Question

Benzene with $\mathrm{Br}$ and ${\mathrm{CONH}}_2$ at the meta position is made to react with $\mathrm{KOBr}$ to give A and the same reactant is made to react with ${\mathrm{LiAlH}}_4$ to give B. Predict products A and B.

• A.

  

• B.

  

• C.

  

• D. None of the above

Answer 1

The correct option is A

Explanation for the correct options:

A) is correct option.

For the formation of product A, the reaction undergoes Hoffmann Bromamide degradation when $3-\mathrm{bromobenzamide}$ react with $\mathrm{KOBr}$ to give$\mathrm{m}-\mathrm{bromoaniline}$ and the same reactant undergoes reduction with ${\mathrm{LiAlH}}_4$ to give $3-\mathrm{bromobenylamine}$.

Explanation for the incorrect options:

B) The product A is produced by Hoffman degradation which means the product has one carbon less than parent carbon. So, $3-\mathrm{bromobenylamine}$ is not formed as product A

C) The reduction does not lead to addition of $\mathrm{Br}$ on ortho position.

Hence, Option A is correct. As the product formed are A - $\mathrm{m}-\mathrm{bromoaniline}$ and B- $3-\mathrm{bromobenylamine}$.