2
$\begingroup$

I was dealing with a question whose answer was justified only when it was assumed that the Finkelstein reaction proceeds through an $\mathrm{S_N2}$ mechanism. So, does it really follow an $\mathrm{S_N2}$ mechanism or is there some other mechanism at play?

$\endgroup$
  • $\begingroup$ What brings you to assume any other mechanism? $\endgroup$ – Jan Apr 19 '16 at 15:43
  • 1
    $\begingroup$ Why do you think it wouldn't proceed through $\mathrm{S_N2}$? $\endgroup$ – SendersReagent Apr 19 '16 at 16:40
  • 1
    $\begingroup$ Consider this reaction: (CH3)3-C-Br + NaI (dry acetone) ---> (CH3)3-C-I I have every reason to believe this reaction will proceed through Sn1 mechanism as there is stearic hindrance to the incoming nucleophile and moreover I'm getting a very stable tertiary carbocation. $\endgroup$ – EuclidAteMyBreakfast Apr 20 '16 at 5:24
  • $\begingroup$ The mechanism should depend on what type of haloalkane you use in the reaction i.e. primary, secondary or tertiary. For primary haloalkanes, the reaction must proceeed through ${S_{N}2}$ mechanism, since the carbocation formed will not be stable, and there is very little steric hindrance while ${S_{N}1}$ mechanism must be used for a tertiary haloalkane sue to stable carbocation formation and steric hindrance. Not really sure about secondary haloalkanes, though. $\endgroup$ – user25546 May 15 '16 at 6:06
2
$\begingroup$

The Finkelstein reaction is a nucleophilic substitution reaction. As such, it generally has the choice of proceeding via the $\mathrm{S_N1}$ or the $\mathrm{S_N2}$ mechanism.

The reaction is typically conducted in acetone; a polar but not very protic solvent. The solvent is used because all the sodium halides except for sodium iodide are not very soluble in it and precipitate out. It does not show any real preference for either $\mathrm{S_N1}$ or $\mathrm{S_N2}$.

Therefore, we should study the nucleophile and leaving group. Iodide is both a good nucleophile and a good leaving group; the former is important as it allows us to assume a viable $\mathrm{S_N2}$ mechanism. The other halides that you are substituting with iodide are not as good leaving groups but still rather okay. This again does nothing really to tilt the scales towards either $\mathrm{S_N1}$ or $\mathrm{S_N2}$.

The final choice we have is that it depends on the substrate. Probably, if you are replacing a primary halide chances are that the reaction will proceed according to the $\mathrm{S_N2}$ mechanism while an $\mathrm{S_N1}$ mechanism should predominate for tertiary halides.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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