Markovnikov's Rule

What the rule says

Take propene, CH₃-CH=CH₂, and add HBr. There are two carbons in the double bond — call them C1 (the terminal CH₂) and C2 (the middle CH bearing a methyl). Either H or Br could end up on either side. Markovnikov's rule predicts the answer in one line:

The H attaches to the carbon of the double bond that already bears more hydrogens.

So propene + HBr gives 2-bromopropane (CH₃-CHBr-CH₃), not 1-bromopropane. The H joined C1 (which already had two H's), and Br joined C2 (which had only one). Markovnikov phrased it that way in 1870 from product ratios alone, with no theory of mechanism. The modern statement points to the underlying reason: HX addition proceeds through a carbocation intermediate, and the more substituted carbocation is more stable, so it forms faster.

The mechanism, in two arrows

Electrophilic addition of HBr to propene runs in two distinct steps:

1. Protonation. The π electrons of the C=C double bond attack H⁺ (from HBr). One arrow goes from the alkene to the proton; a second arrow goes from the H-Br bond to Br. The result is a carbocation on one carbon and Br⁻ floating free.
2. Nucleophilic capture. Br⁻ collapses onto the carbocation. One arrow from a Br⁻ lone pair to the empty orbital. Done.

    H₃C-CH=CH₂  +  H-Br
         |
         | step 1: protonate
         v
    H₃C-CH⁺-CH₃        (2° cation, lower energy)
         vs.
    H₃C-CH₂-CH₂⁺       (1° cation, higher energy — not formed)
         |
         | step 2: Br⁻ captures
         v
    H₃C-CHBr-CH₃        2-bromopropane (Markovnikov)

Both protonation pathways are conceivable, but the rate-determining step is the formation of the cation, and the activation energy for the secondary cation is roughly 12-15 kcal/mol lower than the primary one. The Boltzmann factor on a 13 kcal/mol gap at 25 °C is about 10⁹ — so essentially nothing goes through the primary pathway.

Why the more substituted cation wins

Two effects combine to stabilize a carbocation as you add alkyl substituents:

• Hyperconjugation. Adjacent C-H σ bonds donate density into the empty p orbital. A tertiary cation has up to nine C-H bonds participating; a primary one has at most two. Each interaction is small (≈2-3 kcal/mol) but they add up.
• Inductive donation. Alkyl groups are weakly electron-donating relative to H, so they spread the positive charge over more atoms.

The net stability ordering from gas-phase hydride affinities is roughly:

Cation	Hydride affinity (kcal/mol)	Relative stability
CH₃⁺ (methyl)	314	Worst
CH₃CH₂⁺ (1°)	273	+41 vs methyl
(CH₃)₂CH⁺ (2°)	250	+23 vs 1°
(CH₃)₃C⁺ (3°)	233	+17 vs 2°
Allyl / benzyl	~233	≈ 3° (resonance)

An allylic or benzylic position rivals a tertiary one because resonance delocalizes the charge onto a neighboring π system. This explains why styrene (Ph-CH=CH₂) plus HBr gives PhCHBrCH₃ exclusively — the benzylic cation is dramatically favored over the primary alternative.

Worked example: HBr + propene

Predict the major product of HBr addition to propene at 25 °C in the absence of peroxides.

1. Identify the alkene carbons. C1 = CH₂ (two H's), C2 = CH (one H, one CH₃).
2. Apply Markovnikov. H goes to C1 (more H's already), Br goes to C2.
3. Or equivalently, draw both possible cations:
   • Cation at C1 = primary (CH₃-CH₂-CH₂⁺) — disfavored.
   • Cation at C2 = secondary ((CH₃)₂CH⁺) — favored.
4. Br⁻ adds to the favored cation. Major product: 2-bromopropane, CH₃-CHBr-CH₃.

Experimental ratio: roughly 95:5 in favor of 2-bromopropane in non-polar solvents at room temperature. The 5% minor product is partly genuine attack on the primary cation, partly trace radical chemistry from dissolved oxygen.

Markovnikov vs anti-Markovnikov vs hydroboration

	Markovnikov (HX, ionic)	Anti-Markovnikov (HBr / ROOR, radical)	Hydroboration-oxidation
Reagents	HCl, HBr, HI, H₂O/H⁺	HBr + ROOR (peroxide initiator)	BH₃·THF, then H₂O₂ / NaOH
Mechanism	Two-step ionic (cation)	Radical chain	Concerted (4-center TS)
Where does X go?	More substituted carbon	Less substituted carbon	Less substituted carbon (OH)
Stereochemistry	Mixture (planar cation)	Mixture	syn addition, single face
Rearrangements?	Yes — hydride/methyl shifts common	No (radicals don't rearrange easily)	No
Works for	Any HX	HBr only (HCl, HI fail energetically)	Most alkenes
Typical use	Direct hydrohalogenation	Reversed regiochemistry on demand	Anti-Markovnikov alcohols

The peroxide effect is purely a kinetic switch: peroxides homolyze to give RO·, which abstracts H from HBr to give Br·. Br· then adds to the alkene at the less hindered (terminal) carbon, generating the more stable secondary or tertiary carbon radical. That radical then abstracts H from another HBr, propagating the chain. HCl and HI don't do this efficiently — the HCl bond is too strong to abstract, and HI's I-H bond is so weak that addition is unfavorable in the chain step.

Pitfall: carbocation rearrangements

The two-step Markovnikov mechanism has an Achilles heel: an exposed carbocation is free to rearrange before Br⁻ catches up. A 1,2-hydride shift or 1,2-methyl shift can convert a secondary cation into a tertiary one in essentially zero activation energy if the geometry is right.

The classic textbook trap is 3-methyl-1-butene + HBr:

    (CH₃)₂CH-CH=CH₂  +  HBr
            |
            | protonate (Markovnikov: H to terminal CH₂)
            v
    (CH₃)₂CH-CH⁺-CH₃    (2° cation)
            |
            | 1,2-hydride shift — adjacent C is 3°, more stable
            v
    (CH₃)₂C⁺-CH₂-CH₃    (3° cation)
            |
            | Br⁻ captures
            v
    (CH₃)₂CBr-CH₂-CH₃   2-bromo-2-methylbutane (major)

You expected 2-bromo-3-methylbutane (Markovnikov from the original alkene). You got 2-bromo-2-methylbutane (Markovnikov, then a hydride shift). Always check the cation you draw — if a hydride or methyl shift one bond away gives a more stable cation, expect rearrangement. This is one of the most common exam gotchas in undergraduate organic chemistry.

Real-world relevance

• Petrochemical hydrohalogenation. Industrial production of isopropyl chloride from propene + HCl runs at 30-60 °C with a metal-halide catalyst; the Markovnikov product is the desired one for downstream isopropanol manufacture.
• Acid-catalyzed hydration of isobutylene. (CH₃)₂C=CH₂ + H₂O / H₂SO₄ gives tert-butanol cleanly; the tertiary cation is so stable that essentially no anti-Markovnikov product forms. This is the route used industrially for tert-butanol and, downstream, MTBE.
• Anti-Markovnikov hydration in synthesis. When a chemist needs a primary alcohol from a terminal alkene, hydroboration-oxidation is the textbook answer — pioneered by H. C. Brown (1979 Nobel Prize, in part for this chemistry).
• Drug synthesis. The propranolol synthesis exploits Markovnikov-style epoxide opening to install the secondary alcohol on the correct carbon. Get the regiochemistry wrong and the resulting molecule is biologically inactive.

Variants and special cases

• Halogenation (Br₂, Cl₂). Not a Markovnikov reaction in the usual sense — it goes through a bridged halonium ion, not a free cation. Both carbons get a halogen and the product is anti-stereospecific.
• Hydration via oxymercuration-demercuration. Hg(OAc)₂ / H₂O, then NaBH₄. Markovnikov regiochemistry, but no rearrangements (the mercurinium ion behaves like a halonium ion). Useful when you want Markovnikov hydration without carbocation surprises.
• HF addition. Adds Markovnikov but slowly; HF is a weak acid (pKa ≈ 3) so protonation is rate-limiting.
• Vinyl halides. Adding HBr to CH₂=CHCl follows Markovnikov, with the new Br ending up on the carbon that already has the Cl — the geminal dihalide is favored because the Cl stabilizes the adjacent cation by resonance.
• Allenes (C=C=C). Markovnikov-style protonation gives an allyl cation; the regiochemistry depends on which face protonates and is rarely clean.

Common pitfalls

• Forgetting the peroxide effect. If HBr is run with ROOR, AIBN, or under UV light, the regiochemistry flips. Standard textbook problems specify "HBr, peroxides" or "HBr, ROOR" to flag the radical mechanism.
• Ignoring rearrangements. Always check the cation you're drawing. If a hydride or methyl shift gives a more stable cation, the rearranged product dominates.
• Confusing it with Zaitsev's rule. Markovnikov is for addition. Zaitsev is for elimination. They are not the same rule despite both rewarding "more substituted."
• Assuming cleanliness with Br₂ or Cl₂. Halogens add anti via a halonium intermediate, not via a free cation. Markovnikov regiochemistry is irrelevant — both halogens end up on the alkene.
• Forgetting solvent participation. In water or alcohol solvents, the cation may be captured by solvent before the halide gets there, giving a halohydrin or ether mixture. Use non-nucleophilic solvents (CH₂Cl₂, hexanes) for clean HX addition.