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  • [ 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