about the main events in NMR

spectroscopy for determining structure.

That is, about the chemical shift and about spin-spin

splitting, that allows you to tell how many nearby

nuclei there are.

Today we're going to talk about some complications that

make the spectrum more complicated, and perhaps more

informative, in terms of high order effects, higher-order

effects in spin-spin splitting.

And then about using NMR to study, not structure, but to

study rates.

And we'll do some C-13

labeling stuff there.

So spin-spin splitting with other nuclei besides protons

next door, and we actually saw some of this last time with

deuterium splitting, with C-13.

splitting.

For example, if there's a signal like this and you blow

it way up by 30 times, then you can see that.

What gives these things that are so far apart 126 Hz.

Do you remember this from last--

what are the size of splittings you normally see

for protons that are on adjacent carbons?

They're the order of 7 Hz, right?

Here we have got 126.

What's the big thing that's going on?

What makes coupling so big?

It's C-13-H. In fact, 99% of the sample--

if you haven't specifically put C-13s

in there, are C-12

but about 1% are C-12.

So you see that.

Now here's a paper from 1959, which first noted something

interesting about these C13-H coupling constants, that they

relate to the amount of s-character in the carbon

atomic orbital that's involved in the bonds.

Remember, we said that the information has to be

communicated through the electrons because the action

of magnets through space

if you average over all angles of rotation in a liquid--

goes to zero.

So you communicate through the bonds.

The nice thing about...

that it's so strong for JC13-H, is that

C-13 has just one bond to get to the H. So it's a very short

path to communicate.

Now, but you remember, also, that to do that-- so that it

doesn't average out as you tumble--

you have to have the electrons in an s orbital.

So the hydrogen uses its s orbital, that's all it's got

essentially.

But carbon can use s or p. So the hybridization, as they say

here, relates to the coupling constant.

Because to communicate that electron must

be on the C-13 nucleus.

So this is a table from that paper.

It gives the coupling constant in units of per second.

Those were then changed to be called Hz, nowadays.

So here are a bunch of different compounds and

their JC13-H coupling constants.

If we make a histogram of them, you see that there's a

group between 120 to 140

one between 150 and 170

and another group down around 250.

So there are different groups involved here.

There are three different populations, we could draw the

lines between them.

And then if we look to see what the carbons are, that

bear these hydrogens that show that splitting, we could look

first at the very big ones down there.

And what's special about those carbons that are

splitting the hydrogens?

They're triple bonded.

So sp hybridization

50% s in the bond to hydrogen.

The ones here, the green ones, are in aromatic systems, so

they have sp squared hybridization.

Except for cyclopropane.

Why would it have such a high coupling constant?

Why would it have a lot of s-character

in the bond to hydrogen

the carbon that's in the cyclopropane?

Because its bonds normally are tetrahedral-sp cubed

but when you try to make a smaller angle, more

p-character in the bonds to the other carbons, therefore

more s-character coming out to the hydrogens.

So we can actually see that hybridization change in the

NMR. And then these other are normal methyl groups, I speak

sp cubed hybridized carbons.

So here in NMR, we have direct evidence for this

hybridization stuff we talked about last semester.

So it's sp, sp squared, sp cubed.

Now here's a subtle point.

Perhaps we shouldn't be spending time on it, but it's

so interesting. And I suspect the curious among you may have

wondered about this.

Because we talked about coupling

goes through the bonds.

So from H to C to C to H. How about H to C to H?

That's a shorter path.

So methane, if you're looking at one hydrogen, how many

other hydrogens would you think it would be coupled to?

Three.

And they could be in 1:3:3:1.

So you should see a quartet for a proton of CH4.

But you don't, you just see a single sharp peak.

So where did the splitting go?

And this is related to what we talked about last time, where

a book said the splitting constant is zero.

And I said that's not true, the splitting constant is

high, maybe 10.

But it's not observable.

So why is it, why don't you see quartets for CH4?

OK, so here's one of these spectra-- actually if you go

up here and you can click on that and get taken to the

website of Chem 220 where they have problems. I'll work one

problem later in the class, if we have time.

But there are others you can practice on there.

So this is a compound, and I'll tell you its answer.

It's cinnamic acid.

It's related to the adelhyde related to this acid,

cinnamaldehyde is the smell of cinnamon.

So what do we expect when we look at the proton spectrum--

this is the proton spectrum of it-- what do we expect?

Well, notice that we can draw resonance structures, which

will put positive charge on the carbon with that hydrogen,

so remove electron density from that hydrogen.

Or from that one, or that one, or that one.

So those all should be shifted downfield relative to where

they would otherwise be, relative to the ones that

aren't blue.

When we look at the compound, then, we see these really

complicated peaks in here. We throw up our hands trying

to understand that.

And then we see here two doublets

really nice, sharp, clean things.

So what do you think the two doublets are out here?

Let's ignore that bit for the time being and talk about

these two doublets.

Which protons do you think those come from?

Which one is this?

OK, let me start going around the compound.

The OH is way down further.

We're not showing that.

OK, so I go down.

I come to this one.

What do you think that is?

STUDENT: The one near the COOH.

PROFESSOR: No, that's the one up here.

This is where double bonds normally come.

Then you come here, and that one has a hydrogen on double-

bonded carbon, but it's down here because it's has lost

electron density.

Remember we drew a resonance structure where the blue

ones were more positive.

And then here, we have hydrogens that are on the

benzene ring, shifted down this way because of the ring

shift, the aromatic ring current that we

talked about last time.

And now there's a group of three and a group of two.

These are positive and are shifted down.

That's the group of two.

The group of three here is these two, plus this one.

Which would be shifted down because it's positive, but up

because it's far from this sp squared carbon.

Let's not worry about why that is.

But at any rate, that's the benzene ones.

But that's not what we're going to talk about here.

What we're going to talk about is the clean

signals. Here and here.

Now this is measured with a magnetic field of 7.05 Tesla.

That means it's a 300 MHz spectrometer.

So instead of using parts per million down here, we could

change to Hz if we wanted to, change

to a frequency scale.

And in terms of frequency, if we call zero the middle, then

that's lower-frequency to the top, higher-

frequency to the bottom.

It turns out that the ratio of the chemical shift difference,

delta--measured in Hz, not in parts per million

we want to be comparing apples with apples

here, Hz with Hz--

so the chemical shift from here to here is much bigger

than the coupling constant, J. It's 24.4 times as large, and

that's the pattern we see.

Now suppose we used a smaller magnetic field.

What's going to happen?

The chemical shift in parts per million

will stay the same.

But it'll be the same fraction of a smaller frequency. If we

use a smaller field.

The precession frequency gets smaller.

So if we measure in Hz, as we're doing here, they'll come

closer together

but J will stay the same, because it doesn't

depend on the field.

The magnets are not dependent on the

field, they're coupling.

So what will happen?

These two doublets will move together.

So at 2.89, it turns out the ratio is 10.

Or at point.

0.86 Tesla, it would be 3.

Notice, incidentally, that this would be a much cheaper

spectrometer because it would have only 1/10 as big the

magnetic field as the other one.

But also, those other peaks are going to be in the middle.

Everything's going to be getting on top of one another,

and I'm just showing you what would happen

to this double doublet.

There it is at 0.29.

And finally, if we're 100 times smaller at 0.2 for the ratio

of chemical shift to J, measured in the same units, in

Hz, you get the purple pattern.

Now notice that the J is always the same, the splitting

between the two peaks

but what do you notice between the doublet?

What's curious about this?

I mean, I'm sure you all see.

Some of the peaks are disappearing and others are

getting stronger as they come together.

The outer peaks are getting very weak.

And in the limit where the chemical shift was zero, the

difference-- as in methane--

then the outer peaks disappear altogether and you

just see one peak.

Now notice that that part in the original spectrum was a

mess, because all those hydrogens of the benzene ring

have similar chemical shifts, and they're all coupled with

one another so it's more of this same thing

and all on top of one another, and very complicated. So we'll

forget that.

Let's think about what causes this change.

This is a so-called higher- order effect

not the simple thing, that you have a chemical shift, count

the neighbors and get 1:2:1, 1:3:3:1, 1:4:6:4:1

kind of thing.

So here we're plotting the energy in a big applied field

of different nuclear spin states.

And we're talking about just two nuclei.

So they could both be with the field; that

would be low in energy.

Or one with and the other against, or one against and

the other with.

And then they could both be against the applied field.

So you'd have these different energies, right?

And that's the 100 MHz, that's the energy that it

would take to get from here to here by flipping the red spin,