n- propyl bromide on treating with alcoholic KOH produce

Question

n- propyl bromide on treating with alcoholic KOH produce (A) propane (B) propene (C) propyne (D) propanol

Tardigrade Admin · Accepted Answer

Alcoholic KOH is a dehydrohalogenating reagent, so when n-propyl bromide is treated with alcoholic KOH, propene is obtained. underset textn - propyl bromide CH3CH2CH2Br + alc KOH longrightarrow underset textpropene CH3CH =CH2 +HBr

Q. $n -$ propyl bromide on treating with alcoholic $K O H$ produce

propane

propene

propyne

propanol

Alcoholic $K O H$ is a dehydrohalogenating reagent, so when n-propyl bromide is treated with alcoholic $K O H$ , propene is obtained.

$n - propyl bromide C H_{3} C H_{2} C H_{2} B r + a l c K O H ⟶ propene C H_{3} C H = C H_{2} + H B r$

Q. n− propyl bromide on treating with alcoholic KOH produce