Organic Chemistry I · Lesson 09
Elimination
Elimination is how you build alkenes. You pull an H and a leaving group off neighboring carbons and a double bond snaps into place. The whole art is reading the conditions to know which of the three timings you get, and whether elimination or substitution wins.
Learning Goals
- Explain why elimination reactions are a main route to alkenes.
- Identify the α-carbon, β-carbons, and β-hydrogens in a substrate.
- Describe the three core events of every elimination.
- Draw the E1, E2, and E1cb mechanisms and tell them apart by timing.
- Predict Zaitsev vs Hofmann products from the base.
- Apply anti-periplanar geometry, including trans-diaxial on cyclohexanes.
- Use the master table to choose SN1, SN2, E1, or E2 from substrate and conditions.
Key Terms
Tap any term to see its definition.
Section 1
Why Elimination Matters
Here is the reason to care: elimination is one of the main ways we prepare alkenes. Two reactions cover most of what you will see. Dehydration of an alcohol loses water, turning an alcohol into an alkene. Dehydrohalogenation of an alkyl halide loses HX, turning an alkyl halide into an alkene.
Dehydration: alcohol → alkene + H₂O
Dehydrohalogenation: alkyl halide → alkene + HX
Section 2
What Elimination Means
Elimination means a small molecule is lost during the reaction. In almost every intro case, it is a 1,2-elimination, also called β-elimination, which means the two atoms removed come from adjacent carbons.
The vocabulary here matters, so let me pin it down. The α-carbon is the carbon attached to the leaving group. A β-carbon is a carbon right next to the α-carbon. And a β-hydrogen is a hydrogen on that β-carbon. One group leaves the α-carbon and one hydrogen leaves a β-carbon.
α-carbon holds the LG, β-carbons sit next door
The one-liner for the whole lesson
β-elimination forms an alkene by removing H and a leaving group from neighboring carbons. Hold onto that and every mechanism below is just a variation on the timing.
Section 3
The Three Core Events
Every elimination, no matter the mechanism, is built from the same three events:
- 01Removal of a proton: a base takes the β-H.
- 02Formation of the C=C π bond: those electrons become the new double bond.
- 03Breaking the bond to the leaving group: the LG departs with its electrons.
Proton removal, π bond formation, leaving group loss
The key idea for the next section
The mechanism is decided entirely by when these three events happen. That timing is the difference between E1, E2, and E1cb, so keep these three moves in mind.
Section 4
The Three Elimination Mechanisms
E1: unimolecular elimination
Two steps. First the leaving group leaves and a carbocation forms, then a base removes a β-H and the C=C forms. The rate law is rate = k[substrate], because the slow step only involves the substrate losing its leaving group.
E1: LG leaves first, carbocation forms, then H is removed
E1 is favored by
- ›Stable carbocations (usually tertiary substrates)
- ›Weak bases
- ›Polar protic solvents
- ›Heat
In one phrase: E1 = leaving group leaves first, then the H is removed.
E2: bimolecular elimination
One concerted step. The base removes the β-H, the C=C forms, and the leaving group leaves, all at the same time. The rate law is rate = k[substrate][base], so the base is in the slow step, which is why E2 needs a strong base.
E2: everything happens in one concerted step
E2 is favored by
- ›Strong base
- ›Good leaving group
- ›An available β-H
- ›Anti-periplanar geometry
- ›Secondary or tertiary substrates (or primary with a bulky base)
In one phrase: E2 = base removes H while the leaving group leaves.
E1cb: elimination through the conjugate base
Two steps, but flipped from E1. The base removes the proton first to give a carbanion, then the C=C forms as the leaving group leaves. It is less common in intro courses, but it completes the set of timing possibilities.
E1cb: H leaves first, forming a carbanion
In one phrase: E1cb = H leaves first, then the leaving group leaves.
The timing that separates all three
This is the cleanest way I know to keep them straight: the only thing that changes is which event comes first.
| Mechanism | First event |
|---|---|
| E1 | Leaving group leaves first |
| E2 | Everything happens together |
| E1cb | Proton is removed first |
Section 5
β-Hydrogens
Elimination needs a removable hydrogen on a carbon next to the leaving group. This is the check I never skip, because it can kill the reaction before it starts.
The gate you must pass first
No β-H means no normal elimination. Before I think about mechanism or product, I find the α-carbon, mark every β-carbon, and count the β-hydrogens. If there are none, elimination is off the table.
Mark the α-carbon, the β-carbons, and every β-H
Once you have all the β-hydrogens marked, each different β-H can lead to a different alkene product. Practice mapping every β-H to the alkene it would give, because that is the setup for the next section on which product wins.
Section 6
Regioselectivity: Zaitsev vs Hofmann
When more than one β-H is available, elimination can form more than one constitutional alkene. Which one dominates is the regioselectivity question. The Zaitsev product is the more substituted alkene, and it is usually the major product because more substituted alkenes are usually more stable.
Zaitsev = more substituted, Hofmann = less substituted
But it flips with a bulky base. The Hofmann (anti-Zaitsev) product is the less substituted alkene, and you get it when the base is too bulky to reach the crowded interior β-H, so it grabs the most accessible one instead. tert-Butoxide is the classic example.
A bulky base reaches the easiest β-H, giving Hofmann
The caveat I want you to carry
Zaitsev is a useful rule, not an absolute one. A small strong base usually gives Zaitsev, but a bulky strong base often gives Hofmann. Always check the size of the base before you commit to a product.
Section 7
Stereoselectivity: Trans vs Cis
Even after you know which constitutional alkene forms, there can still be a choice between the cis and trans versions. Elimination usually favors the more stable stereoisomer, and that is normally the trans alkene, because it puts the bulky groups on opposite sides with less steric crowding.
Trans usually beats cis
Quick rule
When both cis and trans alkenes are possible, the trans product is usually favored. Less crowding, more stability.
Section 8
E2 Stereochemistry
E2 is picky about geometry. The β-H and the leaving group usually need to be anti-periplanar, meaning opposite each other in the same plane. That alignment lets the orbitals overlap properly to form the new π bond.
Anti-periplanar: β-H and LG 180° apart
On cyclohexanes this rule becomes very concrete. E2 requires trans-diaxial geometry: the leaving group must be axial, the β-H must also be axial, and the two must be anti to each other. Because only axial positions work, this can decide which alkene actually forms.
Cyclohexane E2 needs the LG and β-H trans-diaxial
Where Lesson 5 pays off
This is your chair and ring-flip skills from Alkanes and Conformations doing real work. If the leaving group is stuck equatorial, the ring has to flip it axial before E2 can happen, and sometimes that changes which product you can even make.
Section 9
E1 vs E2 vs E1cb
Here is the three-way comparison in one place.
| Feature | E1 | E2 | E1cb |
|---|---|---|---|
| Steps | Two | One | Two |
| First event | LG leaves | Simultaneous | H removed |
| Intermediate | Carbocation | None | Carbanion |
| Base strength | Weak base okay | Strong base common | Strong base usually |
| Rearrangements | Possible | No | No |
| Geometry | Less strict | Anti-periplanar | Depends |
| Common products | Zaitsev often | Zaitsev or Hofmann | Stabilized alkenes |
Section 10
Substitution vs Elimination: The Master Table
This is the payoff of the last two lessons, and honestly the single most useful thing on this page. Substitution and elimination are always competing, and the winner comes down to two questions: is the reagent acting as a base, a nucleophile, both, or neither, and what is the substrate (1°, 2°, or 3°). Read the reagent character across the top, the substrate down the side, and the cell tells you the pathway. Read it until you can rebuild it from memory.
Strong base Weak nucleophile | Strong base Strong nucleophile | Weak base Strong nucleophile | Weak base Weak nucleophile | |
|---|---|---|---|---|
| 1° | E2 | SN2 > E2 | SN2 | ✕ |
| 2° | E2 | E2 > SN2 | SN2 | ✕ |
| 3° | E2 | E2 | SN1 | SN1 / E1 |
Primary: a strong base gives E2, a strong nucleophile gives SN2, and when the reagent is both, SN2 edges out E2. A weak base and weak nucleophile is too slow to react usefully.
Secondary: E2 dominates whenever a strong base is present, and E2 stays ahead of SN2 when the reagent is both a strong base and a strong nucleophile. A weak/weak reagent gives no practical reaction, though a secondary alcohol will do E1 with sulfuric acid and heat.
Tertiary: SN2 is impossible, so any strong base gives E2. A weak base with a strong nucleophile gives SN1, and a weak base with a weak nucleophile gives SN1 and E1, depending if the reaction is performed under heat or not (ex. a tertiary alcohol undergoes E1 with sulfuric acid and heat).
A few things I want you to notice as you read it. A strong base always eliminates, E2 on every substrate class, because E2 is not sensitive to steric hindrance. When the reagent is both a strong base and a strong nucleophile, E2 edges out SN2 on primary and secondary, and it is the only pathway left on tertiary. A weak base with a strong nucleophile substitutes, giving SN2 on primary and secondary but SN1 on tertiary. And the weak base, weak nucleophile column is mostly dead: primary and secondary have no practical reaction, and only a tertiary substrate reacts, giving the unimolecular pair SN1 and E1.
About that empty 2° box
You might expect the secondary weak/weak cell to read SN1 / E1, but a plain secondary substrate barely reacts under those conditions, both the bimolecular and unimolecular routes are too slow, which is why it is marked as no practical reaction. The one real exception: a secondary alcohol will undergo E1 with sulfuric acid and heat.
If you want the quick verbal version of the same logic, here it is as a lookup:
| Condition | Favored pathway |
|---|---|
| Strong nucleophile, weak base | SN2 |
| Strong base | E2 |
| Weak nucleophile / base + stable carbocation | SN1 / E1 |
| Heat | Favors elimination |
| Bulky strong base | E2, often Hofmann |
| Tertiary substrate + strong base | E2 |
| Tertiary substrate + weak nucleophile / base | SN1 / E1 |
The rule that overrides mood swings
Heat favors elimination over substitution. When a problem mentions heat or reflux, let that tip a close call toward E1 or E2.
Section 11
The Decision System
When you meet an elimination problem, run these questions in order. This is the exact checklist I walk every time.
Is there a good leaving group?
If not, elimination is unlikely to go at all.
Are there β-hydrogens?
No β-H means no normal elimination, no matter what else is true.
Is the base strong?
A strong base points toward E2.
Can a stable carbocation form?
If yes, E1 becomes possible (think tertiary, weak base, protic solvent, heat).
Is the base bulky?
A bulky base makes the Hofmann (less substituted) product more likely.
Which alkene is most stable?
The more substituted alkene is usually favored, unless a bulky base overrides it.
Is E2 geometry possible?
You need an anti-periplanar β-H and leaving group, or trans-diaxial on a ring.
The cleanest summary I can give you
Elimination = remove H and LG from adjacent carbons to make an alkene. E1 = LG leaves first. E2 = everything happens together. E1cb = H leaves first. Get that, plus the master table, and you own this topic.
Common Mistakes
Watch Out for These
Practice Set
Try each question before opening the answer.
1. Name the two classic elimination reactions and what small molecule each loses.
Answer: Dehydration of an alcohol loses H₂O to give an alkene. Dehydrohalogenation of an alkyl halide loses HX to give an alkene.
2. In a β-elimination, where do the two removed groups come from?
Answer: Adjacent carbons. The leaving group leaves the α-carbon and a hydrogen leaves the β-carbon next to it, forming the C=C between them.
3. What are the three core events of every elimination?
Answer: Removal of the β-proton, formation of the C=C π bond, and breaking of the bond to the leaving group. The mechanism is defined by when those three happen.
4. How do E1, E2, and E1cb differ in timing?
Answer: E1: the leaving group leaves first (carbocation). E2: all three events happen at once (no intermediate). E1cb: the proton is removed first (carbanion).
5. What is the rate law for E2, and what does it tell you?
Answer: rate = k[substrate][base]. The base appears in the rate law, so it is involved in the rate-determining step. That is why E2 needs a strong base.
6. What is the Zaitsev product, and when might you get the Hofmann product instead?
Answer: The Zaitsev product is the more substituted, usually more stable alkene. You get the Hofmann (less substituted) product when the base is bulky, like tert-butoxide, because it grabs the most accessible β-H.
7. What geometry does E2 require, and what does that mean on a cyclohexane?
Answer: The β-H and leaving group must be anti-periplanar (opposite each other in the same plane). On a cyclohexane that means trans-diaxial: both the leaving group and the β-H have to be axial and anti.
8. A secondary substrate reacts with a strong bulky base. Substitution or elimination, and which mechanism?
Answer: Elimination by E2, and likely the Hofmann product. A strong base favors E2, and a bulky base steers toward the less substituted alkene.
9. Why does heat favor elimination?
Answer: Elimination increases the number of molecules (it releases a small molecule), which is entropically favored, and that entropy term becomes more important at higher temperature.
10. A tertiary substrate sits in ethanol with no strong base or nucleophile added. What is likely?
Answer: SN1 and E1 competing. A tertiary substrate forms a stable carbocation, the polar protic solvent supports ionization, and with only a weak nucleophile/base present the unimolecular pathways dominate.
Ready for a bigger set?
Work through the full Elimination practice page: identifying β-hydrogens, drawing E1 and E2 mechanisms, predicting Zaitsev vs Hofmann, applying anti-periplanar geometry, and using the master table to call SN1, SN2, E1, or E2.
Do Elimination PracticeLesson Summary
Elimination removes an H and a leaving group from adjacent carbons to build an alkene. Every elimination is proton removal, π bond formation, and leaving group loss, and the timing sets the mechanism: E1 loses the leaving group first (carbocation), E2 does everything at once (concerted, anti-periplanar), and E1cb removes the proton first (carbanion). You always need a β-H. Zaitsev (more substituted) usually wins, but a bulky base gives Hofmann, and trans usually beats cis. Substitution and elimination compete, and the master table settles it from substrate and conditions, with heat tipping toward elimination.