Q. In the following reaction, $A$ and $B$, respectively are $A\overset{\ HBr\ \ \ \ }{\rightarrow}C_{2}H_{5}Br\overset{\ \ \ B\ \ }{\rightarrow}A$
A) $C_{2}H_{4},\ alc.\ \frac{KOH}{\mathrm{\Delta}}$
B) $C_{2}H_{5}Cl,\ aq.\ \frac{KOH}{\mathrm{\Delta}}$
C) $CH_{3}OH,\ aq.\ \frac{KOH}{\mathrm{\Delta}}$
D) $C_{2}H_{5},\ PBr_{3}$
Explanation
(a)
$H_{2}C = CH_{2}\overset{HBr}{\rightarrow}CH_{3}CH_{2}Br$
$(A)$
$\underset{\mathrm{\Delta}}{\overset{Alc.KOH}{\rightarrow}}H_{2}C = CH_{2}$
$(A)$
Hence, $A = C_{2}H_{4};B = \frac{alc.KOH}{\mathrm{\Delta}}\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.