The optically active alkyl bromide C4H9Br is 2-bromobutane, CH3CH(Br)CH2CH3.
Reaction with ethanolic KOH undergoes dehydrohalogenation to form but-2-ene as the major product [A] (Zaitsev's rule).
But-2-ene [A] reacts with Br2 to form 2,3-dibromobutane [B], CH3CH(Br)CH(Br)CH3.
Compound [B] undergoes double dehydrohalogenation with ethanolic KOH and NaNH2 to give but-2-yne [C], CH3C≡CCH3.
Hydration of but-2-yne [C] with HgSO4 and dilute H2SO4 yields butan-2-one [D], CH3COCH2CH3.
Since compound [D] is a methyl ketone, its functional group is confirmed by the haloform test.
Answer: Haloform test