The reaction this whole page is about: propene + HBr gives 2-bromopropane, not 1-bromopropane. Structures drawn live.
1. Markovnikov's Rule: H Adds to the Carbon That Already Has More Hydrogens
When HX adds to a C=C, H attaches to the carbon with more hydrogens and X to the more substituted carbon.
2. The Rule Is Really About Forming the More Stable Carbocation
The carbon that doesn't get the H becomes a carbocation, and stability rises with substitution (3° > 2° > 1° > methyl).
3. The Full Mechanism: Protonate, Form the Cation, Then Trap It
Protonate the π bond to form the more stable cation, then halide traps the charged carbon.
4. Worked Example: Propene + HBr Gives 2-Bromopropane, Not 1-Bromopropane
The secondary pathway wins overwhelmingly because cation stability, not sterics, decides.
Acid-catalyzed hydration follows the same logic, giving 2-propanol.
5. Watch Out: Carbocations Can Rearrange to Something Even More Stable
A free carbocation will shift a hydride or methyl to reach a more stable cation: 3-methyl-1-butene + HCl rearranges the 2° cation to 3°, so Cl lands on a carbon that was never in the double bond.
Expect rearrangement when a 2° cation sits one shift from a 3° center; carbocation-free reactions (oxymercuration, hydroboration) never rearrange.
6. Anti-Markovnikov: Peroxides and Hydroboration Flip the Outcome
HBr with peroxides goes radical: Br• adds first to give the more stable radical, putting Br on the terminal carbon ("peroxides flip the bromide," HBr only).
With peroxides the radical mechanism puts Br on the terminal carbon — 1-bromopropane.
Hydroboration–oxidation (BH₃·THF, then H₂O₂/NaOH) adds boron concertedly to the less hindered carbon and delivers the anti-Markovnikov alcohol, while oxymercuration gives the Markovnikov alcohol — both without rearrangement.
7. Summary
H adds to the carbon with more hydrogens · the real driver is the more stable carbocation (3° > 2° > 1° > methyl) · protonate → form cation → trap · free cations can rearrange · HBr/peroxides and hydroboration flip to anti-Markovnikov, oxymercuration stays Markovnikov.
Worked example
- The alkene's π electrons grab the proton of H–Br — this is the rate-determining step.
- Protonating the terminal (CH2) carbon puts the + charge on the middle carbon, a 2° carbocation; protonating the other way would give a much less stable 1° cation.
- Bromide then adds to the more-stable 2° cation.
Answer. 2-bromopropane — the H adds to the carbon with more H's, so Br ends up on the more-substituted carbon (Markovnikov), via the more stable carbocation.
Quiz yourself
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The Cl goes on the central (more substituted) carbon, giving 2-chloro-2-methylpropane. Protonating the =CH₂ end builds a tertiary carbocation — the most stable option — so chloride traps that carbon. H goes to the carbon that already had more hydrogens, exactly as Markovnikov's rule predicts.
Because it explains the mechanism instead of just describing a pattern, and it still gives the right answer when the two wordings might diverge — for example when a carbocation rearranges. The empirical rule can't predict rearranged products; the carbocation-stability logic can.
1-Bromopropane (CH₃CH₂CH₂Br). Peroxides switch HBr to a radical mechanism: Br• adds first to give the more stable secondary radical, placing Br on the terminal carbon — the anti-Markovnikov result. "Peroxides flip the bromide."
Hydroboration–oxidation (BH₃·THF, then H₂O₂/NaOH) gives 1-propanol, the anti-Markovnikov alcohol, with no rearrangement. Acid-catalyzed hydration or oxymercuration–demercuration gives 2-propanol, the Markovnikov alcohol.
Draw this on the whiteboard
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.