Difficulty: Intermediate to Advanced | Prerequisites: Functional groups and nomenclature notes, SN1/SN2 substitution notes, carbocation stability basics.
Elimination reactions compete directly with substitution whenever a base or nucleophile encounters a substrate bearing a leaving group. E2 (bimolecular elimination) is a concerted, one-step process requiring anti-periplanar geometry; E1 (unimolecular elimination) goes through the same carbocation intermediate as SN1 and then loses a proton. Being able to predict which of the four pathways (SN1, SN2, E1, E2) dominates for a given substrate, reagent, and solvent is probably the single most important skill tested on an Organic Chemistry I exam. This set of notes also covers alkene and carbocation stability, carbocation rearrangement in mechanisms, and reagent selection.
E2 is a one-step elimination driven by a strong base: the base grabs a proton while the leaving group departs, forming a double bond in a single concerted step. E1 shares its first step with SN1 (leaving group departs to form a carbocation), then a base removes a proton from the carbocation to form the alkene. Both typically give the most substituted alkene (Zaitsev's rule). The four-pathway decision (SN1, SN2, E1, E2) depends on substrate class (1°, 2°, 3°), nucleophile/base strength, and solvent.
E2 (bimolecular elimination)
A concerted, one-step mechanism in which a strong base abstracts a beta-hydrogen while the leaving group departs simultaneously, forming a new C=C double bond. Rate = k[base][substrate]. Think of it as a tug-of-war where the base pulls a proton off one carbon while the leaving group exits from the neighbouring carbon, all at once.
E1 (unimolecular elimination)
A two-step mechanism: (1) the leaving group departs to form a carbocation, (2) a base removes a proton adjacent to the positive carbon, forming the alkene. Rate = k[substrate]. In simple terms, the substrate falls apart first, then the base cleans up.
Anti-periplanar geometry
The spatial requirement for E2: the H being removed and the leaving group must be on opposite sides of the C-C bond (180° dihedral angle). Think of them as pointing in opposite directions when you sight along the bond.
Zaitsev's rule (Saytzeff's rule)
The more substituted alkene is the major product of elimination. In simple terms, the double bond forms toward the more branched side of the molecule because more substituted alkenes are more stable.
Alkene stability
More substituted alkenes are more stable: tetrasubstituted > trisubstituted > disubstituted > monosubstituted. Trans (E) alkenes are more stable than cis (Z) due to reduced steric strain.
Carbocation rearrangement
A carbocation can rearrange via a 1,2-hydride shift or a 1,2-methyl shift to form a more stable carbocation. This happens in SN1 and E1 pathways and can give "unexpected" products.
Hydride shift (1,2-H shift)
A hydrogen atom with its bonding electrons migrates from an adjacent carbon to the carbocation centre, converting a less stable carbocation into a more stable one.
Methyl shift (1,2-alkyl shift)
A methyl (or other alkyl) group migrates with its bonding electrons from an adjacent carbon to the carbocation centre.
E2 is concerted: the base abstracts a beta-hydrogen at the same time as the leaving group departs and the double bond forms.
Requires a strong base: NaOEt, NaOH, NaNH₂, t-BuOK. The stronger and bulkier the base, the more elimination is favoured over substitution.
Anti-periplanar geometry is required: the C-H and C-LG bonds must be 180° apart (anti) for the orbitals to overlap properly. In cyclohexane rings, this means both groups must be axial and trans to each other.
Regioselectivity (Zaitsev): the more substituted alkene is the major product. A bulky base like t-BuOK can shift selectivity toward the less substituted (Hofmann) product, but Zaitsev is the default for standard bases.
Rate law: Rate = k[base][substrate].
Exam key examples:
Problem 3b: a secondary substrate with NaOEt/EtOH. Strong base, so E2 dominates. The major product is the more substituted alkene.
Problem 5c: NaNH₂ gives the fastest E2 elimination because it is the strongest base among the choices (NH₃, NaCN, NaOH, NaNH₂). The amide ion (NH₂⁻) is far more basic than hydroxide.
Problem 6c: a tosylate treated with NaOEt/EtOH. E2 gives the more substituted alkene.
E1 shares its first step with SN1: the leaving group departs to give a carbocation. Then a weak base (often the solvent) removes a proton from a carbon adjacent to the positive charge.
Conditions: weak base or no added base, polar protic solvent, tertiary or secondary substrate with a good leaving group.
Regioselectivity: follows Zaitsev's rule (more substituted alkene is major).
Competes with SN1: under SN1 conditions (tertiary substrate, weak nucleophile, protic solvent), E1 products often appear alongside SN1 products.
Rate law: Rate = k[substrate].
Exam key examples:
Problem 3a: cyclohexanol treated with H₂SO₄. The acid protonates the -OH to make H₂O a good leaving group. H₂O departs (E1 conditions), carbocation forms, loss of a proton gives cyclohexene.
Problem 6a: a secondary alcohol with H₂SO₄. E1 elimination, producing the more substituted alkene.
Problem 6b: a benzylic chloride in CH₃OH. SN1 substitution products (both enantiomers of the methyl ether) plus an E1 elimination product (the alkene).
This is the central skill of the topic. Work through these questions in order:
Is the substrate primary, secondary, or tertiary?
Primary + strong nucleophile → SN2
Primary + strong, bulky base → E2
Tertiary + strong base → E2
Tertiary + weak nucleophile/base, protic solvent → SN1 + E1 (mixture)
Secondary → context-dependent (see next points)
Is the reagent a strong nucleophile or a strong base?
Strong, unhindered nucleophile in polar aprotic solvent → SN2
Strong base (especially bulky) → E2
Weak nucleophile / no nucleophile / just solvent → SN1/E1
What is the solvent?
Polar aprotic (DMF, DMSO, acetone) → favours SN2
Polar protic (H₂O, MeOH, EtOH, AcOH) → favours SN1/E1
Exam key example (Problem 6d): a secondary alcohol treated with HBr. The acid protonates -OH to H₂O (good leaving group). The carbocation forms (SN1 conditions). Br⁻ attacks to give the substitution product. Because the initial carbocation is secondary, a hydride shift can rearrange it to a more stable tertiary carbocation, giving a rearranged product as well.
Problem 4 from the exam tested a full mechanism for the reaction of a tertiary alcohol with HCl:
Step 1: the oxygen lone pair attacks the H of HCl, protonating the -OH to form -OH₂⁺ (a good leaving group).
Step 2: H₂O departs, generating a carbocation.
Step 3: a 1,2-hydride shift converts a secondary carbocation to a more stable tertiary carbocation (or the reverse, depending on the starting structure; here a hydride shift produced a tertiary cation).
Step 4: Cl⁻ attacks the carbocation to give the product.
Key point: always check whether a hydride or methyl shift can produce a more stable carbocation. If it can, assume the rearrangement occurs.
More substituted alkenes are more stable due to hyperconjugation:
Tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted (ethylene)
Trans (E) > cis (Z) for the same degree of substitution, due to reduced steric strain between groups.
Exam key example (Problem 7a): three alkenes ranked from least to most stable. The ordering was A < C < B, where B was the most substituted and A was the least.
More substituted carbocations are more stable:
3° > 2° > 1° > methyl
Resonance-stabilised carbocations (allylic, benzylic) are more stable than their non-conjugated counterparts of the same degree.
Exam key example (Problem 7b): three carbocations ranked from least to most stable. The ordering was C < B < A. The most stabilised (A) had the greatest degree of substitution and/or resonance stabilisation.
Problem 7c: converting a tosylate to an acetate. The best reagent is NaOAc in DMF, an SN2 setup: strong nucleophile (OAc⁻), polar aprotic solvent (DMF), and primary substrate.
Problem 7d: an elimination from a secondary chloride to give an alkene. The best reagent is t-BuOK, a strong, bulky base that favours E2 over substitution.
Students often assume E1 and E2 always give the same product. They usually do (Zaitsev), but E2 requires anti-periplanar geometry, which can restrict which hydrogen is removed, especially in rigid ring systems. E1, going through a free carbocation, has no such geometric constraint.
Forgetting carbocation rearrangement is a common error. If a secondary carbocation can shift to a tertiary one, it will. The product may end up with the functional group on a different carbon than expected.
Students sometimes pick NaOH as the strongest base in a list that includes NaNH₂. Hydroxide (pKa of H₂O ≈ 15.7) is far weaker than amide (pKa of NH₃ ≈ 38). NaNH₂ wins.
Confusing "strong base" with "strong nucleophile." t-BuOK is a strong, bulky base (favours E2) but a poor nucleophile (too bulky for SN2). NaCN is a strong nucleophile but a relatively weak base (favours SN2, not E2).
⚠️ The "predict the mechanism and draw the products" question type (Problem 6 on this exam, worth 24 points) is typically the highest-value question. You must label each product with its mechanism (SN1, SN2, E1, or E2).
⚠️ For elimination reactions, draw only the major alkene product unless the question asks for all products. The major product is the most substituted alkene (Zaitsev's rule).
⚠️ Always check for carbocation rearrangement in any SN1 or E1 pathway. If a hydride or methyl shift can form a more stable carbocation, draw the rearranged product. Some exams give full marks for the rearranged product and partial credit for the non-rearranged one.
⚠️ Pay close attention to stereochemistry. E2 products may have defined (E)/(Z) geometry based on which groups end up on the same or opposite sides of the double bond.
⚠️ Reagent-choice questions ("What is the most appropriate reagent for this transformation?") require you to work backwards: identify the mechanism from the substrate and product, then pick the reagent that drives that mechanism.
True or false: E2 requires anti-periplanar geometry between the leaving group and the beta-hydrogen. True.
Fill in the blank: the E1 rate law is Rate = k[____]. substrate
True or false: t-BuOK favours SN2 over E2. False. t-BuOK is bulky and strongly basic, favouring E2.
Fill in the blank: a 1,2-hydride shift converts a less stable ____ into a more stable one. carbocation
True or false: a trisubstituted alkene is more stable than a disubstituted alkene. True.
Q: A cyclohexanol is treated with H₂SO₄. What mechanism operates, and what is the product?
A: E1 elimination. The acid protonates the -OH to form water (a good leaving group). Water departs to give a carbocation. Loss of a proton from an adjacent carbon produces the alkene (cyclohexene).
Q: A secondary benzylic chloride is dissolved in CH₃OH. What products form, and by what mechanisms?
A: SN1 gives both enantiomers of the methyl ether (racemisation). E1 gives the alkene. All three products may appear because SN1 and E1 share the same carbocation intermediate.
Q: A secondary tosylate is treated with NaOEt in EtOH. What is the major product?
A: E2 elimination. NaOEt is a strong base, the substrate is secondary (E2 favoured over SN2 for secondary substrates with a strong base). The major product is the more substituted alkene (Zaitsev).
Q: A secondary alcohol reacts with HBr. The product has the bromine on a different carbon than the original -OH. Explain.
A: The -OH is protonated to -OH₂⁺, which leaves to form a secondary carbocation. A 1,2-hydride shift converts it to a more stable tertiary carbocation. Br⁻ then attacks the rearranged carbocation, placing the Br on the carbon that was originally adjacent to the -OH.
Q: Which base gives the fastest E2 elimination: NH₃, NaCN, NaOH, or NaNH₂?
A: NaNH₂. The amide ion (NH₂⁻) is the strongest base in the list (conjugate acid NH₃ has pKa ≈ 38). Stronger base = faster E2.
Q: Rank these alkenes by stability: monosubstituted (A), trisubstituted (B), disubstituted (C).
A: A < C < B. More substituted alkenes are more stable due to hyperconjugation.
Q: What reagent converts a primary chloride to an alkene via elimination?
A: t-BuOK (potassium tert-butoxide). It is a strong, bulky base that favours E2 and suppresses SN2.
Elimination connects directly back to substitution: every SN1 condition can also produce E1 products, and every E2-favouring setup might give some SN2 if the nucleophile is small and unhindered. The decision framework here pulls together everything from the substitution notes.
Carbocation rearrangement will return in later topics, including Markovnikov addition to alkenes and the chemistry of carbocation intermediates in terpene biosynthesis. Alkene stability rankings feed directly into predicting the regiochemistry of addition reactions in the next unit.
E1, E2, elimination reaction, Zaitsev rule, Saytzeff rule, anti-periplanar, carbocation rearrangement, hydride shift, methyl shift, alkene stability, carbocation stability, t-BuOK, NaOEt, NaNH₂, strong base vs strong nucleophile, SN1 vs SN2 vs E1 vs E2, reaction prediction, mechanism drawing, organic chemistry I, OChem 1, UMN, University of Minnesota, Exam 3 review