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Q.

n-propyl bromide on treatment with ethanolic KOH produces

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a

1-propanol

b

propyne

c

propene

d

propane

answer is C.

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Detailed Solution

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Dehydrohalogenation

n-propyl bromide on treating with alcoholic KOH produces propene (CH3​−CH=CH2​) which is a elimination reaction.

CH3​−CH2​−CH2​−Br+ethanolic.KOH→CH3​−CH=CH2​+HBr

 

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