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If an organic compound with a chiral center of a specific (R) or (S) configuration, is chlorinated on that carbon center through a radical mechanism, will one product be formed or two? Basically will the compound undergo racemization or not?

For example consider the case of: $\ce{C(CH3)(CH2CH3)(Ph)(H)}$ when monochlorinated, a radical intermediate will form. Now since the radical intermediate appears $\ce{sp^3}$ to me in this case I think that the configuration of compound about the chiral center will remain conserved when Chlorine radical later combines with this intermediate form. Is this true or not? Your analysis and views are appreciated.

(Please ignore effects of steric crowding in the compound, this example is purely for theoretical analysis)

Waylander
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Ashish
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  • I would agree with you, for the same reason that you indicated. – Azamat Jul 17 '21 at 06:53
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    @Ashish https://chemistry.stackexchange.com/questions/34976/what-is-the-geometry-of-an-alkyl-radical?noredirect=1&lq=1 (sort of duplicate...) – Rishi Jul 17 '21 at 08:30
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    From the JEE point of view, the final product would almost certainly be a racemic mixture. The link @Rishi provided says precisely that, how the barrier for inversion in the intermediate radical is extremely low -similar to amine inversion - so it rapidly interchanges between R/S forms, so the final product is also roughly equal in R/S (exactly equal in the theoretical, ideal case). – TRC Jul 17 '21 at 09:34

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