1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

$\ce{EtO-}$ 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).
$\endgroup$
  • $\begingroup$ 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

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.