[hardy5086]

I have been given a few pictures to go off to determine the structure of a compound with a formula of C5H12O. I haven't had to deal with any of these types of things before and it's rather confusing just reading from a book as it's overly technical.
So from the 13C NMR spectra I would say there is 4 distinct chemical environments as there are 4 peaks. I'm guessing the peak at approximately 22 is a primary alkyl group. The following peak at 25 I would guess as a secondary alkyl group. The peak at 42 a tertiary alkyl group. The final peak at 62 an alcohol.
Then the 1H NMR I'm really not sure about. The first peak is a doublet and sits around the value for a shift for a R-CH3 so this is being split by something around it with 1H? I really have no idea. Any help would be great.
Also what is the line on the spectra that continues to increase in steps?

[Borek]

Red line on NMR is integration - it is proportional to number of hydrogens at given shift.

(I am not commenting any further, as at a quick glance I have no idea what compound it is).

[mjc123]

OK, let's see what we've got. 13C tells you there are 4 different types of C, 1H that there are 5 different types of H, in the ratios 2:1:1:2:6. The 6H peak is a doublet, and one of the 1H peaks is a septet - does that suggest anything to you? And there's a strong broad IR band at ca. 3300 cm-1 - what does that tell you?

[hardy5086]

i wasn't sure as I had the question thrown to me with having never had any lectures or information explained on the subject, I was just going off what I had read and many things contradicted something else.
but from what I understand, the peak at 3300 in IR is characteristic of an OH group. now the bit I am unsure on is the 1H NMR, correct me if im wrong, but the doublet containing 6H is 2 CH3 groups, the septet is a CH with the signal being split by 6 different H's, so im assuming it is adjacent the 2 CH3 groups. the quartet at 1.5 is a CH2, the singlet at 2.2 is probably an alcohol, and the triplet at 3.7 is a H-C-O bond?
so putting the molecule together is would look like 3-methyl-1-butanol?

[mjc123]

OK, that's good. The IR tells you you have an alcohol, not an ether. You are right about the CH next to 2 CH₃s, so you have an isopropyl group (CH₃)₂CH-, which also gives a peak at 43 in the MS. As to the other nmr signals, I agree with two of them, but the peak at 3.7 integrates to two H's.
I agree that the molecule is 3-methyl-1-butanol.