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Dehydrohalogenation of Dihalides

n-propylbromi...

Question

n-propylbromide on treating with alcoholic KOH produces

propene

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propane

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propyne

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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.
$C H_{3} C H_{2} C H_{2} B r alc. KOH ⟶ C H_{3} C H_{2} = C H_{2}$
so, correct option is a.

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