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
Solution
Correct 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.
This medium difficulty chemistry question is from the chapter organic compounds containing halogens, covering the topic of dehydrohalogenation. It appeared in the 2025 exam.
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