Reaction of 1-bromobutane with cyanide in ethanol/water to form pentanenitrile

In my textbook it's mentioned that the reaction undergoes $\mathrm{S_N2}$ mechanism. I understand that the carbon attached to the bromine is primary, but both solvents here are polar protic solvents, which favour the $\mathrm{S_N1}$ mechanism. Can anybody explain how to reconcile these?


1 Answer 1


Agreed, $ \ce{Br-} $ is a good leaving group, so one would expect that an SN1 mechanism is not too difficult to have in a polar protic solvent. But do not forget that the mechanism of any reaction is determined by not only the nature of the leaving group, but also the carbon skeleton. A primary carbocation is extremely difficult to obtain, and will not, in general, be formed unless the corresponding anionic complex is extremely stable. For example, the equilibrium $$ \ce{R-CH2-F + SbF5 <=> R-CH2+ + SbF6- }$$ lies predominantly towards the products side because the anion $\ce{SbF6-}$ is an extremely stable system (which is why the mixture $\ce{HF + SbF5}$ is considered a super acid.). Had $\ce{AlBr3}$ been used before adding $\ce{KCN}$ to the solution, I would've gone for SN1 mechanism.

So, firstly, the primary carbocation is a difficulty if you consider an SN1 mechanism.

Moreover, the formation of a carbocation would immediately facilitate rearrangements. Since a primary carbocation is unstable, Wagner-Meerwein rearrangements will take place and your end product would have been a substitution at the second carbon instead of the first (terminal) one.

As for the solvent, it is a mixture of $\ce{Et-OH}$ and $\ce{H2O}$ to facilitate the dissolution of both reactants. Because one of them is an ionic compound, and the other an organic compound, getting them both in the same state can only be achieved by the mixture of an organic solvent and water; the organic solvent being such that it is, itself, miscible with water. A polar protic solvent weakens the ionic bonds by screening, and favours ionisation. This doesn't imply that a polar protic solvent will always completely ionise your solute.

  • $\begingroup$ Would it have been SN1 or SN2 if there was a secondary haloalkane? $\endgroup$
    – Shub
    Oct 6, 2022 at 18:16

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