A) \[\overset{\centerdot }{\mathop{C}}\,{{H}_{2}}C{{H}_{2}}CH{{\left( C{{H}_{3}} \right)}_{2}}\]
B) \[C{{H}_{3}}\underset{\centerdot }{\mathop{C}}\,HCH{{\left( C{{H}_{3}} \right)}_{2}}\]
C) \[C{{H}_{3}}C{{H}_{2}}\underset{\centerdot }{\mathop{C}}\,{{(C{{H}_{3}})}_{2}}\]
D) \[\overset{\centerdot }{\mathop{C}}\,{{H}_{3}}\]
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, \[\overset{\centerdot }{\mathop{\text{C}}}\,{{\text{H}}_{\text{3}}}\] free radical is the least stable.
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