Difficulty: Intermediate | Prerequisites: Lewis structures, functional groups, alkene structure, basic acid-base chemistry
Alkenes react with electrophiles (H–X, X₂, X₂/H₂O, H₂SO₄/H₂O, BH₃, O₃) by breaking their π bond to form new σ bonds. The type of reagent determines the regiochemistry (Markovnikov vs. anti-Markovnikov), the stereochemistry (syn vs. anti addition) and the intermediate (carbocation, bromonium ion or organoborane). If you can predict those three things for each reagent, you can predict the major product of nearly every alkene addition on an exam.
Electrophilic addition
A reaction in which an electrophile (electron-poor species) adds across a π bond, breaking it and forming two new σ bonds. In simple terms, the double bond acts as an electron-rich target that "grabs" an incoming electrophile.
Markovnikov's rule
In the addition of HX to an unsymmetrical alkene, the hydrogen adds to the carbon with more hydrogens already attached (the less substituted carbon), and X adds to the more substituted carbon. Think of it as: "the rich get richer" in hydrogens.
Anti-Markovnikov addition
The opposite regiochemistry: the nucleophilic group (e.g. OH from hydroboration-oxidation) ends up on the less substituted carbon. This happens when the mechanism bypasses a carbocation intermediate.
Carbocation
A positively charged carbon intermediate. Stability order: 3° > 2° > 1° > methyl. In simple terms, the more alkyl groups surrounding the positive carbon, the more stable it is because those groups donate electron density.
Carbocation rearrangement
A 1,2-hydride shift or 1,2-methyl shift that converts a less stable carbocation into a more stable one. This changes where the product ends up and is a common exam trap.
Bromonium ion (halonium ion)
A three-membered ring intermediate formed when Br₂ (or Cl₂) reacts with an alkene. The bromine bridges both carbons of the former double bond, which forces the nucleophile to attack from the opposite face (anti addition).
Halohydrin
The product of X₂/H₂O addition to an alkene: an OH and a halogen on adjacent carbons, added anti to each other. Water acts as the nucleophile that opens the halonium ion.
Hydroboration-oxidation
A two-step sequence (1. BH₃·THF, 2. H₂O₂/NaOH) that adds H and OH across a double bond with syn stereochemistry and anti-Markovnikov regiochemistry.
Ozonolysis
Cleavage of a C=C double bond using O₃, followed by a reductive workup (commonly Me₂S or Zn). The double bond breaks completely, producing carbonyl fragments (aldehydes and/or ketones).
Acid-catalysed hydration
Addition of water across a double bond using H₂SO₄/H₂O. Follows Markovnikov's rule because the mechanism proceeds through a carbocation intermediate.
The π electrons of the alkene attack the proton of HX, forming a carbocation intermediate and releasing X⁻.
The halide ion (X⁻) then attacks the carbocation to give the alkyl halide product.
Regiochemistry follows Markovnikov's rule: H adds to the less substituted carbon, X adds to the more substituted carbon. This is because the more substituted (more stable) carbocation forms preferentially.
Carbocation rearrangements can occur. If the initial carbocation is secondary and a neighbouring carbon would give a tertiary carbocation, a 1,2-hydride shift or 1,2-methyl shift will take place before the halide attacks.
Example from the practice set: an alkene with an allylic bromine treated with HBr. The protonation forms a carbocation that can rearrange, leading to a product where the Br ends up at a position you would not predict without considering the shift.
Stereochemistry: because the carbocation is planar (sp²), the nucleophile can attack from either face. The product is typically a racemic mixture when a new stereocentre forms.
Halogenation (Br₂ or Cl₂, no water): the π bond attacks Br₂, forming a cyclic bromonium ion intermediate. The bromide ion then attacks from the opposite face, giving anti addition of two halogen atoms (a vicinal dihalide).
Halohydrin formation (Br₂/H₂O or Cl₂/H₂O): the same bromonium ion forms, but water (present in large excess) acts as the nucleophile instead of Br⁻.
Water attacks the more substituted carbon of the bromonium ion (the carbon better able to stabilise partial positive charge).
The result is anti addition of OH and Br (or Cl) on adjacent carbons.
Regiochemistry: OH ends up on the more substituted carbon, halogen on the less substituted carbon.
Example from the practice set: a methylcyclohexene treated with Br₂/H₂O gives a bromohydrin with OH on the more substituted carbon and Br on the less substituted carbon, both anti to each other.
Step 1: BH₃ adds across the double bond in a single concerted step (no carbocation). Boron attaches to the less substituted carbon, hydrogen to the more substituted carbon.
Step 2: H₂O₂/NaOH replaces the B with OH, retaining the stereochemistry.
Net result: anti-Markovnikov addition of water with syn stereochemistry.
Anti-Markovnikov: OH ends up on the less substituted carbon.
Syn: H and OH add to the same face of the double bond.
This is the go-to reaction when you need an alcohol on the less substituted carbon. Compare with acid-catalysed hydration, which gives the Markovnikov alcohol.
Example from the practice set: a cyclopentene with a vinyl group treated with BH₃·THF then H₂O₂/NaOH gives syn addition of OH to the less hindered position.
The proton from H₂SO₄ adds to the alkene, forming a carbocation (Markovnikov regiochemistry: proton to the less substituted carbon).
Water attacks the carbocation, then loss of a proton gives the alcohol.
Net result: Markovnikov addition of water. OH ends up on the more substituted carbon.
Because a carbocation forms, rearrangements are possible.
Example from the practice set: a trisubstituted alkene with H₂SO₄/H₂O gives a tertiary alcohol as the major product.
Ozone cleaves the C=C double bond entirely, producing two carbonyl fragments.
Each carbon of the former double bond becomes a C=O.
If the carbon had one hydrogen, it becomes an aldehyde (RCHO).
If it had no hydrogens (fully substituted), it becomes a ketone (R₂CO).
The reductive workup (Me₂S) prevents further oxidation of aldehydes to carboxylic acids.
Example from the practice set: a compound with a carboxylic acid group and a pendant alkene treated with O₃ then Me₂S gives two carbonyl-containing fragments, one retaining the carboxylic acid.
Drawing a mechanism for electrophilic addition typically involves two or three steps:
Step 1 (protonation): draw a curved arrow from the π bond electrons to the electrophilic hydrogen (of HBr, HCl, etc.). A second arrow shows the H–X bond breaking, with both electrons going to X. This forms the carbocation and X⁻.
Step 2 (optional rearrangement): if the carbocation is secondary and can become tertiary, draw a curved arrow showing a 1,2-hydride shift (H with its bonding electrons migrates to the adjacent positive carbon) or a 1,2-methyl shift. The positive charge moves to a more stable position.
Step 3 (nucleophilic attack): draw a curved arrow from a lone pair on X⁻ (or water, or Br⁻) to the carbocation carbon. This forms the new C–X bond and gives the product.
For the practice set's mechanism problem (methylenecyclohexane + HBr/H₂O):
Protonation of the exocyclic alkene places the carbocation on the ring (tertiary position, Markovnikov).
Br⁻ attacks the tertiary carbocation to give the major product (tertiary alkyl bromide).
A ring-expansion rearrangement can also occur: if a secondary carbocation forms initially, a 1,2-alkyl shift can expand the six-membered ring to a seven-membered ring, giving a cycloheptyl bromide as a minor product. This is the second mechanism asked for in the problem.
Reagent | Regiochemistry | Stereochemistry | Intermediate | Product |
|---|---|---|---|---|
HX (HBr, HCl, HI) | Markovnikov | Racemic (if new stereocentre) | Carbocation | Alkyl halide |
X₂ (Br₂, Cl₂) | N/A (symmetric addition) | Anti | Halonium ion | Vicinal dihalide |
X₂/H₂O | OH on more substituted C | Anti | Halonium ion | Halohydrin |
H₂SO₄/H₂O | Markovnikov | Racemic | Carbocation | Alcohol (more sub.) |
BH₃·THF, then H₂O₂/NaOH | Anti-Markovnikov | Syn | Organoborane | Alcohol (less sub.) |
O₃, then Me₂S | Cleaves at C=C | N/A | Ozonide | Aldehydes/ketones |
Halohydrin formation is used industrially to synthesise epoxides (via intramolecular ring closure with base), which are key intermediates in pharmaceuticals and polymers. Hydroboration-oxidation is a workhorse in synthesis labs when you need precise control over where an OH group lands on a carbon chain. Ozonolysis is used in forensic and structural chemistry to determine the position of double bonds in unknown molecules: you cleave the bond, identify the fragments, and work backwards.
Students often assume that a more stable product always forms faster. It does not. Reaction rate is determined by the transition state and intermediate stability, not product stability. A more substituted alkene reacts faster with HBr because it forms a more stable carbocation intermediate, not because the product itself is more stable.
Students frequently confuse the regiochemistry of halohydrin formation: the OH goes on the more substituted carbon (not the halogen). The halogen ends up on the less substituted carbon.
Hydroboration-oxidation and acid-catalysed hydration both add water across a double bond, but they give opposite regiochemistry. Mixing them up is one of the most common errors on exams.
Students sometimes forget that ozonolysis completely cleaves the double bond. It does not leave the carbon skeleton intact the way other addition reactions do.
⚠️ You will almost certainly be asked to predict the major product of at least two different addition reactions. Know the reagent → intermediate → regiochemistry → stereochemistry chain for each.
⚠️ Mechanism questions (curved arrows) appear on nearly every exam. Practise drawing all three steps: protonation, optional rearrangement, nucleophilic capture.
⚠️ Be ready for a rearrangement trap: if the direct carbocation is secondary but a shift would give tertiary, the exam expects you to show the shift.
⚠️ "What reagents would you use to achieve this transformation?" questions test whether you can work backwards from a product. If the OH is on the less substituted carbon, the answer is hydroboration-oxidation, not acid-catalysed hydration.
True or false: Hydroboration-oxidation gives Markovnikov addition of water. Answer: False. It gives anti-Markovnikov addition.
Fill in the blank: The intermediate in Br₂ addition to an alkene is a ______ ion. Answer: Bromonium (halonium).
True or false: Acid-catalysed hydration can involve carbocation rearrangements. Answer: True.
Fill in the blank: Ozonolysis of a trisubstituted alkene followed by Me₂S gives a ______ and an ______. Answer: Ketone and aldehyde.
True or false: In halohydrin formation, the halogen ends up on the more substituted carbon. Answer: False. The OH goes on the more substituted carbon; the halogen goes on the less substituted carbon.
Q: An asymmetric alkene is treated with HCl. Where does the Cl end up, and why?
A: On the more substituted carbon. HCl protonates the double bond to form the more stable (more substituted) carbocation, and Cl⁻ attacks that carbon.
Q: You need to convert 1-methylcyclohexene into trans-2-methylcyclohexanol (OH and methyl anti to each other). Which reagent set would you choose?
A: Br₂/H₂O would give a bromohydrin with anti addition, but the question asks for an alcohol without a halogen. The best route here is an indirect one (e.g. epoxidation then acid-catalysed ring opening). Acid-catalysed hydration would give the Markovnikov alcohol but without stereochemical control. Hydroboration-oxidation gives syn, anti-Markovnikov addition, which is the wrong regiochemistry and wrong stereochemistry for this target.
Q: Draw the products of ozonolysis (O₃ then Me₂S) of 2-methylbut-2-ene.
A: The double bond cleaves between C2 and C3. You get acetone (propan-2-one, from the more substituted side) and acetaldehyde (ethanal, from the less substituted side).
Q: A more substituted alkene reacts faster with HBr. Is this because the product is more stable?
A: No. It reacts faster because it forms a more stable carbocation intermediate. The rate depends on the energy of the transition state leading to that intermediate, not on product stability.
This connects to stereochemistry because many addition reactions create new stereocentres, and whether you get syn or anti addition determines the relative configuration of those centres. It also connects to alcohols and ethers: the products of hydration and hydroboration-oxidation are alcohols, which feed into substitution, elimination and oxidation reactions later in the course. Carbocation rearrangements here are the same ones you will see again in SN1 reactions and Friedel-Crafts alkylation.
Electrophilic addition, alkene reactions, Markovnikov rule, anti-Markovnikov, regioselectivity, stereoselectivity, syn addition, anti addition, carbocation intermediate, bromonium ion, chloronium ion, halonium ion, halohydrin, bromohydrin, chlorohydrin, hydroboration, oxymercuration, acid-catalysed hydration, ozonolysis, reductive ozonolysis, 1,2-hydride shift, 1,2-methyl shift, carbocation rearrangement, ring expansion, alkyl halide, vicinal dihalide, organic chemistry practice problems