Markovnikov's rule and alkene addition reactions
Markovnikov's rule is a consequence, not a law: the electrophile adds so that the more stable carbocation forms, which puts the proton on the carbon that already has more hydrogens. It flips to anti-Markovnikov in exactly two situations — HBr with peroxides (a radical mechanism, no cation) and hydroboration–oxidation (a concerted addition where sterics, not charge, decide).
State the rule properly, then forget the statement
The textbook version — "the rich get richer", the hydrogen adds to the carbon with more hydrogens — is a memory aid for a result, and memory aids fail on unfamiliar reagents. The mechanism behind it never fails.
When H–X adds to an alkene, the π electrons attack the proton first. The proton can land on either alkene carbon, and whichever carbon does not get it becomes a carbocation. So the question is not "where does the H go" but "which carbocation would I rather have" — and the answer is always the more substituted one, because alkyl groups donate electron density into the empty p orbital and hyperconjugation stabilises the charge.
Tertiary beats secondary beats primary, and a resonance-stabilised allylic or benzylic cation beats all of them. Work out which cation is more stable, and the regiochemistry follows automatically.
Propene + HBr, worked
- π electrons attack the proton of HBr. Two choices.
- Proton on C-1 (the CH₂ end) leaves a secondary cation at C-2. Proton on C-2 would leave a primary cation at C-1.
- The secondary cation is far more stable, so that pathway dominates.
- Bromide captures the cation at C-2.
Product: 2-bromopropane. The hydrogen ended up on the carbon that already had two — Markovnikov — but you did not need the rule to get there.
When the rule flips
1. HBr with peroxides (radical addition)
Add a peroxide and the mechanism changes completely. The peroxide homolyses, abstracts H from HBr to give a bromine radical, and that radical — not a proton — adds to the alkene first. Bromine adds to the less substituted carbon because that leaves the more stable, more substituted radical. Radical stability follows the same tertiary > secondary > primary order as carbocations, but the species adding first is different, so the product is reversed.
Propene + HBr/ROOR gives 1-bromopropane. Note that this only works for HBr — not HCl or HI, where the chain propagation steps are energetically unfavourable.
2. Hydroboration–oxidation
BH₃ adds across the double bond in a single concerted step, with boron and hydrogen delivered to the same face at the same time. There is no carbocation, so carbocation stability is irrelevant. Boron — the bigger group — goes to the less hindered, less substituted carbon. Then H₂O₂/NaOH replaces boron with OH with retention.
The overall result is anti-Markovnikov hydration, syn stereochemistry, and no rearrangement. That last point is why hydroboration is the standard answer when a question wants the terminal alcohol from a terminal alkene.
The reagent table
| Reagents | Adds | Regiochemistry | Stereochemistry | Rearranges? |
|---|---|---|---|---|
| HX (HCl, HBr, HI) | H and X | Markovnikov | Not controlled | Yes |
| HBr + peroxides | H and Br | Anti-Markovnikov | Not controlled | No |
| H₂O, H₂SO₄ | H and OH | Markovnikov | Not controlled | Yes |
| 1. Hg(OAc)₂, H₂O 2. NaBH₄ | H and OH | Markovnikov | Anti across the addition step; the NaBH₄ demercuration is not stereospecific | No |
| 1. BH₃·THF 2. H₂O₂, NaOH | H and OH | Anti-Markovnikov | Syn | No |
| X₂ (Br₂, Cl₂) | Two halogens | n/a | Anti (halonium ion) | No |
| X₂ in H₂O | X and OH | OH to the more substituted carbon | Anti | No |
| H₂, Pd/C | Two hydrogens | n/a | Syn | No |
| mCPBA, then H₃O⁺ | Two OH | n/a | Anti | No |
| OsO₄ or cold dilute KMnO₄ | Two OH | n/a | Syn | No |
| 1. O₃ 2. Zn or DMS | Cleaves C=C | n/a | n/a | No |
Read the table as three questions, not thirty facts: does a carbocation form (rearrangement possible, stereochemistry uncontrolled), or a three-membered bridged ion (anti addition), or a concerted delivery from one face (syn addition)?
Bridged intermediates and why anti addition happens
When Br₂ approaches an alkene, the first-formed bromine does not sit on one carbon — it bridges both, forming a cyclic bromonium ion. That bridge blocks the face it sits on, so the incoming bromide must attack from the opposite side. Hence anti stereochemistry, reliably, with no exceptions worth worrying about at this level.
The same logic explains halohydrin formation. Water attacks the bromonium ion from the back, and it attacks the more substituted carbon, because that carbon carries more positive character in the unsymmetrical bridged ion. So OH ends up on the more substituted carbon and Br on the less substituted one, anti to each other.
Watch for rearrangement
Any mechanism that forms a free carbocation can rearrange. 3-methylbut-1-ene plus HCl gives a secondary cation at C-2 with a tertiary carbon next door; a hydride shift converts it to the tertiary cation, and the major product is 2-chloro-2-methylbutane, not the 2-chloro-3-methylbutane you might have drawn. If a question's carbon skeleton makes a shift attractive, it is testing exactly that.
Oxymercuration and hydroboration both exist as the rearrangement-free alternatives to acid-catalysed hydration — one Markovnikov, one anti-Markovnikov.
How Organic Chemistry AI helps here
Addition questions are almost always about a single decision — which intermediate forms — and a solve makes that decision explicit before it states the product, so you can check the intermediate rather than just the answer. Type the reaction with the Addition topic hint, or photograph the scheme. The offline flashcard decks include alkene additions and their stereochemical outcomes if you would rather drill the table than look it up.
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Frequently asked
Why do only peroxides with HBr give anti-Markovnikov addition?
Because the radical chain only propagates favourably for bromine. For HCl the hydrogen-abstraction step is too endothermic, and for HI the halogen-addition step is, so neither sustains a chain. Peroxides plus HCl or HI simply give the ordinary Markovnikov product.
How do I remember which additions are syn and which are anti?
Ask what the intermediate is. A three-membered bridged ring — bromonium, mercurinium, epoxide — blocks one face, so the second group must come from the other side: anti. A reagent delivered from one face in a single step — H₂ on a metal surface, BH₃, OsO₄ — gives syn. Free carbocations control neither.
When should I expect a carbocation rearrangement?
Whenever the mechanism makes a free carbocation and a single hydride or alkyl shift would produce a more stable one — typically a secondary cation with a tertiary carbon adjacent. HX addition and acid-catalysed hydration both go through free cations. Oxymercuration and hydroboration do not.
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