Q. In the reaction, $\ \ \ \ \ \ \ \ \ \ \ \ \ CH_{3}$ $|$ $CH_{3} - C - CH_{2}\ \ \overset{\ \ \ Conc.H_{2}SO_{4\ \ }\ }{\rightarrow}A$ the product $A$ is $|\ \ \ \ \ \ \ |$ $OH\ \ OH$
A) $\ CH_{3}$
$|$
$CH_{3} - C = CH_{2}$
B) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ H\ \ \ \ H$
$|$ $|$
$CH_{3} - C = C - CH_{3}$
C) $CH_{3}$
$\ |$
$CH_{3} - CH_{2} - C = O$
D) $CH_{3}$
$|$
$CH_{3} - CH - CHO$
Explanation
The correct answer is Option D: $CH_{3}$ $|$ $CH_{3} - CH - CHO$.
A detailed explanation for this question is being prepared.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.