When (2​S,3​R)-3-iodobutan-2-ol undergoes a substitution reaction with sodium azide the only organic product from the reaction is (2​S,3​R)-3-azidobutan-2-ol. Give a mechanism for the reaction.

My Attempt

Now I know this is not a normal $\mathrm{S_N1}$ or $\mathrm{S_N2}$ reaction since the stereochemistry remains the same and only one of the enantiomers are formed. I am guessing that at first there is some intramolecular reaction where the iodine atom attacks the other carbon bonded with the hydroxyl group. Then the azide substitutes the iodine atom.


1 Answer 1


I can explain the reason why the configuration does not change by using neighboring group participation of the hydroxyl group. This is the mechanism that I propose:

  1. The hydroxyl group attacks C-3 and $\ce{I-}$ leaves.
  2. Then $\ce{N3-}$, the nucleophile, attacks C-3.
  3. Since two successive inversions occur at C-3, the net result is retention.


As you can see, the configuration has not changed.

@bon brought up a point: The formation of a mixture of products when an epoxide opening takes place. If the nucleophile were to attack C-2 instead,

Attack at alternative carbon

C-3 will still have the S configuration.

As noted by @Jan in the comments, the epoxide intermediate is symmetrical and both epoxide carbons are equivalent. This can be shown by a $C_2$ rotation:

Epoxide symmetry

  • $\begingroup$ Epoxide formation under neutral conditions? Also, one would expect a mixture of products from the epoxide opening. $\endgroup$
    – bon
    Commented Feb 12, 2016 at 16:55
  • $\begingroup$ @bon I am sorry if this is wrong. I read about retention of configuration because of neighboring group participation in jerry march's book. So I thought it would be the reason here. $\endgroup$
    – Aditya Dev
    Commented Feb 12, 2016 at 16:56
  • $\begingroup$ I'm not saying it's definitely wrong but I'm just surprised (and intrigued). $\endgroup$
    – bon
    Commented Feb 12, 2016 at 16:57
  • $\begingroup$ @bon: if the nucleophile were to attack at the third carbon, then the product will also have retention configuration. So Nu attack on 2nd and third carbon will produce the same compound. I just drew the possible products on my book. $\endgroup$
    – Aditya Dev
    Commented Feb 12, 2016 at 16:59
  • 1
    $\begingroup$ The intermediate epoxide is $C_2$ symmetric. This renders the carbons homotopic and means that both attacks give the same product. $\endgroup$
    – Jan
    Commented Feb 13, 2016 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.