Q.$CH_3Br \xrightarrow{KCN} (A) \xrightarrow{H_3O^+} (B) \xrightarrow{PCl_5} (C)$ Product (C) is :

Methyl bromide is converted stepwise via a nitrile substitution, acid hydrolysis, and finally chlorination of the resulting carboxylic acid's $-OH$ by $PCl_5$, ending at acetyl chloride.

Step 1 — formation of A: methyl bromide reacts with potassium cyanide ($KCN$) in a nucleophilic substitution ($S_N2$), where the cyanide ion displaces bromide: $CH_3Br + KCN \rightarrow CH_3CN\ (A,\ \text{methyl cyanide / acetonitrile}) + KBr$

Step 2 — formation of B: the nitrile A undergoes acid-catalysed hydrolysis (via an amide intermediate) with aqueous acid ($H_3O^+$) to give a carboxylic acid: $CH_3CN + 2H_2O \xrightarrow{H_3O^+} CH_3COOH\ (B,\ \text{acetic acid}) + NH_4^+$

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