字幕列表 影片播放 列印英文字幕 >> 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 C 13 NMR shift. We'll talk more about them later on but then we're going to start, 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 equivalence which ties into concepts of symmetry and stereochemistry and confirmational analysis and it's really beautiful, chemical equivalence and so we'll be talking about chemical equivalence 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 stereo chemistry 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 an amplification on chemical equivalents but it's too much to take in on 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 one percent or more specifically 1.1 percent C 13, so most of your molecules, your small molecules don't contain any C 13. For small molecules some of them contain one C 13. Now things get worse. The magnetogyric ratio for C 13 is only a quarter of that of the magnetogyric ratio for a 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 of 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 magnetogyric 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 since for a 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, did I say thousand, million times per second. So that also gives you a quarter as much electricity in the coil. So you've got 1.1 percent and a quarter of a quarter of a quarter which means you're only 1/5,800th as sensitive. Question. >> Is the procession, are you talking about the limiter frequency? >> The limiter frequency, yeah. So if you take a magnet and just spin it in a coil, if you spin it faster you get more voltage. 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 X, Y plane. >> Is that a quarter of all or is that a quarter of what's protons? >> It's a quarter-- compared to protons. So carbon is much less work, much less sensitive technique than proton NMR spectroscopy. Now there are 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 and 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 proton so all the carbon you're going to see, virtually all the carbon you're going to see is called proton-decoupled carbon. That flips the spins of protons rapidly which means the carbon doesn't see them as spin up or spin down so the carbons appear as singlet. Well 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 Overhauser effect and we'll talk more about this as a technique but the basic principle of the Nuclear Overhauser effect is that 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 now days it's easy to collect a carbon NMR spectrum. It typically will take more sample so you can collect the proton NMR sample on strychnine 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 could do it at ten. You could do it at a milligram but it takes a lot more time. And remember if you have one milligram versus ten milligrams it's going to take a hundred times as long to collect the same signal to noise ratio which means if you're in your 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 [inaudible] is three and a half centimeters or requires a three and a half 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 miss 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 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. Aliphatics show up at about 10 to 40 so on that big range of 200 or so ppm that's in the up field region. Carbons next to an electron withdrawing atoms show up down field but it's not quite as pronounced so carbon next to a halogen you might even 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 withdrawing group I'd say 50 to 70 for a carbon next to one oxygen so that's 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 alkine CH. For an alkine carbon it's about 70 to 80 ppms so that kind of stands out. All right, alkenes and aromatics whereas in proton NMR the alkenes show up a little bit more up field, 5 to 6 and the aromatics a little more down field 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 anisole or phenol where you put an oxygen directly on a carbon that carbon might be at 160 parts per million. If you have a very electron donor-- we'll talk more about specific shifts but you could easily if you have high electron donations say an ortho carbon, you could easily be a little below, 115 ppm. All right carbonyls, esters and carboxylic acids you get a little bit of a resonance affect so they show up a little bit less far down field than some of the others let's say about 170 to 180 parts per million. Aldehyde show up further down field, let's say about 190 to 200 ppm and ketones I'll say are RCOR prime for ketones, let's say about 205 to about 220 ppm. Bless you. 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 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 look at when reading a C 13 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 H 1 NMR but with maybe a factor of 20 on the scale. In other words most of what you're going to see on a proton NMR is from zero to ten. Most of what you're going to see in carbon NMR is from zero to 200. Carboxylic acid 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. 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. 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 H1 NMR spectrum of chloro ethane and so my little pigeon sketch of this spectrum of chloro ethane would look something like this. We have a one to three to three to one triplet somewhere down field of three ppm or just a hair down field of three ppm and then we have, I'm sorry, one to three to three to one quartet and then somewhere just hair down field of one we have a one to two to one 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, 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 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 are 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. [ Silence ] All right, before we come to this very important concept of diastereotopicity let's tackle this basic notion of the one to two to one triplet and the one to three to three to one quartet. So we have a triplet and it's one to two to one. It comes from the CH3 and it sees the CH2 and the two 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 one to two to one triplet. For our quartet we have a one to three to three to one ratio. The quartet of course comes from the CH2. The CH2 sees the CH3 and all of their 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 N equal and I'm going to underline equal couplings lead to N plus 1 lines. So if there are four equal couplings you get five lines, those five lines it follows Pascals triangle. You can work out the statistics yourself or you can say that those five lines end up being in a one to four to six to four to one ratio and so forth you basically add the ones above or you work out the statistics. So for example, if we'd go to isopropyl chloride now the CH here is split equally by three methyl groups, so the CH appears as septet 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 a molecule unless you look hard you might not see these lines on the outside, and in fact by the time you get up to great 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 look at a case of what I mean by couplings that may or may not be equal. So imagine the methine group in [inaudible] and specifically this methine group. So this methine group is split by two methyls groups and it's also split by this methine 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 is a multiplet. The OH of alcohols and I know had has already come up on the homework. The OH of the alcohol may or may not couple so the hydrogen of isopropanol, the methine 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 oh, 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 methine 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 Js 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, sometime don't. They're more spherically hindered. The exchange, 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 about 20 millimolar concentration, 10, 20, 30, millimolar concentration and chloroform undergoes photo oxidation to give hydrochloric acid or DCL in the case of CDCL 3 and that will rapidly promote exchange so if I want to minimize exchange I'd typically will pass my chloroform through alumina, not my sample but my chloroform through alumina that I've dried over a flame and that will take out the acid and minimize the amount of water in the air. Okay, so alcohols are kind of a wild card on J coupling. [ Inaudible student comment ] 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 on the sample you will see 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 time scales. One is the time scale for coupling. The other is the time scale for being able to see a peak. Remember I talked about the uncertainty principle before? So in order to see coupling your exchange rate has to be, you have to be able to make the observation such that your line width is on the ordinary 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. A 4 ppm difference is two thousand 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 the hydrogen in the alcohol may or may not couple. If that doesn't couple you can also assume that the proton bond in the alcohol will not couple as well. >> You mean yes. >> If one doesn't couple the other one will. >> Absolutely, coupling is mutual. Now you could envision a circumstance where you had J coupling but quadripolar 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 quadripolar 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. There are real 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 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 the lecture too. 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 equivalents. Chemical equivalents 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 operant, can be exchanged by a symmetry operation or rapid process. [ Silence ] For example, rapid rotation about a bond. All right the other caveat and this ties into something I said last time. Virtually all, you could say all rotations about SP 3, SP 3 carbon bonds are rapid at room temperature. [ Silence ] 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 SP 3 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, 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. 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 major change by reflection and therefore they are chemically equivalent but we're going to get into some cases that 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 ethyl benzene 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. [ Inaudible student comment ] . 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 gadonk 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, 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. It 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 ourselves what the relationship of those two structures is to each other? What are they? >> Diastereomers. >> They're diastereomers so we say that the two hydrogens of the methylene group are diastereotopic and therefore they're not chemically equivalent. [ Silence ] All right, so here's a spectrum of phenylalanine so we can see what's going on. This is a spectrum taken in D2 O with DCL to dissolve it. Deutero-HCl to dissolve it, so all of the OHs and the NHs have exchange 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 seven to eight range. We see our proton connected to the alpha carbon over here just a hair down field at about a 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 and these are expansions here. Each of the beta protons appears 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 position 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 J2 HH. 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 in 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 6 and 8 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 and the example I'll give us is a molecule called acetal. Acetal is acid aldehyde diethylacetal. [ Silence ] 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 OCH2CH3. 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. That'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. >> When you talk about an IR how you get like an internal alkene sometimes you don't see the stretching because there's like that pseudo symmetry. Do you see the same thing with this that if you have a stereo isomer like a stereo center 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 coincidental and in general, the further that they are from the stereo center the more likely they are to be coincident and behave as if they are equivalent. Here we see them as well- defined peaks at very different positions. They're both doublets of quartets. That's a DQ. We'll talk more about the splitting pattern about their being split with one big geminal coupling of about 10 hertz, one two-bond coupling and with three vicinal couplings, three, three bond couplings of about 7 hertz. All right, last thing I want to show you, one last handout while worry on the concept of chemical equivalents. [ Silence ] So here we have the molecule 3 methyl 2 butanol and what I want to point out to you is in 3 methyl 2 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 deutero methyl 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 C 13 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 are 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 a triplet. [ Inaudible student comment ] Well you can discern it if you look carefully because it's not one to three, one to two to one 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 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 could proof 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. ------------------------------23065aa15093--
B2 中高級 化學203.有機光譜學。第10講。13C NMR化學位移。 (Chem 203. Organic Spectroscopy. Lecture 10. 13C NMR Chemical Shifts.) 54 2 Cheng-Hong Liu 發佈於 2021 年 01 月 14 日 更多分享 分享 收藏 回報 影片單字