$A$ hydrocarbon $A$ $(C_{4}H_{8})$ on reaction with $HCl$ gives a compound $B$ $(C_{4}H_{9}Cl)$ which on reaction with $1 \ mol$ of $NH_{3}$ gives compound $C$ $(C_{4}H_{11}N)$. On reacting with $NaNO_{2}$ and $HCl$ followed by treatment with water,compound $C$ yields an optically active compound $D$. The compound $D$ is

• A
  $2-$chlorobutane
• B
  butan$-2-$ol
• C
  butan$-2-$amine
• D
  butane