What are $X$ and $Y$ respectively in the following reaction? $Z$-product $\stackrel{Y}{\longleftarrow} 2$-butyne $\stackrel{X}{\longrightarrow} E$-product

• Correct answer
  A

  $\mathrm{Na} / \mathrm{NH}_3$ (liq. ) and $\mathrm{Pd} / \mathrm{BaSO}_4+\mathrm{H}_2$
• B

  $\mathrm{Ni} / 140^{\circ} \mathrm{C}$ and $\mathrm{Pd} / \mathrm{BaSO}_4+\mathrm{H}_2$
• C

  $\mathrm{Ni} / 140^{\circ} \mathrm{C}$ and $\mathrm{Na} / \mathrm{NH}_3$ (liq. )
• D

  $\mathrm{Pd} / \mathrm{BaSO}_4+\mathrm{H}_2$ and $\mathrm{Na} / \mathrm{NH}_3$ (liq.)

Solution

Hence, reagent $X$ and $Y$ are respectively $\mathrm{Na}$, $\mathrm{NH}_3$ and $\mathrm{Pd} / \mathrm{BaSO}_4+\mathrm{H}_2$.

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