxygen can attack the carbonyl. Now, you could in fact use the C-O or C-N double bond. And, in fact, we’ve seen this before. This, remember, is how urea got made, from ammonia and the corresponding acid. This is just a phenyl derivate of that acid. So it does the same thing. And, in fact, you can do it with all the OHs in the group. So you put that carbamoyl group, as it’s called, on all the oxygens. So that is the chiral stuff that was on the solid that the chromatography was conducted with.

Now, they weren’t able to get very much. They did six injections with chromatography, and were able to get three 3 mg of the dextrorotatory and 4 mg of the levorotatory stuff separated by the rate at which they would go through this chiral column. Now that wasn’t enough for any kind of human testing or anything. You need 20 mg, remember, for a human dose, for just one dose. But what they could do was measure how fast racemization was; because obviously if the stuff racemized immediately, it would be useless to have a single enantiomer. So they found out that the half-life for racemization was an hour at 75°, which they then extrapolated to what it would be at a body temperature to be 100 hours; so plenty of time. It’s not going to racemize just spontaneously. So at least that was not a trouble. So this very first quick way of resolving it allowed them to see that it was maybe worthwhile continuing.

Okay, then they tried reversible formation of a crystalline mandelate ester. Now that red stuff there is mandelic acid, which is a natural substance. So you have that as the pure S enantiomer. And now you want to react that with the omeprazole, with racemic omeprazole. But you do it with an intermediate compound, formaldehyde, such that the formaldehyde, the carbonyl group, gets attacked from both sides, one at a time, by the nitrogen on the omeprazole and the oxygen on mandelic acid, to put a CH₂ group between them. And we’ll talk about that a lot, that kind of thing, when we get onto the chemistry of aldehydes and ketones. But anyhow, they could do that. Now they have what? They have two things. What’s the stereochemical relationship? Because they had racemic omeprazole they had reacted from (S)-mandelic acid. What are the two — you have two compounds, what’s their relationship?

Student: Diastereomers.

Professor Michael McBride: Russell? They’re diastereomers. So they’re different. So you can crystallize them apart. Okay, so they did that. And then the important thing was that it’s a reversible reaction. So they could go back again, adding water, and get the omeprazole back, after the separation. And now they have both the R and the S versions of them; both the d and l, or the dextrorotatory and the levorotatory forms of omeprazole, in separate bottles. And now they have hundreds of milligrams. So it’s not enough to do a large-scale test, but they can do some biological testing. And with it, they found that the R was four times as active as the S, in rats that they studied. But lo and behold, when they got into humans, it turned out it was reversed, that the S was more active than the R. So that’s why the drug is (S)-omeprazole.

Okay but that’s — but you’re not going to do this to make enough to sell to the public, especially because you don’t want to throw away half the material you make; both because you’re polluting the environment and because you’re wasting the money you put into making it. So they developed a chiral catalyst involving titanium. So there’s a titanium bonded to four alcohols; or actually one of them bridges to another titanium. But anyhow, you can do something called ligand exchange. So the electrons in the bond go onto the oxygen. So you have RO- and Ti+; you’ve broken the bond. Now that goes away and a different one can come in, a different alcohol, and form a new bond. So that exchanges the ligands by losing one and putting on another one. Okay? Now you can do that trick with diethyltartrate. So two of the oxygens, the alcohol oxygens of tartaric acid, can exchange for two of the alcohols that are on the titanium, and you get this compound of titanium. Now what you really want is the titanium. That’s what’s the catalyst. So you can lose another RO- and you get this Ti+. And now’s when the catalysis begins, because you bring in a peroxy compound, a hydroperoxide. So instead of an alcohol, there are two oxygens there. And that can do the same trick and combine with the titanium. So you have this compound. But now you have a different functional group there. You have the O-O bond. And what makes the O-O bond reactive? We’ve seen this before. Kevin?

Student: The high nuclear charge gives the low LUMO.

Professor Michael McBride: And what is the low LUMO?

Student: σ*.

Professor Michael McBride: Right. So σ* is going to be reactive there. Okay? And you can bring in a sulfur — right? — with its unshared pairs, and they’ll react, make a bond, break a bond. Right? So you get the sulfur attached to the oxygen, and it’s S+, since it has used its electrons to make that new bond. Now, that thing can come off, the same way the compound above it did; the RO came off, the sulfur compound could come off. And that’s the sulfoxide; remember, that’s the important group in the (S)-omeprazole; that’s what sets things going. Okay, so we’ve got it. And now that compound on the lower left, the titanium compound, is the same as the one on the upper right. So we’ve made a cycle. Okay? So that means we can look at the cycle this way, a catalytic cycle. We bring in ROO-. It combines with the little machine. Then S comes in and RO- comes out. And then S double bond O comes out. And you get the catalyst back again. You can start at the top and go — so you go round and round and round and round, and every time you have the peroxide come in, the sulfide come in, the RO- comes out, and the SO comes out. So essentially the ROO gives one of its Os, the red one, to the sulfur. Right? But it’s catalyzed by this. Now why is it important for their purposes that this thing be involved in catalysis? We looked last time that you can put an oxygen on sulfur, just with peroxide. You don’t need the catalyst to do that, although it’s faster with the catalyst. Did they do it just to gain speed, or can you see something else? What’s special about the catalyst? Lucas?

Student: Well it’s one enantiomer.

Professor Michael McBride: Pardon? It’s one enantiomer, right? It’s a chiral oxidizing agent. The peroxide alone would’ve been an oxidizing agent, could’ve done the trick. But the chiral oxidizing agent means that it will discriminate between the two hands of the product. Right? It could attack one of the unshared pairs here, or it could attack the other. And those are enantiomers of one another. But when it’s combined with this thing, that has tartaric acid on, which makes it chiral and of one hand, then those are diastereomers, when it attacks this one, or when this one does the attacking. Right? So they won’t be 50:50.

Unfortunately, it was 50:50. It just happened to be the same rate. So did they give up? No, they started fiddling. And one thing they did was to add diisopropylethylamine. Why? Who knows? Right? But they discovered that when you do that, then you get what’s called 94% enantiomer excess. That means the desired enantiomer is there at 97%, and there’s 3% of the wrong enantiomer. Right? So there’s an excess of 94% of the enantiomer you want. So now all you do is crystallize it, and it gets purified. You get rid of the 3%, and you have the stuff you want. And now you can sell it to the public as Nexium. So that’s the way you do it practically, right? When you want to make a lot of stuff you make just one.

Chapter 3. Conformational Isomerism: An Introduction [00:15:15]

Okay, so that’s what we’re going to say about configuration. And now we have one “C” to go. We talked about composition, of course, from the time of Lavoisier; about constitution from the time of Couper and Kekulé. But then there are distinctions based on this bonding model. Note that constitution is already involving the bonding model. But something beyond constitution is how things are arranged in space. So configuration is obviously also a question of the model that you make for bonding, and conformation, which is the “C” we haven’t talked about yet. And those last two are stereoisomers, because they have to do with arrangement in space; which constitution, as you remember, doesn’t. So the difference between configuration and conformation, both of which have to do with arrangement in space, is that the configuration, you have to break bonds to go from one isomer to another, and in the conformation you just rotate about single bonds to get from one isomer to the other. And, in general, it’s hard to break bonds and easier to rotate bonds. So typically configurational isomers last longer than conformational isomers; but not always, there can be special cases.

Now all things that you call isomers represent local energy minima. So the molecule can sit there at that geometry — and of course it has to be vibrating, at least zero-point vibration. So the atoms are moving, but they’re in a well; they stay in the same relative locations, more or less, right? They vibrate in there. But then they can get out of that well and go to a completely different well, and vibrate there. So not all the different phases of vibration are considered different molecules; that would be way too much, to call those all isomers of one another. Those are all one isomer, and these are all another isomer, but there’s some barrier to get from one to the other; so you can have one or the other. And the barriers are typically higher for configuration and lower for conformation.

Okay, now we’ve seen that in the nineteenth century stereochemistry was qualitative. It was a question of counting isomers; not measuring something but just counting how many isomers you could get. Very qualitative, right? But conformation, which didn’t come along until much later, which involves rotational isomerism about single bonds, is much more subtle, and it requires quantitative thought about equilibria, rates and energies. It’s not just this qualitative counting of isomers. So notice that this conformational analysis came relatively late to the timeline we’re talking about. All the other stuff we’re talking about happened before — most of the other stuff, with respect to structure — happened before 1900. But it was only in 1950 that conformational analysis began, and then twenty-five years later that people started doing significant, what’s called molecular mechanics modeling, treating molecules as if they were springs and so on. And we’ll talk about that a little bit later.

So remember there you are, and we can trace your lineage back to P.D. Bartlett at Harvard, who was my Ph.D. supervisor, and his Ph.D. supervisor, J.B. Conant. Now, those guys were involved in a special flavor of organic chemistry which influences you. It’s called physical-organic chemistry. It’s interested not so much in making new compounds as in why compounds behave the way they do. So what you’ve suffered through this semester is because of these guys. Right? We talk much more about why things work — how bonding works, how to recognize a functional group — than we do with memorizing reactions that will get you from here to there. Okay, so Conant was one of the founders of this discipline, because he did a Ph.D. with two advisors, one of whom was a physical chemist and one who was an organic chemist. So if we go back a little bit here, you see the blue, the organic chemist, was Kohler, who we’ve talked about, and the red was T.W. Richards, the physical chemist, who himself studied with Wilhelm Ostwald, who together with van’t Hoff and Arrhenius — these three guys here, van’t Hoff, Ostwald and Arrhenius — those are the guys who founded physical chemistry as a discipline. Okay? So we have a lot more physical chemistry in this course than you normally would in an elementary organic course.

Now, if you go a little bit further in the family tree, you get G.N. Lewis as well, whom we’ve of course talked about. Now, in the late nineteenth century, when these guys were founding physical chemistry, organic chemists focused their efforts on molecular structure, on these things having to do with configuration and enantiomers and so on. But physical chemists focused on energy, sometimes to the exclusion of structure. In fact, Ostwald — oh I forgot to bring his book, I was going to bring to show you; but I have a page from it here. He wrote lots and lots of books, Ostwald, who’s like your four times great-chemical-grandfather, or something. This Principles of Chemistry — which he gave the not too modest title An Introduction to all Chemical Textbooks. So this is what, if you read it, then you were ready to read any other chemical textbook. So it was a general view, in 1907. And on page 421, there’s a very long footnote. And if you look at a little bit of this footnote, it says: “Dalton, who developed the law of combining weights on the basis of an hypothesis he had proposed about the composition of matter from atoms, at first took hydrogen as unity, since it had the smallest ‘atomic weight’, i.e., combining weight.” And so there’s the word ‘atom’. And that’s the only place, in 554 pages of this book, that the word atom occurs. It’s in the subordinate clause of a footnote, because, at this time, Ostwald didn’t believe in atoms. He believed that philosophically, since you couldn’t see them, you shouldn’t talk about them; so you should do things that you can deal with. And energy was such a thing, he thought. So physical chemists, these guys who founded physical chemistry, focused on — of course, van’t Hoff is interesting because he’s the guy that had the idea of the tetrahedral carbon. So he didn’t sign off on this, together with Ostwald. But Ostwald was much more influential, he had a much bigger school of students, and so he was — it was only two years later that he finally accepted the existence of atoms. So I’m just drawing for you the contrast between what organic chemists were thinking about and what physical chemists were. And now, to deal with confirmation, we have to become a little physical chemist to see the influence of energy.

So here’s the dedication of Sterling Chemistry Lab in April 1923, and there’s a group of people that we’ll look at. And all of these guys here were students of Ostwald. So he had a very broad influence. The one there is T.W. Richards; got the first American Nobel Prize in Chemistry. Okay? And he, remember is your something-grandfather. Okay? So then Donnan, the next guy — well there’s a guy from Canada, a guy from Britain, and the guy on the left there was a student of Richards, so a grand-student of Ostwald; with his cigar, as he always had. And this guy is Wilder Bancroft, who was at Cornell, and who founded and owned, for that matter, the Journal of Physical Chemistry. And he wrote an obituary of Ostwald, in 1933, in which he said: “Ostwald’s gift for leadership showed itself in the way his pupils regarded him all through their lives. They usually believed what Ostwald said, even when they knew he was not right.” Okay? And no doubt atoms were what he had in mind; that Oswald didn’t believe in atoms, but his students still believed in atoms. Okay, but they all believed in energy.

Now, did anybody walk past here this morning? You recognize this place? Okay, you take a different path. Okay, but have you noticed what’s off on the left there? “Here Stood the House of Josiah Williard Gibbs, Class of 1858, Professor of Mathematical Physics.” So Yale has a really special connection to energy. There’s Gibbs as a freshman, and here he is in later life. He spent his whole life here at Yale, except for one year he went to Germany. So physical chemists were, and still are, quantitative about equilibrium constants and rate constants. So big K is an equilibrium constant; small k is a rate constant. And they’re related to energy. So you can call energy E, or if you’re more particular you can call it H, for enthalpy, or G for Gibbs free energy; and entropy gets into the argument, as we’ll see in a few lectures. So energy determines what can happen at equilibrium; that is, if you have this stuff, is it possible spontaneously to go to this stuff? If you have forever and ever for it to happen, is it conceivable that this stuff will go to this stuff? It can only happen if it’s downhill in energy, in free energy. Okay? And the equilibrium constant is determined by — is related to energy in this way. And here’s something that’s really handy, and you’ll impress subsequent teachers and your roommates and everyone else is you remember this: that if you know the change in energy, in kilocalories per mole, you can say what the equilibrium constant is, because it’s 10^{(-3/4 of whatever that is)}. So suppose the difference between this and this is 4 kcal/mole. What will the equilibrium constant be?

[Students speak over one another]

Student: It’ll be 1000; 10^{(3/4 of 4)}; 3/4 of 4 is 3; 10³ is 1000. So the equilibrium constant is going be 1000. Now whether it’s 1000 or 1/1000 depends on which way you’re going. But obviously it’s going to favor the one downhill. So you have to keep your thinking cap on to get the right direction. But 10^{(-3/4 of ΔH)}, or — yes, 10^(-3/4) is the way to do it. Okay. But you can also say something about how fast it will go. This is just whether it’s conceivable that it will go; the big K. The little k says how fast. Because rates — this is an approximation, this isn’t written on the tablets by the angel or someone. But for practical purposes, for our purposes, the rate in seconds — that is, how many per second — is 10¹³. So that’s very fast; 10¹³ per second. But then it gets cut down by how big the energy barrier is you have to get across. Cutting down is exactly the same kind of thing you have in the big K, 10^{(-3/4 ΔH)}. So you could again — that energy is called activation energy. It’s how high the barrier is that you have to get over. And you can use the same trick of the 3/4ths. So it’s, the rate constant is 10^{(13-(3/4 of whatever the energy))}. So suppose you had a fourty kilocalorie energy barrier you had to get across; suppose you had a forty kilocalorie energy barrier you had to get across. Forty kilocalorie energy barrier, right? How fast would the rate constant be? Becky, why don’t you help me? So the barrier is 40 kcal that you have to get across. So now how am I going to plug that in here?

Student: 10^13th.

Professor Michael McBride: Okay, so 3/4ths of forty I need, right? So what’s that?

Student: Ten — or thirty.

Professor Michael McBride: Thirty. Okay, so it’s 10^(13-30). Okay, what’s thirteen minus thirty?

Student: Negative twenty-seven [should be negative seventeen].

Professor Michael McBride: Okay, so it’s 10^-27 [correction: 10^-17] per second. Do you think that’s very fast, 10^-27 [correction: 10^-17] ? That’s a pretty small number, right? So forget spontaneously, at room temperature, going over a barrier of 40 kcal/mole, for practical purposes. But you have this T here. So if you increase the absolute temperature, then you make that forbidding factor smaller and smaller and smaller, up in the exponent. So if you heat it up, you can make the reaction go. Okay, so anyhow, that’s really helpful, to remember this 3/4ths trick. Okay, now conformation involves rotational isomerization about single bonds. So the question is, how free is single bond rotation? Paternó said it wasn’t free at all; I can count isomers that way. I can have this isomer, and I can have this isomer, and I can have this isomer — right? — for purposes of counting. But van’t Hoff says no, don’t count isomers like that. Because why? Why shouldn’t you count isomers like that?

[Students speak over one another]

Professor Michael McBride: Because they rotate too easily, right? It’s like that in his picture. Right? So you shouldn’t count this and this and this and this. Okay? Now, there are different kinds of models, something like these, and some of them you can rotate easily, but not others. Right? Which one’s realistic? Right? Or this one you can rotate, it’s just not easy to rotate — right? — it sticks, right? Or this one, remember, that Fischer used; which is good because you can bend it, for his purposes, for drawing two-dimensional Fischer projections. But forget rotating that. Right? It could vibrate, but it can’t — you can’t rotate it, it’ll always come back. Right? So the question is, which model is right? Can you rotate or can’t you rotate, and how hard is it to rotate? Because that’s going to determine how fast these things will be.

Okay, now notice that both of these authors drew their pictures with a conformation which we call eclipsed. Now eclipsed means these. It means this, that the ones on the top are right over ones on the bottom. Okay? And the same is true in van’t Hoff’s tetrahedra — right? — right above the others. Obviously you could rotate 60° on the top and everything would be staggered, not directly above. Okay? Notice, incidentally, that from the exam we looked at the Molecule of the Week, for last week, from the American Chemical Society. And notice how it’s drawn. Right? It also is eclipsed. This red one here is directly opposite this red one here. This acid group is directly opposite that hydrogen, and this acid group is directly opposite that hydrogen; it’s eclipsed, right? And here, this CO₂ is directly above that CO₂; it’s eclipsed, right?

Chapter 4. Newman Projections and Nomenclature for Conformations [00:31:37]

Now you need notation and names to talk about these things. And the projection that was invented for this purpose was invented by someone who sat in the seat that some of you are sitting in. I don’t which. Maybe — he probably, in the course of his undergraduate years and also getting his Ph.D., he probably sat in most of the seats in here; or many of them anyhow. But his name was Melvin Newman. You see he graduated from Yale in ‘29 and got his Ph.D. in ‘32, and then he went to Ohio State. And he invented a way of drawing conformations on paper. So there are two carbons, and the convention is, for the Newman projection, that you sight down the carbon-carbon bond. So I’m going to show this, the rotation around this central bond. Okay? So I look right along that bond, and there’s one carbon in front and one carbon behind, and each of them has three things on it. Okay? So the one behind, here, has three hydrogens, and the one in front hides the one behind. Okay? But it has three other groups on it too. Everybody see how he’s doing this; that that shows this? Well it shows it without the methyl groups here. Okay, and he calls that staggered, right? But you can also have them be right on top of one another, like this. Right? And that’s called eclipsed. And that’s the way it was shown on the previous slide by Paternó, by van’t Hoff, and by the ACS operative who made the Molecule of the Week. Okay?

Now that’s not so convenient, because you can’t see the stuff in back. So sometimes people turn it a little bit that way, or sometimes they turn it a little bit that way; by which they don’t mean that it’s not eclipsed, they just mean I want to be able to show both things. Right? Now that’s very cumbersome, and the reason you don’t mind that it’s cumbersome is that things are never eclipsed, they’re always staggered. So all those previous pictures were wrong, including the Molecule of the Week, because conformations tend to be staggered, not eclipsed. Okay, but anyhow you can draw them this way. Okay, and that’s just a conventional way of drawing something that’s eclipsed; because, as you can see, it’s not truly 100% eclipsed. Okay? Now, when you have substituents on those things, you can give different names to different phases of rotation. For example, they could be exactly opposite one another, like that; which if this were a double bond you would call trans, or E. Right? But for a single bond, you look at it with a Newman projection like this, and it’s called anti. Right? Or you can have it rotated like this, and be what’s called gauche. And I’ve never found out why it’s called gauche. Right? But notice something about that. Is this, is it, anti, chiral; is that conformation chiral, or is it superimposable on it? Is there a mirror?

Student: Yes.

Professor Michael McBride: Right, there’s a mirror right here. Right? So it’s its own mirror image. How about gauche, is it chiral; is there a mirror here? There’s a twofold axis here. If I rotate it like this, if I had you close your eyes and did that rotation, you couldn’t tell whether I rotated it or not. Right? It’s like a propeller. But if I take another one like this and make the mirror image of this gauche one, like this — here, right? This is a mirror image. Everybody see? Right? But I can’t superimpose them. Right? If I superimpose the back three, the front ones are not. Right? So these are chiral. So, with respect to gauche, we have enantiomers. Right? That one as well. And one of them we call minus and the other we call plus. Okay? And now if you have — the analog for eclipsed is that you can have something that’s fully eclipsed, like that, where the special groups are right on top of one another. Or it can be like that. Right? So and that can be one way or the other; and again it’s twisted, like a propeller would be. And these are again cumbersome things to draw, with Newman projection. But you don’t worry about their being cumbersome because you don’t have to draw them usually, because they’re not energy minima, they’re not isomers, they’re barriers between going from one to another. Okay? But again, those would be minus and plus.

And if you want to be really pedantic and meet in a committee of the Pure and Applied Chemistry people, the IUPAC, and decide precisely what you’re going to call different things, then you get together and you scratch your head, and for different rotations of the thing in back, with respect to the thing in front, you have these different pedantic names: synperiplanar is the one that we call fully eclipsed; and you can have anticlinal and synclinal and synperiplanar, and plus and minus, and so on. But you never do that really. But it’s there if you need it. Okay, so there’s a notation, Newman projection, and there’s a nomenclature, which can be sort of simple, like on the previous slide, or very complicated, like this, that allows you to discuss these things; to draw and discuss them.

Chapter 5. Understanding the Barrier to Rotate about the C-C Bond of Ethane [00:37:31]

Now, but here’s the practical question. Is the threefold barrier — so if you have ethane, like this, and you rotate this, the energy will go up and down; the energies will be different for different angles of rotation. But obviously there’s a certain symmetry, that this, if it’s all hydrogens, will be the same energy as this, will be the same energy as this. And for the eclipsed, this will be the same as this, will be the same as this. Okay? So there’s a threefold barrier. Now the question is — two, there are two questions. How big is the barrier? So are we going to be able to have isomers? And, what’s the energy minimum?

Now let’s just suppose, together with van’t Hoff and Paternó and the person who made the Molecule of the Week, that the lowest energy — if we plot energy versus phase of rotation here — that the lowest energy is 0°; that’s fully eclipsed, like this. Okay? That’s 0°. And now we’ll start turning it, and the energy may go up, when we go to a staggered, and then down and then up, down, up, down. Right? And then you’re back where you started, once you’ve gone through 360°. Or is there a big barrier? Right? Is the barrier small or big? Now how can you tell? Here’s an interesting thing, a way to tell, right? That if there’s a big barrier, then the things are going to be quantized in it; the energies are being quantized. But the energy is — if the barrier is very, very small, then almost any energy you have is above the barrier, and so it’s not quantized. Remember, when you get above a barrier you can have energy, any energy you want. It’s only when the thing is bounded on both sides that the wave function has to come down and reach zero. Okay? So if the barrier is big, the energy will only be certain amounts, but if the energy’s very small, you can have any energy you want. And that you can tell.

That is, in fact, how it was discovered how big the barrier for rotation was, was by measuring something having to do with energy, with heat. Okay? And here’s — so the question is, is the energy quantized in this triple minimum? We’ve talked about double minima. This is actually a triple minimum. Okay, so they measured in 1936 the absorption of heat by ethane, going down to a very, very low temperature. This starts at 150 Kelvin, but I think they went further than that. So there’s a thing called entropy; and we’ll talk about that in different terms, a little bit — in a week or two, after vacation. Okay? Vacation. [laughter] Okay, so there’s a thing called heat capacity, C_p, which is how much heat you absorb. So entropy is defined, by people who do thermodynamics, in this way: that you sum up, or integrate, starting from zero Kelvin — remember Kelvin? — all the way up to whatever temperature you’re interested in. You sum up how much heat is being absorbed; so C_p times [change in] temperature, that’s how much heat is being absorbed at each temperature, as you go up. As you raise the temperature, more and more heat gets absorbed. But you always divide it by the temperature, as you go up. So if you absorb heat at very low temperature, that makes a much bigger difference than absorbing the same heat at high temperature, in this quantity called entropy. So don’t worry about it, except by this definition for now. And this is very technical and not so very important to what we’re doing. So don’t get tied up in knots about it. But anyhow, if you have larger quantized spacings, then less heat gets absorbed — and I’ll show you that on the next slide — and the heat that gets absorbed gets absorbed at a higher temperature. Right? So both those factors mean that the entropy, at any given temperature, is going to be smaller, because you’ve absorbed less stuff in the numerator, and you’ve absorbed it at a higher temperature; so you have a bigger denominator for all these contributions.

Okay, so the results in 1936 were this. If you’re interested in what the entropy of ethane is, at 298 Kelvin, it turns out experimentally — that’s what they did out at Berkeley — they found it was 54.8 plus or minus 0.2 entropy units. And then they calculated how high that would be, for different heights of this threefold barrier, and the quantization that’s involved. And what they saw was that if the barrier were zero, that number, instead of 54.8, should be 56.4. If the barrier were 0.3, it should be 56.3. If the barrier’s 3.1, it should be 54.6; which is within experimental error of the experiment. Right? So that showed that the barrier was about 3 kcal/mole. So it wasn’t totally free rotation. It was just pretty fast. So suppose the barrier had been 4 kcal/mole, instead of three; just suppose it were four. How fast — what would the rate constant be for the rotation? So the barrier is 4 kcal/mole, that you have to get over. How do I deal with that? Sam?

Student: It’s this —

Professor Michael McBride: There’s a trick, right?

Student: 10 to the —

Professor Michael McBride: 10 to the minus —

Student: -3/4. So 3/4 times four is three.

Professor Michael McBride: Okay, so it’s going to be to the — there’s going to be a -3, up in the exponent. And what else is going to be in the exponent? Remember, for a rate?

Students: Thirteen.

Professor Michael McBride: 10¹³. So it’s 10^(13-3), which is ten to the —

Student: To the tenth.

Professor Michael McBride: 10¹⁰ per second. So it would take 10^-10th seconds for it to rotate. That’s pretty darn easy, at room temperature. Right? Pretty easy. But not perfectly free, and enough to make this difference in the entropy. Now, let me just see what time it is; okay, we have time to do this. Okay, so we’re now going to use that big K, equilibrium constant. So suppose we have something that has — this rotor say — that has a barrier of such that the spacing between successive levels is 1 kcal/mole. Suppose it’s a harmonic oscillator. We already did that. The lowest zero-point level, you got to have that energy, and then 1 kcal above that, 1 kcal, 1 kcal. Now what we’re going to do is see how much energy, extra energy, above zero-point, is there at equilibrium, at any given temperature. So we have some stuff — at equilibrium there’s going to be stuff that has no extra energy, zero-point, a certain amount. But there’ll be a certain amount that has 1 kcal/mole, that’s in that first level. Now how do I know how much is in that first level, at equilibrium? How do I know the ratio of what’s in the top to what’s in the bottom, if the spacing is 1 kcal/mole? Is there a way of doing that? You don’t have to do the numbers, but tell me what formula you would use, to get the equilibrium constant. Katelyn, do you remember? Sophie? How do you get the big K; how does it relate to energy? Anybody remember? We already did the hard one. Eric?

Student: 10^-3/4ths.

Professor Michael McBride: 10^-3/4ths of one. Okay. That’s at room temperature. It’s different at other temperatures. Okay, so we know how much there’s going to be — what fraction of the stuff, of the stuff that’s here, is going to be one kilocalorie. Ah, it’s going to be 10^-3/4th, of this; so something like 1/10th of it. Right? 10^-1 would be 1/10th. Okay? And then there’ll be stuff here, and the amount here, relative to the amount here — this is what we’re doing. So that ratio, the ratio of 0:1 is going to be the same as the ratio of 1:2, because that one is one kilocalorie above this one. And the same for the third to the second, or the fourth to the third. So then we can sum all these up, knowing the ratios, and figure out what percentage there is of each of these. And it turns out that that 10^(-3/4) is 5.6. So there’s going to be 5.6 times as much here as there is here; 5.6 times as much here as there is here; 5.6 times as much here as there is here. And if you translate that into percentages, it’ll be that there’s 81.9% here and 1/5.6, 14.8% there and 2.7, and 0.5%, and 0.1%. But of course this stuff added only — this 14% added only 1 kcal/mole. This one added two. This one added three. This one added four. So you sum all those up and you get how much energy — how much heat was absorbed. This didn’t absorb anything, it’s still in the lowest. This absorbed a certain amount, 14.8 — right? — .148, and then twice that, 0.027, and so on. So you can sum it up and see how much heat would’ve been absorbed by the time you get out there.

Now, if the spacing had been very much smaller, if the spacing had been 0.025 kcal/mole, then we would’ve made — how much is there of this that’s in the lowest level; how much has one unit, 0.025; how much has 0.05, 0.075, and so on? And we do exactly the same trick; although of course you need a spreadsheet to do this with, because it’s tedious. Right? But you would get this. And you see you absorb more heat if you have smaller steps. And now there’s 4.2% here, and 0.8% at exactly 1 kcal/mole. But you have a lot of others in here. So 59% of the population is halfway out to the one; and then 33% is more or less one; and 6.6% is like that; and so on. So anyhow, you absorb more heat. And you absorb it at lower temperature too. Right? So both those things build up the entropy. So that one absorbs 364 kcal/mole more — not kilocalories, small calories — 364 more than the other one. And that is what gives that entropy difference. Okay, so that’s how people knew that there was a barrier to rotation, was by measuring heat capacity. So we’ll stop here, and next time we’ll address the question of whether the — now we know what the barrier is, but we don’t know what the low energy form is, whether it’s eclipsed or staggered.

[end of transcript]

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