We are given:
- Alkyl halide A → C₆H₁₃Cl
- On treatment with alcoholic KOH → gives two isomeric alkenes B and C of formula C₆H₁₂
- Both B and C on hydrogenation → give 2,3-dimethylbutane
Step 1: Analyze the hydrogenation product
2,3-dimethylbutane:
Step 2: Work backward to possible alkenes B and C
To form 2,3-dimethylbutane upon hydrogenation, the alkenes must have a structure that leads to that skeleton.
Two possible isomeric alkenes (B and C) are:
(B):
(C):
Step 3: Determine the alkyl halide A
Both (B) and (C) can be formed by elimination (E2) from a common alkyl halide with structure:
That is:
2-chloro-2,3-dimethylpentane → on elimination, gives both alkenes (B) and (C) depending on base orientation (Saytzeff and Hofmann products).
Final Answers:
- (A) = 2-chloro-2,3-dimethylpentane
- (B) = 2,3-dimethyl-2-pentene
- (C) = 2,3-dimethyl-1-pentene