An alkene $A\left(\mathrm{C}_4 \mathrm{H}_8\right)$ exhibits cis/trans isomerism. $A$ on ozonolysis gives $B$, which when reacted with $\mathrm{NaOH}$ followed by hydroxylamine gave $C$. What are $B$ and $C$ ?

• A

  $\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}==\mathrm{NOH}\end{array}$
• B

  $\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CHCH} \mathrm{CH}_2=\mathrm{NOH}\end{array}$
• C

  $\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}=\mathrm{NNH}_2\end{array}$
• Correct answer
  D

  $\begin{array}{cc}\mathbf{B} & \mathbf{C} \\ \mathrm{CH}_3 \mathrm{CHO} & \mathrm{CH}_3 \mathrm{CH}=\mathrm{CHCH}=\mathrm{NOH}\end{array}$

Solution

Here, $(B)$ is $\mathrm{CH}_3 \mathrm{CHO}$ and $(C)$ is $\mathrm{CH}_3 \mathrm{CH}==\mathrm{CHCH}=\mathrm{NOH}$. (i) On ozonolysis double bond get broken to give $\mathrm{CH}_3 \mathrm{CHO}$. (ii) On reaction with $\mathrm{NaOH}$ followed by hydroxyl amine, we get $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CHCH}=\mathrm{NOH}$

Hence, the correct answer is option: 'D'.