Freshman Organic Chemistry I

CHEM 125a - Lecture 16 - Recognizing Functional Groups

Chapter 1. Why So High, Why So Low? The HOMO/LUMO View of Chemical Reactivity [00:00:00]

Professor Michael McBride: So this is where we were at the end last time, trying to decide when a high HOMO, or a low LUMO, is unusual in its energy; because, as the world is, things like to be at low energy. We’ll talk about that later, why that’s so. But that means that electrons tend to be in the occupied, the low — occupy the low-energy orbitals and not the high-energy orbitals. So that empty orbitals are almost always higher than any occupied orbitals. So if you want to get good energy match, and therefore mixing, and therefore bonding, the occupied orbital that you’re interested in must be unusually high, and the vacant orbital must be unusually low. So the task we have is to recognize what orbitals should be unusually high and what ones should be unusually low, and if we can find the unusual ones, then we’ll know what’s reactive, be able to identify functional groups. Okay, so unusual, compared to what? Compared to normal occupied and vacant orbitals. That is, in organic chemistry, or in most kinds of chemistry actually, carbon-carbon, carbon-hydrogen bonds are occupied orbitals; not unusually high, they’re our standard of usual. And the antibonds, corresponding antibonds are unusual — are the usual vacant orbitals.

So what can make things unusually low or unusually high? When the valence-shell orbitals that, if they bonded, would go down and up, don’t bond with anything, don’t mix with anything, then they’ll be unusual. Second, if the orbitals don’t overlap, then they don’t go down and go up. Third, if you have an unusual atomic orbital. These are made of carbon and hydrogen, going down. But if you start with something that’s unusual, unusually high or unusually low, then the things that come from it will be unusually high or unusually low. And finally, electrical charge. So right at the end last time we were looking at the first example: unmixed valence-shell atomic orbitals. So we saw that the simplest of all acids, H⁺, of course is unusually — if it were occupied, it would be unusually high for an occupied orbital. Right? Because it hasn’t gone down. But it’s vacant, and it’s unusually low, for a vacant orbital, because it hasn’t gone up. And so on with these others: an unshared pair on nitrogen; or a little less so on oxygen. A vacant orbital on boron that’s not bonding with anything is an unusually low LUMO; even though orbitals on boron are not unusually low, because the nuclear charge isn’t very big for something that’s using the second, n=2, quantum level orbitals, valence orbitals. It’s an unusually low nuclear charge for that. So the orbitals are unusually high in energy, not being attracted by more protons. But it’s low for a vacant orbital, if it hasn’t mixed with anything.

Okay, and finally notice that some of these are charged: H⁺; plus means that’s a good place to put an electron, so unusually low; OH^-, unusually high because of the negative charge; CH₃^-, especially high. So this slide also shows the fourth point, that electrical charge makes a difference on how normal occupied and vacant orbitals are. Okay, so these unusually low vacant orbitals are acids, like H⁺. Unusually high occupied orbitals are bases, like OH^-, as we just showed on the previous slide. That means that you can mix the high HOMO of OH^-, with the low LUMO of H⁺, and two electrons, the ones that were on the OH^-, go down in energy, and that makes a bond. And our notation for showing that is to draw a curved arrow, and it gives water.

Now think carefully about what a curved arrow means; it’s not the same as a straight arrow. A curved arrow designates a shift of electron pairs. It doesn’t show that this atom moves from here to here. It shows that a pair of electrons shifts, and that pair of electrons goes from being on the oxygen, to being between oxygen and hydrogen, to form the bond. So you start the curved arrow where the electron pair is at the beginning, and you end the arrow, put its point, where it’s going to be in the product. Right? So it’s not showing that an atom moves from here to there. It’s showing that the electrons that were formerly on oxygen are now between oxygen and hydrogen. The effect of that is, of course, to pull all the hydrogen to the oxygen. But that’s not what’s being shown by the arrow. What’s being shown by the arrow is how the electrons are moving in our picture. Okay, so there’s a case.

Now here’s another base with the same acid, and we can draw a curved arrow there and show making an ammonium ion out of ammonia. Or we could use the same base with a different acid, the low vacant orbital of BH₃, and draw the same curved arrow, and make this anion, make the boron-oxygen bond. Or we could start with completely different ones. We could start with ammonia, not OH^-, not Arrhenius’s base, but ammonia; not Arrhenius acid, H⁺, but BH₃. But it’s exactly the same reaction. The high HOMO mixes with the low LUMO to form a new bond. Okay? So that’s what curved arrows show, and don’t confuse it by trying to draw molecules or atoms moving, and showing a curved arrow. If you want to show them moving, draw some other kind of arrow; draw dotted arrows or a straight arrow or a wiggly arrow, or something. But the curved arrows have a very specific meaning.

Now the second reason that things could be unusually high or low is because they don’t have much overlap. Notice in the first case, it was just an extreme case of that. There was no overlap at all, the orbitals were just there. But even though the orbitals are mixed with something else, if the overlap is poor, they haven’t changed very much, they still look very much like they did originally. Right? So a good example of not having very good overlap is the side-to-side overlap of p orbitals. The π overlap — you remember from looking at those curves of amount of overlap for different orbitals versus distance — the π overlap isn’t very big. So even though the p orbitals start a little higher on carbon than they do on hydrogen, or normal bonds of carbon that have s character in them, they start a little higher, but they don’t mix very much. Right? So the π orbital is unusually high, and the π* antibonding orbital is unusually low, because they didn’t go down or up very much. Okay? Now, which of those do you think is more remarkable, in its highness or lowness? So for this reason, the carbon-carbon double bond could behave either as an acid or as a base. Right? High occupied orbital makes it a base; low vacant orbital makes it an acid. And indeed, it does have both kinds of reactivity. But which do you think is more pronounced? Which would be more familiar, how high the occupied one is, or how low the vacant one is? Lucas, what do you say?

Student: How low the vacant one is.

Professor Michael McBride: And why do you say so?

Student: Because electrons have to sort of — they would likely move into a position of an energy as close as possible to it.

Professor Michael McBride: Yes, but on the other hand, the electrons of the π want to move into something else.

Student: There are lots and lots and lots of electrons going up.

Professor Michael McBride: Well there are lots — there are even more vacant orbitals than there are electrons. Right?

Student: Because you also have starring electrons that will — like in a normal carbon atom.

Professor Michael McBride: Let’s just look at the picture. Yes Catherine?

Student: I think the HOMO is more unusual because it’s farther away from the usual —

Professor Michael McBride: Ah, the HOMO is further above the normal bonds, than the LUMO is below the normal antibonds. Why?

Student: Because the p orbital started —

Professor Michael McBride: Can’t hear very well.

Student: Because the p orbital started a little higher.

Professor Michael McBride: Ah, because they started a little higher. Right? So most reactions that we’ll study, or at least many reactions that we’ll study of the carbon-carbon double bond, it’s reactive because of the high HOMO, not because of the low LUMO; although there are cases where the low LUMO makes it reactive. But it’s this π orbital, here, that makes it particularly remarkable, because they started a little high. Okay, so the high HOMO makes it unusually reactive. Now how about the C=O double bond? Right? So it starts with the same p orbital on the carbon, which has π overlap and therefore not much shifting up and down, with a p orbital of oxygen, which of course is lower than that of carbon, because of the higher nuclear charge. Which of these will be most unusual? Somebody got a suggestion? So what would you think the characteristic reactivity of a C=O double bond — would it be reactive mostly as a high HOMO — that is, as a base — or mostly as a low LUMO? Which is more unusual, compared with what we compare things with? Andrew?

Student: I think it’s a low LUMO.

Professor Michael McBride: Why?

Student: First of all it can’t — it’s further away from its origin.

Professor Michael McBride: And why is it further down, compared to what we usually compare? Why does the π*, C double bond O, why is that orbital so very low? Pardon me, one reason it’s low is because the overlap isn’t very much. So it didn’t go up so much. Right? That’s the same, true with a C-C double bond. But what’s special about the C-O double bond Andrew?

Student: I thought the atoms are different, there’s a —

Professor Michael McBride: Because of the bad energy match, it didn’t go up very much from the carbon, or down very much from the oxygen. But the average is low, that it starts from. So what’s special here is the π is not so unusual, not so far from here. But this one, π*, is very far from here. So what should make a carbonyl especially reactive is its low vacant orbital, because of poor overlap; and that it started with an oxygen, an unusually low orbital, atomic orbital. Okay? We saw another example here. The three-membered carbon ring is unusually reactive, for a carbon-carbon bond. That is, the electrons in that carbon-carbon bond are not shifted down so much, and the antibond is not shifted up so much, as in normal carbon-carbon bonds. Why? What’s special about the bonds in the three-membered ring? Kevin?

Student: Well two of them are bent.

Professor Michael McBride: Ah, they’re bent. And what does that mean?

Student: Greater than 90°.

Professor Michael McBride: Yes, but they’re bent, so what?

Student: It overlaps, there are better overlaps.

Professor Michael McBride: Does it overlap as well? So it doesn’t go up and down as much. Right? So another example of poor overlap. Okay, or you could have an unusual atomic energy orbital in the molecular orbital. For example, how about a C-F single bond? Right? So the overlap can be perfectly good. It’s a regular old σ bond, straight on. Right? But the fluorine starts very low. There’s bad energy match. So the π* isn’t so very different from a vacant orbital on carbon; unusually low as a vacant orbital. So it should act as an acid. Things that have high HOMOs should come up and react with that σ* of C-F. It’s an unusually low LUMO. Can you think of what the flip side of that — how could you have something like this, a carbon bonding to something, that for the same or analogous reason would have an unusually high occupied orbital? What would you want the carbon to bond with? Choose an element.

Student: Lithium.

Professor Michael McBride: Lithium. Why lithium?

Student: Because —

Professor Michael McBride: Because you looked at the PowerPoint?

Student: No.

Professor Michael McBride: [Laughs]

Student: I didn’t look at it.

Professor Michael McBride: Why lithium?

Student: Because its valence electron — it’s easy for the valence electron to be given away.

Professor Michael McBride: Why?

Student: Because it would be as stable as a lithium plus one ion.

Professor Michael McBride: Why? That’s restating the same thing, right? What’s different about lithium? Why are orbitals on lithium, as an atom, unusually high in energy, compared to carbon-hydrogen? Catherine?

Student: Because it has a low nuclear charge.

Professor Michael McBride: Because it has a low nuclear charge. Right? It’s right at the left end of the table. Okay? So its orbitals aren’t very stable because it doesn’t have very high nuclear charge. So it’s a high energy. And therefore it has bad energy match. Or it could’ve been lithium that I drew here. I happened to draw boron, which is, for the same reason, a lower nuclear charge than carbon; therefore higher energy as an atomic orbital. When they mix, what’s special here is the occupied orbital, σ, which is unusually high, because the average started high. Okay.

Chapter 2. Is BH₃ an Acid or a Base? [00:15:22]

So boron. You may remember that in the first slide we showed, boron had a vacant orbital; that p orbital on boron is vacant. Unusually low. Therefore what, acid or base? Unusually low vacant orbitals, which does that make it?

Student: Acid.

Professor Michael McBride: Acid. So BH₃ should be an acid. Right? Well what I just showed you is what Shai? On the previous slide; we’ll go back there.

Student: That it had a big B-H; yes the B-H.

Professor Michael McBride: It has an unusually high HOMO, BH₃. So what does that make it Shai?

Student: It would make it a base but —

Professor Michael McBride: So is it an acid or a base?

Student: Both.

Professor Michael McBride: Wilson?

Student: Both.

Professor Michael McBride: It’s both. And what does that suggest, if the same molecule is both an acid and a base? What do acids react with?

Students: Bases.

Professor Michael McBride: So what if the same molecule is an acid and a base?

[Students speak over one another]

Professor Michael McBride: It’s going to react with itself. Two of the molecules will react with one another. Okay, and that’s why you can’t do a crystal structure of BH₃, because BH₃ becomes B₂H₆. Let’s look at how it happens. So there’s the low LUMO that makes BH₃ an acid. What’s the high HOMO? What’s the high HOMO of BH₃? In localized terms. I don’t mean these big things that go over the whole molecule. It’s the B-H bond, right? Which was poorly matched in energy and so on. So there. Notice that the B-H bonding electrons are big on hydrogen, small on boron. We saw that last lecture. Okay? So they should overlap like this, right? And the vacant orbital on the B should stabilize these high-energy electrons in the B-H bond. But now, you could imagine many different orientations of the top molecule that would allow overlap. Do you know why I chose this particular orientation? Which one acted as the acid and which one is the base? The one on the bottom, the vacant orbital, acted as an acid. The one on the top acted as a base. So what other possibility is there? Yoonjoo?

Student: So then there also — you could reverse the roles of them.

Professor Michael McBride: Ah ha! You could have — so that would make — notice incidentally, what I forgot to mention here, is that there are three nuclei being held together by that pair of electrons now. Originally the top molecule, that pair of electrons, mostly on hydrogen but partly on boron, held the hydrogen and the boron together. Now that same pair of electrons is attracted — is helping form a bond with a boron, the bottom boron. So in fact that’s an unusual kind of bond, because it’s bonding three nuclei together, not just two. It’s doing double, or perhaps triple duty. So we need a new symbol to talk about such bonds, and a reasonable one would be a bond that looks like that. It’s a Y bond. We could make it blue. Okay? But now we have, as Yoonjoo said, the vacant orbital on the top and the high occupied orbital on the bottom, and they can do exactly the same thing. So we have two of these three-center, two-electron bonds. Right? Two electrons holding three atoms together. Right? Or we can draw it that way. Right? And that’s the structure of B₂H₆. Two of the pairs of electrons are each bonding three atoms together instead of two. Any questions about this? Yes Alison?

Student: Is that what the dotted line means for —

Professor Michael McBride: Yes, the dotted just means it’s a little different. You could draw the solid Y if you want to. There’s no really standard notation for that. But it’s clear what it means. It’s just that a pair of electrons is being shared among three nuclei, rather than two, and their electron density is — correspondingly, if you did a difference density map, if you could do an X-ray of this, you’d expect electron density to build up in the middle of all three. Yes Claire?

Student: This may seem like a stupid question but a high HOMO, we’ve talked about as a base, and a low LUMO we’ve talked about as an acid.

Professor Michael McBride: Yes.

Student: And the low LUMO is unoccupied. If you think about it, it’s sort of receiving electrons.

Professor Michael McBride: Right.

Student: But generally, bases are supposed to receive electrons, instead of acids, and —

Professor Michael McBride: No, you got it backwards.

Student: Do I?

Professor Michael McBride: Yes.

Student: Oh okay.

Professor Michael McBride: Because H⁺ is an acid. Right?

Student: Right.

Professor Michael McBride: So it obviously can’t give up electrons. There aren’t any electrons. An acid accepts electrons. Okay? Think about it a little bit in the privacy of your room, and you’ll see that. Okay?

Student: Does the bond between three nuclei only happen —

Professor Michael McBride: I can’t hear very well.

Student: Does the bond between the three nuclei only happen here because of the geometry of the molecule?

Student: Yes. And to get that, you have to have overlap. So if you didn’t have — if the top BH₃ were oriented so that the B-H bond were vertical and the boron was way up at the top, it wouldn’t overlap the other one. So you have to — always to get a bond, you have to have overlap, because otherwise the orbitals don’t mix and you just have the original orbitals, as we’ve seen.

Student: Is this kind of bond very common?

Professor Michael McBride: Pardon me?

Student: Is this kind of bond very common?

Professor Michael McBride: No, it’s not very common, because there are not many really low energy vacant orbitals running around. Right? Boron is a very special case. But a lithium can do the same thing. But lithium doesn’t have energy, orbital energies as low as those of Boron, because it doesn’t have as big a nuclear charge. So boron is particularly good at getting this kind of thing. And this answers the puzzle about Lewis structures that we raised in Lecture two, about how can BH₃ react with BH₃? Right? That’s how it does it, by making three-center bonds.

Now here’s a True and False quiz. On the basis of what you know, is it true that low energy molecular orbitals result in bonding? True or false? I don’t think you trust me anymore. [Laughter] We’ve been saying that when you get those low orbitals — things come together, you get a low orbital — that results in a bond. Lots of overlap. [Laughter] I can’t talk you into it? Good. That’s false. What makes a bond is lowered energy orbitals. It’s when things come together and the electrons get more stable. It’s not how low they are, it’s how much they’re lowered by the coming together, because then pulling apart they have to go back up again. So it’s not whether they’re high or low, it’s whether they get lowered. Okay? Now, compared to what? What do they have to get lower compared to?

Student: By the size.

Professor Michael McBride: Yoonjoo?

Student: So it’s kind of like how in Erwin and Goldilocks you showed the antibonding and bonding. So wouldn’t it be the reactants?

Professor Michael McBride: You can say something in fewer words than that. What do you compare to? When you say energy is lowered, and that makes a bond, what’s it lowered, compared to? Christopher?

Student: The atoms in their standard states.

Professor Michael McBride: Well yes. It wouldn’t necessarily be between atoms, it could be between two molecules, like BH₃ with BH₃. But you’re right. What it’s lowered compared to is the things it was before they came together. Right? Now these things, before they came together, one of them had electrons, one didn’t. Those might’ve been very high. Right? So they came together. The electrons went substantially down, but still aren’t so very low. They could’ve been very low-energy orbitals to begin with, but not gone down very much, because there was bad overlap say. So these, even though they’re lower than the ones we talked about first, would not be so bonding. It’s these that were bonding, because they came down a lot, when the mixing happened. So it’s lowering. Compared to what? Compared to the separated components, before you made this bond. Right? So when things come together, and that results in the electrons going way down in energy, that’s a strong bond. Yes, Chenyu? Pardon me?

Student: How far does it have to go down?

Professor Michael McBride: Well that’s what we’re going to have to learn. That’s a question of lore. Right? And it has to be at least enough to overcome the fact that when they come together other electrons, other orbitals, filled orbitals are overlapping, which is net repulsive. So it may not be that there’s an absolute criterion. Right? It may be that if there are a lot of other things opposing the coming together, filled orbital with filled orbital, then you have to have really enormous going down. So you can’t make a simple answer to that. But we’ll learn as we see examples. Okay.

Chapter 3. HOMO-LUMO Mixing for Reactivity and Resonance: The Cases of HF [00:25:39]

So now, HOMO/LUMO mixing, for reactivity and resonance. So reactivity means between molecules. So far we’ve been talking mostly about atoms coming together and forming a bond. But molecules have high orbitals and low orbitals, as in the case of BH₃. The B-H was a molecular orbital, or not an atomic orbital. Right? So when things come together, orbitals are orbitals. If you have an unusually high energy and an unusually low energy, and they overlap and go down, that makes a bond. So that’s between molecules. But it turns out that what resonance is, is HOMO/LUMO mixing, within a molecule.

Now you might say the molecules have certain molecular orbitals. How can you mix them? Right? The idea is that we made our first analysis on the basis of localized orbitals: σ,σ* here; σ,σ* here; not these big Chladni things that go over the whole thing. But it may be that this σ,σ*, and this σ,σ*, are near one another and overlap. So that when we made our initial analysis, and looked only at this and only with this, we didn’t take into account that this one might interact with this one, and give still lower energy. When that kind of thing is important, that’s when you have to draw other resonance structures. And I’ll show you an example. But first I’m going to show you reactivity, and then we’ll go on to resonance. Okay, now let’s look at the frontier orbitals for H-F. Okay? So it has four valence electron pairs and five valence atomic orbitals; 2s, 2p_xyz on fluorine, and a 1s on hydrogen, and four pairs of electrons. So there are going to be four occupied orbitals. And this is what the lowest orbital looks like. What does it look like?

[Students speak over one another]

Professor Michael McBride: Well it’s a 2s orbital, the 1s being the core on fluorine. But it’s made up of atomic orbitals. Is it exactly a 2s of fluorine? Does it look like sphere on fluorine? Alison, you’re shaking your head.

Student: It’s a little bit distorted.

Professor Michael McBride: And how did it get a little distorted? What did we mix with the F orbital?

Student: The H.

Professor Michael McBride: A little bit of the 1s on hydrogen. It’s mostly the F on fluorine. Why? Because the fluorine’s way down in energy. So the best combination is mostly this. But you can see that it’s a little bit egg shaped, a little bit drawn out toward the hydrogen. Okay, so it’s mostly a 2s of fluorine, but a little bit of 1s on hydrogen. Right? Now here’s the next one. What’s that mostly? Can you see?

[Students speak over one another]

Professor Michael McBride: Tyler, what do you say?

Student: I would say a 2p.

Professor Michael McBride: It looks like a 2p on fluorine. Does it look exactly like a 2p on fluorine, or is it just hard for you to see that it’s not? It’s very similar.

Student: I don’t know, it looks pretty close, but the blue one might be a bit bigger.

Professor Michael McBride: Yes, the blue one is a little bit bigger, because it’s got a little bit of the 1s of hydrogen lowering the energy of the 2p of fluorine. Okay? And now the next two orbitals are these. What are those? Steve, what do you say?

Student: The 2p_y and 2p_z.

Professor Michael McBride: Yes, the 2p_y and 2p_z of fluorine. Is there hydrogen in those too?

Student: Not very much.

Professor Michael McBride: Why not?

Student: Because they don’t overlap with the one —

Professor Michael McBride: Ah, they’re orthogonal. It’s a π versus a σ. There’s no overlap, therefore no mixing. So those are the occupied orbitals. And the 2p’s have the same energy. Okay, and then remember there are going to be five molecular orbitals. And now you make the last one, which is going to have another node, with the leftovers. What’s left over, after we made the occupied orbitals? What’s it mostly left over? We used up the 2p’s of fluorine to make those HOMOs. Those were pure, right? But what was left over from the bottom?

Students: 1s.

Professor Michael McBride: We used very little of 1s on hydrogen, and there’s a little bit of 2p fluorine and 2s fluorine that we didn’t use in the bottom ones. So they’re back in the top. So what we have is mostly a 1s on hydrogen, but a little bit of some kind of sp hybrid on fluorine. Is that clear to everyone? So that’s the vacant orbital. Is it unusual energy, that vacant orbital?

Student: It’s low.

Professor Michael McBride: It’s low. Is it unusually low? Kate, do you have an idea? What kind of criteria do we have for whether an orbital should be unusually low? What do we look for?

Student: We look for overlap and energy match.

Professor Michael McBride: Overlap. Energy match. How about this case, good overlap? Quite good overlap. It’s a hybrid orbital on fluorine pointed right toward a hydrogen. Good overlap. How about the energy match? Kate?

Student: Sure.

Professor Michael McBride: Sure what? Sure it’s good or sure it’s bad? One of them’s hydrogen. What’s the other one?

Student: The other one is fluorine.

Professor Michael McBride: Where’s fluorine, compared to hydrogen?

Student: Fluorine is going to be lower.

Professor Michael McBride: Why?

Student: It has greater nuclear charge.

Professor Michael McBride: Right. Okay, good energy match or bad energy match?

Student: Not great.

Professor Michael McBride: Not very good. So you don’t get much mixing. And you know that already by looking at the picture, because you didn’t mix it very much. It’s almost all 1s of hydrogen. So it’s unusually low for a σ*; it didn’t go up very much from hydrogen. So indeed — now it’s got an unusually low vacant orbital. What does that make it, an acid or a base?

Student: An acid.

Professor Michael McBride: It’s an acid. Are you surprised that H-F is an acid?

Student: No.

Professor Michael McBride: Why? What’s its name?

Student: Hydrofluoric acid.

Professor Michael McBride: Hydrofluoric acid. And that’s what makes it an acid. Arrhenius would say it’s an acid because it gives up H⁺. We say it’s an acid because it’s an unusually low vacant orbital. Right? So notice that those three are made up of three atomic orbitals: the 1s of hydrogen, the 2s of fluorine, and a 2p orbital of fluorine. And they’re in different mixtures in three of these. Usually we’ve looked at just two things, right? There’s one atom, atomic orbital, and another one, and they mix. Right? A pair. Here there are three going in, to give three molecular orbitals. But still you can see quite easily why they should be the way they are; why the bottom ones are almost pure fluorine, and the top one is almost pure hydrogen. Okay. And so that top one is σ*; the LUMO, the unusually low LUMO. Lucas?

Student: How can we be sure that the 1s of the low nuclear-charge hydrogen is going to be that much different from the really high nuclear charge 2s of fluorine?

Professor Michael McBride: Yes, this you have to learn. And I’m going to show you very soon how we can tell that kind of thing. Okay, there’s a different picture, that was made in 1973, of the same thing. These pictures were drawn just a year or two ago, but this is — you can see the same thing. It’s drawn at a different contour level. It was based on a different calculation. But you can see it’s the same thing. And we’ll notice that — how many nodes does this thing have, that are clearly visible?

Students: Two.

Professor Michael McBride: Two nodes, right? Between the H and F. That one is antibonding. Right? When they came together they cancelled in the middle, rather than reinforcing. There’s another node there. But notice that that didn’t have anything to do with the bonding. That was already there in the atomic orbital you started with. So it didn’t have anything to do with lowering. It didn’t have anything to do with whether the thing was bonding or antibonding. The pink node had to do with whether it was bonding. That came when they came together. Right? It’s unfavorable. So it’s better for them to come apart. But if they come apart, you don’t do anything with the blue node. It’s still there, it’s part of the atom. Okay? So you have to recognize that there are two kinds of nodes. There are nodes that were there already, atomic orbital nodes, and there are ones that are associated with the coming together, and that’s what makes something bonding, or antibonding; in this particular case it’s antibonding.

Chapter 4. Comparing HF and CH₃F to Distinguish Molecular Orbital Nodes [00:34:49]

Okay now let’s look, instead of H-F, let’s look at CH₃-F. Sam?

Student: Why is it called a frontier?

Professor Michael McBride: Because you have occupied orbitals, and then you have vacant orbitals, and the ones you’re interested in are the lowest of the vacant and the highest of the occupied, at this border between occupied and vacant orbitals. Okay, so this one has seven valence pairs of electrons. So you’re going to occupy seven orbitals. And here’s what they look like, the seven that are occupied. What’s the very lowest one, mostly?

[Students speak over one another]

Professor Michael McBride: It’s the 2s of fluorine. Right? And what’s the next one? Well it’s C-H bonds, all mixed together. Right? But also a little bit of a p orbital on fluorine. It has the blue, sort of, toward the front, and a little bit of that red behind, as part of the p orbital on fluorine. So it’s a mixture of the C-H bonds and of the p orbital on fluorine. And then we have these others, some of them coming in pairs, and those are the HOMOs, because we have seven occupied. And let’s look at them a little more closely and compare them, each one, with the one beneath it. And I’ll draw another picture too, of that older kind, for making this comparison. And we’re interested in what is it that went together, to make these orbitals, and why is one lower in energy, and the other higher? So what do you see on the top orbital, say this one here, the top left; what’s that made up of? What is it on the left side? Sophie, what would you say the orbital is on the left side of that?

Student: I think it’s 2p over there.

Professor Michael McBride: Right. This part here is a 2p π orbital of fluorine. Now what’s this thing on the right here, the dash bit down here? If you just saw that, without any of the rest of it, what would you say that was?

Students: 1s.

Professor Michael McBride: It’s more than a 1s on hydrogen; that would be spherical. It’s, at a certain contour, the C-H bond. Do you see that? So this is electron density in here, bonding between C and H. And these two on the top are little bits of C-H bonds as well, mixed together. So what this orbital is, is a mixture between some combination of C-H bonds on the right, and the p orbital of fluorine on the left. Now is it a favorable, or an unfavorable combination, of the fluorine orbital with the C-H orbitals? This one up here. Is the interaction between the fluorine orbital and the C-H orbitals bonding or antibonding?

Students: Antibonding.

Professor Michael McBride: Becky, do you have an idea? Are they building up electron density in between, or having a node in between?

Student: A node.

Professor Michael McBride: There’s a node in between. So that’s antibonding. What’s the orbital on the bottom, this one?

Student: Bonding.

Professor Michael McBride: That’s the same components, but it’s the bonding combination. So this is the favorable combination of those, and this is their unfavorable combination. So this lower-energy one is favorable, and the upper energy is unfavorable combination. How about here and here? Can you see what that is? What’s on the left, here, that lump of red and this lump of blue? Eric? Here, this thing, it’s very complicated. Okay, will you agree on that?

Student: Okay.

Professor Michael McBride: But part of it, this part on the left, there’s a red lump in behind and a blue lump in front. You can see it maybe more clearly here. Red behind, and blue in front, surrounding the green fluorine atom. What is that orbital, that atomic orbital?

Student: Probably the p orbital.

Professor Michael McBride: Can’t hear very well.

Student: Probably the p orbital, a different orientation.

Professor Michael McBride: It’s the p orbital on fluorine that’s pointing more or less toward you. Okay? So this is the p orbital of fluorine, that’s mixing with C-H bonds here. Right? In fact, it’s the same thing as this, turned on its side. Okay? So this is the bonding combination, built up between the fluorine and the C-H’s. This is the antibonding one, with a node between the fluorine and the C-H’s. So here we have a node, in this bottom one, but that node came from the atomic orbitals. That’s what made it a p orbital here, on fluorine. That one, you don’t get rid of, if you pull the fluorine away. That doesn’t have anything to do with the bonding, that’s just an atomic orbital node. Right? On the top, you again have the same atomic orbital node, because it’s the same p orbital that’s involved. But what about the top? Sam?

Student: You have another node but —

Professor Michael McBride: There’s another node, that one, and that’s an antibonding node. Right? The electrons in that orbital would get lower in energy if you broke the bond, if the fluorine came away.

So the pink nodes are the ones we’re interested in; the antibonding nodes, not the ones that are just part of the atomic orbital. Okay. Now, we’re getting to Lucas’s question. So this thing down here is made up of fluorine and C-H. The one here is made up of fluorine and C-H. Right? This is the favorable combination, and this is the unfavorable combination. So get it right. So here are the two of them. They come together when the F comes up to the methyl. Right? And you get a favorable one and an unfavorable one. But the fluorine and the C-H’s may not be at the same energy. How do you know which one’s lower? That’s your question. Is the fluorine lower, or is the C-H lower, or are they about the same? Now how are we going to tell? If the fluorine is lower, and they come together, what does the lower one look like, mostly?

Student: Fluorine.

Professor Michael McBride: Fluorine. If the fluorine’s higher, and they come together, what does the low one look like?

Student: C-H.

Professor Michael McBride: Mostly C-H, right? So by looking at how big these are, we can tell which one is lower. So when we look here, we see that here I would say they’re pretty similar, left to right. There’s not much difference between a 2p orbital on fluorine and C-H σ bonds; not much difference. But to the extent they’re different I would say, looking at this, that the C-H is a little bigger here, and the fluorine is a little bigger here. Would you say that? So which is lower, a C-H σ bond or the p orbital of fluorine?

Students: C-H.

Professor Michael McBride: C-H σ bond would be a little lower than fluorine; not much though, pretty similar. So they’re about the same, as we say. But if you have to make a choice, the C-H is a little bit lower in energy; the fluorine’s a little higher. Okay. Now there’s also a vacant orbital, made up with the leftovers here, some of the leftovers. And what’s that? Let’s look at a different picture of it. So this has three nodes that are obvious. There’s one that’s near the fluorine atom, a node, plane, that goes back. The furthest to the left. Everyone see that? Is that an antibonding node, or is that an atomic orbital node?

Student: Atomic orbital.

Professor Michael McBride: That’s part of the fluorine atomic orbital of some hybrid on fluorine. Right? And the same is true at the C-H end. There’s a node that’s an atomic orbital node of the carbon. Right? But what’s important? Between those other two is what’s between the fluorine and the carbon, and the CH_3. And that’s antibonding. So that, the LUMO is a σ* antibond, between C and methyl. And why is it unusually low? The question is whether the LUMO should be unusually low. Why is it unusually low? Is the overlap bad?

Student: It’s fine.

Professor Michael McBride: No, the overlap’s good; the hybrids are pointing right toward one another. But what makes it low in energy? So I’m not giving you specific problems on this, but look over those things and run your brain as to those four different things that make orbitals unusually high or unusually low. Because that’s what you’ll be doing next week when you’re doing these Wikis, each of you, to decide some functional group, why is it functional? What makes it unusually high or unusually low; or maybe neither? Right? Why is the C-F bond, more properly the C-F σ* orbital, the antibonding orbital, why is it unusually low? Compared to what? What do you compare it to?

Student: C-H.

Professor Michael McBride: A C-H bond. How come C-F is lower, for the σ*? Alex?

Student: Energy mismatch.

Professor Michael McBride: Pardon me?

Student: Energy mismatch.

Professor Michael McBride: Yes, the fluorine is really low, their average is very low. So the vacant orbital is unusually low. It didn’t go up much. Right? Where have you seen that before? H-F. The previous slide, H-F, was exactly the same. It’s an acid. Remember Kate, you helped us with that. It’s an acid because of the bad energy match between fluorine and hydrogen. This is the same thing, but it’s the bad energy match between fluorine and carbon. Okay. So CH₃-F is an acid for the same reason that H-F is an acid. Right? There’s the low LUMO of H-F. Right? It’s got that same antibonding node. Shai?

Student: Why is there no energy mismatch between carbon and hydrogen?

Professor Michael McBride: It just happens that it worked out that way. But it’s true. Hydrogen is 1s. That makes it unusually low. But it doesn’t have a very big nuclear charge. Carbon has a higher nuclear charge, but you’re talking 2s and 2p. And it turns out those just cancel out. If that hadn’t been the case, organic chemists — like if boron happened to match hydrogen, then maybe our organic chemistry would be boron-hydrogen, not — there would be borohydrates, not carbohydrates. Right? But that’s the way things are.

Student: So in this picture we can tell that —

Professor Michael McBride: Actually there are other reasons it wouldn’t be boron, because boron has these vacant orbitals and forms B₂H₆ and so on. But it just happens that that cancellation works that way. Lucas?

Student: Just by looking at this picture we can say that CH₃, that there’s poor energy match because the orbitals around fluorine are much smaller than those around CH₃.

Professor Michael McBride: Let’s look at the next slide here. Is it this one? No. We’re going to get to that. If you looked at ones that were very badly mismatched, then the favorable and unfavorable combinations would be very dramatic. Actually, you’ve already seen that in H-F. Okay, we have one minute to start this. Okay, so we’re going to look at how CH₃-F behaves like H-F. Right? Both of them are acids. So first we’ll look at H-F, and next time we’ll go on to CH₃-F. So here’s H-F. So you have to bring — here’s a low vacant orbital. We’ve talked about that ad nauseam. Okay? Now you want to get good overlap. From what direction will another orbital come, in order to get good overlap, without the nuclei getting too close together? Obviously it’ll come from off in the right, where this orbital is big, where you can get a lot of overlap without getting close to the nuclei. So if you had something with a high-energy pair of electrons, it would come and it would overlap, from the right; something like OH^-. And you’d draw a curved arrow, to show those electrons; the high HOMO of OH^-, being stabilized by the vacant orbital of H-F. And how would you draw the curved arrow? Where would it start, where would it end? To show the electrons of OH^- forming a bond with — OH bond?

[Students speak over one another]

Professor Michael McBride: You’d start from the pair of electrons that you’re talking about, and you’d end between H and O. Right? But that means — this is really important — that means you’re putting electrons, putting electron density, into this orbital. Right? What effect does putting electrons in that orbital have on the H-F bond? Okay, the curved arrows, blah-blah; we did that. Sherwin?

Student: It’s like it pushes them out.

Professor Michael McBride: It’s what? What do you call that orbital? What’s the name of the orbital, that we’re putting electrons into?

Student: 1s for —

Professor Michael McBride: It’s σ*. It’s mostly 1s on Hydrogen, but it’s σ*. What does * mean?

Students: Antibonding.

Professor Michael McBride: What does it mean if you put electrons in?

Student: The bond breaks.

Professor Michael McBride: The bond will break. Right? It has that antibonding node. So electrons that go into that will get more stable if the bond breaks. So we’re going to draw another curved arrow, like that. So we make a new bond between H and O, but we lose the bond between H and F. So there’s our product. That’s an acid-base reaction, and it showed H-F acting as an acid; not because it gave H⁺. H⁺ never appears in here. Right? What happened is you make a bond and break a bond to the hydrogen. So it’s an acid-base reaction. And the same thing, we’ll show next time, happens with CH₃-F. Okay.

[end of transcript]

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