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  • >> 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,