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Question

n-propylbromide on treating with alcoholic KOH produces

A
propene
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B
propane
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C
propyne
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D
propanol
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Solution

The correct option is A propene
On dehydrohalogenation with alcoholic KOH , major product will be that alkene which is more substituted and has more number of α-hydrogens.
CH3CH2CH2Bralc. KOHCH3CH2=CH2
so, correct option is a.

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