1. Hofmann bromamide degradation of benzamide

Benzamide is $C_6H_5CONH_2$.

In Hofmann bromamide degradation, an amide on treatment with $Br_2/NaOH$ gives a primary amine with one carbon less:

$RCONH_2 \xrightarrow{Br_2/NaOH} RNH_2$

So,

$C_6H_5CONH_2 \xrightarrow{Hofmann\ degradation} C_6H_5NH_2$

Thus, product $A$ is aniline.

2. Reaction of product A with $CHCl_3$ and $NaOH$

Aniline is a primary amine. Primary amines on heating with chloroform and alcoholic sodium hydroxide give isocyanides. This is the carbylamine reaction:

$RNH_2 + CHCl_3 + 3NaOH \rightarrow RNC + 3NaCl + 3H_2O$

For aniline:

$C_6H_5NH_2 \xrightarrow[\Delta]{CHCl_3,NaOH} C_6H_5NC$

So, product $B$ is phenyl isocyanide (isocyanobenzene).

3. Identify structures of A and B

• $A =$ aniline, $C_6H_5NH_2$
• $B =$ phenyl isocyanide, $C_6H_5NC$

So the correct option is the one showing:

$A = C_6H_5NH_2, \quad B = C_6H_5NC$

4. Comparison with stored correct answer

Stored correct answer: A

Our derived answer is also A.