CHEM 125a: Freshman Organic Chemistry I
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Freshman Organic Chemistry I
CHEM 125a - Lecture 17 - Reaction Analogies and Carbonyl Reactivity
Chapter 1. Similarities among Acid-Base, SN2 Substitution, and E2 Elimination Reactions [00:00:00]
Professor Michael McBride: Okay, so this is what we looked at last time. We looked at an acid-base reaction from a new point of view, from the point of view of bringing up some HOMO to overlap with this unusually low LUMO. Why is it unusually low, the LUMO, σ*, of H-F? Why is it unusually low? Is the overlap bad?
[Students speak over one another]
Professor Michael McBride: Does the hybrid orbital on fluorine and the s orbital on hydrogen, does that have bad overlap, or does it overlap well? Points right at it, good overlap. So that’s not the reason it’s unusual. What’s the reason that the energy of this vacant orbital is unusual? Dana?
Student: The unusual atomic energy of fluorine?
Professor Michael McBride: Right, because fluorine started very low; so that went down, when it mixed with hydrogen. Hydrogen went up, but it didn’t go up very far, because the energy match was bad. So that’s this LUMO, the unfavorable combination of fluorine and hydrogen. Which is it big on, hydrogen or fluorine, the one that went up? Hydrogen, right? The good one is mostly on fluorine; the bad one, σ*, mostly on hydrogen. And it has that node shown as an antibonding node; σ*, the star means antibonding. So in OH-, with its high energy occupied orbital — now wait a second, why is an unshared pair on oxygen high energy? I thought oxygen had a big nuclear charge, which would make electrons stable, not have them be high. Why is OH- high? Obinna, do you have an idea about that? What is it about OH-, the unshared pair that makes the electrons unusually high in energy?
Student: Because it didn’t mix.
Professor Michael McBride: Can’t hear.
Student: It didn’t mix.
Professor Michael McBride: Because it didn’t mix with anything else. Right? It’s not like a bond that went down. We’re comparing it to C-H or C-C σ bonds that went down in energy. So even though oxygen started low, it didn’t go down. Right? So it’s left unusually high. And there’s actually another reason that this one is unusually high. Josh?
Student: Negative charge on —
Professor Michael McBride: Negative charge on oxygen doesn’t make electrons happy. Okay, so it comes along, mixes with that. In the process of putting electrons into that orbital, it makes the bond. But also it makes it antibonding between hydrogen and fluorine. So we draw that second curved arrow that shows the fluorine leaving with the electrons, that it already had most of. The electrons it leaves with — not that you can really label electrons and say this one is this, this one is this. But remember, fluorine mixed with hydrogen went down. It already has most of those electrons that came from hydrogen. Now it has them all when it breaks away. It breaks away as an ion, F-. Okay, and that’s an acid-base reaction; a more general interpretation of acid-base, than just H+ reacting with OH-.
Okay, now let’s generalize a little bit further. So notice that besides creating a new bond, which would happen with H+, mixing the HOMO with this LUMO can break a bond, right? If the LUMO, shown there, has an antibonding node. Putting in electrons means that those electrons will be more stable if fluorine moves away. Right? So you not only make a bond, you also break a bond. So it’s a make-and-break situation; whereas H+ plus OH- is just a make situation.
Okay, now let’s look at F-CH3. Right? And look at the orbital, and I’m going to superimpose on it the corresponding LUMO of F-CH3. Watch it change. The carbon-fluorine bond is longer than the carbon-hydrogen bond. So it moved away. But the shape, especially around the fluorine, is almost identical. I’ll back up and do the change again. Right? So it’s going to have very similar properties to what H-F did. So if I move it over, you see it’s going to be attacked by something coming here to overlap with that orbital. Some high HOMO will mix with the low LUMO. And here we show OH- doing it. And then what will happen? Remember, we have this antibonding node here. What happens when electrons go into that orbital? Kate?
Student: It breaks the bond.
Professor Michael McBride: Kate?
Student: It breaks. The fluorine is going —
Professor Michael McBride: Yes, the fluorine will take electrons and move away, like that. Right? So we get fluoride and methanol. Right? And this reaction, which we’ll study at the beginning of next semester — and we’ll mention a little bit more as we go along this semester, but systematically we’ll talk about it next semester — is called SN2 Substitution. But notice, it’s, from the point of view of orbitals, exactly the same make-and-break situation that you had with OH- reacting with H-F. It’s the same reaction, just the names have been changed. Right? So you could, in fact, have called the reaction of OH- with H-F, you could’ve called that a substitution. It could’ve been SN2 Substitution. The difference would be that the atom on which the substitution takes place, the one that gets attacked and then loses the F-, is hydrogen rather than carbon. So you could call both reactions substitution, or you could call them both acid-base reactions. Right? So that’s the way sciences, as I said last time, progress, by seeing generality, where things were thought to be different before.
Okay, so we have these two reactions, acid-base reaction, SN2 Substitution. What would happen if we had ethyl fluoride, sometimes known as ethanol — improperly labeled, right? — earlier today. Okay, so that’s the LUMO of ethanol [correction: ethyl fluoride]. Now that doesn’t look exactly like the LUMO we just calculated of ethanol [correction: ethyl fluoride]. But remember it was the LUMO+3. Wasn’t it that we found? It was either plus two or plus — it was plus three wasn’t it? The LUMO+3, remember, had this characteristic, that it’s a favorable mixture of two simple pairwise orbitals. That one is what? What’s that orbital, if you just saw that?
Professor Michael McBride: Josh?
Student: It’s the same thing as the other one.
Professor Michael McBride: Ah-ha, it’s the same thing as up there. Right? It’s σ* C-H, instead of C-F. Well it’s certainly σ*, but it’s of H-C, not of F-C. Right? Everybody got that? So but it’s σ*, that’s the important thing. It’s the antibonding combination, σ* C-H. Okay, and what’s the other thing that goes into this LUMO, or actually the LUMO+3? It’s this. What’s that? Right? Angela?
Student: It’s the [inaudible]
Professor Michael McBride: Can you give it a name?
Student: It’s the σ* of [inaudible]
Professor Michael McBride: If you name just two atoms, what is it? Speak up loudly so everybody can hear.
Student: σ* C-F.
Professor Michael McBride: σ* C-F. Everybody see that? Okay, and I say it’s the favorable mixture of these two, because they’re put together so as to be bonding between them. Right? So there were two vacant orbitals, σ* C-F and σ*C-H, and they mixed, and the favorable combination, where they overlapped favorably between those two, is the one we’re looking at. It’s still vacant, it’s an unoccupied molecular orbital. But that’s the one we’re looking at here. Okay, now how should it react? Well that one looks like the top right, except that the colors are changed. Right? So how about doing SN2 substitution? Now Russell cringed when I did that. What’s wrong with doing SN2 substitution?
Student: We have to get a favorable overlap with —
Professor Michael McBride: Ah-ha! what’s hidden here is that bit. It’s going to be hard to overlap with the blue on top, without having an unfavorable overlap with the red on the bottom. So there’s another way you could do it. You could attack here, right? And get a big orbital, a big lobe, all of the same sign. So there you’re going to get good overlap. So might it do that? So there are two possibilities. It could do the same as the top, it could do SN2 substitution, or it could put electrons into this LUMO by attacking from this second arrow position. And how do you know which one it does?
Professor Michael McBride: You do experiment. So this is lore, right? And when we get to studying these reactions systematically, you’ll see that both things can happen. They’re both possibilities, and you can make little tweaks in solvent or base and things like that, what HOMO you’re using, that’ll make it go a little more one way or a little more the other. But that, of course, takes time to do all the experiments and to hear about them. But anyhow, there are these two possibilities.
Now let’s suppose it does the second one. Now how did I set it up so that that would be favorable? I didn’t use the true LUMO of the lowest energy, of the molecule, which I just calculated for you at the beginning of class. What I did was use that LUMO+3, but made it the LUMO. And the way I made it the LUMO was to stretch bonds. I stretched the C-H bond and I stretched the C-F bond. Right? And then if I — in this particular phase of vibration of the atoms, this one is in fact, the LUMO. Okay, now why did I want to show you this? Look what happens if you do that bottom thing. Okay, so we bring OH- up, and the electrons go into this and make a bond between H and O. But there’s a node here, that antibonding node. What does that mean, if you put electrons into this orbital? Lexy, what do you say?
Student: It’ll break.
Professor Michael McBride: It’ll break where?
Professor Michael McBride: Between the hydrogen and the carbon. Okay, so we’re going to — we’ll draw a second curved arrow that shows those electrons pulling out. Okay? Now where do they go? Look here, there’s an atomic orbital node there. That’s not an antibonding node because it’s the node of the p orbitals on the carbons. Everybody see that that’s not antibonding? Because it doesn’t go between the atoms, it goes through the atoms. So, in fact, if you look at that blue and red lobe in the middle, they’re a little bit distorted. But what do they look like? If you look just at the red and blue in the middle? Lucas?
Student: p orbitals.
Professor Michael McBride: Not p. Greek.
Student: Oh π.
Professor Michael McBride: It’s the π orbital, made up of 2p orbitals; the p orbital of the carbon on the top, the p orbital of the carbon on the bottom. And you put electrons in that and you get a double bond between the carbons. Okay? Okay, so there’s a π bonding between the carbons. That means we can draw a second arrow that takes the electrons out of the bond that’s breaking and puts them between the two carbons to make a double bond. Okay? But now there’s another node in this LUMO; there, that antibonding node. What happens with respect to that, when we put electrons in this LUMO? Rick, what do you say?
Student: The C is going to break the bond between the C and the F.
Professor Michael McBride: The bond between C and F will break. How will we draw a curved arrow to show that?
Student: You would draw a curved arrow starting at the C and ending at the —
Professor Michael McBride: Not starting at the C. Which are the electrons that are leaving? You start with the electrons that are changing, right?
Professor Michael McBride: So where are they, in this diagram here? Here, tell me when to stop, when I hit the electrons?
Professor Michael McBride: Not here. That’s the one that’s building up electrons, right?
Professor Michael McBride: These are the electrons that are leaving. And where do I draw the curved arrow to, where are they going? Okay. On to the F. So that’s what happens. When the O-H comes up, you get water, you get ethylene, the double bond, and fluoride. So this is called, as you’ll see — just after we talk about SN2 substitution, we’ll talk about E2 elimination. Right? And that’s what this reaction is. Right? But that’s just an old category. Right? If we want to make it — what it is, is the same kind of HOMO/LUMO interaction, except this time we make two bonds and break two bonds, because of where the nodes were, in the LUMO that was being attacked. Right? There were two antibonding nodes. We break two bonds. Okay, does anybody have a question about this? Do you see how it works? Lucas?
Student: Why did you stretch C-H again?
Student: Why did I do it? Because I knew the answer. I knew that this reaction does this E2 elimination, because when I studied organic chemistry we did E2 elimination and never talked about orbitals at all. Okay? So I knew where I was going. I knew what really happened. So I said, how does that happen, that this is the orbital that actually gets attacked? And it’s, in fact, the LUMO+3. Right? Why do the electrons go into that? And then I saw that in a certain phase of vibration of the molecule, when this bond is long, and when this bond is stretched, then it is the LUMO. So when the OH is attacking, the OH- is attacking, it’s attacking at a time that those bonds are stretched at the same time. Right? And then they break. Okay? It’s a little bit of chicken and the egg as to whether the electrons going in make them stretch, or their stretching allows the electrons to go in. Right? But anyhow, I knew what the answer was, which you wouldn’t have known, if you had just looked at the orbitals for the minimum energy geometry of the molecule.
Chapter 2. The Oxidation of Ammonia by Chlorine in Molecular Orbital Terms [00:15:23]
Okay, now let’s look, let’s generalize this to a complicated, or fairly complicated, reaction scheme, at least by our lights. So three ammonia molecules react with chlorine. That’s called oxidation of ammonia, and it generates hydrazine, the nitrogen-nitrogen bond, or actually hydrazinium — it’s a cation — chloride, and ammonium chloride. That looks pretty complicated. There’s a lot of changes going on. How does it happen? Could we figure it out? I think you can. Okay, so in the starting material I want to find a high HOMO. Where do I find it? Sherwin, can you help us? What do I look — I want to find a high HOMO in this starting material. The starting materials are ammonia and the chlorine molecule. So what about them could give you a high, unusually high-occupied orbital?
Student: The unmixed atomic orbital —
Professor Michael McBride: Say it clearly.
Student: The unmixed atomic orbital in ammonia.
Professor Michael McBride: Yes, the unshared pair of ammonia. Right? N has a bigger nuclear change than C. So you might expect the electrons to be low in energy. But it never mixed with anything, so it didn’t go down to become a σ bond. So, unshared pair on ammonia. I’ll take that. Okay. Now we have to react it with some — and unshared pairs are often denoted n, little n. And I actually don’t know why that is. It’d be fun to find out. Okay, now we need a low LUMO. Anybody got an idea for what a low LUMO might be in this system? What kind of things give rise to low LUMOs?
Student: Electronegative things.
Professor Michael McBride: Sam?
Student: Electronegative —
Professor Michael McBride: Electronegative nuclei. Right? Big nuclear charges. You know, people always say electronegative, but that’s the cart and the horse. Why is something electronegative? That means it’s a good place for electrons to go. Right? It’s because the nuclear charge is big; that’s what’s fundamental. Okay, where do we find that here?
Student: In the chlorine.
Professor Michael McBride: The chlorine. Now which orbital, associated with the chlorines, is going to be unusually low in energy? Does it have a vacant atomic orbital? Bear in mind what’s — chlorine is in the last column, right? So it’s just missing one electron. So a chlorine atom has three unshared pairs and one odd electron, which has come with the other chlorine to make that bond. Right? So after the Cl2 forms, where’s the vacant orbital? Anybody see? We had two chlorines, each with this odd electron. They came together and you got a σ bond. Where’s the low LUMO? Chris?
Student: It’s going be like a 4s atomic orbital.
Professor Michael McBride: No. Kevin?
Professor Michael McBride: σ*, right? They each had one electron. Come together, they went down. But there was a σ* that went up. σ* is usually high energy, but not in this case because, as Sam pointed out, the nuclear charge is big, so it’s low. So σ* on chlorine. Okay? So what kind of reaction are we going to have? If we mix σ* of chlorine with the unshared pair on nitrogen? Have you ever seen a reaction like this before, where you have a σ* mixing with an unshared pair? Katelyn?
Student: On the last slide.
Professor Michael McBride: On the last slide we saw lots of examples. Right? And so Katelyn, what happens? Tell me how to draw a curved arrow here.
Student: The electrons from the unshared — no.
Professor Michael McBride: The electrons on the unshared pair of nitrogen, that’s the high HOMO. Between what?
Student: The two Cl’s?
Professor Michael McBride: Between the nitrogen and the chlorine. They mix with that vacant orbital. But the ones in the chlorine, now that’s an antibonding orbital, σ*, so a chloride leaves. So it’s exactly the same reaction, make and break, that we saw before. Okay? So we get ammonium chloride “chlorammonium” and Cl-, plus/minus, net neutral. We started neutral, we’re still neutral. Okay? And now we have one of the products, the chloride up above we have. Okay, but we also have this molecule. Now it has a low LUMO, so it’s reactive. What is the low LUMO? Yoonjoo?
Student: The N-H bond.
Professor Michael McBride: That’s not what I would’ve said. It turns out to be right, but it’s not what I would’ve said. There are going to be N-H — it’s got to be vacant, so what’s the name of the orbital?
Professor Michael McBride: σ* N-H. Are there any other σ*s here? Russell?
Professor Michael McBride: Cl. Which do you think is going to be lower, σ* N-Cl or σ* N-H? Kevin?
Student: I would guess N-Cl.
Student: I would have too. So the lowest LUMO is for some HOMO. Now what are we going to use for a HOMO? We have to have another reagent for this to react with. Well fortunately that σ*N-H is what we’re actually going to use. Okay? So the high HOMO. We had three ammonias to begin with. So we can use another one; same deal again, right? So the best reaction, I think, would be for the unshared pair to attack the chlorine; at least that’s something that you could imagine. If that happened, what would the product be? If NH3 attacked Cl, it would make an NH3-Cl bond, and lose what? What Russell?
Professor Michael McBride: So what about it?
Student: That’s what it could make there.
Professor Michael McBride: It could happen ‘til the cows come home, but the product is the same as the starting material, and no one cares. It would exchange which nitrogen is attached to the chlorine. But unless you label them with isotopes or something, you’d have no way of knowing. So that probably does happen, or at least I suspect it happens some of the time. Right? But you don’t care. So now we resort to Yoonjoo’s suggestion that we’ll use a different one so we get a different product. Okay? So we’ll attack the N-H bond. Now Corey, tell us how to draw a curved arrow to show this. Where’s the curved arrow going to start?
Student: On the lone pair.
Professor Michael McBride: On the lone pair of the nitrogen. And where will it end?
Student: In between the N and the H.
Professor Michael McBride: In between the N and the H. And what else will happen?
Student: The bond will break.
Professor Michael McBride: Which bond? What’s going to take off with the electrons? Can you see? Nothing. Well, that is, what you lose — it becomes an unshared pair on nitrogen. N was N+, right? Now it gets the electrons that formerly it was sharing with H. So it ends on nitrogen. Right? So there’s the product. And now I’ve highlighted in blue another one of the ultimate products. Right? So now we’ve made ammonium chloride. Right? We’ve made one of the products.
But we need the other one. But we have this product, from the second reaction. Now, what do we have for a low LUMO here? Nate, do you have an idea? You have only σ bonds, and an unshared pair on nitrogen, and three unshared pairs on chlorine. Right? But those are all occupied. What’s a vacant orbital here? What kind of vacant orbitals do you get in molecules that have only σ bonds?
Student: The p orbitals.
Professor Michael McBride: No, the p orbitals are filled with electrons here. Right? So they’re not vacant.
Student: You get a σ* then. You get σ*.
Professor Michael McBride: σ*. Which σ — you have N-H σ* and you have N-Clσ*. Which one do you think is lower?
Student: The last time it was the N-H.
Professor Michael McBride: No, it wasn’t. The one that reacted last time was the N-H, but it wasn’t the lowest. The N-Cl was the lowest. Why didn’t we care last time? Because the product, when that reacted, you got the starting material back. So who cared? Right? So this time it’s the N-Cl. Right? What’s the high HOMO we’re going to use?
Professor Michael McBride: Ah. Wilson?
Student: The NH3, we still —
Professor Michael McBride: Ah, we got a third one up there still. Right? So bring it in. And there are the curved arrows, and there’s the product. Hydrazine — hydrazinium, it’s got a proton and there’s a plus charge — hydrochloride, hydrazinium chloride. Right? So that is the last product. So we’ve done it. So there are three of these reactions that take place, three cycles of make and break. And notice the difference among them, right? It’s always NH3 that attacks. But first it attacks chlorine, then it attacks hydrogen, then it attack nitrogen. But it’s the same reaction we talked about on the previous slide, right? Make and break.
Chapter 3. Reactivity of the Carbonyl Group [00:26:06]
Okay, now we’re going to look at four functional groups: The C double bond O, carbonyl; amide; carboxylic acid; and alkyl lithium. And after that, we’ll have done what we wanted to do about functional groups. Although it will still be in your court to do functional groups, because you have these Wikis, to do the same kind of analysis on the functional groups that we’re doing here, on these four functional groups. And those are due Thursday. And almost all of them are signed up for. There are just a few more to do. So if you want a choice, get there quickly, because I’ll assign them this afternoon, I think, because we’re almost at the end, and I have to make sure everything gets covered.
Okay, so the carbonyl group, C=O double bond. This is probably the most important functional group in organic chemistry. Why is it so important? (a) Because it’s strong, so it gives stable compounds, low energy compounds. So you have a lot of carbonyl groups. But, paradoxically, it’s not only strong, but it reacts easily. So it’s easy to make things from carbonyls. Okay, that sounds contradictory, that it should be both stable and reactive; but it is, and we’ll see why. So let’s look at the shape of the frontier orbitals. There are six valence electron pairs in formaldehyde, shown here from the side. And we could show the molecular orbitals, right?; the Chladni style, plum-pudding molecular orbitals. So the lowest one is 1s, or 2s if you want to count the core. Right? Which is mostly on the oxygen, a little bit on carbon. And then there’s the 2px, 2y and 2pz, as plum pudding orbitals. And then there’s this 3s orbital and this 3dxy orbital.
But let’s look a little more closely at that, from the point of view of pairwise interactions. What things interacted with one another to give these? So let’s do the kind of pairwise mixing we’ve been talking about. There’s the HOMO, an older style picture of the HOMO, the one that’s called 3dxy, on the right. And this is drawn at a — it was based on a different calculation and it’s drawn at a different contour level. But you can see what it — what is it mostly, the HOMO? If you had to give a simple name to it, what would you call it, if you neglected the little tiny pieces on the right? What does it look like? Sherwin?
Student: p orbital on oxygen.
Professor Michael McBride: It’s a p orbital on oxygen, an unshared pair of oxygen, mostly, although it is mixed a little bit with something else. Now, to see what else it’s mixed with, we can go down the list by three steps and look at that orbital, the one that’s called 2py. Because it also has that little bit of oxygen in it, but more of the other stuff. Now let’s look at nodes in these things. There’s nodes that are nodes of atomic orbitals, of the oxygen and of the carbon; they bisect the molecule. Those don’t have to do with bonding. Right? But the one on the top also has an antibonding node. Can you see where that is? Shai?
Student: Between the carbon and the oxygen.
Professor Michael McBride: Between the carbon and the oxygen, there’s an antibonding node. So this is a favorable and an unfavorable combination of, in the front, oxygen, the front left is oxygen, and in the back right is some sort of C-H bond. So those molecular orbitals are made by combining 2p of oxygen with σ C-H; in fact two σ C-Hs. Okay? Now, which one is lower in energy, the unshared pair on oxygen, or σ C-H? Notice that the oxygen has something going for it, which is a big nuclear charge, but the C-H has something going for it, which is the bonding of the C-H, which lowered it from where carbon and hydrogen atomic orbitals would be. Which one is lower, the 2p of oxygen or σ C-H? Which wins? Can you tell by looking at this? Angela?
Student: σ C-His lower because the one on the bottom has less nodes than the one on the top.
Professor Michael McBride: Right. The favorable combination, which looks mostly like the lower one, looks mostly like C-H; and the unfavorable one, the one that went up, looks mostly like O. So clearly the σ C-His lower in energy than the 2p of oxygen. Okay, now let’s look at the LUMO, up on top, in the plum pudding column, right? And here’s another version of that. And you can see what it is. It’s made up of a p orbital on oxygen and a p orbital on carbon. That’s an atomic orbital node. It has nothing to do with the bonding. But there is an antibonding node between them. Right? So this is the π* orbital; not the occupied one but the vacant orbital, the LUMO. Right? There’s poor overlap because it’s π*; that’s why it’s unusually low. And it’s also unusually low compared to C, C-C, because it’s got an oxygen in it, a high nuclear charge. So poor energy match; 2p Ois lower in energy than the 2p of carbon. Okay? And there — we’re going to show this on the next slide, those two — but the top one, you see, is the favorable combination of 2p O and 2pz; the one of which this is the unfavorable combination. But it’s not the HOMO, right? The HOMO is that unshared pair on oxygen, which is the dominant thing in 3dxy.
Okay, so now let’s look at these. There’s the C=O π bonding orbital. Now which one should that be big on? Don’t look at it. Which one should the π — the favorable combination of p on oxygen and p on carbon, which should it be big on? There’s p on oxygen, p on carbon, they mix; π overlap, not too much overlap, so they don’t go so much up and down. But what does this, the bonding one, what does it mostly look like?
Professor Michael McBride: It should look mostly like oxygen. When you look at this picture it doesn’t look that way. That’s because of which contour we drew. Because oxygen holds its electrons closer. So by the time you get out to a certain low — so you get through most of the electron density of oxygen, as you go out one contour by the other of the onion, that’s the oxygen; there are more electrons inside. So indeed it does have more electron density on oxygen than it does on carbon, but it doesn’t show well in this particular contour. Anyhow, that’s the π bonding. And this one is a different unshared pair on Oxygen. And that’s mostly a p-rich, hybrid atomic orbital of oxygen, but it’s got a little bit of the C-H bonding and it’s a favorable combination; there’s a little bit of favorable bonding between there. Okay, but the nodes that you see in this, in neither orbital are the nodes antibonding. They’re always the nodes that go through atoms. They’re part of atomic orbitals. These are bonding orbitals, the occupied ones, as you expect them to be.
Okay, but now let’s look at that LUMO, that vacant orbital, the π* orbital. What direction should a high energy HOMO approach this, in order to mix with it? So it has to get good overlap as it comes in, and not run into the other nuclei, or the other electron density associated with other orbitals on the way in. Okay, well here are three possibilities. You could attack from within the plane, from the oxygen end; from within the plane toward the carbon; or from straight up, halfway between carbon and oxygen. How do those look as ways for an orbital to come in and overlap? Suppose it came from the right. Suppose you had a big ball, like OH-, coming in along that arrow from the right? You going to get good overlap? What would you call that situation?
Professor Michael McBride: Orthogonal, right? So you’re not going to get overlap — how about if you come from the left, the arrow on the left?
Student: It’s also —
Professor Michael McBride: That’s also going to be orthogonal. How about if you come from the top? It’s not precisely orthogonal, but it’s pretty close to orthogonal. You have both positive and negative overlap. So in fact none of those are good. Now here are two other possibilities. You could attack the oxygen on the top, or you could attack the carbon on the top. Which of those do you think looks better?
Professor Michael McBride: Carbon. Why? Angela?
Student: It’s bigger.
Professor Michael McBride: Because it’s bigger. Okay, so I’d say that one’s smaller, so forget that. So come in from the carbon. Okay, so here’s a better drawing of that orbital. And we want to come in actually from that direction, at a certain angle; not 90º, with respect to the oxygen-carbon bond. Now why come in from that angle? Why not come straight down on top?
Student: You get less overlap with the oxygen.
Professor Michael McBride: There are the nodes. Right? This is furthest from the node. So you can overlap the nodes you want to overlap with, without getting near the blue ones. Right? So you should come in at a certain angle. Now, there’s a prediction. Is it true?
Chapter 4. Dunitz and Burgi’s Experimental Results on Carbonyl Attack Trajectory [00:36:11]
So what we’re going to do this is rotate this so the orientation corresponds to a graph, that appeared in a paper by Dunitz and Bürgi in 1983, when they were analyzing this. Now what did they do? They looked at X-ray crystal structures of many, many crystals that had both a nitrogen in it — so the unshared pair — and they had a carbon-oxygen double bond. And they looked to see, in many, many different structures, how those atoms are arranged, because they must be arranged in a low energy way next to one another. Okay? That’s the way they would pack in the crystal.
Okay, so now how is this diagram made? Well notice that the carbon and oxygen are drawn there. So that’s the carbon and oxygen, and they’re made — so all these different crystals, some have the group oriented this way, some this way, some this way, some this way. So you rotate all those crystals so that their carbons are superimposed with one another, in all the structures; the carbons are all there, and the oxygen is along this particular line, which lies in the plane of the screen. Right? And then there are things attached to it, which are R groups; an R group — that is whatever group, hydrogen or carbon or something — attached to the carbon. There’s one in front of the plane and one behind the plane. Okay? So we rotate all these different structures around that carbon-oxygen bond. So those things are all one hiding the other one; the R in front hides the R behind.
And now we look, with that orientation, to see where the nitrogen is nearby. And those dots up on the top show in many different structures, labeled A, B, C, D, E, F, G, H, I — no J — K, L. Okay? In all those different structures where’s the nitrogen? So in structure G, a particular compound, the nitrogen is where the blue dot is on top. The carbon is — all the carbons are always there. That’s the origin, right? Here. The oxygens are along this line. So the oxygen is along that line and at that position, that distance. And the R groups are bent down a little bit from being planar. One is here and the other one is behind. Okay? Okay, but now let’s go G, H, I, K, L. G, H, I, K, L. Do you see what happens? Let’s go back. K, I, H, G. These are five different crystal structures. But look again. So the nitrogen comes down, from one structure to the next. What happens to the R’s, as the nitrogen comes down?
Student: They move away.
Professor Michael McBride: Right, the R’s bend down. So the nitrogen — the carbon, pardon me, in the middle, becomes pyramidal rather than planar. Back up. What happens to the oxygen as we do this? There’s G, H, I, K, L. Right? Okay, so here’s what happened. The nitrogen moves down, the Rs bend down, and the oxygen moves away. Why does the oxygen move away from the carbon, from a theoretical, orbital point of view? Where are you putting electrons? Where are you putting the electrons here, when you put them into this LUMO? Into an orbital that’s antibonding between carbon and oxygen. So the oxygen stretches away from the carbon. Kevin?
Student: On the right side of the screen, the length units, what is that?
Professor Michael McBride: That’s Ångstroms. So it’s zero is the origin; that’s where the carbon is. And there’s one Ångstrom, 1, 2, 3. So they start three — the nitrogen starts three Ångstroms from the carbon, and as it comes in it gets to be within about 1.6 or something. And I forget. I’m guessing that an actual bond between that nitrogen and carbon is probably the order of 1.3 or something; I’m not actually sure it is. So they haven’t quite bonded. But it’s following the pathway of reaction, when you look at a lot of these different structures. Okay? Now that angle, 110º, along the path that they follow, is called the Bürgi-Dunitz Angle; that’s the guys who did the paper, Hans-Beat Bürgi and Jack Dunitz in Switzerland. So there is a bunch of what we might consider snapshots of a reaction actually taking place. And you can see that there’s a well-defined angle from which the nitrogen approaches that LUMO, as it mixes with it.
Now, resonance is intramolecular HOMO/LUMO mixing. We’ve looked here at intermolecular, which is reaction. But next time we’ll go on to this slide and look at the analogous thing that happens within a molecule. Okay.
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