An optically active alkyl bromide C_4H_9Br, reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at 333 K to form compound [D]. The functional group in compound D will be confirmed by :
A. Haloform test
The optically active alkyl bromide C _4 H _9 Br is 2-bromobutane, CH _3 CH ( Br ) CH _2 CH _3. 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 Br _2 to form 2,3-dibromobutane [B], CH _3 CH ( Br ) CH ( Br ) CH _3. Compound [B] undergoes double dehydrohalogenation with ethanolic KOH and NaNH _2 to give but-2-yne [C], CH _3 C CCH _3. Hydration of but-2-yne [C] with HgSO _4 and dilute H _2 SO _4 yields butan-2-one [D], CH _3 COCH _2 CH _3. Since compound [D] is a methyl ketone, its functional group is confirmed by the haloform test. Answer: Haloform test
Related: Chemistry — Hydrocarbons · All PYQ Banks