Freshman Organic Chemistry II

CHEM 125b - Lecture 25 - C-13 and 2D NMR – Electrophilic Aromatic Substitution

Chapter 1. Proton Decoupling [00:00:00]

Professor J. Michael McBride: So, we’ll finish up on NMR spectroscopy, talking about more dynamic things involving decoupling. Then a little bit about C-13 NMR and double labeling involving correlation. And then we’ll get onto another reaction, electrophilic aromatic substitution.

OK, so proton decoupling. Remember that the proton is four times as strong a magnet as the Carbon-13, so it precesses much faster. So it’s easy to do a pulse that will tilt the Carbon-13 net magnetization down into the plane, so that it’ll precess and be an antenna that can tell us its local field, without doing the same thing to the proton. And if we do that, then we see the C-13 NMR spectrum. And there are two peaks, because of the adjacent proton. Right? it can be either up or down, 50:50, or almost exactly 50:50, so you see two peaks.

Now suppose at the same time you’re doing that, when you do this pulse and listen to the C-13 to see what its local field is, you irradiate at 100 MHz. Right? That won’t do anything to the C-13, it’s the wrong frequency. But it’ll now make the proton precess. That is, if we look in the rotating frame, we can see a weak field that’s horizontal, about which the magnetism will precess. So the protons will go down, and then up, and then down, and then up.

And what determines how fast it goes up and down? How strong this field is. The stronger it is, the faster it will precess. And if it’s fast enough, then you won’t see a doublet for the C-13 anymore, because you’ll see the average. Right? So you’ll see a single peak, you’ll proton-decouple the C-13 spectrum and see the C-13 as if the protons weren’t there. OK, so it depends on how much power you put in. Now here’s a C-13 spectrum of a given compound. Where we’re listening to the C-13 and we see all these different local fields, right? But we’re irradiating the proton at the same time, but very, very weakly. This number up there, 40 decibels, is an inverse ratio of how strong that horizontal field is for the protons. So it’s 10⁴ weaker than what you’ll see you later, 10,000 times weaker.

And if we look at the carbon, we see that some of the carbon signals are split and some are not split. The ones that are split are the ones that have hydrogens on them. So if we start from the left, we see there’s one with no hydrogens, then one with one hydrogen, one with one hydrogen. That red one is the solvent, CDCl₃, right? So we’re looking at the carbon of the CDCl₃. But it has a deuterium on it, not a hydrogen, so it’s not split. Then there’s a carbon with two hydrogens, a carbon with three hydrogens, it’s a quartet. Underneath, a carbon with two hydrogens, it’s a triplet. So very low power in the proton, that’s what we see.

Now we’re going to reduce the size of that. And see what happens, is we start increasing the power. So there’s 40 dB, 20 dB, and now things are beginning to change. See how the patterns are changing their multiplicity and their intensity? Because it turns out that when you irradiate the protons, by a complicated mechanism it strengthens the signal of the carbons to which they’re coupled. And that’s called nuclear Overhauser enhancement, as you’ll see.

OK, so now you see the splitting is going away when we get down to two decibels or one decibel. Now we see that each carbon gives just a single peak, we’ve decoupled the protons. So now we have a peak for every carbon. And notice, incidentally, that the CDCl₃ didn’t get stronger as we did this. Why? Because what strengthened the signal was irradiating the protons, and CDCl₃ doesn’t have hydrogens on it. So it doesn’t get strengthened. OK, notice also that the carbon without a hydrogen on it, on the far left here, didn’t get strengthened as much as the others did.

OK, so that’s decoupling, and it can be useful. And it can be useful in several ways. There’s two things, you get rid of the spin-spin splitting, so you see just a peak for each carbon. And also you strengthen the signals when they have nearby protons. And that could be bad if you’re interested in measuring the size of the peaks to get how many carbons there are, because you can’t count properly anymore, because you get different intensities with different neighbors of protons. But it’s good if you want to determine structure and see what hydrogens are how far away, and I’ll show you that shortly.

OK, now bear in mind that carbon-13 is only 1% in natural abundance of carbon. Of course, you can synthesize the molecule with expensive carbon-13-containing precursors and put carbon-13 where are you want it in a molecule, and then the signal from that carbon will be ever so much stronger. If you put in pure carbon C-13 in some position, the signal would be 100 times stronger. In fact, 200 times stronger than without, because when the carbon – . No, pardon me. It would be 100 times stronger, because it would be 100% instead of 1%.

OK, so here’s a compound, neotame, it’s an artificial sweetener. And it’s the proton-decoupled C-13 spectrum. So you see one peak for each carbon. There are 20 carbons, there are 20 peaks. But you can’t go by their intensities. Now, let’s see what we can make of this. It’s proton-decoupled, so there’s one peak per carbon and, being carbon, it’s pretty well spread out. Remember, the range of hydrogen chemical shifts is only about 10 parts per million. The range for carbon is hundreds of parts per million here. 200 parts per million are shown. Now why is there no C-13-C-13 splitting? There’s no proton splitting because it’s decoupled. But why no C-13-C-13 splitting?

Student: The likelihood of two C-13s adjacent in the same molecule is unlikely.

Professor J. Michael McBride: If it’s only 1% natural abundance, and you’re a C-13, so that you’re giving a signal, the chance that your neighbor is C-13 is only 1%, so there would be tiny, tiny peaks from those you don’t see them. OK, so that’s what Chris just said.

OK now, let’s see if we can figure out which carbons these are, that we’re looking at. And of course, this is how it was done empirically. People put in known compounds and figured out what the chemical shifts… what peaks were from one compound to the next. And then they figured out this comes in this range, this comes in this range, this comes in this range. So let us see, knowing what these peaks are.

OK first, there are these ones that are way down close to 200. They’re numbers 1, 13, and 4. So what do 1, 13, and 4 have in common that turns out to make them way far downfield? Ellen, what do you say? What do 1, 13, and 4 have in common that makes them different from the other carbons?

Student: They’re almost like esters. Like, double bond to an O.

Professor J. Michael McBride: Yeah, the double bond to an O. Carbonyl carbons are way downfield. OK, now we’ve got 7 and 8 to 12. So there’s 7 and there’s 8, 9, 10, 11, 12. Pretty clear what those are, right? Those are the ones on a benzene ring, aromatic carbons. OK, then there are ones that are a little bit further down than the others, and those are those blue ones. What do they have in common? Rahul, what do you say they have in common?

Student: Tertiary?

Professor J. Michael McBride: They’re not all tertiary, no. in fact, only two of them have three things associated with them, the others have one and two. Lauren, can you see what the blue ones have in common? In hydrogen, what makes things come a little bit further downfield? Yeah, Amy?

Student: Well, the hydrogen would be like deshielding from an oxygen or something like nitrogen or oxygen.

Professor J. Michael McBride: They all have an electronegative atom next to them, right? Nitrogen or oxygen. So those are ones, a carbon that’s attached to something electronegative. And then, finally, you have the others, which are attached only to carbons. OK, so you could begin to figure out these, and you can get tables of what comes where. And then you can start dealing with unknown compounds and figure out what it’s telling you. But a nice thing about this proton-decoupled spectrum is that there’s just one peak for each carbon, and they’re well spread out. So that it’s good to do this kind of thing. So those are the alkane carbons.

Chapter 2. C-13 NMR: Double Labeling and Lanosterol Biosynthesis [00:04:39]

OK, now the power of correlation. Often when you can measure two things on the same subject simultaneously, you get a lot more information than measuring one, or measuring the other, or measuring them independently. I’ll show you that example with C-13 double labeling and also in two-dimensional, as it’s called, nuclear magnetic resonance.

So first, C-13 double labeling. And we’ll talk about lanosterol biogenesis. In fact, we already talked about it, if you look back at lecture 52. OK, so remember we got squalene from isopentenyl pyrophosphate. Then it curled around, and we had this complicated thing where it zipped up going one way, and then we had all these rearrangements. So as we said at that time, this is the source of cholesterol and the steroid hormones. And it’s a very cute story, how all that rearrangement took place, OK.

But there’s a question whether it’s true. And I said at that time, wait for NMR. So now we’re at NMR, so you can get the answer.

OK, so you can use Carbon-13 labeling to figure this out. So if you feed the plant that makes this stuff Carbon-13-labeled isopentenyl pyrophosphate– in that particular carbon, that substituted carbon– then you can see where that would appear in the squalene. According to the way we linked them together before, it should be those carbons. Then when you curl it around and do all this stuff to make lanosterol, those peaks should be labeled with C-13 and therefore 100 times stronger than the other peaks in that spectrum. And indeed they are. So that supports it. Or you could do it a different way. You could label the methyl group. And if you label the methyl group, it turns out that those methyl groups should be labeled in squalene. And then if you curl it around, it should be those methyl groups. And indeed, they’re 100 times stronger than the other signals.

But much more interesting than this is to do double labeling. So if you label both, you get that. Big deal. That’s just the same information we got before right? Except done in one experiment rather than two. And of course, in a sense this is not as valuable as these two independent label experiments, because you don’t know but what two of those carbons might, in principle, have changed places. But the double labeling is much more important if you do it dilute. Now what do I mean dilute? I’ve been talking as if we put 100% C-13 in here. But suppose we don’t put nearly 100%, suppose we only put 10% in. But suppose that those 10% that are labeled are double labeled. So we prepare a sample that’s 100% labeled in two positions, but then we mix it with 90% percent of unlabeled material and feed it to the organism and get the stuff out.

OK, so what we’re going to do is a dilute double label. Now why do it dilute? Because then, when label goes in, two labels go in. But it’s unlikely that any given squalene will have two different positions double labeled, or two sets of positions double labeled. Because only 10% of the stuff is double labeled. So when a squalene is double labeled, it’ll be labeled in just one pair of positions, a given molecule. Like there. So those two come labeled together. The signal is 10 times stronger because remember, it was 10%. It’s 10 times stronger than it was for the ones that aren’t labeled. But what’s special about it? Can anybody see what’s going to be special about the signal we get from these molecules that look like that? Ayesha, have you got an idea?

Student: They’re going to be the same more or less because they’re just an extra–

Professor J. Michael McBride: They’re going to be the same what?

Student: They’re going to be the same more or less because they’re just extra carbons and there’s no special function.

Professor J. Michael McBride: I didn’t quite understand you, but I was listening for special words and I didn’t hear them. Sebastian, have you got a special word for us?

Student: You get splitting?

Professor J. Michael McBride: Ah, they’ll split each other, because now the C-13s are next to one another. Because they’re not coming in at random 10%, they’re coming in as a pair. So now when we look at the signal from that, it’s going to be a double doublet. The blue and the red one are going to split each other. Now other molecules in the same sample will turn out to have been labeled elsewhere. For example, so that proves that those entered as a unit. Others will have come in there, and those will also be a double doublet, show that they came in as a unit. And if we go over to the other end of the molecule, those will be a double doublet. And those singles will be a double doublet.

 Great. What’s going to be different? Suppose we had that molecule where we had a methide shift. So the carbon got away from its original neighbor. Now it won’t be next to a C-13 and it’ll be just a pair of singlets instead of a double doublet in a proton-decoupled spectrum. And the same thing will be true there.

So this double labeling absolutely proves that these rearrangements took place. So those two, incidentally, are labeled in the same molecule. But they’re not adjacent to one another anymore, so they don’t show splitting. And those are both labeled, the ones that are adjacent to one another, but not in the same molecule. So they don’t do splitting either. So this dilute double label experiment enhances the same 12 peaks as the single label experiments did, the two single label experiments. But only eight of them show spin-spin splitting because their bonds stay intact. So this strongly confirms the story I told you before about the rearrangement scheme.

Chapter 3. 2-D NMR for Protein Structure and Rearrangement Rate [00:19:51]

OK, so this is the power of correlation. And let’s look at a different kind of power of correlation in what’s called two-dimensional NMR spectroscopy. Now here’s a two-dimensional NMR spectrum. And the way it’s collected is rather technical, and we don’t have time to go into it. But it shows that there are two different frequencies being used. And if you go along the diagonal of this plot, those two frequencies are the same. So what you see there, when you have just a certain frequency going in, is the normal spectrum you see along the diagonal. But this is the spectrum of a protein. So it’s got a whole bunch of amino acids linked together like that. And we’re in the range of the spectrum between 6 and 9, where the N-H protons show up In this particular kind of molecule.

So there are a whole bunch of them in this long chain and they all overlap with one another, and it’s very hard to tell what’s going on. But if you don’t have these two frequencies you’re using the same, you’re looking at one while you’re irradiating another one. And that affects the intensity of the one you’re looking at. That’s this nuclear Overhauser enhancement that we talked about, where you irradiate one, other things that it’s coupled to will change their intensity.

So what we’re going to do is plot on the off-diagonal how one frequency is influenced by having other frequencies in there, in its intensity. So that’s called NOE, or nuclear Overhauser enhancement. And it comes, incidentally, by through-space magnetic interaction. Not the stuff like we talked about coupling where it goes through bonds. This works through space. So it works if the two protons you’re looking at– the one you’re looking at and the one you’re tickling to see if it influences this one– if they’re within about 6Å of one another. So as I said, this is mostly these N-H protons. And for one thing, one advantage of having these off-diagonal peaks is they’re much less congested than what’s along the diagonal, than the normal spectrum. So for example, we’re looking here at the NH proton that’s at 7.25 parts per million. But you can see that it interacts with several other protons. It’s within 6Å of signals that are at 8.9, 8.3, 8.25, 7.7.

Now you don’t measure the exact distance, all you know is that it’s close. But this allows you to make a three-dimensional map of the protein. And the way you do it, is first you have to know which NH goes with which R group. So you see that that NH– the one you’re looking at, the red one– is close to that blue one. So it’ll influence it. But the blue one is then close to Hs that are in the R group, so you can tell which R group that that blue H is nearby, and which one the red H is near. So the red is near the blue is near the R. So you know which NH is near which R. And now the way you make a map, then, in this way– Professor Loria, who does this kind of thing, uses this analogy– it’s like making a map of a town if all you have is a telephone and a telephone book. Sometimes there’s a front page that has a map of the town, then that’s easy.

But if you don’t have the map of the town, then what you do is you just call someone at random. Say, “Hi, how are you today? Blah blah blah. Could you tell me who three of your near neighbors are?” And they tell you three of their near neighbors. Then you call those people, and you say, “Hi, can you tell me who three of your near neighbors are?” and they tell you. And then you call the next one, “Who are three of your near neighbors?” So once you know who is near what, then you can see that there will be a map that will allow you to figure out what’s near what. Especially if you have some help, because you know in the case of the protons, they have to fit with normal bond distances, normal bond angles, and so on.

So if you do molecular mechanics together with this knowing what’s near what, you can get a very good three-dimensional structure of the protein. And the nice thing about this, from the point of view of people who are actually working with proteins– what’s the best way of getting the three-dimensional structure of something like a protein or anything?

Student: Crystallogaphy.

Professor J. Michael McBride: X-ray, but you need a crystal. For this, you don’t need a crystal. So you can find the structure that way. So with molecular mechanics constraints, you get the 3D structure without a crystal.

Now here’s a different kind of 2D correlation with NMR. It’s correlation not in space, but in time. So this is the spectrum of a cation, that cation that’s shown at the right. Notice it’s benzene with methyl groups all around it, with one extra methyl group and a positive charge. Now in fact, you can get this in solution as long as you don’t have anything that reacts with cations in there, that reacts with low LUMO, so you need very, very non-basic solvents to do this.

So again, the diagonal running a different way this time– the increase in parts per million is down to the right, rather than what we usually see, increasing to the left. But anyhow, it’s color-coded, according to these peaks that show what they are in the molecule. So the scale is slanted and backwards. So you see that if you look at the resonance structures– that one, that one, that one– you can see that the blue and the green are near where there’s positive charge in the resonance structures. So they’re far downfield, very deshielded compared to the others.

Now, this can rearrange. A methide can shift, so one of the red methyls on the top shifted to the right. So now you’ve changed which methyl groups are in which locations. So now the red one at the top, this one and this one are now C, and will be shifted way downfield. This one will be shifted downfield– what was blue before in the D position, and so on. Then this blue one, which was downfield before is now an A, it’s up here. So we’ve changed the locations, the local frequencies, the local fields. So that’s a methide shift.

But is it a 1,2 methide shift? Because it’s possible that the methyl, instead of going just to its neighbor, would move to the center of the ring, up above the center of the ring and then it could come down anywhere. Wouldn’t have to come down adjacent to where it started. So you could have a 1,anywhere shift. Now you can distinguish that and measure the rates with this NMR spectrum, this two-dimensional NMR spectrum. So again, we’re seeing when one frequency influences another. Now if a given set of protons moves from one position to another, then having been irradiating one influences what happens where it ended up. Because those same protons are now here. So here you see off-diagonal peaks, where one has become another. So we can see here, for example, that the proton that was in position C becomes A. And that’s what happened in the blue, here, when we had that 1,2 rearrangement.

We can also see that D becomes B, and that you can see happened here. And you can see that B becomes C. But you notice there’s not A becomes B, A becomes D, or C becomes D. So you don’t have peaks for some of these patterns that you could have done. Now if you go through a whole sequence of the these, then you can start any place and end any place else. But it takes a longer time. So depending on the time scale of this thing, you can tell which things happen quickly. And what happens quickly is a 1,2 shift, not a 1,anywhere shift. So again, that’s information on mechanism. It comes from two-dimensional NMR spectroscopy. And it’s proton versus proton correlation in time. That’s, incidentally, just another visualization of the same map I showed before, that shows the peaks on the diagonals and the off-diagonal peaks.

Chapter 4. Electrophilic Aromatic Substitution: Substituent Effects [00:39:07]

OK, now we’ve finished with NMR. We’ll refer to it again, but we’ve finished discussing it. And we’ll go on to the next important reaction, which is electrophilic aromatic substitution. We talked about electrophilic things before. Remember, electrophilic addition to an alkene. Now, you might think benzene is an alkene, so you can get electrophilic addition to the double bonds of benzene, but you don’t get addition. You get substitution.

So here’s an example of substituting deuterium for hydrogen. You put benzene and concentrated sulfuric acid that has deuterium in it, and you get deuterium first in one position then, of course, in other positions in the benzene. So you can go all the way to C₆D₆, that’s how you prepare deuterated benzene, if you want to use it for a solvent for something, for example. You can use other electrophiles as well, not just D⁺, you can use NO₂⁺, Br⁺, sulphur⁺, R⁺, and RC=O⁺. All these things can substitute for hydrogen, and we’re going to talk about those.

Now, the mechanism is that the deuterium, D⁺, the deuteron, adds to the ring. But that, notice, changes the aromatic ring into just a regular pentadienyl system. So you lose a lot of resonance stabilization. It happens, but it’s not very stable and it’ll easily come off again, if there’s anything that can take a proton. But if you have a very, very non-nucleophilic, non-basic solvent, nothing to take it off, then it can stay on and in fact it’s observable. You can get spectra of it. And, so in electrophilic addition to alkenes, the next thing that happens is a nucleophile would add, and you’d lose the double bond. But here it’s easier to lose the deuteron or, more interestingly, the proton. Losing the deuteron would take you back to the starting material. But losing the proton, the H, then takes you to the substituted product. And that’s easier than adding a nucleophile, because you get back to the aromatic stabilization, 30 kilocalories. And remember, we just saw a spectrum that looked like that. When you added not H, but CH₃⁺, in this case, to the hexamethylbenzene. So it’s possible to see these things in NMR. So there’s the benzene, the D⁺ comes on, we get that.

Notice that it converts the aromatic ring into a chain, destroying the aromaticity. Let’s look at the orbitals that are involved, just using this simple Hückel program. So benzene, we’ve looked at the orbitals of that on the left, and pentadienyl we looked at before, too, and we see it on the right. Now, the most interesting orbital in pentadienyl is the one that’s indicated in yellow over there. The highest of the low three orbitals, and it’s at 0. And you know it’s at 0, the same energy as a p orbital. How do you know the energy of that molecular orbital, consisting of three p orbitals, is the same energy as a normal p orbital? Remember, if you make a double bond and put the two next to one another, they overlap, one goes down, one goes up, it’s not to same energy.

But this one has the same energy as an isolated p orbital. How do you know that? Notice how many nodes are there in the lowest one, the one that’s about minus two? How many nodes for the π orbitals? One, two, three, four, five p orbitals connected in a chain. How many nodes in the lowest combination?

Student: Zero.

Professor J. Michael McBride: Zero, except for the node of the p orbitals, of course. So zero. Then the next one, the one that’s at about -1?

Student: One.

Professor J. Michael McBride: One. How about the ours?

Student: Two.

Professor J. Michael McBride: OK, and so there are two nodes. Red, white, red on the top. Now, how do we know the energy of that? How much overlap is there?

Student: Zero.

Professor J. Michael McBride: None, they’re not next to one another. So it’s the same as an isolated p orbital at the Hückel approximation. OK, so that one is very interesting, that SOMO. It’s non-bonding in the case where we have a single electron in there, which is what’s shown. So that’s the locus of that odd electron in the free radical. If we have 5 π electrons, the fifth, the one that’s not in a pair, is in those three positions. Now suppose we’re interested in the cation with five p orbitals in a row. So what we’re going to do is just take that electron out. And where we take it out is now going to be a positive charge. So that’s the same as if we put another electron in, in the anion, that would be the locus of the negative charge. And in the cation, the locus of the positive charge. And that’s the same thing we’d get by drawing resonance structures that show those same positions having charges or free radicals in them. So those are the positions of positive charge. And we get the same thing drawing resonance pictures.

Now consider the influence on rate of substituting not benzene, but a benzene that already has some X group on it. Now, a very common thing to have studied with this was nitration, because it’s such an easy reaction to do. It’s a very vigorous reagent, NO₂⁺. You make it by just having a mixture of nitric and sulfuric acid. You see how you get NO₂⁺ from that mixture? It’s quite simple. You get a proton from sulfuric acid, and it protonates the OH, and then you lose water. So you’ve got NO₂⁺. So it could be the thing that comes in, and then a proton comes off from the same position, and you’ve got the substitution. But now, when you have X in there, you can get three different isomers. Ortho, meta, and para, they’re called.

Notice, incidentally, that NO₂⁺ is an interesting molecule because it’s exactly the same as CO₂, except it has one more proton in the nucleus of the central atom. Nitrogen instead of carbon. So CO₂ has a pair of electrons in that double bond, a pair of electrons on the oxygen that are in the y orbitals, pointing up and down. It’s also got x orbitals coming in and out that have the same kind of thing. But if we consider just the y orbitals, that’ll be the LUMO. There are two pairs of electrons, so the one with no nodes has two electrons, the one with one node has two electrons. This is the one with two nodes, no electrons. So that’s the vacant orbital, the LUMO of NO₂ ⁺, just like CO₂.

So NO₂ ⁺ attacks, and we can measure the rate. Now what we’re going to do to measure the rate– we could do careful work with a stopwatch and so on, but there’s a clever way to do it. Which is to compare two molecules by doing them both. Put both in there, and do the reaction for just a little while and then kill it. If you add base, you won’t have the strong acid and the reaction will stop. And now see how much of each one becomes product. So we get relative rates that way. So we’re doing relative rates compared to hydrogen being 1, that is, benzene when X is hydrogen. If X is methyl, it’s 25 times faster. That is, if you have 25 times as much benzene as toluene– the one with methyl on it– you’ll get equal amounts of the products at first, until you’ve used up all the toluene. So it’s 25 times faster. So methyl helps the reaction. OH really helps it, if you use phenol it’s 1,000 times faster at getting the sum of all the products.

Now why are those so fast? Because electron donation eases the formation of the carbon cation intermediates. Or, looked at a different way, the carbon cations are stable because there are other electrons that become stabilized when you get a cation there. For example, the unshared pair on oxygen. OK, now that’s because we have positive charge at those locations. When X is there, that helps out.

Now if you put NO₂ as X, so it’s already there, then it’s 6 times 10^-8. It’s 10^-7, 10 million times slower. And the same if you have trimethyl ammonium with its positive charge. You don’t want to make those cations, if there’s another cation there, or if there’s an NO₂ group there. So those are electron withdrawing, and they retard the formation of cation intermediates.

The most interesting, as many times, is the halogen. The halogen is slower than benzene, but not much slower. Only 30 times slower. And that’s because both things are happening with halogen. It’s withdrawing electrons in the σ bond, but it has an unshared pair that donates back, so that it mostly cancels. Why is NO₂ electron withdrawing when OH is electron donating? O is more electronegative than nitrogen, so you might think that OH should be more electron withdrawing. An unshared pair on nitrogen is higher energy than the unshared pair on oxygen, should be more electron donating. So this seems, at first glance, to be sort of nuts, that OH is 1,000 times faster and NO₂ is 10 million times slower.

But it’s pretty easy to understand when you look at the orbitals that are involved. So in that case of oxygen, there’s this unshared pair on the oxygen, which can overlap with the p orbitals on the benzene ring. It turns out, indeed, that the energy of the oxygen unshared pair is about the same as [gr}pi electrons of a carbon-carbon bond. So that the unshared pair on oxygen has about that energy, about -1 on the scale we’re talking about here. OK, so that’s unusually high, as π are, unusually high. So when you have a vacant orbital next door, those electrons get stabilized. So it’s a high HOMO, a good overlap with phenyl, and therefore OH can donate its unshared pair. OK, that’s good.

How about NO₂? Notice that NO₂ is allylic. It’s got three p orbitals, on each oxygen and on the nitrogen. So it’s going to have a low occupied molecular orbital that gives good overlap with phenyl. So that one is able to donate electrons because it has good overlap, but it’s not willing to share its electrons. Its electrons are very low in energy. They’re not going to be much stabilized by having a vacant orbital next door. So you’re not going to get anything out of that, from the low one.

The next orbital of the NO₂ allylic system has a node in the middle. That one. That one has the energy shown in yellow there, it’s going to be very good. It’s got a good, high HOMO. As high as the p orbital on oxygen, so it should be good at donating electrons, at having its electrons stabilized by the vacant orbital in the benzene ring. But it’s not, why not? Can you see? Why doesn’t this orbital be stabilized by mixing with orbitals that are on the benzene ring? Because there’s no overlap. It doesn’t have any size where it’s adjacent to the benzene ring. So it has no overlap, so it’s willing to give its electrons, but it’s not able because there’s no overlap.

And finally, you have the third of the allylic orbitals, that one. Which is a low LUMO. And it has good overlap with phenyl. So it withdraws electrons, it’s willing to take electrons because it’s unusually low, and it has overlap so it’s able to accept electrons. So that’s why nitro is electron withdrawing when OH is electron donating. it’s because of that node in the middle that makes it unable to give electrons to the benzene ring. So NO₂ is a π accepter, whereas OH is a π donor.

OK now, we just looked at the rates of these reactions, of forming all three products together. But we might be interested in the individual products, ortho, meta, and para. So let’s look at that. In the case of hydrogen, we’ll call each of these 1. And now we’ll do mixtures of hydrogen with others– but now we’ll analyze the ones that already have an X in them– we’ll analyze whether it’s ortho, meta, or para, and how much of each one. Now if it’s methyl, you see that two of them are very fast but one of them is not fast, compared to what it was when only hydrogen is on the ring. How do you understand the two that are fast? Why are those fast when X is methyl? Linda, do you have an idea? When X up here is methyl, this is good, and this is good. 39 and 46. But this one is not any better. Methyl wants to be where the cation is. So it’s especially good here, and especially good here. But not here. There are only hydrogens here, here, and here. And it has the same rate as hydrogen. OK, that’s fine.

If you look at t-butyl, now again it’s fast over here. But it’s not fast here. So why don’t you get this product when you have the R group here, when it’s t-butyl? You get it when you have methyl, but not when it’s t-butyl. Nathan, you got an idea?

Student: Steric hindrance.

Professor J. Michael McBride: It’s steric hindrance, it’s too big. It’s hard to put this extra group here when you have something big there already. OK, so that’s good.

Now there are two things about a substituent, X. Does it make it go faster, or does it make it go slower, is it activating or deactivating? So it’s activating, these are faster than benzene. And how is it directing, what does it direct? Its ortho and para, for the reason we saw. Except that in the t-butyl case, it’s not very much ortho because of steric hindrance, as Nathan told us. OK, so those are electron donating, and like to be ortho/para, other things being equal.

Now here’s one that’s the same size as that neopentyl… or t-butyl group, the carbon with three methyls on it. But notice it’s very much slower. And we saw that before. But it’s selectively slower, it’s slowed much more here, 0.6 times 10^-8, than it is here, 3 times 10^-8. So it’s slow to get product, but when you get product this is the one you get. Meta. You don’t get ortho and para. Especially you don’t get ortho, because it’s sterically hindered. But you see, that’s the same thing. If you want to form these things when X has a plus charge, don’t put it here, and don’t put it here. If you’re going to do it, put it there. Make it as far away as possible. So that’s meta.

And the same thing is true of nitro. It’s slowed down everywhere. But it’s not nearly as slowed down meta as it is here– it’s slowed 10 times more here for ortho and whatever – 300 times, right? Yeah, 300 times slower when it’s para. And if you have an ester group, the same kind of thing is true. Here it’s 10 times slower here than here, and twice slower here than there. So those are deactivating, for the reasons we talked about before, and they’re meta directing.

But again, halogen, is especially interesting. Notice it’s slowed down, it’s deactivating, but it’s ortho/para directing. So it shows both these characters, deactivating and ortho/para. Why? It’s deactivating because of the electron withdrawal, which is σ. But if you’re going to make it, at least make it where the halogen’s unshared pair can be stabilized by the vacant orbital next door.

Chapter 5. Electrophile Activation: Friedel and Crafts [00:46:25]

Now, how can you make the reaction work better, the electrophilic substitution? You can make the electrophile better. So Cl⁺ would be a great nucleophile, but you don’t  buy Cl⁺, you have Cl₂. So what’s the low LUMO here, in Cl₂? I’ve asked this maybe 15 times before in lecture.

Student: σ*.

Professor J. Michael McBride: σ*, OK. But it’s not nearly as low as Cl⁺. One way to make it better is to treat it with a Lewis-acid catalyst, like AlCl₃, which has a vacant orbital on aluminum, which can attack an unshared pair on the chlorine like that. And now you have a minus formal charge on the aluminum and plus on the chlorine. It’s a better leaving group. So now you can do it better. So Cl₂ won’t be nearly as reactive as AlCl₅.

Here, we can compare their LUMOs and see it’s essentially the same LUMO. But it’s lower in the case of AlCl₅, so we can do this reaction and it’ll behave as if it were Cl⁺. If you look at the surface charge, 0.29 positive there—or 29 there, and 38 on the one with AlCl₃. So it’s a better electrophile. So adding a Lewis acid makes these things work better. So the leaving group is then AlCl₄^-, rather than Cl^-. And remember before when we said that the rate of addition to alkenes of bromine is, in some solvents, is second order in bromine. Because that second bromine molecule makes the leaving group Br₃^-, rather than Br^-. It makes the reaction better, exactly the same idea here.

Another way to make the reaction better is to activate the nucleophile. So salicylic acid, which was studied by Kolbe, the same guy who excoriated van’t Hoff about stereochemistry, invented a way to make salicylic acid. So you start with phenol, you react it with CO₂, which has that vacant orbital we talked about, like the one in NO₂⁺. So you could imagine that reaction. Just like NO₂⁺, except that CO₂ isn’t nearly as low as NO₂⁺ with its extra proton. But you can make the nucleophile better. In the previous slide, we saw making the electrophile better. So here we have the unshared pair on oxygen, which helps us out. But we can make it ever so much better by putting base in and making it a higher unshared pair. Now you have a big push to do it, and you get salicylic acid, which where aspirin comes from, from putting CO₂ on phenol.

Now you could imagine the same thing for NH₂. It’s a higher unshared pair than the one on oxygen, normally. So it would do this pushing too, and you could nitrate aniline, aminobenzene, it’s called aniline. And in fact, this was the proposed and, in fact, used in 1944 as a rocket fuel, a mixture of nitric acid and aniline. And there’s a little story I’m going to tell here about a good friend of mine, Dennis Tanner, who was a professor for many years at the University of Alberta. But when he was a post-doc, he was at Stanford. And he’s a very hard worker. And he was working in the lab on the Fourth of July. Not many people were working on the Fourth of July, but several people were, and they passed each other in the hall, and they decided they should have some fun, too.

So they decided that each person would put on a display out in the parking lot. So Dennis had heard about this being a self-igniting rocket fuel, so he thought it would be fun to mix to nitrate aniline, to do the nitration of aniline. So he got concentrated nitric acid and put it in a beaker, and he got aniline and put it in a separatory funnel and hung it above the beaker. And he crouched down and grabbed the thing and twisted it, and started to run. But the instant it hit, before he got more than two paces, there was an enormous fireball that burned off his eyebrows. So indeed, this is a very, very active nitration reaction, as Dennis proved.

But you can tame it. if you put an acetyl group on the aniline, then you lower the energy of the unshared pair. The π* of the carbonyl mixes with it and lowers it. Now you don’t have so much of a push anymore. And now, the nitration still takes place. It’s still ortho/para directing, but it doesn’t burn your eyebrows off. So the acetylation of aniline makes this nitration controllable. So you can start with aniline, react it with acetyl chloride and pyridine, which is often used as a catalyst for putting acetyl groups on, and we’ll talk about that later. And now you can do the nitration. And you see it’s ortho/para directing, not meta directing. So the unshared pair is still doing its work, but not as dramatically, as in Dennis’s case. But then if you really wanted to do the nitration of aniline, now you take the acetyl group back off again. So you have the product, so you tamed the nitration.

Now I just want to introduce the biographical part of this, and we’ll talk about the rest of it next time. So the Friedel-Crafts reaction is one of the most important electrophilic substitution reactions. In fact, it’s one of the most important reactions altogether. And it was named after Charles Friedel and James Mason Crafts. Friedel was a Frenchman. Notice he’s seven years older than Crafts, who was an American. So this is a great example of where America stood in international science in the last part of the 19th century. So these are the publications of James Mason Crafts from 1862, when he was a student, until 1915 when he died.

So some of his papers were on organic chemistry, and they didn’t have Friedel as a co-author. And at different ages, he published a number as a student there, as you’ll see. But then kept publishing. Now here’s his trajectory. He was born in Boston, he studied at… Harvard. But then to learn real science, he went to Europe. First he went to Germany for just a short period. He worked with Bunsen, actually. Then he went to France and worked with Wurtz, the successor of Dumas. And he stayed there for four years. And then he came back, he actually inspected mines in Mexico for a year– and California. Then he went to Cornell– which had just been founded– to be the chemistry professor for a couple of years. Then he went to MIT. And then he was ill, but also tired of not doing research. And he went back to Paris. He went back for just two years, but he stayed for 17 years. And now he was working with his pal, Friedel, whom he’d met as a student in Wurtz’s lab. Then he came back to MIT and, in fact, he was president of MIT for two years around 1900.

Now here are his papers that are not organic chemistry. They’re about thermometry, about physical chemistry. And notice he did most of the publication during this second period in Paris when he was doing only research.

But if you look at his papers with Friedel, there are an enormous number. During this period–when he was in America, he published almost nothing. When he was in Paris, he published a lot, both as a student when he started working with Friedel, and especially after he came back in these 17 years. So he published 100 papers during these years, 99 papers during those years. So it wasn’t possible to do the kind of research in this country that you could do over there. And then in 1877, just after he had gone back to recover and work with his pal Friedel at the School of Mines, they published a really important paper and most of those red publications after that are about this Friedel-Crafts reaction. And they got into it because they had been students together working with Wurtz. And we’ll talk about this then next time.

[end of transcript]

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