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Topic: Hemiaminal reaction  (Read 2198 times)

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Offline orgo814

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Hemiaminal reaction
« on: July 09, 2015, 05:39:29 PM »
Sorry if this picture posts at a different angle I'm having issues with my device. This is a question found in the ACS organic book and the answer is A which is obvious to me. However, in trying to figure out the mechanism in my head I figured the NH2 group would react first into the carbonyl and then one that OH of the original carbonyl was protonated into a good leaving group (such as water) the electrons of N would collapse down forming an imine which would allow the OH to intramolecularly react into that imine forming that product. My issue is that it said the intermolecular OH group would go through a carbocation when reacting to form that cyclic product. I don't see how that would work or if I'm just reading it wrong.

Offline spirochete

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Re: Hemiaminal reaction
« Reply #1 on: July 09, 2015, 07:06:46 PM »
The first few steps I would think would be similar to imine formation. So there would be protonation of the carbonyl oxygen by the pTSOH, nucleophilic attack by the amine, proton transfer to hydroxyl oxygen, then this tetrahedral intermediate would break down to via the lone pair on nitrogen "pushing down" to expel water as a leaving group.

This would form an iminium, that has the general Structure R2C=N+HR . I am thinking it is perhaps this structure they are referring to as a carbocation? This iminium intermediate has a resonance structure with a positive charge on carbon, although it is not contributing as much as the all octet N+ resonance structure.

Offline spirochete

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Re: Hemiaminal reaction
« Reply #2 on: July 09, 2015, 07:35:42 PM »
The other possibility is that you are misreading "oxonium ion" as "carbocation", which would refer to the final intermediate that is deprotonated following intramolecular nuclophilic attack by the tethered OH group.

If you write out exactly what the book says we can help you more.

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