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  • >> So when we were talking about coupling constants,

  • I said sometimes there are some really big differences

  • between coupling constants, distinct differences

  • between coupling constant so like an alkanes,

  • if you have a cis coupling, your J3 H-H is about 10 Hertz,

  • if you have a trans coupling your J3 H-H that's your three

  • bond coupling or your vicinal coupling is about 17 Hertz

  • and by the way I will again caution you

  • to make sure you understand the difference between coupling

  • through a double bond versus coupling between double bonds

  • so a lot of people particularly when they are starting

  • out confuse stereochemistry with confirmation.

  • You can rotate about single bond so even if you have a bond

  • between two single bonds that might be S-cis

  • or S-trans that's always a dynamic equilibrium.

  • It may be heavily weighted toward one way,

  • but that's not stereochemistry that's confirmation and that

  • by mind you has a different Karplus curve then the Karplus

  • curve for an SP2-SP2 carbon pair that's connecting two hydrogens.

  • So anyway this is specifically not on here but let's now talk

  • about cyclohexane which is what I'm going to choose for today's,

  • today's lecture and just remind us of some

  • of the key features I pointed out, previously we talked

  • about the Karplus curve for SP3-SP3 systems and we said

  • that there are some very distinct relationships that are,

  • that in cyclohexane,

  • so 180 degrees gives you a big coupling constant,

  • a 60 degree relationship gives you a small coupling constant.

  • So our axial-axial coupling has a 180 degree dihydro

  • relationship, those of you who saw the previous sophomore class

  • that was just filing out got to see Hanna talking

  • about Newman projections

  • and if you're Newman project you have a 180 degree dihydro angle

  • and typically you have about 8 to 10 Hertz

  • for an axial-axial coupling.

  • If you have an axial equatorial coupling,

  • these are of course all three bond couplings.

  • We're in equatorial-equatorial coupling,

  • you have a 60 degree dihydro-angle

  • and you remember the Karplus curve that I drew out before

  • and so that's about 2 to 3 Hertz for each of those.

  • So what I'm going to do right now is pass out a or,

  • or mention a very real example that just came up in my lab

  • and I thought it would make a cool, cool example for spect

  • and it was a stereochemical problem associated

  • with the conjugate addition to a, an alpha,

  • beta and saturated nitro compound.

  • So we were working with nitrocyclohexene,

  • the nitrocyclohexene is electrophilic

  • at the beta position and we were doing an addition of aniline,

  • I'll write that as pH, NH2 to give the addition product.

  • [ silence ]

  • and so the issue here was whether we have the

  • trans product

  • [ silence ]

  • or the cis product

  • [ silence ]

  • and so of these two stereo-isomers what do we call

  • the relationship between these stereo isomers?

  • Diastereomer, so of these two diastereomers each of them

  • of course is formed as the rasimic and that doesn't matter

  • because you can't tell one enantiomer from another by NMR,

  • the spectrum of a racemic mixture is identical

  • to the spectrum of either

  • in individual enantiomer unless you have some sort

  • of chiral solvent or unless I prepare a chiral derivative.

  • So the only way I could distinguish these two apart

  • would be say to make an amid the two enantiomers apart would be

  • to make an amid where I had a chiral acid brought

  • in to make an amid or to use something

  • like a chiral shift reagent or chiral solvating agent.

  • So the big question is which of these two is formed, the cis,

  • the cis or the trans and I want to pass out the spectrum,

  • we will take a lot and analyze it.

  • [ silence ]

  • And I have a few extras here, and some anyone else?

  • One more back here, great.

  • [ silence ]

  • . Alright, great, so we have here the NMR spectrum

  • and chloroform solution and what I've done here is I've blown

  • up this region, the midfield region

  • so we call this region the downfield region,

  • this real region, the upfield region

  • and I'll call this the midfield region.

  • So our molecule of course is night has a fennel group in it

  • and so I think it's pretty obvious

  • that our fennel group is over here.

  • You might notice for example if you look at the fennel group

  • that we have something that looks like a doublet

  • or an apparent doublet

  • that corresponds to the ortho protons.

  • There may be some metacoupling going on there, we're not going

  • to be talking in any great detail about this,

  • but I just want you to get in the habit

  • of always reading your spectrum

  • so that corresponds to the ortho.

  • We see something that looks like a triplet,

  • that's the same integral, if you notice the height

  • in the integral from here to here is identical

  • from the height from here to here so that triplet

  • or apparent triplet corresponds to the meta protons,

  • there are two of those and then we see something that looks

  • like a triplet over here and again it's either triplet

  • or an apparent triplet

  • or there's some small additional coupling, let's say HM

  • and that corresponds to the power of proton.

  • I don't know if you can see on here, but you certainly can see

  • on yours, do you notice where the ortho-proton

  • and the meta-proton and the, and the para-proton show up?

  • What's their chemical shift roughly?

  • 6.6, 6.7 I said aromatic are typically 7 to 8.

  • Do you know why they're little upfield?

  • Nitrogen donation, so the electrons pushing

  • in from resonance off of the nitrogen.

  • By resonance, we get extra electron density

  • at the ortho-position and the para-position and so

  • that ring is shifted a little bit upfield

  • which is kind of cool.

  • If I made an amid out of that nitrogen

  • which would pull electron density out,

  • these guys would scoot over a little more downfield

  • and here this little guy

  • over here this is your core form [inaudible].

  • Alright, all of this is all of the other cyclohexyl stuff

  • [ silence ]

  • that are not directly involved and then we have three peaks

  • in the midfield region.

  • Remember right now we don't know stereochemistry

  • and we don't yet know what's what.

  • Why do we have three peaks?

  • The NH, so we have these two protons and the NH.

  • Something very interesting happened with this sample

  • and this wasn't anything that I'd planned on,

  • it was simply the day, a day later I went

  • to rerun the sample, same sample

  • and the midfield region looked a little bit different.

  • This is your day old, day old sample,

  • this is the freshly prepared sample.

  • What's going on there?

  • [ silence ]

  • . So we still see it, so James said the nitrogen is swapped

  • with deuterium so we still see it, but now it's broad.

  • What does that tell us?

  • Maybe moist.

  • If it were very, very wet, you see this peak over here,

  • so this is and the sample was nicely sealed

  • so this little peak here at 1.6 is H2O,

  • that's about where you see at 1.6, 1.56 that's very typical

  • for H2O on chloroform, so we don't have a lot of water

  • in there, probably about 10 millimolar, 20 millimolar water

  • and we have lots and lots of sample

  • and what else did I tell you about chloroform?

  • Can generate HCl and a little bit, even a little bit

  • that forms is going to catalyze the acceleration, the exchange

  • of NH's between molecules or between molecules and water

  • so even if you form just a little bit 1 percent for example

  • of I guess would be DCL but the majority of course

  • if it's one percent of your, you know of your molecule,

  • one molar percent majority is still HCl,

  • but what's happening is now your NH is instead of saying attached

  • to the same molecule are getting swapped between molecules

  • and as a result in this case it's also getting swapped

  • with water, but as a result if that proton doesn't stick

  • around for tens of milliseconds or hundreds

  • of milliseconds you can't see the spin as to whether it's been

  • up or spin down, so we lose coupling, it becomes a singlet,

  • in this case a broad singlet.

  • Now let's look at these patterns here in the unexchanged samples,

  • so these are all the midfield.

  • So which ones the, the NH over here?

  • 3.4 ish, the doublet, is the NH and so the other two

  • and we'll talk more about this in a moment, but the other two,

  • one of them changes a little bit in pattern visibly.

  • Alright let's, let's talk now

  • about what we call these patterns.

  • So what do we call these patterns, we saw it last time.

  • It's a triplet of doublets, you have two big coupling constants

  • and one small coupling constant.

  • What do we call this pattern here?

  • A quartet of doublet, so you're having roughly 1:3:3:1 ratio

  • for little doublets and if you wanted to be fussy

  • about it you could say well you know my quartet isn't quite

  • perfect, I can see these two aren't quite the same

  • or they're ones litter fatter than the other you know

  • because all for coupling constants may not be exactly the

  • same, but they're the same within a Hertz

  • or half a Hertz basically that within the line within the limit

  • of digital resolution.

  • So if you didn't like to analyze it is a QD you could call it an

  • apparent QD, I'm fine with either.

  • I think either of these accurately reflects what you see

  • and here if you said you know I kind of sort

  • of see a little humpy thing over here and over there,

  • you could call it an apparent so I'll write or apparent TD,

  • I'm happy with either analysis on this.

  • Alright, so and you'll notice this peak simplifies the quartet

  • of doublets simplifies which makes sense

  • because that must've been the proton that was coupled

  • to the NH group because we lose that coupling so that quartet

  • of doublet now has become a TD or an apparent TD.

  • Well it's not deuterium,

  • remember even though you're getting a minuscule amount

  • of DCL, let me put it in numbers.

  • This sample is 50 millimolar roughly,

  • we have may be generated 0.5 millimolar DCL,

  • but that DCL is enough to catalyze the exchange

  • so now this NH isn't sticking

  • around on this molecule every millisecond this NH is bumping

  • into another molecule and trading partners

  • or every millisecond this molecule is bumping

  • into NH a little bit of the H2O

  • in the sample and trading partners.

  • Typically when I prepare sample like this I pass the solvent

  • through flame or furnace dried alumina to remove traces

  • of water and DCL first if I want a nice spectrum

  • or DMSO-D6 is another good solvent

  • because if hydrogen bonds nicely and typically keeps your OHs

  • in place by stabilizing them, so carboxylic acids that are hard

  • to see in chloroform often show up well in DMSL.

  • Question?

  • >> [inaudible]

  • >> And the water, so the water could help the exchange,

  • but apparently not enough that it actually you know

  • that proton actually does stay in place very nicely

  • and it's just when you get some acid in the sample.

  • >> Yes so I was saying that the double, the double,

  • the NH doubling [inaudible].

  • >> A singlet so we would call this a broad singlet, BRS

  • >> But then the QD [inaudible].

  • >> So okay, great question.

  • The QD is becoming a TD and remember how we teetered

  • on the edge here between being a triplet of doublets

  • and you can kind of see a little bit of humpy stuff,

  • all the 3 J's aren't exactly alike,

  • they're all different sorts of protons

  • with their own specific dihedrals coupling to it.

  • So there are some differences and so depending on the shimming

  • of the instrument, one day one time it looks like a TD

  • and the other time this looks like a DDD with two very,

  • very similar coupling constants so we can just

  • at this point see eight lines here and so

  • at this point just due to chance of the shimming,

  • we're able to resolve the two coupling constants that differ

  • by about a Hertz so becomes a DDD that's,

  • this is purely, purely coincident.

  • If I had this with slightly less good shimming it would have

  • worked exactly like that.

  • Alright, what I want us to do now that we thought

  • about this a little bit is

  • to extract the stereochemistry from here.

  • So we by figuring already actually happen to know then

  • that this proton here corresponds to this one,