NMR spectroscopy, talking

about more dynamic things involving decoupling.

Then a little bit about C-13 NMR and double labeling

involving correlation.

And then we'll get onto another reaction,

electrophilic aromatic substitution.

OK, so proton decoupling.

Remember that the proton is four times as strong a magnet

as the Carbon-13, so it precesses much faster.

So it's easy to do a pulse that will tilt the Carbon-13

net magnetization down into the plane, so that it'll

precess and be an antenna that can tell us its local field

without doing the same thing to the proton.

And if we do that, then we see the C-13 NMR spectrum.

And there are two peaks, because of

the adjacent proton.

Right? it can be either up or down, 50:50

or almost exactly 50:50

so you see two peaks.

Now suppose at the same time you're doing that, when you do

this pulse and listen to the C-13 to see what its local

field is, you irradiate at 100 MHz.

Right?

That won't do anything to the C-13,

it's the wrong frequency.

But it'll now make the proton precess.

That is, if we look in the rotating frame, we can see a

weak field that's horizontal, about which the

magnetism will precess.

So the protons will go down, and then up, and then down,

and then up.

And what determines how fast it goes up and down?

How strong this field is.

The stronger it is, the faster it will precess.

And if it's fast enough, then you won't see a doublet for

the C-13 anymore, because you'll see the average.

Right?

So you'll see a single peak, you'll proton-decouple the

C-13 spectrum and see the C-13 as if the

protons weren't there.

OK, so it depends on how much power you put in.

Now here's a C-13 spectrum of a given compound.

Where we're listening to the C-13 and we see all these

different local fields, right?

But we're irradiating the proton at the same time, but

very, very weakly.

This number up there, 40 decibels, is an inverse ratio

of how strong that horizontal field is for the protons.

So it's 10-4th weaker than what you'll see you

later, 10,000 times weaker.

And if we look at the carbon, we see that some of the carbon

signals are split and some are not split.

The ones that are split are the ones that have

hydrogens on them.

So if we start from the left, we see there's one with no

hydrogens, then one with one hydrogen,

one with one hydrogen.

That red one is the solvent

CDCl3, right?

So we're looking at the carbon of the CDCl3.

But it has a deuterium on it, not a

hydrogen, so it's not split.

Then there's a carbon with two hydrogens, a carbon with three

hydrogens, it's a quartet.

Underneath, a carbon with two hydrogens, it's a triplet.

So very low power in the proton, that's what we see.

Now we're going to reduce the size of that.

And see what happens, is we start increasing the power.

So there's 40 dB, 20 dB,

and now things are beginning to change.

See how the patterns are changing their multiplicity

and their intensity?

Because it turns out that when you irradiate the protons by a

complicated mechanism it strengthens the signal of the

carbons to which they're coupled.

And that's called nuclear Overhauser enhancement, as

you'll see.

OK. So now you see the splitting is going away when

we get down to two decibels or one decibel.

Now we see that each carbon gives just a single peak, we've

decoupled the protons.

So now we have a peak for every carbon.

And notice, incidentally, that the CDCl3 didn't get stronger

as we did this.

Why?

Because what strengthened the signal was irradiating the

protons and CDCl3 doesn't have hydrogens on it.

So it doesn't get strengthened.

OK, notice also that the carbon without a hydrogen on

it, on the far left here, didn't get strengthened as

much as the others did.

OK, so that's decoupling, and it can be useful.

And it can be useful in several ways.

There's two things, you get rid of the spin-spin splitting

so you see just a peak for each carbon.

And also you strengthen the signals when

they have nearby protons.

And that could be bad if you're interested in measuring

the size of the peaks to get how many carbons there are,

because you can't count properly anymore because you

get different intensities with different

neighbors of protons.

But it's good if you want to determine structure and see

what hydrogens are how far away, and I'll

show you that shortly.

OK, now bear in mind that carbon-13 is only 1% in

natural abundance of carbon.

Of course, you can synthesize the molecule with expensive

carbon-13-containing precursors and put carbon-13

where are you want it in a molecule, and then the signal

from that carbon will be ever so much stronger.

If you put in pure carbon C-13 in some position, the signal

would be 100 times stronger.

In fact, 200 times stronger than without.

No, pardon me.

It would be 100 times stronger, because it would be

100% instead of 1%.

OK, so here's a compound, neotame, it's

an artificial sweetener.

And it's the proton-decoupled C-13 spectrum.

So you see one peak for each carbon.

There are 20 carbons, there are 20 peaks.

But you can't go by their intensities.

Now, let's see what we can make of this.

It's proton-decoupled, so there's one peak per carbon

and, being carbon, it's pretty well spread out.

Remember, the range of hydrogen chemical shifts is

only about 10 parts per million.

The range for carbon is hundreds of

parts per million here.

200 parts per million are shown.

Now why is there no C-13-C-13 splitting?

There's no proton splitting because it's decoupled.

But why no C-13-C-13 splitting?

STUDENT: The likelihood of two C-13s adjacent in the same

molecule is unlikely.

PROFESSOR: If it's only 1% natural abundance, and you're

a C-13 so that you're giving a signal, the chance that your

neighbor is C-13 is only 1%, so there would be tiny, tiny

peaks from those

you don't see them.

OK, so that's what Chris just said.

OK now, let's see if we can figure out which carbons these

are, that we're looking at.

And of course, this is how it was done empirically.

People put in known compounds and figured out what the

chemical shifts... what peaks

were from one compound to the next.

And then they figured out this comes in this range, this

comes in this range, this comes in this range.

So let us see, knowing what these peaks are.

OK first, there are these ones that are way

down close to 200.

They're numbers 1, 13, and 4.

So what do 1, 13, and 4 have in common that turns out to

make them way far downfield?

Ellen, what do you say?

What do 1, 13, and 4 have in common that makes

them different from the other carbons?

STUDENT: They're almost like esters.

Like, double bonding to an O.

PROFESSOR: Yeah, the double bond to an O. Carbonyl carbons

are way downfield.

OK, now we've got 7 and 8 to 12.

So there's 7 and there's 8, 9, 10, 11, 12.

Pretty clear what those are, right?

Those are the ones on a benzene

ring, aromatic carbons.

OK, then there are ones that are a little bit further down

than the others, and those are those blue ones.

What do they have in common?

Rahul, what do you say they have in common?

STUDENT: Tertiary?

PROFESSOR: They're not all tertiary, no. In fact, only

two of them have three things associated with them, the

others have one and two.

Lauren, can you see what the blue ones have in common?

In hydrogen, what makes things come a

little bit further downfield?

Yeah Amy?

STUDENT: Well, the hydrogen would be like deshielding from

an oxygen or something, like nitrogen or oxygen.

PROFESSOR: They all have an electronegative atom next to

them, right?

Nitrogen or oxygen.

So those are ones. A carbon that's attached to something

electronegative.

And then, finally, you have the others, which are attached

only to carbons.

OK, so you could begin to figure out these, and you can

get tables of what comes where.

And then you can start dealing with unknown compounds and

figure out what it's telling you.

But a nice thing about this proton-decoupled spectrum is

that there's just one peak for each carbon, and they're well

spread out.

So that it's good to do this kind of thing.

So those are the alkane carbons.

OK, now the power of correlation.

Often when you can measure two things on the same subject

simultaneously, you get a lot more information than

measuring one, or measuring the other, or measuring them

independently.

I'll show you that example with C-13 double labeling and

also in two-dimensional.

As it's called, nuclear magnetic resonance.

So first, C-13 double labeling.

And we'll talk about lanosterol biogenesis.

In fact, we already talked about it, if you look back at

lecture 52.

OK, so remember we got squalene from isopentenyl

pyrophosphate.

Then it curled around and we had this complicated thing

where it zipped up going one way, and then we had all these

rearrangements.

So as we said at that time, this is the source of

cholesterol and the steroid hormones.

And it's a very cute story, how all that rearrangement

took place

OK. But there's a question whether it's true.

And I said at that time, wait for NMR. So now we're at NMR,

so you can get the answer.

OK, so you can use Carbon-13-labeling to figure

this out.

So if you feed the plant that makes this stuff Carbon-13

labeled isopentenyl pyrophosphate--

in that particular carbon, that substituted carbon--

then you can see where that would appear in the squalene.

According to the way we linked them together before, it

should be those carbons.

Then when you curl it around and do all this stuff to make

lanosterol, those peaks should be labeled with C-13 and

therefore 100 times stronger than the other

peaks in that spectrum.

And indeed they are.

So that supports it.

Or you could do it a different way.

You could label the methyl group.

And if you label the methyl group, it turns out that those

methyl groups should be labeled in squalene.

And then if you curl it around, it should be those

methyl groups.

And indeed, they're 100 times stronger

than the other signals.

But much more interesting than this is to do double labeling.

So if you label both, you get that.

Big deal.

That's just the same information

we got before right?

Except done in one experiment rather than two.