Freshman Organic Chemistry I

CHEM 125a - Lecture 33 - Conformational Energy and Molecular Mechanics

Chapter 1. The 1918 Ernst Mohr Illustrations of Cyclohexane [00:00:00]

Professor Michael McBride: Okay, so as you remember, before we ended, we were talking about Sachse and how he explained everything in more detail than people wanted to hear about. Okay? And he knew exactly the story; that you had chairs of cyclohexane, and you had boat, and you could interconvert them by rotation about single bonds. But it was only after the Braggs had determined the X-ray structure of diamond, and Mohr, not equally importantly but very importantly, had drawn clear pictures of it, like this, that people understood what Sachse had been talking about twenty-eight years earlier. Okay? Because if we look in the middle there, we see that chair form of cyclohexane. Okay? We call it a chair. People like to give sort of silly, trivial names to things. Scientists are probably sort of geeks, actually; but that’s okay. So there’s a chair — right? — that looks like this chair. And that chair has a particular property, which is you can fold the back down and fold the legs up. Okay? But you can do the same thing, to do the ring flip, as we call it, in chair cyclohexane, by counter-rotation of two parallel bonds. So take these two bonds there, rotate it that way, and there rotate that way; counter-rotation, right? One rotates clockwise, the other counter-clockwise. And you rotate group four, at the bottom right, so that it goes up. Everybody see how the rotation does that? And I can do it with a model here. Right? So I rotate around this bond and this bond, hold this part here, and I rotate, and it goes up. Okay?

Now, so the product, after rotating that up, is a different conformation, which we call boat cyclohexane, or which people call boat cyclohexane, because it looks a little like a boat. Okay? And people who got very imaginative named various bonds on this; like the bowsprit, or the flagpole. But, in fact, this isn’t really a conformation at all, because it’s not a minimum of energy, it’s a maximum of energy. You get lower energy — this is the boat — you get lower energy by twisting a little bit, like that. Right? So the boat is actually not a minimum of energy. Usually we reserve the name “conformation” for isomers that are minima in energy. Right? They can vibrate, but they’re at a minimum of energy. That’s not true of the boat. But the boat is what Sachse made his picture of, and it’s easy to think about. So we often talk about the boat, even if it isn’t true; if it really wants to twist a little bit. We’ll discuss that a little bit more later, why it wants to twist. But anyhow, there’s a boat.

Now, if you then did the same trick to the blue bonds, on the other side — that is, counter-rotate so those go in and down, rather than in and up, as the red ones did on the left — it goes down like that. And now you see what you have is a chair where everything that was up is down, and everything that was down is up. Okay? So we started with a boat here. I can flip it like that to make a capsized boat; or a chair, start with a chair. Flip it to make a capsized boat. Flip it again and I have another chair. But notice what happened. Here all these bonds that are pointing vertical, or down and vertical, are black, and the others are silver. Right? Everybody see that? But after I do this, they’ve changed; the silver ones point up and down, and the black ones point out. Right? So that’s changed the environment. That’s interchanged the environment of this one and this one, of the black and the metal one — right? — by doing that so-called ring flip.

Okay, now you should learn how to draw chair cyclohexanes. That’s a very popular discipline among organic chemists, and it shows that you understand what’s going on with the conformation, if you draw it right. And if you don’t draw it right, it shows that you don’t understand. So let’s see how — people say, “Oh, I’m not an artist. I can’t really draw it right.” But if you understand it, you can draw it right. The only thing you have to be capable of doing is drawing things that are parallel to one another. That’s not too big a challenge. Okay, so notice that the carbon-carbon bonds are parallel in pairs. Right? So the red ones are parallel, the blue ones are parallel, and the green ones are parallel. Okay? Now that means when you draw the frame it looks like this, so that opposite ones are parallel to one another. And now the only challenge is to put the hydrogens on, or whatever the substituents are. Right? And for that purpose it’s worth knowing — noticing the symmetry of the ring. Right? There’s a three-fold axis of symmetry, vertical, as they’ve drawn it here. Right? So you can rotate 120°, a third of a circle around that axis, and you can’t tell that it happened; it’s symmetry. Okay? So notice that some of the bonds are parallel; are called axial because they’re parallel. But some of them are parallel up, and the intervening ones are parallel down, to keep the carbons looking tetrahedral.

Okay, so six of the hydrogen bonds are parallel to the axis of symmetry. So even if you drew the six-membered ring in some cockeyed direction, so that its mean plane was like this, or like this, or like this, or like this, still you can see where that axis points and make the axial bonds parallel to that axis. Okay, now you have the problem of drawing the last bond, which completes the tetrahedron. Notice that the equatorial bonds as they’re called — because it’s sort of like the equator, relative to the axis — they’re not strictly horizontal. Right? They go a little bit opposite the way the red bond went on the same carbon. And notice in particular that they’re parallel to the next-adjacent carbon-carbon bonds, which are shown in blue here. Okay? So they make a sort of an N, or a Z, with the next adjacent carbon-carbon bond. Okay?

Now, what o’clock — if we start from that carbon at the back left — what direction, starting from the carbon, do I draw a line, to draw the proper orientation of the hydrogen, at that carbon? You know, one o’clock, two o’clock, three o’clock, eight o’clock, seven o’clock, nine o’clock. What o’clock? Think about it a second, and then I’ll ask you. Oh, we can do it like an auction. I’ll go one, two, three, four, five, and you raise your hand when I get the right one. Okay, twelve o’clock? one o’clock? two o’clock? three o’clock? four o’clock? five o’clock? six o’clock? seven o’clock? eight o’clock? nine o’clock? ten o’clock? eleven o’clock? Okay, the answer is eight o’clock. Right? Well how do I know that? Because it’s parallel to the next-adjacent C-C bond; there. Right? See the Z? So it’s parallel to the next-adjacent C-C bond.

Okay, now how about there, what o’clock? one o’clock? two o’clock? three o’clock? four o’clock? five o’clock? Okay. It’s about 1:30, right? — sort of like one o’clock — because it’s parallel to the next-adjacent C-C bonds, which are pointing anti-parallel, right?; parallel, but in the opposite direction from their nearest carbon. Okay, try here. One o’clock? two o’clock? three o’clock? four o’clock? five o’clock? Ah, you’re getting better, right? About four o’clock; 3:30 maybe. Okay, how about here? One, o’clock? two o’clock? three o’clock? Okay, two o’clock. Okay, how about here? One? Two? Three? Ah, I got some votes for one. Now I’m going to come back and ask what’s wrong with one o’clock? Okay? There’s what it is. It’s actually seven o’clock. How is that related to one o’clock? It’s anti-parallel, right? If you did one o’clock, it would have made a U with the next-adjacent bond, rather than an N or a Z. Do you see that, see where the mistake was? Okay, so if you understand this, if you understand that the hydrogen is anti to the next-adjacent C-C bond — right? — then you won’t draw it wrong, you won’t draw a U, which would be eclipsed. Right? You’d draw it anti, which makes a Z or an N. Okay, so if you understand it, you’ll draw it. So it wouldn’t be surprising if that were on some test sometime, to draw a cyclohexane sometime.

Chapter 2. The Invention of Conformational Analysis [00:09:41]

Okay, now this got interesting because in the 1940s and ’50s, synthetic organic chemists were very interested in steroid hormones; in the ’30s, ’40s and ’50s. Right? And it turned out that you had these two alcohols, which were called β and α. And beta is the one that came up, toward the viewer, and alpha is the one that goes back into the paper. But of course, that depends on how you drew the molecule. If you drew the molecule upside down, it’d be just the opposite. But synthetic organic chemists, who were interested in this kind of thing, weren’t bothered about this, because they always drew it the same way. Right? So they knew what they meant by up and down. But that’s like names that don’t really tell you what the — don’t allow a novice to know what the structure is, to talk α and β. People still talk about α and β substituents, the ones that point out of the paper and into the paper, in steroids. But you have to know the lore. You have to know how people always draw the rings, in order to know what comes out and what comes in. So it’s like cis and trans, a little bit, not knowing what’s cis to what.

Baeyer, remember, said it didn’t make a difference, because there was only one cyclohexane carboxylic acid. Right? Because if you had axial and equatorial ones, the rings could flip and they would interconvert and it’d all be the same. Right? So, but it’s interesting that the β and the α isomers of these alcohols have different reactivity. They are different. Right? They behave differently. They don’t interconvert. Right? And you can see why that would be so. What’s the relation between the β and the α? Are they mirror images? Are their environments identical? Enantiotopic? Diastereotopic? There’s an easy way to recognize this. They exist at a chiral center, because going around one carbon and going around the other are different paths; they encounter different things as you move along. Right? So it’s a chiral center. But there are other chiral centers in the molecule. And when you have two chiral centers, then changing just one doesn’t give the mirror image. It gives an epimer, not an enantiomer. So they are diastereotopic. Right?

Now, for these kinds of problems, Derek H.R. Barton invented what was called “conformational analysis”, in 1950. He was interested in these steroid hormones, as many people were. Right? So conventionally these six-membered rings were labeled A, B, C, and then the five-membered ring D, so people could know what they were talking about. Right? And Barton redrew ring A. And notice that the β, α is configurationally diastereotopic. You’d have to break bonds to make the α into β. So it doesn’t surprise us that they’re different. Right? But let’s look at just how they’re different. So this is the picture, from his paper in 1950, that Barton drew to show that ring A; and the bracket there, that I’ve drawn in, shows where it would go on to ring B. I’ve truncated his picture. Now I want to ask you a question. So β goes up, relative to the mean plane of the ring, and α goes down. And he labeled these things e or p, for equatorial — that’s what we still call it, as I told you, the one that points more or less in the plane of the ring. And he called the others p, for polar, as if it were a globe and had an axis, the pole. But we call it now axial. So that’s changed since he invented this. But I think you’re probably sophisticated enough now, having had this practice, to recognize an error in what he drew. Do you see what’s wrong with his pictures? Marty, what do you say? Are the equatorial bonds, do they make Z’s with the carbon frame? That is, are they anti to the next adjacent carbon-carbon bond? Take an equatorial one, like this one here. Is this carbon-carbon bond anti-parallel, at the correct o’clock, relative to this bond?

Student: It needs to be more [inaudible].

Professor Michael McBride: Can’t hear very well.

Student: It needs to be more like eleven-ish.

Professor Michael McBride: No, this one, notice, is exactly horizontal, and this one is exactly horizontal. So they’re perfectly anti. Or take this one. Right? Here’s its bond, and here’s the next adjacent C-C bond, one to ten. And they’re exactly antiparallel. Okay? So that’s anti too. So the equatorial ones are fine. How about the axial ones? Do they look good? Catherine, what do you say? What does axial mean? What’s axial about it? Maria?

Student: The molecule can be rotated around the axis.

Professor Michael McBride: Okay, what direction does that axis point, that you can rotate that six-membered ring around it and get the same thing? What o’clock does it point, from the center of the ring?

Student: So it should point like eleven o’clock.

Professor Michael McBride: Yeah, about eleven o’clock. There are the sets of three carbons that are related by that, and the axis goes at about eleven o’clock. What does that mean Maria, about the axial bonds?

Student: That they should also be pointing up.

Professor Michael McBride: They should be parallel to that axial direction. Right? So they really should be this direction — right? — parallel to that axis. So the very guy who invented it, in the very paper in which he did the invention, drew them wrong. That’s not too surprising. Right? But it’s interesting to note. So you don’t have to feel bad when you draw it wrong the first few times. By the time we get to an exam, you’ll know how to draw it right; better than Barton did, when he published his paper. Okay, he got the Nobel Prize, in 1969, for development of the concept of conformation and its applications in chemistry. So he didn’t get the Nobel Prize for drawing the axial bonds wrong. Right? What he got the Nobel Prize was for the application, showing how important this was in chemistry; that those axial and equatorial groups — hydrogens in this case; but they could be other groups — have different chemistry. Right? And notice the date, 1950. All these things about stereochemistry were happening within plus or minus a few years of that. We’ve already talked about Bijvoet determining the absolute configuration of tartaric acid; Newman figuring out how to draw his projections, to show conformation; the Cahn-Ingold-Prelog rules; and, as we’ll see shortly, the idea of molecular mechanics. Okay?

Now, what made Baeyer say everything was the same, although he didn’t know it, was that the ring would flip, interconverting axial and equatorial, as we just looked. But can the ring — if the ring can flip in this one, it would also interconvert axial and equatorial — but can the ring flip? That’s the question. So let’s think a little bit more about how the ring-flip works. Notice that during a ring-flip, as in the top left there, what’s equatorial becomes axial. So you look on the top left, there’s an equatorial bond, and in the right that becomes an axial bond; well actually you’d call it a — what is it? — flagpole, in the boat. Right? But when you do the other flip of the blue one, that one that’s purple is there. Right? So the one that was equatorial in the chair on the top left, after a flip becomes axial. And we already saw that here, that the ones that are equatorial become axial when you do the double ring flip. Okay? Now, and by the same token, the ones that are axial become equatorial. So if you start on the top left with those two green ones, which are anti to one another, axial — notice they point a little to the right because the axis of — the symmetry axis is not straight up and down. Right? After the first part of the flip to the boat looks like that; and the second part of the flip, they become equatorial, and gauche to one another, where they were anti originally. Right?

So now if you have fused rings — that is, two six-membered rings that share a bond, as on the bottom there — those two green ones can be part of the second ring; gauche to one another, as all the carbons are around the cyclohexane ring in its chair form. So they’re gauche, and that’s fine. But if you tried to flip it, those two green ones, with respect to the front ring, would become axial, as on the top left. And if they pointed axial, you couldn’t possibly complete the ring. Right? So here’s two chairs, here and here. Now if I could try to flip this one up — I could flip this one down probably; there, down. Right? So I got it almost into a boat. But I can’t even make it into a boat, let alone flip the other ring. And the reason is that these two are tying these two — while I’m trying to flip this one, these two are holding this one and won’t let it go. They can’t become axial to one another. Right? Just because the carbons can’t reach. Okay, so in what’s called — decalin means decahydro, ten hydrogens, on naphthalene; naphthalene is like benzene, except two of them. Right? So this is decalin, or ten hydrogens on naphthalene. And this one is called trans-decalin. Right? These two hydrogens are trans to one another. And in that one, I can’t flip the rings. If I had cis-decalin, like this, then it’s possible to flip the rings; which I could do like this. Let’s see. Well ho, ho, ho. I think I did it there. Yeah, I did. Okay.

So you can flip the ring if it’s cis, but not if it’s trans. And the point is that you can’t take two equatorial things and make them both axial in a ring, because those carbons get too far apart to be bridged by the rest of the ring. But if they’re gauche to one another, then they can be gauche to one another on the other side and it can flip. It’s fun to play. You know, we used to require people to buy models. Now you can do things with computers that make it that it’s probably not worth your money to buy the models. But it’s still a lot of fun, and you learn something with your hands. It’s like when people have trouble in elementary school with arithmetic; the teacher gives them toothpicks to count with and so on. It’s a higher level than that, but it’s the same thing; you learn a lot with your hands. So it’s fun to play with these. In fact, let me pass around here simplified models that are a chair. But you can do this trick of changing one into the other by rotating. And feeling it is fun. Right? So that you have the opportunity to feel it, I’ll pass these around. There you go. There you go. Okay, the ring flip is impossible for trans-decalin. Okay, but so gauche because gauche is okay within the second ring of decalin, but not anti. Okay, so and you try with models if you’re skeptical. If you want to come afterwards and fiddle with the decalins, you’re free.

Chapter 3. Conformational Animations of Ethane, Propane, and Butane [00:22:25]

Okay now, nowadays people, much more often, encounter these things on animations. But this is a nice webpage that we have permission to use in the course; I mean, you could use it on your own anyhow, on the net if you wanted to. But it’s a cautionary tale because just because it looks nice and works, doesn’t mean it gives the right answers. So let me show you what I mean. Okay, so if you go to this website, you can click there to get it. And it’s a nice tutorial about conformation. So they show here, for example, ethane, and if you click ‘Play’ — let’s see, I think I’ve got it animated here — you can step from one position to the next and see what the shape looks like. Not that ethane is so very exciting, but you can see it go down and up, as it goes from eclipsed to staggered. And you can click on a point and the model will turn, to show you what it’s like. I’ll give an example of this later. And there’s the staggered and there’s the eclipsed. That’s the barrier that goes across. And you see that that barrier is 5.2 kJ/mole, because they use — they’re modern and they use joules, whereas American organic chemists tend to use kilocalories. But anyhow, if you multiply that by 0.239, you get it in kilocalories, as what we’ve been talking about, which would say that the barrier is 1.24 kcal/mol. How big is the barrier in ethane? Does anybody remember from before Thanksgiving? Dana?

Student: Three.

Professor Michael McBride: It’s three. Right? Just because they give a fancy chart with numbers on it, and just because it’s been calculated by some quantum mechanical program that they got access to and could use, doesn’t mean the numbers are going to be right. And they’re not, they’re off by a factor of more than two. Right? So don’t believe everything you see. That’s maybe the primary lesson that this whole unit tells you. Okay, it should be 2.9. So let the buyer beware in situations like that. If you don’t pay anything for it, you probably get about that. But the pictures are still very nice, and the general shape is correct. So here’s the rotation in propane, and it says the barrier is 5 kcal/mole. And you know it’s something like 3.4, 3.3, instead. Here’s butane, and again you can animate that and see the various staggered conformations and the fully eclipsed one. And we know the gauche is supposed to be 0.9 kcal/mole; which is not what this says. And forget the scale on the left.

We can use that 0.9 to tell how much of the gauche there should be at equilibrium. And let’s rehearse this again. Remember, it’s 10^{(3/4 ΔH)},in kilocalories. So 3/4thsof 0.9 is 0.68. So it means the ratio of gauche to anti should be 1:4.7 (4.7 is 10^0.68). Okay? So 1:5, about. But it depends on what you’re talking about, because that gauche is in fact chiral. Let’s see, I didn’t have it here. Well you can — I do it with my hands. Okay, so here’s anti. Okay? Here’s gauche. Right? But gauche is chiral. It could also be this. Right? It could be one hand, or the other; the mirror image. It’s like a propeller of gauche. Right? So when I say gauche, what do I mean? Do I mean gauche+, or do I mean gauche-, or do I mean both of them, taken together? Because obviously, if I include — if I say gauche and mean both of them — right? — then I have to multiply this by two, and the ratio is 1:2.4, instead of 1:4.7. Okay? So there has to be this statistical factor taken into account as well, when you use a collective name like gauche. Okay, the eclipsed is 3.4 kcal/mole. That will tell how fast anti goes to gauche. So the difference in the well heights, tells how much gauche there is at the equilibrium, and the barrier height tells how fast you go, from one to the other. So how fast do you go from anti to gauche? Do you remember how to do that? Anybody remember? Chenyu, do you remember how to do it?

Student: How to do what?

Professor Michael McBride: How to find out how fast something is, if you know how big the barrier is you have to go across?

Student: I’m not sure.

Professor Michael McBride: No, but you’ll know it for the final.

Student: I’m sure.

Professor Michael McBride: Okay? It’s 10^13th/second; pretty fast. But then slowed down by that same kind of equilibrium constant, 10^{(-3/4 of how big the barrier is)}. Now, if the barrier is 3.4 — say it was four, say the barrier’s four; it’s about that. 10(^{3/4ths of 4)} is 10^3rd. Right? So it’s slowed down by 10³. It’s 10^13th/second times 10^-3. Right? So it’s 10^10th per second. Okay, there it is. Oops, I did it wrong. What did I do wrong here?

Student: You said four instead of 3/4ths.

Professor Michael McBride: Pardon me?

Student: You made it four instead of 3/4ths.

Professor Michael McBride: That’s not the difference though. I think I’ve got a typo here. That can happen. Okay, anyhow, I think — subject to thinking while I’m not on my feet — that this should be about 10^10th per second. Okay? Anyhow, you see how you would use it. Right? 10^13th times — whoops, oh I forgot to plug; no, I did plug it in. Oh, it’s okay. Okay, so there we go. Okay? And the fully eclipsed barrier is about 4.4 kcal/mole. But that’s hard to access experimentally. Why? Because experimentally you try to tell how fast one thing goes to another. Right? But it’s possible to get from anti to gauche without going over that barrier, and it’s possible to go from gauche to anti without going over that barrier. To go from gauche to gauche, do you have — notice that 360° is the same as 0°. Right? So you can go from this gauche to this gauche, by going over that barrier. Right? So can you measure that rate and see what that barrier is? No, because there’s an easier way to go. Instead of going like this, to get from one gauche to the other, you just go like this. Right? So there’s an easier way to do it. So you can’t measure the rate and know what that barrier is. That barrier is more or less irrelevant experimentally.

Okay, or here’s the ring flip in cyclohexane. Now let me — I’ll use this one to animate it, to show you how. You can try this program yourself. Okay, so here we go. And there’s the thing doing its ring-flip. And it’s nice because you can grab it and rotate it around and see it from different points of view as it does its ring-flip. So there’s flipping one end down, and then flipping the other end up, to go from one chair to the other. So you can fiddle with this. Or you can get models and try them too. Okay or, as I said, you can go — we could stop that — and you can go over here and see what it looks like at the halfway point. Now notice that that is a little bit like a boat. But it’s a twisted boat, as I said, because the actual boat is up here at the top. In fact, even at the top it’s not — oh, pardon me, I said the wrong thing. It’s not — up at the top, it’s sort of a half chair, half boat. Right? This top left carbon is still in the sort of chair-like form, or these five carbons are. But this one is halfway bent up, toward a boat. Okay?

Anyhow, you can fiddle with those things. Oh let’s see — what do I need to do here? Go back here. Okay. So there’s the chair conformer, there’s the flexible or twisted-boat conformer. Now that one is really fun. I’m going to do it here, and you people who have this can do it. Notice — did you notice when you played with this, that the chair is sort of rigid. If you try to twist it, it’s hard. Has everybody felt it, to feel that? But if you get into the boat form, like that, then it’s quite flexible. You can do this, to your heart’s content. Did you feel that? After you’ve passed it across, pass it back so everybody can feel how — this is great, like the orb. Right? But the chair is quite rigid. Right? And as it goes from chair to boat, it clicks; like click. See it click? Okay. So there’s a barrier; that’s the point. But in the flexible form there’s not a barrier, it just smoothly rotates. Okay, so there’s that barrier of 11 kcal/mol. So it takes some time to get back and forth. Okay, and here’s the flexible one, and you can animate that and watch it flex. Okay? So that’s the twist-boat form.

Chapter 4. Molecular Mechanics as an Alternative to Quantum Mechanics [00:32:06]

Okay, now we’re going to talk about the shape, strain energy, and molecular mechanics. The point is to talk about molecular mechanics. How do you get these energies? That particular animation used a quantum mechanical calculation. It didn’t do a very good job on the energies, as we just showed. But is there an easier way to get these energies? And there is, and it’s called molecular mechanics. Essentially it’s just using Hooke’s Law for the model, to calculate strain energies. So we already saw the torsional energy of ethane at 3 kcal/mole — a threefold barrier, three minima, as you go across — and the conformational energy of butane. And it’s fun to remember those numbers: 0.9 for anti to gauche; 3.4 barrier between them, and then maybe 4.4, but who knows, for the other one. Okay, now remember 1950 is when all these things were happening. So in 1946 there was a paper by Frank Westheimer and Joseph Mayer from the University of Chicago, about using mechanics to calculate the energies of conformations. The particular thing they were interested in is when you have two benzene rings hooked together, so that they can rotate like this. Right? They can’t rotate freely if you have things here that are sticking out, that would run into one another, as you tried to rotate them. Right? So you could get one that’s twisted this way and one that’s twisted this way, in a substituted biphenyl. And you can resolve them, and get optically active forms — one that’s twisted to the right, one that’s twisted to the left — even though you don’t have a carbon that has four different things on it; because it can’t rotate. Right?

But the question is, how big should that barrier be? And Westheimer, in fact, is your great-uncle. Right? Because he also studied with Conant and Kohler, at Harvard; the same as my teacher, Bartlett, did. Okay? So he was interested in physical-organic chemistry and did this. Incidentally, he’s also the guy that figured out which hydrogen you pull off on ethanol; you know, that enzymes choose between the pro-R and the pro-S hydrogens. So anyhow, Westheimer did this in 1946 and figured if you have a bromine here and a bromine here, that when it’s flat, and achiral, that’s the transition state between being twisted one way and being twisted the other; that — whoops, sorry — that these two things would run into one another. That would mean the bonds might bend back a little bit or something; how much energy would that cost? So he tried to do it by what’s called molecular mechanics. So these programs calculate the energy, and they can minimize it by adjusting angles to get the minimum energy, by treating these molecules as if they were mechanical. And to achieve useful precision they require a very large set of empirical force constants; that is, how strong are the various springs? And you adjust these arbitrarily on some test molecules, in order to get the best values, and then try to apply it to something else.

You can also do it nowadays by reliable — unlike the ones we just talked about — quantum mechanical calculations. But I want to show you how many parameters there are, when you do molecular mechanics. Okay, so this is so-called MM2 — that’s a particular molecular mechanics scheme — and these are parameters. So there are sixty-six different atom types; fourteen different types of carbon, depending on what it’s bonded to. Okay, so here are some of — here are the fourteen different types of carbon. The carbon in an alkane, in an alkene, in an alkyne. That carbon in a carbon cation. The carbon in a carbonyl group. The carbon in a carboxylate; and so on. And these have different van der Waals radii; which is one of the things you have to put in. Okay? Now, then you have these various kinds of atoms hooked together, and you have a certain strength for stretching the spring. That is to say you need what the equilibrium bond distance is, the minimum energy bond distance, and how strong the force constant is, for stretching that spring. So notice there are 138 different bond stretchings. These are just the ones that involve where the first atom is of type one. Okay?

So you need to find out what all those force constants are. Okay. Then you can bend bonds, and there are 624 different strengths of springs for bending different types of bonds. Forty-one, shown here, involve alkane carbon, alkane carbon, and then some other group; it could be another alkane carbon. That’s the very top one; one, one one is three alkane carbons. So that’s the force constant for bending the bond. But then there are different equilibrium bond angles; all close to 109.5 — remember, that’s the tetrahedral angle — but a little bit different, according to whether that particular carbon has two other R groups on it, or an R group and an H, or two Hs, in addition to the other two groups you’re talking about actually bending the bond. So you get — so there are — so then, of course, you can twist bonds also. That’s what conformation involves, after all. And there are 1494 different bond-twisting, and for each bond-twisting you have three parameters here. For example, V3 — also V2, V1 — for each of those 1500 whatever number it was of bonds. I’ll just show you what that means. V3 means a threefold barrier. And it says its height is 0.093 kcal — yeah, kcal/mole. So here it is. Right? It’s maximum when it’s eclipsed, it’s minimum when it’s staggered, maximum eclipsed, minimum staggered. So that’s a threefold barrier, for rotating. But there’s also, for a particular kind of set of three bonds, there’s a twofold barrier, which looks like that. Right? And that one, you see, is 0.27 kilocalories high. And there’s also a one-fold barrier, like that, which is, for that particular one, alkane, alkane, alkane, alkane, 0.2 kilocalories high. Now you add all these together, and that is the torsional contribution to the energy in butane; or anything where you have alkane, alkane, alkane, alkane. Right?

Now we know what it really looks like in butane. It looks like that. The scale is much bigger. So this is a rather minor contribution. Now why is it that the anti is the most stable, according to this scheme? It’s because it’s — the van der Waals repulsion is least when these two carbons are as far apart from one another, rather than being eclipsed with one another. Right? But that then is biased — you calculate that van der Waals repulsion, which depends on the radii of various things. Right? And you then tweak it by adding this to it, as well. Right? So you can see that these are pretty complicated — I mean, for a computer it’s not a problem. It’s a lot easier than quantum mechanics, finding all these curvatures and stuff. But it’s still a pretty intricate thing and an enormous numbers of parameters. After simplification, the so called MM3 scheme has more than 2000 arbitrarily adjustable parameters that you have to fiddle out by knowing a lot of different experimental results. And what would I have to make the numbers to make that right? And then it has to be also right with this one, this one, this one. So obviously you need at least 2000 different molecules to determine 2000 parameters, or 2000 different measurements at least.

Chapter 5. Assigning Strain to Estimate Energy in Bonds [00:40:13]

So it’s a very highly parameterized system. Contrast it with quantum mechanics, where there are no arbitrary parameters; all you have is the particle masses, their charges, and Planck’s constant. Right? So there’s nothing fundamentally correct about molecular mechanics. It’s just a very complicated scheme that’s been tweaked to try to give good results for the molecules that it’s tweaked on the basis of. But that doesn’t necessarily mean it’ll work for anything else. But it does work pretty well. And the nice thing about it is by looking at the calculations, you can figure out why things happen the way they do. For example, here’s ideal cyclohexane, the way we’ve been drawing it. And we can look at the various kinds of strain that there are in that. First we take an ideal cyclohexane; which means it has ideal bond lengths, bond angles, and is staggered, like this. Okay? And there’s no strain for stretch, bend, or what’s called stretch-bend. Why is there a term called stretch-bend? Because when you stretch a bond — it might be easier to bend, or conceivably harder to bend when it’s stretched. So you have to put another term in for that. Okay, but there is torsional energy, because there are gauche interactions here, like that one. But there are six of them, as you go around the ring. So they’re not anti. So there’s torsional energy.

And then there’s what’s called “non-1,4 van der Waals energy”; which in fact is favorable, it’s attractive. Okay. And here you see an example: one, two, three four, five, those atoms, atom one and atom five there, are five atoms apart. Right? So it’s non-1,4; it’s 1,5. And they’re at a distance where their interaction is attractive, according to van der Waals energy. But you can have 1,4 van der Waals energy, as here. There’s one, two, three four; and that, you see, costs you — Its strain of 6.3 kcal/mol — not that particular one, but sum them all up. Okay? There’s a very bad one here, you can see, between one and eight. Those are rather close together. Okay? So there’s this much strain in that. Now, what a molecular mechanics program can do is adjust the geometry — twist bonds, stretch them, bend them and so on — to minimize the energy. Okay? And here’s what happens. Whereas the total strain energy before deformation is almost 8 kcal/mole, after — let’s see, what did I do? There. After you’ve minimized it, those are the energies. It falls to 6.56. So it gets a kilocalorie 0.3 better. Okay? Now notice what happened. It stretched the bonds a little bit. It bent the bonds a little bit. Right? Let’s see exactly what it did, by looking at the model. So there’s the — it’s going to stretch and flatten the ring slightly, to reduce that bad van der Waals repulsion. Notice the 1,4 van der Waals before was 6.32, and it’s cut down to 4.68. So watch what happens. So this is — did you see it? Let’s back up. See, it flattens and stretches, just a bit, to get — to reduce those van der Waals repulsions. So this way I told you to draw cyclohexane, to get it the same way Mohr did, is the way organic chemists do it. But it’s not quite right. Actually the rings flatten a little bit and the things that should be axial spread out just a little bit.

Okay, fine, so some things get better, some things get worse; overall it gets a little better. And it would be very hard to do that in your head. Right? Molecular mechanics is good to do that. Okay, now there you notice is a gauche butane within the cyclohexane. And we know how much gauche butane costs, compared to anti butane. How much strain is there in gauche butane? Remember gauche versus anti, I said that was worth remembering. It’s 0.9 kcal/mol. Okay, but in fact, in the whole thing, there are six gauche butanes, because every bond is part of a gauche — is the central bond of a gauche butane. So if you had six gauche butanes, that would be a strain of six times 0.9; 5.4 kcal/mol. Okay? And in fact that’s rather close to 6.56. So that actually is way over-simplified. But it’s a good mnemonic device for remembering how big it is. It’s about six gauche butanes. Or suppose you had axial methylcyclohexane. Now you have much, much worse van der Waals interactions — right? — 6 and 8 kcal/mol of strain. But what’s going to happen if I let it relax, if I run molecular mechanics and let it minimize its energy? Can you guess what’s going to happen? How will the structure change? Virginia?

Student: The last methyl group.

Professor Michael McBride: Yeah, that methyl group on the top is really in trouble — right? — because of those non-1,4 van der Waals repulsion. But if you bend it back to the right a little bit, you’ll reduce that. So here’s what happens, it relaxes like that. And now notice that the non-1,4 van der Waals went from being bad, by 6 kcal/mol, to being good by 1.3, overall, summing them all up. Right? So it went from sixteen kilocalories of strain to only nine, or from seventeen to nine. Okay. And notice that here there are two more gauche butanes, in axial methylcyclohexane. There were already six in the cyclohexane ring. Now there are two more. So you’d guess — whoops I went too fast — you’d guess eight times 0.9; 7.2. It’s 8.6. It’s in the right ballpark. So it’s roughly what you’d expect for eight gauche butanes. Okay?

Now, if you go from axial to equatorial, then it turns out that energy difference is 1.8 kcal/mole. Equatorial is 1.8 better than axial. Does that surprise you, that it’s 1.8? Could you have guessed that it would be 1.8? Notice when it was axial, there were two gauche butane interactions. When it’s equatorial, those two gauche butanes become anti butanes. And a gauche to anti is 0.9. There are two of them. So 1.8. Right? Two gauches become two antis. So that’s axial to equatorial. And, in fact, you could put other groups there. Like instead of methyl group, it could be chlorine or bromine or ethyl or something like that. And for any of these, if you can measure the amount of equatorial and axial, you can get a measure of how big that group is. Right? So that, these so-called A values, the difference between energy when it’s axial and when it’s equatorial, is a nice rough measure of the effective group size. So we’ll stop here, for now, and continue next time.

[end of transcript]

Back to Top