Alkene Addition Reactions: Markovnikov and Anti-Markovnikov Worked Examples

Alkene addition is one of the highest-frequency topic areas on orgo exams. Hydrohalogenation, hydration, halogenation, hydroboration-oxidation, and epoxidation show up on every first-semester organic chemistry exam because they test multiple skills at once: carbocation stability, regioselectivity, stereochemistry, and mechanism drawing. Mastering alkene addition builds the foundation for E/Z/R/S logic, SN1/SN2/E1/E2 reasoning, and later aromatic electrophilic substitution.

The unifying principle: an alkene has a π bond that acts as a nucleophile. Electrophiles attack the π bond, producing a carbocation intermediate (or a cyclic bromonium ion for halogen additions, or a concerted transition state for hydroboration). The carbocation — or its analog — determines the product selectivity.

Markovnikov's rule, in its classical form, states: when HX adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already on it, and X attaches to the carbon with fewer hydrogens. Modern restatement: the electrophile (H+) adds to produce the more stable carbocation intermediate.

Carbocation stability ranking: 3° > 2° > 1° > methyl. Tertiary cations are most stable because alkyl groups donate electron density through hyperconjugation and inductive effect. When HX adds to an unsymmetrical alkene, protonation occurs at the position that generates the more stable carbocation.

Example: 2-methyl-2-butene + HCl - Two possible cations: tertiary (at C2) or secondary (at C3) - Tertiary cation is more stable — H adds to C3, Cl adds to C2 - Product: 2-chloro-2-methylbutane (the Markovnikov product)

The mechanism is two-step: 1. H+ adds to the less-substituted carbon, forming the more stable carbocation 2. Cl- attacks the carbocation, forming the C-Cl bond

Rearrangements: if the initial carbocation can rearrange to a more stable one (via hydride or methyl shift), it often does. This is a common exam trap. For example, (CH3)2CHCH=CH2 + HBr produces the rearranged tertiary carbocation via hydride shift, giving 2-bromo-2-methylbutane as the major product rather than the expected 2-bromo-3-methylbutane.

When you see a secondary alkene carbocation next to a tertiary carbon, check for rearrangement.

Problem 1: Predict the product of 1-methylcyclohexene + HBr.

Step 1: Identify alkene carbons and possible cation locations. - Alkene: between C1 (bearing methyl) and C2 of cyclohexene - Cation at C1 (tertiary, more substituted): bearing methyl + 2 ring carbons → tertiary - Cation at C2 (secondary): bearing 2 ring carbons only → secondary

Step 2: Markovnikov → H+ adds to C2, generating the tertiary cation at C1.

Step 3: Br- attacks the tertiary cation at C1.

Product: 1-bromo-1-methylcyclohexane.

Note on stereochemistry: since addition proceeds through a planar carbocation, both syn and anti additions are possible. The product is a racemic mixture if a new stereocenter forms. In this problem, C1 becomes a stereocenter (bearing CH3, Br, and two different ring substituents), so the product is racemic.

Problem 2: Predict the product of 3,3-dimethyl-1-butene + HCl.

Step 1: Identify possible cations. - CH2=CH-C(CH3)3 + H+ → two possibilities - Cation at C1 (primary): highly unstable - Cation at C2 (secondary next to quaternary C): secondary, potentially rearrangeable

Step 2: The secondary cation at C2 can undergo a 1,2-methyl shift from the adjacent quaternary carbon to give a tertiary cation at C3.

Step 3: Cl- attacks the rearranged tertiary cation.

Product (major): 2-chloro-2,3-dimethylbutane (rearranged) Product (minor): 3-chloro-2,2-dimethylbutane (unrearranged)

This is a classic rearrangement exam problem. Always check for cation rearrangement possibilities when the initial cation is secondary or primary and a tertiary cation can form via 1,2-hydride or 1,2-methyl shift.

Acid-catalyzed hydration adds water across the alkene using H2SO4 or H3PO4 as the acid catalyst. The overall transformation: alkene + H2O → alcohol. Regioselectivity follows Markovnikov's rule — the OH ends up on the more substituted carbon.

Mechanism (3 steps): 1. H+ adds to the less-substituted alkene carbon, forming the more stable carbocation. 2. Water (as nucleophile) attacks the carbocation, forming an oxonium ion (-OH2+). 3. Deprotonation of the oxonium by base (water or HSO4-) gives the final alcohol.

Worked example: 2-methyl-2-butene + H2O / H2SO4.

Step 1: Protonation at C3 (less substituted carbon) gives tertiary cation at C2. Step 2: Water attacks C2, forming -OH2+. Step 3: Deprotonation gives 2-methyl-2-butanol.

Product: 2-methyl-2-butanol (Markovnikov alcohol).

Key features: - Carbocation intermediate can rearrange, same as hydrohalogenation - No stereospecificity — racemic if a new stereocenter forms - Reversible — high water concentration drives forward; equilibrium governs yield - Generally produces the most substituted alcohol

Acid-catalyzed hydration is practical for synthesizing tertiary alcohols from the corresponding alkenes. For secondary and primary alcohols, other methods (like hydroboration-oxidation) often give cleaner results.

Hydroboration-oxidation adds water across an alkene with anti-Markovnikov regiochemistry — the OH ends up on the less-substituted carbon. This is a two-step reaction:

Step 1 (hydroboration): BH3 (or BH3-THF) adds across the alkene in a concerted, syn fashion. Boron bonds to the less-substituted carbon; H bonds to the more-substituted carbon.

Step 2 (oxidation): treatment with H2O2 and NaOH replaces the C-B bond with C-OH, retaining configuration at the carbon.

Overall: alkene + H2O → anti-Markovnikov alcohol, with syn stereochemistry.

Why anti-Markovnikov? In the hydroboration step, boron is electrophilic but the transition state is concerted. Boron has partial positive character and prefers the less-substituted carbon (less steric hindrance). H, having partial negative character in this context, goes to the more-substituted carbon. This pattern reverses the regiochemistry compared to HX addition.

Worked example: 1-methylcyclohexene + BH3-THF, then H2O2/NaOH.

Step 1: B adds to C2 (less substituted), H adds to C1 (more substituted — bearing the methyl). Syn addition. Step 2: Oxidation replaces C-B with C-OH, retaining syn stereochemistry.

Product: trans-2-methylcyclohexanol (trans because H and OH came from the same face; syn addition means the methyl and OH are on adjacent carbons but on opposite faces when you account for the existing methyl).

Actually in this case the existing methyl is on C1, hydroboration adds H to C1 (no change — already had H) and B (later OH) to C2. Syn relative positioning of H (at C1, new) and B (at C2) is on the same face. After oxidation, the OH replaces B with retention. Net: OH and H (at C1) are cis; but the existing methyl at C1 forces the stereochemistry: the OH at C2 ends up trans to the methyl at C1 due to the geometry of syn addition.

Key features: - Anti-Markovnikov regiochemistry - Syn stereochemistry (both H and OH add to the same face) - No carbocation intermediate → no rearrangements possible - Concerted mechanism - Excellent for synthesizing primary and secondary alcohols

This is the go-to method when you need a specific anti-Markovnikov alcohol or need to avoid rearrangements.

Halogenation adds X2 (Br2 or Cl2) across the alkene with anti stereochemistry.

Mechanism: 1. Alkene π bond attacks Br2 (or Cl2), displacing Br- and forming a cyclic bromonium ion (3-membered ring with bridging Br+). 2. Br- attacks from the opposite face, opening the ring. Backside attack gives anti addition.

Product: vicinal dihalide with anti (trans) configuration.

Example: cyclopentene + Br2 → trans-1,2-dibromocyclopentane.

Anti stereochemistry is a characteristic feature. If the bromonium ion is symmetric, Br- attacks either carbon equally, giving racemic anti product. If the bromonium ion is unsymmetric (e.g., from unequal alkene substitution), Br- preferentially attacks the more substituted carbon — this matters for halohydrin regiochemistry.

Halohydrin formation: when X2 addition occurs in water (H2O as solvent), water opens the halonium ion instead of halide, giving a halohydrin (one X, one OH on adjacent carbons).

Mechanism: 1. Alkene attacks Br2 or Cl2, forming halonium. 2. Water attacks the more substituted carbon (where more positive charge is localized). 3. Deprotonation gives the halohydrin.

Result: Markovnikov regiochemistry (OH on more substituted carbon, X on less substituted), anti stereochemistry.

Example: propene + Br2/H2O → 1-bromo-2-propanol. - Br bonds to C1 (less substituted) - OH bonds to C2 (more substituted) - Anti configuration

This is a useful orgo synthesis for making halohydrins, which are often precursors to epoxides via intramolecular SN2.

Epoxidation with meta-chloroperoxybenzoic acid (mCPBA) adds an oxygen atom across the alkene, forming a three-membered epoxide ring. The reaction is concerted and syn.

Mechanism: the oxygen of the peroxy acid transfers to the alkene face in a single step through a 'butterfly' transition state. Both new C-O bonds form simultaneously and on the same face.

Result: syn stereochemistry (the two new C-O bonds are on the same face of the original alkene).

Example: cis-2-butene + mCPBA → cis-2,3-epoxybutane (both C-O bonds on same face; the methyls retain their original cis relationship). Example: trans-2-butene + mCPBA → trans-2,3-epoxybutane (methyls retain trans).

Key features: - Syn addition: both C-O bonds on same face - Alkene geometry (cis or trans) preserved in product - No rearrangement possible (concerted mechanism) - Produces an epoxide, which is a useful reactive intermediate (opens with various nucleophiles under acidic or basic conditions)

Epoxide opening: once formed, epoxides can be opened by nucleophiles. Under acidic conditions, protonation activates the more substituted carbon for attack (Markovnikov-like). Under basic conditions, nucleophiles attack the less substituted carbon (SN2-like). The stereochemistry of epoxide opening is always anti because of the SN2-like backside attack.

For synthesis problems, choose the reaction based on desired product:

Want Markovnikov alcohol (OH on more substituted carbon): - Acid-catalyzed hydration: H2O, H2SO4 or H3PO4 - Oxymercuration-demercuration: Hg(OAc)2/H2O, then NaBH4 (no rearrangement, unlike acid-catalyzed)

Want anti-Markovnikov alcohol (OH on less substituted carbon): - Hydroboration-oxidation: BH3-THF, then H2O2/NaOH - Syn stereochemistry

Want Markovnikov alkyl halide: - HX addition (HBr, HCl, HI) - Watch for carbocation rearrangement - Generally anti stereochemistry dominates, but since carbocation is planar, can get mixed stereochemistry

Want anti-Markovnikov alkyl halide: - HBr + peroxides (radical addition; ROOR as initiator) - Radical mechanism produces anti-Markovnikov product - Does NOT work for HCl or HI

Want vicinal dihalide (both X on adjacent carbons): - X2 addition (Br2 or Cl2) - Anti stereochemistry

Want halohydrin (one X, one OH): - X2/H2O - Markovnikov-type regiochemistry (OH on more substituted C) - Anti stereochemistry

Want epoxide: - mCPBA - Syn addition - Alkene geometry preserved

This table is exam gold. Memorize the pattern: regiochemistry, stereochemistry, rearrangement possibility. Then the specific reagents match reliably to specific transformations.