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Question

Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH.
(I) CH3H|C|CH3CH2Br
(II) CH3CH2Br
(III) CH3CH2CH2Br

A
I>II>III
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B
III>II>I
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C
II>III>I
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D
I>III>II
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Solution

The correct option is D I>III>II
More the number of β- substituents (alkyl groups), more stable alkene it will form on β- elimination and more will be the reactivity.
Hence, the decreasing order of the rate of β-elimination reaction with alcoholic KOH is
I2βsubstituents>III1βsubstituents>II0βsubstituents.

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