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. eg,\[C{{H}_{4}}\frac{B{{r}_{2}}}{hv}C{{H}_{3}}Br\] Mechanism: Initiation: \[Br-Br\xrightarrow{hv}B{{r}^{.}}+B{{r}^{.}}\] Propagation: \[C{{H}_{4}}+B{{r}^{.}}\xrightarrow{{}}CH_{3}^{.}+HBr\] \[CH_{3}^{.}+Br-Br\xrightarrow{{}}C{{H}_{3}}Br+B{{r}^{.}}\] Termination: \[B{{r}^{.}}+B{{r}^{.}}\xrightarrow{{}}B{{r}_{2}}\]
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