An optically active alkyl halide C4H9Br [A] reacts with hot KOH dissolved in ethanol and forms alkene [B] as major product which reacts with bromine to give dibromide [C]. The compound [C] is converted into a gas [D] upon reacting with alcoholic NaNH2. During hydration 18 gram of water is added to 1 mole of gas [D] on warming with mercuric sulphate and dilute acid at 333 K to form compound [E]. The IUPAC name of compound [E] is :
- Butan-2-ol
- But-2-yne
- Butan-2-one
- Butan-1-al
Answer
(3) Butan-2-one
Solution
The only optically active C4H9Br is 2-bromobutane: [A] = CH3- *CHBr-CH2-CH3
[A] alc. KOH, [B] = CH3-CH=CH-CH3 (but-2-ene, Saytzeff product)
[B] Br2 [C] = CH3-CHBr-CHBr-CH3 (2,3-dibromobutane)
[C] alc. NaNH2 [D] = CH3-C C-CH3 (but-2-yne, gas) by double dehydrohalogenation
[D] [333 K] H2O, Hg2+/H+ CH3-C(OH)=CH-CH3 (enol) CH3-CO-CH2-CH3
[E] = butan-2-one
Answer: (3)