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Asked By :  Herry harvar
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An alkyl halide a of molecular formula c6h13cl on treatment

An alkyl halide (A) of molecular formula C6H13Cl on treatment with alcoholic KOH gives two isomeric alkenes (B) and (C) of molecular formula C6H12. Both alkenes on hydrogenation give 2, 3-dimethylbutane. Write the structures of (A), (B) and (C).




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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:

  • Structure:

    CH₃–CH(CH₃)–CH(CH₃)–CH₃\text{CH₃–CH(CH₃)–CH(CH₃)–CH₃}
  • It’s a branched alkane, meaning the alkenes (B and C) must be branched alkenes.


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):

CH₃–C(CH₃)=CH–CH(CH₃)–CH₃(2,3-dimethyl-2-pentene)\text{CH₃–C(CH₃)=CH–CH(CH₃)–CH₃} \quad \text{(2,3-dimethyl-2-pentene)}

(C):

CH₃–CH=C(CH₃)–CH(CH₃)–CH₃(2,3-dimethyl-1-pentene)\text{CH₃–CH=C(CH₃)–CH(CH₃)–CH₃} \quad \text{(2,3-dimethyl-1-pentene)}


Step 3: Determine the alkyl halide A

Both (B) and (C) can be formed by elimination (E2) from a common alkyl halide with structure:

CH₃–CH(Cl)–CH(CH₃)–CH(CH₃)–CH₃\boxed{\text{CH₃–CH(Cl)–CH(CH₃)–CH(CH₃)–CH₃}}

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



Answered By

Sam

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