Hoffmann bromamide degradation of benzamide gives product $A$,which upon heating with $CHCl_{3}$ and $NaOH$ gives product $B$. The structures of $A$ and $B$ are

• A
  $A = \text{p-bromoaniline}, B = \text{p-bromophenyl isocyanide}$
• B
  $A = \text{aniline}, B = \text{phenyl isocyanide}$
• C
  $A = \text{aniline}, B = \text{o-formylaniline}$
• D
  $A = \text{benzamide}, B = \text{o-formylbenzamide}$