An optically active alkyl bromide C _4 H _9 Br, reacts with...

The optically active alkyl bromide $\text{C}_4\text{H}_9\text{Br}$ is $2$-bromobutane, $\text{CH}_3\text{CH}(\text{Br})\text{CH}_2\text{CH}_3$.

Reaction with ethanolic $\text{KOH}$ undergoes dehydrohalogenation to form but-$2$-ene as the major product [A] (Zaitsev's rule).

But-$2$-ene [A] reacts with $\text{Br}_2$ to form $2,3$-dibromobutane [B], $\text{CH}_3\text{CH}(\text{Br})\text{CH}(\text{Br})\text{CH}_3$.

Compound [B] undergoes double dehydrohalogenation with ethanolic $\text{KOH}$ and $\text{NaNH}_2$ to give but-$2$-yne [C], $\text{CH}_3\text{C}\equiv\text{CCH}_3$.

Hydration of but-$2$-yne [C] with $\text{HgSO}_4$ and dilute $\text{H}_2\text{SO}_4$ yields butan-$2$-one [D], $\text{CH}_3\text{COCH}_2\text{CH}_3$.

Since compound [D] is a methyl ketone, its functional group is confirmed by the haloform test.

Answer: Haloform test