June 19, 2019, 11:51:59 AM
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Topic: H-NMR of paracetamol/acetaminophen  (Read 203 times)

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Offline Peter4524

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H-NMR of paracetamol/acetaminophen
« on: June 02, 2019, 11:12:45 AM »
How do i determine which equivalent atoms show up downfield? I've given the different equivalent atoms letters.
I already know where they'll turn up but what im looking for is an explanation as to why they turn up where they do.

I know the reason for the methyl group turning up first, but im not quite certain on the other groups.



Heres the spectrum. There's two peaks that's presumably DMSO and an impurity but thats to be ignored.
How come group E and B turn up where they do?

Does B turn up to be the most downfield due to the near presence of the benzene ring and the oxygen?
Both E and B are quite close to each other in terms of ppm and i don't really know why E turns up most downfield. Is my initial theory right?

Also I'm not quite on the explanation for C and D either.
Heres what i got so far: C is close to both oxygen and the nitrogen which in turn are more electronegative than the other side, thus it will deshield the C group and cause it to show up further downfield than the D group.

Thanks in advance.

Offline Peter4524

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Re: H-NMR of paracetamol/acetaminophen
« Reply #1 on: June 06, 2019, 11:27:56 AM »
Figured it out

Offline wildfyr

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Re: H-NMR of paracetamol/acetaminophen
« Reply #2 on: June 06, 2019, 11:50:06 AM »
I'm sorry Peter, i meant to sit down and work this out myself and forgot about it!

Offline Peter4524

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Re: H-NMR of paracetamol/acetaminophen
« Reply #3 on: June 06, 2019, 11:50:54 AM »
I'm sorry Peter, i meant to sit down and work this out myself and forgot about it!

It's alright, i appreciate the thought  :)

Offline Babcock_Hall

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Re: H-NMR of paracetamol/acetaminophen
« Reply #4 on: June 06, 2019, 06:28:48 PM »
When explaining the shifts in the presence of a heteroatom bound to an aromatic ring, I find it helpful to think about minor resonance forms.

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