If I understood correctly, having a protic solvent increases chance of E1 mechanism. But I can not rationalize this because carbocation intermediate is a positively charged molecule, and by surrounding that with a lot more positive protons will make the whole system in an unstable state. Hence it should make it less likely for E1 to occur. Is my understanding wrong or did I miss some important fact?


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First of all, a protic solvent does not mean a highly acidic solvent; it mainly means a solvent whose molecules contain protons that can reasonably be expected to act as an acid. Take for example methanol: its $\mathrm pK_\mathrm a$ is somewhere around $15$, so calling the hydroxy proton acidic is a stretch by most definitions but there are sufficiently basic compounds that can abstract said proton.

Secondly, the question of how a protic solvent affects the E2 versus E1 ratio has less to do with the intermediate cation and more to do with the leaving group. In an E1 reaction, the leaving group dissociates from the main molecule first taking the bonding pair with it. That typically means this leaving group is now anionic or has a somewhat nucleophilic lone pair on it. The protons of the solvent can ‘block’ that lone pair from reattacking the intermediate cation, preventing the backwards reaction and increasing the odds of proceeding. The resulting cation can then be stabilised by the polar bit of a polar protic solvent: the more negatively polarised part of the solvent molecules will interact favourably with the cation.

In an E2 reaction, the base must interact with the starting material for both the starting material’s proton and the leaving group to be displaced in one step. In this case, having a protic solvent which is capable of protonating the base (reducing its basicity) is harmful meaning the reaction will proceed slower. In E1 reactions, weaker bases can be used that are not able to deprotonate the solvent molecule so this pathway is not much of a problem.

  • $\begingroup$ @DDD4C4U I did go into solvation effects stabilising the E1 intermediate. $\endgroup$
    – Jan
    Commented Jul 30, 2020 at 6:12

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