In my textbook, the following statement is given :

Attempts to prepare optically active iodides by nucleophilic displacement on optically active bromides using the iodide ion normally produces a racemic mixture of iodoalkanes.

The reason given is that this is due to the following equilibrium:

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My doubts:

  1. $\ce{I-}$ is a stronger nucleophile than $\ce{Br-}$. Then, why does the backward reaction even occur?

  2. Is the reaction SN2?

  3. Does this happen for other types of nucleophilic displacements also?

    Book: Arihant Organic Chemistry (DPP Problems)

  • 1
    $\begingroup$ This suggests the reaction has some SN1 character. Formation of a carbocation leads to loss of stereochemical integrity. $\endgroup$ – Waylander Mar 11 '18 at 17:06

Yes the backward reaction is an sn2 reaction.and I'm sure that you have the knowledge of iodide being nucleophilic catalyst-it is a very good nucleophile as well as a good leaving group.that is why the backward reaction is not an illogical one.you can't generalize chemistry.so for this reaction it occurs but it does not mean that it will happen in others also.


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