When I first learned about substitution reactions, I started wondering something that I never ended up figuring out:

Suppose you have a strong nucleophile which is also a good leaving group, such as I$^{-}$. Can such a group substitute itself using the SN1 mechanism? If so, would that mean if such a group were bonded to a chiral carbon, an enantiomerically pure solution could spontaneously racemize?

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    $\begingroup$ Very short answer: Yes. Possibly also via SN2. $\endgroup$ Commented Jul 6, 2015 at 1:28
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    $\begingroup$ Has spontaneous racemization been observed? $\endgroup$ Commented Jul 6, 2015 at 1:36
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    $\begingroup$ see this and this $\endgroup$
    – ron
    Commented Jul 6, 2015 at 1:41
  • $\begingroup$ The reaction you refer to has been done and racemisation observed, however I don't have the reference $\endgroup$
    – Beerhunter
    Commented Apr 2, 2016 at 13:07

1 Answer 1


Yes - a nucleophile can replace itself in a substitution reaction. Under the SN1 mechanism, once the carbocation is formed, it becomes prone to attack from any negatively charged species, including the one that left it in the first place. Likewise, in an SN2 reaction, there is the possibility of a nucleophile replacing itself. Given enough time, I would expect alkyl halides to become racemic in solutions that contain the halide anion; however, substitution reactions remain stereo-selective. This is because when you are preforming a substitution reaction, you normally use a better leaving group than the nucleophile that you are adding. Because of this, the nucleophile in solution will preferentially replace the initial leaving group over its resulting product. This is because the rate laws for both SN1 and SN2 reactions depend on the concentration of the alkyl group - implying that an alkyl group with a better leaving group will react faster than one with a weaker leaving group. In SN1, the rate determining step is the leaving group dissociating from the carbocation, which happens quicker with a better leaving group. In an SN2 reaction, the rate determining step involves both the nucleophile attacking and the leaving group leaving. This - once again - means that the reaction will proceed quicker with a better leaving group. Given enough time, your product should racemize; however, your substitution reaction should be done before this can have a major effect on your yield.

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    $\begingroup$ Iodine is quite a good nucleophile for saturated carbon atoms because of its softness. a good nucleophile is a poor leaving group - this statement only really applies to reactions with hard electrophiles like carbonyl groups. $\endgroup$
    – bon
    Commented May 9, 2016 at 9:22
  • $\begingroup$ @NielsKornerup Can you clarify what you mean when you say, "Because of this, the nucleophile in solution will preferentially replace the initial leaving group over its resulting product"? $\endgroup$ Commented May 10, 2016 at 0:43
  • $\begingroup$ I modified my answer. Hopefully this helps! $\endgroup$ Commented May 10, 2016 at 2:28

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