字幕列表 影片播放 列印英文字幕 [ Silence ] >> -- and get good at recognizing and identifying the multiplets and extracting a couplet in constants. So the things that we are talking about and the question that was raised: Is that coupling is really 700 Hertz? What makes it 700 Hertz? So let's talk about factors that J depends upon. And I've tried to sort of group out maybe five factors and give us some points. So let's say one is the magnetogyric ratios of the nuclei involved. [ Writing On Board ] And - so remember how when we started talking about magnetogyric ratio, I gave you two examples? We talked about the C13 satellites of chloroform and protiochloroform, right? So when you see the H1 anomer of CDCl3, of course you don't see any of the deuterium chloroform that's in there; but you see that 0.2% or 0.1% of protiochloroform. And you see a peak that's very big at 7.26. But then, symmetrically disposed about it, you see these two satellites. And so this is your CHCl3, or more specifically your C12HCl3 peak. But then your satellites come from your 1% -- 1.1% -- of C13 CDCl3 - CHCl3. And the spacing of those two lines, the distance here between these two, is 208 Hertz. And that distance happens to be 1.6. Is 1.6 the - is 1 over 6.6 the distance that we see in the C13 anomer of CDCl3, right. In CDCl3, you do, indeed, see the C13, the 1% C13. And you see this one-to-one-to-one triplet centered at 77 ppm. And the distance between the lines here is 32 Hertz. And that ratio of 208 to 32 is equal to 1 over 6.5, which is the magnetogyric ratio of proton over the magnetogyric ratio of a deuteron -- meaning that when everything else is equal, if you have a magnetogyric ratio nucleus with a magnetogyric ratio that's big, you get big coupling. If you have a nucleus with a magnetogyric ratio that's small, you get small coupling. For heteronuclei, we said that fluorine, for example, has a big magnetogyric ratio. So fluorine is going to have big coupling. We said that carbon has a relatively small magnetogyric ratio, so carbon in general is going to have small coupling. So the gamma of the nucleis [phonetic] is important. The number of bonds: So generally -- generally, generally -- we see coupling through one or two or three bonds, sometimes four or five; that's called long-range coupling. So if we have some system w, x, y, z -- and we look at it if we are talking about coupling through one bond, coupling through two bonds, or coupling through three bonds -- and I need to make some wildly general generalizations, I could say our J1's are on the order. And I - you can just sort of look at the appendix I handed out. And if you want to keep a number in your head, I'd say - let's say 30 to 300 Hertz. Kind of one-bond couplings are big. They are on the order of 100 Hertz give or take, you know, a factor of half a log unit. We saw you can have a 700-Hertz coupling in the phosphorus. But generally, yeah, like 100 Hertz would be a good number to keep in mind. J2's are generally on the order of 0 to 20 Hertz. And again, I'll give you examples that fall out of there. But if you want to keep a number in mind, you know, 10 Hertz, 20 Hertz, something like that. And J3's are generally on the order of 0 to 20 Hertz. So in other words, one-bond coupling is huge; two-bond coupling, three-bond coupling is smaller. All right, other factors involved: geometry, particularly in three-bond couplings where good overlap leads to big coupling. [ Writing On Board ] Remember, coupling is carried by electrons in bonds, where a nucleus feels it - another nucleus through polarizing the pair of electrons making up the bonds so that the one with its spin up is near the one with its - nearer to the nucleus that spin down and so that travels along the line. And if you have good orbit overlap -- for example, an anti-periplanar relationship or a syn-periplanar relationship -- you'll have a big coupling constant. If you have bad overlap -- for example, a 90-degree dihedral angle -- you'll have essentially no coupling constants, no - zero-coupling constant. And if you are sort of at a low angle like a gauche angle -- you know, something close to 90 degrees, 60 or 120 -- you have still relatively poor overlap; so your coupling constants are small. That was what we were talking about the other day with cyclohexane, where I said, "Yeah, for a tran - for an axial coupling on the order of 10 Hertz, 8 to 10 Hertz." For axial equators, that's axial-axial. For axial equatorial, or equatorial-equatorial where your dihedral angle is about 60 degrees, yeah, 2 or 3 Hertz. All right. So geometry also matters. Hybridization. [ Writing On Board ] And a general rule is, more S character leads to bigger J's. [ Writing On Board ] And other factors: electronegativity. [ Writing On Board ] You know, hybridization is like Y in trans-alkenes. We see a J of 17 Hertz, typically, versus in a one - in a diaxial interaction on a 1-A axial-axial coupling on a cyclohexane, which also has 180-degree dihedral angle just like a trans-alkene; we only see a coupling constant on the order of 10 Hertz. In other words, you have more S character in an sp2 bond and in a bond involving an sp2 carbon and a bond involving an sp3 carbon. So you have a bigger coupling constant. So the other - another factor, let's say - I'll just say other factors such as electronegativity. [ Writing On Board ] So for example, a hydrogen that's next to an oxygen that's on a carbon with an oxygen will have slightly smaller J values, particularly if there are two oxygens on there. All right. Let's - let me give you some typical J values, and then we'll look at some examples. And I don't want -- [ Erasing Board ] And I want us to have a - you - I mean these sort of generalizations that I just put up are pretty useful. But unless you are doing a project with particular nuclei, you may not end up keeping all the numbers in your head. But let's just look at sort of some typical examples, and these are actually in the appendix here. So let's take a look at a four-O alkane. So your J2HF -- and this is all in the appendix -- is really, really big. It's, like, 44 to 81 Hertz, so even bigger than a sort of generic two-bonds coupling that I talked about. Fluorine has a large magnetogyric ratio, so you have got very big coupling. Your three-bond HF, your three-bond coupling, is on the order of 3 to 25 Hertz. [ Writing On Board ] And fluorine is so, so good at coupling, right? So this is two-bond coupling. This is three-bond coupling. Fluorine is so good at coupling that you can even sometimes see a little four-bond coupling: J4HF is on the order of, let's say, 0 to 4 Hertz. [ Writing On Board ] [ Inaudible Class Question ] Part of it's the polarization, but part of it is the orbitals that fluorine is used - using in bonding. So if you have a hydrogen, right, we talked about two-bond and three-bond H-ings [phonetic] coupling. Right? So a typical two-bond H-H coupling might be -- I said -- 14 Hertz, sort of, in an unperturbed system. And a typical three-bond H-H coupling I said, "Let's use 7 Hertz as a typical number." So that hydrogen is contributing a 1s orbital. But the fluorine is going to be contributing orbitals that are going out further, because it's using the second shell in coupling. So we are in the 2 - you know, sp3 technically. So your coupling is going to be bigger with fluorine. So I don't think it's just - and it probably involves some electric. Yeah, actually, you know, it would involve electronegativity as well, because it is going to be pulling those electrons in really tight and that's going to give more interaction of the nuclei with the electrons. So yeah, I guess it's both of those. Other questions or thoughts on this? [ Inaudible Class Question ] Of one over the other? >> Yes. >> No. I mean think about your Newman projection of a perfect chair cyclohexane. If I draw it as a Newman projection -- [ Writing On Board ] -- I'll just draw half of the cyclohexane. So this is axial. This is axial. This is equatorial. This is equatorial. And in a perfect chair cyclohexane, this dihedral angle is 60 degrees; this dihedral angle is 60 degrees; and this dihedral is 60 degrees. So it really should be about the same unless you flatten the ring out or do something to perturb it. [ Inaudible Class Question ] Ah, OK. Do you ever see coupling that's essentially through space? One can, so outside of the realm of covalent chemistry that we are seeing. You can, for example, see coupling or cross-hydrogen bonds and other situations where things are held into close proximity with each other. So for example, you can do ing-15 [phonetic] coupling across DNA, for example, from one nuclear base to another through the hydrogen bonds. But just for a normal ordinary organic molecule where, you know, it was in some confirmation, not that I know of any examples. [ Inaudible Class Question ] Nothing where the molecules aren't locked together. I mean I suppose you could come up with some example where you take some poor molecule and bang two methyl groups into it so hard that they are banging in at van der Walls radius. And you might. I don't know the answer. That's the sort of thing that one would try experimentally. Other questions? Benzenes are interesting with fluorobenzenes. And you will have -- I think it's on this coming Monday's homework set -- you'll have some fluorobenzenes. And it's a little bit counterintuitive. The problems are pretty easy. They are out of Chapter -- what is it? -- 5? 4? What's the one that I gave you to read that - the coupling involving other nuclei? Six. OK. Anyway, so keep this in mind. So your ortho coupling, your J3HF, is on the order of 6 to 10 Hertz. And so that's kind of what you'd expect. And what's sort of surprising: I mean in the case of hydrogens, you have meta coupling, but it's usually on the order of a couple of Hertz -- you know, 2 or 3 Hertz, 1 to 3 Hertz. In the case of fluorine, your meta coupling is bigger than you might otherwise expect. It's on the order of 5 to 6 Hertz. And then because we know that fluorine is very good at coupling, you even see some para coupling, and it's on the order of 2 Hertz. And these are all kind of approximate. [ Pause ] Yeah, I mean the good news on all of this -- that Appendix F is just such -- [ Pause ] -- such a treasure trove. And so the good news is Appendix F puts a lot of these data right at your fingertips. And the only reason I'm talking about this is because I think hearing it once is sort of helping you see where to look it up and so forth. Let me just point out also the numbers that I had given you before and just put this into some context. So when we started talking about NMR and spin-active nuclei, I mentioned the magnetogyric ratios. And just let me put these up here again. Magnetogyric ratio for a proton is big. Magnetogyric ratio for fluorine is big. Magnetogyric ratio for phosphorus: 30 - 31. I guess I'll even put up F19 and H1 -- is 10,840. And the magnetogyric ratio for C13 is 6-7 - 6,728. So let me tell you what I was talking about when I was saying you can correlate things roughly with magnetogyric ratios. So for example, JCF - J1CF is pretty darn big. It's at the high end of what you would typically expect for couplings. It's on the order of 200 to 300 Hertz. J2CF is also pretty big. It's at the order of 20 to 40 Hertz. And J3CF is on the order of 10 Hertz. And then phosphorus: As was already pointed out, one-bond phosphorus-hydrogen coupling is huge. It's on the order of 700 Hertz. Two-bond PH coupling is on the order of 10 Hertz. And three-bond PH coupling is on the order of 20 Hertz. Certainly these aren't numbers to keep in your head. These aren't numbers to keep in your head. These are just, sort of, numbers that have seen once. [ Pause ] [ Inaudible Class Question ] Often J3'sds are bigger than J2's. In carbon-hydrogen coupling, often J3 to hydrogen, often three-bond coupling, is a little bigger than J2. Sometimes J2 is bigger. Remember, of course, one case you can actually have an anti-periplanar relationship: So if you think about it, in -- [ Writing On Board ] In J2 you can - in J3 you can get a nice zigzag relationship. In J2 you have got, you know, a different relationship. These electrons aren't as directly overlapped. And J2 often depends on hybridization in geometry. I mean a striking example: So you look at alkenes, right? And this coupling is on the order of 0 to 2 Hertz, the geminal coupling. And then you look at your vicinal couplings, and it's on the order of 17 Hertz. And even the cis coupling is on the order of 10 Hertz. So here is another example where J2 is much - is smaller than J3. [ Pause ] [ Inaudible Class Question ] These ones. Yeah, carbon fluorine. So we have now just had this tremendous traipse through all of these weird and wild couplings -- and sort of in the abstract in the sense that I've listed some numbers. So let's now look at some real compounds. And these are just compounds that I pulled out of Aldrich, and for that matter compounds that you might well - or types of groups that you might well encounter. And the first thing I'll start with is triethyl phosphites. And you would encounter this - last night we talked about Horner-Wadsworth-Emmons type reagent. So if you made a Horner-Wadsworth-Emmons reagent, you will encounter - or a Wittig reagent, or a phosphonium precursor to a Wittig reagent, you will suddenly find yourself encountering fluorine protons and fluorine-carbon coupling as get to know your molecule. So let's take a look at an example. And I love these Aldrich spectra because you can just pull all sorts of spectra out and say, "All right, what would I get?" So I pulled this out of www.sial.com, and then I just blew things up. OK, so let's start with - we have a 300-megahertz proton spectrum. All the Aldrich spectra are 300 megahertz H1, and 75 megahertz C13 spectra. So triethyl phosphite, we see something that looks like a triplet for the methyl group and then something that looks a little more complicated over here. How do we describe this pattern? [ Inaudible Class Comment ] "Quintet" would be a good place. And so we could call this a quintet or - what was that? [ Inaudible Class Comment ] Quintet, I like that -- so, quintet. I like even better "apparent." [ Writing On Board ] Because I have taken this peak, and I have picked it up and blown it up here. And what do we see? [ Inaudible Class Comment ] Low shoulders. So we know it's not a perfect quintet. We don't have five - four couplings that are exactly the same. Can we recognize a pattern hiding under here? [ Inaudible Class Comment ] Triplet - a triplet. Now that -- [ Inaudible Class Comment ] Doublet of quartets. Watch this. So doublet of quartets is a pair of quartets: 1, 3, 3, one for this shoulder. And then for this shoulder 1, 3, 3, 1. Do you see that? 1, 3, 3, 1; 1, 3, 3, 1. So it's a doublet of quintets. [ Writing On Board ] >> Quartets. >> Quartets -- I'm sorry. [ Writing On Board ] So if we want, we could extract both J's. The smallest J -- the small J, the J associated with the quartet -- is going to be the distance from the last line to the next to the last line, or the first line to the second line. And I handily put a scale on here. And I read that distance as 7 Hertz. [ Writing On Board ] And the big J is going to be the distance from this line to the - from the first line - the second from the last line to the last line, or from the first line to the third line. And I can't exactly read where that line is, but it looks like it's about a Hertz more than here. So let's call that 8 Hertz. [ Writing On Board ] So we could characterize this as a DQ -- use Q for quartet, quint for quintet. J equals 8, 7 Hertz. [ Pause ] So at 7 Hertz, that's just our - and a quartet, that's just our coupling to the methyl group, right, because we have P-O-C-H-H-C-H-H-H. So 7 Hertz is just our J3HH. [ Writing On Board ] And 8 Hertz is our J3PH through one, two, three bonds. [ Writing On Board ] [ Inaudible Class Question ] Those are just approximations. I - yeah - trying to give you sort of general numbers to ballpark it. So one-bond couplings? On the order of 100 Hertz. Two-bond, three-bond couplings? Yeah, on the order of 10 Hertz. You know, sometimes bigger, sometimes smaller. It depends on gamma; it depends on other factors; it depends on hybridization. We can do the same thing with carbon here. And even though our two peaks in the C13 anomer are really small here, they conveniently give us a peak printout up over here. So those two constitute a pair, and those two constitute a pair. Each of them is a doublet. The doublet for this carbon we can deal with as 63.64 -- I want to calculate the coupling constant -- minus 63.56; those are just the two values in the top of the table in ppm -- is 0.08 ppm. And we have 0.08 ppm times 75 Hertz per ppm is equal to 6 Hertz. So that's our J1P - that's our J2PC. [ Writing On Board ] [ Pause ] And we can figure out our J3PC from this other one. So we have 16.19 in the up left-hand - upper - your - right-hand corner and 16.10 as the position of the two lines. And that difference is 0.00 - 0.09, times 75 is equal to 7 Hertz. And so that's our J3PC. [ Writing On Board ] [ Pause ] Thoughts? Questions? [ Silence ] [ Inaudible Class Question ] Ah. In the carbon anomer. So in the upper right-hand corner, they list the two line positions. There is -- [ Inaudible Class Question ] In the proton anomer. You mean how we see the doublet? >> Yes. >> So OK. I can't do this well with my hands because my hands aren't in 1-to-3 to 3-to-1 ratio. But OK, imagine my hand is a quartet. And if we have - so the distance between my hands is the big J. And if the distance was one finger's worth, we would see a perfect quintet. If it - this is a bad - oh. I got a thumb there. OK. If it were 1 - I got to get rid of my thumbs, chop 'em off. OK. If it were one finger's worth, we would have 7 Hertz spacing between the fingers; and then it would be split into five lines, all equally spaced, at 7 Hertz apart. But they are a little bit further apart, so we see a quartet and another quartet. And the distance between lines 1 and 2 is the J of the quartet; and the distance between lines 1 and 3 -- or in this case, 1 and 2, and 1 and 3 -- is the distance of the doublet. [ Inaudible Class Question ] Interesting question. So do you notice these little jaggies right on the edge of the peak? Do you see how the peaks aren't smooth? So remember I was talking about digital resolution? So you take a spectrum that's, let's say, 16,000-point slide. And you have those 16,000 points over a spectral width that's, let's say, 4,000 Hertz here. And so your digital resolution is 1/4 Hertz. So you think of your spectrum as a smooth curve. But what it really is, is a series of points that been - have been splined together. And because we may be missing a point right on top of the peak, you end up with something not looking completely symmetrical. [ Inaudible Class Question ] That's just an artifact of this. As a matter of fact, I would say if I went and told the spectrometer by increasing the acquisition time, told it to have more digital resolution, we would probably see something that was better resolved. [ Inaudible Class Question ] You just type in - there are two ways. So you'd type in number of points, and you make the number bigger. Or you make the acquisition time longer. And the number of points in the acquisition time in the spectral width are all intimately linked. So you can basically collect data for 5 seconds instead of 3 seconds to increase your number of points, or you can tell it more points. [ Inaudible Class Question ] So that would be the J between the methylene and the methyl group. >> Oh, OK. OK. >> So let's take a look at a fluorine anomer or a - rather, a fluorinated compound. And maybe I'll just show you some highlights. So I grabbed from Aldrich the spectrum of fluoropentane. And so let me write out the structure here, just so you can see it. All right. What do we call that? [ Inaudible Class Comment ] It's a doublet of triplets. [ Writing On Board ] And what's doing the splitting? What's giving the doublet part of the splitting? Which proton is it, first of all? Which protons are? [ Inaudible Class Comment ] One alpha to the fluorine. [ Writing On Board ] And so this is giving a two-bond JHF. And we can - and what else is giving the splitting? [ Inaudible Class Comment ] The methylene. So we have this methylene is coupled to the adjacent methylene. And so that's giving the triplet part of the splitting. And you can read off the two-bond JHF. What value do we get here? [ Inaudible Class Comment ] Yeah, somewhere around 48, 47. So your two-bond coupling, I think I got 47. [ Writing On Board ] You notice this multiplet over here? It's not so neat. Remember I talked about non-first-order coupling, when you've got things lumped on top of each other you've got virtual coupling? So the next couple of methylenes are lumped on top of each other, and you've got coupling here and coupling to the fluorine. So it's not so pretty. But if you pull it out and you expand it, you can pick out your J3HF even if you can't quite pick out what the multiplet is. If it's just sort a generic multiplet, you can measure that distance there. Here it looks like it's about 24 Hertz. So our J3HF is 24 Hertz. [ Inaudible Class Comment ] What's that? [ Inaudible Class Comment ] And that fits into the model. You can do the same thing with the carbons. And if you look at the carbon spectrum here, you'll notice a big, big, bigly [phonetic] separated doublet over here. What does that correspond to? >> J1. >> J1, you see. Yeah. So that's the carbon that's directly alpha to the fluorine. And you have got a J1CF. And if I want to calculate the J1CF, what do I do? [ Pause ] [ Inaudible Class Comment ] And what numbers do I subtract to multiply? [ Inaudible Class Comment ] 85 minus 83 -- or more specifically, 85.26 minus 83.09, times 75; and that gives us 167 Hertz. So that's a J1CF coupling. And I won't work through it, but you'll notice that your next peak here is split into a doublet. You can see the pair of lines over here, and that corresponds to these two. That's our J2CF. And then the next one is split, and that corresponds to our J3CF coupling. So you could pull all of this data out of your spectra. [ Inaudible Class Question ] Answer on that blackboard. Take your carbon anomer and go ahead and run it at 75. Or I guess for our department we have carbon at 100, carbon at 125, and carbon at 150. So you could just go ahead, run it at 200, and see if the two lines are now at the same ppm. If each line is at the same ppm, their distance in Hertz has changed and they correspond to two singlets. If, on the other hand, their positions in ppm have changed and they have moved toward each other but they are still centered at the same position in ppm and their distance in Hertz apart is the same, then they correspond into a doublet. [ Pause ] All right. Let's have some fun with carbon-carbon and carbon-hydrogen coupling now. [ Erasing board ] All right: CH coupling. [ Writing On Board ] So sp3 - so I'm talking right now about one-bond coupling; we'll start with one bond, CH coupling. One-bond CH coupling is important because all of your heteronuclear techniques and your DEPT technique rely on carbon-hydrogen coupling, one-bond or in some cases two-bond and three-bond coupling. So a typical S - remember I said how hybridization matters, and the more S character you have for everything else being equal, the bigger the J if everything else is equal. All right. So typical J1CH for an sp3 carbon is about 125 Hertz. So for example, if you look at ethylene - if you look at ethane, your value is 124.9 Hertz. If you look at cyclohexane -- and these are all in your appendix table as well -- it's 123.0 Hertz. I'm just pulling out some highlights from one of the appendices I've handed you. sp2: more S character -- 25% S character in sp3, 33% in sp2. And so your J1CH goes up proportionately; it's about 160 Hertz. So you look at ethylene, and it's 156 Hertz - 156.2 Hertz. And these are all in one of the appendices that I've passed out to you. You look at benzene, it's 159.0 Hertz. All right. Now where does this - where else does this become interesting? Where else does it become important? Yeah. [ Inaudible Class Comment ] Beautiful. Yes. So C13 anomer: You will invariably run it proton decoupled. So you won't see proton coupling. And yet you will still see people use the term to refer to a DEPT spectrum where methylene is referred to as T for a triplet. And there's history there. And the history is that before they had DEPT and other techniques, they would run what's called off-resonance decoupling -- partial decoupling. And so you would get a methylene. And you would see it as a triplet because it would be falling out most of the couplings so you wouldn't see the two- and three-bond coupling -- which is really horrendous because it peaks - when you talk about two- and three-bond coupling, you can imagine how - what your carbons are like. The methyl group in ethanol is spilt into a triplet of - into a quartet of triplets. And the methylene group in ethanol is split into a triplet of quartets. So these are - you know, you'll have huge, huge splittings in full-link [phonetic] or proton-decoupled carbon. But you will still see a methylene referred as parentheses T from that old, old thing. So yeah. All the carbon anomer we are going to run is proton decoupled. You do see - remember I said in your C13 satellites of course you see the reverse because you are not carbon decoupling when you are running a proton anomer. And for the most part it doesn't matter because 99% of your carbon is Carbon 12. But if you look closely in your methyl groups, you can see little satellites and another short singlet. [ Inaudible Class Question ] You could, indeed. One of - the problem with X nucleus - so the question was, could you decouple fluorine so you didn't get fluorine coupling. And the short answer is yes. One of the problems is the amount of power you put in has to - depends on your spectral width. And so if you have the width of fluorine -- which is very wide; it's 200 or 300 ppm -- you've got to put in a huge amount of power to irradiate all of the fluorines. And basically that turns your anomer into a giant microwave, and so you'll cook your sample. So it actually isn't so simple. It's easy. Protons have a narrow spectral range; it's only 12 ppm. Carbons, on the other hand, have about 200, 240 ppm. So you would have to put in a lot of power to carbon decouple. You could do specific decoupling experiments where you irradiate at one specific frequency, and that's another way you could do it. But it's hard to blast an entire wide spectral width. [ Inaudible Class Comment ] Yeah, but normally it's proton decoupling. It's - proton is the easy one. [ Inaudible Class Question ] Negative - OK. Negative and positive, for the most part, don't mean anything in terms of what you will observe except in the phase-sensitive [phonetic] spectrum. But a positive J value means that if one nucleus is spin up, the other feels a stronger magnetic field. The other - but because you are going through electrons and you are polarizing the electrons in the first bond, which are polarizing the electrons in the second bond, you may end up with polarization so that if one proton is - one nucleus is spin up, you end up with the other feeling a weaker magnetic field. So that's what a negative J value means. But in terms of the doublet, you'll see the same doublet. All right. Let's try a little bit more; and then I want to show you one really, really, really cool example. All right. It kind of makes sense that if you start to change the amount of S character in the PH bonds, you are going to end up changing the coupling constant. So for example, if you go to cyclobutane, you use more p orbitals in making the carbon-carbon bond framework, you use more P character, so you have more S character for the CH bonds. So your J1CH is bigger. So your J1CH for cyclobutane is 134 Hertz instead of 125 Hertz. For cyclopropane, your J1CH is 161 Hertz. All right. So this is sp3 and sp2. Now you come to sp, though, and things end up really, really tricky. So sp, your J1CH, now you have twice as much S character - your J1CH is twice as big as an sp3. It's 250 Hertz. Now that's tricky, because a lot of the experiments you do depend upon using an average value. So when you do a DEPT experiment, there is a delay in there that corresponds to 1 over the J1CH coupling. That's how it ends up working and picking out whether something is a methylene or a methine. We'll talk more about it when we talk about complex pulse sequences. But if the J value is twice as big, everything can get mixed up. The practical implications are that acetylene in DEPT. So for example, anything where you have a methine on the acetylene may be completely messed up. [ Writing On Board ] Opposite, I'll say, of what is expected. [ Writing On Board ] So I'll say DEPT. And later on we are going to talk about HMQC. And so because your J's are opposite, you may see things that you do not expect there. All right. So last example I want to give is incredibly wild, and I pulled this out because I thought, "This is fun." And I thought, "This is fun, and this is cool." [ Pause ] Because it gives us every sort of heteronuclear coupling you could imagine, and all sorts of bonds coupling. And I found this in that Aldrich collection of spectra, and I thought, "Hey, we got to take a look at this, because it's fun and it's cool." All right. So this is nice, a topically labeled Horner-Wadsworth-Emmons reagent. It would be something that if you wanted to do an isotopic labeling experiment and put C13 into your molecule in specific places, you could put it in. But it's got an absolutely funky proton and carbon spectra. So just to orient ourself: Remember the Horner-Wadsworth-Emmons reagent is a phosphonate ester, and you have a regular ester in typical Horner-Wadsworth-Emmons ester. Both of them have ethyls. And we won't worry about the ethyl groups other than to say this is OCH2, CH3 from both parts of the molecules, and this is the OCH2, CH3. But what I would like to worry about is what we see here in the middle. Now what we see here in the middle is that CH2. [ Writing On Board ] Let's think about what's going on here. So that CH2 is being split. It's being split by the carbons to which the hydrogens are attached, which are C13. It's being split by the phosphorus through a two-bond PH coupling, and it's being split by the C13 of the carbonyl by a two-bond CH coupling. So what pattern do you observe with three distinct coupling constants? [ Inaudible Class Comment ] Doublet of doublet of doublets. And that's exactly what you'd see over here. It's a DDD. You can pick out your smallest coupling at this distance. Your next coupling is this distance. And your biggest coupling is this distance over here. So I get that this distance is 7 Hertz. This distance is 22 Hertz. And this distance, which corresponds to the biggest J, is either the distance between 1 and 4 or 1 and 5; but here it's clearly the distance between 1 and 5. I get that it's 130 Hertz. So it's a DDD. J equals 130, 22, and 7 Hertz. And that corresponds to a J1P. J1CH equals 130 Hertz. J2PH equals 22 Hertz. And J2CH is equal to 7 Hertz. And that's pretty cool. [ Pause ] Now there's even more cool stuff embedded in this if you look the C13 anomer. And if you remember, this thing is isotopically labeled. So most of the carbons are present only at 1% abundance, but then that central methylene and carbonyl are present at 100% or 99% isotopic abundance. So you get to see something that you never, never see in regular carbon anomers except in what's called an inadequate experiment, which we'll talk about at the very end of this - the quarter -- which is carbon-carbon coupling. And if you look at the spectrum, now we see some neat stuff. So this is the proton decouple. Wrap your head around it. This is a proton-decoupled carbon anomer. But of course you still see carbon-carbon coupling and carbon-phosphorus coupling. So you look at this peak over here, and you look at the lines over here, and that peak is ADD. That's your carbonyl. So that's your carbonyl. It's a doublet of doublets. And doing the same thing we did before of taking 1 minus 2 and 1 minus 3, we can extract our J values. So our DD here for the carbonyl ends up being J is equal to - let's see, I get 59 Hertz. And I get 6 Hertz for analyzing that DD. And if I do the same thing for this carbon here, and I look at this carbon -- and we can see it over here as this very, very, very, very nice, very pretty doublet of doublets with this CH2. And I do the same thing here, and I get 134 Hertz. And I get 59 Hertz. [ Writing On Board ] All right. So let's figure this out, because now we have a puzzle problem. We have three different J's, and we have three different relationships here. What does the 59 Hertz have to, have to, have to, correspond to? [ Inaudible Class Comment ] It has to correspond to the carbon-carbon coupling, because carbon - because coupling is mutual. So your J1CC is equal to 59 Hertz. Now this guy is a doublet of doublets with 59 Hertz and 6 Hertz. So what does the 6 Hertz correspond to? [ Inaudible Class Comment ] Two-bond phosphorus-carbon coupling. So your J2PC is equal to 6 Hertz. And the last coupling we observe is 134 Hertz. What's the 100 - for the methylene here, what's that coupling? [ Inaudible Class Comment ] 130 - it must be 134, yeah. [ Inaudible Class Comment ] That's your one-bond carbon-phosphorus coupling. That's your - right. One bond, J1 - whoops. Did - yeah, right. That's your J1PC is equal to 134 Hertz. All right. Last thing -- stretch your imagination. I don't have a phosphorus anomer. But imagine I had a proton-decoupled P31 anomer spectrum. What would the peak for phosphorus look like? One peak? Everyone agree? [ Inaudible Class Comments ] Splitting pattern. What would its splitting pattern be? [ Inaudible Class Comment ] A doublet of doublets. And what would the J's be? [ Inaudible Class Comment ] Six and 134. So if we took a proton-decoupled phosphorus anomer, you'd expect to see a pattern that looks -- what is this, Richard Nixon here? -- [ Laughter ] -- pattern that looks like this: a doublet of doublets with a really big J for the one-bond coupling and a much smaller cut pattern for the two-bond - the CPPP coupling. All right. Well I think I've taken enough of your time today. So that sort of wraps up, and that'll set you in good stead to attack this next homework set that has all sorts of cool coupling sets. You had a question? >> Yes. ------------------------------0be8e63a723f--
B2 中高級 化學203.有機光譜學。第15講.涉及其他原子核的偶合 (Chem 203. Organic Spectroscopy. Lecture 15. Coupling Involving Other Nuclei) 60 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