Q. In the following reaction, X and Y respectively are $C_{2}H_{5}OH\overset{\ \ \ \ KMnO_{4}/H^{+}\ \ }{\rightarrow}X\underset{H_{2}{SO}_{4}/\mathrm{\Delta}}{\overset{\ \ \ \ \ \ \ \ \ \ \ Y\ \ \ \ \ \ \ \ \ \ \ }{\rightarrow}}CH_{3}CO_{2}C_{2}H_{5}$
A) $CH_{3}OH,\ C_{2}H_{5}OH$
B) $CH_{3}CHO,\ CH_{3}OH$
C) $CH_{3}CO_{2}H,\ C_{2}H_{5}OH$
D) $C_{2}H_{4},\ CH_{3}CO_{2}H$
Explanation
(c)
$C_{2}H_{5}OH + \lbrack O\rbrack\overset{{KMnO}_{4}/H_{4}}{\rightarrow}CH_{3}COOH\underset{H_{2}{SO}_{4}/\mathrm{\Delta}}{\overset{{\ \ \ \ C}_{2}H_{5}OH(Y)\ \ \ }{\rightarrow}}CH_{3}COOC_{2}H_{5}$
ethanol (X) (esterification) ethyl ethanoate
ethanoic acid
Hence, $X = CH_{3}COOH$
$Y = C_{2}H_{5}OH$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.