00:03
This is the answer to chapter 20, problem number 61 from the mcmurray organic chemistry textbook.
00:10
This problem asks us about crotonic acid versus metacrylic acid.
00:20
And so we're given two nmr spectrums, and we're asked to decide which spectrum corresponds to which molecule.
00:29
So there's a lot about these molecules that are similar.
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They're both carboxylic acids.
00:36
We're going to get four signals in each of these proton nmr spectrums.
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But it's actually pretty easy to distinguish which of these two is which.
00:47
And in order to do that, we just need to look at the signal from the methyl groups.
00:53
And so in both cases, the methyl groups give a signal.
01:01
Right about 1 .9 parts per million.
01:06
So we have 1 .91 parts per million in part a, in part a, spectrum a, and then we have 1 .93 in spectrum b.
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And so what we need to look at is the splitting patterns here.
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So in part a, our methyl group signal is split into a doublet.
01:36
In part b, we just have a singlet.
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And so with that in mind, what we can do is think about what is splitting or not splitting the methyl groups here.
01:49
So when we look at crotonic acid, i'm going to draw this proton in green.
01:54
So we have one proton here on the double bond.
01:59
And then the protons that are actually giving rise to this methyl group signal are going to be right here, these three.
02:11
And so again, this is the signal at about 1 .9 ppm.
02:21
And so here we can draw the same thing for the metacrylic acid...