00:01
In this problem, we have our molecular formula and we have two different sets of nmr data.
00:10
So our first isomer has a double it at 1 .2 and it represents six hydrogens.
00:22
We have a septet, so that's seven peaks, for one hydrogen at three.
00:36
And then we have a singlet at 4 .1, and that's for two hydrogens.
00:45
So let's look at our degrees of unsaturation.
00:49
This will tell us the number of rings and pie bonds.
00:52
So we multiply the number of carbons by two, add two.
00:57
We subtract the number of hydrogens and the number of halogens and divide by two.
01:08
So here we get one.
01:10
And based off our data, it's likely that we have a carbonyl, so double bond between carbon and oxygen.
01:25
So these two sets of data, we typically see when we have two ch3 groups coming off of a carbon and one hydrogen.
01:37
So this is this hydrogen, and these are for these hydrogens.
01:43
And this splitting patterns make sense.
01:46
This hydrogen has six neighbors, and these hydrogens have one neighbor...