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What Are E1 and E2 Elimination Reactions?

Elimination reactions remove a hydrogen and a leaving group from adjacent carbons, forming a double bond. E1 (unimolecular) proceeds through a carbocation, while E2 (bimolecular) is a concerted, one-step process. Both compete with substitution reactions under certain conditions.

Short answer

E1 is a two-step mechanism via a carbocation intermediate (first-order kinetics, favored by polar protic solvents and weak bases). E2 is a one-step, concerted mechanism (second-order kinetics, favored by strong bases and polar aprotic solvents). Zaitsev's rule predicts the major alkene product.

E2 Mechanism: One-Step Elimination

1
Base attacks H
Strong base removes β-hydrogen in backside orientation.
↓
2
C-H and C-X bonds break
Simultaneous bond breaking and π-bond formation.
↓
3
Alkene forms
New C=C double bond replaces C-H and C-X bonds.

Step-by-step worked examples

Predict the major alkene from 2-bromo-2-methylpropane (tert-butyl bromide) with KOH in ethanol.

Substrate: tertiary alkyl halide (stable carbocation)
Base: KOH (moderate strength)
Solvent: ethanol (polar protic)
E1 dominates. Carbocation forms, H removed from most substituted position (Zaitsev's rule).
Product: 2-methylpropene (isobutylene), (CH₃)₂C=CH₂

What is the major product when 1-bromo-3-methylbutane reacts with tert-BuOK in DMSO?

Substrate: primary alkyl halide
Base: tert-BuOK (strong, bulky)
Solvent: DMSO (polar aprotic)
E2 dominates (strong base, aprotic solvent).
Zaitsev's rule: H removed from more substituted β-carbon.
Product: 3-methylbut-1-ene (major) and 3-methylbut-2-ene (minor)

With 2-bromobutane and KOH in EtOH, which elimination product is favored?

Substrate: secondary alkyl halide (competing mechanisms)
Conditions favor E1 (polar protic, weak base)
Zaitsev's rule: H from most substituted β-carbon
Product: but-2-ene (CH₃CH=CHCH₃) major, but-1-ene minor

Flashcards

Quick quiz

Q1.Which conditions favor E2 elimination?

Correct answer: B. E2 requires a strong base (fast H removal) and aprotic solvent (no carbocation stabilization).

Q2.Zaitsev's rule predicts…

Correct answer: B. Zaitsev = more stable, more substituted C=C is major.

Q3.E1 kinetics are…

Correct answer: C. E1 rate = k[RX] (first-order) — carbocation formation is rate-limiting, independent of base.

Q4.Which substrate favors E1?

Correct answer: C. 3° alkyl halides form the most stable carbocation → E1 preferred.

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Common mistakes

✗

E2 and E1 have the same rate law. — Correct: E1 is first-order (rate = k[RX]), E2 is second-order (rate = k[RX][base]).

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Hofmann's rule is the same as Zaitsev's rule. — Correct: Zaitsev = major is more substituted alkene. Hofmann = major is less substituted (with bulky bases).

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E1 always gives only one alkene product. — Correct: E1 can give multiple products if there are β-hydrogens on different carbons (Zaitsev dominates).

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Strong bases favor carbocation formation. — Correct: Strong bases prevent carbocation formation by quickly removing H (E2); they favor elimination over substitution.

FAQ

What is an elimination reaction?

Removal of a hydrogen and leaving group from adjacent carbons to form a C=C double bond.

Why is E2 called 'concerted'?

Because H removal and C-X bond breaking happen simultaneously in one step, not via an intermediate.

What is Hofmann's rule?

With very bulky bases, the less substituted alkene can become major (steric hindrance blocks Zaitsev path).

Can substitution and elimination happen together?

Yes — SN1/E1 compete via the same carbocation intermediate; SN2/E2 both require strong Nu/base but follow different paths.