Source: Organic Chemistry, The Ohio State University (Baldwin)
Tags: E1, E2, elimination, Zaitsev's rule, Hofmann product, alkene stability, carbocation rearrangement, hydride shift, antiperiplanar, LDA, base, concerted mechanism
Difficulty: Intermediate | Prerequisites: SN1 mechanism (Part 1 of these notes), SN2 mechanism (Chapter 6), acid-base chemistry
Big picture: Elimination reactions compete with substitution reactions. Whenever you have a substrate with a leaving group, there are (in principle) four possible pathways: SN1, SN2, E1, and E2. E1 shares its first step with SN1 (carbocation formation), so the two always compete under the same conditions, with heat pushing the balance towards E1. E2 is a completely different, concerted mechanism that competes with SN2 when a strong base is present. Knowing both elimination mechanisms, and when each dominates, is essential for predicting the products of any reaction in this chapter.
E1 is a two-step elimination that forms the same carbocation as SN1, then loses a proton to give an alkene. It competes with SN1 and is favoured by heat. E2 is a one-step, concerted elimination requiring a strong base and an antiperiplanar arrangement of the H and the leaving group. E2 competes with SN2. Both eliminations follow Zaitsev's rule (most substituted alkene is the major product), unless a bulky base is used, which gives the Hofmann (least substituted) product.
E1 (elimination, unimolecular)
A two-step elimination mechanism. The leaving group departs first to form a carbocation (same as SN1 step 1), then a base removes a proton from a carbon adjacent to the cation to form the alkene. In simple terms, the molecule falls apart, then a base plucks off a hydrogen.
E2 (elimination, bimolecular)
A one-step, concerted elimination. The base removes a proton at the same time as the leaving group departs and the double bond forms. Everything happens in a single transition state. Think of it as a coordinated departure: the base pulls, the leaving group leaves, and the pi bond snaps into place simultaneously.
Zaitsev's rule (Saytzev's rule)
The more substituted alkene is the major product of elimination. In simple terms, the double bond forms preferentially towards the side with more alkyl groups, because that alkene is more stable.
Hofmann product
The less substituted alkene, formed preferentially when a bulky base is used. The base is too large to reach the more hindered hydrogen, so it takes the more accessible one instead.
Antiperiplanar
A geometric relationship where two bonds are on opposite sides of a carbon-carbon bond and in the same plane (180° dihedral angle). E2 requires the C–H bond and the C–LG bond to be antiperiplanar. Think of it as "anti and flat," like two people standing back-to-back.
Alkene substitution classification
Alkenes are classified by how many carbon groups are attached to the double-bond carbons: monosubstituted (1), disubstituted (2), trisubstituted (3), tetrasubstituted (4). More substituted = more stable.
Carbocation rearrangement
A structural change in a carbocation intermediate, either by a hydride shift (H moves) or an alkyl shift (carbon group moves), to produce a more stable cation. This happens only when a more stable carbocation can be reached, and it is not reversible under reaction conditions.
Hydride shift
Migration of a hydrogen atom (with its bonding electrons) from an adjacent carbon to the positively charged carbon. A 1,2-hydride shift is the most common type.
LDA (lithium diisopropylamide)
A very strong, very bulky, non-nucleophilic base. Because of its size, LDA gives the Hofmann (least substituted) product in E2 reactions. It is too bulky for SN2.
E1 shares its first step with SN1. The difference is what happens to the carbocation:
Step 1 (RDS): The leaving group departs to form a carbocation. Identical to SN1.
Step 2: A base (often the solvent) removes a proton from a carbon next to (beta to) the positively charged carbon. The electrons from the broken C–H bond form the new C=C double bond.
The rate law is Rate = k[substrate], same as SN1 (unimolecular).
Because E1 and SN1 share the same first step, they always compete. The product mixture typically contains both the substitution product and the elimination product. Heat (Δ) favours E1 over SN1, because elimination produces more molecules (higher entropy), which is favoured at higher temperatures.
Any hydrogen on a carbon adjacent to the carbocation (a beta hydrogen) can, in principle, be removed. But:
The major product follows Zaitsev's rule: the base removes the hydrogen that gives the more substituted (more stable) alkene
If two different beta hydrogens are available, the more substituted alkene is major and the less substituted is minor
If both sides are equivalent, only one product forms
E1 goes through a carbocation, so there is no stereochemical requirement for the H and LG. Any beta hydrogen can be removed regardless of its spatial relationship to the leaving group.
The reaction uses reversible arrows (equilibrium arrows) for the carbocation formation step
Like SN1, E1 prefers polar protic solvents
E1 occurs above room temperature (typically 25–75 °C range)
Watch for E/Z alkenes in the products. When the elimination can produce both cis and trans isomers, trans (E) is generally major because it is more stable.
Understanding alkene stability is essential for predicting the major elimination product.
Classification by substitution:
Terminal (the double bond is at the end of the chain): monosubstituted
Internal (the double bond is within the chain): disubstituted, trisubstituted, or tetrasubstituted
Stability order: monosubstituted < disubstituted < trisubstituted < tetrasubstituted
More substituted alkenes are more stable because alkyl groups are electron-donating (through hyperconjugation), which stabilises the electron-rich pi bond.
Cis vs. trans:
Cis (Z): substituents on the same face of the double bond. Less stable, higher energy (steric strain between groups on the same side). German: zusammen = together.
Trans (E): substituents on opposite faces. More stable, lower energy. German: entgegen = opposite.
Whenever a carbocation forms (in SN1 or E1), you must ask: is this the most stable carbocation possible, or can it rearrange?
Rearrangements happen through:
Hydride shifts (1,2-H shift): A hydrogen migrates from an adjacent carbon to the cation carbon, moving the positive charge to a more substituted position
Alkyl shifts (1,2-alkyl shift): An alkyl group (or ring bond) migrates in the same way
Rules for rearrangements:
Shifts occur only when a more stable carbocation results (e.g. 2° to 3°)
Any adjacent group can shift: H, CH₃, or part of a ring
Rearrangements are not reversible under these conditions
No more than two shifts per molecule on an exam (a practical rule for problem-solving)
Example pattern: A secondary alkyl halide loses its leaving group to form a 2° carbocation. A hydride on the adjacent carbon shifts to convert it to a 3° carbocation. The nucleophile then attacks the rearranged cation, giving a product with a different connectivity than you would predict from a simple, direct substitution.
When you see an SN1 or E1 product whose structure does not match a direct substitution or elimination at the original carbon, a rearrangement has occurred. Always check for this.
Rearrangements also affect elimination products. If the carbocation rearranges before the base removes a proton, the resulting alkene may form in a different position than expected. Compare "expected" (no rearrangement) vs. "observed" (with rearrangement) products.
E2 is fundamentally different from E1:
Concerted and bimolecular: everything happens in one step. Rate = k[substrate][base].
A strong base removes a beta hydrogen at the same time as the leaving group departs and the pi bond forms.
There is no carbocation intermediate, so there are no rearrangements in E2.
Key features of E2:
Competes with SN2 (both require a strong nucleophile/base)
Rate = k[electrophile][base]
Basicity matters more than nucleophilicity for driving E2
Prefers polar aprotic solvents (like SN2)
E2 has a strict geometric requirement: the C–H bond being broken and the C–LG bond must be antiperiplanar (180° dihedral angle, coplanar, on opposite sides).
This matters because the orbitals involved need to be aligned for the concerted mechanism to work. The electrons from the C–H bond flow into what becomes the pi bond, while the leaving group departs from the opposite side.
Consequences of the antiperiplanar requirement:
You may need to rotate the molecule (Newman projection) to find the conformation where H and LG are anti
In cyclohexane rings, both the H and the LG must be axial and on opposite sides (trans-diaxial). If the LG is equatorial, a chair flip is needed to put it axial before E2 can occur.
Some substrates have only one beta hydrogen that is antiperiplanar to the LG, which locks in a single elimination product regardless of Zaitsev's rule
Small bases (HO⁻, RO⁻, H₂N⁻): follow Zaitsev's rule, giving the most substituted alkene as major product
Bulky bases (t-BuO⁻, LDA): give the Hofmann product (least substituted alkene), because steric hindrance prevents the base from reaching the more substituted position
LDA (lithium diisopropylamide) is the classic bulky, strong, non-nucleophilic base. It reliably gives E2 with the least substituted alkene. On cyclohexane substrates treated with LDA, the antiperiplanar requirement still applies: the LG must be axial, and the eliminated H must be the axial one anti to it.
E1 Rate Law: Rate = k[substrate]
E2 Rate Law: Rate = k[substrate][base]
Alkene stability ranking: monosubstituted < disubstituted < trisubstituted < tetrasubstituted
Carbocation stability ranking: methyl < 1° < 2° < 3°
Antiperiplanar geometry: H and LG at 180° dihedral, both axial in cyclohexane chairs
Elimination reactions are how synthetic chemists install double bonds in target molecules. Controlling whether you get Zaitsev or Hofmann products by choosing the right base is a standard strategy in organic synthesis. Carbocation rearrangements, while sometimes inconvenient, are also exploited deliberately in terpene biosynthesis, where enzymes guide rearrangements to build complex ring systems from simple precursors.
"E1 and E2 are just two versions of the same reaction." They are mechanistically completely different. E1 is stepwise (carbocation intermediate, no geometric requirement). E2 is concerted (no intermediate, strict antiperiplanar requirement). They compete with different partners: E1 with SN1, E2 with SN2.
"The most substituted alkene is always the major product." Only when a non-bulky base is used. Bulky bases (t-BuO⁻, LDA) flip this to give the least substituted (Hofmann) product.
"E2 can happen with any orientation of H and LG." The antiperiplanar requirement is absolute. If no beta hydrogen is antiperiplanar to the leaving group in a given conformation, that conformation cannot undergo E2. A chair flip or bond rotation may be needed.
"Carbocation rearrangements happen in E2." They do not. E2 has no carbocation intermediate, so there is nothing to rearrange. Rearrangements occur only in E1 and SN1.
⚠️ When given an SN1/E1 problem, always check for carbocation rearrangements. If a more stable cation is one shift away, it will rearrange.
⚠️ E2 antiperiplanar requirement: exam problems often use cyclohexanes where you must do a chair flip to get the LG axial, then identify which axial H is anti to it.
⚠️ Know the distinction between Zaitsev (most substituted, small base) and Hofmann (least substituted, bulky base) products.
⚠️ Heat favours elimination over substitution in SN1/E1 competition. This is a common exam condition to watch for.
⚠️ LDA problems: LDA is always E2, always Hofmann product, never SN2 (too bulky to be a nucleophile).
True or False: E1 and SN1 share the same rate-determining step.
Fill in the blank: E2 requires the H and the leaving group to be in a(n) _______ arrangement.
True or False: Bulky bases give the more substituted alkene as the major product.
Fill in the blank: A 2° carbocation can rearrange to a 3° carbocation via a _______ shift.
True or False: E2 reactions can involve carbocation rearrangements.
Q: A tertiary alkyl bromide is heated in methanol with no strong base added. What products do you expect?
A: Both SN1 (methyl ether) and E1 (alkene) products. Heat favours E1 over SN1, so the alkene is the major product. Check for carbocation rearrangements before drawing either product.
Q: What is the antiperiplanar requirement, and why does E2 need it?
A: The C–H bond and C–LG bond must be at a 180° dihedral angle (anti and coplanar). This alignment is needed because E2 is concerted: the base removes H, the pi bond forms, and the LG leaves all in one step. The orbital overlap required for the new pi bond demands this geometry.
Q: Draw the major E2 product of 2-bromobutane with NaOH.
A: NaOH is a strong, small base, so Zaitsev's rule applies. The major product is 2-butene (the more substituted alkene), with trans-2-butene preferred over cis-2-butene.
Q: Same substrate (2-bromobutane) with LDA. What is the major product?
A: LDA is a strong, bulky base. The Hofmann product (least substituted alkene) is major: 1-butene.
Q: A secondary substrate undergoes E1. The initially formed 2° carbocation is adjacent to a quaternary carbon bearing a methyl group. What might happen?
A: A methyl shift (1,2-alkyl shift) could convert the 2° carbocation to a 3° carbocation if that produces a more stable intermediate. If so, the elimination product forms from the rearranged cation, and the double bond may appear in an unexpected position.
E1 and E2 connect directly back to SN1 and SN2 (Chapters 6 and 7), because every reaction problem asks which of the four pathways is favoured. The alkene products of elimination reactions are the starting materials for the addition reactions you will study next (Chapters 8 and beyond: hydrohalogenation, hydration, halogenation). Carbocation rearrangements also appear in acid-catalysed reactions and biosynthetic pathways later in the course.
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