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Topic: Formation of a Benzaldehyde  (Read 3843 times)

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Offline AlphaScent

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Formation of a Benzaldehyde
« on: January 02, 2015, 12:52:51 PM »
I was given a problem and am a bit confused as to exactly how it proceeds.

I am trying to study for an upcoming academic semester.

I am given 1,2-dimethoxybenzene, commonly referred to as veratrole. Treating with n-butyl lithium and them dimethylformamide gives a benzaldehyde with the formylation happening alpha to one of the methoxy groups.  It the process lithiation and then nucleophilic attack by the aromatic ring?  Then hydroylsis of the hemiaminal to the aldehyde??

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Offline Rutherford

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Re: Formation of a Benzaldehyde
« Reply #1 on: January 02, 2015, 02:04:33 PM »
The amine is more basic than the hydroxyl group, hence the formed ammonium ion will be expelled by the oxygen lone pair.

Offline AlphaScent

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Re: Formation of a Benzaldehyde
« Reply #2 on: January 02, 2015, 02:16:22 PM »
Yes, my fault.  Where I have the hemiaminal written, the hydroxy should not have a proton but be an ion pair with lithium.  The proton should actually be on the nitrogen (forming a ammonium ion) and is then expelled through the formation off the carbonyl.

No hydrolysis.  Thanks for the clarity. 
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Offline discodermolide

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Re: Formation of a Benzaldehyde
« Reply #3 on: January 02, 2015, 07:35:55 PM »
Is this not an example of directed ortho-metallation as defined by Viktor Snieckus?
http://en.wikipedia.org/wiki/Directed_ortho_metalation
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Offline AlphaScent

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Re: Formation of a Benzaldehyde
« Reply #4 on: January 02, 2015, 08:08:21 PM »
Thank you so much!!
I am trying to get ahead for an upcoming design class. 

If you're not part of the solution, then you're part of the precipitate

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