In a reaction with a secondary leaving group, why does using $\ce{NaI}$ with acetone as a solvent lead to an $\mathrm{S_N2}$ reaction mechanism instead of $\mathrm{S_N1}$? I thought that halogens were weak nucleophiles, and only really led to $\mathrm{S_N1}$ reactions?

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    $\begingroup$ Students seem to like hard and fast rules like ‘$\ce{\text{iodide} -> \text{weak nucleophile} -> \mathrm{S_N1}}$’ where reality allows for grey tones, in between scenarios and others. $\endgroup$ – Jan Dec 18 '18 at 1:36
  • $\begingroup$ @Jan could you explain why this is not a hard and fast rule then? In most of the sources I read about this, they all just say NaI is good example of an SN2 reaction, so it seems like that is the case in most scenarios, and I'm just trying to understand why. $\endgroup$ – Ben Dec 18 '18 at 1:49
  • $\begingroup$ If it’s a nucleophile, it’s capable of $\mathrm{S_N1}$ and $\mathrm{S_N2}$ pathways. The ratio of $\mathrm{S_N1}$ to $\mathrm{S_N2}$ depends on factors such as the leaving group, steric influence of the electrophile, the solvent, the temperature and the strength of the nucleophile. That’s all there is to it. Saying ‘x will only do y because z’ is wrong most of the time. $\endgroup$ – Jan Dec 18 '18 at 1:54
  • $\begingroup$ No, iodide is an amazing nucleophile. It also also happens to be a great leaving group, so substituting for iodide and having it stay on the product is hard. $\endgroup$ – Zhe Jan 18 '19 at 16:36

As Jan wrote, there isn't a hard and fast rule encompassing every mechanism, so there may not be a definitive answer to the question. Nevertheless, I found an interesting site that talked a little bit about the Finkelstein reaction:

$$\ce{NaI + RX -> NaX + RI}\qquad\ce{X} = \ce{Cl}, \ce{Br}$$

This is in acetone solvent (I don't know how to format it along the arrow).

If you go by Henry Rzepa's blog, the solvation of sodium halides by acetone leads to a solvent shell that isn't well "fit" for a chloride or bromide anion. The writer discusses how the gaps between solvated sodium ions is just large enough to fit an iodide anion, which means the anion has minimal distance to the sodium atoms around it, and therefore a minimized potential energy. Bromide and chloride are too small, leaving a gap of empty space around them, and the distance to sodium cations is larger than optimal. It becomes energetically more favorable for the sodium and halide ions to simply fall out of solution than arrange themselves in a solvated cage. This drives the reaction forward in accordance with Le Chatelier's principle.

While this relates to the solubility of the halide salts, maybe it can also serve to speculate on the mechanism of the Finkelstein reaction. In the vicinity of the solvated $\ce{NaI}$ cage, there isn't a favorable space to release bromide or chloride into, so the iodine attack is forced to proceed through $\mathrm{S_N2}$.

This is far from rigorous research, so I'd take it with a grain of "salt" as a correct answer, but thought it might be interesting to think about.

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NaI in acetone is the reagent used in Finkelstien reaction. Alkyl halides are treated with the reagent to give alkyl iodides. The important aspect of this reaction is that the reaction is driven forward because of the lower solubility of the products NaCl and NaBr in acetone even though I- is a weak nucleophile.

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    $\begingroup$ This in itself doesn’t explain why one nucleophilic mechanism should dominate over another. $\endgroup$ – Jan Dec 18 '18 at 5:28

Note : When does SN2 attack occur. Let's analyse this and you will get your answer in end.

First of all the steric hindrance around the carbon atom should be least i.e if RX is 1° (primary) then SN2 attack always (mostly) occurs.

If RX is 3° (tertiary) then SN1 attack always (mostly) occurs, as steric hindrance is very high.

Now if RX is 2° (secondary), the mechanism can be SN1 or SN2, depending on the Nucleophile and the solvent. If solvent is aprotic and nucleophile is fairly good (strength): like DMF, acetone etc., then SN2 attack is favoured (I am not explaining why it occurs, but you can refer to the mechanism) If solvent is protic and nucleophile is not so strong: then SN1 mechanism is favoured.

In your question, $\ce{I-}$ (iodide ion) is mild nucleophile and acetone is aprotic solvent. Thus SN2 mechanism is favoured more.

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