Difficulty: Intermediate | Prerequisites: SN1/SN2 mechanisms, carbocation stability, acid-base chemistry, stereochemistry basics (Ch. 1-7 and substitution notes)
Elimination reactions are the other half of the substitution-elimination coin. Where substitution swaps one group for another, elimination removes a leaving group and a neighbouring hydrogen to form a pi bond (usually an alkene). E1 and E2 compete directly with SN1 and SN2, and exam questions regularly ask you to predict which pathway dominates.
Once you have formed an alkene, the next question is what you can do with it. Alkenes are electron-rich functional groups that undergo addition reactions: HX, hydroboration-oxidation, oxymercuration-demercuration, hydrogenation, and radical addition. Each gives a different regiochemistry and stereochemistry, and picking the right reagent to get the right product is central to CHEM 237.
E2 elimination is a one-step process requiring a strong base and anti-periplanar geometry between the leaving group and the hydrogen being removed. E1 elimination goes through the same carbocation intermediate as SN1, then loses a proton to form the more substituted alkene (Zaitsev's rule). Alkene addition reactions reverse the process: they add atoms across the double bond, with regiochemistry and stereochemistry that depend on the reagent and mechanism.
E2 (elimination, bimolecular)
A one-step elimination where a strong base removes a beta-hydrogen at the same time the leaving group departs. Rate = k[substrate][base]. Think of it as a tug-of-war: the base pulls a hydrogen while the leaving group walks away, and the electrons left behind form a new pi bond.
E1 (elimination, unimolecular)
A two-step elimination: the leaving group departs to form a carbocation (same as SN1 step 1), then a base removes a proton from the carbocation to form the alkene. Rate = k[substrate].
Anti-periplanar geometry
The spatial arrangement required for E2. The H being removed and the leaving group must be on opposite sides of the C-C bond, at a dihedral angle of 180°. In simple terms, they need to be lined up pointing in opposite directions for the electrons to flow properly.
Zaitsev's rule
The more substituted alkene is the major product (the thermodynamic product). In simple terms, the double bond forms toward the side with more alkyl groups.
Markovnikov's rule
In the addition of HX to an alkene, the hydrogen adds to the less substituted carbon and the halide adds to the more substituted carbon. This follows from the preference for forming the more stable carbocation intermediate.
Anti-Markovnikov addition
The opposite regiochemistry: H adds to the more substituted carbon, and the other group adds to the less substituted carbon. This occurs in hydroboration-oxidation and in radical addition of HBr (with peroxides).
Hydroboration-oxidation
A two-step sequence: (1) BH₃ adds across the double bond with syn addition, placing boron on the less substituted carbon. (2) H₂O₂/NaOH oxidises the C-B bond to a C-OH bond with retention of configuration. The net result is anti-Markovnikov, syn addition of water.
Oxymercuration-demercuration
A two-step sequence: (1) Hg(OAc)₂ and a nucleophilic solvent (water or alcohol) add across the double bond with Markovnikov regiochemistry. (2) NaBH₄ removes the mercury. The net result is Markovnikov addition of water (or an alcohol) without carbocation rearrangements.
Syn addition
Both new groups add to the same face of the double bond. Hydroboration and catalytic hydrogenation are syn additions.
Anti addition
The two new groups add to opposite faces of the double bond. Halogenation (Br₂) is an anti addition.
Mechanism
A strong base abstracts a beta-hydrogen while the leaving group departs simultaneously.
The C-H bond, the C-C bond between the alpha and beta carbons, and the C-LG bond must all be in the same plane (anti-periplanar).
This is a concerted, one-step process. Rate = k[substrate][base].
Requirements
A strong base is essential. Common E2 bases: NaOH, KOH, NaOEt, KOtBu.
KOtBu (potassium tert-butoxide) is a bulky, strong base that favours E2 over SN2 because it is too sterically hindered to act as a nucleophile.
The beta-hydrogen and the leaving group must be anti-periplanar. In cyclohexane systems, this means they must both be axial and trans to each other.
Regiochemistry
Zaitsev's rule: the more substituted alkene is the major product when a small base is used.
Hofmann's rule: a bulky base (e.g. KOtBu) can favour the less substituted alkene, because it cannot easily reach the more hindered beta-hydrogen.
Stereochemistry of the product
When E2 produces a disubstituted alkene, the geometry (E vs Z) is determined by the anti-periplanar arrangement of the starting material.
Example from the exam: a substrate with Br, H, CH₃, and D (deuterium) at specific positions. KOtBu (strong, bulky base) causes E2. The anti-periplanar requirement dictates which hydrogen (or deuterium) is removed, and the product must show where the deuterium ends up.
Mechanism
Step 1: the leaving group departs to form a carbocation (identical to SN1 step 1).
Step 2: any base (even a weak one like H₂O) removes a proton from the beta-carbon adjacent to the carbocation, forming the alkene.
Rate = k[substrate]. The base does not appear in the rate law.
When E1 occurs
E1 competes with SN1 under the same conditions: tertiary substrates, weak nucleophiles/bases, polar protic solvents, and heat.
Higher temperature generally favours elimination over substitution (both E1 over SN1 and E2 over SN2) because elimination has a higher activation entropy.
Regiochemistry
Zaitsev's rule applies: the more substituted alkene is the major product.
Because a carbocation forms, rearrangements are possible before the proton is removed.
Key difference from E2
E1 does not require anti-periplanar geometry, because the leaving group has already departed before the proton is removed.
E1 can give a mixture of E and Z products (and often does), since the carbocation is planar and the base can remove a proton from either side.
Addition of HX (HCl, HBr, HI) to alkenes
Follows Markovnikov's rule: H adds to the less substituted carbon, X adds to the more substituted carbon.
Mechanism: the alkene pi electrons attack the H of HX (forming the more stable carbocation), then X⁻ attacks the carbocation.
This is mechanistically the reverse of E1 elimination. The two reactions exist in equilibrium.
Example from the exam: HCl adding to 2-methylpropene gives 2-chloro-2-methylpropane (the Markovnikov product), because the tertiary carbocation is more stable than the primary.
Hydroboration-oxidation (1. BH₃; 2. H₂O₂, NaOH)
Anti-Markovnikov, syn addition of water.
BH₃ adds to the less substituted carbon (anti-Markovnikov regiochemistry) because of steric preference.
The addition is syn: both B and H add to the same face of the double bond.
Oxidation with H₂O₂/NaOH replaces B with OH, retaining the stereochemistry.
Example from the exam: hydroboration-oxidation of a cyclic alkene gives the alcohol on the less substituted carbon with syn stereochemistry.
Oxymercuration-demercuration (1. Hg(OAc)₂, solvent; 2. NaBH₄)
Markovnikov addition of water (when solvent is H₂O) or an alcohol (when solvent is an alcohol such as CH₃OH).
No carbocation rearrangements, because the intermediate is a mercurinium ion (a three-membered ring with Hg), not an open carbocation.
Example from the exam: oxymercuration with Hg(OAc)₂ in CH₃OH followed by NaBH₄ gives the Markovnikov ether product (OCH₃ on the more substituted carbon).
Catalytic hydrogenation (H₂, Pd or Pt)
Syn addition of H₂ across the double bond.
Both hydrogens add to the same face of the alkene.
No change in the carbon skeleton, no carbocation intermediate.
Example from the exam: hydrogenation of a cyclopropane-containing alkene with H₂/Pd reduces only the alkene double bond. The stereochemistry of the new C-H bonds reflects syn addition.
Hydrogenation does not break C-Cl or C-Br bonds and does not change the configuration at existing stereocentres unless the double bond itself is at the stereocentre.
Radical addition of HBr (HBr with peroxides)
Anti-Markovnikov addition of HBr only. This does not work with HCl or HI.
Mechanism: peroxides generate radicals. A bromine radical adds to the less substituted carbon (forming the more stable radical intermediate). Then H is transferred from HBr.
Example from the exam: 2-methylpropene with HBr and peroxides gives 1-bromo-2-methylpropane (Br on the less substituted carbon).
Free-radical halogenation (Br₂, hv)
A distinct reaction from the additions above: this is substitution of a C-H bond, not addition to a double bond.
Br₂/hv selectively brominates the most substituted C-H (most stable radical intermediate), giving the tertiary bromide as the major product.
Example from the exam: free-radical bromination of 2-methylbutane gives predominantly 2-bromo-2-methylbutane.
Adding HX to an alkene and E1 elimination of the corresponding alkyl halide are the forward and reverse of the same equilibrium. You can shift the equilibrium using Le Chatelier's principle.
To increase the yield of the alkyl halide (addition direction)
Add excess HX (increase reactant concentration).
Lower the temperature (addition is exothermic, so cooling favours the product side).
Remove the alkyl halide as it forms.
To increase the yield of the alkene (elimination direction)
Raise the temperature (drives elimination, the endothermic direction).
Remove the alkene as it forms (e.g. distillation, since alkenes are often more volatile).
Dilute the HX or remove it.
E2 rate law
rate = k [substrate] [base]
E1 rate law
rate = k [substrate]
Alkene addition reagent summary
Reagent | Regiochemistry | Stereochemistry | Product |
|---|---|---|---|
HX | Markovnikov | Not controlled | Alkyl halide |
1) BH₃; 2) H₂O₂/NaOH | Anti-Markovnikov | Syn | Alcohol |
1) Hg(OAc)₂, H₂O; 2) NaBH₄ | Markovnikov | Not controlled | Alcohol |
1) Hg(OAc)₂, ROH; 2) NaBH₄ | Markovnikov | Not controlled | Ether |
H₂, Pd | N/A (both H) | Syn | Alkane |
HBr, peroxides | Anti-Markovnikov | Not controlled | Alkyl bromide |
Br₂ | N/A (both Br) | Anti | Dihalide |
Students often assume Markovnikov's rule is arbitrary. It follows directly from carbocation stability: the proton adds to give the more stable (more substituted) carbocation.
Students frequently confuse hydroboration-oxidation with oxymercuration-demercuration. Both produce alcohols, but they give opposite regiochemistry: hydroboration gives anti-Markovnikov, oxymercuration gives Markovnikov.
Students sometimes think peroxides change the regiochemistry of all HX additions. They only affect HBr. HCl and HI still follow Markovnikov regiochemistry even with peroxides present.
Students often forget that E2 requires anti-periplanar geometry, leading to incorrect elimination products, particularly in cyclohexane systems where axial/equatorial positions matter.
⚠️ Predicting the correct alkene addition product for each reagent set is heavily tested. Know the reagent summary table cold.
⚠️ Drawing curved-arrow mechanisms for E1 and SN1 (which share the same first step) is a common question. Show the carbocation intermediate explicitly.
⚠️ E2 with deuterium-labelled substrates tests whether you understand anti-periplanar geometry. You must show where D ends up in the product.
⚠️ Le Chatelier's principle applied to the HX addition/E1 equilibrium is directly tested: know how to shift the equilibrium in either direction.
⚠️ Free-radical bromination (Br₂/hv) is a substitution reaction, not an addition. The exam includes it alongside alkene reactions to test whether you recognise the difference.
True or False: E2 elimination requires the H and the leaving group to be syn-periplanar.
Fill in the blank: Hydroboration-oxidation gives ______ (Markovnikov/anti-Markovnikov) addition of water.
True or False: Radical addition of HBr with peroxides also works for HCl and HI.
Fill in the blank: The more substituted alkene is the major E2 product according to ______ rule.
True or False: Oxymercuration-demercuration avoids carbocation rearrangements.
Answers: 1. False (anti-periplanar, 180°). 2. Anti-Markovnikov. 3. False (only HBr). 4. Zaitsev's. 5. True.
Q: Draw a curved-arrow mechanism for the addition of HCl to 2-methylpropene to give the Markovnikov alkyl chloride product.
A: Step 1: the pi electrons of the alkene attack the H of HCl, forming a bond to the less substituted carbon and generating a tertiary carbocation on the more substituted carbon. Cl⁻ departs. Step 2: Cl⁻ attacks the tertiary carbocation to give 2-chloro-2-methylpropane.
Q: Why is the Markovnikov product the major product in HCl addition to 2-methylpropene rather than a different isomer?
A: The proton adds to the terminal (less substituted) carbon, generating the more stable tertiary carbocation. Adding the proton to the internal carbon would give a less stable primary carbocation. Because the tertiary carbocation is lower in energy, it forms preferentially, and the chloride then attacks that carbon.
Q: Suggest one method to increase the yield of the alkyl chloride from the HCl/alkene equilibrium, and one method to increase the yield of the alkene.
A: To favour the alkyl chloride: add excess HCl or lower the temperature (Le Chatelier). To favour the alkene: raise the temperature or remove the alkene as it forms by distillation.
Q: What is the major product of hydroboration-oxidation of cyclobutene? Include stereochemistry.
A: The product is cyclobutanol with syn stereochemistry. BH₃ and H both add to the same face. Since cyclobutene is symmetric, the alcohol forms on the less hindered face, and both enantiomers are produced equally (write "and enantiomer").
Q: What is the product of Hg(OAc)₂ in CH₃OH followed by NaBH₄ on propene?
A: 2-Methoxypropane (the Markovnikov ether). The OCH₃ group ends up on the more substituted carbon (C-2), with no rearrangement.
Elimination reactions connect directly to substitution (SN1/SN2): the same conditions that promote one often compete with the other, and exam questions frequently ask you to predict the dominant pathway.
Alkene addition reactions are the reverse of elimination, forming a conceptual loop. Understanding both directions deepens your grasp of thermodynamics and kinetics. These reactions also preview later material in CHEM 238 and 239, where nucleophilic addition to carbonyls and electrophilic aromatic substitution follow similar logic of electron-rich species attacking electron-poor sites.
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