Question

Which gives more than one amides on treatment with $N{H}_{2}OH$ followed by $PC{l}_5$?

• A.
• B. $HCHO$
• C.
• D.

Answer 1

The correct option is D

Unsymmetrical carbonyl give more than one oximes on reacting with hydroxylamine, hence more than one amides in the subsequent Beckmann rearrangement is formed.
Here, option (d) has the unsymmetrical carbonyl compound.
Hence, it is the correct answer.