A) \[{{C}_{6}}{{H}_{5}}-C{{H}_{2}}-\overset{\bullet }{\mathop{C}}\,{{H}_{2}}\]
B) \[C{{H}_{3}}\overset{\bullet }{\mathop{C}}\,{{H}_{2}}\]
C) \[{{C}_{6}}{{H}_{5}}-\overset{\bullet }{\mathop{C}}\,H-C{{H}_{3}}\]
D) \[C{{H}_{3}}-\overset{\bullet }{\mathop{C}}\,H-C{{H}_{3}}\]
E) \[C{{H}_{3}}-\overset{\bullet }{\mathop{C}}\,{{H}_{2}}-\overset{\bullet }{\mathop{C}}\,{{H}_{2}}\]
Correct Answer: C
Solution :
The benzylic free radicals are more stable than alkyl free radicals because of the delocalization of odd or unpaired electron on the benzene ring. Further, secondary free radical is more stable than primary free radical. Therefore, \[{{C}_{6}}{{H}_{5}}-\overset{\bullet }{\mathop{C}}\,H-C{{H}_{3}}\] is more stable than \[{{C}_{6}}{{H}_{5}}-C{{H}_{2}}-\overset{\bullet }{\mathop{C}}\,{{H}_{2}}\]
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