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Question

Hoffmann degradation of m-bromobenzamide gives:

A
aniline
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B
m - bromoaniline
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C
bromobenzene
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D
m - bromoethyl benzene
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Solution

The correct option is C m - bromoaniline
The Hoffmann's bromomide reaction is as follows :
RCONH2+Br2+KOHRNH2
Here, the R group is m-bromobenzene group
So, m-bromoaniline is the product.
71517_24106_ans_db7c4f2dfdd148c1bdf150cebf3c9dc9.gif

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