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Question

Identify $$X$$ and $$Y$$ in the following reaction.
$$H_ 2\underset { \underset {\displaystyle Br }{ | }  }{ C } -\underset { \underset { \displaystyle Br }{ | }  }{ CH_2}+KOH\xrightarrow [  ]{ alcohol } X\xrightarrow [  ]{ NaNH_ 2 } Y$$


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 $$X: CH_2=CHBr, \,\, Y: CH\equiv CH$$
The reactant is symmetrical alkanes. On reacting with alcoholic KOH, dehydrohalogenation reaction will occur. It will give monobromoalkene.
$$Br{H}_{2}C-C{H}_{2}Br + KOH \xrightarrow{alcohol} C{H}_{2}=CHBr$$
$$C{H}_{2}=CHBr$$ on reaction with $$NaN{H}_{2}$$ releases sodium bromide and ammonia and forms ethyne.
$$C{H}_{2}=CHBr \xrightarrow{NaN{H}_{2}} CH \equiv CH + NaBr + N{H}_{3}$$

Chemistry

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