Lecture 10. 13C NMR Chemical Shifts
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Nowick, James
This video is part of a 28-lecture graduate-level course titled "Organic Spectroscopy" taught at UC Irvine by Professor James S. Nowick. The course covers infrared (IR) spectroscopy, mass spectrometry, and nuclear magnetic resonance (NMR) spectroscopy, the latter of which is the main focus. Topics covered in the NMR spectroscopy part of the course include chemical shifts, spin-spin coupling, dynamic effects in NMR spectroscopy, and 2D NMR spectroscopy (COSY, HMQC, HMBC, TOCSY, NOESY, ROESY).
 All right. Today we're going to talk about two different things. We talked about proton chemical shifts and today I want to talk briefly about C13 NMR shifts. We'll talk more about them later on. But then we're going to start and we're going to spend a reasonable amount of time talking about spin-spin coupling. And in order to understand this we really have to understand the concept of chemical equivalents which ties into concepts of symmetry and stereochemistry and conformational analysis and it's really beautiful chemical equivalents. And so we'll be talking about chemical equivalents and spin-spin coupling. We're actually going to be spending a good deal of time because there's a lot to understand and as you can already see from the problems people are saying, hey what's going on here? And these problems, these very simple molecules have all sorts of cool issues of spin-spin coupling and all sorts of cool issues of stereochemistry. So we're actually going to spend a number of lectures on them. Next time we're going to develop a concept called magnetic equivalents which is different which is an amplification on chemical equivalents but it's too much to take in in one and then we're going to spend a couple of times talking about details of spin-spin coupling. All right, what do I want to say about carbon NMR spectroscopy first of all? All right, if proton NMR was difficult because you have a very small population between your alpha and beta states, carbon NMR is even worse. And first you know that carbon 12 doesn't have an NMR spectrum, it's not active, it doesn't have a magnetic dipole and we only have 1% or more specifically 1.1% C13. So most of your molecules for small molecules don't contain any C13. For small molecules some of them contain 1C13. Now things get worse, the magnetogyrich ratio for C13 is only a quarter that of the magnetogyrich ratio for proton. Now remember what the implications are of that. That means that you get roughly a quarter, I'll use tilde to say approximately a quarter the Boltzmann difference, say difference in Boltzmann distribution in alpha and beta states. So already we've got fewer nuclei that can flip up. To put it another way, a 500 megahertz spectrometer gives you a carbon spectrum at 127.5 megahertz, a hair over a quarter because the magnetogyrich ratio is a hair over a quarter. Things get worse than that. Your dipole is only a quarter as strong. Guess what? If you want to generate an electric current like a generator, you want a big hunking magnet to spin in a coil. Proton is a little magnet but a carbon is a tiny magnet because it's got a quarter of the magnetic dipole. So you're damned again. And then you further get damned because the procession rate is also a quarter. This for proton at that same 117,500 gauss magnet, you're processing at 500 times per second. For carbon, you're only processing at 125 million times, I say 1,000 million times per second. So that also gives you a quarter as much electricity in the coil. So you've got 1.1% and a quarter of a quarter of a quarter which means you're only 1,500 as sensitive. Question? The lemur of frequency. So if you take a magnet and just spin it in a coil, if you spin it faster, you're going to get more voltage. You get, if you spin it twice as fast, you get twice as much voltage. If you take a magnet that's twice as big, you get twice as much voltage. And so for all of these reasons, and you've got, in addition to a smaller magnet, you've got fewer of them because you've got, even if you give a 90 degree pulse, you get only a quarter of the magnetic dipole from having only a quarter as many nuclei going down into the XY point. Is that a quarter of all, is that a quarter of what, of proton? It's a, of, compared to protons. So carbon is a much less work, a much less sensitive technique than proton and a more spectroscopy. Now there are a few redeeming features. So one thing that's redeeming is we typically do proton decoupling. So normally a carbon would be split by all of the protons. So for example, the carbon in ethanol would be split into a quartet in the carbon in the methyl group of ethanol, would be split into a quartet by the three hydrogens that are attached to it, and then it would be further split by the hydrogens over on the methylene carbon. But what we do is we irradiate the protons. So all the carbon you're going to see, virtually all the carbon you'll see, is called proton decoupled carbon. That flips the spins of the protons rapidly, which means the carbon doesn't see them as spin up or spin down, so your carbons appear as a singlet. So that's good because that means all your carbon signal is gathered in one peak. So that gives you more intensity. Now the other thing is when you do that, so that leads to singlets, which is good, sharper, bigger, and the other thing it leads to is what's called the nuclear overhouser effect. And we'll talk more about this as a technique, but the basic principle of the nuclear overhouser effect is by perturbing the alpha and beta states of the protons, you end up enhancing the difference in Boltzmann population between the alpha and beta states of the carbon. So that too gives you a bigger signal. So all of this leads to a better signal than you would otherwise get in a proton non-decoupled carbon. All right, anyway, suffice it to say nowadays it's easy to collect a carbon NMR spectrum. It typically will take more sample. So you can collect the proton NMR sample on Strick-9 and use a milligram of material or even tenths of a milligram. For a carbon, you might want to put 30 megs in the NMR tube. If you have a chance, you know, you can do it at 10. You could do it at a milligram, but it takes a lot more time. And remember, if you have one milligram versus 10 milligrams, it's going to take 100 times as long to collect the same signal to noise ratio, which means if you're in the research lab and you have some sample, it actually makes sense if you're trying to collect a carbon spectrum to weigh your sample or at least be cognizant of how much you put in your NMR tube because you want to get your spectrum quickly and you want to get good signal to noise ratio. It also makes sense when you're filling your NMR tube to only fill it with the appropriate amount for the coil. The coil on our Brookers is 3.5 centimeters or requires a 3.5 centimeter high sample. That's a half a mil. So if you dissolve your sample, don't dissolve it in a mil and a half. Don't try to be clever and dissolve it in.3 mils because then you mess up the shims on the spectrum because you get flux lines at the end of the sample. So anyway, that's the way to get a good spectrum. All right. I want to talk about where the peaks show up. So carbon NMR spectrum, the carbon NMR spectrum has a big range typically from about zero to about 200 or 200 and change parts per million, 220, 240 parts per million. Allophatic show up at about 10 to 40. So on that big range of 200 or so ppm, that's in the upfield region. Carbons next to an electron with drawing atoms show up downfield, but it's not quite as pronounced. So carbon next to a halogen, you might even sort of see it in this range. Carbon next to a nitrogen, it's going to be sort of at the end of that range. But by the time you're next to an oxygen is an electron with drawing group, I'd say 50 to 70 for a carbon next to one oxygen. So that sort of stands out. Alkines aren't that common, but remember I said there's about two and a half parts per million, 2.2 parts per million maybe for a typical alkyne CH. For an alkyne carbon, it's about 70 to 80 ppm. So that kind of stands out. All right, alkenes and aromatics, whereas in proton NMR, the alkene show up a little more upfield, 5 to 6, and the aromatics a little more downfield, 7 to 8. Alkenes and aromatics are all sort of lumped in at about 110 to 150. Beyond about, and of course these are all typical values. If you put in oxygen on an aromatic like anisol or phenol where you put in oxygen directly on a carbon, that carbon might be at 160 parts per million. If you have a very electron, we'll talk more about specific shifts, but you could easily have, if you have high electron donations, say an ortho carbon, you could easily be below 110 ppm. All right, carbonyls, esters, and carboxylic acids, you get a little bit of a resonance effect so they show up a little bit less for a downfield than some of the others. Let's say about 170 to 180 parts per million. Aldehydes show up a little further downfield, let's say about 190 to 200 ppm. And ketones, I'll say R-C-O-R prime for ketones, let's say about 205 to about 220 ppm. I just want to give you one little handout. By the way, if you ever miss your handouts or misplace them, I put them up on the web with the video part of the course so you can always go ahead and download the handouts. All right, so just as I had my, does anyone else need one? Just as I had my little pigeon drawing of my take of what to look at in reading a proton NMR spectrum, I have my little pigeon drawing of what to take, what to look at when reading a C13 NMR spectrum. In other words, this type of region is aliphatics, here you have a carbon next to a significantly withdrawing group like an oxygen, here's your alkene aromatics, here's your esters and acids, then you go to aldehydes and ketones. So to a large extent, it's sort of like H1NMR but with maybe a factor of 20 on the scale. In other words, most of what you're going to see in a proton NMR is from 0 to 10, most of what you're going to see in a carbon NMR is from 0 to 200, carboxylic acids will be further out, ketones will be further out but that kind of gives you my read on things. All right, this should give you the basics to start to use carbon NMR in helping to analyze some of the homework sets that we're going to get. So one thing that you can get is a reading of things like carbonyl groups in there and another thing you'll be able to do is count up and see how many different types of atoms because whereas in proton NMR you may have overlapping resonances, most often because the carbon spectrum is more dispersed and peaks typically show up as singlets, most often you will be able to see one peak for each type of carbon. All right, I want to talk now about spin-spin coupling. We're splitting. And if I had to give you a very, very general way of thinking about it, the way I'll describe it and we're going to amplify on this in today's talk, the way that I would describe it is that protons that peaks are split by adjacent protons that are different. And we'll later on amplify on the concept of adjacent talking about two bond coupling, three bond coupling and long range coupling. But today what I'd like to amplify on is the concept of same and specifically different. So let's start with an example that's very, very, very intuitive, very, very freshman, sophomore rather. Chemistry, let's take the H1NMR spectrum of chloroethane. And so my little pigeon sketch of the spectrum of chloroethane would look something like this. We have a 1 to 3 to 3 to 1 triplet, somewhere downfield of 3 ppm or just a hair downfield of 3 ppm. And then we have a, I'm sorry, 1 to 3 to 3 to 1 quartet. And then somewhere just a hair downfield of 1, we have a 1 to 2 to 1 triplet. The triplet comes, of course, from the CH3. And I can say that in this particular example, CH3s are split by the adjacent CH2s. And I can say because the three hydrogens of, well, let's talk about the CH2s, the CH2s are split by the CH3. The three protons of the CH3 group are the same. So the CH3 does not split each other. So I'll say the CH3 protons do not split each other. And then I'll say in this particular case, the CH2s do not split each other. In other words, in this particular case, the two hydrogens of the methylene group are the same so they don't split each other. But more in general, I'll give this as an exception. So if your compound has a chiral center in it or the protons or otherwise diastereotopic, then they will split each other. And so this except is something that we're going to be playing with in today's lecture. Thank you. All right, before we come to this very important concept of diastereotepicity, let's tackle this basic notion of the 1 to 2 to 1 triplet and the 1 to 3 to 3 to 1 quartet. So we have a triplet and it's 1 to 2 to 1. It comes from the CH3 and it sees the CH2 and the 2 protons act as little magnets in each molecule. And those little magnets can be spin up or one can be spin up and one can be spin down or they both can be spin down. So you get three lines. There's one way for them both to be spin up. There are two ways for them to be one spin up and one spin down and there's one way for them to be spin down. And so that gives rise to our 1 to 2 to 1 triplet. For our quartet, we have a 1 to 3 to 3 to 1 ratio. The quartet of course comes from the CH2. The CH2 sees the CH3 and all of the protons can be spin up and some of the molecules there are. Or two can be spin up and one can be spin down and there are three different ways that that can occur. Or two can be spin down and one can be spin up. And again, there are three ways that can occur. Or they all can be spin down. And we can generalize this idea to say if there are in equal and I'm going to underline equal couplings lead to in plus 1 lines. So if there are four equal couplings, you get five lines, those five lines, it follows Pascal's triangle. You can work out the statistics yourself or you can say that those five lines end up being in a 1 to 4 to 6 to 4 to 1 ratio and so forth. You basically add the ones above or you work out the statistics. So for example, if we go to isopropyl chloride, now the CH here is split equally by three methyl groups. So the CH appears as a subtet and the ratio of the lines are 1 to 6 to 15 to 20 to 15 to 6 to 1. And because the quartet, because the septet is going to be very small compared to triplets and quartets in the molecule, unless you look hard, you might not see these lines on the outside. And in fact, by the time you get up to very big multiplets like septets and so forth, octets, often people will just report them as a multiplet in reporting an NMR spectrum. All right, let's take a case, a look at a case of what I mean by couplings that may or may not be equal. So imagine the methane group in veiling. And specifically this methane group. So this methane group is split by two methyl groups. And it's also split by this, this methane group. And so depending on whether the couplings are equal, that is whether the coupling constants are equal, it may be an octet. Or it may be more complicated. And I can tell you from experience, in this case, it would be more complicated. It would be in this case a doublet of septets, which you would probably report as a multiplet. The OH of alcohols, and I know this has already come up on the homework. The OH of the alcohol may or may not couple. So the hydrogen of isopropanol, the methane of isopropanol might appear as a septet, or it might be more complicated. In order to couple, this hydrogen has to stick around in one place. It has to stay attached to that proton for a time that's roughly one over the coupling constant. In other words, it has to stay attached since coupling constants are typically on the order of about 7 hertz. It has to stay attached for hundreds of milliseconds or longer. If that hydrogen exchanges on the order of tens of milliseconds, this methane is not going to see it as either spin up or spin down. It will see an average, and you won't get coupling. If this hydrogen doesn't exchange, then we'll be in the situation here, and it will see it either as spin up or as spin down, and you will have splitting either as an octet or as a doublet of septets, or septet of doublets depending on the relative J's, but we'll get into that later. Typically, primary alcohols often exchange rapidly. So an alcohol at the end of a CH2 chain, like you may have in the homework problem, usually will exchange rapidly. Secondary alcohols sometimes do, sometimes don't. They're more sterically hindered. The rate of exchange is going to depend on a number of factors, including the concentration of the sample because the molecules can exchange by colliding. It will depend on the amount of water in the sample. There's invariably adventitious water in your NMR tube at about 20 millimolar concentration, 10, 20, 30 millimolar concentration, and chloroform undergoes photo oxidation to give hydrochloric acid, or DCL in the case of CDCl3, and that will rapidly promote exchange. So if I want to minimize exchange, I typically will pass my chloroform through alumina, not my sample, but my chloroform through alumina that I've dried over a flame. That will take out the acid and minimize the amount of water in there. Okay, so alcohols are kind of a wild card on J coupling. So the proton is exchanging. Do you still see that proton shifts? Ah, do you still see the proton? In general, if the proton is exchanging with other molecules, you will see it, and it will be there. If the proton is exchanging with water, and you have a little bit of water in the sample, you will see it, it and the water peak at the weighted average of the two. If it is exchanging with water, and there's a lot of water, then it will become part of the water peak. Also, there are two different timescales. One is the timescale for coupling. The other is the timescale for being able to see a peak. Remember I talked about the uncertainty principle before? So in order to see coupling, your line width has to be, your exchange rate has to be, you have to be able to make the observation such that your line width is on the order of better than 7 hertz. In order to not have swapping with water, which is very far away, so water is typically 1.6 parts per million in chloroform. Remember I said an alcohol can be 1 to 5 parts per million. Let's take 4 ppm. Those are hundreds or even thousands of hertz away, right? A 4 ppm difference is 2,000 hertz. So in order to not have exchange with water where you can see the alcohol peak, it has to stay attached for tens of milliseconds rather than hundreds of milliseconds. Carboxylic acids are real bad boys in this regard because they do exchange rapidly, which is one of the reasons the peak typically broadens out in chloroform. Alcohols often will stay attached. Secondary amines are a big pain in the ass. Secondary amines are almost impossible to see. You have base catalyzed exchange mechanisms. Primary amines tend to be a little bit better behaved. Aromatic amines are well behaved, so I'm talking aliphatic amines. But if you're working on an alkaloid project or some other project and you have a secondary amine, you really are in trouble. So you said earlier, like, speed, that I didn't have the carbon adjacent to the alcohol layer may not couple. If that does not couple, can you also assume that the proton on the alcohol will not couple as well? Ex- you mean, yes, it's you. If one doesn't couple, the other one will. Absolutely. And coupling is mutual. Now, you could envision a circumstance where you had J coupling but quadrupolar broadening. So you could have a hydrogen on a nitrogen that could be broadened to a point where it was a broad singlet, but the hydrogen on the carbon could be split by it. And there the broadening would be quadrupolar. But yes, typically. So again, let's talk in rules rather than exceptions now. Typically, it's going to be seen both ways. Sometimes what will happen is you may, for example, if you have a very complex coupling pattern, if you do have coupling, you may just see this guy as a complex multiplet and then see this guy as a doublet. Why? Because this guy is splitting this guy, but it's such complex splitting that you can't actually discern what the splitting pattern is. On the other hand, it's the same coupling constant, but you only have one coupling and you split into a doublet. Other questions. These are good. These are really important. This is really understanding the real stuff you're going to see in the trenches. What about the methyl groups? The methyl groups won't couple to this. Not will not because it's far away. In general, coupling is going to be through two or three bonds, although on the case of double bonds and triple bonds, you can get four bond couplings. There's a very good appendix in the back of your book. I forget whether it's appendix A or appendix F at the back of Silverstein. What was the appendix? Somebody looked it up. Anyway, that is a very good place to start. But in general, unless you have long-range coupling, we'll get to that in a lecture or two, F. Unless you have long-range coupling, in general, coupling is going to be through two or three bonds. All right. Let us now tackle the concept of chemical equivalence. Chemical equivalence is the first level of sameness that I was talking about. Remember I said that protons only couple to other protons if they are different? Two protons are the same, are chemically equivalent, if they exchange by a symmetry operate, can be exchanged by a symmetry operation or rapid process. For example, rapid rotation about a bond. All right. The other caveat, and this ties into something I said last time, virtually all, you can say all rotations about SP3, SP3 carbon bonds are rapid at room temperature. I'll say virtually always rapid at room temperature. That's a hint. If you see two hydrogens splitting each other, and they're both on SP3 carbons attached to each other, chances are you need to think deeper. And it's not, oh, there's some slow process. All right. One corollary to this is chemically equivalent protons have the same chemical shift. So this thing here is kind of like the golden rule of splitting, you know, the basic do unto others as you want others to do unto you. If you keep this in mind, you're going to be very, very well set on all of the details. All right. Let's talk about specific way of thinking about what's the same and what's different. One way that I like to do it, you can do it by a couple of ways. If you're good with symmetry, you can just go ahead and say, all right, do we have a symmetry operation that interchanges these two hydrogens, say in diphenylmethane? You can say they interchange by reflection and therefore they are chemically equivalent. But we're going to get into some cases that at least the first time you see them are tricky. And another way to do it is just to perform a little thought experiment and say, all right, what is their topological relationship to each other? And the way you do this thought experiment is you replace one proton by a deuterium and the other proton by a deuterium and then you ask yourself, what is the relationship between these two molecules? And in the case of this particular thought experiment, these two are the same. And since they're the same, topologically we call those two protons homotopic and homotopic protons are chemically equivalent. Let's try another molecule. Obviously, obviously, obviously these are trivial. So let's take instead of diphenylmethane, let's take ethylbenzene. And again, we'll consider the methylene group. The two hydrogen groups of the methylene group still exchange by reflection. So they meet the criterion there. They're chemically equivalent. Yeah? You can't tell them apart. So they have the same chemical shift. And two protons that are chemically equivalent are the same and don't split each other. So chemically equivalent protons don't split each other. Now, later on we're going to talk about a concept called magnetic equivalents. And what you'll see is that although chemically equivalent protons, well actually I'm going to hold off on this statement here. Right now I'm just going to leave it as there are two types of sameness. And chemically equivalent protons don't split each other if they're also magnetically equivalent. But we're going to get to that later. For now, the simple rule is chemically equivalent protons are the same in a certain way. All right, let's try our little gadanke experiment here. We replace one by a deuterium. And we replace the other by a deuterium. And now these two molecules, are they the same? No, what are they? Enantiomers. Enantiomers. So now they are enantiomers, which means topologically the protons are enantiotopic. And enantiotopic protons are also chemically equivalent. All right, so now let's come to a molecule where we have a chiral center in the molecule. And since we've been building on this idea of phenyl groups, I'm going to give us the molecule phenylalanine. Doesn't matter whether I draw it racemic, which means we have two enantiomers or whether I have one molecule. Okay, these two protons are now no longer exchangeable by reflection. And no matter how rapidly you rotate, they still are not the same. So these two protons now, the methylene protons, are not chemically equivalent. And let's try our thought experiment here again, because sometimes things can get more complicated than you might anticipate. And so we'll take one, replace it with a deuterium, and we'll take the other and replace it by a deuterium. And we'll ask ourself what the relationship of those two structures is to each other. What are they? They're diastereomers. So we say that the two hydrogens of the methylene group are diastereotopic. And therefore they are not chemically equivalent. All right, so here's a spectrum of phenylalanine, so we can see what's going on. This is a spectrum taken in D2O with DCL to dissolve it, deuterohCl to dissolve it. So all of the OHs and the NHs have exchanged with deuterium and are exchanging rapidly, and for all intents and purposes, don't J couple. So we see a peak for our HOD, and then we see the rest of it are carbons attached to hydrogen. We see our phenyl group over here in the 7 to 8 range. We see our proton connected to the alpha carbon over here, just a hair down field at about 4, about 4.3 parts per million. And then we see our beta protons, the ones that are next to the phenyl, and we see them as two peaks. Each of the beta protons appears at, and these are expansions here, each of the beta protons appears at distinct positions. They're not the same. They're not chemically equivalent. They could be coincident, meaning appearing at the same position, but they are not, and so they appear at different positions. And what's more, each of them is a doublet of doublets, or DD. Why? Because each of the protons splits the other one and is split by the alpha. So we have both a J2HH. That's a two bond coupling since you go one, two bond to get from these two beta protons to each other. And we have a three bond coupling between the alpha proton and each of the beta protons. And the alpha proton is also a doublet of doublets. So each proton is a doublet of doublets. The alpha proton is a doublet of doublets because it's split by the two beta protons, and it's split with different coupling constants. The beta protons are coupling to the alpha proton, probably with about six and eight hertz coupling constants, and we'll be talking more about these in a moment or in a future lecture. And the beta protons are coupling to each other with about a 13 or 14 hertz coupling constant. All right. Let's talk about why we have to keep our heads attached to us. So I'm going to give us another example. The example I'll give us is a molecule called acetal. Acetal is acetaldehyde diethylacetal. All right. So what do we have here? We have a couple of ethyl groups, and we have a methyl group. The CH3s of the ethyl group appear here as a triplet, so this is OCH2 CH3. That looks pretty reasonable. We have another methyl group. It appears as a doublet. That's our CHCH3. That looks pretty reasonable. We have our methine. It's the one that's attached to two oxygens, so it's shifted down field. It's at 4.6 parts per million. It's a quartet because it's split by the methyl group. And finally, we come to the CH2s. And the CH2s show up as two peaks. And it takes a moment to wrap your head around this. And to wrap your head around this, what you have to realize is that these two protons are not chemically equivalent. You can think of this a couple of ways. They don't exchange by symmetry operation. If you reflect through the plane, you don't get the same thing. Another way to think about it, and I think the easiest way to keep your head on straight is to think about it rigorously. You imagine replacing one of them with a deuterium, replacing the other of them with a deuterium. And the molecules that you get are diastereomers. In other words, these two protons are diastereotopic. Okay. We talked about in IR how if you have like an internal outcheam, sometimes you don't see the scratching because there's like that pseudo symmetry. Do you see the same thing with this that if you have like a stereosenter or like a stereocenter, but it's like really far away? Absolutely. Great question. Just because two protons are topologically diastereotopic doesn't mean that they won't be coincident. And in general, the further that they are from the stereocenter, the more likely they are to be coincident and behave as if they are, clearly. Here we see them as well-defined peaks at very different positions. They're both doublets of quartets. That's a DQ. Oops. DQ. We'll talk more about the splitting pattern, but they're being split with one big geminal coupling of about 10 hertz, one two bond coupling. And with three visceral couplings, three three bond couplings of about 7 hertz. All right, last thing I want to show you, one last handout while we're on the concept of chemical equivalence. So here we have the molecule three methyl two butanol. And what I want to point out to you is in three methyl two butanol, the two methyls are not the same. They're diastereotopic. You can think of this a couple of ways. One way is to think about, remember the OH constitutes a chiral center, and you can do a little thought experiment and say all right, I'll envision replacing one of the methyl groups or the other methyl group with a deuteromethyl group. Or the other way is just to think about the topology. But when you look at it, you say all right, how many peaks do we see in the C13 NMR? One, two, three, four, five. There are five types of carbon because there are five non-chemically equivalent carbons. If you look at the proton NMR, it's a little confusing at first until you think about what's going on. It's not hard to realize that this methyl group that's beta to an oxygen gives rise to this doublet. Now we have two more methyl groups in the molecule, and each of those methyl groups is next to a hydrogen. Each of those methyl groups appears as a doublet, but the doublets are very close. In other words, we have a doublet and we have another doublet and so it looks like the triplet. Can you discern that because it's not the 1 to 3 along? What's that? Can you discern that because? Well, you can discern it if you look carefully because it's not 1 to 3, 1 to 2 to 1, but it's confusing. So it's two doublets, but let me tell you a secret. This is a 300 megahertz spectrum. Remember I said the coupling constant is fixed in hertz. The position is fixed in PPM. So my two doublets would look like this at 300 and it would look like a triplet. If I went to 500 megahertz, the two doublets would now separate and I'd see a doublet and a doublet and that would be one way you can prove it. All right, I think that's what I want to say for today about chemical equivalents. We'll pick up next time talking about magnetic equivalents.
 All right.  Today we're going to talk about two different things.  We talked about proton chemical shifts and today I want to talk briefly about C13 NMR  shifts.  We'll talk more about them later on.  But then we're going to start and we're going to spend a reasonable amount of time talking  about spin-spin coupling.  And in order to understand this we really have to understand the concept of chemical equivalents  which ties into concepts of symmetry and stereochemistry and conformational analysis and it's really  beautiful chemical equivalents.  And so we'll be talking about chemical equivalents and spin-spin coupling.  We're actually going to be spending a good deal of time because there's a lot to understand  and as you can already see from the problems people are saying, hey what's going on here?  And these problems, these very simple molecules have all sorts of cool issues of spin-spin  coupling and all sorts of cool issues of stereochemistry.  So we're actually going to spend a number of lectures on them.  Next time we're going to develop a concept called magnetic equivalents which is different  which is an amplification on chemical equivalents but it's too much to take in in one and then  we're going to spend a couple of times talking about details of spin-spin coupling.  All right, what do I want to say about carbon NMR spectroscopy first of all?  All right, if proton NMR was difficult because you have a very small population between your  alpha and beta states, carbon NMR is even worse.  And first you know that carbon 12 doesn't have an NMR spectrum, it's not active, it  doesn't have a magnetic dipole and we only have 1% or more specifically 1.1% C13.  So most of your molecules for small molecules don't contain any C13.  For small molecules some of them contain 1C13.  Now things get worse, the magnetogyrich ratio for C13 is only a quarter that of the magnetogyrich  ratio for proton.  Now remember what the implications are of that.  That means that you get roughly a quarter, I'll use tilde to say approximately a quarter  the Boltzmann difference, say difference in Boltzmann distribution in alpha and beta states.  So already we've got fewer nuclei that can flip up.  To put it another way, a 500 megahertz spectrometer gives you a carbon spectrum at 127.5 megahertz,  a hair over a quarter because the magnetogyrich ratio is a hair over a quarter.  Things get worse than that.  Your dipole is only a quarter as strong.  Guess what?  If you want to generate an electric current like a generator, you want a big hunking magnet  to spin in a coil.  Proton is a little magnet but a carbon is a tiny magnet because it's got a quarter of  the magnetic dipole.  So you're damned again.  And then you further get damned because the procession rate is also a quarter.  This for proton at that same 117,500 gauss magnet, you're processing at 500 times per  second.  For carbon, you're only processing at 125 million times, I say 1,000 million times per  second.  So that also gives you a quarter as much electricity in the coil.  So you've got 1.1% and a quarter of a quarter of a quarter which means you're only 1,500  as sensitive.  Question?  The lemur of frequency.  So if you take a magnet and just spin it in a coil, if you spin it faster, you're going  to get more voltage.  You get, if you spin it twice as fast, you get twice as much voltage.  If you take a magnet that's twice as big, you get twice as much voltage.  And so for all of these reasons, and you've got, in addition to a smaller magnet, you've  got fewer of them because you've got, even if you give a 90 degree pulse, you get only  a quarter of the magnetic dipole from having only a quarter as many nuclei going down into  the XY point.  Is that a quarter of all, is that a quarter of what, of proton?  It's a, of, compared to protons.  So carbon is a much less work, a much less sensitive technique than proton and a more  spectroscopy.  Now there are a few redeeming features.  So one thing that's redeeming is we typically do proton decoupling.  So normally a carbon would be split by all of the protons.  So for example, the carbon in ethanol would be split into a quartet in the carbon in the  methyl group of ethanol, would be split into a quartet by the three hydrogens that are  attached to it, and then it would be further split by the hydrogens over on the methylene  carbon.  But what we do is we irradiate the protons.  So all the carbon you're going to see, virtually all the carbon you'll see, is called proton  decoupled carbon.  That flips the spins of the protons rapidly, which means the carbon doesn't see them as  spin up or spin down, so your carbons appear as a singlet.  So that's good because that means all your carbon signal is gathered in one peak.  So that gives you more intensity.  Now the other thing is when you do that, so that leads to singlets, which is good, sharper,  bigger, and the other thing it leads to is what's called the nuclear overhouser effect.  And we'll talk more about this as a technique, but the basic principle of the nuclear overhouser  effect is by perturbing the alpha and beta states of the protons, you end up enhancing  the difference in Boltzmann population between the alpha and beta states of the carbon.  So that too gives you a bigger signal.  So all of this leads to a better signal than you would otherwise get in a proton non-decoupled  carbon.  All right, anyway, suffice it to say nowadays it's easy to collect a carbon NMR spectrum.  It typically will take more sample.  So you can collect the proton NMR sample on Strick-9 and use a milligram of material or  even tenths of a milligram.  For a carbon, you might want to put 30 megs in the NMR tube.  If you have a chance, you know, you can do it at 10.  You could do it at a milligram, but it takes a lot more time.  And remember, if you have one milligram versus 10 milligrams, it's going to take 100 times  as long to collect the same signal to noise ratio, which means if you're in the research  lab and you have some sample, it actually makes sense if you're trying to collect a  carbon spectrum to weigh your sample or at least be cognizant of how much you put in  your NMR tube because you want to get your spectrum quickly and you want to get good  signal to noise ratio.  It also makes sense when you're filling your NMR tube to only fill it with the appropriate  amount for the coil.  The coil on our Brookers is 3.5 centimeters or requires a 3.5 centimeter high sample.  That's a half a mil.  So if you dissolve your sample, don't dissolve it in a mil and a half.  Don't try to be clever and dissolve it in.3 mils because then you mess up the shims on  the spectrum because you get flux lines at the end of the sample.  So anyway, that's the way to get a good spectrum.  All right.  I want to talk about where the peaks show up.  So carbon NMR spectrum, the carbon NMR spectrum has a big range typically from about zero  to about 200 or 200 and change parts per million, 220, 240 parts per million.  Allophatic show up at about 10 to 40.  So on that big range of 200 or so ppm, that's in the upfield region.  Carbons next to an electron with drawing atoms show up downfield, but it's not quite as pronounced.  So carbon next to a halogen, you might even sort of see it in this range.  Carbon next to a nitrogen, it's going to be sort of at the end of that range.  But by the time you're next to an oxygen is an electron with drawing group, I'd say  50 to 70 for a carbon next to one oxygen.  So that sort of stands out.  Alkines aren't that common, but remember I said there's about two and a half parts per  million, 2.2 parts per million maybe for a typical alkyne CH.  For an alkyne carbon, it's about 70 to 80 ppm.  So that kind of stands out.  All right, alkenes and aromatics, whereas in proton NMR, the alkene show up a little  more upfield, 5 to 6, and the aromatics a little more downfield, 7 to 8.  Alkenes and aromatics are all sort of lumped in at about 110 to 150.  Beyond about, and of course these are all typical values.  If you put in oxygen on an aromatic like anisol or phenol where you put in oxygen directly  on a carbon, that carbon might be at 160 parts per million.  If you have a very electron, we'll talk more about specific shifts, but you could easily  have, if you have high electron donations, say an ortho carbon, you could easily be below  110 ppm.  All right, carbonyls, esters, and carboxylic acids, you get a little bit of a resonance  effect so they show up a little bit less for a downfield than some of the others.  Let's say about 170 to 180 parts per million.  Aldehydes show up a little further downfield, let's say about 190 to 200 ppm.  And ketones, I'll say R-C-O-R prime for ketones, let's say about 205 to about 220 ppm.  I just want to give you one little handout.  By the way, if you ever miss your handouts or misplace them, I put them up on the web  with the video part of the course so you can always go ahead and download the handouts.  All right, so just as I had my, does anyone else need one?  Just as I had my little pigeon drawing of my take of what to look at in reading a proton  NMR spectrum, I have my little pigeon drawing of what to take, what to look at when reading  a C13 NMR spectrum.  In other words, this type of region is aliphatics, here you have a carbon next to a significantly  withdrawing group like an oxygen, here's your alkene aromatics, here's your esters and acids,  then you go to aldehydes and ketones.  So to a large extent, it's sort of like H1NMR but with maybe a factor of 20 on the scale.  In other words, most of what you're going to see in a proton NMR is from 0 to 10, most  of what you're going to see in a carbon NMR is from 0 to 200, carboxylic acids will be  further out, ketones will be further out but that kind of gives you my read on things.  All right, this should give you the basics to start to use carbon NMR in helping to analyze  some of the homework sets that we're going to get.  So one thing that you can get is a reading of things like carbonyl groups in there and  another thing you'll be able to do is count up and see how many different types of atoms  because whereas in proton NMR you may have overlapping resonances, most often because  the carbon spectrum is more dispersed and peaks typically show up as singlets, most  often you will be able to see one peak for each type of carbon.  All right, I want to talk now about spin-spin coupling.  We're splitting.  And if I had to give you a very, very general way of thinking about it, the way I'll describe  it and we're going to amplify on this in today's talk, the way that I would describe it is that  protons that peaks are split by adjacent protons that are different.  And we'll later on amplify on the concept of adjacent talking about two bond coupling,  three bond coupling and long range coupling.  But today what I'd like to amplify on is the concept of same and specifically different.  So let's start with an example that's very, very, very intuitive, very, very freshman,  sophomore rather.  Chemistry, let's take the H1NMR spectrum of chloroethane.  And so my little pigeon sketch of the spectrum of chloroethane would look something like  this.  We have a 1 to 3 to 3 to 1 triplet, somewhere downfield of 3 ppm or just a hair downfield  of 3 ppm.  And then we have a, I'm sorry, 1 to 3 to 3 to 1 quartet.  And then somewhere just a hair downfield of 1, we have a 1 to 2 to 1 triplet.  The triplet comes, of course, from the CH3.  And I can say that in this particular example, CH3s are split by the adjacent CH2s.  And I can say because the three hydrogens of, well, let's talk about the CH2s, the CH2s  are split by the CH3.  The three protons of the CH3 group are the same.  So the CH3 does not split each other.  So I'll say the CH3 protons do not split each other.  And then I'll say in this particular case, the CH2s do not split each other.  In other words, in this particular case, the two hydrogens of the methylene group are the  same so they don't split each other.  But more in general, I'll give this as an exception.  So if your compound has a chiral center in it or the protons or otherwise diastereotopic,  then they will split each other.  And so this except is something that we're going to be playing with in today's lecture.  Thank you.  All right, before we come to this very important concept  of diastereotepicity, let's tackle this basic notion  of the 1 to 2 to 1 triplet and the 1 to 3 to 3 to 1 quartet.  So we have a triplet and it's 1 to 2 to 1.  It comes from the CH3 and it sees the CH2 and the 2 protons act  as little magnets in each molecule.  And those little magnets can be spin up or one can be spin up  and one can be spin down or they both can be spin down.  So you get three lines.  There's one way for them both to be spin up.  There are two ways for them to be one spin up and one spin down  and there's one way for them to be spin down.  And so that gives rise to our 1 to 2 to 1 triplet.  For our quartet, we have a 1 to 3 to 3 to 1 ratio.  The quartet of course comes from the CH2.  The CH2 sees the CH3 and all of the protons can be spin up  and some of the molecules there are.  Or two can be spin up and one can be spin down  and there are three different ways that that can occur.  Or two can be spin down and one can be spin up.  And again, there are three ways that can occur.  Or they all can be spin down.  And we can generalize this idea to say if there are in equal  and I'm going to underline equal couplings lead to in plus 1 lines.  So if there are four equal couplings, you get five lines,  those five lines, it follows Pascal's triangle.  You can work out the statistics yourself or you can say  that those five lines end up being in a 1 to 4 to 6 to 4  to 1 ratio and so forth.  You basically add the ones above or you work out the statistics.  So for example, if we go to isopropyl chloride,  now the CH here is split equally by three methyl groups.  So the CH appears as a subtet and the ratio of the lines are 1  to 6 to 15 to 20 to 15 to 6 to 1.  And because the quartet, because the septet is going  to be very small compared to triplets and quartets  in the molecule, unless you look hard,  you might not see these lines on the outside.  And in fact, by the time you get up to very big multiplets  like septets and so forth, octets,  often people will just report them as a multiplet  in reporting an NMR spectrum.  All right, let's take a case, a look at a case of what I mean  by couplings that may or may not be equal.  So imagine the methane group in veiling.  And specifically this methane group.  So this methane group is split by two methyl groups.  And it's also split by this, this methane group.  And so depending on whether the couplings are equal,  that is whether the coupling constants are equal,  it may be an octet.  Or it may be more complicated.  And I can tell you from experience, in this case,  it would be more complicated.  It would be in this case a doublet of septets,  which you would probably report as a multiplet.  The OH of alcohols, and I know this has already come  up on the homework.  The OH of the alcohol may or may not couple.  So the hydrogen of isopropanol, the methane  of isopropanol might appear as a septet,  or it might be more complicated.  In order to couple, this hydrogen has to stick  around in one place.  It has to stay attached to that proton for a time  that's roughly one over the coupling constant.  In other words, it has to stay attached  since coupling constants are typically  on the order of about 7 hertz.  It has to stay attached for hundreds  of milliseconds or longer.  If that hydrogen exchanges on the order of tens  of milliseconds, this methane is not going to see it  as either spin up or spin down.  It will see an average, and you won't get coupling.  If this hydrogen doesn't exchange, then we'll be  in the situation here, and it will see it either as spin  up or as spin down, and you will have splitting either  as an octet or as a doublet of septets,  or septet of doublets depending on the relative J's,  but we'll get into that later.  Typically, primary alcohols often exchange rapidly.  So an alcohol at the end of a CH2 chain,  like you may have in the homework problem,  usually will exchange rapidly.  Secondary alcohols sometimes do, sometimes don't.  They're more sterically hindered.  The rate of exchange is going to depend  on a number of factors, including the concentration  of the sample because the molecules can exchange  by colliding.  It will depend on the amount of water in the sample.  There's invariably adventitious water in your NMR tube  at about 20 millimolar concentration,  10, 20, 30 millimolar concentration,  and chloroform undergoes photo oxidation  to give hydrochloric acid, or DCL in the case of CDCl3,  and that will rapidly promote exchange.  So if I want to minimize exchange,  I typically will pass my chloroform through alumina,  not my sample, but my chloroform through alumina  that I've dried over a flame.  That will take out the acid and minimize the amount  of water in there.  Okay, so alcohols are kind of a wild card on J coupling.  So the proton is exchanging.  Do you still see that proton shifts?  Ah, do you still see the proton?  In general, if the proton is exchanging  with other molecules, you will see it, and it will be there.  If the proton is exchanging with water, and you have a little  bit of water in the sample, you will see it,  it and the water peak at the weighted average of the two.  If it is exchanging with water, and there's a lot of water,  then it will become part of the water peak.  Also, there are two different timescales.  One is the timescale for coupling.  The other is the timescale for being able to see a peak.  Remember I talked about the uncertainty principle before?  So in order to see coupling, your line width has to be,  your exchange rate has to be, you have to be able  to make the observation such that your line width is  on the order of better than 7 hertz.  In order to not have swapping with water,  which is very far away, so water is typically 1.6 parts per  million in chloroform.  Remember I said an alcohol can be 1 to 5 parts per million.  Let's take 4 ppm.  Those are hundreds or even thousands of hertz away, right?  A 4 ppm difference is 2,000 hertz.  So in order to not have exchange with water  where you can see the alcohol peak, it has to stay attached  for tens of milliseconds rather than hundreds of milliseconds.  Carboxylic acids are real bad boys in this regard  because they do exchange rapidly, which is one  of the reasons the peak typically broadens out in chloroform.  Alcohols often will stay attached.  Secondary amines are a big pain in the ass.  Secondary amines are almost impossible to see.  You have base catalyzed exchange mechanisms.  Primary amines tend to be a little bit better behaved.  Aromatic amines are well behaved,  so I'm talking aliphatic amines.  But if you're working on an alkaloid project  or some other project and you have a secondary amine,  you really are in trouble.  So you said earlier, like, speed, that I didn't have  the carbon adjacent to the alcohol layer may not couple.  If that does not couple, can you also assume that the proton  on the alcohol will not couple as well?  Ex- you mean, yes, it's you.  If one doesn't couple, the other one will.  Absolutely.  And coupling is mutual.  Now, you could envision a circumstance  where you had J coupling but quadrupolar broadening.  So you could have a hydrogen on a nitrogen that could be broadened  to a point where it was a broad singlet, but the hydrogen  on the carbon could be split by it.  And there the broadening would be quadrupolar.  But yes, typically.  So again, let's talk in rules rather than exceptions now.  Typically, it's going to be seen both ways.  Sometimes what will happen is you may, for example,  if you have a very complex coupling pattern,  if you do have coupling, you may just see this guy  as a complex multiplet and then see this guy as a doublet.  Why? Because this guy is splitting this guy,  but it's such complex splitting  that you can't actually discern what the splitting pattern is.  On the other hand, it's the same coupling constant,  but you only have one coupling and you split into a doublet.  Other questions.  These are good.  These are really important.  This is really understanding the real stuff you're going  to see in the trenches.  What about the methyl groups?  The methyl groups won't couple to this.  Not will not because it's far away.  In general, coupling is going to be through two  or three bonds, although on the case of double bonds  and triple bonds, you can get four bond couplings.  There's a very good appendix in the back of your book.  I forget whether it's appendix A or appendix F at the back  of Silverstein.  What was the appendix?  Somebody looked it up.  Anyway, that is a very good place to start.  But in general, unless you have long-range coupling,  we'll get to that in a lecture or two, F.  Unless you have long-range coupling,  in general, coupling is going to be through two or three bonds.  All right.  Let us now tackle the concept of chemical equivalence.  Chemical equivalence is the first level of sameness  that I was talking about.  Remember I said that protons only couple  to other protons if they are different?  Two protons are the same, are chemically equivalent,  if they exchange by a symmetry operate,  can be exchanged by a symmetry operation or rapid process.  For example, rapid rotation about a bond.  All right.  The other caveat, and this ties  into something I said last time, virtually all,  you can say all rotations about SP3,  SP3 carbon bonds are rapid at room temperature.  I'll say virtually always rapid at room temperature.  That's a hint.  If you see two hydrogens splitting each other,  and they're both on SP3 carbons attached to each other,  chances are you need to think deeper.  And it's not, oh, there's some slow process.  All right.  One corollary to this is chemically equivalent protons  have the same chemical shift.  So this thing here is kind of like the golden rule of splitting,  you know, the basic do unto others  as you want others to do unto you.  If you keep this in mind, you're going to be very,  very well set on all of the details.  All right.  Let's talk about specific way of thinking  about what's the same and what's different.  One way that I like to do it, you can do it by a couple of ways.  If you're good with symmetry, you can just go ahead and say,  all right, do we have a symmetry operation  that interchanges these two hydrogens, say in diphenylmethane?  You can say they interchange by reflection  and therefore they are chemically equivalent.  But we're going to get into some cases  that at least the first time you see them are tricky.  And another way to do it is just  to perform a little thought experiment and say, all right,  what is their topological relationship to each other?  And the way you do this thought experiment is you replace one  proton by a deuterium and the other proton by a deuterium  and then you ask yourself, what is the relationship  between these two molecules?  And in the case of this particular thought experiment,  these two are the same.  And since they're the same,  topologically we call those two protons homotopic  and homotopic protons are chemically equivalent.  Let's try another molecule.  Obviously, obviously, obviously these are trivial.  So let's take instead of diphenylmethane,  let's take ethylbenzene.  And again, we'll consider the methylene group.  The two hydrogen groups of the methylene group still exchange  by reflection.  So they meet the criterion there.  They're chemically equivalent.  Yeah?  You can't tell them apart.  So they have the same chemical shift.  And two protons that are chemically equivalent are the  same and don't split each other.  So chemically equivalent protons don't split each other.  Now, later on we're going to talk  about a concept called magnetic equivalents.  And what you'll see is that although chemically equivalent  protons, well actually I'm going to hold off  on this statement here.  Right now I'm just going to leave it  as there are two types of sameness.  And chemically equivalent protons don't split each other  if they're also magnetically equivalent.  But we're going to get to that later.  For now, the simple rule is chemically equivalent protons  are the same in a certain way.  All right, let's try our little gadanke experiment here.  We replace one by a deuterium.  And we replace the other by a deuterium.  And now these two molecules, are they the same?  No, what are they?  Enantiomers.  Enantiomers.  So now they are enantiomers,  which means topologically the protons are enantiotopic.  And enantiotopic protons are also chemically equivalent.  All right, so now let's come to a molecule  where we have a chiral center in the molecule.  And since we've been building on this idea of phenyl groups,  I'm going to give us the molecule phenylalanine.  Doesn't matter whether I draw it racemic,  which means we have two enantiomers or whether I have one molecule.  Okay, these two protons are now no longer exchangeable by reflection.  And no matter how rapidly you rotate, they still are not the same.  So these two protons now, the methylene protons,  are not chemically equivalent.  And let's try our thought experiment here again,  because sometimes things can get more complicated  than you might anticipate.  And so we'll take one, replace it with a deuterium,  and we'll take the other and replace it by a deuterium.  And we'll ask ourself what the relationship  of those two structures is to each other.  What are they?  They're diastereomers.  So we say that the two hydrogens  of the methylene group are diastereotopic.  And therefore they are not chemically equivalent.  All right, so here's a spectrum of phenylalanine,  so we can see what's going on.  This is a spectrum taken in D2O with DCL to dissolve it,  deuterohCl to dissolve it.  So all of the OHs and the NHs have exchanged with deuterium  and are exchanging rapidly,  and for all intents and purposes, don't J couple.  So we see a peak for our HOD,  and then we see the rest of it are carbons attached  to hydrogen.  We see our phenyl group over here in the 7 to 8 range.  We see our proton connected to the alpha carbon over here,  just a hair down field at about 4,  about 4.3 parts per million.  And then we see our beta protons, the ones that are next  to the phenyl, and we see them as two peaks.  Each of the beta protons appears at,  and these are expansions here,  each of the beta protons appears at distinct positions.  They're not the same.  They're not chemically equivalent.  They could be coincident, meaning appearing  at the same position, but they are not, and so they appear  at different positions.  And what's more, each of them is a doublet of doublets,  or DD. Why?  Because each of the protons splits the other one  and is split by the alpha.  So we have both a J2HH.  That's a two bond coupling since you go one, two bond to get  from these two beta protons to each other.  And we have a three bond coupling between the alpha proton  and each of the beta protons.  And the alpha proton is also a doublet of doublets.  So each proton is a doublet of doublets.  The alpha proton is a doublet of doublets  because it's split by the two beta protons,  and it's split with different coupling constants.  The beta protons are coupling to the alpha proton,  probably with about six and eight hertz coupling constants,  and we'll be talking more about these in a moment  or in a future lecture.  And the beta protons are coupling to each other  with about a 13 or 14 hertz coupling constant.  All right.  Let's talk about why we have to keep our heads attached to us.  So I'm going to give us another example.  The example I'll give us is a molecule called acetal.  Acetal is acetaldehyde diethylacetal.  All right.  So what do we have here?  We have a couple of ethyl groups, and we have a methyl group.  The CH3s of the ethyl group appear here as a triplet,  so this is OCH2 CH3.  That looks pretty reasonable.  We have another methyl group.  It appears as a doublet.  That's our CHCH3.  That looks pretty reasonable.  We have our methine.  It's the one that's attached to two oxygens,  so it's shifted down field.  It's at 4.6 parts per million.  It's a quartet because it's split by the methyl group.  And finally, we come to the CH2s.  And the CH2s show up as two peaks.  And it takes a moment to wrap your head around this.  And to wrap your head around this, what you have to realize is  that these two protons are not chemically equivalent.  You can think of this a couple of ways.  They don't exchange by symmetry operation.  If you reflect through the plane, you don't get the same thing.  Another way to think about it, and I think the easiest way  to keep your head on straight is to think about it rigorously.  You imagine replacing one of them with a deuterium,  replacing the other of them with a deuterium.  And the molecules that you get are diastereomers.  In other words, these two protons are diastereotopic.  Okay.  We talked about in IR how if you have like an internal outcheam,  sometimes you don't see the scratching  because there's like that pseudo symmetry.  Do you see the same thing with this that if you have like a  stereosenter or like a stereocenter,  but it's like really far away?  Absolutely.  Great question.  Just because two protons are topologically diastereotopic  doesn't mean that they won't be coincident.  And in general, the further that they are  from the stereocenter, the more likely they are  to be coincident and behave as if they are, clearly.  Here we see them as well-defined peaks  at very different positions.  They're both doublets of quartets.  That's a DQ.  Oops. DQ. We'll talk more about the splitting pattern,  but they're being split with one big geminal coupling  of about 10 hertz, one two bond coupling.  And with three visceral couplings,  three three bond couplings of about 7 hertz.  All right, last thing I want to show you,  one last handout while we're  on the concept of chemical equivalence.  So here we have the molecule three methyl two butanol.  And what I want to point out to you is  in three methyl two butanol,  the two methyls are not the same.  They're diastereotopic.  You can think of this a couple of ways.  One way is to think about, remember the OH constitutes  a chiral center, and you can do a little thought experiment  and say all right, I'll envision replacing one  of the methyl groups or the other methyl group  with a deuteromethyl group.  Or the other way is just to think about the topology.  But when you look at it, you say all right,  how many peaks do we see in the C13 NMR?  One, two, three, four, five.  There are five types of carbon  because there are five non-chemically equivalent  carbons.  If you look at the proton NMR, it's a little confusing  at first until you think about what's going on.  It's not hard to realize that this methyl group that's beta  to an oxygen gives rise to this doublet.  Now we have two more methyl groups in the molecule,  and each of those methyl groups is next to a hydrogen.  Each of those methyl groups appears as a doublet,  but the doublets are very close.  In other words, we have a doublet and we have another doublet  and so it looks like the triplet.  Can you discern that because it's not the 1 to 3 along?  What's that?  Can you discern that because?  Well, you can discern it if you look carefully  because it's not 1 to 3, 1 to 2 to 1, but it's confusing.  So it's two doublets, but let me tell you a secret.  This is a 300 megahertz spectrum.  Remember I said the coupling constant is fixed in hertz.  The position is fixed in PPM.  So my two doublets would look like this at 300  and it would look like a triplet.  If I went to 500 megahertz, the two doublets would now separate  and I'd see a doublet and a doublet  and that would be one way you can prove it.  All right, I think that's what I want to say for today  about chemical equivalents.  We'll pick up next time talking about magnetic equivalents.
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