The base plays a central role in elimination reactions because it is responsible for removing a proton (hydrogen atom) from the β-carbon, which allows a double bond to form. Here’s how it works for different types of elimination:
1. In E2 Reactions
- The base directly removes a β-hydrogen while the leaving group departs in a single, concerted step.
- A strong base is required to pull off the proton efficiently.
- The strength and size of the base can affect:
- Rate of reaction (stronger base → faster E2)
- Regioselectivity (bulky bases often give the less substituted Hofmann product, smaller bases favor Zaitsev product)
- Examples of bases: hydroxide, alkoxides, tert-butoxide, LDA.
2. In E1 Reactions
- The base removes a proton after the leaving group has left and a carbocation forms.
- A weak base is usually sufficient because the carbocation is already reactive.
- The base’s role is mostly to complete the elimination, forming the alkene.
- Examples of bases: water, alcohols, acetate ion.
3. Factors Affected by the Base
- Strength: Strong bases favor E2; weak bases favor E1.
- Bulkiness: Bulky bases hinder substitution (SN2) and often favor elimination.
- Solvent compatibility: Some bases work better in polar aprotic solvents (for E2).
In summary:
- The base is the key agent that removes a proton, allowing the formation of the double bond.
- Strong bases → favor E2 elimination.
- Weak bases → can support E1 elimination after carbocation formation.
- Base size and strength also influence which alkene forms (more or less substituted).