Dec 14, 2009
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What is the product of this reaction? (all c=o is carbonyl)
C6H5CH2(C=0)NH2 -------------(Br2/NaOCH3, CH3OH) -------->

answer is

C6H5CH2NH(C=0)OCH3

what kinds of reaction is this?
 

MrBeans

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Nov 6, 2009
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Although I'm not an organic expert, it looks kind of like a Fischer Esterification to me.
 
Dec 14, 2009
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but there is no leaving group. As you can see, amine is still there and switch its position to left.
 

Maygyver

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but there is no leaving group. As you can see, amine is still there and switch its position to left.
Would you be able to draw out the product. I'm having trouble figuring out how to draw it out. I feel like the alkoxide would just replace the amide and create an ester.
 

MrBeans

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Would you be able to draw out the product. I'm having trouble figuring out how to draw it out. I feel like the alkoxide would just replace the amide and create an ester.
I agree, it's a little confusing for me too
 
Jun 14, 2009
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It's a Hoffman Rearrangement with a twist. Instead of water, a methoxy group acts as the nucleophile in the second-to-last step, preventing decarboxylation.
 

Maygyver

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It's a Hoffman Rearrangement with a twist. Instead of water, a methoxy group acts as the nucleophile in the second-to-last step, preventing decarboxylation.
What is the product? I am seeing this as a nitrogen with a carbonyl and other things attached to it. :-\
 
Jun 14, 2009
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What is the product? I am seeing this as a nitrogen with a carbonyl and other things attached to it. :-\
I'm going to abbreviate the Benzyl group as R-

RC(=O)NH2 > RNHC(=O)OCH3
Basically the amino group shifts positions, turning the primary amide into a secondary amide.
 

sfoksn

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I see that it is Hofmann Rearrangement, but how does the carboxyl group on the N arise in the product?