Potential energy profile

ReactantsInt.Products
via 2° cation (major) via 1° (minor)
Ready
HBr approaches
HBr is a polar molecule: Br is more electronegative, so H is δ⁺ (the electrophile) and Br is δ⁻. Propene is unsymmetrical.
Step 1 — slow
π bond attacks δ⁺H
The π electrons attack δ⁺H. H adds to the CH₂ carbon (the one with more H). The H–Br bond breaks heterolytically, giving Br⁻.
Intermediate
Secondary carbocation
The + charge sits on the middle carbon, stabilised by two electron-donating alkyl groups. This is why the major route is taken. Simulation paused.
Step 2 — fast
Br⁻ attacks carbocation
Br⁻ uses a lone pair to attack the planar carbocation. A new C–Br bond forms on the middle carbon.
Product
2-bromopropane (major)
Br ends up on the more substituted carbon, which is Markovnikov addition. 1-bromopropane is the minor product.
Speed: 1.2× Ready

Markovnikov's rule — the two possible routes

When HBr adds to an unsymmetrical alkene, the H can add to either carbon of the C=C. The two routes go through different carbocations, and the more stable carbocation forms faster (lower activation energy), so its product dominates.

H adds to the CH2 carbonMajor
  • CH3–CH=CH2  +  H–Br
  • H joins the carbon that already has 2 H
  • CH3–CH⁺–CH3   secondary (2°) carbocation, 2 alkyl groups
  • Br⁻ attacks → CH3–CHBr–CH3   2-bromopropane
H adds to the CH carbonMinor
  • CH3–CH=CH2  +  H–Br
  • H joins the carbon that has 1 H
  • CH3–CH2–CH2⁺   primary (1°) carbocation, only 1 alkyl group
  • Br⁻ attacks → CH3–CH2–CH2Br   1-bromopropane
Carbocation stability
methyl< 1°< 2°< 3° Alkyl groups are electron-donating (+I) and disperse the positive charge.

HBr is already polar

Unlike Br₂, HBr does not need to be polarised by the π cloud. Br is more electronegative than H, so the H–Br bond carries a permanent dipole: H is δ⁺ and acts as the electrophile.

Step 1 — heterolytic fission

π electrons attack δ⁺H. The H–Br bond breaks heterolytically, both electrons going to Br and forming Br⁻. This is the slow, rate-determining step because a carbocation must be made.

Step 2 — nucleophilic attack

Br⁻ acts as a nucleophile, using a lone pair to attack the positively charged carbon. This step is fast, so it is the stability of the carbocation that decides the product.

Stating the rule

In the addition of HX to an unsymmetrical alkene, the hydrogen adds to the carbon of the C=C that already carries the greater number of hydrogen atoms. The real reason is carbocation stability, not the rule itself, so always explain it that way in an answer.