Markovnikov / anti-Markovnikov — OChem Notes

Add HBr to propene, and the bromine could in principle end up on either carbon of the double bond. In practice nearly all of it lands on the middle carbon. Addition reactions showed why. This page restates that reason on propene, follows it into cases where it still holds, and then shows two ways to put the new group on the other carbon instead. Together, these reactions let you choose which product you make.

First, the word: regiochemistry

Propene has two different alkene carbons, so adding HBr could give 2-bromopropane or 1-bromopropane. Which one forms is a question of regiochemistry, which Addition reactions introduced: which atom a reagent ends up bonded to when more than one position is open. A reaction that strongly prefers one position is regioselective.

Regiochemistry is only about position. How the new groups are arranged in space is a separate question, called stereochemistry. A reaction can control one without the other. HBr addition is highly regioselective but does nothing to control stereochemistry. Bromination is the reverse: Br₂ puts one bromine on each alkene carbon, so position is never in question, but the two add anti, on opposite faces, as shown in Addition reactions.

Markovnikov's rule, restated mechanistically

Go back to propene and HBr. In the first step, the pi bond takes the proton from H–Br and bromide leaves. The proton can bond to either alkene carbon, and whichever carbon does not get it is left with the positive charge.

So the two choices give two different cations. If the proton goes to C1, the CH₂ end, the charge sits on C2, which has two alkyl groups beside it: a secondary (2°) cation. If the proton goes to C2, the charge sits on C1, which has only one alkyl group beside it: a primary (1°) cation. By the 3° > 2° > 1° > methyl order from Carbocations, the secondary cation is the more stable. The reaction goes through it, bromide bonds to C2, and the product is 2-bromopropane.

The proton can land on either end of the double bond. The highlighted group is the carbon that took it, and the other carbon is left with the positive charge.

Addition reactions named this preference Markovnikov's rule. Its classic wording is "the hydrogen adds to the alkene carbon that already has more hydrogens." For propene it gives the right answer, but it does not say why. The reason is that the proton adds where it leaves the more stable carbocation. The nucleophile then bonds where the cation is, so the halogen or OH ends up on the more substituted carbon.

The more stable cation wins because it forms faster. Protonation is the slow step, so the product comes from whichever route has the lower barrier. Making a carbocation is uphill, so by the Hammond postulate from Energy diagrams & the Hammond postulate the transition state looks like the cation it leads to. Whatever stabilizes the cation stabilizes that transition state too. Markovnikov selectivity is therefore decided at the transition state of the first step, not by how stable the final product is.

Styrene, PhCH=CH₂, is a benzene ring carrying a CH=CH₂ group. Ph, short for phenyl, means a benzene ring used as a substituent. HBr puts its proton on the CH₂ end of styrene. The cation left behind is benzylic: its charged carbon sits next to the ring, and resonance spreads the charge onto three ring carbons, as Carbocations showed. Bromide bonds to that carbon, next to the ring. Here the old wording happens to agree, but the reason is resonance: a cation stabilized by resonance beats one stabilized only by alkyl groups.

The cation argument also settles cases the "more hydrogens" wording cannot. In 1-phenylpropene, PhCH=CHCH₃, each alkene carbon carries one hydrogen, so "more hydrogens" cannot choose. The cation argument can. A proton on the carbon away from the ring leaves a benzylic cation, while a proton on the carbon next to the ring leaves an ordinary secondary cation. The benzylic cation wins, so the bromine again bonds to the carbon next to the ring.

The highlighted CH₃ is the carbon that took the proton, and δ+ marks the carbons that share the charge. Right: each alkene carbon has one H, but only one choice gives a benzylic cation.

Hydroboration–oxidation: a different mechanism

Start again with propene, and compare two ways to add water across it. Aqueous acid, H₃O⁺, gives propan-2-ol, with the OH on the middle carbon, as in Addition reactions. Treating propene first with borane, BH₃, and then with hydrogen peroxide and sodium hydroxide gives propan-1-ol, with the OH on the end carbon. Both routes add H and OH, but the OH lands on opposite ends. This second route is hydroboration–oxidation, and its result is called anti-Markovnikov: the OH goes to the less substituted carbon.

Follow the OH: on the middle carbon at the left, on the end carbon at the right.

The difference comes from the mechanism. BH₃ is a Lewis acid: boron has only six valence electrons and an empty p orbital. In the lab it is used as a complex with THF (tetrahydrofuran, a cyclic ether used as the solvent), whose oxygen lends a lone pair to boron's empty orbital, or with dimethyl sulfide, (CH₃)₂S. BH₃ adds to the alkene in one concerted step, meaning every bond change happens at once, with no intermediate. Boron bonds to one alkene carbon while one of its hydrogens moves to the other, and no carbocation ever forms.

With no cation, cation stability cannot set the regiochemistry. Three effects decide it instead, and all three come from one picture, the four-center transition state, in which boron, hydrogen and the two alkene carbons are joined in one four-membered ring.

Dashed lines are bonds partly made or partly broken. δ− marks the electron density boron is gaining from the pi bond.

In that transition state, the pi electrons start flowing into boron's empty orbital before the new bonds are complete. The flow is uneven. It leaves a partial positive charge on the alkene carbon further from boron, and a partial positive charge, like a full one, is best carried by the more substituted carbon. So boron bonds to the less substituted carbon.

The other two effects agree with it. The less substituted carbon is also the less crowded one, so the bulky boron group fits there more easily. And hydrogen is slightly more electronegative than boron, so the B–H hydrogen carries a partial negative charge and moves to the carbon with the partial positive charge. All three point boron to the less substituted carbon.

The four-membered ring also fixes the stereochemistry. Boron and hydrogen are still bonded to each other as they bond to the alkene, so they must arrive on the same face. This is syn addition, the opposite of the anti addition of bromine. This is not a preference; it follows from the closed ring.

The second step, with H₂O₂ and NaOH, replaces the C–B bond with a C–OH bond. The OH takes exactly the place boron held, on the same carbon and on the same face, which chemists call retention of configuration. The net result is OH on the less substituted carbon, added syn with the new H.

Hydroboration–oxidation never rearranges, because it never makes a carbocation to shift. That makes it useful in three ways. It gives the anti-Markovnikov alcohol. It gives an unrearranged alcohol from an alkene whose cation would shift. And on a ring, its syn addition puts both new groups on the same face, so the reaction gives one predictable diastereomer.

Worked example — one alkene, two alcohols

Take 1-methylcyclohexene. C1 carries the methyl group, and C2, its partner in the double bond, carries one hydrogen. Run each hydration.

H₃O⁺: the proton goes to C2, which leaves a tertiary cation at C1. Water bonds there. The product is 1-methylcyclohexan-1-ol, the Markovnikov alcohol, with the OH on the more substituted carbon.

BH₃, then H₂O₂/NaOH: boron bonds to C2, the less substituted carbon, and hydrogen bonds to C1, both from the same face. Oxidation then puts the OH exactly where boron was. The product is 2-methylcyclohexan-1-ol. Its numbers differ from the alkene's because in the alcohol's name the OH carbon becomes C1.

Now read the stereochemistry carefully, because this is where students lose the mark. "Syn" describes the two groups that were added: the new H on the methyl-bearing carbon and the new OH on its neighbor sit on the same face. The methyl was there before the reaction and is not part of that relationship. The new H took the front face of the methyl-bearing carbon, so the methyl must sit on the back. The OH (front) and the methyl (back) are therefore on opposite faces, and the product is trans-2-methylcyclohexan-1-ol.

So syn addition does not mean a cis product. It means the two new groups are cis to each other, and whether the product is named cis or trans depends on what the ring already carried.

The trans alcohol is chiral, and it forms as a racemic mixture, because borane attacks the top and bottom faces of the flat alkene equally often. One starting material has given two different positions, and only one route controls which face the groups use.

Left, the Markovnikov alcohol. Right, the anti-Markovnikov alcohol, where the highlighted H and OH are the two groups added.

Radical HBr addition: anti-Markovnikov by another route

Add HBr to propene with a trace of a peroxide present, and the product changes from 2-bromopropane to 1-bromopropane. A peroxide, ROOR, is a compound with a weak O–O bond, which breaks to give two RO· radicals. These switch HBr addition from the ionic route to a radical chain, the same kind of chain as in Radical halogenation. The table lays it out in the same three stages, the form exams ask you to reproduce. Every arrow in the mechanism is a fishhook, a single-barbed arrow that moves one electron.

Stage	Step	What it does
Initiation	RO–OR → 2 RO·	Gentle heat or light breaks the weak O–O bond. This happens only now and then; each event starts a long chain.
Initiation	RO· + H–Br → RO–H + Br·	Makes the chain carrier, Br·, so the cycle can start.
Propagation 1	Br· + C=C → ·C–C–Br	Bromine adds to the less substituted carbon, leaving the more stable radical.
Propagation 2	·C–C–Br + H–Br → H–C–C–Br + Br·	The radical takes a hydrogen and releases a new Br·, which starts propagation 1 again.
Termination	Br· + Br·, or two carbon radicals combining	Two radicals meet and the chain stops. This is rare, because radicals are scarce.

The two propagation steps are the reaction. Together they use up one alkene and one HBr and give back the bromine radical, so a single initiation can convert thousands of molecules. Notice which species reaches the alkene first: the bromine radical, not H⁺. That one difference explains the change in regiochemistry.

Every arrow has one barb. Follow the Br· made at the end of propagation 2: it is the one that starts propagation 1 again.

The bromine radical adds to whichever carbon leaves the more stable radical behind. Radical stability follows the same 3° > 2° > 1° order as carbocation stability, for the same hyperconjugation reason. With propene, bromine adds to C1, the CH₂ end, and leaves a secondary radical on C2. That radical then takes a hydrogen from another HBr.

The net result is Br on the less substituted carbon and H on the more substituted one, the reverse of ionic HBr addition. The carbon radical is nearly flat, so H–Br can reach either face of it, and a single new stereocenter formed from an achiral alkene comes out racemic.

This peroxide effect works only with HBr. HCl and HI ignore peroxides and add by the ordinary ionic Markovnikov route. The reason lies in the speed of the two propagation steps. The ionic addition is always running alongside the chain, so the chain gives the product only if both of its steps are fast. An uphill step is slow. In alkane bromination (Radical halogenation), a +11 kcal/mol step still works: it slows the chain, but nothing else competes for the reactants. Here the ionic route competes, and a chain with an uphill step loses to it.

You can estimate each step's ΔH from bond strengths, as the energy of the bonds broken minus the energy of the bonds made, with the same arithmetic as for alkane halogenation. Propagation 1 breaks the pi bond (about 65 kcal/mol) and makes a C–X bond. Propagation 2 breaks the H–X bond and makes a C–H bond (about 98 kcal/mol). The table runs the numbers for each acid.

Acid	Propagation 1: X· adds	Propagation 2: H taken from H–X	Does the chain win?
HBr	C–Br about 70, so ΔH ≈ −5	H–Br about 87, so ΔH ≈ −11	Yes: both steps are mildly downhill and fast
HCl	C–Cl about 84, so ΔH ≈ −19	H–Cl about 103, so ΔH ≈ +5	No: propagation 2 is uphill and slow, so the ionic route wins
HI	C–I about 56, so ΔH ≈ +9	H–I about 71, so ΔH ≈ −27	No: propagation 1 is uphill and slow, so the ionic route wins

Exams like to swap the two failures. Down the group, both the C–X bond and the H–X bond get weaker. Iodine bonds too weakly to carbon, so its radical cannot add to the alkene fast enough. Chlorine holds its own hydrogen too tightly, so the carbon radical cannot take it fast enough.

The five routes side by side

The table collects the five ways this chapter adds H–Br or H–OH to an alkene. "Either face" means that a single new stereocenter formed from an achiral alkene comes out racemic. Hydroboration is racemic too, but its H and OH always share a face.

Conditions	Intermediate	Regiochemistry	Faces used
HBr	carbocation	Markovnikov	either face
HBr, ROOR	radical	anti-Markovnikov	either face
H₃O⁺	carbocation	Markovnikov (OH)	either face
Hg(OAc)₂ / NaBH₄	mercurinium bridge	Markovnikov (OH)	either face
BH₃ / H₂O₂	none (concerted)	anti-Markovnikov (OH)	syn (same face)

What carries forward

Read the table as two groups. Three routes run through a cation, or the cation-like mercurinium bridge, and put the halogen or OH on the more substituted carbon. Two avoid a cation, one through a radical and one with no intermediate at all, and put it on the less substituted carbon. Choosing a reagent means choosing one of these two groups. If the substrate's cation would shift, it also means choosing whether a cation may form at all.

The idea that changing the pathway changes the product comes back later in the course. Kinetic and thermodynamic enolates in Enolate Chemistry, 1,2- versus 1,4-addition to dienes in Conjugation & Pericyclic Reactions, and ortho/para versus meta directors in Aromatic Chemistry all give different products from one starting material. Whenever two sets of conditions do that, ask what differs along the way.