Arrange the following alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH. (I) CH3−H|C|CH3−CH2Br (II) CH3−CH2−Br (III) CH3−CH2−CH2−Br
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 DI>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 →