$\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N} \stackrel{\mathrm{X}}{\longrightarrow}…

$\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N} \stackrel{\mathrm{X}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}$
The compound $X$ is
  1. $\mathrm{SnCl}_{2} / \mathrm{HCl} / \mathrm{H}_{2} \mathrm{O}$, boil
  2. $\mathrm{H}_{2} / \mathrm{Pd}-\mathrm{BaSO}_{4}$
  3. $\mathrm{LiAIH}_{4} /$ ether
  4. $\mathrm{NaBH}_{4} /$ ether $/ \mathrm{H}_{3} \mathrm{O}^{+}$

Solution

It is Stephen's reaction.
$\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \equiv \mathrm{N} \stackrel{\mathrm{SnCl}_{2} / \mathrm{HCl}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{NH} \cdot \mathrm{HCl} \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}+\mathrm{NH}_{4} \mathrm{C}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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