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Question

The dehydrohalogenation of neopentyl bromide with alcoholic KOH mainly gives


A
2-methyl-1-butene
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B

2-methyl-2-butene

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C
2, 2-dimethyl-1-butene
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D
2-butene
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Solution

The correct option is B

2-methyl-2-butene



In this reaction 1 carbonium ion is formed which rearranges to form 3 carbonium ion from which base obstruct proton. Hence 2-methyl-2-butene is formed as a main product

1 carbonium less stable


2-Methyl-2 Butene



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