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  • Arrange the following alkyl halides in decreasing order of the rate of β– elimination reaction with alcoholic KOH.

Arrange the following alkyl halides in decreasing order of the rate of \beta – elimination reaction with alcoholic KOH.

 (B) CH_{3}-CH_{2}-Br   (C) CH_{3}-CH_{2}-CH_{2}-Br

(i) A > B > C
(ii) C > B > A
(iii) B > C > A
(iv) A > C > B

Answers (1)

The answer is the option (iv) A > C > B

Explanation: When alkyl halides are heated with alc. KOH, an alkene is formed by eliminating one molecule of halogen acid. Hydrogen is eliminated from the beta carbon atom. Order of reactivity is  3^{\circ} > 2^{\circ} > 1^{\circ}, hence option (iv). The rate of reaction is determined by the nature of alkyl groups.

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