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 :
- Haloform test
- Lucas test
- Silver mirror test
- Benedict test
Answer
(1) Haloform test
Solution
Optically active C4H9Br: 2-bromobutane, CH3CH2 *CHBrCH3
alc. KOH [A] CH3CH=CHCH3 (but-2-ene, Saytzeff product)
Br2 [B] CH3CHBr-CHBrCH3 (2,3-dibromobutane)
alc. KOH, NaNH2 [C] CH3C CCH3 (but-2-yne)
[333 K]H2O, Hg2+/H2SO4 [D] CH3COCH2CH3 (butan-2-one)
D is a methyl ketone (CH3CO- group) ⇒ confirmed by the haloform (iodoform) test. Lucas test is for alcohols; silver mirror and Benedict tests are for aldehydes.
Answer: (1)