Q. $n$-propyl bromide on treating with alcoholic $KOH$ produces
A) Propane
B) Propene
C) Propyne
D) Propanol
Explanation
(b)
Alcoholic $KOH$ is a dehydrohalogenating reagent, so when $n$-propyl bromide is treated with alcoholic $KOH$, propene is obtained.
$CH_{3}CH_{2}CH_{2}Br + alc\ \ KOH$
$n$-propyl bromide
$\rightarrow CH_{3}CH = CH_{2} + HBr$
propene
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.