CHEM 125a: Freshman Organic Chemistry I

Lecture 27

 - Communicating Molecular Structure in Diagrams and Words


It is important that chemists agree on notation and nomenclature in order to communicate molecular constitution and configuration. It is best when a diagram is as faithful as possible to the 3-dimensional shape of a molecule, but the conventional Fischer projection, which has been indispensable in understanding sugar configurations for over a century, involves highly distorted bonds. Ambiguity in diagrams or words has led to multibillion-dollar patent disputes involving popular drugs. International agreements provide descriptive, unambiguous, unique, systematic “IUPAC” names that are reasonably convenient for most organic molecules of modest molecular weight.

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Freshman Organic Chemistry I

CHEM 125a - Lecture 27 - Communicating Molecular Structure in Diagrams and Words

Chapter 1. The Development of the Fischer Projections [00:00:00]

Professor Michael McBride: We’re talking about facts, ideas and words, in relation to stereochemistry. And the words have to be generalized, in the way we communicate things, to include models; things like this, right? [Technical Adjustments] Okay, so models, but also what you can draw on a piece of paper that will convey stereochemical structure to people. So the standard we use in drawing is wedges, to mean they’re coming out at you, in the direction that they expand, and dashes mean they’re going back into the board. Although if all you draw is dashes, it’s not immediately obvious which end is going back, if it’s a dash. Right? But if one end has nothing on it, it’s clear, that it’s going back from the carbon that’s shown. So you have to use your stereochemical intuition sometimes to interpret that. Although if you draw a wedge, it’s clear which way it’s coming. And sometimes people draw wedges with dots, meaning they’re going back in. But there are many ways that people make mistakes in drawing this. So it needs a little practice. That one’s okay, and clearly corresponds to a tetrahedron with 109° degree bond angles, shown like this — right? — the blue one’s coming out at you, the black- the yellow ones going back, green and red, up and down. Okay, so that drawing is fine. What’s wrong with this drawing? Sherwin?

Student: It couldn’t mean like a [inaudible]. The upper bond and the lower bond are —

Professor Michael McBride: I can’t hear — I still can’t hear.

Student: Oh, the upper bond and the lower bond are [inaudible].

Professor Michael McBride: What are the bond angles? How about between the solid lines? 109.5°?

Student: 90°?.

Professor Michael McBride: Between the solid lines, not the wedges?

Student: Oh, 180°.

Professor Michael McBride: 180°. How about the other angles?

Student: 90°.

Professor Michael McBride: 90°, right? This is a square plane, drawn that way. The dash has to go the other way, and the lines that are in the plane of the board, or the piece of paper, have to be at 109°. Okay, so forget that one. I assure you that we will get these on exams, because we always have in the past; but this is warning. Okay, so that one’s planar, forget that. How about this one? There I’ve got the wedges drawn the right way. Everything else right? No, the solid lines in the plane are 180°. That one’s bad obviously. Not many people draw it that way. Okay? Ah, there’s one that looks pretty good. Everybody agree? Any complaints? Lucas, what’s the complaint?

Student: An acute angle there.

Professor Michael McBride: Pardon me?

Student: An acute angle, over there.

Professor Michael McBride: Ah, the angles between the normal lines and the wedge are all less than 90°. Well one of them is bigger than 90°, the one to the top. But the one to the bottom is less than 90°, if the normal line is in the plane of the screen. Right? That’s an acute angle. Right? So the one on the top left is the one you want to go for. Okay? Now how about this? Do those look good? See any problems? Kate, what do you say? Looks fine to you. Tell me what the angle is here, between this bond and this one. Where would this bond point if it were 90° from this one, which is in the plane of the board?

Student: Straight out.

Professor Michael McBride: It would point straight out. Now if at 90° it would be coming straight out from that point. How about there? Is this angle greater than or less than 90°? Here’s the carbon, here’s the one that’s in the plane of the board. Right? It’s going like that. Here’s the one — that’s 90°, right? Suppose I bend it up like this, which is what the wedge shows, now what’s the angle?

Student: It’s acute.

Professor Michael McBride: It’s acute, right? So these are not right, right? The same is true at the bottom. But this structure is okay because, by convention, it doesn’t mean to show the exact angle. It just means to show that it’s going relatively out, right? But it wouldn’t be — the solid — these three lines here, this one, this one, this one, the ones that are not wedges and not dashes, could not be in the plane of the board. They’d be 120° angles, or something like that. They couldn’t be 109°, right? But in order to draw this, in a finite amount of time, when you’re writing on paper — people understand what you mean, in this case — but if you draw that one, which means it’s planar, people will be very much offended. Right? Okay, so this again, at a certain level, is lore. You have to be criticized before you find out where you’re making mistakes. Okay? But anyhow, it’s important to be sensitive about the fact that carbon is tetrahedral.

Now, these things are hard to do in three dimensions. And the Fischer projection was invented for that purpose in 1891. And the reason for its invention was interesting. He wrote:

“With the help of Friedländer’s convenient rubber models, one can construct molecules of right-handed tartaric acid, left-handed tartaric acid, and inactive tartaric acid and lay them in the plane of the paper so that the four carbon atoms are in a straight line and the attached hydrogens and hydroxyls lie above the plane of the paper.”

Now, here’s the idea, that Friedländer’s models used rubber tubing to connect these things. Right? So drawing this in two dimensions and making clear what you mean in three is not easy. But he says what you’re going to do is take the four carbons that are in a chain — this is tartaric acid, here’s CO2, carbon with an OH, carbon with an OH and a carbon with O2 on it, right? So what he wants to do is to lay all four carbons in the plane of the paper, which you can’t do. Right? But if the bonds are made of rubber, then you can bend the bonds — right? — and then you can make them all four, in the plane of the paper. Right? But what you have to understand is that in this convention you draw the four carbons vertical in a line; not horizontal. And then you understand that the bonds between them are severely bent. Right? So that the curious thing is that when you draw a Fischer projection, as is most convenient for sugars, for discussing the configuration of sugars, the bond, the four bonds for this carbon are understood that these two, the horizontal ones, come out toward you, as if they were wedges, and the other two from this carbon go back into the board. Okay?

This is his paper where he originally proposed it. Nowadays you wouldn’t use dots here, which help make it clear that it’s going back, you’d just draw lines. But understand that vertical lines mean they’re going back into the board, and horizontal lines mean they’re coming out at you. Okay, so that’s the Fischer convention, that he invented in 1891. So this bond here is a very funny bond, because in bonding from this carbon to this carbon, it’s going back into the board from this carbon, but also back into the board from that carbon, because it’s bent. Okay? And it was because Friedländer made his models with rubber tubing that it occurred to Fischer that you could do this. And then all you have to do is draw straight lines — you don’t have to do wedges and so on — and just put things in the right place.

So here he could draw right-handed tartaric acid, left-handed tartaric acid, and inactive or meso, as we call it now, tartaric acid. So the attached hydrogens and hydroxyls lie above the plane of the paper — right? — and the other carbons, vertical bonds, go back in from both of the atoms that they attach. Okay, so there are the wedges. And if you rotated that 90°, it would look like that. Very highly distorted, but you can do it with rubber and be clear about what the configuration is — unambiguous. Okay, they go back from both carbons, and it was because you had rubber models that you could do that. Okay, so here are those models. And notice that you can rotate in-plane by 180°, and it means the same thing. So if I have this and it’s bent like this, and I rotate it like that, 180°, it obviously means the same thing. Right? I can rotate a Fischer projection by 180° and it’s still the same model. But I can’t rotate it by 180° — or by 90° this way. Why not? If I write it this way it won’t mean the same thing. Why not? Shai?

Student: Because a number of horizontal bonds are going in.

Professor Michael McBride: Yeah, the horizontal bonds are going back in, and the vertical ones are coming out at you. You’ve reflected it in a mirror, each carbon. So you could rotate it 180°, but you can’t rotate it 90°. Okay? So there, we’ll rotate that 180°. That’s still fine. And in fact what do you notice about it, other than the fact that the symbols are written backwards?

Student: The same thing.

Professor Michael McBride: It’s the same thing it was above. Right? So rotating it 180° doesn’t change it. Okay? How about the next one, if you rotate it 180°? Same as above? Yes. How about the last one? Okay. That one is now the mirror image. If you slide it up, like that, you can see that there would be a mirror there. Everybody with me? So that molecule is its own mirror image, whereas the first two — and remember you can draw the mirror any direction and always get the same mirror image; xy, yz, xz, they all give the same mirror image, just differently oriented. Another way to draw the mirror one, that’s easier, is to draw it right through the middle of the molecule horizontally, and then the top half is the mirror image of the bottom. Right? So it’s obviously not a chiral molecule. It’s superimposable on its mirror image, whereas the — and that one’s called meso, the name then; that was called meso-tartaric acid and the name was generalized to mean any molecule which is its own mirror image is “meso”. Okay? And notice that these terms name relationships. They don’t name an absolute molecule, these stereochemical terms, the ones like this. What’s the relationship between those two? Are they identical, or are they just plain different, or are they different in the special way that they’re mirror images of one another; the two on the top left?

Student: They’re just different.

Professor Michael McBride: Okay? Can you rotate them and make them be on top of one another? No. If you rotate them, they become the same thing they were originally, right? That’s what we already did. Are they mirror images of one another? Right. There’s a vertical mirror between them, right? So those are non-superimposable mirror images. And the name given to things like that is “enantiomer”. So the one on the left, the top left, is the enantiomer of the one that’s in the middle. Caitlyn, what’s the question?

Student: Okay. It’s just the notation I guess. If the OOH were on the other side.

Professor Michael McBride: Oh no, but that’s — but all that means is the grouping C double bond OOH. That doesn’t have any arrangement in space when it’s just written with letters that way. Okay? That was just because I wanted to make strictly the rotation. Right? It’s where the bonds point that makes a difference. Okay? Now, how about the relationship between the middle and the right? Are those identical; are they superimposable? Elizabeth, what do you think?

Student: No.

Professor Michael McBride: No you can’t move those, or rotate them any way, and make them on top of one another. Are they mirror images?

Student: No.

Professor Michael McBride: So what are they?

Student: Diastereomers.

Professor Michael McBride: They’re just plain different. They’re diastereomers. So the one in the middle is the enantiomer of the one on the left but the diastereomer of the one on the right. So you don’t say it’s a diastereomer, or it’s an enantiomer. You say it’s a diastereomer of this and an enantiomer of that. Right? So these name relationships. Okay? Okay, how about those two, the first and the last? What’s the proper relationship there? Maria?

Student: Diastereomer.

Professor Michael McBride: Right, diastereomer. Good, so we got the idea. Okay now, how many isomers do you have? Obviously if you have one chiral center you have right-handed and left-handed, if there are four different things on it. So there are two isomers. But suppose you have several chiral centers. That’s not so obvious then how many you have. So if you have n stereogenic centers, as they’re called, and each could be right or left, then you’d imagine the number of permutations would be 2n. Right? That would be expected. But that’s not true because of meso compounds. And we’ll illustrate that with a quote from van’t Hoff’s brochure, in 1877, where he said: “Next we consider a symmetrical formula” (where carbon, the first carbon, has one, two, three; the next one has two fours, and the last one has one, two, three, different substituents; and also the one where the two in the middle are different, four and five on the middle carbon.) “As is easily conceived from the foregoing discussion, they lead to only three isomers.” Because you have this possibility of meso, that we just were talking about. Although this is with three centers possibly in the second one.

Chapter 2. Diastereomers and Enantiomers in van’t Hoff’s Brochure [00:16:27]

Now, Baeyer and Fischer were — Baeyer, a student of Kekulé, was the teacher of Fischer. And they got together at a resort on the shore of the Mediterranean, in Italy, and were trying to figure out this particular question of how many isomers you have here. And they had read van’t Hoff’s thing. So they knew there were supposed to be three isomers. So they were sitting at breakfast and they had bread rolls and they had toothpicks. So they started making models of these and trying to count how many isomers. But they ran out of bread rolls, presumably, and couldn’t answer the questions. They were completely flummoxed by trying to understand what van’t Hoff was saying. So they gave up on using bread rolls. And that is what prompted Fischer to go back and figure out how to make Fischer projections. Okay?

Now let’s look at that particular case using Fischer projections. So we have these three carbons that might have four different things on them. The one in the middle is a little bit tough to decide, because the ones on — as to whether it has four different things. Obviously it has H, OH, and a carbon. But those carbons that are attached are identical constitutionally. They have the same nature and sequence of bonds, things attached to them, but they might or might not be the same from the point of view of stereochemistry. So it’s not so obvious how this is going to work out. But it turns out that it’s really easy, if you use Fisher projections. And we’ll try it here. So here are the five carbons in a row, the top and the bottom one being COOHs, and you have the three carbons in the middle, each of which has an H and an OH on it, but they could be either way. So at first glance it looks like 23, or eight possible isomers. So let’s check them out, instead of using bread rolls, using Fischer projections.

So we’ll draw eight Fischer projections, and try to be systematic and draw all the possibilities, and then count them. Okay, so first of all take the bottom carbon, and we’ll put in the top row all the OHs to the right, and in the bottom row all the OHs to the left. And then we’ll do the analogous thing with the second row, except we’ll make the first two to the right, the next two to the left, the next two to the right, the next two the left. Okay? And finally what are we going to do on the top row? Can you see how we’re going to — so that we get all eight? We’ll go right, left, right, left, right, left, right, left. Okay, and now we’ve got all the possibilities. So now we have to go through them one by one and see if they’re unique, or how they are related to the others, if they’re related. Are they diastereomers? Are they enantiomers? Are they identical? And we’re going to use these Fischer projections to do it. Okay, so we’ve got the first one. Okay? And now we look at the second one and compare it to the first one. Is it the same?

Student: No.

Professor Michael McBride: No. Are they superimposable?

Student: No.

Professor Michael McBride: No, that would make them the same. Are they mirror images?

Student: No.

Professor Michael McBride: No. So two is different, right? They’re diastereomers. Okay, now about how the next one, three? Is it like any of the previous ones, or the mirror image of any of the previous ones?

Student: No.

Professor Michael McBride: No, three is unique. Okay, now how about four? Does it look like any of the preceding ones? Wilson?

Student: It’s a rotation of two.

Professor Michael McBride: Ah, you can rotate it 180°; which you can allow, right? That’ll keep the bonds pointing back, in the convention, right? And it’ll look like two. So that one actually is two. So we’ll blank it out. Now how about the first one in the bottom row here? Does it look like one? No. Does it look like three? No. Does it look like two?

Student: [Inaudible]

Professor Michael McBride: That’s not so easy, right? But it’s not superimposable. If you rotate it 180°, about the axis into the screen, it still doesn’t look like two. Right? So it’s different. Four. How about the next one? Virginia, you got an idea? Does this one here look like any of the previous ones, if we rotate it?

Student: Well, it’s the mirror image of three.

Professor Michael McBride: Can’t hear.

Student: Yeah, it looks like three.

Professor Michael McBride: Ah ha, it’s like three. If we rotate it 180° it’s like three. So forget that one. How about the next one? Does it look like anybody we’ve already seen?

Student: Four.

Professor Michael McBride: Four. Right, if you rotate it 180° it’s four. Okay. And the last one?

Student: One.

Professor Michael McBride: Pardon me? Lexy, what do you say?

Student: It’s the same as one.

Professor Michael McBride: It’s one, if you rotate it 180°. Okay. So we’ve got these four that are different. Now what are the relationships? Okay, are any of them meso? Do any of them have mirror planes, so that their own — they’re their own mirror image. Angela?

Student: One.

Professor Michael McBride: One has a mirror image. The top is the mirror image of the bottom. There’s the mirror. Any others?

Student: Three.

Professor Michael McBride: Three. Okay, but not two or four. Anything else that you want to note about two and four?

[Students speak over one another]

Professor Michael McBride: Sophie?

Student: They’re mirror images.

Professor Michael McBride: They’re mirror images of one another. Okay? So what we have are two meso isomers — they’re diastereomers of one another, right? — and a pair of enantiomers. Each of those enantiomers is the enantiomer of one another, and the diastereomer of either of the previous ones, one and three. Okay, so with Fischer projections he could easily settle this question. And that became very important in the chemistry of sugars, where Fischer was deeply involved, as we’ll see next semester.

Chapter 3. Notation Ambiguities and Multibillion Dollar Pharmaceutical Disputes [00:23:15]

Okay, now here’s Halichondrin B, which comes from a marine sponge; as it says here. Okay? Now, you don’t get very much of this out of a sponge. Right? But it was found that it was useful in anticancer therapy. So the hope was you could make it and then see if it will work as a good drug. So it was found that you don’t need the whole molecule. If you cut it apart, to have only this portion we’ve seen here, then that part is just as good, medically, as the whole molecule. So that simplifies the synthesis problem quite a bit. But it’s still non-trivial. So there’s the active fragment of Halichondrin B. And SAR means Structure Activity Relationship. They take off this, take off that, trim it here and so on, change one thing to another thing, and see if you can get better activity than you get from just this thing itself; and especially looking to changes that would make it easier to synthesize. Okay? So they found out that this one, which is called E7389, that model, is about as easy as you can get to synthesize and still have really good activity. But it’s not easy to synthesize. And you have the question that you have to get the right diastereomer because the other diastereomers are different. Now how big a problem is that going to be? Well it depends on how many stereogenic centers there are; how many carbons can generate two isomers, by being right or left-handed. So let’s start up here at the left, and I’ll go along this chain, like that, and you stop me when we get to a carbon that’s chiral, that’s stereogenic. Okay? So I’ll start here, at the top left, and I’m going across. Stop me when I get to one. Oh, you didn’t tell me.

Student: That one.

Professor Michael McBride: That one looks — well you say oh that has only three things on it.

[Students speak over one another]

Professor Michael McBride: You have to have four different things, to be stereogenic. Does anybody think that one is stereogenic? Chenyu?

Student: It left the hydrogen out.

Professor Michael McBride: Can’t hear.

Student: It doesn’t have the hydrogen in it.

Professor Michael McBride: Ah, it doesn’t show the hydrogen; the hydrogens aren’t being shown. So there are, in fact, four different things on that third atom — nitrogen, carbon, but the second carbon is stereogenic. Okay? So there’s one. Now let’s keep going. How about here, that one? No, that one has two hydrogens. How about this one?

Students: Yes.

Professor Michael McBride: That’s stereogenic. How about this one?

Students: Yes.

Professor Michael McBride: Yeah. How about this one?

Students: Yes.

Professor Michael McBride: Okay? How about this one?

Students: Yes.

Professor Michael McBride: Okay, how about this one? This one? This one?

Students: Yes.

Professor Michael McBride: Ah, that’s got four bonds already, no hydrogens there. Okay, that one’s not. There? Here? Okay there. There? There? There? Yeah, that’s got four. There?

Students: Yes.

Professor Michael McBride: Yeah. There?

Students: Yes.

Professor Michael McBride: Yeah, four different things, because you have to count the hydrogen. Here? Yeah, the hydrogen also. That one shows the hydrogen. Okay. Here? Yeah. Here? Here, no. Here? Well you say there are two oxygens. Right? Are they the same?

Student: No.

Professor Michael McBride: No, they’re attached to different things. Right? So that one too. Okay, here? No, no. Here?

Student: Yes.

Professor Michael McBride: How many think so? Yeah, right, that one. The next one? This one?

Student: No.

Professor Michael McBride: Yes or no here?

Students: No.

Professor Michael McBride: No. Here?

Students: Yes.

Professor Michael McBride: Yes. Here?

Students: No.

Professor Michael McBride: Speak up.

Students: No. Yes. No. Yes. No. Yes.

Professor Michael McBride: And there’s one more carbon here.

Students: No.

Professor Michael McBride: No. Okay. Well done. There are nineteen stereogenic centers. That means there are 219 possible isomers. Will any of them be meso? Are you going to get any mirror images within this molecule? No, it’s completely unsymmetrical, right? It’s not like tartaric acid. So there are going to be something like — something greater than half a million configurational isomers. So it’s not just a question of putting the right groups in the right place to get constitution. You have to put them stereochemically correct, to get one out of million, right? So this is a problem. But they figured out how to do it. And this, I just got this, this morning, off the web — pardon me. It’s a — it’ll say so at the top — but this is the current report on clinical testing of this; so testing in humans. Some of them are in — like if you look at the top one, it’s in phase three. In phase one they test for toxicity with humans; obviously if it’s toxic, you can’t use it as a drug. The next one they test “does it work?” And the third they test whether it’s better than current things for doing it, and what the proper dose should be and so on.

So this is phase three trials, and they’re recruiting people to — I’m sorry, I got — pressed the wrong button — they’re recruiting subjects for it. They already have 1100 people in it. It started in June 2006, and this was updated last May, and it’s to — it doesn’t have a completion date given there. This one has a completion date of March 2010. But in fact this is the objective, to see what the quality of life is; tumor response rate; duration of response; how many survive one, two and three years, in this breast cancer test. And it’s being compared with another drug, as it says there. So here’s what they’re comparing in the two different things. Some people do this, some people do this. So the drug, this drug is injected 1.4 mg/m2, IV infusion over two to five minutes, on days one and eight, every twenty-one days. And it’s being compared with this drug, which has 2.5 g/m2/day, administered orally twice daily in two equal doses, blah, blah, blah. Okay, so that’s the test that’s being undergone now. And the estimated primary completion date is September 2011. It’s been going since June 2006. So it’s going to be five years that they’re doing this test, with 1100 people.

So these are really, really expensive things. And you notice it says it’s being — maybe it was on a previous slide — that it’s being funded by industry. Because they hope that it’ll work and then they’ll be able to recoup their costs and make a profit on the basis of selling the drug afterwards. Okay. If you go to this NIH website,, you find that they have 64,268 trials in 158 countries going as of today. So these are very, very long, expensive things, to try to get a drug that will work. But making that stuff was really something. But if they get a hit, then the company is on easy street, for a while at least. Like Lipitor. So here’s a Lipitor pill, which is the world’s best selling drug. In 2004 it sold almost eleven billion dollars. Right? But it raised a 10 billion dollar problem in stereochemical notation, that I’ll show to you here. So this is a news report from 2005, dateline New York.

“Pfizer Inc. won a significant victory on Wednesday when a British judge upheld a key patent covering its blockbuster cholesterol drug Lipitor in the United Kingdom. But the medication still faces a similar yet more important case in the United States.” (And that was decided in 2006, in the same way, in the U.S. Court of Appeals.) Okay, so, “Judge Nicholas Pumfrey” (in the High Court in London) “upheld the patent covering atorvastatin, Lipitor’s active ingredient, but ruled that another patent was invalid. Indian pharmaceutical company Ranbaxy Laboratories Ltd.” (a generic chemical company) “had challenged both patents, and was joined by Britain’s Arrow Generics Ltd. against the second patent that was ruled invalid. Pfizer said the decision upholding the exclusivity of the patent covering atorvastatin until November 2011 was an important victory for scientists.”

And perhaps also for stockholders and executives of Pfizer. Okay? Now, what is patented? Here’s the patent that was under discussion. This is the U.S. version. There was also a British version. Okay, so it’s patent number 4,681,893, from 1987. Okay? So the important part of a patent is the claims at the end, what’s to be protected. And it said, “I claim a compound of structural formula 1.” So this compound, okay? And notice that it has codes in here; like X can be any of those groups, linking the right and the left; R1 can be any of those groups; R2 and R3 can be any of these groups. Right? Still going. R4 can be any of these groups. Or it can be a hydroxyl — instead of being this kind of ester, which is called a lactone, it forms a ring — it could be the acid that you’d get from that by hydrolyzing it, or a salt of that acid. Okay, so all those things are covered in that patent. How many molecules are covered in the patent? If you made all the permutations of those groups, you’d find that it’s greater than the number of protons, neutrons, and electrons in the solar system. So it’s obviously impossible to make all the molecules. You’d need quite a few solar systems to have enough particles to do it with. Okay? But the patent is supposed to protect that.

Okay, now. So here — but what is patented and what was being discussed was not what those groups are. That wasn’t in contention. It’s known that Lipitor is this stuff: the calcium salt, and it has those particular R groups on it. But that wasn’t in question, the fact that it covered so many compounds. What was in question was what the stereochemistry is. Because Lipitor, the drug, is a single enantiomer. Right? Not its mirror image and not a mixture, not a racemate. Okay? But the question is, did this patent cover the single enantiomer? Which is what the picture shows; although it doesn’t show it very well. Why not? If you were drawing this structure, on the top right, would you draw it that way? How about that carbon? Does it show its tetrahedron well?

Student: No.

Professor Michael McBride: It certainly doesn’t show 109° angles, but it’s — and this dash should be up here, to be even close. How about that carbon? That’s even worse, if you were trying to really show the angle, because that one’s 180°. But at least it’s unambiguous about that versus its mirror image, even though it’s not very good. Right? It’s not ambiguous. So the structure drawn in the patent was a single enantiomer, and it turned out to be the one that is Lipitor. Okay? But the text of the patent didn’t do anything. It never resolved it and made it into one enantiomer. It just talked about it generically. And what it would be making is the racemate, not the single enantiomer. So does the patent cover Lipitor or doesn’t it, was the question for the judge. Okay, so this is his opinion. He said:

“In the ‘633 patent,” (when they talk about patents they just talk about the last three digits, not the big number) “it is absolutely clear from context throughout that formula (1) is being used to denote a racemate.” (So what they talked about and what they prepared, tested and so on was both enantiomers. Okay?) “In my judgment, every time the skilled person sees formula 1 or formula X,”

Now, skilled person has a very special meaning. It means someone who is “skilled in the art”, that knows about pharmaceutical chemistry and so on, knows what these compounds are, how to make them and that kind of thing, but is not creative. Because what you can patent is original creation. So the skilled person knows everything but can’t figure anything out; if you see what I mean, can’t devise anything new. Okay so the skilled person sees these formulas.

“He will see them with eyes that tell him that in that racemate,” (which is described in the text) “there is a single enantiomer that is the effective compound,” (So even though what’s described is both, the person that isn’t very swift, but is knowledgeable, knows that within that there’ll be two mirror images, and one of them will do the trick, and not the other one. So you don’t have to be creative to know that.) and he will know “that he can resolve the racemate using conventional techniques.” (He doesn’t have to invent something new, in order to get from that mixture, the one that will be the active one. Okay, so that’s not creative. Okay?) “When one comes to claim 1, which echoes the purpose of the invention with its conventional reference to pharmaceutically acceptable salts, he will, in my judgment, continue to see the formulae in this light.” (So the salt’s the same deal, right?) “In my view, the claim covers the racemate and the individual enantiomers.”

So the person without any creativity knows that yes, the patent described the racemate, but there’ll be one in there that I can get, and that’ll be the useful drug. And it doesn’t take anything great to get there. We’ll talk about that soon, how you would do that; why it’s well-known. Okay? So this is good news for Pfizer, right? Their patent holds. They put out a press release saying this is a victory for science. Okay? But it’s in a sense a pyrrhic victory, because they had a later patent, on the single enantiomer. Right? And now that patent is invalid. Because if the first patent covers it, you can’t just every three years come up and say, “I patent this again,” “I patent this again,” I patent this again,” and keep getting another fifteen years of exclusivity. Right? So once they were covered by the first patent, the second patent is invalid. Why do they care, if they’re already covered by the first patent? Isn’t that good news? Dana?

Student: Their fifteen years of exclusivity will run out.

Professor Michael McBride: Ah, they’re going to run out. So this means that three years earlier their patent is going to run out — right? — at four or five billion dollars a year. That adds up. Okay?

Chapter 4. The IUPAC and the Standardization of Molecular Nomenclature [00:39:26]

Now, so drug — so there are conventions for how you name things, including stereochemistry. In fact, there was a stamp issued in Switzerland in 1992 commemorating the 100th year of the Geneva Convention, where chemists got together, once they knew stuff about how things were connected. Then they had to agree on how to describe it, on what the nomenclature and the notation should be. So the International Union of Pure & Applied Chemistry continues this. And you can go to these websites and learn about what the rules are for notation. So they had to devise rules for what you want to do when you make a name for a compound. So what is maybe the first thing you want to do? Nate?

Student: You need to —

Professor Michael McBride: No, not you.

Student: The other Nate?

Professor Michael McBride: Well let’s see. Let’s try somebody else. Andrew? No, no. Kate? Kate?

Student: [inaudible].

Professor Michael McBride: Kate, what was the first rule?

Student: They want the name to discuss composition, like which atoms are in it.

Professor Michael McBride: Well — I didn’t mean you anyhow. So what’s the first rule about names?

[Students speak over one another]

Professor Michael McBride: John? Not you. [Laughter] Other Kate?

Student: They should describe only one.

Professor Michael McBride: Ah, they should be unique. There should only be one compound that has that name. You shouldn’t say, “Take this,” and they take the wrong John or the wrong Kate or the wrong Andrew. Right? So the very first thing should — well it should be — Kate, you were right first. It has to be clearly descriptive of what the compound is. And so it has to tell composition, constitution, configuration and conformation maybe, if you care, i.e., stereochemistry. And it can be like this. E7389 was the name of that compound. Forget trying to name it systematically, according to rules. Right? E is for Eisai, the name of the pharmaceutical company, and this is their 7,389th compound. Right? But at least everybody knows what that means. Right? Or you can look it up. But this is what we were just talking about. It has to be unambiguous. Right? One name must mean one structure, not some other structure as well. Okay, so unfortunately “amide” doesn’t fit that bill, because it means the anion of an amine, but it also means that functional group we’re talking about. It’s the same name. You have to figure out from context which one you’re talking about.

It also has to be unique. One structure is one name. So everybody will have the same name for the compound. Right? Which is more important, that it be unambiguous or that it be unique? Like you could have two names for the same compound; like bromobutane and butyl bromide. Right? And if you put a number in, you would know which — where the bromine was. Okay? Or bromoethane and ethyl bromide. Two different names, right? So the names aren’t unique. Right? But they’re unambiguous. You know what compounds you’re talking about. Why would you care whether you have just one agreed-on name by the people who get together in the International Union of Pure & Applied Chemistry? Why would care? What’s wrong with having two names? Anybody see any disadvantage to having different names for the same compound? Sam? I could’ve called Sam too. Yeah, go ahead.

Student: It’s hard to communicate about the stuff.

Professor Michael McBride: Not if everybody knows what you’re talking about. Shai?

Student: Well like if something has a lot of names, eventually you’re going to get to a point where someone doesn’t know. You can say — you’re going to say a name and they’re not going to know what you’re referring to.

Professor Michael McBride: Or if you have to look it up, you have to choose the right name to look it up. So this was more important, when you didn’t have computers to do the work for you, when you had to look in an index and get — and be looking for the right name that everyone agreed, and the person who made up the index entered it there. Nowadays you can draw a structure often on a screen, and the computer will figure out what the official name is and then look it up. But anyhow, that — so it’s not as important that it be unique as that it be unambiguous. But if you’re looking things up, it’s very nice if you know that everybody will have named it a certain way. So you can also index things according to their composition, CHN, or using computer graphics, as we say. And, if possible, the name should be manageable. Right? It should be possible to write it fairly quickly. It should be easy to figure out; short, if possible, and pronounceable, if possible, so that you can talk about it.

Now let’s look at systematic constitutional nomenclature, for this sort of intermediately complex compound. Compared to these others we’ve looked it’s not much. Okay, so the first thing you do, these people decided, when you get together, is decide what the main chain of carbons is. Because that will provide the root with a Greek designation that says what the number of things is. So we have to find a carbon chain, and we’ll find the longest carbon chain. We could take this one, or we could take that one. Which one is longer? The one, it could be an octane or it could be a nonane. Which one should we choose?

Students: Nonane.

Professor Michael McBride: Well we could’ve decided you take the shortest one. But that’s sort of silly. There’d be a zillion shortest ones; or short ones, right? Or even shortest. The shortest is one carbon I guess. But there’s only one longest one, right? Wrong. Because there’s also that one. Same length. So now we have to decide when we’re going to take the longest; how are we going to choose, if there are two of them? Well what you choose by is the number of substituents. Okay? Because that one has one, one, three, four, five substituents, and this one has four substituents. Which will you choose, the one with four or the one with five? You have to make the rule up. five or four?

Student: Five.

Student: Four.

Professor Michael McBride: Somebody said four. Why? Sophie you said four?

Student: Less naming to do.

Professor Michael McBride: There’d be less naming to do. Ah ha. But in fact they chose the most substituents, and you’ll see why shortly. Okay, so you choose the most substituents. So you choose the one on the top, not the one in the middle. Now why? Sophie’s not happy with this. But you’ll see soon. Now you have to number the carbon atoms, to say where the substituents are. Now, we can start from either end. Right? If we start one, we could go one, two, three with green and continue; or one, two, three with red, from the other end, and continue. Now how are we going to decide which way to go? Kevin?

Student: Right. So you want in this case the Cl and the Br to have the lowest — to be associated with the lowest number of carbons on the chain.

Professor Michael McBride: Okay, there are two ways to do it. You could make it so the overall sum of the numbers you’re going to get — because you’ll get different numbers, according to which end you start from. You could sum them all up and see which one gives the lowest sum and choose that one; or the highest sum you could do. Or you could see which one gives the lowest number at the first number. So this, if you chose the red, you’d have 2-chloro. If you chose the green, you’d have 3-bromo. Right? And the other numbers would all be higher. Which one do you want to do, use the lowest sum of all the numbers, or the lowest number at the first difference? Lucas?

Student: Lowest number at the first difference.

Professor Michael McBride: Why?

Student: So you can actually build it as you go, instead of just saying —

Professor Michael McBride: Yeah, so you don’t have to do all the work of going through the whole thing and adding them all up and so on. You just go ‘til you get a difference and when you get a difference that’s it. That’s much quicker. Okay? So we’ll go to the first difference, lowest number at the first difference. And now we have to say — now we know we have the chain and its numbers, and now we say where the various substituents are. But first we have to name the substitutents. Now in the top you can see we have methyl, bromo, methyl, ethyl and chloro. Now had we followed Sophie, we would have to name that one, and that’s not as easy to name. So the fewer substituents you have, the harder they are to name, because they get more complicated. Right? So that’s why you choose to have the most substituents, so they’ll be the easiest to name. You have to do more but it’s easier. That particular one is called 1-chloroethyl. But why have to name complicated things if you can name simple things? Okay? So you choose the most substituents in order to get simpler names.

Then you alphabetize them. You don’t put them on by order of their number, you alphabetize the names of the groups, and you count. So that compound is 7-bromo- (b is first) 2-chloro- (c) 3-ethyl-6,7-dimethylnonane. Now you might say dimethyl should be alphabetized by d, not by m. But that’s not the way it’s done. It’s done by the group name and then the prefix that tells the number doesn’t enter into the alphabetizing, usually; although some people do it the other way. Okay, so that’s the question about the d. Okay.

Now, if you go to this website you can get help with nomenclature. Here’s a compound that it goes through. And the minute you look at this and see the name at the bottom, the compound name, 4,5-dichloro-2,4-chloro-2-hydroxymethyl-5-oxo-hexyl-cyclohexane-1-carboxylic acid, you say “Wow!” And aren’t you glad there are computers that can do this for you? But you can go through it and figure out why they did it that way. But that’s sort of a parlor game. It’s not fundamental chemistry. Okay? But there’s some very useful non-systematic names for simple groups: isobutane, isopentane, neopentane. So you can have an isobutyl group or an isopentyl group or a neopentyl group, isopropyl, secondary-butyl, tertiary-butyl, neopentyl. So these names you just have to learn. Right? They’re not systematic. And there’s a nomenclature drill available on the course website. So it’s good to know these so we can talk to one another. But in principle the systematic name is the way to go. Okay, that’s it. But we haven’t gotten to stereochemistry yet. That’ll be tomorrow, or Monday.

[end of transcript]

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