Question
When HBr is added to propene (CH₃–CH=CH₂), what is the major product? How does the answer change when the reaction is carried out in the presence of peroxide (ROOR)?
This exact comparison — with and without peroxide — was tested in JEE Main 2024 and appears frequently in NEET organic chemistry sections.
Solution — Step by Step
Propene is CH₃–CH=CH₂. The two carbons of the double bond are C2 (attached to CH₃, so more substituted) and C1 (the terminal carbon, less substituted). HBr will add across this double bond — the question is which carbon gets H and which gets Br.
Markovnikov’s Rule says: the H goes to the carbon that already has more hydrogens (the less substituted carbon), and the Br goes to the more substituted carbon. Why? Because the carbocation intermediate at C2 (secondary carbocation) is more stable than one at C1 (primary carbocation). The reaction goes through the more stable intermediate.
Major product: 2-bromopropane
Step 1: H⁺ from HBr attacks the double bond. It adds to C1 (terminal), generating a secondary carbocation at C2.
Step 2: Br⁻ attacks the carbocation at C2.
Result: Br ends up on C2 — the more substituted carbon. This is Markovnikov addition.
When peroxide (ROOR) is present, the mechanism switches to a free radical chain reaction. The Br• radical (not H⁺) adds first to C1, generating the more stable secondary radical at C2.
Major product: 1-bromopropane (anti-Markovnikov)
The peroxide effect is specific to HBr. HCl is too strong (Cl• addition is exothermic but the chain propagation is unfavorable energetically), and HI’s iodine radical is too stable to react with the alkene. This selectivity is a standard MCQ trap — peroxide effect ≠ anti-Markovnikov for all HX.
Why This Works
The key is understanding which intermediate is more stable. In ionic addition, the reaction goes through a carbocation. Secondary carbocations (positive charge on a carbon with two alkyl groups) are more stable than primary ones because alkyl groups push electron density toward the positive charge through hyperconjugation and induction.
In radical addition, the same logic applies to radicals instead of carbocations. A secondary radical is more stable than a primary radical, so the Br• radical still ends up generating the more stable intermediate — but now Br is at C1, not C2. The product is the opposite regioisomer.
Both Markovnikov and anti-Markovnikov follow the same underlying principle: the reaction prefers the more stable intermediate. The mechanism determines which intermediate forms first.
Alternative Method — Using the “Stability of Intermediate” Shortcut
Instead of memorizing Markovnikov’s Rule as a statement, train yourself to draw the two possible intermediates and pick the stable one:
| Condition | Intermediate type | Which carbon? | Product |
|---|---|---|---|
| No peroxide | Carbocation | C2 (2° carbocation) | 2-bromopropane |
| With peroxide | Free radical | C2 (2° radical) | 1-bromopropane |
Notice that in both cases, the more stable intermediate forms at C2. But the timing of Br addition differs — ionic gives Br at C2, radical gives Br at C1 (since Br• added first to C1 to generate the radical at C2).
Shortcut for MCQs: “No peroxide → Markovnikov → Br on more substituted carbon. Peroxide → anti-Markovnikov → Br on less substituted carbon.” Write this once on your rough sheet before starting organic chemistry MCQs — it saves 30 seconds per question.
Common Mistake
The most common error: students apply the peroxide effect to HCl or HI by mistake. Questions often give you CH₂=CH₂ + HCl with peroxide and ask for the product — the answer is still the Markovnikov product (chloroethane), because the peroxide effect does not work with HCl or HI, only HBr. If you get this wrong in JEE Main, you lose a mark and give a free mark to students who specifically memorized this exception.
A second trap: confusing which product is Markovnikov vs anti-Markovnikov. Remember — Markovnikov = Br on the more substituted carbon = 2-bromopropane. Anti-Markovnikov = Br on the terminal carbon = 1-bromopropane.