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