E1 vs E2 Elimination

The setup

Both reactions take an alkyl halide (R-X) — or a tosylate, mesylate, or any substrate with a decent leaving group — and convert it to an alkene by removing a proton from a carbon adjacent to the leaving group (the β-carbon).

     H  X                           H + X⁻
     |  |       elimination          |
   — C—C —     ───────────►         C=C
     |  |                            |
                                   (alkene)

The α-carbon is the one bearing X. The β-carbon is one bond away. Eliminating across the α-β pair forms the new π bond. If multiple β-carbons exist, the regiochemistry of the product depends on which β-H gets removed.

E1: stepwise via a carbocation

Step 1 (slow, rate-determining): the leaving group ionizes off, leaving a carbocation behind. The polar protic solvent (water, ethanol) stabilizes both the cation and the departing X⁻ via hydrogen bonding.

      CH₃                     CH₃
       |                       |
   CH₃-C-Br        ────►   CH₃-C⁺   +   Br⁻
       |                       |
      CH₃                     CH₃        (slow, k₁)

Step 2 (fast): a weak base — usually the solvent itself — removes a β-H. The two electrons in the C-H bond become the second bond of the alkene π system.

      CH₃                                    CH₃
       |                                      \
   CH₃-C⁺           ────►                      C=CH₂   +   H₃O⁺
       |   ↑                                  /
      CH₂— H                                CH₃
            \  (base, e.g. H₂O)                              (fast, k₂)

Because step 1 is rate-limiting and the base concentration doesn't matter for it, the rate law is rate = k[R-X], independent of base. Anything that stabilizes the carbocation (tertiary > secondary >> primary; allylic and benzylic >> alkyl; polar protic solvents) accelerates E1.

E2: one concerted step

The base attacks the β-H, the C-H bond breaks, the C-X bond breaks, and the C=C π bond forms — all in one transition state.

          B⁻                              BH
           ↓
       H—Cβ                              Cβ
           |        ────►                ‖    +   X⁻   +   B-H
        Cα—X                             Cα

(anti-periplanar:                       (alkene + leaving group + protonated base
 H, X dihedral = 180°)                   in one step)

The transition state has partial bonds everywhere — base···H is forming, C-H is breaking, C-X is breaking, C=C is forming. Because two molecules (R-X and base) come together in the rate-limiting step, the rate law is rate = k[R-X][base].

The constraint that defines E2: the breaking C-H bond and the breaking C-X bond must lie in the same plane. The molecule must rotate so that H and X are anti-periplanar (180° dihedral) before the reaction proceeds. Syn-periplanar (0° dihedral) elimination exists but is ~40 kJ mol⁻¹ higher in energy and only matters in rigid cyclic systems where anti is geometrically impossible.

E1 vs E2 vs E1cb side-by-side

	E1	E2	E1cb
Steps	2 (LG leaves, then β-H plucked)	1 (concerted)	2 (β-H plucked, then LG leaves)
Rate law	k[R-X]	k[R-X][base]	k[R-X][base] (slow LG step)
Intermediate	Carbocation	None (one TS)	Carbanion / enolate
Substrate preference	3° > 2° >> 1° (cation stability)	3° > 2° > 1° (steric & thermo)	Acidic β-H needed (β to C=O, NO₂, CN)
Leaving-group requirement	Excellent (I⁻, Br⁻, OTs)	Good (Cl⁻, Br⁻, I⁻, OTs)	Often poor (OH, OR, F)
Base needed	Weak (solvent)	Strong	Strong
Solvent	Polar protic (EtOH, H₂O)	Polar aprotic OK; protic OK	Polar protic, often hot
Geometry constraint	None (free cation)	Anti-periplanar required	Carbanion-LG planarity
Regiochemistry	Zaitsev (thermodynamic)	Zaitsev with small base; Hofmann with bulky base	Hofmann (kinetic, reflects acidity of β-H)
Stereochem of alkene	Mix (cation lost stereoinfo)	Specific E or Z from anti requirement	Often more E (extended TS)
Competing reaction	SN1	SN2	Reverse aldol; nothing if substrate locked

The cleanest test in the lab is to double the base concentration and watch the rate. If it doubles, you have E2. If it doesn't change, you have E1.

Zaitsev vs Hofmann: a worked example

2-bromobutane has two β-carbons: C1 (terminal, three β-H) and C3 (internal, two β-H). E2 elimination can take an H from either, giving either 1-butene or (E)/(Z)-2-butene.

          CH₃     Br
            \    /
            CH—CH       (2-bromobutane)
           /    \
         H₂C     CH₃

Path A: take H from C3   →   CH₃-CH=CH-CH₃     (2-butene, more substituted)
Path B: take H from C1   →   CH₂=CH-CH₂-CH₃    (1-butene, less substituted)

The product distribution depends on the base:

Base	Steric size	2-butene (Zaitsev)	1-butene (Hofmann)	Notes
NaOEt in EtOH	Small	~80 %	~20 %	Reaches internal β-H easily; thermodynamic product wins
NaOH in H₂O/EtOH	Small	~75 %	~25 %	Similar to NaOEt
NaOMe in MeOH	Small	~80 %	~20 %	Similar to NaOEt
KOtBu in tBuOH	Bulky	~30 %	~70 %	Can't squeeze in; takes accessible terminal H
LDA	Very bulky	~10 %	~90 %	Even more selective for Hofmann

The crossover happens at potassium tert-butoxide, which is bulky enough to prefer the unobstructed terminal β-H. This is the canonical Hofmann-vs-Zaitsev choice.

Why E2 fixes alkene geometry

The anti-periplanar requirement makes E2 stereospecific. Take 2-bromo-2,3-diphenylbutane: the two diastereomers (meso and racemic) give different alkenes.

(2R,3S)-meso isomer:  E2 → (E)-2,3-diphenyl-2-butene
(2R,3R) or (2S,3S):   E2 → (Z)-2,3-diphenyl-2-butene

Each diastereomer can only achieve anti-periplanarity in one way; that one way fixes which substituents end up cis on the resulting alkene. Run the same starting materials through E1, and the carbocation intermediate scrambles the stereoinformation, giving a mix of E and Z. This stereospecificity is one of the cleanest tools for distinguishing E1 from E2 experimentally.

Cyclohexane: when E2 is locked

In a cyclohexane ring, anti-periplanar means both H and X must be trans-diaxial. Equatorial groups can never participate in E2 — the geometry is impossible.

cis-1-bromo-2-methylcyclohexane          trans-1-bromo-2-methylcyclohexane

   Br is axial in one chair flip;          Br is axial only when methyl is also
   no axial H on C2 to eliminate            axial. In that chair, an axial H on
   to give the methyl-substituted          C6 sits anti-periplanar to Br.
   alkene; reaction must use C6 H.

   Result: 1-methylcyclohexene (3°)        Result: 3-methylcyclohexene (Hofmann)
           dominates                        only — the Zaitsev product is
                                            geometrically forbidden.

In other words: cyclohexane stereochemistry overrides Zaitsev. The classic exam example is menthyl chloride, which gives almost exclusively the unsubstituted alkene because the more substituted β-H is equatorial and inaccessible.

Isotope labeling distinguishes E1 from E2

Replace β-hydrogens with deuterium. E2 breaks β-C-H in the rate-determining step → k_H/k_D ≈ 6 (primary kinetic isotope effect). E1 breaks C-H only in the fast second step → k_H/k_D ≈ 1.0-1.5. The cleanest mechanistic test on a given substrate.

Variants of elimination

• E1cb — base removes the β-H first to give a carbanion, then the leaving group departs. Important when the β-H is acidic (β to a carbonyl, nitro, sulfone) and the leaving group is poor. Classical aldol-condensation dehydration is E1cb.
• Hofmann elimination — quaternary ammonium salt (R-N⁺R'₃) is heated with hydroxide. The poor amine leaving group and the highly basic OH⁻ favor Hofmann regiochemistry: the least-substituted alkene forms.
• Cope elimination — N-oxide pyrolysis, syn-periplanar by geometric necessity (intramolecular). Useful in steroid synthesis.
• Ester / xanthate pyrolysis — purely syn elimination through a six-membered transition state; makes alkenes without a strong base.

Pitfalls

• Forgetting the SN/E competition. Most E1 substrates also do SN1, most E2 substrates also do SN2. The answer "what does the reaction give" is almost always a mixture; only the dominant product is reported in textbooks.
• Assuming Zaitsev applies to bulky bases. Potassium tert-butoxide flips the regiochemistry to Hofmann. Always check the steric profile of the base.
• Anti-periplanar in rings. Forgetting that cyclohexane geometry forces axial-axial elimination produces wildly wrong predictions. Sketch the chair before you push arrows.
• Carbocation rearrangement in E1. A secondary cation can shift to a tertiary cation by a 1,2-hydride or 1,2-methyl shift before β-H removal — the product alkene often has different connectivity than the substrate suggests. Bridgehead substrates won't eliminate at all (Bredt's rule).