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Chemistry Forums for Students => Analytical Chemistry Forum => Topic started by: Peter4524 on June 02, 2019, 11:12:45 AM

Title: H-NMR of paracetamol/acetaminophen
Post by: Peter4524 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.
(https://cdn.discordapp.com/attachments/194160897734737920/584759299281256460/Unavngivet.png)
(https://cdn.discordapp.com/attachments/194160897734737920/584759296550764584/Untitled-1.png)

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.
Title: Re: H-NMR of paracetamol/acetaminophen
Post by: Peter4524 on June 06, 2019, 11:27:56 AM: Figured it out
Title: Re: H-NMR of paracetamol/acetaminophen
Post by: wildfyr 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!
Title: Re: H-NMR of paracetamol/acetaminophen
Post by: Peter4524 on June 06, 2019, 11:50:54 AM: Quote from: wildfyr 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!

It's alright, i appreciate the thought :)
Title: Re: H-NMR of paracetamol/acetaminophen
Post by: Babcock_Hall 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.