00:01
In this problem, we're looking at proton nmr data and formulas.
00:06
We want to give compounds that are going to match these descriptions.
00:10
We have c2h6o.
00:13
First, we'll find degrees of unsaturation.
00:16
This will give us number of pi bonds in rings.
00:18
We take two times the number of carbons.
00:21
We add two, subtract the number of hydrogens, and divide by two.
00:25
We get zero.
00:26
We have no double bonds or pi bonds.
00:31
What if here, oxygen is in the middle, and then there's a ch3 group on each side? with this, if we think about our splitting patterns, we have n plus one, where n is number of neighboring hydrogens on the next carbon.
00:49
Both of these n would be zero, so there would be a singlet, and we have perfect symmetry.
00:54
It would show up in one peak.
00:55
This would be a reasonable structure.
01:00
Then, we have c3h7co, and we have a doublet and a septet.
01:14
Again, i'll find degrees of unsaturation, number of carbons times two, add two, subtract number of hydrogens, subtract number of halogens, divide by two, and again, we get zero.
01:29
When we see a septet, that usually means we have a carbon that's going to have two ch3 groups and one hydrogen.
01:43
This has six neighboring hydrogens, so that gives us seven peaks.
01:48
Then, our hydrogens here are covalent, so they would show up in one peak.
01:55
There's only one neighboring hydrogen, so we get a doublet.
01:59
This is going to be your structure.
02:03
Then, we have c4h8cl2o, and we have two triplets...