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Question

What is the product when CH₃CH₂CH₂CHBrCH₃ reacts with alcoholic KOH ?

Options

Choose one · Correct answer highlighted

Explanation

Alcoholic KOH promotes elimination (E₂) in CH₃CH₂CH₂CHBrCH₃ , forming CH₃CH₂CH=CHCH₃ (2-pentene) per Zaitsev's rule.