E2 Reactions

Updated: Jan 11, 2020

Elimination reactions are helpful in many situations, especially in synthesis when you have an alkane and need an alkene. Similar to substitution reactions, eliminations produce similar products but depend on different factors. Most notably, E1 reactions have a carbocation intermediate while E2 reactions do not (for more information on E1 reactions, see “SN1/E1 Reactions”).

While SN2 reactions require a nucleophile, E2 reactions require a base. This base will pull off a b-hydrogen which opens up two electron. These electrons form another bond between the carbons, creating a double bond, and expelling the leaving group in order for the second carbon to maintain octet:

Note: Only the b-hydrogen shown.

In order for E2 to occur, the hydrogen and the leaving group must be antiperiplanar. This just means that the hydrogen and leaving group have to be on the same plane, but in opposite directions, forming a “Z” shape with the two carbons involved. With linear molecules this usually isn’t a problem, as sigma bonds can freely rotate to accommodate. However, ring structure can prove to be problematic. In the case of cyclohexane, E2 can only happen if the hydrogen and leaving group are both in the axial position on adjacent carbons. If one (or both) are equatorial, E2 can’t happen. It may help to build a molecule with your modeling kit if you’re having trouble visualizing this.

Elimination reactions also introduce another rule: Zaitsev’s Rule. This is similar to Markovnikov’s Rule in that the more substituted option will be the major product, and the less substituted option the minor product. And just like Markovnikov’s Rule, there is an anti-Zaitsev’s exception: tert-butoxide. tBuO- is too bulky to get close to the more substituted carbon, so instead it pulls a hydrogen off from the less substituted b-carbon, causing the less substituted option to be the major product.

Zaitzev's Rule

Anti-Zaitzev's Rule

Predicting E2

Unlike SN2, E2 only depends on the base present. A strong base is required to pull off a b-hydrogen, so if you don’t have a strong base you cannot proceed through E2. Additionally, if you have a strong base, SN1/E1 can’t occur, so with a strong base you know you’ll be doing either SN2 or E2.


1. Non-antiperiplanar b-hydrogens: E2 requires the b-hydrogen and leaving group to be antiperiplanar, if they aren’t then E2 can’t occur. This is most problematic in ring structures due to the rigidity of the molecule.

2. tBuO-: Again here’s tert-butoxide causing problems. When you see it, circle it, and remember that this molecule causes the major product to be the less substituted option—anti-Zaitsev’s Rule.

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