from your homework, and I want to do this

as a demonstration regardless

of whether you've already solved this problem.

In fact, I hope that you've solved this problem

because what I'd like you to do is to be able

to think really deeply about it, and show you my perspective

on structure solving and in particular we're going

to be talking about using HMBC as a very focused tool

to help solve structures and specifically for the problem

of putting the pieces together.

You've seen on the templates that I've given you,

I've had this tremendous emphasis, or at least I've tried

to have this tremendous emphasis on thinking your way

through the problem, reading the spectrum, getting the formula,

writing out fragments, jotting down what you know,

jotting down what you don't know if possible, and then being able

to ask focused questions.

HMBC is an incredibly data rich technique,

it also has ambiguities to it because we get two bond

and three bond couplings.

So you've got a huge amount of information, a huge amount

of data and what I'd like us to do today is to learn how

to use it as a focus tool.

So I want to zip through the problem from Silverstein

at the beginning and then focus on the HMBC angle.

So this is problem 8.43 from Silverstein, and I want to jot

down my -- you know I've worked this problem,

just like you folks, and I want to jot down my impressions

on working the problem.

So I look at this problem and I generally start with --

I have -- I mean what's the most useful thing you can get?

Molecular formula and functional group.

That's sort of the most general stuff, and we'll be able

to get molecular formula in just a moment

because we'll actually see all the hydrogens

and all the carbons so we basically will get them,

get the oxygen's by difference.

So before we start with the NMR spectra though,

looking at the IR spectrum, I see what looks like an OH group,

I see what looks like a carbonyl.

At 1728, it's a little high for a typical ketone, right.

A typical ketone is a little but lower, it might be an ester.

I've been emphasizing it's very hard, in any sort of level

of complexity, to get CO single bonds stretches associated

with esters.

It's probably hiding right here at 1188.

You sort of get a hint at it, but we'll see in a moment,

there's some other hints there pointing--

pointing towards ester.

We probably have an alkene -- that's probably an alkene CH,

it could be an aromatic, we'll see in a moment it's not.

That's probably an alkene CC bond.

All right, mass spec it's 200 molecular weight

in the EI mass spectrum.

What I'm going to do before we --

before we actually get the formulas,

so maybe at this point I'll just jot down sort of a question mark

on the IR spectrum -- ester is sort of my thinking on this.

All right, at this point I want to go and look

at the proton NMR, look at the carbon NMR, and I've really,

really, really been emphasizing this notion

of keeping the resonances separate from the atoms.

In other words, we know resonance is something you can

see, we're going to be assigning those resonances

to the structure as we build the structure,

and so we have a common language, I go ahead

and I simply letter the peaks A, B, C --

it's hard staring into the light -- D, E, F, G, H, I, J,

and we number the carbon resonances, and I'll go 1, 2,

3 -- that's our chloroform -- 4, 5, 6, 7, 8, 9, 10, 11.

I like to do a good job of working the integrals.

My philosophy, every particular integral has experimental errors

associated with it.

The best way that you can do it if you know the number

of hydrogens in the molecule or you know multiple hydrogens,

is add up a bunch and divide by the number of hydrogens.

If you don't or you're in a hurry, often you can get it

by dissection with a ruler.

These integrals are a little hard to read,

they're starting right in the baseline; they're not offset.

You can use a ruler and draw straight lines,

you can slap a grid on it.

I'm just using my ever, ever useful grid, and if I'm looking,

I start to see a ratio that's just a hair under 6,

a hair under 6, a hair under 3, hair under 3, hair under 3.

This one's just a little bit --

actually this just a hair under 6, hair under 9,

so basically we're talking about like 2 point --

2.8 units, these happen to be tenths of an inch per hydrogen.

This guy is interesting, he's coming up a little bit,

this one right at 1.6 ppm, so he's coming up over .6,

and I'll tell you about that in a second,

and then we see one that's just a hair under .9,

and another that's just a hair under .9.

So you pretty much, by inspection, can go ahead

and get -- and again it's really hard staring in here --

2H for A, 2H for B, 1H for C, 1H for D, 1H for E, 1H for F.

[ Inaudible ]

2H for F, thank you, 3H for G. Now H is interesting, 1.6 --

this is in chloroform solution, and chloroform,

water typically shows up at about 1.6 parts per million,

and so you can kind of see the water peak right over here.

So you have this multiplate that's reasonably symmetrical,

and then you'll have a little bit

of water off on this side here.

I would go in if I were at my own spectrometer --

I'd go in, and zoom in and get a better look and see,

but I think you can see the integral just gives a little

extra kick up for the water.

So H is 2H, I is 3 -- 3H, J is 3H.

So in other words, our molecular formula has a total

of H20 in it.

Now it's certainly possible, if I had an amine or something,

that the amine NH would not show up or an alcohol NH.

So I'm not immediately locking it in,

and this is really important,

you have to be keeping your wits about you.

Secondary amine NH is aliphatic.

Secondary NH's are terribly hard to see.

Alcohols can be broad.

Alcohols can be with the water peak.

Carboxylic acids can be broad as well, and can be disappeared,

particularly if you're in chloroform

versus DMSO due to exchange.

If we look at the carbon NMR here, peak number one is

at about 170, just a hair down of 170, about 173 ppm

where you would expect for an ester,

not where you'd expect for a ketone.

We've got a couple of peaks over here, if I want to go ahead

at this point, I'll say, one is a quat, two is a quat,

looking at the depth, three is a CH2, four is a quat,

five is a CH2, six is a CH2, seven and eight are all CH2's,

and then nine, 10, and 11 are all CH3's.

So basically if we do this, we see we have C11H20.

If we have symmetry in a molecule, which you've gotten

in a couple of homework problems this week,

of course you might end up with a count that's lower,

if your carbons aren't all different,

or you could have overlapping peaks,

but symmetry's a typical reason.

If you have a plain phenyl group, you're going

to see four carbon peaks, but you're going to get six carbons.

If you have a ring like a cycle propane ring,

and there's no stereo center in the molecule,

you may if you have one point of attachment have two methylene's

that are the same, but if I look here, I'm suspecting an alcohol,

I'm suspecting an ester.

I have 200, if I take away C12, C11,

and H20 from that, I get O3.

So I'm reasonably happy at this point,

I think I have a molecular formula.

Now, the next thing I do

with the molecular formula is I'll start to --

it's nice to have a scorecard here, so I might, for example,

over here, jot down some thoughts; ester, alcohol,

alkene as some of the group that we're seeing in the molecule.

If I want, I can calculate degrees of unsaturation,

another very useful thing.

Remember, I don't do it by formula,

I just say, C11would be H24.

I'm at H20 so we have two degrees of unsaturation.

So I can help keep my wits about me.

That also tells me if I have a carbonyl,

and I have an alkene I know I have no rings in the molecule.

All right, now the next place --

any questions or thoughts at this point?

>> So it doesn't matter [inaudible]?

>> Does it?

Let's see.

>> Maybe I'm doing my math -- I'm doing my math wrong.

>> And this is actually a good point in checking yourself.

I have an incredible ability to screw up arithmetic on my feet

which means go ahead and double check yours.

>> I think you're actually right I calculated the formulas wrong.

>> All right.

>> [Inaudible] C could and D would be the OH.

So we've got some interesting points here you can already see.

So if you're reasonably experienced,

I'm going to bet anything

that we have a stereo center in the molecule.

Now, even though I haven't measured the coupling constants,

I could measure the coupling constants,

they very conveniently give me a scale here

and you get a peak print out, but very conveniently,

a typical aliphatic CH coupling constant is about 7 hertz.

A methyl group next to a methylene is really very typical

for this, you'll -- a methyl group next to just

about anything is going to give you about 7 hertz.

So immediately if I'm eye-balling this,

even if I don't know I'm at 600 megahertz,

and I see this triplet here that's obviously a CH3 next

to a CH3 next to a CH2.

In fact I can write this as one of my fragments here

because I think we're pretty obvious at that point for this

and this here, but if I'm looking at that, I see okay,

these guys are leaning into each other, we have an AB pattern,

it's obviously a big coupling constant.

This separation is at least twice the separation

in the triplet, so it's like a 14 hertz

or 15 hertz coupling constant.

That is very typical for a germinal coupling.

So in the back of my mind I'm thinking stereo center,

something with the methylene that's diastereotpic.

If I have a stereo center, every methylene

in the molecule is going to be diastereotopic,

but usually the ones that are further

from the stereo center behave as if things are coincident.

So we look here, we have what looks like an ethyl group,

it wouldn't have surprised me

to see a more complicated coupling pattern

for this ethyl group, but it doesn't necessarily surprise me

to see a less coupling constant.

I'm pretty [inaudible] less complicated coupling pattern

just to see a plain old quartet.

I'm pretty certain I have an ethyl ester at this point.

Pretty confident I have an ethyl ester because that sure looks

like an OCH2CH3 off of an ethyl ester,

but it could be something else that's shifting it downfield.

All right, at this point it's time to get analytical,

and where I'd like to go next is actually the HMQC rather

than the COZ.

As I was saying in discussion the other day,

the problem is you're drowning in data in the COZ quite often,

and a lot of the data isn't particularly useful.

You'll have two diastereotopic protons, and they'll both couple

to another two diastereotopic protons or even to one proton,

and you're getting more information than you need

because you'll get the diastereotopic protons are

coupling with each other, okay, big whoop.

You'll get that each one is coupling to one of the protons,

all right that gives you some new information,

but that's redundant.

If you have two diastereotopic protons coupled

to two diastereotopic protons, now you've got lots and lots