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  • >> 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.