CHEM 125a: Freshman Organic Chemistry I

Lecture 29

 - Preparing Single Enantiomers and the Mechanism of Optical Rotation


Within a lecture on biological resolution, the synthesis of single enantiomers, and the naming and 3D visualization of omeprazole, Professor Laurence Barron of the University of Glasgow delivers a guest lecture on the subject of how chiral molecules rotate polarized light. Mixing wave functions by coordinated application of light’s perpendicular electric and magnetic fields shifts electrons along a helix that can be right- or left-handed, but so many mixings are involved, and their magnitudes are so subtle, that predicting net optical rotation in practical cases is rarely simple.

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Freshman Organic Chemistry I

CHEM 125a - Lecture 29 - Preparing Single Enantiomers and the Mechanism of Optical Rotation

Chapter 1. Introduction: Challenges in Isolating Enantiomers Despite Optical Activity [00:00:00]

Professor Michael McBride: So you remember Pasteur was very fond of racemic and tartaric acid. So when he’d travel around he would look for samples of tartaric acid. There are very detailed accounts that he published of these travels. I haven’t found this one, so I’m not sure it’s true. It may be apocryphal, but it was told to me by the guy who taught me organic chemistry. So I pass it onto you, for what it’s worth, because it makes a good point. He said that Pasteur was in Alsace, the wine-producing region in northeast France, and in the town of Thann he found in a pharmacy a bottle of racemic acid that was moldy. So remember racemic acid’s the 50:50 mixture. So, since he liked racemic acid — and in fact it was becoming rare. It’s a not a very common product; mostly you get the handed stuff. So he took it back and cleaned it up, got rid of the mold that was growing on it. And what he found was that after he had purified it, it was the unnatural, the left-handed, not the one you normally get. Can you see an explanation for that? Had the apothecary mislabeled it because he thought he could get a higher price for racemic than for tartaric acid? Or might he have been innocent? Zack?

Student: Was it the mold by any chance?

Professor Michael McBride: What did the mold do?

Student: Might have converted one into all.

Professor Michael McBride: Didn’t convert.

Student: It didn’t?

Professor Michael McBride: The mold ate the natural stuff, converted it into something completely different, leaving the unnatural one. Okay? Remember the smell of the carvones? Natural things, enzymes and so on, can distinguish between the two hands, and naturally the stuff that normally eats tartaric acid is going to eat normal tartaric acid, not the unusual one. So it left the unusual one. The penicillium glaucum had eaten the R,R. Okay, so that’s one way of doing resolution, to get rid of one enantiomer from a racemic mixture, leaving the other one, because diastereomeric reactions have different rates. It’s a right hand shaking a right hand versus a right hand shaking a left hand; they’re just different. So one will go faster than the other. So if you react a racemate with some chiral reagent — of course you have to use only one hand of that other thing you add in there — but then you’ll react with one more than the other. It could be even a catalyst so that it doesn’t get consumed; an enzyme, for example. But that’s not Nature’s way, because it’s very inefficient to make both and then destroy one. What nature does is prepare only one enantiomer. And you can do that, prepare only one, by starting with something that’s already a single enantiomer and building on that. Or you can use a reagent, which is resolved, that will — because diastereomeric rates are different — will tend to produce one of the two enantiomers.

So here’s an example of this molecule we’ve been talking about, Eisai-7389, with its nineteen asymmetric centers, that’s made artificially, commercially. And this is where those nineteen centers came from. Five of them came from starting materials that they bought as a single hand. But the rest were all made. One of them was done by chromatography. They have chromatography columns about this big around, and this high, with chiral stuff inside. So you pass your thing through. One hand goes through a little faster than the other, because of its diastereomeric interaction with the packing; one comes through quicker. And they actually do that for one of the centers. But the others are all done by reactions that preferentially give one center rather than the other. Three of them are ones that everybody knew which one they would give. The other ones they just had to hope, and fiddle around with different reactions until they found one that would do the trick. So anyhow, they were able to generate these nineteen stereocenters.

Now we’re left with a problem. That incidentally is all we’re going to have for the exam on Friday. Okay? So now we’re on to other stuff that will be covered on the final exam, and in particular return to this question about the tartaric acids. Remember there’s d-(+) and l-(-). Could you have l-(+) and d-(-), of a different compound? Why not? What does d or l mean, in this context?

Student: [inaudible].

Professor Michael McBride: Andrew?

Student: It reflects light to the left.

Professor Michael McBride: It means which way it rotates the light. Right? And the plus means the same thing, as d. So it’s redundant, this particular nomenclature. But there’s a question mark because when — Fischer just guessed, remember? And it could easily have been the opposite. So which way is it? Right? So if we knew, if we knew how optical activity worked, then by measuring the optical activity we’d know which of those structures is right. Okay? But we don’t know that; at least I don’t know that. Now there’s a lot of knowledge about this, but it’s a very, very tough problem. Fortunately there’s a book by Laurence Barron called Molecular Light Scattering and Optical Activity, which goes into this stuff. And I see Professor Vaccaro —

Professor Laurence Barron: I’m here.

Professor Michael McBride: Oh, so we have a copy of the book. That’s great. But even better than that, we have Laurence Barron.

Professor Laurence Barron: Or a copy of Laurence Barron.

Professor Michael McBride: Or a copy of Laurence Barron. [Laughter] Perhaps it’s the mirror image of Laurence Barron. So he’s going to tell us how this works.

[Technical instructions]

Chapter 2. Barron: A Sketch of Lord Kelvin and Chirality [00:06:09]

Professor Laurence Barron: Well good morning everybody. I was both delighted and dismayed, at the same time, when Professor McBride asked me to try and — well to address you this morning. Delighted to have the opportunity to talk science to some of America’s brightest budding young scientists, but also dismayed that he asked me to try and explain the molecular origin of optical rotation, optical activity. It’s a very subtle, difficult, delicate problem, that’s exercised some of the finest minds in physics and chemistry for the last hundred years. But anyway, let’s see how it goes. Towards the end I think I may be presenting some stuff that’s sort of beyond the boundaries of your current knowledge. But anyway, it can at least pass in front of your eyes. So chirality then means, as you well know, right- or left-handedness. It pervades much of modern science; from the physics of elementary particles, through organic stereochemistry, to the structure and behavior of the molecules of life; with a lot more besides. It comes up in what’s called nonlinear optics, involving intense lasers; nanotechnology; materials, electrical engineering; pharmaceuticals; astrobiology; and origin of life. So it’s a very important theme in modern science.

Now first of all though I’ll tell you a little bit about Lord Kelvin. He was the first person to introduce the word chirality into science. He was professor of natural philosophy in Glasgow — which is my home university — through most of his career; well all of his career. He was one of the giants of physics of the nineteenth century. He’s best known for inventing the absolute Kelvin temperature scale. Now he was originally, his original name was William Thomson, but then he became famous and became Sir William Thomson. Then he became even more famous, so they made him a Lord. And when you’re made a Lord in the U.K., you choose your title from someplace that’s dear to your heart, maybe your home area. He took his title from the name of the River Kelvin, which runs through the University Park in Glasgow. So whenever you use the absolute temperature scale now in the future, you can picture this idyllic scene.

Anyway, so he was the first to introduce the word chirality into science. And here’s his definition, which you’ll be familiar with: “I call any geometrical figure or group of points chiral, and say that it has chirality if its image in a plane mirror, ideally realized, cannot be brought into coincidence with itself.” That was in his Baltimore Lectures. So he’s just emphasizing the non-superimposabity of the mirror image, the enantiomers of a chiral molecule. But of course the whole subject started earlier with the wonderful work of Louis Pasteur, who showed mirror-image chiral molecules show optical rotation of equal magnitude but opposite sign; which was an epoch-making discovery. So you’ve come across then this, the fundamental manifestation of optical activity, which is natural optical rotation. You put linearly polarized light beam into a sample — say a sample of an isotropic collection of chiral molecules, like a sugar solution — and it will come out the other side with the plane of polarization rotated through some angle. And if you put in the mirror image version, you’ll get an equal but opposite sense of optical rotation.

Now it’s not to be confused with something called magnetic optical rotation, the Faraday effect. I’m just mentioning this to you. You probably haven’t come across the Faraday effect yet, but you may do later on in your studies, or in your professional life. So I’ll mention the Faraday effect. Faraday discovered, in 1846, that achiral samples — no natural optical activity there — if you apply a static magnetic field, parallel to the light beam, that will induce an optical rotation. And if you put in — if you reverse the direction of the magnetic field, relative to the light beam, you’ll get an equal and opposite sense of optical rotation. It would even work say for a sample of water, for instance. Any material will show a Faraday effect. But this has been a source of much confusion, in fact, to scientists.

Now Lord Kelvin was on the ball here. He knew all about it. He made a statement here. He said, “The magnetic rotation has neither right-handed nor left-handed quality.” (That is to say no chirality; it’s got nothing to do with chirality.) “This was perfectly understood by Faraday and made clear in his writings, yet even to the present day we frequently find the chiral rotation and the magnetic rotation of the plane of polarization classed together in a manner against which Faraday’s original description contains ample warning.” Well Lord Kelvin would be turning in his grave today, 100 years later, because you still see papers which involve the Faraday effect, in some way or other, and in the introduction they talk grandly of “inducing chirality with a magnetic field.” That is completely wrong. Just for the record — probably beyond your knowledge at the moment — chiral phenomena, like natural optical rotation, they’re characterized by what’s called time-even pseudoscalar observables. A pseudoscalar is a number that changes sign under reflection or inversion; we’ll leave it at that. So that’s for the record. But well it turns out the essential symmetry characteristics of natural and magnetic optical rotation are completely different and you need different sorts of molecular quantum states, different characteristics to support them. But we won’t pursue that.

Chapter 3. Natural and Magnetic Optical Rotation [00:12:35]

Right, now back to natural optical rotation. So now we bring in circularly polarized light. So in order to detect molecular chirality, you must have some sort of chiral probe. Well what we’re using here is right and left-circularly polarized light beams. They are actually mirror-image chiral systems. So they can be used as chiral probes. So here’s a representation of a right-circularly polarized light beam. Now you know that light involves electromagnetic oscillations in space, and usually you just think of the oscillating electric vector of a light wave, and if it’s linearly polarized, it’s oscillating in one plane. There’s actually also an oscillating magnetic field vector that oscillates perpendicular to the electric. We’ll come back to that later. We’re just looking at the oscillating electric vector here. So if it’s linearly polarized, or plane polarized, it’s just oscillating in a plane; but circularly polarized light, as well as an oscillation in this direction. You also have an electric [magnetic] vector oscillating perpendicular but 90° out of phase. And what happens is you get this circular polarization.

Now this picture here, this represents the instantaneous electric vectors at different points in space, in the direction of propagation, along the z direction. And so here we are, that’s just showing — that’s just connected the ends of the instantaneous electric vectors. And there it is for left-circularly polarized. Anyway, you can see that at the very least those are chiral, those are helical, and those are mirror-image chiral systems. Now here we go. This is some extra fancy stuff that Professor McBride added to my [laughter] presentation. He doesn’t like — it was too simple and static. Anyway, so there we are. So now if you look at the wave, if you just look at the electric vector through — in a fixed plane, as the light wave is propagating, there’s the electric vector, and it will rotate in the fixed plane. So rotating clockwise, that defines right-circularly polarized light. And here we go, rotating counter-clockwise; that’s left-circularly polarized light.

Okay, so we now have a chiral optical probe, circularly polarized light. Now chiral molecules respond slightly differently to right and left-circularly polarized light. I mean, an extreme example is say in the world of engineering. You can’t fit a left-handed nut onto a right-handed bolt. That’s an extreme example of different chiral interactions. Well it’s obviously much more delicate than that here. But the point is right and left-circularly polarized light interact just slightly differently with a molecule of a given chirality. Anyway, there’s a differential absorption of right and left-circularly polarized light. That corresponds to a phenomenon called circular dichroism, which is the basis of a widely used form of spectroscopy used to study chiral molecules. But now, what gives rise to optical rotation is a difference in refractive index towards — of right and left-circularly polarized light beams.

Now linearly polarized light, you can describe it as a coherent superposition of right and left-circularly polarized waves of equal amplitude. Coherent means they’re in-phase with each other. Rather than being random, they’re oscillating in-phase with some fixed phase relationship. So, for example, now here’s a linearly polarized light beam. But you can decompose it into a superposition of a left-circular and a right-circular component. You see when those are at the top they’re reinforcing, and you’ve got the maximum linearly polarized vector up here. As they come away, left and right, they will tend to increasingly cancel, and that will decrease. And when they’re this way and this way, you’ll have zero electric field vector there, and as you come down it will increase again. So you can decompose a linearly polarized light into a coherent superposition of left and right waves. Now, refractive index corresponds to velocity through a medium. So if there’s a difference in refractive index for right and left-circularly polarized light beams, that means there’s a slight difference in the velocity of the right and left-circularly polarized light beams going through the medium. So the phase relations between the two contrarotating electric vectors will change. And you can easily see, this will give you a rotation in the plane of polarization. You see, if there’s a difference in velocity, then at some instant this vector, this electric vectors of the left component will be here, and the right component here. And if you take the resultant, you see, it’s no longer where it was. So this is a simple picture of how optical rotation develops, in terms of different refractive indices of the right and left components.

Now there’s a picture in Atkins’ Physical Chemistry. He tries to illustrate this there. He’s got the linearly polarized beam coming through, and you’ve got the — he’s broken it down into the left and right, and he’s saying the two velocities are slightly different, and that gives you the resultant optical rotation. You can easily develop an expression for the angle of rotation as a function of the difference in refractive indices for left and right-circularly polarized light beams. And it’s also — it’s a function of the path length. Obviously the longer the path length, the more rotation will develop. And there’s the wavelength there in the denominator. In fact, I mean, this is the secret; this path length, that’s the secret of how you get a measurable rotation. Because this is an incredibly tiny effect. If it was just a single molecule event you’re looking at, the polarization changes, you wouldn’t see anything, they’re so tiny. But you can build up this rotation over long path lengths, centimeters or even meters. In fact, if you go to Google, Google Images, and just Google ‘circularly polarized light,’ you’ll find lots of sites there which describe polarized light, and they provide beautiful simulations, animations of this effect here. I didn’t want to download any and try and show them here, but I would encourage you to go to that. But go and look in particular at this site here, That comes from an institute of enzymology in Budapest. They’ve got some beautiful simulations of this optical rotation process and other more exotic polarization effects, as light propagates through matter.

Chapter 4. Understanding Optical Activity via the Carbonyl Chromophore [00:20:50]

Now, let’s try and — so that’s just, what would you say, a phenomenological description. And in fact that’s where most physical chemistry textbooks stop. And in fact Atkins’ Physical Chemistry stops at that point, and then just there’s just sort of hand waving saying okay, right and left-circularly polarized light interacts slightly differently with the chiral molecule; and they leave it at that. They don’t attempt to try and give you a picture in molecular detail. But Professor McBride wanted me to try and attempt this. So just as a start, this is a simple picture, a simple scattering picture of optical rotation. Now a circularly polarized light wave ‘bouncing’ from one group to the other, as it scatters from a simple two-group chiral structure, will sample the chirality. So here we have a simple two-group chiral molecular structure. So we’ve got two achiral groups, held in a rigid, twisted arrangement. I think that’s left-handed I’ve got there. And you can break down — if you look at most chiral molecules, just look at the bonds, you can often break them down. You can see these sorts of two-group structures throughout. Anyway, so a particular model of optical rotation is that the light wave, it bounces from one group to the other, before coming off and getting involved in generating the optical rotation phenomenon. But you can see, it’s sampling the chirality as it bounces from one to the other. So if the light beam is right-circularly polarized, that bouncing process will have a slightly different amplitude, as we say, from if it’s left-circularly polarized. So that gives you a simple picture. It’s worth mentioning that this is the basis of something called the “dynamic coupling model of optical activity” that was developed by somebody called John Kirkwood, who was chairman of this department in the 1950s.

Well now we really have to [laughter], we have to grasp the nettle at this point, because you just — it’s hopeless messing around with these simple models. [Laughter] They don’t get you anywhere as regards prediction; you know, relating structure to sign and magnitude of optical activity. You have to go to the quantum-mechanics. So now this was the expression for the optical rotation, in terms of the refractive index difference for left and right-circularly polarized light. Now you can develop this using something called quantum-mechanical perturbation theory, and you develop an expression for the rotation angle in terms of this incomprehensible looking stuff. But let me just try and give you a feel for what it’s telling us. The heart of it is this so-called rotational strength, which involves this so-called scalar product of an electric dipole and a magnetic dipole transition moment. So what we’ve got here is there’s the ground state of the molecule n; j is some excited electronic state; and μ is the electric dipole operator that’s connecting the ground to the excited state. And the light wave is interacting, is coupling with this electric dipole operator, and it’s driving the transition. And it’s this — these electric dipole transitions, like this, that’s behind conventional spectroscopy. You’ve done probably UV and infrared spectroscopy.

But now what we have in addition is the same transition, but now brought about through m; that’s called a magnetic dipole operator, and that’s activated by this oscillating magnetic component of the light wave. And so you have this so-called scalar product. So μm, that would be μxmx + μymy + μznz. Some of you have probably done vectors. Maybe others haven’t. But okay. So that’s the heart of it. But this is a very important feature here. We’re summing over all excited states, j, all excited states. So the whole plethora of excited states of a chiral molecule come in here, and some can give you — one particular excited state can give you a positive contribution to the optical rotation; another one a negative. You know? So it’s a very subtle problem, and you have to consider them all carefully. So optical activity ultimately originates in interference between electric and magnetic dipole transitions during the light scattering process. That’s at the heart of it.

Now, let me just show you how this works out. I can now give you a chemical, an interpretation, in terms of a real and highly important system in organic chemistry: the carbonyl chromophore. I believe you have come across the carbonyl chromophore. A lot of important organic molecules contain this carbonyl chromophore. It gives rise to a transition in the near ultraviolet at about 290 nanometers, and it’s widely used in physical organic chemistry. It’s called the n to π* transition. Now the carbonyl group itself — here it is — that’s not chiral, that’s got a plane of symmetry. So by itself it’s not going — there’s going to be no optical activity there. But you see, in this particular molecule here, 3-methylcyclohexanone, that molecule overall is chiral. There’s a chiral center there. So that carbonyl group is experiencing a chiral perturbation, a through space perturbation from the rest of the chiral molecule, and that induces chirality into the electronic transitions of the carbonyl group, and induces optical rotation and circular dichroism. Well let’s look at this in a bit more detail, this famous n to π* transition. So here we’ve got the σ bonding orbital of the carbonyl group. There’s the π bonding orbital, made up of two — of the on carbon and on oxygen. And here is a lone — sorry, here is a py orbital. So those are px orbitals. That’s the py orbital on the oxygen. Feed electrons in, and so you’ve got, in the ground state, you’ve got two there, two there. And you’ve got two electrons in the py orbital. So those are lone-pair electrons. So the lowest order, the lowest transition now is the py to π*, often called the n to π* transition. You’re promoting one electron from the py orbital up to the π* orbital; and that’s the origin of this transition. Well let’s — ah now.

Okay, now this transition, it happens to be fully magnetic dipole-allowed but completely electric dipole-forbidden. What happens is electric dipole character is induced by mixing — well I’m giving the example of a higher oxygen dyzorbital into the π* orbital. Now Professor McBride has been messing around with my presentation here. Let’s see what he’s done. So here’s the n orbital, the py orbital of the carbonyl group. There’s the π*. Ah, now here we go. So this is now — but wait — yeah, so here you can see, in that n to π* transition, there’s a net rotation of charge, a rotation of charge. And that’s the essence of the magnetic dipole-allowed character. Magnetic dipole transitions involve a rotation of charge.

Professor Michael McBride: And the reason we messed about with them in this way is we’ve seen the mixing p orbitals changing the orientation, causing them to rotate that way.

Professor Laurence Barron: Right. So what’s — ah okay, now here we go again. Now here’s a dyz orbital. Now one could consider a whole loads of other possible orbitals to mix in. But this is the simplest one, just to illustrate the idea here. So there’s a dyz orbital. Now going from — you see n to dyz, if you add those orbitals together, you tend to get a displacement of charge in the z direction. So now, you see, you have a combination of a rotation of charge with a displacement of charge, and rotation plus translation gives helicity. And here we go. There. So by mixing in a little bit of this dxz orbital, you get this — which is electric dipole-allowed — you get this helicity in the transition.

Now, in fact, the rotation of charge generates a magnetic dipole perpendicular to the plane of rotation, and it’s pointing that way. So you’ll get an mz component, a component of the magnetic moment along the z axis. Here you can see immediately you’ve got a component of the electric dipole along the z axis. So that generates a μzmz component, from this scalar product of μ and m, in the rotational strength. So this is sort of making a meal of it here now. So we’re just putting down these so-called quantum-mechanical matrix elements. So here we’ve got n to π*, with a little bit of that mixed in, and that’s fully magnetic dipole-allowed. Then we have n to π*, which is forbidden electric dipole, but mixing that in gives you a little bit of electric dipole character. And so you now get a non-zero contribution to the rotational strength.

However, you wouldn’t want to go on and actually calculate the optical rotation of the carbonyl chromophore, in some situation from this. Because, as I said, there’s many other excited electronic states you could have also considered, which may be giving opposite contributions to the rotational strength. You have to sum them all, and to do that you now have to go to modern ab initio quantum-chemistry, quantum-chemical calculations. There’s wonderful programs out now, from Gaussian and Dalton; you can calculate all sorts of molecular properties, with quite good — very good accuracy, in many cases. But this now turns it all into a black box procedure. You just feed in appropriate atomic orbitals and press buttons and turn handles, whatever you do for these calculations, and out will pop some physical quantity. So you can calculate this whole thing, ab initio, and taking in however many excited states in the sum are necessary. And you can calculate the sign and magnitude of the optical rotation, for a given absolute configuration. So you would feed into the calculation whether it’s the S or the R absolute stereochemistry. So you’d put that in, and that will determine the sign of the optical rotation that comes out. And so — I mean, in this particular case, this small molecule, the S absolute configuration, goes with the plus rotation and the R goes with the minus. So from the calculation then you can relate the sign of the optical rotation to the absolute stereochemistry.

But you wouldn’t want to stake your life on it, because it’s not completely reliable. It usually gives the right answer, for smaller molecules, but not always. And, for example, if this was — as you know, and I think you’ll hear more after my talk here. Many drugs now are chiral, and drug companies like to market now single enantiomer versions of the drug. So it’s a tremendously important problem for them to know, with absolute certainty, the absolute configuration of the particular enantiomer they’re marketing, because if they make a mistake and specify the wrong absolute configuration in their patent, they can literally lose billions of dollars. Anyway, there’s the famous — the importance of chirality in drugs is exemplified by the famous thalidomide case. But I won’t elaborate that any more. I think Professor McBride is going to tell you some more of that in more detail. I should say that the cornerstone, the definitive method for determining absolute configuration has, for some years now, been something called the Bijvoet method of anomalous X-ray scattering, which again Professor — have you told them about that?

Professor Michael McBride: Yes.

Professor Laurence Barron: Yeah. Yeah, that’s the definitive method. But it’s sort of cheating because you’re actually seeing — you’re seeing it through X-ray eyes. But even then, you can occasionally make a mistake. But I should just mention also, there are some newer optical activity techniques involving something called vibrational optical activity. Here we’ve been looking at optical activity in electronic transitions, using visible light. But in recent years newer methods, measuring optical activity in vibrational transitions, have come along. And these are actually comparable with X-ray, anomalous X-ray scattering, for reliability of absolute configuration. That’s because these calculations are much more reliable for these vibrational optical activity phenomena than electronic ones. Well that’s about it. That’s all I would like to say now. So thank you for listening.


Professor Michael McBride: Now that you’ve heard from the authority, do you have any questions about it? You are unwontedly quiet, like you usually are. Any questions? Yes?

Professor David Spiegel: I’m just wondering if the magnitude of the Faraday rotation can ever interfere with the measurement of optical rotation in the chiral material? That is to say, if one situates a polarimeter in proximity to an NMR spectrometer, for example?

Professor Laurence Barron: No. No, you wouldn’t need to worry about that at all. No. You need a reasonable strength of magnetic fields just applied directly through the sample. Yeah.

Professor Michael McBride: Lucas?

Student: Just if I’m understanding. The n to π* is magnetically dipole-allowed, electric dipole-forbidden, and that’s why then it goes to the xz as well, in order to get the electric dipole…

Professor Laurence Barron: The electric, yeah.

Student: …in there, that contribution.

Professor Laurence Barron: Yes.

Student: For my knowledge.


Professor Michael McBride: Okay, thanks again Laurence.

Professor Laurence Barron: My pleasure.

Professor Michael McBride: Very good.


Professor Michael McBride: Thank you.

Laurence Barron: Okay.

Professor Michael McBride: So yeah, this is just a plug. This afternoon — the reason Professor Barron is here is he’s the Tetelman lecturer in Jonathan Edwards College. So he’s giving a talk for the history majors and so on. But it’ll be very interesting for anyone. And I think since we’re into chirality, you would enjoy this, particularly this afternoon, in Davies Auditorium at 5:00. But there’s the question, who cares? No offense. But who cares? Why do we care about chirality? Well Professor Barron hinted at it. Living things care, because they’re chiral. Right? So which one they react with. Okay? The Food and Drug Administration cares, for the same reason, with respect to medications you might be taking. Drug companies care a lot, and their lawyers. And the US Patent Office cares a lot. Right? Which has generated a thing called a “chiral switch.” Most drugs that used to be developed were developed as racemates, because it was difficult to separate the two hands. But if your patent runs out on the racemate, and you can resolve it and now sell just one of them, and if it’s better, and the FDA will approve it, and you can convince people that that’s the one they should buy, then you can go back to not having to compete with generic drug companies anymore and charge five dollars a pill instead of fifty cents a pill. Okay?

So this so-called chiral switch, to go from racemic to a single enantiomer, is a big movement nowadays. For example, consider this pain reliever. Let’s figure out what it is. Do you remember what that group is: four carbons, in that sort of Mercedes structure? Do you know what that group’s called? This, what we’re doing for the rest of the lecture here out is actually review for the exam on Friday. You’re not specifically responsible for it, but it’s review. So you know that group?

Student: Isobutyl.

Professor Michael McBride: Isobutyl. Okay, and what acid is that, with three carbons, do you know? The first fatty acid?

Student: Propionic.

Professor Michael McBride: Propionic acid. And in the middle we have —

Student: Phenol?

Professor Michael McBride: Phenyl group. Okay? So what’s the name of the drug?

Student: Ibuprofen.

Professor Michael McBride: Sherwin?

Student: Ibuprofen.

Professor Michael McBride: Ibu-pro-fen. Advil or Motrin. Okay, now in this case you have a chiral center there — right? — because there’s a hydrogen on there that we don’t see. And the S-form is a pain reliever; it’s said to be so anyhow. And the R is inactive. Right? But it’s marketed as a racemate. Right? One might try doing a chiral switch and selling only the S. But the trouble is that it very quickly racemizes inside you. So it wouldn’t be doing any good, right? Because you’d do all the work of selling the S and then it would become R immediately, when you ate it. And here’s another one. This is the one that Professor Barron referred to, which is a sedative; thalidomide. And there’s the chiral center in that one, because there’s an H on there. So the S-form is a sedative, but the R-form, at least it’s said, is a teratogen; which means it makes monsters. [Laughter] And it’s not so funny, because it was sold as a racemate from 1957 to ‘62 — never in the U.S. because the FDA didn’t — they were lucky and never approved it. But in Europe it caused 10,000 birth defects; children born without arms, legs, things like that. So it was a tragedy. But this one also undergoes in vivo racemization. Okay?

You can find rate constants for these. It’s curious that the rate of S going to R, and the rate of R going to S, are not the same. There’s got to be more to it than that, that one has to be more stable than the other, of these mirror images; if that’s true. It may be that the rates aren’t exactly true. But if so, they must be bound to something that makes — that’s chiral — that makes one of them more stable than the other. Anyhow, that’s how fast they go back and forth. And this is how fast they get eliminated from the body; one gets eliminated much faster than the other. And if you put these things together, you can see how the concentration should vary with time, if it’s going according to that rate. So the blue one, the one that’s good for you, quickly drops off. The bad one, the teratogen, grows quickly. So you have maybe two hours of twenty-four hours where you got a lot more of the good one than the bad one. So essentially this drug is completely off-limits, especially for any women that could conceivably, under any circumstances, be pregnant. Okay? So but it’s a wonderful drug for things like leprosy and so on. David here is an MD, as well as a professor of chemistry, and you probably know more about that. But it’s a wonderful drug for certain things. Isn’t that right?

Professor David Spiegel: That’s right, that’s right. It’s still actually in common usage for anxiety disorders.

Professor Michael McBride: Yeah, but there are all sorts of warnings, in letters this big, about if you have a chance of getting pregnant, stay away from this baby. Right?

Professor David Spiegel: In fact, if you’re a male patient taking it, and your wife is pregnant or likely to become pregnant, you’re encouraged not to take it.

Professor Michael McBride: Wow.

Professor David Spiegel: Because of possible contamination.

Professor Michael McBride: Wow. Okay, so now here’s a drug. We’re going to look at the name of this one. It’s a really whopper of a name, right? So let’s just use this as a practice about nomenclature. Okay, so that thing there is called imidazole, and the 1H tells where — which one has a hydrogen on it. Right? Okay, so that’s position one. And that is the benzene ring; so benz. So it’s benzimidazole, that group on the right. And you number it. Remember it was 1H; there’s an H on the nitrogen one there. And you go around the ring and number in that conventional way. Okay, now but then you notice that there’s something on the number five, and there’s something on the number two carbon of that ring. On five, there’s a methoxy group, which appears first among all the things in this — named here; not the thing that’s on two. Why is the methoxy group first? Do you remember what the rule is?

Student: [inaudible].

Professor Michael McBride: How do you arrange the substituent groups? Angela?

Student: Alphabetically.

Professor Michael McBride: Alphabetically; m is going to be the first one. Okay, so 5-methoxy, and then two is sulfinyl; that S with an O on it is called sulfinyl group. Okay, so it’s 2-sulfinyl. But now there are all sorts of curly brackets and square brackets and so on, to tell what’s attached to that. Okay, so attached to that is methyl group; the methyl is substituted. So it’s methyl sulfinyl. Okay, that’s the curly — the square brackets. Okay, now what’s on the methyl? There’s a pyridine group; the benzene with a nitrogen, is pyridine. And it’s substituted on the methyl at its own two position; the two position of the pyridine is what’s attached to the methyl. So it’s 2-pyridinyl, the end. But now on that, in the four position, is methoxy. And methoxy comes alphabetically before methyl; not before d, but before methyl, and you don’t count the di. Right? So there’s the name of that compound.

And this stuff is a drug. It’s a gastric proton pump inhibitor. So it treats acid reflux disease. And it’s the world’s largest selling drug in the Year 2000; 6.2 billion dollars. And it’s called omeprazole, or Prilosec. And you’ve seen probably, some of you at least, have seen boxes of Prilosec. I hope you don’t have to take it, like I do occasionally. This is called OTC. We’re going to be talking about that on Monday. Okay, so there it is. Now get your glasses up. Because can you see this? Can you see any sort of three-dimensions here? What’s in front, or what’s behind? Can anybody tell? You have any luck in seeing it in three-dimensions? Some people can’t see it, but most of you, about 95% of you, should be able to see. It’s not a really high quality three-dimensions. And because the computer — the projector doesn’t do it exactly right, for the colors. So one eye sees only one and one sees only the other. Anybody see? Okay, now what is chiral here? Can you see why this thing is chiral? There’s no carbon with four different things on it. But what makes it handed? Can you see? In fact, there’s no group that has four different things on it. Any suggestions? Lucas?

Student: Sulfur.

Professor Michael McBride: Sulfur. Sulfur has an unshared pair, an oxygen, and two different R groups. And which is pointing out toward you? Can you see it enough in three-dimensions, with the glasses, to see that, to see which is pointing out?

Student: The unshared pair.

Professor Michael McBride: Yeah, it’s the unshared pair that’s pointing out. So if you use your thumbs and recognize that the unshared pair has the lowest atomic number — zero, right? — then you can do it. And you’ll find out that that particular one drawn there is the S enantiomer. Okay? So it’s known — omeprazole, Prilosec, Prilosec OTC, are the racemate. But this one that’s drawn here is the S isomer, and it’s called what? Esomeprazole; that’s the name of it, right? Or Nexium. Right? And it’s the S isomer, right? So this is the product of one of these chiral switches, where for a long time the stuff was marketed as a racemate and now they market it as a single enantiomer, Nexium. And we tried this one last time, so we’re not going to spend time on it right now. That’s what the class looks like when they’re doing this. It’s fun for me to see you. [Laughter] Okay?

But there are other ways of doing the stereoviewing. You can use a pair of periscopes, like this. Right? And the way that works of course — well the point is, for each eye to see a different image. So if you want to try those and look at this after class, feel free. Or borrow them if you wish. Right? So what the eyes perceive is a superposition in the middle. It’s like this. There are two pictures. But if you can do it with the glasses, you can do it probably just by looking at this picture, if you have a little while. But we don’t have the little while now. The right eye will see that, the left eye will see that, and you see, in the middle — there see, something in the same position but in slightly different projections. So the image in the middle of the three seems to be in 3-D. Okay.

Now we don’t have time to go — we’re going to do a lot of curved arrow stuff, about how reactions happen in making omeprazole; and then, even more interesting, in the action of omeprazole that stops the stomach acid. But we’ll have to wait ‘til after the exam for that. Okay? Thanks again to Professor Barron.


[end of transcript]

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