JEE Main · 2004 · 24 Apr 2004 · Offline · Haloalkanes and Haloarenes

Acetyl bromide meets the Grignard reagent twice over. The first equivalent of $\mathrm{CH}{}_3\mathrm{MgI}$ adds to the acyl carbon and, with loss of bromide, gives a ketone, acetone; the excess reagent then adds again to that ketone to build a tertiary alkoxide. Working up with saturated ammonium chloride protonates it to the tertiary alcohol, so overall $\mathrm{CH}{}_3\mathrm{COBr}+2\mathrm{CH}{}_3\mathrm{MgI}\to (\mathrm{CH}{}_3){}_3\mathrm{COH}$, which is 2-methyl-2-propanol.