Ethyl bromide gives ethylene when reacted with

Ethyl alcohol

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Dilute

H_{2} S O_{4}

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Aqueous KOH

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Alcoholic KOH

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Solution

The correct option is D Alcoholic KOH
$C H_{3} - C H_{2} - B r + K O H (a l c) \to C H_{2} = C H_{2} E t h e n e + K B r + H_{2} O$

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The final product formed when ethyl bromide reacts with excess of alcoholic KOH

C H \equiv C H H g S O 4 - --- \to H_{2} S O_{4} (X) L i A l H_{4} - --- \to (Y) P B r_{3} - -- \to (Z)

Z

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