Q. An unknown compound ‘D ’ first oxidised to aldehyde and then acetic acid by a dilute solution of $K_{2}Cr_{2}O_{7}$ and $H_{2}SO_{4}$. The compound ‘D ’ is
A) $CH_{3}OH$
B) $C_{2}H_{5}OH$
C) $CH_{3}CH_{2}COOH$
D) $CH_{3}{CH}_{2}CHO$
Explanation
(b)
$D\overset{Oxidation\ \ \ \ }{\rightarrow}aldehyde\underset{H_{2}SO_{4}}{\overset{\ \ \ K_{2}{Cr}_{2}O_{7}\ \ }{\rightarrow}}CH_{3}COOH$
acetic acid
$1{^\circ}\$alcohol on oxidation gives aldehyde having same number of carbon and aldehyde on oxidation gives acid having same number of carbon atoms.
It means, D will be alcohol having two carbon atoms that is $C_{2}H_{5}OH$(ethyl alcohol) and the alcohol on oxidation will give $CH_{3}CHO$ (acetaldehyde)
$C_{2}H_{5}OH\overset{\ \ \ \lbrack O\rbrack\ \ \ }{\rightarrow}CH_{3}CHO\overset{\ \ \lbrack O\rbrack\ \ \ }{\rightarrow}CH_{3}COOH$
(D)
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.