A) \[\overset{\bullet }{\mathop{C}}\,{{H}_{2}}C{{H}_{2}}CH{{(C{{H}_{3}})}_{2}}\]
B) \[C{{H}_{3}}\underset{\bullet }{\mathop{C}}\,HCH{{(C{{H}_{3}})}_{2}}\]
C) \[C{{H}_{3}}C{{H}_{2}}\underset{\bullet }{\mathop{C}}\,{{(C{{H}_{3}})}_{2}}\]
D) \[\overset{\bullet }{\mathop{C}}\,{{H}_{3}}\]
Correct Answer: D
Solution :
The stability of free radical increases by increasing the number of alkyl groups because the power of delocalisation of odd electron increases. Therefore,\[\overset{\bullet }{\mathop{C}}\,{{H}_{3}}\]free radical is the least stable.
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