The final product formed when ethyl bromide is treated with an excess of alcoholic $K O H$ is

Ethylene

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Ethane

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Ethyne

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Vinyl bromide

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Solution

The correct option is A Ethylene
Alkyl halide on reacting with alcoholic KOH gives 1, 2-elimination product.
Ethyl bromide undergo 1, 2- elimination to form ethylene.
${CH}_{3} - {CH}_{2} - Br + KOH \to {CH}_{2} = {CH}_{2} Ethylene + KBr + H_{2} O$
This reaction is known as dehydrohalogenation.

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