Elimination Vs Substitution
2-Bromobutane on treatment with alcoholic KOH mainly gives:
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing halogens
- Topic
- elimination vs substitution
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
But-1-ene and But-2-ene
Alcoholic KOH favors beta-elimination (E2) in haloalkanes, removing HBr to form alkenes. Thus 2-bromobutane yields a mixture of but-1-ene and but-2-ene, with but-2-ene usually major by Saytzeff rule. Formation of alcohol is favored with aqueous KOH, not alcoholic KOH.
This medium difficulty chemistry question is from the chapter organic compounds containing halogens, covering the topic of elimination vs substitution. It appeared in the 2025 exam.
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