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Question

Identify X and Y in the following reaction.
H2C|Br−CH2|Br+KOHalcohol−−−−→XNaNH2−−−−−→Y

A
X:CH3CHBr,Y:CH2=CH2
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B
X:CH2OHCH2OH,Y:CH2=CH2
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C
X:CH2=CHBr,Y:CHCH
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D
X:CHCBr,Y:CHCH
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Solution

The correct option is C X:CH2=CHBr,Y:CHCH
The reactant is symmetrical alkanes. On reacting with alcoholic KOH, dehydrohalogenation reaction will occur. It will give monobromoalkene.
BrH2CCH2Br+KOHalcohol−−−CH2=CHBr
CH2=CHBr on reaction with NaNH2 releases sodium bromide and ammonia and forms ethyne.
CH2=CHBrNaNH2−−−−CHCH+NaBr+NH3

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