Question
Show the mechanism of electrophilic addition of HBr to propene (CH₃CH=CH₂). State which product is the major product and why, with reference to Markovnikov’s rule.
Solution — Step by Step
HBr is a polar molecule. In the presence of the π electrons of the alkene, the HBr bond polarises further:
The electrophile is (the proton). The alkene’s π electrons act as the nucleophile, attacking the electrophile. This is electrophilic addition (AdE) — the alkene adds across the double bond.
The π electrons of propene attack the hydrogen of HBr. The H–Br bond breaks heterolytically (both electrons go to Br).
Two possible carbocations form depending on which carbon receives the proton:
Option A — Proton adds to C1 (CH₂ end):
CH₃–CH=CH₂ + H⁺ → CH₃–CH⁺–CH₃ (secondary carbocation)Option B — Proton adds to C2 (CH end):
CH₃–CH=CH₂ + H⁺ → CH₃–CH₂–CH₂⁺ (primary carbocation)The secondary carbocation (Option A) is more stable than the primary carbocation (Option B), because alkyl groups donate electron density by induction and hyperconjugation, dispersing the positive charge.
Since the more stable carbocation forms more easily (lower activation energy for its formation), Option A is the major pathway.
The bromide ion (), generated in Step 1, attacks the carbocation:
CH₃–CH⁺–CH₃ + Br⁻ → CH₃–CHBr–CH₃This gives 2-bromopropane as the major product.
The minor product would be 1-bromopropane (CH₃CH₂CH₂Br), formed via the less stable primary carbocation.
Markovnikov’s Rule (modern version): In addition of HX to an unsymmetrical alkene, the proton (H⁺) attaches to the carbon bearing more hydrogen atoms, while the halide (X⁻) attaches to the carbon bearing fewer hydrogen atoms.
In propene, C1 (=CH₂) has 2 H’s; C2 (=CH−) has 1 H. By Markovnikov’s rule, H adds to C1 and Br adds to C2.
Major product: 2-bromopropane (CH₃CHBrCH₃)
Markovnikov’s rule is not a coincidence — it is a consequence of carbocation stability. The H always adds to give the more stable carbocation (more substituted = more stable), which then reacts faster. So Markovnikov’s rule is really a paraphrase of “the reaction goes through the more stable intermediate.”
Stability order of carbocations: 3° > 2° > 1° > methyl
Why This Works
Electrophilic addition is driven by the nucleophilicity of the π bond and the electrophilicity of HBr. The π electrons are above and below the molecular plane — exposed and accessible to electrophiles.
The two-step mechanism (first carbocation, then halide attack) is slower than the concerted SN2 mechanism for saturated systems, because forming a carbocation intermediate takes energy. This is why alkenes don’t react with non-electrophilic reagents under normal conditions.
Alternative Method
For peroxide-initiated (anti-Markovnikov) addition of HBr, a radical mechanism operates:
- Br• (radical) adds to the less hindered carbon (C1) to give the more stable radical at C2
- H abstraction gives 1-bromopropane
This gives 1-bromopropane — the opposite regiochemistry. Key: peroxide addition reverses the product; it doesn’t apply to HCl or HI.
Common Mistake
Students write the carbocation as positively charged on the wrong carbon. In propene + H⁺ via Markovnikov addition: H⁺ adds to C1, giving a secondary carbocation at C2. Writing the charge at C1 (giving a primary carbocation at C1) would be the anti-Markovnikov intermediate and the wrong major product. Always draw the complete carbocation structure before showing bromide attack.
JEE Main tests this with slightly more complex alkenes — cycloalkenes, disubstituted alkenes, or involving rearrangements. The strategy is always the same: (1) identify which H placement gives the more stable carbocation, (2) that’s the major product. If a 3° carbocation is available, check for 1,2-hydride or 1,2-methide shifts that could convert a 2° to 3°.