The increasing order of stability of the following free radicals is

  • Option 1)

    (CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H< (C_{6}H_{5})_{3}\dot{C}

  • Option 2)

    (C_{6}H_5)_3\dot{C}< (C_{6}H_5)_2\dot{C}H< (CH_3)_3\dot{C}< (CH_3)_2\dot{C}H

  • Option 3)

    (C_{6}H_5)_2\dot{C}H< (C_{6}H_5)_3\dot{C}< (CH_3)_3\dot{C}< (CH_3)_2\dot{C}H

  • Option 4)

    (CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H

 

Answers (2)

As we learnt in

Stability of alkyl free radical -

Electron donating group stabilises alkyl free radicals and withdrawing group destabilises alkyl free radicals.

- wherein

 

 

Stability of alkyl free radical due to resonance -

More the no of resonating structure more is the stability.

- wherein

 

 Free radical is stabilised due to resonance, hyperconjugation and inductive effect. In given problem the order of stability of free radical is 

(CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H< (C_{6}H_{5})_{3}\dot{C}


Option 1)

(CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H< (C_{6}H_{5})_{3}\dot{C}

This option is correct

Option 2)

(C_{6}H_5)_3\dot{C}< (C_{6}H_5)_2\dot{C}H< (CH_3)_3\dot{C}< (CH_3)_2\dot{C}H

This option is incorrect

Option 3)

(C_{6}H_5)_2\dot{C}H< (C_{6}H_5)_3\dot{C}< (CH_3)_3\dot{C}< (CH_3)_2\dot{C}H

This option is incorrect

Option 4)

(CH_{3})_{2}\dot{C}H< (CH_{3})_{3}\dot{C}< (C_{6}H_{5})_{3}\dot{C}< (C_{6}H_{5})_{2}\dot{C}H

This option is incorrect

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