Mechanism comparison
E1 and E2 Elimination: How Each Mechanism Works
Elimination reactions form alkenes by removing a hydrogen (from the beta carbon) and a leaving group (from the alpha carbon) to generate a pi bond. The two elimination mechanisms — E1 and E2 — differ in exactly the same fundamental way as their substitution counterparts SN1 and SN2: E2 is concerted (one step), and E1 is stepwise (two steps through a carbocation intermediate).
This mechanistic difference has profound consequences for which substrates and conditions favor each pathway, what products form, whether rearrangements are possible, and what stereochemical requirements must be satisfied.
Rate law
Rate = k[substrate][base] — bimolecular
Geometry requirement
Anti-periplanar H and LG required (180°)
Base required
Strong base (KOH, NaOEt, KOtBu, LDA)
Substrate
Primary, secondary, or tertiary
Rearrangements
Never — no carbocation intermediate
Steps
Two steps (ionization → deprotonation)
Rate law
Rate = k[substrate] — unimolecular
Geometry requirement
None — carbocation is planar
Base required
Weak base acceptable (solvent, water)
Substrate
Tertiary > secondary; primary impossible
Rearrangements
Possible via 1,2-hydride or methyl shifts
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The Shared Carbocation: E1 and SN1
E1 and SN1 share the exact same rate-determining step — ionization of the leaving group to form a carbocation. The reaction then branches: if a nucleophile attacks carbon → SN1. If a base removes a beta proton → E1. Higher temperatures shift the ratio toward E1.
Regiochemistry
Zaitsev vs Hofmann Product: Which Alkene Forms?
When a substrate has more than one type of beta hydrogen, elimination can produce multiple different alkene products. Which alkene is the major product depends critically on the size of the base.
Zaitsev rule states that the major product is the most substituted alkene. Non-bulky strong bases like NaOH, KOH, NaOEt favor the Zaitsev product. The Hofmann product is the least substituted alkene, favored by bulky bases like KOtBu.
🌿 Zaitsev Product
Alkene formed
Most substituted (most stable)
Base used
Small, strong base: KOH, NaOEt, NaOMe
Why it forms
Small base accesses more hindered beta-H freely; most stable product preferred
Classic example
2-bromobutane + NaOEt → but-2-ene (major)
🔀 Hofmann Product
Alkene formed
Least substituted (kinetically accessible)
Base used
Bulky strong base: KOtBu, LDA
Why it forms
Bulky base cannot reach hindered internal beta-H; removes accessible terminal H
Classic example
2-bromobutane + KOtBu → but-1-ene (major)
Base guide
Which Base Gives Which Product
| Base / Reagent | Reaction Type | Product (2° substrate) | Notes |
|---|
| KOtBu / tBuOH | E2 | Hofmann | Bulky base — definitive Hofmann reagent |
| KOH / EtOH (heat) | E2 | Zaitsev | Classic E2 conditions |
| NaOEt / EtOH | E2 | Zaitsev | Non-bulky alkoxide |
| NaOH / H₂O (heat) | E2 | Zaitsev | Also gives some SN2 |
| NaOH / DMSO | SN2 | Substitution | Polar aprotic → SN2 |
| LDA / THF | E2 | Hofmann | Extremely bulky and strong |
| H₂O / heat (3° only) | E1 | Zaitsev | Weak base — E1 for tertiary |
| EtOH / heat (3° only) | E1 | Zaitsev | Solvolysis — competes with SN1 |
| DBU / CH₂Cl₂ | E2 | Hofmann | Bulky, non-nucleophilic |
Geometry requirement
The Anti-Periplanar Requirement for E2
E2 requires the H and the leaving group to be anti-periplanar: positioned at exactly 180° to each other. This geometry is required because the C–H and C–LG bonds must both overlap with the developing pi bond in the transition state simultaneously.
✓ Anti-Periplanar — E2 Allowed
H — C — C — LG
180° dihedral
H and LG are anti
H and LG on opposite sides — backside overlap enables simultaneous bond breaking and pi bond formation.
✗ Syn or Gauche — E2 Blocked
H
|
C — C — LG
0–120° dihedral
H and LG on the same side — orbital overlap insufficient for concerted E2.
🪑
Cyclohexane E2: Both Must Be Axial
In cyclohexane systems, the anti-periplanar requirement means H and LG must both be in axial positions (trans-diaxial). This is heavily exam-tested: for trans-4-tert-butylcyclohexyl bromide, Br is equatorial and E2 is extremely slow. For the cis isomer, Br is already axial and E2 proceeds readily.
Competition reactions
E2 vs SN2 and E1 vs SN1: When They Compete
Every alkyl halide with a beta hydrogen can undergo both substitution and elimination. Strong, non-bulky nucleophiles in polar aprotic solvents favor SN2. Strong, bulky bases favor E2. Higher temperatures always shift toward elimination.
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Quick Decision Guide
Strong base + primary + polar aprotic → SN2
Strong bulky base (KOtBu) + any substrate → E2
Strong base + secondary + protic → E2 + some SN2
Strong base + tertiary → E2 exclusively
Weak base + tertiary + polar protic + heat → E1 + SN1
For a complete treatment, see the SN1 and SN2 solver and the SN1 vs SN2 guide.
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Related solvers and guides
Frequently asked questions
E1 and E2 Elimination Solver — FAQ
What is the difference between E1 and E2? +
E2 is concerted (one step, Rate = k[substrate][base]). E1 is two steps through a carbocation (Rate = k[substrate]). E2 requires strong base and anti-periplanar geometry. E1 requires only a weak base and has no geometry requirement.
What is Zaitsev rule and when does it apply? +
Zaitsev rule: the major product is the most substituted alkene. Applies with non-bulky strong bases (NaOH, KOH, NaOEt). Does NOT apply with bulky bases (KOtBu, LDA) which give the Hofmann product instead.
What is the anti-periplanar requirement for E2? +
E2 requires H and LG at exactly 180 degrees. In cyclohexane systems, both must be axial (trans-diaxial). If the molecule cannot achieve this geometry, E2 is prevented.
When does E2 compete with SN2? +
E2 and SN2 compete with strong base/nucleophile. Bulky base = E2. Non-bulky nucleophile + polar aprotic = SN2. Higher temperature favors E2. Tertiary substrates: SN2 impossible, strong base gives exclusively E2.
Can E1 give carbocation rearrangements? +
Yes. E1 forms a carbocation intermediate that can rearrange via 1,2-hydride or methyl shift before the base removes the beta proton. E2 cannot rearrange (no carbocation).
What base gives the Hofmann product? +
Bulky bases: KOtBu, LDA, DBU. They cannot reach the hindered internal beta-H, so they remove the accessible terminal H, giving the less-substituted (Hofmann) alkene.