Dehydrohalogenation
Mediumchemistry
Dehydrohalogenation of 2-bromobutane with alcoholic KOH gives:
Select the correct option:
Solution
Incorrect! Answer:
2-Butene as major product
Dehydrohalogenation is a β-elimination reaction.
- Reagent: Alcoholic KOH favors elimination over substitution.
- Reaction: CH3−CH(Br)−CH2−CH3+KOH(alc)→Alkene+KBr+H2O
- There are two types of β-hydrogens available (on C1 and C3).
- Saytzeff's Rule: The major product is the more highly substituted alkene (the one with the greater number of alkyl groups attached to the double-bonded carbons).
- Product: 2-Butene (CH3−CH=CH−CH3) is more stable and forms the major product (~80%), while 1-Butene is the minor product.
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About This Question
- Subject
- chemistry
- Chapter
- organic compounds containing halogens
- Topic
- dehydrohalogenation
- Difficulty
- Medium
- Year
- 2025
This medium difficulty chemistry question is from the chapter organic compounds containing halogens, covering the topic of dehydrohalogenation. It appeared in the 2025 exam. Practice this and similar questions to strengthen your understanding of organic compounds containing halogens concepts.
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