A) \[{{(C{{H}_{3}})}_{2}}\overset{\bullet }{\mathop{C}}\,H<{{(C{{H}_{3}})}_{3}}\overset{\bullet }{\mathop{C}}\,<{{({{C}_{6}}{{H}_{5}})}_{2}}\overset{\bullet }{\mathop{C}}\,H\] \[<{{({{C}_{6}}{{H}_{5}})}_{3}}\overset{\bullet }{\mathop{C}}\,\]
B) \[{{({{C}_{6}}{{H}_{5}})}_{3}}\overset{\bullet }{\mathop{C}}\,<{{({{C}_{6}}{{H}_{5}})}_{2}}\overset{\bullet }{\mathop{C}}\,H<{{(C{{H}_{3}})}_{3}}\overset{\bullet }{\mathop{C}}\,\] \[<{{(C{{H}_{3}})}_{2}}\overset{\bullet }{\mathop{C}}\,H\]
C) \[{{({{C}_{6}}{{H}_{5}})}_{2}}\overset{\bullet }{\mathop{C}}\,H<{{({{C}_{6}}{{H}_{5}})}_{3}}\overset{\bullet }{\mathop{C}}\,<{{(C{{H}_{3}})}_{3}}\overset{\bullet }{\mathop{C}}\,\] \[<{{(C{{H}_{3}})}_{2}}\overset{\bullet }{\mathop{C}}\,H\]
D) \[{{(C{{H}_{3}})}_{2}}\overset{\bullet }{\mathop{C}}\,H<{{(C{{H}_{3}})}_{3}}\overset{\bullet }{\mathop{C}}\,<{{({{C}_{6}}{{H}_{5}})}_{3}}\overset{\bullet }{\mathop{C}}\,\] \[<{{({{C}_{6}}{{H}_{5}})}_{2}}\overset{\bullet }{\mathop{C}}\,H\]
Correct Answer: A
Solution :
Free radicals stability \[{{C}_{6}}{{H}_{5}}-\underset{\begin{align} & | \\ & {{C}_{6}}{{H}_{5}} \\ & Highly\text{ }stable\text{ }by \\ & delocalisation \\ \end{align}}{\overset{\bullet }{\mathop{C}}}\,-{{C}_{6}}{{H}_{5}}>{{C}_{6}}{{H}_{5}}-\underset{\begin{smallmatrix} | \\ {{C}_{6}}{{H}_{5}} \end{smallmatrix}}{\overset{\bullet }{\mathop{CH}}}\,\] \[>C{{H}_{3}}-\underset{\begin{align} & | \\ & C{{H}_{3}} \\ & 9-hyperconjugative \\ & hydrogens\text{ }and+l\,effect \\ \end{align}}{\overset{\bullet }{\mathop{C}}}\,-C{{H}_{3}}>{{H}_{3}}C\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\bullet }{\mathop{C}}}\,-H\]
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