Freshman Organic Chemistry II

CHEM 125b - Lecture 23 - Diamagnetic Anisotropy and Spin-Spin Splitting

Chapter 1. Diamagnetic Anisotropy and Aromaticity [00:00:00]

Professor  McBride: OK, so we’re talking about NMR spectroscopy. And today we’ll talk about the tools that you use to identify– or the aspects of NMR spectroscopy that allow you to assign structures to molecules. So the chemical shift, which involves diamagnetic anisotropy, the word that you will impress your friends with. And spin-spin coupling, and maybe at the end we’ll get to dynamics, if not, next time. 

So, as we said last time, the precession frequency of the proton in a molecule depends on the magnetic field they see, which is mostly this enormous applied magnetic field. But there are also little fields in the molecule that make a difference to give this effective field. In particular, the orbiting of electrons, the net orbiting of electrons, because they orbit both ways and mostly cancel. But a little bit of net orbiting gives rise to what’s called the chemical shift, which can change the field by 12 parts per million, roughly for protons, sometimes a little more, 200 parts per million for carbon, because there are more electrons around carbon.

But there’s another magnet too. Bear in mind that the magnetic aspect of nuclei is not influenced by electric stuff. So that a proton, the nucleus in a molecule, is held in position in x, y, z by the electron density around it. That’s the bonds we talk about. But the orientation of the magnet is not at all, or is rarely, affected by that. It’s affected only by magnetic things, not by electric charges. It’s just moving electric charges that generate magnetism that does it. So that it’s as if the proton is on a perfect ball bearing or something. If this is the molecule, and the molecule rotates, the spin keeps pointing the same way. It doesn’t rotate with the molecule. We’ll talk about that a little more in just a second.

OK, so the things that are magnetic are the circulation of electrons and nearby magnetic nuclei. They’re obviously magnetic too. So if there’s more than one magnetic nucleus, then you get what’s called spin-spin splitting, as we’ll discuss. And notice that the units are different from that. We talk about the units of spin-spin splitting in terms of Hertz. That is, frequency. That is, energy, not parts per million, not a fractional unit. And you’ll see why.

So first the chemical shift, which is due to this electron orbiting and involves diamagnetic anisotropy. We saw last time, this idea that when you have higher electron density around the proton, you get more of this diamagnetic shielding, and the peaks are shifted to the right. And when you have less electron density, as in the carboxylic acid, they’re shifted to the left. And that made some sense. 

But there were things that didn’t make sense, like acetylene, where you’d think that the sp hybridization of the carbon would withdraw electrons from the hydrogens and shift it downfield. Compared to the double bond, it shifts it upfield. And furthermore, as you’ll see, the same thing applies to benzene. It’s shifted downfield, compared to other hydrogens on carbon-carbon double bonds. And the reason is this thing called diamagnetic anisotropy, which we will now discuss.

So you get diamagnetism from orbiting electrons. So suppose that red dot is the nucleus, and we’re in an applied field. And there are electrons that are circulating around the nucleus, and they create– the name “diamagnetism” means it opposes the applied field– so they reduce the field that’s seen by the nucleus. And that’s the source of these splittings we were talking– these shifts that we were talking about.

But now we’re talking here about electrons orbiting around this nucleus, the one we’re looking at. There are lots of other electrons in the molecule orbiting other nuclei. What effect do they have? So suppose we’re interested in the field at the red nucleus, but there’s some other nucleus, the blue one, and electrons are circulating around it, too. Now, what effect is that going to have on the field seen that the red nucleus? Well, we can look at the lines of force, and see it has the same kind of effect, but not as strong, because it depends on 1/r³.

So as you move away, the effect gets smaller. That’s not surprising, that you can see the lines of force get further away as you move further up. OK, so the effect is weaker from the same number of orbiting electrons on another nucleus. That’s fine. But suppose that the nucleus we’re studying is fixed, relative to the other nuclei by bonds, by a bond like that, or other bonds– it could be connected in a complicated way to the blue nucleus– and suppose that the molecule rotates. So it’s not always in the same field.

So we can look at it like this. Here is the– can you see this here? OK, so here is the nucleus we’re interested in, and here’s the other nucleus about which things are orbiting. And as we go like this, sometimes we’re in a region where the field’s pointing up, sometimes in a region pointing down. In fact, I have it the wrong way here, because I’m going to use this same diagram for another purpose later. But you can see that we’re going to have to average, if the molecule is tumbling rapidly, to see what the influence of electrons here are when we’re over here.

So let’s try to do the average. When it was up, when the bond was pointing up, we saw that the field pointed down. How about here, when the bond is down? Which way does the field point from the blue circulation? Debby? Look at the lines of force, which way are they pointing, down or up, where the red nucleus is?

Student: Down?

Professor McBride: Down. OK, so it’s the same. Good. That’s going to reinforce it. But suppose it’s turned at 90° there. Keep us going Debby. Now which direction to those lines point?

Student: Up.

Professor McBride: Up. And how about if we rotate to the other side? 

Student: Up.

Professor  McBride: And how about if we’re at 45, or actually, a little bit less than 45° there?

Student: 45°.

Professor  McBride: Ah! They’re pointing sort of down to the right. And if we’re over there, they’re pointing down to the left. And if they’re down here, down at the bottom, they’re pointing down to the right or the left, depending on where we are. So we have to average all of these. Some of them are bigger than others. They’re pointing different directions. 

But notice that if we average over a whole circle of rotation, then the ones that are at the top and at the bottom, within the scope of those double-headed yellow arrows, those are all net down. Some are down very strongly. Some are down very weakly. And the right and the left ones cancel one another, in the right to left. So it’s net down at the top, net down at the bottom also.

But on the sides, it’s net up. Everybody with me? Now, which one’s going to win? The ups or the downs?

Students: Downs.

Professor  McBride: Why do you say the downs?

Student: More of them.

Professor  McBride: First, there are more of them. And second, they can be stronger. In fact, this turns out, up at the top, the lines of force are twice as close together here as they are out here. There’s a factor of 2. So clearly the downs have it. So if you average around the circle, you’re going to get net down. However, that’s not the way you average. Can anybody see why you don’t average around a circle? When I look at the effect of this, the fields over here, on this, why don’t I just average around a circle?

Student: It’s a sphere.

Professor  McBride: Because the molecule is not held in a plane. It can go this way, too. So we have to average over a sphere, not around a circle. And notice that when you average over a sphere, you get much more area around the equator than you do up at the poles, going around a sphere. So now that’s going to be an advantage for the ones that are pointing up, a much heavier weight in the averaging. And it turns out that the net average over a sphere is 0. And that’s great, because it means you don’t have to worry about it. 

You can’t get an effect from the circulation someplace else. You’re only interested in the ones that are circulating about you. That obviously simplifies our life greatly, except for diamagnetic anisotropy. So let’s see what that is. So we can ignore the electrons on other atoms. Good news. Unless the amount of orbiting depends on the orientation of the molecule, and then it’s not always the same orbiting for different orientations. 

So let’s just suppose that we’re up there. And suppose when the molecule is oriented vertically that way, when our nucleus is above these circulating electrons, suppose there’s less orbiting for some reason in that orientation. So we’re going to make that pink thing thinner, and that will have the effect of making the magnetic field less. The lines of force are now further apart. And that means we don’t have as big a field up on top. Everybody see?

So if it averaged to 0 before, it will average up now. Everybody with me? Because we’ve cut down on the amounts that are going up, when the molecule’s in this orientation. If, when it’s down in the other orientation, you have the full complement of circulation, then the ups are going to win. And this is what’s called diamagnetic anisotropy. Diamagnetic is this circulation that causes a field opposed to the applied field, and is driven by the applied field. The reason it’s there is because you put the applied field on. And if you put more applied field on, you get more of it.

But in a fixed field, you have this diamagnetism. But it’s not isotropic. Isotropic means the same in all directions. If it’s different in different directions, then it doesn’t cancel anymore. So diamagnetic anisotropy means that you’re in trouble, that you have to consider the electrons on other nuclei as well. 

So let’s consider that in the case of benzene. So we have these π orbitals overlapping in benzene. And then when you put a magnetic field on in that direction, you can cause circulation of those electrons, net circulation. Now that creates the field that’s opposed to the applied field in the middle of the ring, but reinforcing the applied field out where the protons are. Everybody with me on that? So it’s reinforcing the applied field. However– so it’s like that– but if you turn the molecule edge on, then you don’t have the path anymore, where they can circulate, because they’re driven by the big field. They circulate around the big field, and now the path isn’t there to go on.

So it turns it off, or at least reduces it, when you turn the molecule that way. So this is exactly a situation of diamagnetic anisotropy, where you have the effect for some orientations and not others, and now there’ll be a net average field, due to the circulation in these p orbitals of carbon, as seen by the hydrogens, which is what you’re looking at in your NMR spectroscopy. So it’s a net deshielding. The orientation shown here is the one that dominates, where the field is big, where the circulation is big. It’s not big when you turn it the other way.

So this one wins, and this one, notice, reinforces the applied field at the position of the nuclei we’re interested in. So that means that they get shifted downfield. So that’s why in benzene, the protons are further downfield, that way, then you would have expected just from their being in an sp² bond.

Now, this then provided, once this was appreciated… NMR came along. It came along at the same time that people were interested in Hückel aromaticity, 4n+2. And it was a happy coincidence. Because in a molecule like this one here, which has 14 π electrons, is that 4n+2? It’s 4 x 3 + 2. So this should be like benzene. 4 x 3 + 2. And this is the NMR spectrum in the normal region of that compound. However, what comes at 0? Remember what comes at 0? 

Student: TMS.

Professor  McBride: That’s TMS, our reference peak. And what comes at 7.2?

Student: CDCl₃.

Professor  McBride:  That turns out to be the solvent, chloroform. So TMS, and chloroform, there are no peaks for our compound. The reason is we have to look out of the normal range.

So if you look for those protons, you find them way down there, further down from where benzene usually comes, which is in the 7 to 8 region. And that’s exactly the same region, that they’re out beside the plane of this thing, the electrons are circulating around that loop when the molecule is flat like this. And it feels exactly the same kind of effect as in benzene, except more so. But there are methyl groups there, too. Where are they? Do we have to go still further downfield? Where are the methyl groups?

So there’s a picture of the molecule taken from an x-ray crystal structure, and you see the circulation would be around this path. Where are the methyl groups? The methyl groups are in the middle, where the lines of force are going the other direction. And you find them up here. So that’s a really dramatic example of this diamagnetic anisotropy.

Now, once people got this particular bit in their mouth, they started looking at a lot of other cases. For example, this molecule that we talked about, which, remember, had that CH₂ bridge across the middle, and then 10 π electrons, 4 x 2 + 2. And again, you see them shifted downfield when they’re outside, and upfield when they’re inside. Now, they’re not shifted as far. Why not shifted as far? There’s not as much ring current. There are not as many electrons, but the other reason is that that was distorted, so the overlap isn’t as good and you don’t get as much circulation. But still you can see that it is benzene-like. It is, in this context, aromatic, because the CH₂ is shifted up above TMS. Very unusual.

Now, that’s 14 π electrons. Suppose you treat it with potassium. Potassium easily gives up electrons. It reduces the compound. In fact, you can put two electrons into it. And for this purpose, we’re going to condense the scale, same spectrum but I just squeezed the scale in for reasons that will be obvious. 

So there the sort of super aromatic hydrogens shifted downfield to about 9 ppm, and the ones on the inside shifted way up. But if you put two more electrons in, and make the dianion, like that, now it’s not aromatic anymore. It’s 4n instead of 4n + 2. And look at its NMR spectrum. They exactly shift. So aromatic goes one way, Hückel, antiaromatic goes the other way. So there were lots and lots of examples done like this during the 1960s and early ’70s.

So the CH₃ signals shift downfield by 26 ppm, despite adding– remember, there are more electrons now. We put two more electrons into the system. They should shield it, but in fact, it’s deshielded by 26 ppm, really dramatic. So it’s an antiaromatic dianion. 

Now how about acetylene? Now, acetylene has these p orbitals that.. we’re writing it vertical here. There are two p orbitals. And remember, when we talked about what the electrons look like if you have p orbitals equally occupied? If you have three p orbitals, each with two electrons in it, and look at the total electron density, what does it look like? It’s a sphere. It’s not lumpy. And two give a cylinder.

So this thing has a cylinder around the carbon-carbon triple bond. So there’s a path in which electrons can circulate. So we get this diamagnetic current when we put a field on that, and get these fields. How is this different from the case in aromaticity? Do you see? When it was benzene, it could circulate when the molecule was flat and the hydrogens were out here. Now it circulates when the molecule’s oriented this way and the hydrogens are above the middle. So the lines of force go exactly the opposite direction. But as in benzene, when you turn the molecule, you don’t have the path for circulation.

So it’s diamagnetically anisotropic. It goes one way and not the other. This is the way that wins. And the nuclei of acetylene lie above the orbiting path where there is ring current. So the applied field is diminished and the signal shifts upfield rather downfield. So that’s why, when you made it sp rather than sp², it didn’t go downfield because of the electronegativity, it went upfield because of diamagnetic anisotropy. And it’s just the opposite of what you see in benzene, where the protons are beside the circulating path rather than on its axis. 

But here’s a warning. This is a handy picture and it’s been taught for 50 years now, I guess, close to. But it might be nonsense. And Professor Wiberg has done calculations and observations that show that it doesn’t work for C-13. Whether it works for protons, I’m not sure. But at any rate, it’s a really handy picture to remember who shifts which direction. And it has sort of the ring of truth about it, but it might not be supported by really careful calculations of the kind Professor Wiberg does.

So that’s the idea of diamagnetic anisotropy and what gives rise to chemical shifts, these things we talk about. Pulling electrons away, mostly it’s the electrons on the nucleus in question that are important, and as you pull the electrons away you get a shift downfield. Pardon me, downfield is that direction for you. OK, good. But in special circumstances, you have this diamagnetic anisotropy, and then you have to think about it in the terms we just talked about.

Chapter 2. Multiplicity in Spin-Spin Splitting [00:20:50] 

But there’s also the local fields that are due to other nuclei, and this is called spin-spin splitting. For example, this is taken– and again, I will recommend to you this Chem 220 website, which has NMR problems. They give you a spectrum and say, what is it? And then the nice thing is they show a spectrum like this, they allow you to go in and look in detail at the peaks and see what their integrals are, and also gives a detailed explanation. So there’s an answer key. So that’s a handy place to go. But this is the first problem in that set. 

Now, you see there are three kinds of protons in this. It turns out– I’ll give you the answer first– that it’s ethyl acetate. So there’ll be two groups of three protons, and one group of two protons. So obviously, these are the three and that’s the two. But let’s zoom in, as this program allows you to do, as that website allows you to do, and when you look at that triplet, when you look at that peak, you see it’s, in fact, three peaks. And they’re in the ratio 1:2:1 in their intensity. Now, why is that? 

Those are the CH₃ groups of the ethyl group, CH₃ protons of the methyl group. But adjacent to it, there are two other protons in the vicinity. And you can have those oriented, each of those can be oriented either with or against the field. So there are two of them. Each can be with or against the field. So there are four different ways to do it, 2². So there are different molecules which have different orientations of next-door protons, of those protons.

Now let’s think about those. They could both be down. They could both be against the field, and then we get this peak. Or they could both be with the field, and then we get this peak. So that when you think you have just one kind of molecule there, in fact, you have what you might call– nobody calls them that, I think– spin isomers. They differ only in how these neighboring protons have their spin oriented. But there are different molecules in there.

But they are so similar an energy, the magnetic interaction of the protons with the field is so small in energy, that the Boltzmann constant is 1, equal amounts of the two. I mean, there’s a tiny, tiny difference, but not that you could measure by the intensity of the lines. So they’re equally abundant, so the yellow peak is just as big as the blue peak, or green. Now what’s that? That’s when one of them is up and the other is down.

And it’s just as likely as the others, so it’s 1:1:1, but you can also have one down and the other up, the opposite direction. So that’s why it’s 1:2:1, because all four of those are equally likely, and two of them come in the middle where they cancel each other in their effect. OK, so that’s a triplet.

Now how about this thing here? That’s a quartet. And we can figure out how that works, too. That’s the CH₂’s. Why are they shifted downfield? Why do they come farther to the left from where the methyl group was? It’s a chemical shift. Derek?

Student: The oxygen.

Professor McBride: Because the oxygen’s withdrawing electrons. OK, so they’re shifted down there. It integrates for two. That’s how many protons are being involved. So it’s that CH₂ group, but it’s a quartet. Why? Well, let’s look at the neighbors. Now there are three hydrogens on the next-door carbon. So there are going to be eight different spin isomers that differ in the spins then of the adjacent hydrogen nuclei. 

So let’s think about it. There are the ones where they all oppose the field. They’ll be up at the right. The ones where they’re all with the field. That’ll be at the left. And then you can have one down, or two down… or [rather] a second one down, or a third one down. But they’re all the same, all have one down.

So they come to that peak. And then there’s one of them up, a different one up, a different one up. So it’s 1:3:3:1.

OK, so if we have three neighbors, we get 1:3:3:1. If we have two neighbors, we get 1:2:1. What if you have only one neighbor? What pattern should you see? Po-Yi, if you had just one neighboring proton, and you’re looking at this one here, what will the signal of this one you’re looking at look like?

Student: 1:1

Professor  McBride: Right. Some molecules will be like that, others will be like that, it’ll be 1:1. So if it’s just 1, it’ll be 1:1. Now, what if it’s 4? Anybody know what that one’s going to be?

Student: 1, 4, 6, 4, 1.

Professor  McBride: 1, 4, 6, 4, 1. How did you get that?

Student: Pascal’s triangle.

Professor  McBride: Right. It’s Pascal’s triangle, the binomial coefficients. So it looks like we can do this for anything, right? We’re in good shape.

OK, now, how big is this splitting? It’s measured in energy units, in Hertz. And that’s called J. And they’re equal. You expect those to be equal, because both of them involve the same change in the neighbor, flipping one proton. And at the same time, those are all the same, too. But this is exactly what you should say, because the influence of this magnet on this magnet has got to be the same as the influence of this magnet on this magnet, if they’re the same kind of magnet, all protons. 

So that’s great. And notice, it won’t depend on how big the magnetic field is, because this is just the interaction between two magnets, which have fixed magnitudes. It was different for the orbiting, because the orbiting was created by the magnetic field. When you had more magnetic field, you got more orbiting. So the chemical shift is measured in fractional units, what fraction of the big field does it shift at some parts per million.

But this is measured in Hertz, in energy, because it doesn’t depend on the applied field. So it’s independent of B₀, versus the chemical shift, which, as we know, does depend on B₀. The orbiting is driven by B₀. 

Okay, now let’s look at another one. This is ethanol, a spectrum measured over the break by Phillip for us. Now, so we have OH, CH₂, and CH₃. Let’s see if we can figure out all these peaks. Which one is that, furthest to the right? Which of the protons in ethanol is that peak that’s furthest to the right, which turns out to be a triplet. What? 

Student: It’s on the methyl.

Professor  McBride: It’s on the methyl group, because the other ones have oxygen withdrawing, so it has the highest chemical shift– or the lowest number of chemical shift, the least shifted down from 0. And it’s a triplet. Does that make sense to you, that it should be three lines? Kate, what do you say? 

Student: Yes.

Professor  McBride: Why?

Student: Integration.

Professor  McBride: No, it’s three protons, but the signal is not a single peak three times as big as the others. It’s three peaks, 1:2:1. What causes that splitting? Do you remember? It’s not chemical shift. It’s called spin-spin splitting. What’s the other spin that’s splitting your signal? It’s the protons on the adjacent carbon. And there are two of them, and you can orient it in 2² ways, 4 ways, one, bum, bum. It’s what we just talked about. So it’s a triplet, that’s what you expect. So that’s the splitting of the CH₃ by the CH₂ protons. 

OK, now let’s look down at the other end. Oh, incidentally, the chemical shift positions as measured on the spectrometer are that, but actually, we should measure it not in parts per million, where the differences is 0.018 ppm. We should measure it in Hertz, so the Hertz units are 7.2 Hz. We multiply by the frequency of the spectrometer, which it says here is 400 MHz. Okay, so it’s 7.2 Hz.

Ok, now we look down there and we see a triplet. What’s that due to? Alex?

Student: That OH.

Professor  McBride: Yeah, why? Why is it the OH?

Student: It’s a triplet, so– 

Professor  McBride: So it’s got two neighbors. And it’s shifted all the way downfield, because the oxygen, electron withdrawing, it would appear. So there, that’s the CH₂ splitting the other direction. Now let’s compare those two. They’re not the same size. The interaction between the OH and the CH₂ is smaller than the interaction between the CH₃ and the CH₂. That one is 5.1 Hz. 

Now, what’s this thing here? What’s that due to? Matt?

Student: The protons on the CH₂. 

Professor  McBride: So that’s the CH₂. And what’s it split by? 

Student: The H on the OH, and the H’s on the CH₃.

Professor  McBride: OK, so how many H’s altogether? 

Student: Four.

Professor  McBride: And what do you expect the intensity to be?

Student: 1:4:6:4:1 

Professor  McBride: 1:4:6:4:1, so let’s blow it up. Is it a 1:4:6:4:1 quintet? Not quite. Now what is going on? Notice that those two interactions are not the same. One is 7.2 Hz and the other’s 5.1. So if we look at those, there we see a quartet. That’s split by how many?

Student: Three. 

Professor  McBride: That’s split by three. That’s the CH₃ splitting. But in fact, there are two quartets. So that’s the smaller splitting due to the OH. Now, had they been the same, then the separation of the three– the separation of the two quartets would have been– let’s see– instead of being like this, it would be like this. And then it would be 1:4:6:4:1, but it doesn’t look like that. It’s actually what we call a doublet of quartets, or a quartet of doublets. OK, so that’s a little bit different.

OK, now there are some other tiny peaks here. What’s this one due to? Any thoughts?

Student: Water.

Professor  McBride: What?

Student: Water.

Professor  McBride: Good idea. It could be water, but it’s not. The solvent is DMSO-d₆. But unless you would spend an absolute fortune, you have some DMSO-d₅ in there that has one proton. So those few molecules of the solvent that have hydrogen on them look like that. So you see a signal from that. OK, that’s fine. And that’s what the signal looks like.

Now, there are no other hydrogens in there. What’s causing spin-spin splitting? Deuterium is also magnetic. It’s not a very strong magnet, so it doesn’t have a very big splitting. These splittings are very small. And furthermore, not only is it a weaker magnet than a proton, so that it’s only 1.8 Hz, it can be oriented three ways in the field. Remember, a proton can be with or against, but a deuterium can be with, against, or across. So now there are three ways of putting it.

So if you have two of them with three ways, it turns out what you get instead is a 1:2:3:2:1 quintet. And you can work that out for yourself. If a single one splits it 1:1:1, because there are three orientations, what do two do? So you can work that out. And if you figure that out then you’ll understand the math we’ve been doing here. OK, now, here’s another peak. Matt? What’s this peak? Why did I ask you? 

Student: Is that water?

Professor  McBride: That’s water. OK, good. But, look over here.

Student: Just ignore this. [laughter] 

Professor  McBride: What are those? Blow it up. This bit here, these little ones, are due to an artifact of the instrument. Don’t worry about those. But these are real, here and there. So they’re triplets, right? Like what’s in the middle of them is a triplet. And there’s a 7.2 Hz splitting, which is exactly what you had here. So it looks like it’s being split by the CH₂ next door. That’s this splitting here. We’re looking at something that has to do with this CH₃. It’s being split by that. But there’s something else that’s splitting it. What else could be splitting it?

There’s another magnetic nucleus. It can’t be this H out here, because that’s so far away it would have a minuscule effect. This is an enormous effect, this doublet splitting. Any ideas? It’s sort of related to this one, actually. There’s another nucleus there that’s magnetic, but only 1% of the time. These are only half a percent as big as this is. So the total of this and this is just 1%. There’s 1% of some other magnetic nucleus in there.

Student: Carbon. 

Professor  McBride: Carbon. It’s carbon-13. And that has an enormous interaction with the protons, 125 Hz. That’s much closer to the protons. OK, that’s 1.1% of carbon in natural abundance. 

And notice one other thing that we’re going to address later– probably we won’t get to it today– which is that these are slightly asymmetric. This one’s just slightly bigger than that one. And the green ones here are slightly bigger than the pink ones. And this one is slightly bigger than that one. And we’ll get to that, too. So we’ve looked at this in pretty much detail.

Chapter 3. Tumbling, Orbitals, and the Magnitude of J [00:36:05] 

Now, what determines the strength of the spin-spin splitting, how many Hertz it’s going to be? So we’re going to talk about isotropic J-coupling between protons. And it’s not a through-space magnetic interaction. It’s mediated by bonding electrons. Because, that’s what this was actually made for, here are the lines of force created by this nucleus, and we’re looking at this one. 

So it can be like this, or it can be like that. So we get a doublet when we’re looking at this proton over here, because this one could be either this or this way. But if we follow its lines of force, they can either oppose the applied field, or they can help the applied field. And what do we get if the molecule is tumbling in three-dimensional space for the average of that? 

Student: Zero.

Professor  McBride: Zero, because now you’re not going to get diamagnetic anisotropy, because this magnet has nothing to do with the circulation of electrons, and where the paths are. It’s just the nucleus itself. And they’re on perfect ball bearings, so there’s no special stuff going on. So it’ll absolutely cancel to 0.

But then how in the world do we get interaction between the two? How do we get splitting if the through-space interaction of magnets won’t do the trick? It goes through bonds. And let’s look at this. So let’s look at the magnitude of some of these coupling constants, as they’re called, J. For adjacent positions in benzene, it’s 6 to 8 Hz. For next adjacent positions, it’s 1 to 3 Hz. And for remote positions, it’s 0 to 1 Hz. So what does it look like it depends on? Any thoughts? 

Student: The position of the H’s. 

Professor McBride: Yeah, it depends on position, obviously, but in what way?

Student: The closer they are the–

Professor McBride: The closer they are, the bigger the interaction. Can I see the hands of those who are surprised? OK, now let’s look at these. That’s 1.85. This is 2.38. That’s 3.07 Å. And here are their coupling constants. What do you think your theory, Derek? 

Student: It’s not holding up right here. 

Professor McBride: What?

Student: It’s not holding up right here. 

Professor McBride: It’s exactly wrong. The closest ones have the smallest splitting, and the farthest ones have the biggest splitting. But we know that it shouldn’t be distance, because we’ve just seen that it averages to 0. So it must be something else. It’s not spatial proximity. The through-space interaction of dipoles averaged to 0 on tumbling.

Now let’s look at ethane here, and look at those two protons, top-right and bottom-left, and look at this particular molecular orbital, which turns out to be HOMO - 3. There are lots of others we could look at, but this one is good to look at for reasons you’ll see. Now, when there are two electrons that are going to be in this orbital, and when one of them is on this nucleus, the one, let’s say– the electron, remember, it’s a magnet, too, when it’s pointing up– it’s going to be in that orbital. There’s going to be one electron pointing up, one electron pointing down.

So suppose the up one happens to be here. Then the down one can’t be up there. The down one can only be over here. That is, only the down electron is available to be here. It can’t be the blue one, when the blue one is up there. So the electrons provide a way to communicate between the nuclei. Like, suppose that this electron likes to interact with that nucleus when that nucleus is up, let’s say. Then in those particular molecules where this nucleus is up, this nucleus prefers to be down. And vice versa. The nuclei would be lower in energy when they’re antiparallel than when they’re parallel.

So it’s as if the nuclei were interacting with one another. But in fact, one is interacting with one electron, the other is interacting with the other electron. But the two electrons are in the same orbital, so they interact with each other. One has to be up, and the other down. So it’s this indirect interaction through the bonds. 

Now, that has an interesting implication. So in tumbling molecules, nuclear spins communicate not through space, but through paired electrons on the nuclei. Because if that other electron wasn’t right on the nuclei– on the nucleus– then as it orbits, it’ll cancel. There’s only one place that electron can be where rotating the molecule won’t cause it to cancel. What’s that place? On the nucleus.

When the electron is on the nucleus, tumbling won’t average it. So what orbital does it have to be in to be on the nucleus. It has to be an s orbital. So it’s only the s orbital electrons that allow this communication to take place. So the J, the magnitude, depends on the s orbital content of the various molecular orbitals. Not just the one I showed, but others as well. 

Why is this one the anti– or trans– why is that bigger than this one? Could it be that you have more overlap here than you do here? So the molecular orbital is better for communicating between these two than for communicating between these two. So we have to decide which of these orbital overlaps looks better to you. The one on the right or the one on the left? If you had to guess which one is better overlap, what would you guess? Or you wouldn’t guess.

Suppose I asked it on an exam, and you had a 50-50 chance of getting some credit. Or maybe it’s 60-40, if your intuition is good. Which one gives better overlap, this or this? Right or left? How many say right? How many say left? There are no lefts. You think this one’s better overlap? OK, let’s see. 

OK, so which gives better overlap? Now, the way we can tell is to dissect it into the atomic orbitals that are part of those orbitals, so examine the overlap of the components. So notice that this one is a combination of a p orbital here and an sp hybrid pointing that way. This is a combination of a p orbital here and an sp hybrid pointing that way, and corresponding over there. And let’s take them apart so we can consider them separately. So notice that these are the same on both sides, so we can forget that. That’s going to be the same overlap. 

But now the question is, which is the better overlap, this or that? Let’s vote again, partial credit here, your chance to redeem yourself. How many think this one is better overlap? Notice that means that, if it’s true, that means this is better overlap. Because these are the same. And how many think that’s better overlap? OK, not many people think that, and for obvious reasons.

But let’s dissect it a little more. So let’s just look at this one where the difference lies. And this, you notice, is a combination of these p orbitals and these s orbitals. And this is a combination of these p orbitals and these s orbitals. Everybody with me? Now let’s see which ones have better overlap. OK, so this is good s-s overlap, and that one’s bad, antibonding. And the p orbitals good and bad. Shall we vote again? How many people think this one’s better? Good. OK, you’re thinking.

What haven’t you thought about though? This s can overlap with this p. And this s can overlap with that p. And the same thing over there. So here are two bad overlaps between s and p. And here are two good overlaps between s and p. So now the question is, which is more important, s with p, or s with s and p with p? And if you had a photographic memory, you at hark back to lecture 12 [of last semester], where we talked about the overlap of carbon hybrids, and saw that, curiously, the sp overlap is bigger than either s-s or p-p.

So in fact, this is the dominant one. And that one has better overlap. Isn’t that cute? So the backside overlap is really counterintuitive, that this is better overlap than that. And it really doesn’t look like it in simple pictures. I absolutely agree with you.

OK, and here’s that thing from lecture 12 last semester. There’s s with s, at the distance of a carbon-carbon single, double, triple bond. Here’s p with p, in the σ overlap. And here’s s with p. So s with p wins. Good.

OK, so now let’s look at something else here. This is a table that you can find in the book, and many books say this kind of thing, but there are some errors here, which is what I want to show you. I mean, overall it’s worthwhile, but there are a couple errors. 

It says, “Identical hydrogens do not couple.” That is absolutely not true. They do interact with one another, but the coupling is invisible. So in fact, the magnitude of the coupling is not 0, it’s 10 to 20 Hz. But you can’t see it, because there’s no handle for the radio frequency. The light can’t cause a transition between the peaks [correction: levels] that would show that splitting. And in frame 26, that we won’t get till next time, we’ll show that.

One other one here. This one, it says that the coupling is 6 to 8, and that’s fine. But it depends on this angle, because remember, the overlap of this bond with this bond, which is what we just talked about, is going to be bigger when it’s anti than when it’s eclipsed, and something different when it’s like this. So it’s going to depend on this angle. And in fact, it can be between 2 and 13 Hz, depending on the conformation.

Now, of course, if it’s spinning around rapidly, it’ll average. Spinning around the internal bond. So it’s 13 Hz when it’s like that. It’s 2 Hz when they’re perpendicular. And it’s 11 Hz when they’re syn. That’s just like what we saw last time, bigger like this than like this. When they’re gauche, which is one of the more likely ways to be, then it’s 7 Hz, which is that. And that’s also the average of all these others. 

And notice that if you have a torsional angle– if it’s spinning, then you average to 6 to 8. But if it’s fixed in some molecule, then you can measure the angle by measuring J, so that’s another way to get structure. 

Let’s see what time we got here. Let me do this one last– no, we’ll do this next time. OK.

[end of transcript]

Back to Top