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# Proton NMR practice 2

More practice determining the structure of a molecule from the molecular formula, hydrogen deficiency index, and proton NMR spectrum. Uses example of ether. Created by Jay.

## Want to join the conversation?

• how would you do this with out integration numbers?
• Without lines of integration, solving NMR problems is slightly harder. Usually, without the lines of integration, they'll ask you which spectrum corresponds to which molecule.

Lines of integration can be pretty important in cases where you have multiple stereoisomers but you have one formula, like in here.
• how did you get the figures of 48.7 or 48.4 for example
• These are the areas of the peaks (in arbitrary units).
Usually, the NMR machine prints out these values for you.
• What if you have 4 protons? Would that be C-C and 2 Hydrogens on each Carbon?
(1 vote)
• It could be -CH₂-CH₂-CH₂-. The central CH₂ would be split into a quintet by the four protons on the outside.
• What happens and what should I do if I have an impure H-nmr scan. Or even a mix. I have a about 50/50% mix of methanol and 1-propanol and now have to find the integral ratio.
(1 vote)
• Here's one way to do it.
I'm guessing that your spectrum has peaks at about δ 3.6 t, (EtCH₂O, A), 3.3 s (CH₃-O, B), 1.6 m (MeCH₂CH₂, D), 0.9 t (CH3CH2, E) plus an OH peak that could be anywhere. I will assume it's at δ 2.5 (C), but it might overlap with the peaks at 3.6 and/or3.3.
Each compound contributes to the OH signal.

Assume that your integrals in arbitrary units are
A:B:C:D:E = 0:0
Then A:D:E =80:0= 2:2:3. This corresponds to –OCH₂CH₂CH₃.
Also, B:E = 0. This corresponds to the CH₃ groups in MeOH and PrOH.
% MeOH = 180/(180+120)×100% =67 % CH₃OH.
• How would this NMR be different if there was a carbonyl group in the molecule?
(1 vote)
• If the second compound had a carbonyl group, the formula would be C₄H₈O.
The two possibilities are butanal (CH₃CH₂CH₂CHO) and butanone (CH₃CH₂COCH₃).
For each of these compounds, I would predict the spectra to be as follows.

Butanal
δ 9.8 (1H, t), 2.3 (2H, td), 1.7 (2H, sex), 0.9 (3H, t)

Butanone
δ 2.4 (2H, q), 2.1 (3H, s), 1.0 (3H, t)
(1 vote)
• Why does it peak at 2 and not the other positions for C5H10O2?
(1 vote)
• This might seem really stupid, but why is there no jump at the 3 position?
(1 vote)
• There is no signal at 3 ppm, so you do not see anything.
(1 vote)
• thank you for your amazing explanation
can you tell us why did you start with the carbonyl?
why didn't you start with other parts of the puzzle

thank you
(1 vote)
• Jay started with the carbonyl because the carbon was bonded to a a methyl group which caps off that end of the molecule. You can continue building off a methyl so you only have to worry about building off the other side of the carbonyl.

Additionally it's the most complete fragment we could have initially generated from the spectrum.

Hope that helps.
(1 vote)
• Sir please explain about spinning concept in NMR
(1 vote)
• For the first example (green protons; ), there are actually 7 peaks (there's a tiny peak on the far right that he ignored), why is that?