Q. Which of the following reactions proceeds $via$ secondary free radical?
A) $CH_{3} - CH = CH_{2}\overset{\ \ \ \ HBr\ \ }{\rightarrow}CH_{3} - CH - CH_{3}$
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Br
B) $CH_{3} - CH = CH_{2}\underset{UV\ light}{\overset{\ \ \ \ \ \ \ \ \ \ \ HBr\ \ \ \ \ }{\rightarrow}}CH_{3} - CH_{2} - CH_{2}Br$
C) $C_{6}H_{6}\overset{Br_{2}/FeBr_{3}}{\rightarrow}C_{6}H_{5}Br$
D) $C_{6}H_{6}\ \underset{UV\ ligh}{\overset{\ \ \ \ \ \ \ \ Br_{2}\ \ \ \ \ \ \ \ \ }{\rightarrow}}CH_{3} - CH_{2}Br$
Explanation
The correct answer is Option B: $CH_{3} - CH = CH_{2}\underset{UV\ light}{\overset{\ \ \ \ \ \ \ \ \ \ \ HBr\ \ \ \ \ }{\rightarrow}}CH_{3} - CH_{2} - CH_{2}Br$.
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.