Q. A gas formed by the action of alcoholic $KOH$ on ethyl iodide, decolourises alkaline$\ KMnO_{4}$. The gas is
A) $C_{2}H_{6}$
B) $CH_{4}$
C) $C_{2}H_{2}$
D) $C_{2}H_{4}$
Explanation
(d)
Ethylene is formed by dehydrohalogenation of alkyl halide in presence of alcoholic $KOH$. Ethylene decolourise alkaline $KMnO_{4}$ due to get oxidized by it.
$CH_{3} - CH_{2}I\overset{Alc.KOH\ \ \ }{\rightarrow}CH_{2} = CH_{2}$
ethylene
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.