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  • $\begingroup$ It has to do with an (equimolar) Racemic mixture being formed! Note that a racemic mixture still has chiral atoms but cannot contribute to any net rotation of the plane polarized light making it optically inactive as a combined product $\endgroup$
    – Utkarsh Sharma
    Commented May 22, 2020 at 17:03
  • $\begingroup$ chemistry.stackexchange.com/questions/132885/… $\endgroup$
    – user94580
    Commented Jun 6, 2020 at 6:16
  • $\begingroup$ This question does not make sense. At the beginning of the reaction, the rate of loss of radioactivity is the rate of both types of substitution. But the rate of loss of optical activity is tempered by the fact that only one of the two mechanisms produces more of the other enantiomer to progress towards racemization. By that logic, how could the rate of loss of optical activity be greater than the loss of radioactivity? $\endgroup$
    – Zhe
    Commented Jun 7, 2020 at 1:36
  • $\begingroup$ @Zhe As I have written in my answer, rate of loss of optical activity is proportional to extent of racemisation. If I take a substrate with optical activity having only one enantiomer, then initially, there is no racemisation. Suppose 10% of substrate reacts and undergoes inversion, then the solution will be 20% racemic since a racemic mixture consists of equimolar amounts of two enantiomers. So rate of loss of optical activity is twice of rate of reaction for SN2 mechanism. $\endgroup$
    – Indian135
    Commented Jun 12, 2020 at 1:32
  • $\begingroup$ "Suppose 10% of substrate reacts and undergoes inversion, then the solution will be 20% racemic" That statement is unfortunately not correct. As soon as any of the initial starting material reacts, you generate the other enantiomer which reacts at the same rate as the initial enantiomer. $\endgroup$
    – Zhe
    Commented Jun 12, 2020 at 3:11