3. The 1H NMR and the IR spectra of compound A are shown...

3. The 1H NMR and the IR spectra of compound A are shown below. In the mass spectrum the parent ion appears at m/z = 164, the base peak at m/z = 121, and there are few other fragments. a) (5 points) What is the molecular formula of A? __________ b) (10 points) What is the structure of A? Structure: s, 1H d, 2H d, 2H t, 2H m, 2H t, 3H 1691 cm-1 1601 cm-1

Transcript

00:01 All right, for a spectroscopy question, there's always something you're a little better at than others, and just focus on the things that work the best for you.

00:10 If you really like mass spec, you can subtract these out and get that you are 43 different, so you're probably going to have a propyl group that's getting popped off.

00:20 For the structure of a, we've got a dead giveaway that we have an aldehyde.

00:28 We know that we have to have benzene because it lives in 7, because we've got two groups that are identical.

00:38 We have to have two substitutions.

00:41 So our benzene has to look like this.

00:49 Ch2, ch2 and ch3.

00:53 Well, looking at the ch3 is a triplet.

01:01 So chances are is connected to a ch2 that will give it the triplet.

01:05 So looking out the two triplets, we've got one that is a triplet, well that has to be at least a quartet if it's connected to the ch2.

01:17 So let's look at the other one.

01:20 We see that it has one, two, three, four, five, six.

01:24 So it's got five buddies around it.

01:27 So if the other ch2 was attached to it, then these ones are the highlighted ch2.

01:36 All right, let's look at the last one.

01:42 So it's a triplet, so it should have two buddies on it.

01:45 And that's exactly what this highlighted ch2 has.

01:49 So chances are we're going to have this n -progroup coming off from our benzene.

02:05 Looking at the ir, well, again, we see that there's something at 1601.

02:10 That's give away that's benzene.

02:12 And we've got a carbonyl group...

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