Acetonitrile on reduction with sodium in ethanol gives A, which...

Acetonitrile ($\text{CH}_3\text{CN}$) on reduction with sodium and ethanol undergoes Mendius reduction to form ethylamine (A).

$\text{CH}_3\text{CN} + 4[\text{H}] \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{CH}_3\text{CH}_2\text{NH}_2$

Ethylamine (A) is a primary aliphatic amine. It reacts with nitrous acid (generated in situ from $\text{NaNO}_2$ and dil. $\text{HCl}$) to form an unstable aliphatic diazonium salt, which immediately decomposes to yield ethyl alcohol (B) and nitrogen gas.

$\text{CH}_3\text{CH}_2\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \rightarrow [\text{CH}_3\text{CH}_2\text{N}_2^{+}\text{Cl}^{-}] \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CH}_2\text{OH} + \text{N}_2 \uparrow + \text{HCl}$

Thus, product B is $\text{CH}_3\text{-CH}_2\text{-OH}$.

Answer: $\text{CH}_3\text{-CH}_2\text{-OH}$