I was reading about Substitution and Elimination Reactions and I came across the following on MasterOrganicChemistry:

Polar protic solvents tend to favor elimination (E2) over substitution (SN2). Polar aprotic solvents tend to favor substitution (SN2) relative to elimination (E2).

I understand the first part. But I have trouble understanding the part with the polar aprotic solvents.

Consider the following example (from the same website):

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Why is E2 not favoured? I can understand why polar aprotic solvents favour SN2:

A polar aprotic solvent like acetone or dimethylformamide preferentially solvates cations, leaving an almost "bare" nucleophile. This increases its nucleophilicity

But I could find no explanation as to why E2 does not happen because of the solvent. What keeps the nucleophile (cyanide anion here) from abstracting a beta hydrogen?

  • $\begingroup$ E2 does happen but product formed is little compared to SN2 in polar aprotic solvent $\endgroup$
    – knoftrix
    May 7 '19 at 17:36
  • $\begingroup$ I have searched incessantly, but no avail. $\endgroup$ May 7 '19 at 18:51

Competition between substitution and elimination is one of the many points of confusion for introductory organic students.
In general, they are taught that there are four factors that will determine the outcome of a reaction where there may be competition between substitution or elimination. They are:

  • Type alkyl halide (methyl, $1^o$, $2^o$ or $3^o$)
  • The strength/basicity of the nucleophile
  • The solvent
  • The temperature

I think that many students over-emphasize the impact of any one of these factors. It is more useful to look at the overall impact of the factors and evaluate the reaction as a whole.

Let's examine the case in question:

  • Primary alkyl halide - generally favours substitution
  • The nucleophile is $\ce{^-CN}$ which is a good nucleophile but a weak base.
  • Polar aprotic solvent
  • Low temp.

The big factor here is not the solvent. All of the factors point toward substitution, especially the alkyl halide and the nucleophile. Remember that elimination reactions are essentially acid-base in nature. If the nucleophile is not a strong base, elimination reactions are less likely.


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