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How does inductive effect affect the stability of free radical?

I can't seem to understand whether the carbon with the odd electron is deficient or not.

That is, whether a +I group, like alkyl, will stabilize it or not? I know that an adjacent alkyl group like $\ce{-CH3}$ will anyway stabilize it by hyperconjugation.

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    $\begingroup$ Generally, we consider radicals to be electron deficient species. $\endgroup$
    – Zhe
    Commented Aug 16, 2017 at 18:55
  • $\begingroup$ +I groups stabilize free radical and -I groups destabilize free radical. $\endgroup$ Commented Dec 26, 2023 at 5:11
  • $\begingroup$ Related: chemistry.stackexchange.com/q/164828 $\endgroup$
    – Karsten
    Commented Dec 26, 2023 at 13:25

1 Answer 1

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TL;DR - Free radicals, which are carbon intermediates with an extra electron, are stabilized by electron-donating groups and destabilized by electron-withdrawing groups.


Explanation:

  1. Stability Factors: Free radicals, being electron-deficient, benefit from neighboring atoms providing electron density. The more alkyl groups present on the carbon, the greater the stability, following the order: $\mathrm{\dot{C}H_3 < 1^\circ < 2^\circ < 3^\circ}$.

    Stability order of free radicals


  1. Electron Configuration: Analyzing the electrons in an alkyl free radical reveals three bonding pairs and one unpaired electron. This results in a trigonal pyramidal geometry due to hybridization to an $\mathrm{sp^3}$ state.

    Carbon free radical resonance


  1. Delocalization (Resonance): Similar to carbocations, delocalization or resonance plays a role in stabilizing free radicals.

Sources: NCERT, class notes, and Master Organic Chemistry.

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