An alkene $\mathrm{X}$ with formula $\mathrm{C}_4 \mathrm{H}_8$ does not exhibit geometrical isomerism. In the conversion of $\mathrm{X}$ to $\mathrm{Y}$, the correct sequence of reagents A and B used are ( $\mathrm{Y}$. gives iodoform test)

• A

  $\mathrm{BH}_3, \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{-}, \quad \mathrm{PCC}$
• B

  $\mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}, \quad \mathrm{ZnCl}_2 / \mathrm{HCl}$
• Correct answer
  C

  $\mathrm{H}_2 \mathrm{O} / \mathrm{H}^{+}, \quad \mathrm{Cu} / 573 \mathrm{~K}$
• D

  $\mathrm{BH}_3, \mathrm{H}_2 \mathrm{O}_2 / \mathrm{OH}^{-}, \quad \mathrm{Cu} / 573 \mathrm{~K}$

Solution

Since ' $\mathrm{X}$ ' does not exhibit geometrical isomerism, two groups on the doubly-bonded carbon atoms will be identical for each carbon.
$$ \text { or } \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}=\mathrm{CH}_2 $$ $\mathrm{Y}=$ Gives iodoform test so it must be a methyl alcohol or methyl ketone.

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