Organic chemistry I · Elimination
E1 vs E2: how elimination reactions make alkenes
Both E1 and E2 remove a hydrogen and a leaving group from neighboring carbons to create a double bond. The difference is timing: E2 happens in one concerted step; E1 passes through a carbocation.
Start with the carbon skeleton
Call the carbon bearing the leaving group the alpha carbon. An adjacent carbon is a beta carbon, and a hydrogen attached there is a beta hydrogen. A normal beta elimination needs a suitable beta hydrogen. Mark every candidate before predicting an alkene; different beta carbons can give different products.
| Feature | E1 | E2 |
|---|---|---|
| Steps | Ionization, then proton removal | One concerted step |
| Typical rate law | Rate = k[substrate] | Rate = k[substrate][base] |
| Base | A weak base can work | Usually a strong base |
| Carbocation shifts | Possible | No carbocation intermediate |
E2: three arrows, one step
A base removes the beta proton while the C–H electrons form a pi bond and the leaving group takes the C–leaving-group bond pair. Draw all three arrows together. The base is involved in the elementary reaction, explaining the usual second-order rate law.
Hydroxide and alkoxides are common strong bases. Secondary and tertiary alkyl halides often undergo E2 with them. Primary substrates often give competing SN2, so a strong, bulky base such as tert-butoxide can help favor elimination by making attack at carbon harder.
Anti-periplanar means a specific arrangement
For the usual E2 pathway, the C–H and C–leaving-group bonds are anti-periplanar: in one plane, pointing opposite ways, with a 180° dihedral angle. Rotate an open-chain single bond to check this. The arrangement lets the orbitals form the new pi bond efficiently.
In a cyclohexane chair, this normally requires a trans-diaxial hydrogen and leaving group. A ring flip may be necessary, and some beta sites lack an eligible axial hydrogen. Geometry can override a preference for the more substituted alkene. Anti elimination is strongly preferred; less favorable syn elimination can occur in special constrained systems.
E1: form a carbocation first
First, the leaving group departs with its bond electrons. Then a base, often solvent, removes a beta proton; that C–H bond pair forms the pi bond. Ordinary E1 is most plausible when the carbocation is stabilized, especially for tertiary or suitable resonance-stabilized substrates. An ordinary primary carbocation is unfavorable.
A carbocation may undergo a 1,2-hydride or alkyl shift when a favorable pathway is available. Check for rearrangement before drawing the final alkene. E2 has no carbocation to rearrange. E1 also shares its intermediate with SN1, so substitution commonly accompanies elimination.
Use conditions and geometry to choose products
Polar protic solvents such as water and alcohols can help ionization by stabilizing ions, supporting E1/SN1 when the substrate is suitable. Polar aprotic solvents often make anionic reagents more reactive, but do not automatically mean E2: SN2 may compete. E2 can also occur in protic solvents. Judge the substrate and base together.
The Zaitsev product has the more substituted double bond and often predominates with a small base or in E1. A strong bulky base can favor the less substituted Hofmann product by accessing a less crowded beta hydrogen. Neither rule is absolute: accessible geometry, conjugation, and substrate structure matter. Higher temperature often favors elimination relative to substitution, but heat alone does not identify the mechanism.
Frequently asked questions
Does E2 always give the E alkene?
No. The available anti conformations and starting stereochemistry control the outcome. Do not turn a common preference for a less crowded E alkene into a universal rule.
Does E1 need an anti hydrogen and leaving group?
No fixed anti relationship to the original leaving group is required because it has already left. The C–H bond still needs suitable overlap with the carbocation's empty p orbital during alkene formation.
Can I identify E1 or E2 from the product alone?
Usually not. Both can make the same alkene. Use the reagents, substrate, rate information if supplied, and stereochemistry; a rearranged product is evidence for an intermediate that can rearrange.
Textbook checks and further reading
OpenStax Organic Chemistry: E2 electron flow and geometry, the E1 mechanism, and reactivity trends.