E1 & E2 Elimination Mechanisms

When Elimination Competes: E1 & E2

The carbocation formed in S_N1 can also lose a β-hydrogen to a base and form an alkene instead — this competing pathway is an elimination reaction. Commonly, a weak nucleophile acts as the base in this pathway. Since it follows the same first-order kinetics as S_N1, it's called unimolecular elimination, abbreviated E1 (E stands for elimination and 1 for unimolecular).

Mixtures of substitution and elimination products result under both S_N1 and E1 conditions; higher temperature increases the elimination percentage. On an unsymmetrical alkyl halide, E1 gives a mixture of elimination products along with a minor amount of S_N1 substitution product. The more substituted alkene (major product) is the Zaitsev product; the least substituted is the Hofmann product.

Recall: no S_N2 occurs on a tertiary alkyl halide (steric hindrance), and a base and nucleophile are chemically the same species. So when the nucleophile can't approach a tertiary carbon, it behaves as a base instead and abstracts a proton from the β-carbon.

This reaction follows second-order kinetics, similar to S_N2, and is called bimolecular elimination, abbreviated E2 (E stands for elimination and 2 for bimolecular).

E2 Reaction Mechanism

The bimolecular elimination mechanism occurs in a single step, with three changes happening simultaneously:

• Abstraction of a proton by the base
• Conversion of sp³ to sp²
• Departure of the leaving group

For a smooth E2 reaction, the β-hydrogen and leaving group need an anti-periplanar arrangement. E2 is more regioselective than E1. Like E1, E2 also gives a mixture of products and the Zaitsev product is always predominant — but while the alkene mixture in E1 is not controlled, in E2 it is possible to control the Hofmann vs. Zaitsev product distribution by using a highly sterically hindered base, a bulkier substrate, and a larger leaving group. Increasing bulkiness of the substrate, increased leaving-group size, and sterically hindered bases all favor the Hofmann product.

Stereoselective vs. stereospecific E2

If the alkyl halide has two equivalent β-positions, there's no regioselectivity issue, but the base may abstract either β-proton, giving two stereoisomeric products in unequal amounts (stereoselective). If the alkyl halide has only one β-proton, only one stereoisomeric product is possible (stereospecific) — provided the leaving group and that β-proton are anti-periplanar.

If the leaving group and β-proton aren't already anti-periplanar, the molecule will rotate into that arrangement before the elimination reaction can occur.