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Topic: C4H10O C-NMR  (Read 212 times)

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Offline sharbeldam

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C4H10O C-NMR
« on: July 02, 2020, 08:14:44 AM »
this graph shows C-NMR spectroscopy of C4H10O, I Tried everything to predict the molecule, nothing works!
there is a shift at 130 which should be a double bond, but there is only 3 signals, so the double bonded carbons need to be somehow symmetrical, how do i do this??
« Last Edit: July 02, 2020, 09:12:20 AM by sharbeldam »
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Offline Borek

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Re: C4H10O C-NMR
« Reply #1 on: July 02, 2020, 09:03:46 AM »
What is degree of unsaturation of the molecule?
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Offline sharbeldam

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Re: C4H10O C-NMR
« Reply #2 on: July 02, 2020, 09:11:57 AM »
I'm not given that.
And if you are asking me, I dont know. How can it contain any double bonds with that formula, its not possible
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Online Babcock_Hall

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Re: C4H10O C-NMR
« Reply #3 on: July 02, 2020, 09:57:10 AM »
Your last reply is paradoxical.  You don't need to be given the number of degrees of unsaturation; you can calculate it, as apparently you did.  I agree with your conclusion.

Offline sharbeldam

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Re: C4H10O C-NMR
« Reply #4 on: July 02, 2020, 10:14:57 AM »
Yes i agree, but still, what molecule would that be? is there something wrong with the question? what is the signal at 130 if not double bond?? it's really urgent or i would keep trying. Can you tell me the answer?
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Online Babcock_Hall

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Re: C4H10O C-NMR
« Reply #5 on: July 02, 2020, 12:17:49 PM »
I was not able to generate a structure.

Offline Marko

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Re: C4H10O C-NMR
« Reply #6 on: July 02, 2020, 12:24:30 PM »
Another way of looking at it: a molecule with formula [chem]C4H10O[/chem] must have a C-O bond. In other words, it must have a peak somewhere between let's say 50 and 80 ppm.

This, in addition to the observations regarding the peak at 130 ppm and degree of unsaturation, can only lead to the conclusion that the formula given must be wrong.

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