n-Pentylbromide on treatment with ethanolic potassium hydroxide produces:

Pent-2-ene

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Pent-1-ene

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Pentyne

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Pentanol

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Solution

The correct option is B Pent-1-ene
Alkyl halide on reacting with alcoholic KOH undergo elimination via E2 mechanism.
n-Pentylbromide undergo 1, 2- elimination to form Pent-1-ene.
${CH}_{3} {CH}_{2} {CH}_{2} {CH}_{2} {CH}_{2} Br n - P e n t y l b r o m i d e Ethanolic KOH - -------- \to {CH}_{3} {CH}_{2} {CH}_{2} CH P e n t - 1 - e n e = {CH}_{2} + KBr + H_{2} O$
Hence, (b) is the correct answer.

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