字幕列表 影片播放 列印英文字幕 >> I want to spend the next 3 lectures, 3 classes, talking really closely about first order coupling and the reason is that there's so much to be gained by deeply understanding NMR spectra. As I said, a lot of what one is going to be doing is asking specific questions about stereochemistry and being able to ask those questions is intimately linked to understanding what's going on. Also just in general for solving structures, being able to read spectra, really read them at a level that goes beyond the level of sophomore organic chemistry, involves intimately understanding [inaudible]. So we're going to take a relatively slow path through this. In fact, we're going through the midterm exam only have 1-d spectra on our exam so that we really focus on understanding things. So I want to start by kind of making the bridge between last time's lecture where we talked about magnetic equivalents and we talked about non-first order systems and so last time was sort of the bad and today is going to be the good. So the bad is that I said a lot of the rules that you learned in simple sophomore organic chemistry really were oversimplifications. There are very few systems that truly behave in the way that you learned they should behave. These are the first order systems. So first order systems are anything like AX systems, AMX systems, A2MX. In other words, anything where coupled protons, protons within a spin system are far apart in chemical shift and if you do have 2 protons that are chemically equivalent like we have in A2MX system that those protons are both chemically equivalent and magnetically equivalent. We divided this and separated it from non-first order systems. [ Writing on board ] And these are systems in which you either have magnetically inequivalent protons that are chemically equivalent or you have protons that are similar in chemical shifts. For example, a non-magnetically equivalent protons we saw, for example, A, A prime, X, X prime systems and we talked about just how ugly those systems could be. Those were like the phthalate [phonetic] system where I said no matter how far apart, no matter how high a magnetic field you look at dioctyl phthalate or ortho dichlorobenzene is never going to get better than this complex pattern of lines and then I said we have other systems like AB systems where the protons are similar in chemical shift and ones that are related to this, for example, ABX systems. The good news about many of these types of systems is that many of these non-first order systems behave very much like first order and that you can start to apply some type of simple, rational understanding to them, which is more than I can say for an AA prime system, XX prime system or an AA prime BB prime system. Now sometimes these systems will look like first order, which is great because sometimes you can analyze these types of systems as first order and many times you can, but what I tried to show you last time was how there are ones that simply defy simple reduction. >> So what do you use X or Bs based on the distance or the separation of chemical shift not the actual distance between them? >> So let me show you exactly and let's take the AB system because I think this is a great starting point and what's nice is the AB system is going to be an archetype for many sorts of systems that although they're not first order we can apply first order analysis to and we can start to see the distortions that occur. So, a pure AX system is one in which you have a doublet so it's 2 hydrogens that are J coupled. Again, that's going to be the whole spin system so I'll just put on XX and YY to represent some other nuclei that aren't going to couple and not, of course, something with a hydrogen on it where it's J coupling. So you would have a doublet and then a big, big span between it and then another doublet. This little squiggly is just a break, break in the spectra. If those 2 doublets are far apart in chemical shift, then you're going to see them each as a simple 1 to 1 doublet. Now as the distance between them becomes smaller. In other words, either you have different substitutents that instead of having them be very far apart they're closer together in PPM or you simply went to a lower field spectrometer, now you start to see a distortion that we would call an AB pattern where the inner line, and so now instead of saying these are effectively very, very far apart now I'm saying they are far apart like so. In other words, this means, you know, 1 here and 1 way over there. Okay, now the typical way in which one characterizes this is the distance between these line sis the J value, the distance between these doublets and technically one takes not the dead center of the doublet but the weighted average because technically with a multiplet the position of the multiplet is not at its average but at its weighted average. In other words, since this line is a little bit bigger we take the center as just a hair over. It's the weighted average. In other words, if this line is 4 times as, if these lines are in a 4 to 3 ratio and they're separated by .07 PPM, we'd say, all right, you're .4 of the way over there; just a little hair. So, if we call this distance Delta nu, typically if Delta nu over J is much, much greater than 10. [ Writing on board ] We're in the situation like this and if Delta nu over J is less than or equal to 10 and those are approximations, then we're sort of into this AB situation. By Delta nu I mean the difference in position in hertz. So in other words, let's say the center of this line was at 7.30 PPM and the center of this line was at 7.10 PPM and let's just say here that our J value is let's say R, what will work out? What will work out well? Let's say that our J value equals 17 hertz. Now, imagine for a moment you're on a very low field spectrometer. Imagine you're on a 100 megahertz spectrometer what's Delta nu at that point? [ Pause ] 730 hertz. Everyone agree? >> Delta [inaudible] 20 hertz. >> Delta, 20 hertz. So at 20 hertz these guys would be hugely close together. In fact, we'd have a situation that looked. [ Pause ] Like this. At this point Delta nu actually will be just a hair further apart because it's the weighted average. I'm going to shift it over just a hair. I'll make the outer lines just a little bit bigger. This would be a situation where Delta nu over J is very small where Delta nu is about 20 hertz and J is about 17 hertz. If we had the same system at 500 megahertz, what would the difference in, what would Delta nu be for 500 megahertz? [ Pause ] A hundred hertz, right? So at 500 hertz, 500 megahertz, Delta nu is equal to 100 hertz. So you'll look at this situation and at 500 megahertz you'd be more like this and 100 megahertz you'd be like this. So this is your AB pattern and if they were ever closer they'd be like what I sketched out before where the inner line would be huge and the outer line would be very tiny. What? That might, it would be like a 60 megahertz spectrometer like one of the freshmen or sophomore and we actually have like a 100 or maybe it's 60 in the sophomore lab it would be like this or imagine the situation that instead of having substituents that put these apart at .2 PPM, imagine they were separated by .1 PPM, but the main thing to keep in mind is for any given doublet no matter what the center of this peak whether I looked at it at a 500 megahertz spectrometer or at a 100 megahertz the center of this peak is going to be 7.30 and the center of this peak, again, weighted average center is 7.10. Now 17 hertz is more characteristic of a trans alteen [phonetic], which was actually what I was doing when I was drawing this. For something like this we'd be more like about 7 hertz for a J value. Thoughts or questions at this point? [ Inaudible question ] Okay, will the center move, so, if you improve the equipment? So here we've gone from this is our 100 megahertz, this is our 500 megahertz and the point is the center of this peak for this whatever hypothetical compound this is, the center of this peak is always at 7.3 PPM whether I'm at 500 megahertz or 100 megahertz, but the distance between the peaks because the number of hertz per PPM is much smaller at 100 than at 500, the distance between the peaks here is very far, it's 100 hertz apart or relatively far, and over here it's only 20 hertz apart. [ Inaudible question ] The inner one? The closer they are together the more they tent into each other and that really is the difference between the AX. [ Inaudible question ] The center is related to the ratio of the bigger. [ Inaudible question ] Absolutely. Absolutely. Well, here the bigger one is at 7.29 PPM or 7.29 PPM and the smaller one is at 7.31 PPM and by here we've got these 2 lines and one of them is at 7.2 PPM and the other is at 7. whatever the number is. Now what's valuable about looking at AB pattern and understand it is it really becomes an archetype for all sorts of systems that behave very near to first order. So we were talking before about phenylalanine and I guess the example I gave when we were talking about spin systems was acetyl phenylalanine methyl amide. [ Pause ] And I pointed out that we had 1, so this is like a spectrum in chloroform solution so I'll say NCDCL3, and we decided that we had 1 spin system over here and the multiplicity of this proton of the NH is a doublet because it's split by 1 coupling partner. Each of these protons they're non-chemically equivalent so they split each other but they're going to be similar in chemical shift, they're similar in the environment so they'll both be at about 2 and a half, 3 parts per million. Why do I say about 3 parts per million? Well, they're off of a phenyl group so if we were methyl group off of a phenyl I'd say 2 parts per million. It's an ethylene so that pushes it to like 2 and a half parts per million. They're beta to a couple of electron withdrawing groups, they're beta to a nitrogen, they're beta to a carbonyl, so then it's going to shift them downfield by about another half a PPM. So we'd expect them to both be at about 3 parts per million but probably not to be on top of each other. So each of these is going to show up at as a DD and that DD is going to be part of what looks like an ABX pattern because this is part of an ABMX system. M is something that's far apart from either A and B and C and so forth and X. So we have 1 proton that's going to be way downfield and nitrogen protons are typically at about 7 parts per million. One proton that's going to be moderately downfield because it's next to an electron withdrawing group and it's alpha to a carbonyl and beta to a phenyl group so this is going to be about 4 and a half parts per million. Then these guys that are both going to be close to 3 parts per million. So we have far apart from this and H far apart from alpha and the alpha is far apart from the beta. So this guy here is going to be split by 3 different protons. So he's going to be a DDD if all of the Js are different or a TD, and we'll talk more about these or DT, if 2 of the Js are the same. Or a quartet if all 3 Js are the same within the limits of experimental error. [ Pause ] So the one I really want to draw our attention to then is these 2 hydrogens here because now this type of AB pattern really can serve as an archetype for more complex patterns that are non-first order but are close to first order. So an ABX pattern is something where you have the AB pattern in which each line is further split. So imagine this type of pattern here with some level of separation, but now with each of these 2 lines split into a doublet. So what you see is line line, line line, and then line line, line line for these 2 protons. [ Pause ] This is for the ABX system. So this is what we're seeing right at about 3 PPM. [ Writing on board ] Which of them is which? [ Inaudible question ] Okay, so stereochemically one of these protons is pro R and the other proton is pro S. They're diastereotopic protons. If we can go ahead and so, for example, if I replace this with a deuterium in the thought experiment, then the ranking of the carbons, the ranking of the 4 constituents on here becomes highest rank, next rank, next ranked and so that would become an S center. So this is a pro S proton and this proton is pro R. If I knew the geometry here, for example, if I knew the phenyl group preferred to point in one way or another or I could by Nuclear Overhauser Effect experiments and the like detect certain proximities, then I would be able to get an experimental correlation or predicted correlation based on say proximity to anisotropic groups of 1 proton and 1 peak. So right now I don't know which is which, but with additional experiments -- this is obviously just a sketch -- but with additional experiments in context, yes, you can figure out with diastereotopic proton is which. Other thoughts and questions? [ Inaudible question ] Do they ever look the same height? Great question. So, right now I've made a sketch for a situation in which these are relatively near to each other. In other words, maybe they're separated by a tenth of a PPM. So if they're separated by a tenth of a PPM, the JAB here is going to be about 14 hertz. At 500 megahertz that would be a separation if they're separated by a tenth of a PPM about 50 hertz. So Delta nu over J would be about 50 to 14 about 3, but if they were very, very far apart, if something held these in very different magnetic environments, then you would see the outer lines getting bigger. [ Writing on board ] And the peaks becoming more like a regular doublet of doublets. So if they were further apart, it would look more like this, these guys would be bigger. They would be tenting into each other less. Other questions? These are really important and this is one of the reasons I'm going really slow over this. [ Inaudible question ] Coupling, coupling is always going to be mutual. So if we call this, let's name our peaks 1, 2, 3 and 4 and let's name our peaks 1 prime, 2 prime, 3 prime and 4 prime. So the JAB is going to be 1 minus 3 and 2 minus 4. They will within the limits of experimental error be the same and JAB here will be the same within the limits of experimental error. It will be 1 prime minus 3 prime and 2 prime minus 4 prime. In this case since it's an ABX pattern, the coupling with the other proton so the coupling with the remote partner, we'll call it AX, JAX equals 1 minus 2 and 3 minus 4 and, again, those will be the same within experimental error. So let's say for a moment this is 14 hertz that's also 14 hertz. Let's say for a moment that this is 6 hertz, actually it looks the way I've draw it it looks more like about 9 hertz so let's say this is 9 hertz, that's going to be 9 hertz within the limits of experimental error. Here this distance will also be 14 hertz as will that distance. This distance the way I've drawn it looks like it's about 12 hertz and that looks like it's about 12 hertz. [ Inaudible question ] Ah, where? Oh, no, no, no. One minus 2 will not equal to 2 minus 3. So over here JBX equals 1 prime minus 2 prime and 3 prime minus 4 prime and, yeah, if you look at this pattern and you draw a splitting tree diagram, we split into a doublet so that's our big J and then each of those lines is further split with a small j and so you get this pattern of line line, line line, and if I call my lines 1, 2, 3 and 4, 1 minus 2 is the small j, 3 minus 4 is the small j, 1 minus 3 is the big J, 2 minus 4 is the big J. [ Inaudible question ] From the AB coupling, the geminal coupling in this particular case, and the source of the small j is from the coupling to, in this case, this nucleus over here. [ Inaudible question ] That's going -- beautiful question -- that's going to depend on the geometrical relationship on the Karplus curve so typically you will not have exactly the same coupling between 1 of the 2 protons, let's say the pro S and this proton versus the other proton and this proton. So rather than my, let me put up some real data. You'll see exactly the same thing but at least it will be a nice, nice chance for you to have a real spectrum and I know I passed this out before, but we didn't look as deeply at this spectrum. Now let's look at it with a fresh pair of eyes. Let's look at it more deeply. So I didn't use this exactly compound. I just grabbed this right off of the Aldridge [phonetic] webpage and remember you can go to sial.com, www.sial.com, to get yourself lots of spectra. It's a great way to check out your ideas and your understanding of things. So here we see a real compound. I've shown you this before. This is phenylalanine in D20. [ Pause ] So unlike the example that I, the hypothetical example I gave in chloroform, this one doesn't have an amide here it has an amine here and it has a carboxylic acid. In D20, those exchange and so this becomes ND2 and you essentially see no coupling and don't see it. Actually there's DCL here so this really becomes ND3 plus and you don't see coupling. This becomes D. So this system here what remains of the 2 methylenes and the methine, so this is an ABX system and so you see coupling here and this is your phenyls and this is your HOD and this is your alpha proton and these are your beta protons. So this is a very real example of what I've sketched out and you'll notice the distance between these 2 lines does, indeed, match the distance between those 2 lines. In other words, the 1 to 3 or 2 to 4 distance does match the 1 prime to 3 prime or 2 prime to 4 prime distance, and you'll also notice that the 2 couplings with the alpha proton are a little bit different, a little bit different from each other. So it looks like if I had to eyeball it here that our coupling here, this distance between 1 and 2 or 3 and 4 is about 6 hertz and the distance here, the distance between 1 prime and 2 prime or 3 prime and 4 prime is about 8 hertz. So you'll notice now the alpha proton is split into a doublet of doublets so each of these is a DD ABX pattern, DD ABX pattern, and then you'll notice that our alpha proton is a DD and the doublet reflects the 2 different couplings. In other words, the distance between lines 1 and 2 or 3 and 4 corresponds to this coupling, to this 6 hertz coupling, and the distance between lines 1 and 3 and 2 and 4 corresponds to this coupling to the 8 hertz coupling. [ Pause ] Thoughts or questions? [ Inaudible question ] Well, great question. So the question is if the alpha proton is a doublet of doublets shouldn't it be leaning a lot more? You notice these guys are really tenting into each other and this one is just barely tenting in. So now look this is a 300 megahertz spectrum. So the distance between these 2 looks like it's about a tenth of a PPM. So they're separated by about maybe it's two-tenths. So, the distance between these 2 is about 60 hertz because it's 300 megahertz so it's 300 hertz per PPM. So they're separated by about 60 hertz and the J value is about 14 hertz so that's the case where Delta nu over 12 is about 4 or 5 whereas here the difference between the alpha proton and the beta protons is about 1 hertz, about 1 PPM, about 300 hertz, and the J value is 6 and 8. So this is a case remember I said the big difference between the AB and AX type of situation this is a case where Delta nu over J is very big, 300 versus 6 or 8 is a factor of well over 10. So they're effectively far apart and you get very little tenting inward. Other thoughts? [ Inaudible question ] So great question. So this is in D20 the most hydrogens on heteroatoms exchange and so most hydrogens on heteroatoms. As a matter of fact, I will say I can give you exceptions but I will say hydrogens on nitrogen, oxygen are replaced with D. So they get replaced with deuterium. Deuterium shows up in completely, shows up not 500 megahertz but at about 80 hertz, 80 megahertz. So they don't show up in the same spectrum and the J values are so small that for intents and purposes you don't see coupling plus they're exchanging very quickly. [ Inaudible question ] Even without DCL they will exchange. Because of DCL the amine is protenated so the amine is an ammonium group and because of DCL it dissolves whereas phenylalanine in just pure water wouldn't be nearly as soluble. All right so I've started to hint that different types of coupling relationships have different coupling constants and what I'd like to do at this point is to talk about typical coupling constants and see how we can use them to enhance our understanding. So, this coupling constants generally if you needed just one number to keep in your head you could keep 7 hertz or 6, 6 to 8 hertz. Let's say we're talking SP3 to SP3 right now and if I needed a number, actually, I'm going to put this as a general CH to CH. I'll show you double bonds in a second but if you need 1 number to keep in your head, 6-8 hertz or 7 hertz is a great number to keep in your head without confirmational preference. I'll say without a confirmational biases. What do I mean by a confirmational bias? Well, if there's a strong, strongly held relationship, for example, if 2 hydrogens are locked in an antiperiplanar relationship so here we have 180-degree dihedral angle, now our J value is going to be bigger. It's going to be about 8 to 10 hertz. So, for example, if you have 2 axial protons. So I'll put this as JAX, AX, that would be a typical example for 8-10 hertz where you're locked into an axial relationship. If you have something locked into an equatorial relationship where now you have axial equatorial or equatorial, now you're talking about a 60-degree dihedral angle and so a typical JAX equatorial or J equatorial equatorial is on the order of 2. I'll put little tildas, little squiggles here just to indicate that that's approximate 2 to 3 hertz. [ Inaudible question ] This is based exactly off of the Karplus curve. So a general way of thinking about coupling is that coupling comes from interaction of the nuclei with electrons in the bond that polarize the next bond, that polarize the next bond. If at one extreme you have 180, a 90-degree dihedral relationship between those 2 bonds, you get no overlap of orbitals. If at the other extreme you have 180, you get very good overlap and antiperiplanar relationship and at 0 you also get a good overlap and so you see very large coupling constants at 180 or at 0 and very small at 90 or 60. So, Karplus relationship is basically a relationship between theta, your dihedral angle, and J, and as sort of a general relationship would be if we go from 0 to 90 to 180 that we go at 0 it's about 8 hertz. We have kind of a cosine wave going down at 90 to a minimum and up to about 10 at 180 degrees. [ Pause ] Now this is sort of for general kind of plain vanilla, it's modulated, the coupling constants are modulated by electronegativity and hybridization. In general, electronegative substituents lead to a smaller coupling constant. In general, if you've got an SP2, SP2 bond between the 2 atoms like a double bond, you have bigger J values. So let me show you a couple of examples. So as I said, first number to keep in your head is about 7 hertz, but now if you want to think about some oddball situations, you can think of like an aldehyde where now you have an electronegative oxygen and you have an SP2 carbon here and aldehydes are very funny in that your coupling constant is on the order of 2 to 3 hertz so that's sort of unusual. Alkenes, it's good the keep. These are numbers that really are worth keeping at your fingertips. For a CIS alkene we're talking typically on the order of say 7 to 12 hertz with let's say 10 hertz being typical. For a trans alkene we're talking maybe 12 to 18 hertz or 14 to 18 hertz. Let's say 14 to 18 hertz with maybe 17 hertz being typical. So these are all examples of vicinal couplings. One more example falling right in that sort of general 7 hertz range. Let's take on a benzene as an example. On a benzene as an example, we're talking maybe for ortho coupling maybe 6 to10 hertz with maybe 8 hertz or 7 hertz typical. [ Pause ] Two bond couplings tend to show more variation than 3 bond couplings. So, for example, if you have 2 carbons on a methylene group, on an SP3 hybridized carbon with different substituents on the carbon you can see anything from say, oh, 5 to 20 hertz depending on the electronegativity if it's just sort of carbons on here maybe 14 hertz is typical. [ Pause ] If you have an SP2 carbon on a double bond, you're talking maybe 0 to 2 hertz. Maybe 1 hertz being typical. [ Pause ] So all of these are examples of vicinal and geminal coupling. In other words, 2-bond and 3-bond coupling. These are J2 HH couplings, these are J3 HH. If you have anything more than that, if you have greater than or equal to 4 bond coupling, we're talking long-range coupling. So, for example, J4 HH would be an example of long-range coupling. Normally in saturated systems, you don't see coupling, but if you have a system where you have certain geometrical relationships, then you may see it. So, we're talking about say a carbon not with its neighbor but with a carbon 1 over. So where does this come up? Usually if you have intervening double bonds. So, for example, [inaudible] systems. We've talked a little bit about this in discussion section. Typically let's say 0 to 3 hertz depending on the geometrical relationship. Metacoupling on a benzene same type of thing. Let's say 1 to 3 hertz. [ Pause ] The only real situation that you can actually see a visible splitting where you have just SP3 carbons is if you have a locked relationship what is sometimes called W coupling. So usually you need a locked relationship where you have a series of antiperiplanar bonds. This occurs, for example, in the norbornene ring system. They call it W coupling because you have a W-like relationship. So say these 2 hydrogens on a norbornene you can see you have the series of antiperiplanar relationships that make a W. Norbornene actually is loaded with W coupling so you have another W relationship across the ring like this and there's even a third geometrical W relationship hiding in the molecule like so. [ Pause ] So I want to pass out, there's 1 table that's really useful in your book, and I want to pass it out just because the stuff that's in the back of your book is so much better when it's passed in front of your eyes rather than simply waiting there in the back of the book undiscovered and uncared about. So this is the Appendix F I mentioned before and I just want to show you how many good things are hiding in this 1 little appendix here. So everything we've talked about and more is hiding in this appendix. So we have our [inaudible] coupling and this is all going to come up on homeworks. If you're wondering what your typical [inaudible] couplings are, you'll find answers here. If you're wondering what happens if you have a double bond next to a double gone, you'll find answers here. If you're wondering, we've already seen in the homework we've already seen coupling across acetylenes and so how many bond couplings is that? [ Inaudible question ] Four-bond coupling, right? So 1, 2, 3, 4 bond coupling, but if you ask as many people do when they come up with some of the homework assignments here, well, can you couple further? Can you get 5 bond coupling? The answer is right here waiting, waiting for you to read about it. Right over here typically you see a little coupling across there. Coupling on pyridines is going to come up won the homework and already I think we see that very soon. Now, all of this is given in more detail in Prech [phonetic]. So Prech has wonderful examples for pyridines, for thiophenes, for all these types of systems off of real compounds and off of typical examples, but here distilled into really 1 little appendix and really 1 page of the appendix is so much different good stuff that's going to help you out with some of the problems that you're working on. So, okay, I think this is where I'd like to wrap it up today. We're going to talk more about first order splitting next time and we're going to walk through some examples of doublets of doublets and triplets of doublets and doublets of triplets and doublets of doublets of doublets and that'll prepare you for a workbook assignment that comes later on. ------------------------------bb79b27d2c62--
B2 中高級 化學203.有機光譜學。第十二講.系統中的耦合分析 (Chem 203. Organic Spectroscopy. Lecture 12. Coupling Analysis in Systems) 36 3 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