Question
How do we predict whether an elimination reaction proceeds via E1 or E2? What factors determine the mechanism?
Solution — Step by Step
In E2, a strong base abstracts a beta-hydrogen at the same time as the leaving group departs. Both events happen simultaneously — no intermediate.
Key requirement: the beta-hydrogen and leaving group must be anti-periplanar (180-degree dihedral angle). This geometric requirement is critical and often tested in JEE Advanced.
In E1, the leaving group departs first (forming a carbocation), then the base removes a beta-hydrogen in a separate step.
E1 follows the same first step as SN1 — carbocation formation. So E1 and SN1 are always in competition when conditions favour carbocation formation.
Strong base (like , , ) + any substrate = E2
Weak base / no base + tertiary substrate + heat = E1
Bulky strong base (like tert-butoxide, DBU) = E2 even more strongly (bulky bases cannot act as nucleophiles for SN2, so elimination dominates)
Temperature: higher temperature always favours elimination over substitution (for both E1 and E2).
graph TD
A{Strong base present?} -->|Yes| B{Bulky base?}
A -->|No / Weak base| C{Tertiary substrate?}
B -->|Yes: t-BuOK| D[E2 dominant]
B -->|No: NaOH, NaOEt| E{Substrate type?}
E -->|1-degree| F[SN2 competes with E2]
E -->|2-degree / 3-degree| D
C -->|Yes + Heat| G[E1 + SN1 compete]
C -->|No| H[SN1 dominant]Why This Works
The core logic is simple: base strength determines E1 vs E2, just as nucleophile strength determines SN1 vs SN2.
- Strong base = E2 (the base actively pulls the proton — it does not wait for carbocation formation)
- Weak/no base = E1 (the substrate ionises first, then whatever base is around removes the proton)
E1 and SN1 share the same carbocation intermediate, so they always compete. E2 and SN2 compete when the base is also a good nucleophile (like ). Temperature tilts the competition: heat favours elimination because it has a larger entropy of activation ( — one molecule becomes two).
Alternative Method
A rapid exam-ready decision table:
| Factor | Favours E2 | Favours E1 |
|---|---|---|
| Base | Strong, concentrated | Weak or absent |
| Substrate | Any (even 1-degree) | 3-degree only |
| Solvent | Polar aprotic | Polar protic |
| Temperature | High | High |
| Kinetics | Bimolecular | Unimolecular |
| Stereochemistry | Anti-periplanar required | No geometric requirement |
Common Mistake
Students forget the anti-periplanar requirement in E2. In cyclic systems (like cyclohexane derivatives), the beta-H and leaving group must both be in axial positions (trans-diaxial) for E2 to occur. If the only available beta-H is equatorial, E2 cannot happen from that hydrogen. JEE Advanced 2021 tested this with a substituted cyclohexane — many students got it wrong because they did not draw the chair conformation.