enter image description here Give the main product and reaction type: SN1, SN2, E1 , E2.

As the 1-bromohexane is primary and the nucleophile a strong unhindered base, the reaction should be a SN2 reaction. However, the solvent $\ce{EtOH}$ is a polar protic solvent which favours SN1/E1 reactions. In addition, the temperature of the reaction is quite high and high temperatures favour the formation of elimination reactions, thus the reaction should be E1. But an E1 doesn't require a strong base, or else an E2 reaction would take place at the beginning. So does that make this reaction an E2 reaction? Am I thinking correctly on how to consider all the different parts of the reaction?


$\ce{EtO-}$ is a strong base, as well as a strong nucleophile, and the temperature is not too high. Therefore a mixture of products is most likely to be formed, where substitution will lead to the major and elimination will lead to the minor product.
Also, the reactant is a primary halide. So, in substitution, SN2 will be favoured, whereas in elimination E2 will be favoured.
So, final product will be,

  • Ethoxy hexane (via SN2) (major)
  • Hex-1-ene (via E2) (minor). (Maybe, a very little amount of hex-2-ene may also form by very very small fraction of E1).
  • $\begingroup$ Why is Sn2 major? I'm having difficult time understanding how a polar protic solvent can result in a Sn2 reaction. Isn't elimination reactions favoured in polar protic solvents? $\endgroup$ – J.Se Feb 12 '18 at 20:51
  • 1
    $\begingroup$ @J.Se take a look at this, will try to post a bit more detailed explanation. $\endgroup$ – bonCodigo Feb 13 '18 at 0:44
  • $\begingroup$ As a general rule, carbocations do not form in the presence of strong bases. This means that $\ce{S_N1}$ or $\ce{E1}$ mechanisms are ruled out immediately. It is important to look at all of the factors that are at play in the reaction: the alkyl halide, the nucleophile/base, the solvent and the temperature. It is the cumulative effect of these that determines the major product, not any one of them alone. $\endgroup$ – Michael Lautman May 8 at 14:55

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