May 25, 2024, 01:07:20 AM
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Topic: I would like some help understanding this mechanism (acetaminophen lab).  (Read 4181 times)

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Offline thescepticalchymist

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I need to explain this mechanism for a lab report:


I understand most of it, but I’m not sure why, in the third step, that particular oxygen (is there a special name for it?) deprotonates the nitrogen.  Why is that oxygen a better nucleophile than the carbonyl oxygens?  I just started learning reactions with anhydrides so I’m kind of new to this stuff.

Offline OrganicDan96

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you are missing a methyl group and a few charges there. and you have an arrow which doesn't do anything

Offline thescepticalchymist

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you are missing a methyl group and a few charges there. and you have an arrow which doesn't do anything

I didn't draw this mechanism out myself; I got it from ResearchGate...  Where is the methyl supposed to be?  And I don't see any missing charges.

Offline wildfyr

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The carbon next to nitrogen in step 3 should have a methyl on it instead of being a C=O. It should have 4 bonds, C-O-C, C-OH, C-N, and C-CH3.

The missing charge is in step 4.

Offline rolnor

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Should not the aminogroup be protonated if the reaction is run under acidic conditions?

Offline wildfyr

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Good point rolnor! This reaction should not be run under acidic conditions. I looked up a student lab version of it and its run without acid or base catalyst.

http://rene.souty.free.fr/IMG/pdf/ParacetamolProtocolFLORIDacetaminophen.pdf

I would do it neutral too as well. Acid will protonate the amine and make it a bad nucleophile, and base would deprotonate the phenol and make it compete with the aniline to attack acetic anhydride.

For OP, I think the mechanism you posted is undependable, and you not consult it. There are better examples to be found in textbooks. Just to start you out, aniline can attack the anhydride on its own, without any of the carbonyls needing to be protonated to boost its electrophilicity.

Offline thescepticalchymist

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Here is a mechanism in neutral conditions, and my description below the figure:

https://imgur.com/gallery/MyXts

Have I gotten it right so far?  Also, what exactly is deprotonating the nitrogen in the second step?  Is it H2O?  Because in our lab, we put water in the reaction mixture with p-aminophenol and acetic anhydride.

Offline Babcock_Hall

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Speaking only for myself, when I see an arrow such as the one concerning the loss of a proton by nitrogen, I interpret that as a proton lost to solvent.  Some reactions are subject to general-base catalysis, but not all reactions are sped up in this way.  The  concept of general acid or general base catalysis is not something that I have seen taught in introductory organic classes, but it might come up in more advanced classes.

Offline wildfyr

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I'm much happier with that mechanism.

I think the proton is taken up by the released ethanoate. Whether it is concerted or not is a matter of opinion. The carbonyl reforming is what kicks off the ethanoate, not the deprotonation of the amine.

Offline thescepticalchymist

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I think the proton is taken up by the released ethanoate. Whether it is concerted or not is a matter of opinion. The carbonyl reforming is what kicks off the ethanoate, not the deprotonation of the amine.

Oh, of course, that makes total sense.  Thanks.

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