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The major product formed when 2-bromobutane is treated with alcoholic KOH, is
- a)2-butanol
- b)trans-2-butene
- c)1-butanol
- d)1-butene
Correct answer is option 'B'. Can you explain this answer?
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The major product formed when 2-bromobutane is treated with alcoholic ...
Reaction Mechanism:
The reaction involves the elimination of a bromine atom and a hydrogen atom from 2-bromobutane to form an alkene. The alcoholic KOH acts as a strong base, abstracting the hydrogen atom from the beta-carbon to form an alkoxide ion. The alkoxide ion then eliminates a bromide ion from the alpha-carbon to form the alkene.
Step 1: Formation of alkoxide ion
2-bromobutane is treated with alcoholic KOH to form an alkoxide ion.
2-bromobutane + KOH → 2-butanol + KBr
Step 2: Elimination of bromide ion
The alkoxide ion acts as a strong base, abstracting a hydrogen atom from the beta-carbon to form an alkene.
2-butanol + KOH → trans-2-butene + H2O + KBr
The major product formed is trans-2-butene. This is because the elimination of a hydrogen atom and a bromine atom from the beta- and alpha-carbons respectively leads to the formation of the trans isomer.
Conclusion:
The treatment of 2-bromobutane with alcoholic KOH results in the formation of trans-2-butene as the major product. This is due to the elimination of a hydrogen atom and a bromine atom from the beta- and alpha-carbons respectively, leading to the formation of the trans isomer.
The reaction involves the elimination of a bromine atom and a hydrogen atom from 2-bromobutane to form an alkene. The alcoholic KOH acts as a strong base, abstracting the hydrogen atom from the beta-carbon to form an alkoxide ion. The alkoxide ion then eliminates a bromide ion from the alpha-carbon to form the alkene.
Step 1: Formation of alkoxide ion
2-bromobutane is treated with alcoholic KOH to form an alkoxide ion.
2-bromobutane + KOH → 2-butanol + KBr
Step 2: Elimination of bromide ion
The alkoxide ion acts as a strong base, abstracting a hydrogen atom from the beta-carbon to form an alkene.
2-butanol + KOH → trans-2-butene + H2O + KBr
The major product formed is trans-2-butene. This is because the elimination of a hydrogen atom and a bromine atom from the beta- and alpha-carbons respectively leads to the formation of the trans isomer.
Conclusion:
The treatment of 2-bromobutane with alcoholic KOH results in the formation of trans-2-butene as the major product. This is due to the elimination of a hydrogen atom and a bromine atom from the beta- and alpha-carbons respectively, leading to the formation of the trans isomer.
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The major product formed when 2-bromobutane is treated with alcoholic KOH, isa)2-butanolb)trans-2-butenec)1-butanold)1-buteneCorrect answer is option 'B'. Can you explain this answer?
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