Q.
The treatment of
with $NaIO_{4}$ or boiling $KMnO_{4}$ produces
$KMnO_{4}$ produces
A) $CH_{3}COCH_{3} + CH_{3}COOH$
B) $CH_{3}COCH_{3} + CH_{3}CHO$
C) $CH_{3}CHO + CO_{2}$
D) $CH_{3}COCH_{3}$ only
Explanation
(a)
$CH_{3}C = CH_{3}\underset{\mathrm{\Delta}}{\overset{{KMnO}_{4}}{\rightarrow}}CH_{3}COCH_{3} + CH_{3}COOH$
$\ \ \ \ \ \ \ \ |$
$\ \ \ \ \ CH_{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.