Like SN1 vs SN2, elimination is either stepwise through a carbocation (E1) or concerted (E2) — here 2-bromopropane loses HBr to propene.
Base takes a β-H, C–Br breaks, a C=C forms.
1. Both E1 and E2 Form an Alkene by Losing a β-Hydrogen and a Leaving Group
Both lose a leaving group from the α-carbon and an H from a β-carbon to form C=C; only the timing differs.
2. E2 Is One Concerted Step; E1 Goes Through a Carbocation in Two Steps
E2 pulls the β-H as C–LG breaks (one step); E1 loses the leaving group first, then a base takes a β-H.
3. Their Rate Laws Differ: E2 Is Second-Order, E1 Is First-Order
E2 is rate = k[substrate][base]; E1 is rate = k[substrate] — so adding base speeds E2 but not E1.
4. E2 Demands an Anti-Periplanar β-Hydrogen and Leaving Group
E2 needs the β-H and leaving group 180° apart (both axial on a ring), so it is stereospecific; E1's planar cation is not.
5. Strong or Bulky Bases and Heat Favor E2; Weak Bases and Polar Protic Solvents Favor E1
E2 wins with a strong base and heat; E1 wins with a weak base, polar protic solvent, and a tertiary substrate.
Strong base on a tertiary halide → E2 to isobutylene.
6. Regiochemistry: Zaitsev Usually Wins, but a Bulky Base Gives the Hofmann Alkene
Usually the more-substituted Zaitsev alkene wins, but a bulky base (KOtBu) can't reach the internal H and gives the less-substituted Hofmann alkene.
7. E1 Can Rearrange Because It Has a Carbocation; E2 Cannot
An E1 cation can undergo a 1,2-hydride or alkyl shift to a more stable cation, so a scrambled skeleton signals E1; concerted E2 never rearranges.
8. Summary
E2: concerted · strong base · anti-periplanar · second-order · stereospecific · no rearrangement | E1: stepwise cation · weak base + polar protic · first-order · tertiary · can rearrange (with SN1) | both Zaitsev except bulky base → Hofmann.
Worked example
- Ethoxide is a strong base and a poor bulky-ish nucleophile; with a 2° substrate this favours E2 (concerted, second-order).
- E2 needs a β-hydrogen antiperiplanar to the C–Br bond; the base removes it as bromide leaves — one step.
- Two β-carbons are available. Zaitsev's rule: the more-substituted (more stable) alkene dominates, so eliminate toward the internal carbon.
Answer. (E)-but-2-ene as the major product, by an E2 mechanism (Zaitsev, more-substituted alkene).
Quiz yourself
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E2. Its rate law is rate = k[substrate][base], so the base concentration appears in the rate. E1's slow step is just ionization of the substrate (rate = k[substrate]), so adding more base would not change its rate.
E2 needs the β-hydrogen and the leaving group to be anti-periplanar — on a ring, both must be axial. If the leaving group is locked equatorial (no adjacent axial H), there is no anti-periplanar geometry available and the concerted E2 cannot proceed.
1-Butene, the less-substituted alkene — the Hofmann product. The bulky tert-butoxide can't reach the crowded internal β-hydrogen, so it removes a terminal one. A small base would instead give the Zaitsev product, 2-butene.
A carbocation was involved, so it was E1 (with SN1 competing). Only the stepwise pathway forms a cation that can undergo a 1,2-hydride or alkyl shift before losing a proton. Concerted E2 has no intermediate and never rearranges.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.