A) carbanions
B) carbocations
C) carbenes
D) free radicals
Correct Answer: D
Solution :
Bromination of alkanes in the presence of sunlight involves the formation of free radical. e.g., \[C{{H}_{4}}\xrightarrow[hv]{B{{r}_{2}}}C{{H}_{3}}Br\] Mechanism: Initiation: \[Br-Br\xrightarrow[{}]{hv}B{{r}^{\bullet }}+B{{r}^{\bullet }}\] Propagation: \[C{{H}_{4}}+B{{r}^{\bullet }}\xrightarrow{{}}CH_{3}^{\bullet }+HBr\] \[CH_{3}^{\bullet }+Br-Br\xrightarrow{{}}C{{H}_{3}}Br+B{{r}^{\bullet }}\] Termination: \[B{{r}^{\bullet }}+B{{r}^{\bullet }}\xrightarrow{{}}B{{r}_{2}}\]
You need to login to perform this action.
You will be redirected in
3 sec
