Difficulty: Intermediate | Prerequisites: SN1/SN2 mechanisms, carbocation stability, acid-base chemistry.
Elimination reactions are the other major pathway available to alkyl halides (alongside substitution). Instead of a nucleophile replacing the leaving group, a base removes a proton from a carbon adjacent to the leaving group. The leaving group departs, and a double bond (alkene) forms.
Elimination and substitution compete with each other constantly. Whether you get substitution or elimination depends on the substrate, the reagent (nucleophile vs base), and the conditions. This is why these topics are taught together and why exam questions so often ask you to predict which pathway wins.
You should be comfortable with SN1/SN2 before tackling E1/E2, because the decision framework at the end of these notes ties all four together.
Elimination reactions remove a proton and a leaving group from adjacent carbons to form an alkene. E1 goes through a carbocation (two steps, favoured by weak bases and 3° substrates). E2 is concerted (one step, favoured by strong bases and 2°/3° substrates). Choosing between SN1, SN2, E1, and E2 is the core skill tested on exams.
Elimination reaction
A reaction in which a leaving group and a proton on an adjacent carbon are both removed, forming a new pi bond (double bond). In simple terms, instead of swapping groups, the molecule loses two groups and gains a double bond.
E1 (elimination, unimolecular)
A two-step elimination where the leaving group departs first (forming a carbocation), and then a base removes a proton from an adjacent carbon to form the alkene. The rate depends only on the substrate.
E2 (elimination, bimolecular)
A one-step (concerted) elimination where the base removes a proton at the same time as the leaving group departs. The rate depends on both the substrate and the base.
Beta carbon (β-carbon)
The carbon adjacent to the one bearing the leaving group. The proton removed in elimination comes from this carbon.
Zaitsev's rule (Saytzeff's rule)
The more substituted alkene is the major product in most elimination reactions. In simple terms, the double bond forms preferentially toward the side with more alkyl groups.
Anti-periplanar geometry
The spatial arrangement required for E2: the proton being removed and the leaving group must be on opposite sides of the molecule (180° dihedral angle). Think of it as the H and the leaving group pointing in exactly opposite directions.
Base
A species that accepts a proton. In elimination, the base pulls a proton off the β-carbon. Strong bases (e.g. HO⁻, RO⁻, NaH) favour E2; weak bases favour E1.
Substrate
2° and 3° alkyl halides
Mirrors SN1 in substrate preference (both need a stable carbocation)
Base/nucleophile
Weak bases (e.g. H₂O, alcohols)
E1 often competes with SN1 under the same conditions; the product mixture typically contains both substitution and elimination products
Solvent
Polar protic solvents (same as SN1)
Mechanism (two steps)
The leaving group departs, forming a carbocation intermediate. This is the slow, rate-determining step (identical to the first step of SN1).
A base removes a proton from the β-carbon. Electrons from the C-H bond form the new C=C double bond.
Product prediction
Zaitsev's rule applies: the more substituted alkene is the major product
Carbocation rearrangements (hydride shifts, methyl shifts) can occur before the proton is removed, leading to unexpected products
Rate law
Rate = k[substrate]
Unimolecular, same as SN1
Key relationship to SN1
E1 and SN1 share the same first step (carbocation formation). Once the carbocation forms, it is a competition: if a nucleophile attacks the carbocation, you get SN1; if a base removes an adjacent proton, you get E1. Higher temperatures generally favour elimination.
Substrate
2° and 3° alkyl halides
1° substrates can undergo E2 with a strong, bulky base (e.g. tert-butoxide)
Base
Strong bases required (e.g. NaOH, NaOEt, KOtBu)
Bulky strong bases (e.g. KOtBu) especially favour E2 over SN2 because they are too large to act as nucleophiles
Solvent
Not as critical as for substitution, but polar aprotic solvents are common
Mechanism (one step, concerted)
The base removes a proton from the β-carbon at the same time as the leaving group departs. The C-H bond breaks, the C=C bond forms, and the C-LG bond breaks, all in a single step.
The H and the leaving group must be anti-periplanar (180° dihedral) for the orbitals to overlap properly.
Product prediction
Zaitsev's rule typically applies: the more substituted alkene is favoured
Exception: bulky bases (e.g. KOtBu) favour the less substituted (Hofmann) product due to steric effects
E/Z geometry matters for the product and is determined by which groups end up on the same or opposite sides of the double bond
Rate law
Rate = k[substrate][base]
Bimolecular
Key relationship to SN2
E2 and SN2 compete when you have a strong nucleophile/base with a 2° substrate. If the reagent attacks carbon, you get SN2. If it attacks a β-hydrogen, you get E2. Bulky bases, higher temperatures, and more substituted substrates all push toward E2.
Feature | E1 | E2 |
|---|---|---|
Steps | Two (carbocation intermediate) | One (concerted) |
Substrate | 2° and 3° | 2° and 3° (1° with bulky base) |
Base | Weak | Strong |
Solvent | Polar protic | Less critical (often polar aprotic) |
Rate law | Rate = k[substrate] | Rate = k[substrate][base] |
Carbocation | Yes (can rearrange) | No |
Geometry requirement | None | Anti-periplanar H and LG |
Product rule | Zaitsev (more substituted) | Zaitsev (Hofmann with bulky base) |
Competes with | SN1 | SN2 |
When you are given a reaction and asked to predict the mechanism, work through these four factors in order of priority.
1. Structure of the electrophile (substrate)
This is the most important factor.
Methyl or 1°: SN2 (or E2 if a strong, bulky base is used)
2°: ambiguous, depends on the other factors below
3°: SN1 or E1 (weak nucleophile/base) or E2 (strong base). SN2 is essentially impossible at 3°.
2. Identity of the leaving group
Better leaving groups make the reaction faster, regardless of mechanism.
Order: I⁻ > Br⁻ > Cl⁻ ≈ H₂O > F⁻ > CH₃CO⁻ > HO⁻
HO⁻ and F⁻ are poor leaving groups; reactions with these are slow or do not proceed without activation
3. Identity of the nucleophile/base
Strong nucleophile + not bulky: favours SN2
Strong base + bulky: favours E2
Weak nucleophile/base: favours SN1/E1 (which one dominates depends on temperature; heat favours elimination)
4. Reaction conditions (solvent)
Polar protic (H₂O, MeOH, EtOH): favours SN1/E1
Polar aprotic (DMSO, DMF, acetone): favours SN2/E2
Quick decision summary
Substrate | Reagent | Solvent | Likely mechanism |
|---|---|---|---|
Methyl / 1° | Strong nucleophile | Polar aprotic | SN2 |
Methyl / 1° | Strong bulky base | Any | E2 |
2° | Strong nucleophile | Polar aprotic | SN2 |
2° | Strong bulky base | Any | E2 |
2° | Weak nucleophile | Polar protic | SN1/E1 mixture |
3° | Weak nucleophile | Polar protic | SN1 (+ some E1) |
3° | Strong base | Any | E2 |
Worked example from the source notes
An alkoxide nucleophile (strong nucleophile) reacts with a substrate in DMSO (polar aprotic solvent). The leaving group (a halogen) is displaced directly, with no carbocation step. This is SN2. The product shows inversion of stereochemistry.
Students often forget that E1 and SN1 compete under the same conditions. When you predict SN1, you should also expect some E1 product. Exams sometimes ask for both.
Students often think E2 requires a 3° substrate. It does not. E2 can occur at 1° substrates if the base is strong and bulky (e.g. KOtBu).
Students sometimes apply Zaitsev's rule blindly. With bulky bases, the less substituted (Hofmann) product is actually favoured because the base cannot access the more hindered β-hydrogen.
Students frequently overlook the anti-periplanar requirement for E2. If the H and the leaving group cannot achieve a 180° dihedral angle (common in rigid ring systems), E2 may not occur even when the other conditions favour it.
⚠️ The four-way decision (SN1 vs SN2 vs E1 vs E2) is the single most commonly tested skill in this part of the course.
⚠️ Drawing the correct elimination product requires applying Zaitsev's rule (or Hofmann with bulky bases) and assigning E/Z geometry.
⚠️ Questions about 2° substrates are intentionally ambiguous. Be ready to justify your mechanism choice with reasoning about the nucleophile/base, solvent, and temperature.
⚠️ Carbocation rearrangements in E1 (and SN1) are a favourite trick on exams. If a more stable carbocation is accessible via a hydride or methyl shift, assume it will form.
True or false: E1 reactions form a carbocation intermediate.
Fill in the blank: E2 requires the H and the leaving group to be in a(n) ______ arrangement.
True or false: Zaitsev's rule always predicts the major product in E2.
Fill in the blank: E1 competes most directly with ______ because they share the same first step.
True or false: A strong, bulky base with a 3° substrate will most likely give E2.
Answers: 1. True. 2. Anti-periplanar. 3. False (bulky bases give Hofmann product). 4. SN1. 5. True.
Q: A 3° alkyl bromide is treated with NaOH (a strong base) in ethanol. What mechanism is most likely, and what product do you expect?
A: E2. The substrate is 3° (SN2 is impossible due to steric hindrance), and the base is strong, which rules out E1. The major product is the more substituted alkene (Zaitsev's rule).
Q: A 3° alkyl bromide is dissolved in methanol with no added base. What mechanism(s) are likely?
A: SN1 and E1. Methanol is a weak nucleophile/base and a polar protic solvent. Both pathways go through the same carbocation intermediate. You would expect a mixture of the substitution product (an ether) and the elimination product (an alkene).
Q: Why does a bulky base like potassium tert-butoxide favour E2 over SN2?
A: The tert-butoxide is too sterically large to attack the electrophilic carbon directly (which would give SN2). Instead, it can only reach the more exposed β-hydrogens, so it acts as a base and gives elimination.
Q: A 1° alkyl bromide is treated with KOtBu. Predict the mechanism and the product.
A: E2. Even though the substrate is primary (which normally favours SN2), the bulky base cannot perform backside attack. The product is the terminal alkene (Hofmann product), because the bulky base preferentially removes the less hindered proton.
Q: List the four factors, in order of priority, for determining which mechanism will occur.
A: 1. Structure of the electrophile (methyl, 1°, 2°, 3°). 2. Identity of the leaving group. 3. Identity of the nucleophile/base (strong vs weak, bulky vs compact). 4. Reaction conditions (solvent: polar protic vs polar aprotic).
Elimination is the reverse of electrophilic addition to alkenes. Understanding both directions gives you a complete picture of alkene chemistry.
The competition between substitution and elimination recurs in alcohol chemistry, where you convert alcohols to alkyl halides or alkenes by choosing reagents that favour one pathway over the other.
Carbocation rearrangements in E1 (and SN1) connect to the broader topic of carbocation chemistry, which also appears in electrophilic aromatic substitution and in the biosynthesis of terpenes.
Elimination reaction, E1, E2, alkene formation, dehydrohalogenation, Zaitsev's rule, Saytzeff's rule, Hofmann product, anti-periplanar, concerted elimination, carbocation rearrangement, hydride shift, methyl shift, SN1 vs SN2 vs E1 vs E2, reaction decision framework, strong base, weak base, bulky base, KOtBu, potassium tert-butoxide, polar protic, polar aprotic, leaving group ability, organic chemistry elimination, beta elimination, rate law bimolecular, rate law unimolecular.