In a lab, a common experiment is to perform SN2 reactions involving a primary alkyl bromide and sodium iodide in acetone (polar aprotic solvent).

The result is that the iodide ion is a good nucleophile, so the iodide will displace the bromide from the primary alkyl bromide, and sodium bromide will precipitate from the solution.

However, without the presence of the sodium cations, would the reaction still take place?

I'm somewhat confused, because iodide is both a better leaving group and a better nucleophile than bromide, so what is the determining factor in the reaction?

Is the strength of the nucleophile more important or the ability of the leaving group be displaced?

  • $\begingroup$ There aren't any silver cations. NaBr will precipitate. $\endgroup$
    – EJC
    Commented Aug 31, 2016 at 22:59
  • $\begingroup$ I meant sodium. Changed to reflect that. $\endgroup$ Commented Aug 31, 2016 at 23:05
  • $\begingroup$ In aprotic solvents bromide is a stronger nucleophile than iodide. $\endgroup$
    – EJC
    Commented Aug 31, 2016 at 23:13
  • 1
    $\begingroup$ @CayetanoGonçalves Please, see this. $\endgroup$ Commented Aug 31, 2016 at 23:36
  • $\begingroup$ @ChemistryHelpCenter thank you. That was my next question if it was being driven by Le Châtlier's principle $\endgroup$ Commented Aug 31, 2016 at 23:37

1 Answer 1


This is very famous "Finkelstein Reaction" that involved the exchange of one halogen to another through SN2 mechanism. The driving force for the reaction is the solubility of salts (NaI, NaBr, KF) under equilibrium. There are different factor that effect the equilibrium position of a reaction, firstly, nucleophilicity of anion, presence of good leaving group, and stabilization of anion in a given solvent.

first part of your question, I don't think with the absence of sodium cation the reaction would take place, because the driving force is the precipitation of salts of NaBr and NaCl in acetone.

In the second part of your question, you are right iodide is better leaving group compared to Bromide because iodide is weaker base compared to bromoide. Apparently, iodide can't replace bromine but here the reaction is driven toward products by "mass action" due to the precipitation of the insoluble salt as mentioned above.


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