An optically active alkyl bromide C₄H₉Br, 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✓ Correct
- (B)Lucas test
- (C)Silver mirror test
- (D)Benedict test
Answer: (A)
Solution
The optically active is 2-bromobutane; ethanolic KOH eliminates it to but-2-ene (A), which adds bromine to give 2,3-dibromobutane (B). Successive dehydrohalogenation with ethanolic KOH and then sodamide removes two molecules of HBr to give but-2-yne (C). Markovnikov hydration of but-2-yne with at 333 K yields butan-2-one (D). Being a methyl ketone, butan-2-one is confirmed by the haloform test.
JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Single correct option