CHEM 125a: Freshman Organic Chemistry I
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Freshman Organic Chemistry I
CHEM 125a - Lecture 28 - Stereochemical Nomenclature; Racemization and Resolution
Chapter 1. Ambiguity in the D/L Genealogical Designations [00:00:00]
Professor Michael McBride: Okay, today in the Times there was a report of a clinical trial, which event, the kind of thing we talked about on Friday. This was of a statin, which is like Lipitor. It’s a different one, it’s Crestor. But they just reported that a study of 18,000 people — so a very big clinical trial — seemed to show that this, that it helped; it reduced the risk of heart attack by a half for people who didn’t have high cholesterol, but had another risk, this C-reactive protein, CRP. And it said in the story this morning: “Like many clinical trials, the Jupiter Study was sponsored by a pharmaceutical company, in this case AstraZeneca. It makes the drug in the trial, rosuvastatin, which is sold as Crestor.” And then it says: “Dr. Ridker, a co-inventor of a CRP test, said he first sought federal financing for the study and was turned down. He and other scientists interviewed for this article, except for Dr. Nabel, Dr. Gardner and Dr. Wolfe, have consulted for or received research money from stain makers.” I guess I should say that I’ve occasionally gotten money from statin makers too, but not for this kind of thing.
Okay, so that’s interesting and relevant to what we’re talking about. But specifically what we’re talking about now is nomenclature. And last time we talked about constitutional nomenclature, the rules people made up for naming the nature and sequence of bonds. But since arrangement in space is also important, you have to be able to have names for stereochemistry. And right now we’re interested in configurational stereochemistry, the kind that can only be changed by breaking bonds — right? — whereas, as we’ll see soon, conformational changes don’t require breaking bonds, so typically are much easier to do.
So configurational isomers include the tartaric acids that we’ve been speaking about, that have these physical properties. One has a unique melting point. The other two are the same. But those two differ by the sign of the optical rotation of polarized light. So the first one is meso. Now how do we name these things? One way is just to name them according to the phenomenon, to what you observe. So sometimes those two tartaric acids that are enantiomers — mirror images of one another but not superimposable — sometimes they’re referred to as small d, or small l, which stands for dextro, or levo, and means right or left. And all it means is the direction they rotate the plane of polarized light, which you observe experimentally. So there’s nothing more arcane about that. Or you can say plus or minus, meaning that they rotate light to the right or rotate light to the left. And those are absolutely the same thing, right? Dextro and plus; levo and minus. Sometimes people use both names, but they’re redundant. They both mean it rotates light one way or the other. So it just describes the phenomenon.
So there’s nothing ambiguous about that. It’s a perfectly good name. But it doesn’t tell you what model you should make to show it, which mirror image is which. All you know is which way it rotates light. So that’s phenomenological. But there can be other names that have to do with these Fischer projections we draw, as to exactly which one is which. So that one is clear enough. That one has a mirror image — is its own mirror image. There’s a mirror that goes through the middle of it. So the top is the mirror image of the bottom. It’s a mirror image of itself. And that one’s clearly the meso. Right? But how about the others? One of them will look one way and one the other. But is there any fundamental thing that we can understand that would tell us which will rotate light to the right and which will rotate light to the left? And I’ll tell you that I can’t say that, and most people I know — in fact almost everybody I know — can’t say that. But there are a few people who think they can say that, and we’ll hear about one of them soon, on Wednesday. But anyhow it’s a question. Is it like — should it be those for the right and the left-handed, the d and the l, or those? Which one is which?
Well Fischer decided that we just have to have to names for this and we have to be able to draw pictures, so let’s make a guess as to which way it is. And he devised a new system of naming which doesn’t depend on the phenomenon but depends on genealogy; that is, synthesis. You take a compound like glyceraldehyde, which is small d and +. What does that mean, d and +?
Student: The same thing.
Professor Michael McBride: For this? There’s a particular bottle of glyceraldehyde, which has that constitution. Right? And we’re going to label it d-(+). And what does that mean? Zack?
Student: Does it rotate light to the right?
Professor Michael McBride: That it rotates light to the right. But we don’t know exactly how we should draw it — right? — the structure or its mirror image. So, and d and + here, remember, are redundant. They both just mean it rotates light to the right. But Fischer went further. He defined that particular glyceraldehyde as capital D. There’s a different between capital D and small d. Capital D is the name he used, and that big D-glyceraldehyde, he guessed — the one that rotates light to the right — he guessed had that structure, which when drawn with a Fischer projection is that structure. So that’s just a guess. So he said this now gives us a starting point. This glyceraldehyde, which I’m going to call capital D, and I’m guessing has that structure — I might be 100% wrong, right? It might be the other one. But let’s just make this guess and then we’ll be able to talk about everything. So what could you talk about? You could do a multi-step synthesis in which the carbons connected by the arrows are the same carbon. That is, you redo the top half of the molecule by doing chemical reactions, the part that’s in green. So CHO becomes that much more complicated thing. But you leave intact that carbon on the bottom. So it’s the same in tartaric acid as it was in D-glyceraldehyde, whatever that was. And we’ll guess it’s this one, with the OH on the right. Is everybody with me, with the guess he made? So now if you can make that particular tartaric acid isomer from D, capital D, glyceraldehyde, then you say that that’s D-tartaric acid, because you can make it from D-glyceraldehyde. Any questions about that? So it’s just what you make from what. If you can make it from a D-compound, you call it a D-compound, and the mirror image would be the L-compound, capital L. Claire?
Student: Is he just swapping the things in the green squares? The arrows are a little confusing.
Professor Michael McBride: The swap — the thing that’s in the green square on the bottom becomes the stuff in the green square on the top. Obviously you have to add another carbon and you have to add a couple of hydrogens and some more oxygens. So you have to do some chemical transformations. But the important thing is that the thing on the top becomes the thing on the top, and the carbons connected by the arrow are precisely the same carbon.
Professor Michael McBride: Okay? So that one you know because you know where it came from by genealogy. So that tartaric acid you’ll call capital D, according to Fischer. But do you see a weakness in this nomenclature scheme? Yeah, Sherwin?
Student: It’s dependent on the structure, your knowledge of the structure that he showed.
Professor Michael McBride: Well of course it could be 100% wrong. Everything could be the mirror image of what we thought it was. Right? That’s clearly a problem, and that was well recognized by Fischer. But maybe I didn’t understand your question — did I?
Student: Yeah, that’s what I was saying.
Professor Michael McBride: Okay. Does anybody see another weakness? Sam?
Student: Capital D isn’t very descriptive.
Professor Michael McBride: It’s certainly not descriptive in structure, except — unless you know what d-glyceraldehyde is, then it’s descriptive. But it’s not descriptive alone. It’s only relative, right? It’s only relative to the d-glyceraldehyde. And you have to know how that relation works. For example, someone else might go into a lab and they want to prepare tartaric acid from d-glyceraldehyde. What might the other person do?
Student: Prepare it a different way.
Professor Michael McBride: Lucas, I can’t hear.
Student: Prepare it a different way.
Professor Michael McBride: Prepare it a different way. So somebody else might put that carbon into the one on the top and change the brown ones. Right? There could be other transformations you do that would change the carbon on the bottom to the carbon on the bottom, and the carbon on the top is now the same as the one on glyceraldehyde. Right? In which case it would be exactly — that one on the left would be named capital D, if you made it that way. Everybody see? The genealogy would just be different. Okay? So there’s a fundamental problem here, that the capital D / capital L designation is ambiguous, without having some detailed synthesis recipe, or some agreement that people say we’ll always make things this way rather than another way. Right? So it’s not — doesn’t have meaning on its own. You have to know exactly how you made the transformation. So that’s a pretty cumbersome way to do names. But that’s what was done for sixty years. Right? Those were the names, capital D and capital L; for sugars, for example, for amino acids. Okay? And it’s relative, and it’s based on a guess.
Chapter 2. The Discovery of Tartaric Acid’s Atomic Arrangement and Notation by Priority [00:10:15]
But in 1950 things changed. Because in 1950 a man named Bijvoet, an X-ray crystallographer at Utrecht — in fact working in the van’t Hoff Laboratory, the same building in which van’t Hoff did the things in 1874 — Bijvoet determined the crystal structure of sodium rubidium salt — remember, there are two carboxylic acids in tartaric acid. So the sodium rubidium salt of d (small d)-L-tartrate. What does the small d mean? Andrew?
Student: It reflects the light of the —
Professor Michael McBride: It rotates the plane of polarized light, or twists it to the right. And what does the capital L mean? Allison?
Student: Previously we said it rotated it to the left.
Professor Michael McBride: I couldn’t hear very well.
Student: That previously we assumed that it would rotate it to the left.
Professor Michael McBride: No, that’s not true. Or maybe I didn’t understand “previously,” what you mean.
Student: Like well, the model we were just talking about with Fischer.
Professor Michael McBride: Right, it’s what we talked about with Fischer, that he was able to make this tartaric acid from l-glyceraldehyde, from the enantiomer of what we talked about before. So it was related by that particular synthesis to l-glyceraldehyde. So there’s no conflict between saying d and L. Right? It rotates light to the right — that’s the small one — but it’s genealogically related to l-glyceraldehyde. Okay? But he was able to use a very special kind of X-ray technique, called anomalous dispersion, which we don’t have time to talk about right now. But it was able to show exactly which way the atoms were oriented. And this is the picture he drew. Right? So clearly on the top is C, two Os, right?; a carboxylate group. On the bottom a carboxylate group. The middle two carbons each have an H and an O on them. The O is stippled, so you see what it is. And it’s clear the way you draw it what’s coming out — the way he drew it, what’s coming out toward you and what’s going back in.
Now here’s what Fischer had guessed, sixty years before, for L-tartrate, what he called L-tartrate. Was Fischer right or wrong? How about the top carbon that has OH on it, the HCOH? Is it the same as in the model that Bijvoet published? Does it agree or not? Yes, it agrees. The OH is coming out toward you and to the right, same as in Fischer’s picture of the top carbon. Everybody got that? How about the bottom carbon? The bottom carbon, Fischer has it going to the left and Bijvoet has it going to the right; this one here. Right? That’s going to the right from this carbon. Fischer’s drawing it to the left. Do they disagree?
Professor Michael McBride: Who said something? Do they agree or disagree?
Professor Michael McBride: Seth, what do you say? What does Fischer mean by this bond here, the CH bond? Which is it, going out toward you, the H, or back in? Seth? The Fischer projection, which is shown here. It’s got vertical bonds and horizontal bonds. Which one is coming out toward you, from the carbon? Okay. Somebody else? Yeah, Kate?
Student: In a Fischer projection the bonds are going into the board.
Professor Michael McBride: Which bonds?
Student: The carbon-carbon bonds.
Professor Michael McBride: The carbon-carbon bonds are going into the board; that’s Fischer’s convention, yeah.
Student: Because he twisted them so that everything else would be coming out of the board.
Professor Michael McBride: Yeah, so these two are coming out of the board. How about in Bijvoet’s picture? For the second carbon — here’s the first one, here’s the second one.
Student: The OHCs are going into the board.
Professor Michael McBride: Ah ha. So, to make it like Fischer, you have to rotate it so they’re coming out of the board. So in fact they do agree. Right? It’s just that Bijvoet’s is made realistically, and Fischer’s was made with these rubber tubing bonds and you had to bend it to get that way. But it’s the same configuration. So Fischer was right. So these structures that had been drawn for sixty years were drawn correctly. Right? They could’ve — it was a 50:50 chance that it could’ve been the other way around. But thank God he was right, so we don’t have to go back and correct everything, and know whether things were written before 1950 or after 1950. Yeah?
Student: But don’t you think that if you rotate that molecule, isn’t the OH that’s coming, that’s like each — his drawing coming out of the board, doesn’t that go back in?
Professor Michael McBride: Okay, if you rotate this molecule by 180° around the vertical axis, then this carbon — with respect to the carbon we’re interested in here — this carbon would be going back into the board and this carbon would be going back into the board. That’s the way these are. Right? But in the process of that rotation, this OH, that was back and to the right, after 180° rotation about the vertical axis, will be out in front and to the left; which is the way it’s drawn in the Fischer Projection.
Student: So the other OH group would be going back in?
Professor Michael McBride: Yeah, but you don’t care. This is just to show the configuration, right? When you do that rotation, yeah, these will do the other. But the point is that Fischer’s structure was right, with respect to each carbon, about what’s going which way, which tetrahedron it is. Okay, so that’s fine. And then Bijvoet wrote, or in the same paper he wrote: “The question of nomenclature is beyond the scope of our investigation.” (He’s just doing the X-ray to find out which way it is.) “The problem of nomenclature now concerns given configurations” (now you know which one it is you want to show) “and requires a notation which denotes these configurations in an unambiguous and if possible self-explanatory way.” So you don’t need to know how it was synthesized or anything. Now you know how it really is — right? — what name are you going to give to it so people can know, just from the name, which way it is.
Now, this is related to the idea of naming configurations involving double bonds. So we’re going to first show you how you do that — which actually came second — and then we’ll go back to see how they named the configuration at a tetrahedral carbon. So malic acid, from apples, has that structure shown, and if you heat it you lose water and get a double bond. But you get two isomers, maleic and fumaric acids. And you’ll remember we showed that before, that van’t Hoff made those with his models. And one of them has the two COHs near one another and one has them farther from one another. And we can draw them that way as well; near on the left and far on the right. And if you heat these molecules further, the one on the left can lose water and form an anhydride, but not the one on the right. Why not the one on the right? It’s pretty obvious. Russell?
Student: They’re too far apart.
Professor Michael McBride: They’re too far apart. So you know which one they’re close together and which one they’re far apart. So that experiment proves which is which. It’s not as hard as the case of the right and left-hand, where you had to do this very special X-ray technique. So it was known that maleic acid was the one that can form an anhydride, and therefore had them on the same side. And that was called cis, meaning on this side of, and the other one was called trans, across. So that was a perfectly good nomenclature for those two isomers of the dehydrated malic acid. Maleic was cis and fumaric acid was trans. So these names were used for a very long time, but they weren’t really good names. And the reason, you can see, by looking at this molecule. Is this one, would you call it cis or trans? Same side or opposite side? Andrew, what do you say?
Student: I don’t know.
Professor Michael McBride: Next Andrew. This is our nomenclature problem.
Student: I think trans.
Professor Michael McBride: Would you call it trans, you say?
Student: I would probably call it —
Professor Michael McBride: This one here.
Student: Okay, that one, I would call it cis.
Professor Michael McBride: Why?
Student: Because of the CH3’s.
Professor Michael McBride: Because the CH3’s are next to one another. So you might call it cis. On the other hand, if you wanted to say how this end is related to the acid, you’d say it’s trans, right? So that means it’s — as an absolute nomenclature, it’s very hard to generalize this because you have to know which one you’re picking, to say it’s cis or trans. It’s fine to talk about the relative configuration, to say the methyl on the left is cis to the methyl on the right, or to say the methyl on the left is trans to the COOH. That’s fine. But as to whether to say the molecule itself is cis or trans, you have to know which one you’re talking about, on the right. So that’s another thing where you’re going to have to have some convention or something. And the way people decided to do it was this. You assign the groups at each end, take one end and then the other end of the double bond, and assign priority. So you say we’re going to give higher priority to one group or the other, and that’s the one we’re going to use for the name. So on what basis might you assign a priority to a group; to say which group has higher priority, CH3 or H? Which one do you think should have higher priority?
Professor Michael McBride: CH3, okay. How about on this end, CH3 or COOH, which should have higher priority?
Professor Michael McBride: Why?
Student: It’s heavier.
Professor Michael McBride: It’s heavier. Okay. Now let’s think about how that is going to work out. So we’ll assign them by atomic number; or atomic weight, if we have two that are different isotopes of one another. Right? But we won’t sum up the atomic numbers of all the groups that are in that; that is, we won’t sum up C and O, and O and H, and C and 3 Hs. Why not sum them all up and then see which one has higher total atomic number? What’s wrong with that? Perfectly well-defined.
Student: It’s tedious.
Professor Michael McBride: It’s tedious, it’s cumbersome, to have to add it all up. So what rule are you going to do?
Student: Find a difference.
Professor Michael McBride: You’ll go until you find a difference, go out until you find a difference. Once you find a difference you stop, so you don’t have to go everywhere. Okay. So at the first difference; the same as we said with numbering last time, right? Okay, so on the left, in each case, you have a carbon compared to a hydrogen as the thing that’s immediately attached. Carbon has higher priority, higher atomic number. On the right, it begins with a tie. Incidentally, we’re going to have to deal with double bonds, COOH, at the bottom. Right? The way you do it, just by convention, there’s nothing — God didn’t write this on the tablets, right? It’s — you just pretend that there are two of those atoms, when there’s a double bond; just pretend.
Okay, so now we compare the top with the bottom. If we go to the first step, it’s carbon versus carbon. It’s a tie. So we’re going to have to go further. When we go to the next level we see on the left it’s oxygen versus hydrogen. Right? And oxygen is higher. We don’t have to go any further, we’ve found a difference. So forget the rest of it. Everybody with me? Okay, so we might say the one on the left is trans and the one on the right is cis. However, people, for sixty years, or longer in this case, have been using the names cis and trans to mean one thing or another, and sometimes we’ll agree with them and sometimes we don’t; which is going to cause great confusion as to whether you’re using a pre-1950 name or a post-1950 name, and at least half the compounds are going to be wrong. Right? So the names trans and cis have been polluted by previous usage. So we’re going to have to have new names. Yeah Lucas?
Student: You don’t really have — in the second part of the double bond, as a substituent, and that itself is like cis or trans.
Professor Michael McBride: You’d mirror all — you’d duplicate all those as well.
Professor Michael McBride: Okay. Okay, so this one was used with a German root, because the guys from Switzerland were doing this, entgegen, which means opposed. And the one on the right, so it’s E when they’re trans, when they’re opposite one another. And when they’re adjacent to one another it’s called together, zusammen, Z. I’ve always thought this is a real pity that it wasn’t the other way around. So I hope this doesn’t confuse you. Because these are on the same side and these are on opposite sides. Right? But it’s just backwards from that. So just remember that it’s unfortunate. Okay? So Z means together and E means apart. Okay, that’s the name we use for there.
Okay, now notice that in assigning priority, as we said, just to belabor this is a little bit, you proceed one shell at a time. So we saw a tie when it was carbon versus carbon, and when it was three oxygens versus three hydrogens, it was obvious that the top wins. But now let’s go to this case. Okay, now carbon versus carbon, that’s a tie. We go out further, three oxygens beat a carbon and two hydrogens. Right? But there are chlorines on the bottom. Right? The chlorine is really high priority. Right? But it’s irrelevant, because the decision has already been made. So you respect earlier decisions. Right? You only go as far as you need to go, to see a difference. Okay, I think we’ve said that enough.
Chapter 3. The Cahn-Ingold-Prelog Priority Scheme [00:24:49]
Now these are the guys that invented the scheme to name handedness. It’s called the CIP Priority Scheme. And the C is R.S. Cahn, and the I is C.K. Ingold — both of them are from Britain — and the P is Vladimir Prelog, who’s in Switzerland, although he’s a native of Yugoslavia. Now, these were — many people, particularly synthetic chemists, would consider Robert Robinson and R.B. Woodward the greatest organic chemists of the 20th century. They both thought they were probably the greatest individual one. Right? So I love this picture, taken by Jack Roberts of them at a seminar at MIT around this same time, because they sort of aren’t looking at one other. Right? What’s Robinson looking at? Robinson is looking at Ingold, whom he really despised, and vice-versa. All right? There was no love lost between Ingold and Robinson. And this is illustrated by an account that Prelog gave in his autobiography. He encountered Robinson in the airport in Zurich. Robinson was on his way to a meeting in Israel.
Robinson: “Hello Katchalsky. What are you doing here in Zurich?”
Prelog: “Excuse me, Sir Robert, I am only Prelog; I live here.” (He got a Nobel Prize too, Prelog.) “I am only Prelog and I live here.”
Robinson: “You know Prelog, your and Ingold’s configurational notation is all wrong.”
Prelog: “Sir Robert, it can’t be wrong. It is just a convention. You either accept it or not.”
Robinson: “Well then if it’s not wrong, it’s absolutely unnecessary.”
Okay? So anyhow, the point of this slide is that it’s a convention. There’s not right and wrong about it. It’s that they propose rules, and those are the rules that people have adopted. And that’s how you give a name to the absolute configuration of a stereogenic carbon. And for fun, for Wednesday, you can try this exercise, going back to this medieval manuscript, and using the figures in it to devise your own convention to describe chirality. So you can put yourself in the position of Cahn, Ingold and Prelog. Okay, so here’s the Cahn-Ingold-Prelog R/S nomenclature scheme for stereogenic centers. First you have to decide on each of these carbons — which are mirror images, you see, of one another — which atoms have which priority. So let’s start on the left. What’s the highest priority for a substituent on the central carbon? There are four groups. Which has the highest priority?
Professor Michael McBride: Which one?
Professor Michael McBride: Zack?
Student: I thought it was OH.
Professor Michael McBride: OH. Yeah O is the highest atomic number. What’s the lowest?
Professor Michael McBride: H, right? There’s also a D there. It’s the same atomic number; hydrogen and deuterium. Which one should be higher?
Professor Michael McBride: Deuterium’s heavier. If it’s isotopes, you take the heavier one. Okay, so here’s one, four, three, two. Right? That’s the priorities; and analogously on the right. And now we have to decide whether they’re — which one to call right and which one to call left. And the way you do it, one way of doing it, is to take a thing like this where I have these red, yellow, green, blue; red, orange, yellow, green, blue, right; the order of the spectrum, red, yellow, green, blue. And I make a spiral that connects them, like that. And now, is that conventionally what you call a right-handed or a left-handed spiral? You know, if you look at it, if I turned it like this, it would move this way. Right? So screws are called right-handed and left-handed screws. And the reason is that the — you know, toe-bone connected to the foot-bone kind of thing. The way your arm is put together, for right-handed people it’s easier to drive a screw like this. Right? So that’s conventionally called a right-handed screw. So this you would call right-handed. Okay?
Now, if you can’t remember from how to turn a screwdriver, you can look at — you can do this trick with your hand. You put your thumb along the direction of the lowest priority; hydrogen, in this case. So it’s coming out of the board, right? And you curl your fingers. Right? And you notice it goes one, two, three, and then four. So that one on the left is left-handed. Everybody see how I did that? Curl your fingers, they go one, two, three, four. Okay? So that’s a left-handed one. The one on the right, if you do it this way, you — so that’s. And you call it, not left, you call it S, for the Latin sinister. So it’s not going to get confused with D and L and so on, that have already been used. It’s a new way of doing it. So that one’s S. And the one on the right, if I put my thumb so it goes back toward the H, and curl my fingers, I go from one to two to three. Okay? So that one is called R for rectus; Latin for right. Okay?
Now there’s another — there are lots of tricks to do this. If you don’t like doing it that way, you can do it this way. Pretend that this is a steering wheel of a car, and the H is going back in. Right? So here I’m — this one on the right — I’m sitting here driving a car like this and the H is going away from me. Everybody got that? And I notice that I go one, two, three. Right? That’s going to turn the car to the right. Everybody see that? So, and this one will turn the car to the left, if I turn the wheel to go from O to C to D. Okay? So that one’s a left turn and that one’s a right turn. So whatever works for you is fine. And I can assure you that on some test I’m going to look out and see people with their papers there and so on, and they’re going to be going like this. Right? Because it’s a handy way to do it. [Laughs] Handy. [Laughter] Okay, so let’s try it on tartaric acid. Here, let’s do the top carbon of tartaric acid. Okay? So we’ve got to decide on the priority of the groups here. Here’s one group, another group, another group, another group. Which is the highest priority?
Professor Michael McBride: OH is the highest, right? O is the highest. Now we got two carbons that are tied. Which one’s going to win, top or bottom?
Professor Michael McBride: Top you say Kate? Why?
Student: A tie.
Student: Because it has — well we can pretend three oxygens, whereas the bottom one has a hydrogen, a carbon, and an oxygen.
Professor Michael McBride: And only one oxygen, right? So the one on the top has — not more oxygens, but it’s got the oxygens nearer. Okay? So we don’t have to go as far. So this is going to be one, two, three, four. Okay, now let me see your hands operating. Put a thumb along the H, the direction the H goes, and curl your hands one, right, top, bottom, one, two, three, and tell me whether that’s right or left-handed? Okay? How many think left? How many think right? This is a democracy, the rights have it; they’re also correct. Okay, that one’s one, four, two, three. Okay, that, it’s right-handed. And the bottom one’s also right-handed. Why didn’t I have to take a lot of time to have you do this, to do the bottom one? Why does the bottom one follow, if you know the top one? Because you can rotate the thing 180° and it’s the same, and that will change the top into the bottom. So whatever the top is, the bottom’s the same thing. So the name of this is, in parentheses, (2R, — carbon two is R; carbon three is also R), then dash (-), 2,3-dihydroxybutanedioic acid.
Okay, so now we have a scheme that does that. And notice, this is the gate into the new Chemistry building, just up Prospect Street, and it’s got things that relate to the different branches of chemistry on it. And this one is those carbon tetrahedra with the spiral on it that says which handedness it is. There are both right and left-handed ones there. Okay.
Chapter 4. Racemization and Epimerization [00:34:12]
Now, racemization. If you start with a molecule, which maybe you’ve gotten from nature, that’s all one hand, like lactic acid, from milk, (R)-lactic acid, there are processes which convert it into a 50:50 mixture of R and S. For example, suppose you had a base and it could pull off a hydrogen. What hydrogen would it pull off from this molecule, a base? Any idea? What is the most reactive group here?
[Students speak over one another]
Professor Michael McBride: Or another way of saying it, this is a high HOMO. What’s the low LUMO going to be? Well there’s several low LUMOS. There’s the CO double bond, π*. That’s a good one, right? However, it won’t get you anyplace. You can put it on, come off again. Okay? Or there’s σ* OH here, but there’s also σ* OH here. But this one’s better because it’s next to the π*. Right? So this — that’s why it’s called an acid. You already knew that. And you can lose the H+ and get down to this. Okay, and that’ll go on off, on off, on off, a long time. That’s fine, nothing special there. But once in a long while — that’s easy to do — once in a long while you might pull off the wrong hydrogen; you might pull off this one. Right? So that’s not nearly as easy. But when it happens it’s interesting, for two reasons. One is, it’s not as bad as you might’ve thought, because it’s adjacent to this π*, which means that the vacant orbital here can stabilize that high HOMO. Right? Which means you can draw that resonance structure, with a double bond there. So it’s hard, but it’s not as hard as it might be.
Once in awhile that’ll happen, and usually it’ll go back again. Right? But what’s especially interesting — well just as a footnote, it’s possible to pull both of them off. But that would be very unusual, to make a di-anion, which is high in energy, because you’re putting so many electrons close to one another. Okay? But what’s relevant for our purposes is that that molecule is planar. So it’s not handed. This one had a chiral carbon in it, but this one doesn’t have an asymmetric carbon, a stereogenic carbon. So this one, the one in the middle, is neither right nor left. So if you go from right to something that isn’t right or left, you can come back, put the hydrogen on. The hydrogen comes from the front of the screen, goes on to where the minus charge is, and you’re back to the left. But the hydrogen could’ve come from the back of the screen, at the same carbon, and that would give the one on the right, and would be (S)-lactic acid. So after a long time of heating, with base, you can take one that’s all R and have it become 50:50, R and S. And that process is called racemization. Why would it be called racemization? Because it makes a racemic mixture; a 50:50 mixture from something that was pure. Okay, that’s interesting.
But what may be more interesting is — well this is just a footnote on that. If you here, to go from (R,R)-tartaric acid, to meso-tartaric acid, you change just one such hydrogen. The one here has been taken off and put back on the other side, to give that one. So that’s not racemization, you’re not going to a 50:50 mixture of the two hands, right? So it has a different name. It’s called epimerization. That’s just for vocabulary. But what’s more interesting than racemization is the reverse, to start with a 50:50 mixture and go to all one. And that’s the main subject for the rest of what we’re going to be talking about, before the exam.
Okay, so you start with (R,S) and you separate it into R and S, in separate vials. So you have just the one you want. Okay, one way we’ve already seen of doing that was Pasteur. A conglomerate is a mixture of crystals where each crystal is all one hand or all the other hand; but it’s a 50:50 mixture of such crystals. So there’s no net handedness to it. Right? But then if you, if you have a situation like that — notice you wouldn’t have that, if the crystals had a right-handed molecule, and then a left-handed, and then a right-handed, and then a left-handed; you couldn’t pick it apart because you can’t separate one molecule from another. Right? But if all crystals are right-handed, and some crystals are all left-handed — right? — then in principle you can pick them apart. And that’s precisely what Pasteur did when he noticed it. Right? So you can separate them. And there are tricks that can help you do it even better, or more easily, when the crystals — when the things crystallize as all right-handed or all left-handed. One is to start with molecules in solution and have only one form crystallize and have the other one stay in solution. Then all you have to do is filter it. Okay? And how can you do that? It’s because — and chiral-resolved poison is another one, which will keep one form from growing, so that only one form crystallizes.
How do you do that? Here’s a crystal which is in solution. And suppose it’s exactly at equilibrium; so it neither dissolves nor crystallizes. It’s exactly at equilibrium, it’s saturated. Okay, now the interior molecules of the crystal are very stable, because they have exactly the right environment. Right? They’re more stable than the ones in solution. The ones on the surface are not so stable, because they don’t have the neighbors — all the neighbors they need. Right? But the average molecule is exactly the same as in solution. So it just sits there, it’s at equilibrium with solution. Now, so that’s the situation we have. That’s the temperature and concentration, so that this crystal is exactly at equilibrium with solution. Now suppose that you had a smaller crystal. Would it also be at equilibrium with solution; if it were smaller? Or do you see any difference? The smaller one has a higher fraction of molecules on the surface. Okay? What does that mean? The molecules on the surface are less stable than the ones inside. If the average molecule here is precisely the same as the ones in solution, so it just sits there, how about this one? The average molecule is less stable than the ones in solution. So what will it do?
Student: It’ll dissolve.
Professor Michael McBride: It’ll dissolve. Okay, so a small one dissolves. This one just sits here. So that’ll go away. And so as time goes on it gets smaller and smaller, the faces move in. How about this one, what will it do?
Student: It’ll grow.
Professor Michael McBride: It’ll grow, because it’s more stable in solution. Right? Bingo. So that one grows, that small one shrinks. And this one in the middle is called the — is metastable. We defined it as being stable, as just sitting there. Right? But it’s metastable. If it gets any smaller it’ll shrink. If it gets any larger it’ll grow. Right? So you have to get a crystal that size, called the “critical size” before the crystal will grow. Right? So this suggests a way of getting only one form to grow. You have a solution, that’s racemic, 50:50. Right? But there are no crystals in there that are big enough to grow. Can you see what you can do? You add a little fairy dust, right? Powdered crystal of one of the two forms, but have them be big enough to grow. So only those will grow. And then you can filter it and you have just that one. And that’s done industrially, as a way of getting just one form, to separate the two forms; a means of resolution. Okay, and the other way is to add a poison that will absorb on one and keep it from growing. Okay, so that’s one way that’s actually used industrially, as well as in the lab. Another way is to form temporary diastereomers. Because remember, the right and the left are mirror images. So almost all their properties are the same; like their solubility. Okay? But suppose you put something else with it. Like here, let’s shake hands Josh. Here, does that feel normal?
Professor Michael McBride: It feels weird, right?
Student: Very weird.
Professor Michael McBride: That feels good. So right with right is different from right with left. Right? It’s a different property. Okay? So if what I did was make diastereomers by adding something else to these molecules, which itself is only one hand, like right, now I’ll have things that reacted that are right with right, and things that are reacted right with left. And what’s the relationship between those new things? Are they enantiomers? Right with right, right with left, and now I’m comparing RR with RS. What’s the relation? Are they mirror images?
Professor Michael McBride: No, they’re not mirror images. They’re just different. They’re different solubility, different boiling point — everything will be different — different reactivity. Right? So on the basis of that I can separate them. But if I did it in such a way that now I can remove that thing I added, now after the separation I can go back and have the original things separated. Okay? So temporary diastereomers. One way to do it is chromatography. So the stuff that I put in the chromatography column can be all one hand. I put the new stuff on — that’s the packing of the column. I now run stuff through it. One absorbs more strongly than the other, doesn’t move as fast. So one will come out the bottom of the column faster. That’s one way of making temporary diastereomers; or making a compound with a chiral-resolved mate.
Okay, for example, this allenic compound that we saw before that proved van’t Hoff was right, this is how it was actually resolved, how it was separated. So it says at the top they used alkaloids. An alkaloid is an organic base, isolated from plants, and the plants made only one hand of it. And so you use it to make a diastereomeric salt with a racemic mixture of right and left-handed acids. So now you have two salts — right-right salt and right-left salt — and they’ll have different solubilities, and you can crystallize. So here’s what happened. There’s brucine; that’s one of these alkaloids. And you see it has all those centers, stereogenic centers. So it’s chiral, just one hand, because you got it from nature. And now you mix 4.7 grams of the acid and 5.2 grams of brucine, which means that there is about a 1:1 ratio. So you have both salts there. And now you crystallize, you get a solid salt, and it weighed 4 grams, which means you got 42% yield — right? — about half of the stuff out. And that stuff then you — and you recrystallize it, to make sure it’s pure, that salt you got out. The melting point didn’t change, it was already pure. Liberate the acid by shaking it with hydrochloric acid and ether. Now you’ve got the acid; and it melted 145 to 146, and its rotation was +29.5°. And it recrystallized and didn’t change it. It was pure. And it turns out that the other one melts 144 to 146°. And so it’s the same melting point essentially, but exactly the opposite rotation. So they manage the separation by making salts that were diastereomeric — so they had different solubility — and then take the salts apart again. And Pasteur figured a way of doing this, in this story, which I suspect is not a true story.
Now I’m going to stop here. But we’re going to have — mostly what’s on Wednesday’s lecture is not going to be on the exam. But there’s a little more here that I’m probably going to make available for the exam; that I’ll talk about on Wednesday. And I’ll have to think over just how much more it’s going to be. Not too much more probably. Okay? So we’ll see you on Wednesday. Most of what’s on Wednesday won’t be on the exam.
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