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Organic Chemistry I · Alkene reactions

Markovnikov vs anti-Markovnikov: practice problems for alkene additions

When something adds across an unsymmetrical C=C, the two pieces can land two ways round. Markovnikov's rule tells you which way, but only if you know why it works. Once you see the carbocation behind it, the anti-Markovnikov exceptions stop being a list to memorize and start making sense.

Jump to: the rule, properly · reagent table · worked examples · reagent sorter · 11 practice problems · FAQ

The rule, said properly

The version you were probably taught: the H goes on the carbon that already has more H's. That works for HX and H₂O, but it breaks the moment the electrophile isn't H. The version that always works:

Markovnikov addition: the electrophile adds to the alkene carbon that leaves the more stable carbocation on the other carbon, and the nucleophile ends up on that more substituted carbon.

Carbocation stability runs tertiary > secondary > primary, and a cation next to a benzene ring (benzylic) or a C=C (allylic) is extra stable because the charge is shared by resonance. The faster pathway goes through the more stable cation, so that's where the nucleophile bonds.

Markovnikov: H⁺ adds first

Propene + HBr. H⁺ can add to C1, giving a secondary cation on C2, or to C2, giving a primary cation on C1. The secondary cation forms much faster, so Br⁻ bonds to C2.

Product: 2-bromopropane.

Anti-Markovnikov: Br• adds first

Propene + HBr with peroxides (ROOR). Now a bromine radical adds first. It adds to C1 to leave the more stable secondary radical on C2, which then takes an H from HBr.

Product: 1-bromopropane.

Same principle both times, the more stable intermediate wins. It's just that in the radical reaction the piece that adds first is Br, not H, so the Br lands on the less substituted carbon. This only works for HBr. HCl and HI still add Markovnikov even if peroxides are present.

Hydroboration is the other big anti-Markovnikov reaction, for a different reason. B and H add in one step, with no cation. Boron, the bulky end, bonds to the less hindered, less substituted carbon, and oxidation then swaps the boron for OH in the same spot.

Every addition reagent at a glance

Regiochemistry, stereochemistry and rearrangements for alkene additions
ReagentAddsWhere the group goesFacesRearranges?
HCl, HBr, HIH, XMarkovnikov: X on the more substituted CBoth (no control)Yes
HBr, ROOR (heat or light)H, BrAnti-Markovnikov: Br on the less substituted CBoth (no control)No
H₂O, H₂SO₄ (cat.)H, OHMarkovnikovBothYes
1. Hg(OAc)₂, H₂O 2. NaBH₄H, OHMarkovnikovNot controlled overallNo
1. BH₃·THF 2. H₂O₂, NaOHH, OHAnti-Markovnikov: OH on the less substituted CSynNo
Br₂, H₂O (or NBS, H₂O)Br, OHOH on the more substituted C, Br on the otherAntiNo
H₂, Pd/C · Br₂ · OsO₄ · mCPBATwo identical ends (H/H, Br/Br, OH/OH, one O)No regiochemistry questionSyn, anti, syn, synNo

Two shortcuts help. If a reaction goes through a free carbocation (HX, acid-catalyzed hydration), expect Markovnikov and watch for a hydride or methyl shift. If it doesn't (oxymercuration, halohydrin formation through a bridged ion, hydroboration, radical HBr), there are no rearrangements.

Worked examples

Example 1: One alkene, two HBr conditions

2-Methylbut-2-ene, (CH₃)₂C=CHCH₃, with HBr alone, then with HBr and peroxides.

  1. Find the two alkene carbons. C2 carries two methyls (it's the more substituted end). C3 carries one H and one methyl.
  2. HBr alone. H⁺ adds to C3, leaving a tertiary cation on C2. Br⁻ bonds to C2: 2-bromo-2-methylbutane.
  3. HBr, ROOR. Br• adds to C3, leaving a tertiary radical on C2, which takes an H from HBr. Br ends up on C3: 2-bromo-3-methylbutane.

Peroxides flip where the Br goes, because they change which piece adds first.

Example 2: When the cation moves (and when it can't)

3-Methylbut-1-ene, CH₂=CH–CH(CH₃)₂, with HCl, and then with oxymercuration.

  1. HCl. H⁺ adds to C1, giving a secondary cation on C2. Next door, C3 is a CH with two methyls. Its H can shift over with its electrons (a 1,2-hydride shift), leaving a more stable tertiary cation on C3.
  2. Chloride traps both cations, so you get a mixture: 2-chloro-3-methylbutane (unrearranged) and 2-chloro-2-methylbutane (rearranged). Roughly half of each forms.
  3. 1. Hg(OAc)₂, H₂O 2. NaBH₄. The mercury forms a bridged ion, not a free cation, so nothing shifts. Water attacks the more substituted carbon (C2): 3-methylbutan-2-ol only.

Free carbocation, possible shift. Bridged ion, no shift. That's why oxymercuration is the Markovnikov hydration you can trust.

Example 3: Styrene and the benzylic cation

Styrene (vinylbenzene), C₆H₅–CH=CH₂, with HBr.

  1. H⁺ on the CH₂ end leaves a cation on the carbon attached to the ring. That cation is benzylic: the ring shares the positive charge by resonance.
  2. Br⁻ bonds there: (1-bromoethyl)benzene, as a racemic mix since the cation is flat.
  3. With peroxides, Br• adds to the CH₂ end to leave the benzylic radical, so you get (2-bromoethyl)benzene.

"More substituted" really means "more stable cation." Benzylic beats plain secondary.

Example 4: Halohydrin from 1-methylcyclopentene

1-Methylcyclopentene with Br₂ in water.

  1. Br₂ forms a three-membered bromonium ion across the double bond.
  2. The C–Br bond to the more substituted carbon (C1, with the methyl) is weaker, and C1 carries more of the positive charge, so water attacks C1, from the face opposite the bromine.
  3. Loss of H⁺ gives the alcohol. OH on C1, Br on C2, on opposite faces: trans-2-bromo-1-methylcyclopentan-1-ol (OH and Br trans, racemic).

Here neither added piece is H, which is why "H goes where the H's are" fails. Ask instead where the positive charge is better supported.

Want more alkene problems like these? OrgoDrill's alkenes chapter has every addition reaction, and review brings back the ones you miss until they're automatic.

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Reagent sorter

For each reagent added to an unsymmetrical alkene like propene, choose where the non-hydrogen piece (the Br, Cl, I or OH) ends up. Then press Check all. Watch the traps.

Markovnikov, anti-Markovnikov, or no regiochemistry question?

Show the answers
  • HBr: Markovnikov. H⁺ adds first to make the more stable cation, and Br⁻ bonds to the more substituted carbon.
  • HBr, ROOR (peroxides), heat: Anti-Markovnikov. With peroxides present the addition runs as a radical chain, and Br• adds first, to the less substituted carbon, so Br ends up there.
  • HCl, ROOR (peroxides): Markovnikov. The trap: the peroxide effect only works for HBr. HCl still adds through a cation, so Cl goes Markovnikov.
  • H₂O, H₂SO₄ (catalytic): Markovnikov. Acid-catalyzed hydration goes through a carbocation, so OH lands on the more substituted carbon.
  • 1. BH₃·THF 2. H₂O₂, NaOH: Anti-Markovnikov. Boron adds to the less substituted carbon, and oxidation puts OH right where the boron was.
  • 1. Hg(OAc)₂, H₂O 2. NaBH₄: Markovnikov. Water opens the mercurinium ion at the more substituted carbon, so OH is Markovnikov, with no rearrangements.
  • Br₂, H₂O: Markovnikov. Water attacks the bromonium ion at the more substituted carbon, so OH goes Markovnikov and Br goes to the other carbon.
  • H₂, Pd/C: No regiochemistry question. Both atoms added are H, so there is nothing to place. Only the faces (syn) matter.
  • Br₂ in CH₂Cl₂: No regiochemistry question. Both ends get a Br, so there is no regiochemistry question. The two Br atoms add anti.
  • HI: Markovnikov. HI adds through a cation like HCl and HBr, and peroxides wouldn't change that either.

Markovnikov practice problems

Pick an answer and press Check answer. Every wrong choice tells you what it would take to get that product.

Problem 1 · PropenePropene reacts with HBr (no peroxides). What is the major product?

Show solution

2-Bromopropane. The secondary cation on C2 forms much faster than the primary cation on C1, so bromide ends up on C2.

Problem 2 · PeroxidesPropene reacts with HBr in the presence of a peroxide (ROOR) and light. What is the major product?

Show solution

1-Bromopropane. RO• takes H from HBr to make Br•. Br• adds to the CH₂ end, leaving a secondary radical on C2. That radical takes H from HBr and regenerates Br•. Br is on C1, the anti-Markovnikov position.

Problem 3 · The HCl trapPropene reacts with HCl in the presence of a peroxide. What is the major product?

Show solution

2-Chloropropane. Only HBr adds anti-Markovnikov with peroxides. With HCl and HI, one of the radical chain steps is uphill in energy, so the chain doesn't run and the normal cation mechanism wins and the product is Markovnikov.

Problem 4 · Acid-catalyzed hydration1-Methylcyclohexene is stirred with water and a little H₂SO₄. What is the major product?

Show solution

1-Methylcyclohexan-1-ol. Protonating C2 leaves a tertiary cation on C1, which water traps. Loss of H⁺ from the oxonium ion gives the tertiary alcohol. C1 isn't a stereocenter here, so there's no stereochemistry to show.

Problem 5 · No rearrangement3,3-Dimethylbut-1-ene is treated with 1. Hg(OAc)₂, H₂O, then 2. NaBH₄. What is the product?

Show solution

3,3-Dimethylbutan-2-ol. The bridged mercurinium ion never becomes a free carbocation, so the methyl shift that plagues acid-catalyzed hydration can't happen. Water attacks C2, and NaBH₄ swaps the HgOAc on C1 for H.

Problem 6 · Hydroboration2-Methylbut-2-ene is treated with 1. BH₃·THF, then 2. H₂O₂, NaOH. What is the product?

Show solution

3-Methylbutan-2-ol (racemic). In (CH₃)₂C=CHCH₃, C2 has two methyls and C3 has one H. Boron adds to C3, H to C2, then oxidation replaces boron with OH. The product is (CH₃)₂CH–CH(OH)–CH₃.

Problem 7 · StyreneStyrene (C₆H₅CH=CH₂) reacts with HBr, no peroxides. What is the major product?

Show solution

(1-Bromoethyl)benzene, racemic. The benzylic cation is far more stable than a primary cation on the end carbon, so the H goes on CH₂ and Br on the carbon next to the ring.

Problem 8 · Halohydrin1-Methylcyclopentene reacts with Br₂ in water. What is the major product?

Show solution

trans-2-Bromo-1-methylcyclopentan-1-ol (OH and Br trans, racemic). Br₂ forms the bromonium ion. Water attacks C1, the more substituted carbon, from the side opposite Br, then loses H⁺. OH on C1, Br on C2, anti.

Problem 9 · Rearrangement3-Methylbut-1-ene reacts with HCl. Which statement about the products is right?

Show solution

Both, by a 1,2-hydride shift. H⁺ adds to C1 to give a secondary cation on C2. Chloride can trap it (2-chloro-3-methylbutane), or the H on C3 moves over with its electrons to make a tertiary cation on C3, which chloride traps as 2-chloro-2-methylbutane. Roughly equal amounts form.

Problem 10 · Pick the reagentWhich reagents turn hex-1-ene into hexan-1-ol?

Show solution

Hydroboration–oxidation. It's the one direct way to put OH on the less substituted carbon of an alkene. Both hydration methods give the Markovnikov alcohol, hexan-2-ol.

Problem 11 · Alkynes tooHex-1-yne is treated with H₂O, H₂SO₄ and HgSO₄. What is the product?

Show solution

Hexan-2-one. Markovnikov applies to alkynes too: the OH goes on the more substituted carbon (C2). That gives an enol, CH₂=C(OH)–C₄H₉, which tautomerizes to the methyl ketone.

Frequently asked questions

What is Markovnikov's rule in simple terms?

When an unsymmetrical reagent like HBr or H₂O adds to an alkene through a carbocation, the H goes on the carbon that already has more hydrogens, and the other group goes on the more substituted carbon. The real reason is that this route goes through the more stable carbocation.

Which reactions are anti-Markovnikov?

The two you need for Orgo 1 are HBr with peroxides (a radical reaction that puts Br on the less substituted carbon) and hydroboration–oxidation (BH₃ then H₂O₂ and NaOH, which puts OH on the less substituted carbon). For alkynes, hydroboration with a bulky borane turns a terminal alkyne into an aldehyde.

Why does the peroxide effect only work with HBr?

The radical chain has two steps that both need to be downhill in energy. With HBr they are. With HCl, the step where the carbon radical takes H from H–Cl is uphill because the H–Cl bond is strong. With HI, the step where an iodine radical adds to the alkene is uphill. So HCl and HI add the normal, Markovnikov way.

Does Markovnikov's rule apply to symmetrical alkenes?

No. If both alkene carbons carry the same groups, as in but-2-ene or cyclohexene, both ways of adding give the same product, so there's nothing to decide. The same goes for reagents with two identical ends, like H₂ or Br₂.

Sources

Regiochemistry questions show up on every alkene exam. Practice them with the rest of Orgo I and II.

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