$A$ 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

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