This question is confusing and I can’t seem to find a answer to this. The question is what solvent promotes E2 and E1 reactions? My teacher says that methanol as solvent with strong base favors E2 over Sn2. But she never explained why. Also she keeps using methanol as the example which is confusing. Can anyone please elaborate on this?


1 Answer 1


This website gives good rules of thumb: http://www.masterorganicchemistry.com/2012/12/04/deciding-sn1sn2e1e2-the-solvent/

And so does this website: http://www.mhhe.com/physsci/chemistry/carey/student/olc/ch08substitutionvselimination.html

In addition, according to page 583 of "Physical Organic Chemistry" by Anslyn and Dougherty:

Almost all bases are also nucleophiles, and hence we expect competition between elimination and substitutions. In both $\ce{S_N2}$ and $\ce{E\text{2}}$ reactions, the nucleophile or base reacts in a single rate-determining step with the reactant. In both $\ce{S_N1}$ and $\ce{E\text{1}}$ reactions, the nucleophile or base reacts in a step after the rate-determining heterolysis...

An $\ce{E\text{2}}$ reaction requires the addition of a base, and can be performed in solvents of low or high ionizing power. Since most strong bases are also good nucleophiles, $\ce{S_N2}$ reactions are important competitors. However, elimination will dominate if the carbon with the leaving group is not susceptible to nucleophilic attack, such as a tertiary R group. Examples of this trends are given in Table 10.6.

$\ce{E\text{1}}$ reactions involve carbenium ion intermediates, and therefore are facilitated by all the factors that stabilize carbenium ions. These are the same factors that facilitate $\ce{S_N1}$ reactions. Strongly ionizing solvents and substitution of electron donating groups on the carbon undergoing heterolysis are necessary.

In highly ionizing solvents and with R groups that readily form carbenium ions, the ratio of substitution to elimination is typically independent of the leaving group. This evidence supports the notion that the substitution and elimination products are formed b branching from a common intermediate, and therefore the two reactions share a common rate-determining step. In contrast, in solvents of lower ionizing power, the ratio of substitution to elimination products does depend on leaving group, and indication that the two reactions do not share a common intermediate. Instead, contact ion pairs are formed, and either the leaving group or the solvent can remove the proton to give elimination products. The formation of a contact ion pair also has an influence on the stereochemistry of the elimination.

Table 10.6

Reactant            Solvent   Base      Elimination
Tert-Butyl Bromide  Ethanol   NaOEt     100.00%
Tert-Butyl Bromide  Acetone   NBu4Cl    96.00%
isopropyl Bromide   Ethanol   NaOEt     75.00%
isopropyl Bromide   Acetone   NBu4Cl    0.00%
Ethyl Bromide       Ethanol   NaOEt     9.00%
Ethyl Bromide       Acetone   NBu4Cl    0.00%

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