Question Detail

CHM 102 Practice Questions Objective Question
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Question

The dehydrohalogenation of 2-bromobutane with alcoholic KOH gives mainly
Options
A 2-Butene
Correct Answer
B 2-Butyne
C 1-Butene
D 1-Butyne
Correct Answer

Option A is the correct answer.

Detailed Explanation

Dehydrohalogenation of 2-bromobutane (CH3CHBrCH2CH3) follows Zaitsev's rule, giving the more substituted alkene as the major product. The major product is 2-butene (CH3CH=CHCH3), which is more substituted than 1-butene.

Hint

Which alkene is more substituted according to Zaitsev's rule?

Question Info

Course: CHM 102
Topic: Practice Questions
Type: Objective
Difficulty: Not specified
Question 1096 of 1999
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