CHEM 125a: Freshman Organic Chemistry I

Lecture 18

 - Amide, Carboxylic Acid and Alkyl Lithium



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Freshman Organic Chemistry I

CHEM 125a - Lecture 18 - Amide, Carboxylic Acid and Alkyl Lithium

Chapter 1. Resonance of the Amide Group [00:00:00]

Professor Michael McBride: Okay, so we’re at the conclusion of our theoretical study of bonds and reactivity, then we’ll go on to something completely different, and we’re ending by talking about reactivity in terms of four functional groups, and you’re doing your own functional groups, remember, for tomorrow. Okay? So we already talked about carbonyl last time. And now we’re going to go on to amide, carboxylic acid and alkyl lithium, as other functional groups. And then we’ll do something just completely different, for the next quarter of the semester.

Okay, so resonance is HOMO/LUMO mixing that happens within a molecule. The molecule, of course, knows all about it. It only becomes “resonance” because we draw a picture that’s too simple, and we need to modify the picture, right? And that modification is called resonance. And a good example of that is the amide functional group. Now when I include an amide on the first exam where people are supposed to identify functional groups, circle them and name them, it’s inevitable that some people will circle the C double-bond O and say that’s a carbonyl group, and some people will circle the nitrogen and say that’s an amine. But an amide is not an amine, and it’s not a ketone, or a carbonyl group. It’s its own thing. And the reason is that those interact with one another in a way that makes them no longer like an amine, and no longer like a carbonyl group. That’s resonance.

So we’ll look at it here in terms of two different pictures of the molecule. One is as a carbonyl amine, and the other is as an amide, which if you’re into Lewis structures you draw with a double-headed arrow and show two resonance structures. And in the carbonyl and amine, you identify a HOMO that makes the thing a base, and a LUMO that should make it an acid. But resonance can be thought of as a reaction already within the molecule, between the HOMO and the LUMO. And it’s a make-and-break situation, just like we saw between molecules last time, where the unshared pair makes a new bond — in this case the second of a double bond — and something leaves from the carbon that’s being attacked. The thing that leaves is the second bond of a double bond. So it’s another one of these make-and-break situations. But in this case, it’s a correction to our naïve, localized picture that we drew at the beginning.

Okay, so let’s see about this. First we have the naïve prediction, what that should say about the characteristics of the molecule, and then what the experimental observation is, which requires that we use something more complicated, the resonance structures or thinking in terms of molecular orbitals. Okay, so the amines and carbonyls are perfectly stable molecules in terms of being low energy. They don’t just do something. But what’s associated with the idea of resonance structures is the thing is more stable than you would’ve predicted if you didn’t consider the resonance structure. So you say that’s more stable. And you can see why it would be more stable, because the unshared pair is mixing with a vacant orbital and becoming lower in energy. In fact, the observation is that it’s more stable by about 16 kcal/mole, which is about a fourth as much as a carbon-nitrogen bond is worth. So it’s a significant amount of energy. Now, if you have a C-N single bond, that’s longer than a double bond, and longer than a partial C-N double bond that you’d have for the average of the resonance structures on the right. And in fact that bond is shorter than normal single nitrogen-carbon bonds, by about 0.14 Ångstroms. Right? On the left you have a short double bond for C double-bond O. But on the right it’s a partial single bond. So it should be longer, not held as tightly. And in fact it is longer than other C-O double bonds, by about 0.03 Ångstroms. Okay?

We’ve seen before, when we were talking about XH3, why nitrogen tends to be pyramidal, so as to give s-character to the unshared pair of electrons. But on the right, if it’s going to be part of a double bond it has to be planar. And we can see why that is. So here’s the π* LUMO, of the C double-bond O, and here in front is the unshared pair on nitrogen, in a p orbital, which overlaps and delocalizes, mixes, with the vacant LUMO. Right? That means that you have a partial double bond for the C-N, and a partial single bond for the C=O. How is it that putting in electrons makes it a single bond rather than a double bond? Why would putting in electrons weaken a bond? Alex? Can’t hear.

Student: Antibonding.

Professor Michael McBride: Because it’s an antibonding orbital, the π* of C-O. Okay? Now, so it’s planar, indeed — and I’ll show you pictures that prove it — because it gives the best overlap. If there were an sp hybrid on the nitrogen, the s part of it wouldn’t overlap with the . So a pure p orbital gives the best overlap. So that’s why it’s planar, from the molecular orbital point of view. Now, if you have a single bond, it’s easy for things to rotate around it, because it’s cylindrically symmetric. But we know that a π bond is not cylindrically symmetric; it has that oval shape. So there’s a barrier to rotation about the central bond of a carbon-carbon double bond. Because when you rotate the nitrogen in front, like this, by 90º, you’ve wrecked the overlap that allowed those electrons to be stabilized. Right? In fact, once you’ve done this, there’s no reason for the nitrogen to be sp2 hybridized anymore, because the unshared pair is not trying to overlap with the carbon p orbital behind it. Right? So it might as well hybridize and become sp2 the way it normally does. And this structure, the halfway-rotated structure, turns out to be 16 kilocalories less stable than the one where you do have the overlap. So twisting it has a barrier of 16 kcal/mole. Have you ever seen that number before? Zack?

Student: Yes, at the top there’s one.

Professor Michael McBride: That’s what we saw at the beginning. It’s more stable when those electrons can be stabilized, as compared to an amine and a carbonyl that are not interacting. Because when you turn it 90º, they indeed are not interacting, because they’re orthogonal. Okay, an amine, as we said, should be basic, and a carbonyl should be acidic because of the low LUMO, the π* C=O. But if they mix with one another, then the HOMO is lower and the LUMO is higher. So they’re not reactive anymore; they’re not as unusual in their highness and their lowness. Right? So it turns out that an amide is relatively unreactive; relative to what you’d expect for an amine and a carbonyl. Now do you know where you encounter amides all the time?

[Students speak over one another]

Professor Michael McBride: You can touch an amide, with no problem. How? Sophie?

Student: Part of amino acids.

Professor Michael McBride: Pardon me?

Student: They’re amino acids.

Professor Michael McBride: Amino acids, which make proteins. Amino acids aren’t amides. But proteins, which are made of amino acids, an amine reacting with a carbonyl, those are amides. And you want them to be stable. Why? You don’t want them to be reactive, like a carbonyl and an amine.

Student: [inaudible]

Professor Michael McBride: It’s so your skin works. Otherwise, in lab if you — what does a base — if you feel sodium hydroxide — you’ve done this — what does it feel like?

Student: It’s slippery.

Professor Michael McBride: It’s slippery. If amides weren’t unusually stable, then when you felt it, your skin would go away. Right? So it’s a good thing. In fact — and the last thing is also important. So they’re opposing dipoles, mostly, right? Carbon to oxygen is plus towards carbon, minus toward oxygen. Carbon nitrogen, plus toward carbon, minus toward nitrogen. But mostly those two are opposed to one another, and a little bit they add. So you get a net displacement of positive/negative charge in the molecule, from left to right, positive to negative. Okay? That’s what would happen if it were a carbonyl and amine. But it’s not a carbonyl and amine. It’s strongly dipolar, not cancelled, and it’s in this direction, perpendicular to the original direction. Because you take electrons that were on nitrogen in the amine, and you shift them into the carbonyl group. So it’s as if — the magnitude of that effect, is as if you transferred one-third of an electron from nitrogen to oxygen. If you did that, you’d get as much displacement of charge, on average, as there actually is. Right?

Now, it turns out that all these latter properties are crucial for structural biology. We already saw that if you want skin to work, it has to not be as reactive. But these other things are important too. The geometric things; that nitrogen has to be planar and that there’s a barrier to rotation is very important. And also the dipolarity; the direction of the dipole is very important. And we’ll show that. So here’s an amide bond. Okay, the nitrogen is blue, and the oxygen is red. You see the double bond. But you shift electrons from the nitrogen to the carbonyl group, and that is the π* — pardon me, that’s the HOMO, not the π*, it’s the HOMO — the π electron pair, which has shifted from nitrogen and shared with the C double bond O. Now, that sharing, as we just said on the previous slide, creates an electric dipole. So the electrons are shifted in the direction shown. And that’s crucial. Because here is a picture of a helix, an α-helix of a protein, and I outline the backbone of it there. Okay? So several turns of an α-helix, which has a lot of amide bonds in it. And if we look at it, we see there’s an amide, there’s another amide, there’s another amide, all more or less on the top of this helix. What’s interesting about those? Andrew, what’s interesting about the set of them? How are they lined up? Do you think it’s favorable or unfavorable, the way they’re lined up: plus, minus; plus, minus; plus, minus, in a row?

Student: Favorable.

Professor Michael McBride: They’re all lined up so as to be attractive to another. So that stabilizes this structure. In fact, another way of talking about it is to say, as shown here by these δs, that the oxygen is partially minus, and that the hydrogen attached to the nitrogen is partially plus. So getting this minus near that plus is favorable. What do you call that? Anybody know? That is a hydrogen bond. Right? So you have — it’s stabilized by electrostatic interaction, these hydrogen bonds. Not only there, but there are other ones too, like here, a chain of them there, and another chain down here at the bottom. So this helix is stabilized by the fact that the dipole points the way it did. If it pointed in the perpendicular direction, you wouldn’t get this stability. Okay? And it’s also stabilized by the geometry that’s favored, the local planarity, the fact that you don’t twist around the C-N double bond, the partial double bond. So here’s carbonyl, here’s the C-N, and that’s — although it’s written as a single bond, it’s partially a double bond. So you don’t twist about it. So this atom, this atom, this atom and this atom are all in the same plane. Right? And you see another such group here, back behind, where here is the bond in question. So this atom, this atom, this atom, this atom are all in the same plane. Or here at the bottom you see it, where you’re looking almost in the line — in the plane. Right? And what does that mean? It means that the backbone is not as floppy, because you take away some of the possibilities of twisting it. So it helps it take this shape. So both the dipole, and the geometrical freezing of some of the single bonds, means that this structure is particularly stable, and therefore very prevalent in biological macromolecules. So that makes a lot of difference, that amides are amides and not amine carbonyls.

Chapter 2. Acidity of Carboxylic Acids [00:14:17]

Okay, now how about carboxylic acids? That’s the next functional group. So compared with what are these acidic? Compared to alcohols. Because you might think that, in the same way that an amide is an amine ketone, you might think that a carboxylic acid is an alcohol ketone. But it’s not. Right? They interact with one another, in much the same way. Now, if you had just a regular old alcohol, it can dissociate as an acid. It’s not very strong as an acid. You don’t burn your skin with it. Right? So it has a pKa of sixteen; the dissociation constant is 10-16. That’s not very impressive. But for a carboxylic acid, the reason it’s called an acid, is because its pKa is much lower, it’s more acidic. The equilibrium constant is 10-5. So it’s acidic, it’s not an alcohol. It’s 1011 times stronger than an alcohol. Right? Now why? Well just like an amide, you have an unshared pair on the O-H oxygen, which could be stabilized by the π* orbital of the C double-bond O. Now, if that stabilizes the material on the left, how should it affect the equilibrium constant? Should it be more or less acidic? That is, because of this interaction, because of this resonance structure we could say, because this pair of electrons is stabilized by the carbonyl, there’s a factor that’s stabilizing this, which doesn’t exist in the alcohol. So you’re stabilizing the material on the left of the equilibrium. Would that tend to make it more acidic, or tend to make it less acidic, if it’s stable on the left?

Student: Probably more.

Professor Michael McBride: Right? So you have — here’s the stuff on the left, here’s the stuff on the right. It’s uphill in energy. No matter what you’re talking about, it’s hard to dissociate as an acid. Right? So I make this more stable, which is what I just showed. There’s resonance here that stabilizes this HOMO/LUMO mixing within the molecule. Does that make it more or less easy to dissociate? Harder or easier?

Student: Harder.

Professor Michael McBride: Harder. Right? It’s further uphill. So this seems backwards. The analog of the stabilization we saw in an amide that makes your skin not dissolve, has the opposite direction. It predicts that the dissociation would be more uphill, where in fact it’s less. Why? Because if we look at the product, you also have HOMO/LUMO interaction, between an unshared pair on this oxygen and the carbonyl. And that will stabilize it on the right and make it easier to dissociate. Which should be more important, the amount of extra stabilization on the left? Here’s what it was for an alcohol, originally. We stabilize it on the left and we stabilize it on the right. Which is more stabilized, the carboxylate or the original alcohol - the carboxylic acid. What do you know from the fact that it’s more acidic; which must have gone down further?

Student: That one.

Professor Michael McBride: This one must’ve gone down further. Can you see why? Why would this one be more stabilized by HOMO/LUMO mixing, that is, this be more important, than it is here, with that? Lucas?

Student: Symmetry.

Professor Michael McBride: No. Becky?

Student: Because of bad charge separation.

Professor Michael McBride: Okay, there’s bad charge separation here. Here, on the right, you don’t separate charge. That may have been part of what Lucas was trying to say. But there’s another way you can say it, just in terms of HOMO/LUMO mixing. The language you’re using is the language of resonance, right? That you don’t want to separate charge. I want to see why, in terms of molecular orbital energies, or atomic orbital energies, that you get more stabilization here. Alex?

Student: There’s an electron pair that moves into an antibonding —

Professor Michael McBride: Right, but that’s true in both cases. There’s an electron pair here that mixes with a vacant orbital. There’s an electron pair here that mixes with a vacant orbital. Why is it more effective here than it is here? Claire?

Student: Just guessing, but if the resonance, for the two structures on the left, when the resonance comes down to the O-H bond it’s anti-bonding. But with the other one, because the H isn’t there, it’s actually making it more stable.

Professor Michael McBride: I’d have to think about that awhile. But I want you to try again, in terms of the — what makes HOMO/LUMO mixings good? What makes them important?

Student: If there’s a higher one over.

Professor Michael McBride: Overlap.

Student: Overlap.

Professor Michael McBride: That’s the same in both cases. It’s the same oxygen reacting — mixing with the same double bond. And what else?

Student: If there’s a high HOMO —

Professor Michael McBride: If it’s a high HOMO. The HOMO here is an unshared pair on oxygen.

Student: Right.

Professor Michael McBride: The HOMO here is an unshared pair on oxygen. Are they the same?

Student: No, because the HOMO on the left is taken up b hydrogen.

Professor Michael McBride: Right, okay. So the one on the left has another proton, which this one doesn’t. That means that the electrons here are more stable, lower in energy. And this is a higher HOMO, which is associated (another way of saying the same thing) with the fact that that’s a negative charge. When you have a negative charge, it’s a bad place to put electrons. They are high in energy. Another way of saying it, and the way you said it, which is probably a better way of saying it, is that the proton is gone, so it’s not stable. So you expect more HOMO/LUMO mixing here because of better energy match than on the left. Therefore, the observation, that it’s 1011 times stronger as an acid. So it’s a higher HOMO, because of that negative charge, or the absence of a proton. Okay, so there’s our carboxylic acid.

Chapter 3. The Aggregation of Alkyl Lithium [00:20:46]

And finally the last functional group we wanted to talk about today, which is a really complicated one. It starts very simple, but you’ll see it gets pretty baroque pretty soon here. So it’s methyl lithium. And the chance of your figuring all this stuff out ahead of time is about zero. Right? So don’t be depressed if this looks like it’s getting very complicated. Remember, there are three approaches here. One is to draw lines, à la Lewis, to understand how things come together. Another one is to use a computer to generate molecular orbitals; to break the total electron distribution into orbitals and look at them one at a time. And having seen what the computer produces, some of those orbitals are interpretable; as we’ve seen, that they’re antibonding in a certain region, bonding in a certain region, and so on. But the chances of your competing with the computer in drawing accurate molecular orbitals — to the extent that molecular orbitals are real — the chance of your competing with the computer is not very good. So don’t expect that you can exactly draw them. But if you see the one that the computer has drawn, at least in some cases you can make some guesses about it. You can see how many nodes it has; whether there are nuclei in the middle of lobes, which would make that a stable orbital; where it’s bonding; where it’s antibonding. Those kinds of things you can analyze, even though you can’t produce them yourself. So don’t think that you’re going to be able to. Right?

But then there’s another point of view, which is to look at localized pairs, bonds and antibonds, and how they’ll come together. Right? And that you can do. That’s the intermediate between using just Lewis and doing the full molecular orbital kind of thing. This is something you can do in your head, and that’s what we’re going to show, these different points of view with this one. So here’s methyl lithium. Now what will make methyl lithium reactive? What’s a high HOMO of methyl lithium? Marty? What makes HOMOs — what makes for a high HOMO? What kinds of things do you think about?

Student: [inaudible]

Professor Michael McBride: Can’t hear very well.

Student: Like what the outlines on the slide were before, unshared pairs and —

Professor Michael McBride: Just with a pair of atoms. I’m not trying to make like our slide before where you’re doing all the complicated stuff. But just you have two atoms and they come together. Will they give a very high HOMO? What determines how far they go up and down, the HOMO and the LUMO? What’s one thing that determines it?

Student: Oh, the overlap.

Professor Michael McBride: Overlap. Okay. So here are two atoms that are coming right together, to a single bond distance. So good overlap. Let’s suppose it’s the same as all overlaps. Okay? What else? What is it besides overlap that determines how far they go?

Student: Energy difference.

Professor Michael McBride: The energy match. Right? Now what’s special about methyl lithium? Methyl’s not going to be funny — that’s our standard thing — but lithium is funny. Where is it on the periodic table? Maybe that’s the problem we’re having.

Student: It’s on the —

Professor Michael McBride: On the far left, which means it doesn’t have a very big nuclear charge. So does that mean electrons are going to like being there or not like being there?

Student: They’re going to not like being there.

Professor Michael McBride: Because it’s not much nuclear charge. So it’s going to be a bad energy match, high energy. So there’s going to be a high HOMO. Right? The HOMO here, from the carbon, doesn’t go down very much, because of the bad energy match. Okay? So there’s an unusually high HOMO. What will it look like? Where will it be, on the carbon or on the lithium?

Student: The HOMO?

Professor Michael McBride: The HOMO.

Student: It’ll be on the first one.

Professor Michael McBride: Is it going to be symmetrical, or is it going to be more Carbon, or going to be more Lithium?

Student: More carbon because —

Professor Michael McBride: More on the carbon. That’s the lower one. Okay? So if we want to look at the HOMO of this thing, it’s going to be mostly on carbon, not so much on lithium, and an unusually high HOMO because of the lithium. And there it is, that’s the HOMO. Marty, you’re going to finish us here by saying what orbital does it look like?

Student: It looks like —

Professor Michael McBride: I’ll back up. There it is and there’s the orbital.

Student: 2 p-ish.

Professor Michael McBride: It’s a 2p orbital on carbon, mostly. Right? With just a little bit of lithium, 1s, or 2s, pardon me, on it. But just a teeny bit of lithium. And it’s antibonding between the lithium and the carbon. It’s the — or no, no, wait, it must be a p orbital on lithium, because it’s the favorable one, it’s the bonding one, right? We see the blue of lithium, but there’s a little red on the other side that we don’t see, without making a net, which I didn’t do. Okay, so that’s the HOMO. It’s a σ. Now what will the LUMO be? Kate, what do you say? What do you think for a LUMO, of methyl lithium?

Student: It’s going to be a p orbital.

Professor Michael McBride: Can’t hear very well.

Student: Lithium orbital —

Professor Michael McBride: Yes, lithium has a lot of vacant orbitals on the atom. Right? It comes in with only one valence electron. Only one of its four valence orbitals is occupied. So it’s got, in fact, three vacant orbitals. So that’s going to be a LUMO. There’s the LUMO, mostly on the lithium. Okay? Now, but the LUMO+1 is this orbital, which I’ve drawn in a funny way, at a relatively high electron density contour, making the things very small. Because if I made them bigger it goes way off the top of the sheet and off the bottom of the sheet, because this orbital isn’t held very tightly by lithium, it sticks way out. And it turns out to be out of the range that gets plotted by Spartan. So you get funny looking holes in the orbital. So I chose this particular density. But anyhow, you see it’s a p orbital, that’s different signs, top and bottom. Now, so you have a — is this an unusually low LUMO?

Monique, what do you say? Let me tell you something. I’d say lithium has a small nuclear charge, not a very good place for electrons to be. Right? Which means that it’s going to be a very high orbital energy. Can you see any contrary things that say it might be unusually low?

Student: I would think it would be high.

Professor Michael McBride: Can’t hear very well.

Student: I would think it would be high.

Professor Michael McBride: Because of the low nuclear charge?

Student: Yes.

Professor Michael McBride: Compared to what? What are we comparing with, to say whether orbitals are high or low?

Student: To a Carbon or H.

Professor Michael McBride: The carbon-carbon, or carbon-Hydrogen σ and σ* bonds. Right? So clearly enough the carbon and hydrogen start lower in energy than the lithium does; but what? But they have big overlap. So σ* is quite high, much higher than lithium, because it’s an atomic orbital, it hasn’t mixed with anything, this one. Okay, so methyl lithium has an unusually low vacant orbital and an unusually high occupied orbital. And what do you expect?

[Students speak over one another]

Professor Michael McBride: You ever seen a case like this before where the same molecule was —

Student: BH3.

Professor Michael McBride: BH3, for exactly the same reason, right? Except more so, because lithium is more so than boron, being further to the left in the periodic table. So we can put two of these things, above, up and down from one another, like this, and have the HOMO of one be stabilized by the LUMO of the other. Kevin?

Student: Sorry, could you please explain again why it has a low LUMO?

Professor Michael McBride: Yes. Because it’s an atomic orbital. It’s not a molecular orbital, it’s not a bonding orbital. It didn’t mix with anything. The things we’re comparing with are not carbon and hydrogen, which are in fact lower than lithium. It’s the σ and σ* bonds of carbon and hydrogen which — and this, the LUMO here is higher than that of lithium — lithium’s unusually low — because it’s still an atomic orbital, it didn’t mix with anything. Okay? So there’s stabilization of the σ by the π. But at the same time we can go the other way and the HOMO on the top can be stabilized by the LUMO on the bottom. Same as in BH3, giving B2H6. Okay, so these things dimerize. They come together, HOMO of one stabilizing the LUMO of the other, and vice-versa. They move together, and you get a thing that looks like that, very much like B2H6. Right? So you have two vacant lithium orbitals, stabilizing the unshared pair of carbon. You can think of it that way. And it also goes the other way, the unshared pair on that carbon is stabilized by two vacant lithiums. Okay? So these are three-center two-electron bonds; just as in B2H6. Okay now, but remember, lithium has two other vacant orbitals, each atom. We’ve only used one here. Right? So they can combine to give this. This is actually the LUMO of that dimer. Right? Those are in the plane of the four-membered ring.

Right? And there are other vacant orbitals too, like this π one, which is in and out from that four-membered ring, made up of p orbitals on the lithium. That’s the LUMO+1, of that dimer. Okay, now I’m going to rotate this. Now, pay attention here because the next picture’s a little confusing. I’m going to rotate this thing so that we’re looking just a little bit — we’re looking mostly in the plane of the molecule but twisted a little bit, so we can still see the top of the molecule. And we’re going to twist it such that the red lobe in back is behind the red lobe in front. So what we’re going to do is rotate around that axis and shift it a little bit toward the center of the picture. So it’s that same orbital. And there it is now. But it’s a little confusing in the middle because there’s one red lobe in the back and one in the front. Okay? We could draw it down — okay, so that’s what it is. But we’re going to drop an axis that goes right through the middle of that parallelogram. Okay? And now along that axis we’re going to bring up another molecule. But it’s going to be rotated 90º. So those are the same two orbitals, LUMO+1 in both cases. The one on the bottom is rotated 90º around the axis I drew. Okay? Now will those things interact favorably with one another? First, they’re orthogonal. And second, even if they mixed you wouldn’t care. Why?

[Students speak over one another]

Professor Michael McBride: Because they’re vacant. So what we want to do is look at the bottom, not at LUMO+1. We want to look at the HOMO of that one, that dimer, which looks like this. And now you see something neat, that the blue in front overlaps the blue vacant one, and the red in back overlaps the behind red vacant one. So the electrons in the HOMO on the bottom get stabilized, by overlapping with the LUMO+1 on the top. And, of course, by the same token, the HOMO of the one on top overlaps with the LUMO of the one on the bottom, like that. So we already had two molecules come together to form dimers. Now we’re having two dimers come together to form a tetramer. Yoonjoo?

Student: Why are we using LUMO+1 on the right?

Professor Michael McBride: Ah, because that’s the one that does the overlap. The actual LUMO doesn’t overlap this one on the bottom. Okay? Lucas?

Student: Does that make the LUMO+1 essentially the new LUMO, because it’s more stabilized?

Professor Michael McBride: Well, but it mixes with this one. So the LUMO+1 and this one interact. Here’s the HOMO.

Student: And they make it better —

Professor Michael McBride: LUMO+1. So it contributes to this one. Right? When the HOMO here, of the dimer, mixed with the LUMO+1, of the other dimer, you’ve got a mixture. This one looks mostly like the HOMO of this dimer. But it’s a little bit helped out by looking like the LUMO of this one. There’s a new vacant orbital, but it’s still higher. You don’t care about that one anymore.

Okay, so now we can take two of these and bring them together. That’s a bonding, the orbitals we just showed. And you get a cube. Right? So here’s the one on the bottom, one, two, three, four; and here’s the one on the top, one, two, three, four. And they’ve been turned 90º, to line up these orbitals, and they come together and the electrons get more stable. Now that used, on every lithium, another one of its vacant orbitals. Right? But what Angela?

Student: There’s another one.

Professor Michael McBride: There’s another one. Okay. So the HOMO now turns out to be this. And you can understand that, the biggest part of it, by seeing that it’s a four-center, two-electron bond. Mostly it’s a p orbital, or a hybrid orbital of the carbon, an unshared pair of carbon, but it’s stabilized by three vacant orbitals of lithium. So three vacant lithium atomic orbitals stabilized the unshared pair of carbon. But, as Angela points out here, there’s another one. That’s still a vacant orbital on lithium. That’s on the top lithium. There are other ones on the other three lithiums. So we need more electrons — or it’s capable of stabilizing more electrons. Where do you get them? Well you might think you bring another cube up. But in fact there’s something easier to do, because this is done in solution, and the solvent is dimethyl ether, as shown here, or some other ether, which has an unshared pair of electrons, and it comes along and gets stabilized by that. But at the same time the other three oxygens, the other orange balls — the other three lithiums I should say — stabilize oxygens of other ethers. So you have four ether molecules associated with a tetramer of methyl lithium. And this is about as complex as we’re ever going to get. Right?

So I think you would’ve had a hard time, at the outset, predicting this. But having seen it, I think you can understand it. That’s what we’re aiming for at this stage. Because as we go on and get lore, that this reacts with this and gives this, what I want you to be able to do is to look at it and say, ah ha, that’s what happened, that’s why that reacted that way. In some cases you’ll be able to predict. But in other cases you’ll be able to understand and correlate it with what you know. So if you put more ether in, if you have a lot of excess ether, then it rips these things apart. So it can fly apart and you get a bunch of ethers on every lithium. All right?

And in fact this is a further development of organic chemistry, beyond what we usually do in an elementary course, which is to understand the non-bonded interactions. Because often you wouldn’t consider that oxygen-lithium to be a real bond. It’s just sort of a weak solvation thing; although truly it is a bond. Okay? So these so-called non-bonded interactions and the effects of solvent are a vital part of lore, what solvent you choose for a given reaction. For example, if you want to facilitate ionization, you use certain solvents and not other solvents. And we’ll talk about that a little bit. But I just want to show you what there is out there, beyond the bonding that we’ve already talked about. But we’ve really taken theory pretty far here, in the first half of the first semester of the course.

Now, I said we would vote. And I forgot to bring a piece of paper. I have this piece of paper, but I’m not going to use it for that, because this is the thank you note. So if you want to — if you haven’t signed it — why don’t you pass it around? Because there are a few people who haven’t signed it. Let’s pass it down, back and forth, down the aisles. I think most people have signed it, but some haven’t. You don’t have to. And you also don’t have to contribute. But if you want to be — have a sense of satisfaction of having contributed to the class gift, you can leave a small coin on the table over there. Okay, but the other business we have to transact is I’m going to give a review session before the exam. But it’s not obvious when we should give it. So I’ve decided that the easiest thing to do is to see how many people cannot — of course we want people not only that can, but would come. Right? If you could come but wouldn’t, you’re irrelevant. So the question is, how many people cannot — or would come, other times, but cannot come at the time shown? And the one that has the fewest will win. That make sense? Okay, so how many people cannot come Thursday evening, 8:00? That’s too many. How many at 8:30? Here we have about, my judge is about fifteen. How many cannot come Friday afternoon?

Student: Lab.

Professor Michael McBride: Whoa! Okay. How many people cannot come Saturday, early? That’s a lot. How many Saturday at 10:00? How many Saturday at 2:00? How many Sunday, 1:00 to 3:00?

Students: Too early.

Professor Michael McBride: Well bear in mind that there’s also the TA session Sunday evening. I think it’s better if I do it two days before the exam, rather than the day before. But on the other hand, none of these earlier times can get — they’re always at least fifteen people. Well let me see about option two. That was the one Thursday at 8:30. How about if it were 9:00? How many could not come if it was at 9:00?

Student: Wait, which Thursday?

Professor Michael McBride: Thursday evening, tomorrow evening. That actually is — and let’s see Sunday again, Sunday, sometime early afternoon. There are actually more who can’t come on Sunday than Thursday. So I’m sorry, but it’ll be Thursday. It could be 8:30 or 9:00. Does it make a difference? Is there somebody who’d come if we started at 9:00?

Student: Is Saturday one?

Professor Michael McBride: Pardon me?

Student: Is Saturday one?

Professor Michael McBride: Saturday what?

Student: Is Saturday bad? Like how bad was that?

Professor Michael McBride: Saturday was pretty bad. But let me see, Saturday 9:00? Yes, that’s worse. Saturday 10:00? Saturday 2:00? Saturday 8:00 p.m.?

[Students speak over one another]

Professor Michael McBride: So it’ll be tomorrow night at 8:30, and I’ll send you an email, once I got a room. Okay, let’s continue.

Chapter 4. Why Wouldn’t Past Organic Chemists Be Surprised? [00:41:21]

Okay, so to summarize the first half of the semester. We’ve seen amazing modern tools that show us molecules, the real thing, at the level of Ångstroms and at the time scale of picoseconds, 10-12 seconds. We’ve seen scanning probe microscopy that can feel atoms, and X-ray diffraction that can see them, and can see bonds, with difference density. We’ve seen spectroscopy, IR and ESR. And next semester we’ll talk about nuclear magnetic resonance. So there are amazing tools, not only experimental tools, but theoretical tools. This stuff about quantum mechanics so that — and people believe quantum mechanics enough that they talk about doing calculations on a computer as “experiments,” rather than just things you do in the lab. So that’s amazing. Right? And it all has developed fairly recently, say since the time — most of the important things we’ve talked about have developed since I was in graduate school. But what’s amazing is that organic chemists were not at all surprised when they were finally able to see these things. Because they already knew what they looked like. So it wasn’t surprising. It was satisfying. And there were a few little queer things that are understood better now than they were before, certainly. Right? And things like being able to generalize about what makes a functional group, that’s completely new. Right? But fundamentally what molecules look like and how atoms behave and that this reacts with this, that was already known.

So how could they possibly know it when they didn’t have these tools? And that’s going to be our chore for the next quarter of the semester, is to figure out how they knew; how did organic chemistry really develop? And a reason for this is that if you know how it developed, then you understand the language much better. Because the language we use isn’t the language of quantum mechanics usually. It’s the language that developed during this long gestation period. And it started, as we said before, in the seventeenth century. And we’ve gone on this, and we’ve talked now about science, the scientific method and force laws, and how we finally found out what the force law was for atoms. Right? And in the twentieth century, this whole theoretical, experimental stuff came together and we saw that bonds were shared electrons, and we could see them, and that quantum mechanics could help us to understand them. But we’ve left out this middle period, starting with the end of the eighteenth century, with Lavoisier discovering oxidation, and ending with the discovery of quantum mechanics. Because this is the period during which most of organic chemistry developed, and we have to see how it happened. Okay?

So this, if you’d been here 107 years ago, and been graduating, you would’ve been in this picture, with your fancy boots and so on. Right? This is the Computer Science Building. You walk up now — it used to be the Chemistry Lab, on the right. You know where it is? Watson Lab down there. Okay. So let’s look at some, see what they understood at the time. They chalked things on the bricks, to show off their knowledge. So here’s CH4 methane; obviously they knew that. And here’s the reaction of sulfuric acid with zinc, which gives hydrogen; they knew that. And here, what do you think that crazy thing is in the middle?

Student: A benzene ring.

Professor Michael McBride: That’s a benzene ring. And remember how it looks like what they confirmed at about the same time? The clairvoyant people saw that the valences all went to the center. Okay? And down here you see another thing. Those are probably aromatic rings, with the double bonds not shown. But an interesting thing that’s shown, the Greek letters, α and β there. Those are symbols that are still used to denote different positions that you can put substituents in the ring. Okay, this is inside the Chemistry Lab, which was where SSS is now. And you see things like this, right? Which came from that lab I’m sure. There was a big box of them floating around here, and I rescued one of them a long time ago. The others got thrown out. There. And what’s that second arrow? There’s another important tool.

Student: Bunsen burner.

Professor Michael McBride: The Bunsen burner. Right? Bunsen invented the burner about 1860, or something like that, or ‘70. Okay, here, there’s one that looks just like this, right? Actually, this isn’t that old, because it says on it “West Germany.”


Professor Michael McBride: Okay, there, what do we see here, the white thing?

Student: Mortar and pestle.

Professor Michael McBride: Mortar and pestle; they could grind things up. Right? And there, what’s that?

Student: It’s a burette.

Professor Michael McBride: It’s a burette drying upside down. Notice, that’s the only quantitative tool. So were there any other quantitative tools in this picture, that they were showing off?

Student: Well there’s somebody who has it on the side.

Professor Michael McBride: Ah ha, there were burettes. Anything else that will give you a number?

Student: A scale.

Student: There’s a scale.

Students: Oh.

Professor Michael McBride: But that’s just a crude balance, right? There had analytical balances, but they weren’t portable, they couldn’t schlep them out to Prospect Street, for the picture. Okay, but that was it. Okay? This guy here. [Laughter] Looked like that; C. Mahlon Kline, Class of 1901S. And you know his name because GlaxoSmithKline. And he gave money for the Kline Biology Tower, and the Kline Chemistry Lab; and the Kline Geology Lab too. But you could also measure angles on crystals. Right? But that’s about it. Somehow, with these tools, they figured out where atoms were, in molecules. Isn’t that crazy? So if you look at the front of the building, you see down here — he’s wrapped in stuff now because they’re painting him — Benjamin Silliman, Yale’s first scientist; first chemist anyhow. And he’s holding something. Can you see what he’s holding?

Student: A crystal.

Professor Michael McBride: A big, whopping crystal. Right? That’s where science was at. That’s what he’s mostly known for, is mineralogical things. Okay. But here, on these plaques, is something else. This is the names of the people who actually developed all this stuff, during the nineteenth century. So it starts in the seventeenth century, actually, with Boyle and the gas laws. But that was way before — notice, he was born 120 years before the time we’re talking about, Lavoisier. So things happened, but not a lot, for our purposes, during that period. And it really got underway with the second person. These are in chronological order. The second one was Lavoisier, and that’s where we’re going to start. And then Gay-Lussac, Berzelius, Faraday. We’re going to talk about these people. Okay? And here is you, and your cousins, who did all this stuff. Because we have a family tree here.

So there’s you, and me. And I studied quantum mechanics as a graduate student with E.B. Wilson — right? — who was a post-doc with Linus Pauling at Caltech, who did X-ray and proteins. He’s the first guy that saw the helices of the proteins. Okay? And another post-doc of Pauling was Jack Dunitz, who we’ve talked about a lot. So Dunitz, who we’ve talked about a lot, is your great-uncle, academically. [Laughter] But if we go back — oh, and here are things — notice how many of these things, the twentieth century things we’ve talked about. Notice that boldface is things that are experimental, and normal face is theoretical things. Shared-pair bonds, mesomerism, which is resonance. Electron pair shifts, the curved arrows. Quantum Theory. Reaction mechanisms; we already talked about ammonia being oxidized by chlorine, for example. IR spectroscopy we talked about. We haven’t talked about — well we’ve talked a little about polymers, not much. We haven’t done mass spectroscopy. Well we did actually the problem set in the first thing, right? Conformational analysis we’ll do at the end of the semester. X-ray we’ve talked about. NMR, next semester. Molecular orbital methods, we’ve talked about a lot. This is what we’ve been talking about. That we’ll do at the end of the semester. Non-covalent structures we just talked about a little bit. Scanning probe microscopy is quite recent, and so on. So we’ve actually talked about a lot of this stuff already, the more recent stuff. But what we want to talk about now is the earlier stuff, how people discovered things in the nineteenth century. So these are the guys that did it. And in particular we’re starting with Lavoisier. But Lavoisier was not the first altogether. But I’m going to have to wait ‘til next time to tell you who was first.

[end of transcript]

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