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Topic: 1H NMR Question?  (Read 5247 times)

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Offline caliheat

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1H NMR Question?
« on: June 07, 2013, 06:50:43 PM »
Hi everyone, I have a molecular formula C11H14O2 and the following 1H NMR data:
1.0 (3H)
1.5 (2H)
1.7 (2H)
4.4 (2H)
7.4-8.1 (5H)

I was thinking it should be something like this: (butyl-benzoate)
because the structure fits, except for the benzyl group with 3 equivalent H groups. My question is, how can you get a 5 equivalent proton (5H) group? I can't think of any structure and I can't find it online. Thanks!

Offline Corribus

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Re: 1H NMR Question?
« Reply #1 on: June 07, 2013, 07:04:01 PM »
The range of values for the 5H peak suggest they're not all equivalent and that it's multiple peaks overlapping.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline caliheat

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Re: 1H NMR Question?
« Reply #2 on: June 07, 2013, 07:16:35 PM »
Oh ok, so then that group should be the benzene ring since there are 5 protons (non-equivalent)and the signals lie in the 6.0-9.0 Aromatic hydrogen range?

Offline Corribus

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Re: 1H NMR Question?
« Reply #3 on: June 07, 2013, 07:24:43 PM »
Yes
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline caliheat

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Re: 1H NMR Question?
« Reply #4 on: June 07, 2013, 07:44:30 PM »
Great thanks, I was just confused because I thought that the (nH) (where n is an integer) group meant that they were all equivalent, but now I will check if there are multiple values within the range.

There is one more I am having trouble with. The molecular formula is C6H14O. The 13C NMR data is given:
3 peaks, one at ~10ppm, one at ~20ppm and one at ~70ppm.

I know that means there are 3 non-equivalent carbon groups but I can't find the proper structure. The degree of unsaturation is ((2*6+2)-14)/2 = 0 which means it only has single bonds. The closest I have gotten is this: (diisopropyl ether), but due to symmetry I believe it only has 2 non-equivalent C groups. I have also found structures with 4 non-equivalent C groups, but not 3 :(.

Offline Corribus

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Re: 1H NMR Question?
« Reply #5 on: June 07, 2013, 09:27:23 PM »
Great thanks, I was just confused because I thought that the (nH) (where n is an integer) group meant that they were all equivalent, but now I will check if there are multiple values within the range.
The n refers to the integration value.  If it's a single peak, this will usually mean the number of equivalent protons.  However if the integration is over a collection of overlapping peaks, then they aren't necessarily equivalent.

Quote
I know that means there are 3 non-equivalent carbon groups but I can't find the proper structure. The degree of unsaturation is ((2*6+2)-14)/2 = 0 which means it only has single bonds. The closest I have gotten is this: (diisopropyl ether), but due to symmetry I believe it only has 2 non-equivalent C groups. I have also found structures with 4 non-equivalent C groups, but not 3 :(.
I think ether might be the right direction.  Can you think of another simple one that fits the formula?
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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