Compound X, $C X_{4} H X_{8} O$ , gives a crystalline bisulphite adduct, does not reduce Tollens' reagent, and gives a yellow precipitate with $I X_{2}$ and $NaOH$ . X is treated with (i) $CH X_{3} MgBr$ , dry ether; (ii) $H X_{3} O X^{+}$ . The product is

Answer: C

Bisulphite adduct plus a negative Tollens test gives a ketone, and a positive iodoform test gives a methyl ketone, so X is butan-2-one

CH X_{3} - CO - CH X_{2} CH X_{3}

. Adding

CH X_{3} X^{-}

to the carbonyl and hydrolysing gives

CH X_{3} - C (OH) (CH X_{3}) - CH X_{2} CH X_{3}

, 2-methylbutan-2-ol.