I am quite confused about the competition of $\mathrm{E2}$ and $\mathrm{S_N2}$ reactions. My lecturers say $\mathrm{E2}$ and $\mathrm{S_N2}$ reactions are not competing reactions but most websites say they are. Please I would like to be clarified on whether $\mathrm{E2}$ and $\mathrm{S_N2}$ reactions are competing reactions, and if not, why?

  • $\begingroup$ If the question has been sufficiently answered, please accept it! $\endgroup$ – HappyFeet Me Apr 5 '20 at 7:44

Yes, E2 and SN2 are competing reactions. Keep in mind, E2 reactions require a base whereas SN2 reactions require a nucleophile. Nucleophilicity corresponds to the ability to donate a lone pair of electrons ‒ here, to a carbon bearing leaving group ‒ which is a kinetic property (time does matter). Basicity corresponds to accepting a hydrogen, a thermodynamic property determined in the state of equilibrium (where time does not matter).

1.Both properties, nucleophilicity and basicity of a reagent may be in competition. For example, a sterically hindered compound which potentially could act as a nucleophile by the presence of a free electron pair will be considered a poor nucleophile and instead a good base. Examples of sterically hindered bases are 2,6-lutidine (mild), potassium tert-butoxide, LDA (strong). The small methanolate, $\ce{CH3O-}$, on the contrary, may act simultaneously as nucleophile as well as a base.

2.Secondly, for SN2 reaction to take place, the substrate (the compound on which the nucleophile will react) should not be crowded, since it increases the energy in the transition state and the reaction becomes less favourable. In these cases, E2 can take place, provided the nucleophile here actually acts as a good base and the product formed is stable by hyperconjugation, resonance, etc.

3.Also, temperature is an important factor. If the temperature is high, E2 will be favoured owing to the thermodynamics of the reaction.


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.