Subtitles section Play video Print subtitles >> Today's lecture is basically going to show you that a lot of the rules that you learned in sophomore organic chemistry like the N plus 1 rule are simplifications and we'll be seeing a lot of examples that look really weird because I want to show you where these simplifications break down and then next time we'll get to more sort of the typical rules of splitting. So as I said, I wanted to begin with this notion of magnetic equivalents and let me begin with a definition. We'll say magnetic equivalents and let's say that 2 protons or nuclei, in other words, we're normally talking about proton NMR but, of course, it could be fluorines or phosphorescence or whatever are magnetically equivalent. [ Writing on board ] If they are chemically equivalent. [ Writing on board ] And remember 2 nuclei that were chemically equivalent were nuclei that were interchangeable by a symmetry operation or a rapid process like rotation about a bond. So we saw lots of examples of chemically equivalent nuclei by symmetry and then we saw how when you have, for example, a chiral center, methylene group is no longer chemically equivalent. So chemically equivalent. So there is a subset of chemically equivalent and they have the same geometrical relationship to all other nuclei in the spin system. [ Writing on board ] So this brings up one other concept and that's the question of what is the spin system and the spin system is just a complete set and I'll underline complete meaning all of the nuclei. So complete set of nuclei in which members are coupled. [ Writing on board ] All right this is, let's start with the easy idea, the spin system, and let's just do this by example. So, for example, if we have ethylproplether, CH3CH20, CH2, CH2, CH3, we have in the molecule 2 spin systems. We have 1 spin sustain comprising the ethyl group and another spin system comprising the propyl group. So in other words, the ethyl group is a set of nuclei. Obviously we're talking about the protons since for all intents and purposes there are no C13s in this fragment. So we have these hydrogens and these hydrogens and at least one member is coupled to every other member. They make up a set together. In the propyl group, we have the methyl group, the methylene group and the other methlyene group and there's coupling among them. In other words, the CH3 is coupled to this CH2, the hydrogens of the CH2 are coupled to each other, the hydrogens or the CH3 are coupled but that doesn't count. The hydrogens of the CH2 are coupled to the next CH2 and what's important. So each of these is complete meaning it takes in all the coupled nuclei. What's important is we don't have coupling between the ethyl spin system and the propyl spin system. So they kind of have 2 separate sets so we can consider the ethyl, we can consider the propyl and there's no interaction between them. Let's try another example. Let's take a acetyl phenylalanine and we'll take the methyl, the mid. [ Writing on board ] So what are the spin systems in this molecule? [ Writing on board ] So you have the 2 methyls on the end and the benzo group. [ Pause ] For all intents and purposes the benzylic protons are not coupled to the phenyl so what would you do for the spin system here? [ Inaudible question ] Separately so I'm going to revise this, okay, so we have the phenyl that's going to be 1 spin system. What do we do here in the middle? >> Couple those 2 together. >> Alpha and beta. What about the NH? [ Inaudible question ] We saw an example where I said you have an ND20 and you exchange so it's deuterium there, which although it has a spin for all intents and purposes you can discount but what about this NH? Is that going to be J coupled? Remember, amides are different than alcohols. Alcohols exchange amides and, remember, I said alcohols can exchange or cannot amides on the laboratory timescale if I throw them in D20 will exchange but on the NMR timescale that NH stays there and we're not doing this in D20 so what should I do with this middle part of the molecule? So that all becomes a spin system and what about the very end of the molecule? [ Inaudible question ] Okay, good. So these guys interact. Now, I'll tell you right now so we have 4 spin systems in the molecule and we have the methyl group, we have the NH, the alpha and the beta protons and we have the phenyl group and then we have the methyl and mid group. So forced in systems in a molecule and we can look at each of these separately. I'll tell you there's a miniscule like undetectably small and I'll show you how to see it if you squint right later on coupling between these hydrogens and the benzogroup, but for all intents and purposes you can say that the phenyl group is not coupled over here. So for all intents and purposes we have this is an isolated spin system, a phenyl group, the alpha, beta and NH, the methyl and methyl NH. Other thoughts? [ Pause ] [ Inaudible question ] Yeah, yeah, and so in chloroform solution, in D20, these would eventually wash out but in chloroform solution what we'd see for this NH is a doublet. It would be a little bit broad. This one is going to have 3 coupling partners in chloroform solution where this has an exchange or in DMSO. So we'd see either a doublet of doublets of doublets or a triplet of doublets or doublet of triplets and we'll talk more about that if you're not familiar with those terms. This NH is going to have 3 coupling partners in chloroform solution where this hasn't exchanged or in DSMO. So we see either a doublet of doublet of doublets or triplet of doublets or doublet of triplets and we'll talk more about that if you're not familiar with those terms. This NH would appear as a quartet and chances are it would be broadened out a little bit. Remember, I mentioned this nitrogen quadrapolar coupling? So couple of ways this can appear so it can appear as a 1 to 3, 3 to 1 quartet slightly broadened. It can appear just due to this nitrogen quadrapolar broadening as an envelope that encompasses the whole thing. Or it can appear as something where if you don't see the wings of the quartet and you just see a little dip you might say, oh, it looks like a doublet to me. So depending on the quadrapolar broadening from the nitrogen this methyl group in turn is not going to have significant quadrapolar effects; it's going to be split into a nice doublet. So this will be a quartet or broad quartet or something that looks like a single. If it's very broad, the methyl group will be a doublet. [ Pause ] All right let's now tack all this notion of the same geometrical relationship. So, let's look at this molecule. Let's take 2, 6 dichloro 1 tert butyl benzene. So as far as chemical equivalents goes, we have 2 types of protons. We have the proton that's powered to the tert butyl group and the proton that's meta to the tert butyl group. So these 3 constitute a spin system. Chlorines don't count; they're quadrapolar nuclei and essentially not spin active. The tert butyl group is magnetically isolated; it's its own spin system. So we look at this and we say, all right, we have 2 protons that are the same as far as chemical equivalents; they're interchangeable by a symmetry operation. Now we ask this geometrical question. Do they have the same relationship to all other nuclei in the spin system? This hydrogen says, oh, look, I'm ortho to this hydrogen and this hydrogen says, oh, look, I'm ortho to it also. So, these 2 are magnetically equivalent as well as chemically equivalent. [ Pause ] Now there's a way of naming systems where you have different types of protons and we'll give a different letter to each type of non-chemically equivalent proton. So, for example, we'll use letters like A and B and C and M and X and Y if you need to. The general idea is if the protons are close in chemical shift we'll use letters that are right next to each other in alphabet, As and Bs and Cs. If they're far apart in chemical shift, we'll use letters that are far apart in the alphabet. Letters like A and X or A and M and X. So depending on whether these protons are close in chemical shift to the center proton or whether they're fall in chemical shift, we'll either call this an A to B spin system or an A to X spin system. Now, technically only ones where they're far apart are truly first order, but even if they're close there's some very regular patterns that you can see. If they're far apart in chemical shift and by far apart what I mean is the separation of the peak centers in hertz is many, many times the coupling constant. So like a typical ortho coupling constant is about 7 hertz, so if the peaks are far apart like 10 times as far apart like 70 hertz or 100 hertz or 200 hertz apart, then they will end up being As and Xs. Now, remember, at 500 PPM, 1 PPM is 500 hertz. So, in other words, if these guys are about two-tenths of a PPM apart, you know, three-tenths of a PPM apart, we would call this an A to X spin system. What we'd expect would be to see a doublet for the 2 on the outside because they're being split by the 1 in the middle and a triplet and so I'll just draw a little squiggly to indicate these are far apart in the spectrum and a triplet like so for the center hydrogen. I guess technically the triplet would be shorter than the doublet so I'll make the doublet a little bigger. If they're close together, and I'm going to actually start in just a moment with the archetypical AB system, if they're closer together, what you will see so if it is, indeed, A to B, what you'll see is a slight tenting inward depending on how far. In other words, the lines of the doublet instead of being equal in height will become unequal in height with the bigger line toward its J coupling partner and the lines of the triplet will be similarly distorted so that the inner line is a little bigger than the outer line. I always like to think of these as sort of tenting in toward each other and that would be what it would like as an A to B system. [ Pause ] Let's try another example and I'll take difluoromethane. [ Writing on board ] Remember, fluorine is spin active, spin of a half, it's magnetic gyro ratio [phonetic] is about 90% of that of a proton. So it shows up a million miles away whereas your protons are resonating at 500 megahertz, your fluorine is resonating at 470 or 460 megahertz. So they are far, far, far away from each other, but they're J coupled to each other so collectively the hydrogens and the fluorines constitute a spin set, a spin system. If I want to remember my geometry because we're going to ask what type of spin system it is, it's a tetrahedral molecule. So the geometrical relationship of this hydrogen to the fluorine is the same as the geometrical relationship of this hydrogen to the fluorine. In other words, we would call this spin system an A2, X2 spin system. We have 2 hydrogens that are chemically equivalent, they're interchangeable by a symmetry operation, and 2 fluorines that are chemically equivalent, they're interchangeable by a symmetry operation reflection, and the hydrogens if you test everyone has the same geometrical relationship to all other nuclei in the spin system.