Subtitles section Play video Print subtitles >> I want to continue our discussion of first order analysis and analysis of couple in first order and your first order systems. And, as I said, the good news is that even though very few of the systems that we deal with are truly first order, many of the systems we deal with can be analyzed as if they were first order, realizing that sometimes you will see some deviation. So, today we're going to look at multi-plates and look at trying to really understand them, and to extract coupling constants from them. And so, we've already talked about multi-plates where you have all of the same type of coupling like a quartet or a septet and now I want to go and talk about multi-plates where we have coupling where you have different coupling constants, you know, to different types of protons. As I said, if you have different types of protons but the coupling constants are the same, so you have a proton that's coupled to a methane on one side, and a methylene on the other side, and all three of those coupling constants are the same, no big deal. It's going to be a quartet. You'll see it as a quartet, you can analyze it as a quartet, you can call it a quartet, and that's great. But, now we'll look at some first order coupling analysis where we have different coupling constants. And, I'll start with a simple example. Remember last time we were talking about typical coupling concepts and I said, well you know most of your vicinal couplings, most of your three bond couplings, are about seven hertz but, there's some that fall out of that range, and I said one of the categories that you have are alkenes. So, let's take a moment to think about alkenes and we'll think about tert-butyl, I guess technically you'd call this 3, 3, dimethyl, 1 butane, so we'll talk about tert-butyl ethylene here. So, in alkenes your cist couplings are on the order of about 10 hertz. I think I gave you a range of numbers and I said if you want to keep one number in your head, 10 is a good number to keep in your head. So, let's say we would expect about 10 hertz. Trans couplings are typically very different than cist couplings. I gave you a range of numbers from about 14 to 18 and I said if you want to keep one number in your head, let's keep 17 hertz as a typical coupling for trans. And, vicinal couplings are all over the map. I said for SP3 a sort of normal value might be 14 hertz. SP2 tends to be a lot smaller, just a couple of hertz, so if you wanted to keep one number as an expectation, I said zero to 2 hertz. Let's say about 1 hertz. And, let's take a look at how these residences, we would expect in [inaudible]. So, HA is going to be split by HB and HC, and HC is going to split with about a 17 hertz coupling constant, and HB is going to split it with about a 10 hertz coupling constant. To put it another way, in the different molecules, HA is going to see some molecules where HB and HC are both spin-up. Some molecules in which one is spin up and one is spin down. Some in which the other is spin up and the other is spin down, and some in which both are spin down. Now, in the case of a simple triplet where you have the same coupling constant, if one is spin up and one is spin down, or the other is spin up and their swapped, it doesn't make a difference. But, if your coupling concepts are different, then you're going to see different magnetic environments for those 2 molecules, and the result is that you're going to for HA, and indeed we will see for all the others that a doublet of doublets were DD. We always name our species by way of the first coupling constant gives the first name. The big coupling constant gives the first name, and the small coupling constant gives the last name. In this case of course it's moot because its name is doublet of doublets. But as we see of course when we get to triplet of doublets, and doublet of triplets it's going to matter. So, one way to conceptualize these split interactions is as a splitting diagram. And so we can say that HA is going to be split with a big coupling of 17 hertz, and so I'll just remind us that this is 17 hertz. And each of those lines is going to be further split with a coupling of 10 hertz, and I'm trying my best here to be proportional. And so we would expect, we would expect a pattern somewhat like this, four lines with a Lorentzian line shape. Like so. And if I call these lines, 1, 2, 3 and 4 and we see such a pattern, we see a doublet of doublets, our big J is always going to be 1 minus 3, and that's going to be the same as 2 minus 4 within the limits of experimental error, because there is experimental error in peak positions because of things like digital resolution which says although your NMR spectrum is depicted as a series of smooth curves, each curve is actually composed of a series of data points and the separation of those data points depends on your sweep with, and your acquisition time, typically you have about 30,000 to 40,000 about 32,000 to say 50,000 real data points divided over the entire width of the spectrum. So, if your spectrum width is 14 parts per million, and your 500 megahertz, so your spectra width is 500 hertz per PPM, your sweep width is 7,000 hertz, and if you have a 30,000 point spectrum that means, each of your points is going to be separated by a few tenths of a hertz, right? By 7,000 divided by the number of points, by let's say 32,000 real points. That would be called the 64K data set because in the Fourier Transform you collect 64,000 real -- you collect 64,000 points in the time domain, and when you do a Fourier Transform that converts you to 32,000 real points in the frequency domain and 32,000 imaginary points in the frequency domain. So, that spectrum is going to have 7,000 divided by 30,000 is what about a quarter, about .25 or so hertz, right? Yeah, about .25, .4 hertz digital resolution which means your points are separated by a few tenths of a hertz. Now for that very reason, when we report our spectral observations you don't want to report your coupling constants to better than a tenth of a hertz because by the time you're at a hundredth of a hertz, your numbers are insignificant. I mean think about it, your Lorentzian Line width is on the order of a hertz due to the uncertainty principle and errors and shimming. It's usually about 1.2 hertz. You have a digital resolution of about 1.2 hertz. So, that means you can determine the peak position to a couple of tenths of a hertz. So, what I typically do to get the best accuracy is I take the position 1 and position 3, and I subtract them, and I take position 2 and sub 4 and I subtract those, and then I average them. And in this case here, let's say I get 17.0 hertz, and I report this as a DD 17.0 10.0 hertz. Small J, similarly is going to be 1 minus 2 and 3 minus four, and I would take those and average them and get about 10 hertz. You will find on many of the homework problems as you go along and we get to the more advanced part of the course, I include a peak print out. If you want on the PDF, you can simply highlight that peak printout, paste it into excel, split your data from using the text to columns command, and then simply in your excel spreadsheet just take 1 minus 3 and 2 minus 4 and 1 minus 2 and 3 minus 4 and average them appropriately, and get your coupling constant out without having to resort to pencil and paper calculations. All right, so this is AJ. Let's take a look at what we'd expect for HB. So, HB is going to be split by HA with a coupling constant of 10 hertz and it's going to be split by HC with a coupling constant of 1 hertz. So, it too will be a doublet of doublets and I will try to draw everything proportionally on my black boards here. And so here we have our 10 hertz, and here we have our 1 hertz, and we get a pattern that looks like so. HC we'd also expect to be a doublet of doublets, and here we'd expect it to be -- have coupling constants of about 17 hertz and 1 hertz, so again I will draw my little splitting diagram. And that would be the doublet of doublets that we'd observe. So, how does that sound? Does that make sense? >> So, I have a question [inaudible]. >> Ah. Okay, the distance from 1 to 4. So, look we've moved apart 17 hertz, and then I've moved 5 hertz out here, and 5 hertz out here. [ Inaudible Question ] What? [ Inaudible ] Would be, I'm sorry 1.5 to 3.5? >> Yeah, 15 hertz over the [inaudible]. >> From 1 to 4? So, 1 to 4 is going to be 17 plus 10 hertz, which is 27 hertz, and actually you've hit upon something that's a super, super point. The difference between the first line in a multi-plate and the last line is the sum of all of the Js. Now, by all of the Js I mean with all of their multiplicities built in. And, this is what came up when we were talking about that problem in discussion section and we were trying to figure out how many hertz the spectrometer was, and I said well, let's assume you have a typical triplet. Let's assume a typical coupling is 3 hertz. Remember the problem, the question was what was the field strength of the spectrometer and we got 200 and something or 350, and we have one peak that looked like this, and another peak that was a sextet that looked like this. And so one way to do this problem was to measure this distance and say oh, that's about 7 hertz, but you're measuring it with your ruler and it's not so accurate. So, another way to do this is to say all right, so we'll measure this distance, that's going to be 14 hertz. Why? Because a triplet is a proton that's split by two 7 hertz couplings. So, 2 times 7 is 14, and then I say well the way I did it which got a little more accuracy was to look at the sextet. And since I said all of the Js are about 7 hertz, this distance here from outer line to outer line, corresponds to the sum of the 5 Js, all of which are the same that are coupling it. So, this corresponds to 35 hertz. And what's important about this then, is that the distance between the outer lines becomes a check sum. What do I mean? I mean that if you have analyzed your multi-plate correctly, and you understand it correctly, and you've all of the Js correctly, then you can go ahead and add up all of those J's, and of course if one of those Js is corresponding to a triplet, you add it in twice, and that's going to correspond to the distance between the first and the last. And, if you haven't gotten it right, it won't come out right. And what's also important is let's say a multi-plate is a little bit broad, and this came up with somebody in the [inaudible] research group, who's looking at confirmations of oxocarbenium ions, and couldn't exactly see the size of the multi-plate but he darn well could see whether the multi-plate was roughly 7 hertz wide, or whether the multi-plate was roughly 15 hertz wide. And that was able to tell him whether his proton was axial, or equatorial, on a cyclohexane type ring, on a pyran ring, and hence the stereochemistry and the confirmation. So, this is why this is extremely important. So, thoughts or questions? All right, I handed out before, but I realize not everyone brings the paper. I handed out before a two-sided thing, so if you had the handout from last time that had the phenylalanine on it, you don't need to grab a new one. But, I decided to make extras because I realize not everybody is prepared. Actually, if we can sweep the extras over to this side of the room now. That'll take care of a lot of stuff. All right, everyone have one or the other sort of copy? So, let's go ahead and let's take a look,