Question
When HBr is added to propene (CH₃–CH=CH₂), two different products are possible depending on the reaction conditions. Write the major product when:
(a) HBr is added in the absence of peroxide (b) HBr is added in the presence of peroxide (H₂O₂ or ROOR)
Explain the mechanism in each case.
Solution — Step by Step
Propene is unsymmetrical — the two carbons of the double bond are not equivalent. C1 (=CH₂) carries two hydrogens; C2 (CH₃–CH=) carries only one hydrogen and the methyl group. This asymmetry is what makes the question meaningful.
Without peroxide, HBr adds via ionic mechanism (electrophilic addition). The H⁺ from HBr attacks the double bond first.
H⁺ can go to either C1 or C2. If H⁺ goes to C1, we get a secondary carbocation at C2 (more stable — 2 alkyl groups). If H⁺ goes to C2, we get a primary carbocation at C1 (less stable — 1 alkyl group).
The reaction goes through the more stable secondary carbocation, so Br⁻ attacks C2.
Major product: CH₃–CHBr–CH₃ (2-bromopropane)
Peroxides (ROOR) generate free radicals under heat or light. This completely changes the mechanism to free radical addition.
The bromine radical (Br•) attacks the double bond first — not H•. Br• is the chain-carrying species because the Br–Br bond is weaker than H–H, making Br• easier to generate.
Br• attacks C1 (the less substituted carbon) because this gives a secondary radical at C2, which is more stable than a primary radical at C1.
Major product: CH₃–CH₂–CH₂Br (1-bromopropane)
| Condition | Mechanism | Br goes to | Product |
|---|---|---|---|
| No peroxide | Ionic (electrophilic) | More substituted C (C2) | 2-bromopropane |
| With peroxide | Free radical | Less substituted C (C1) | 1-bromopropane |
The product is literally the opposite in each case — that’s why this is called peroxide effect or Kharasch effect.
Why This Works
In ionic addition, the attacking species is H⁺ (electrophile). The driving force is carbocation stability — the intermediate that forms first must be stable enough to exist long enough for Br⁻ to attack. Secondary carbocations (two alkyl groups donating electrons) are more stable than primary ones, so H always goes where it makes the carbocation more stable. Markownikoff’s rule is really just “H goes to the carbon that can best stabilise the positive charge that results.”
In free radical addition, Br• attacks first (not H•). The radical intermediate that forms must be stable. A secondary radical (at C2) is more stable than a primary radical (at C1), so Br• goes to C1 — placing the radical on C2. Then H• from HBr completes the addition by attacking C2.
The peroxide doesn’t add to the product. It only triggers the chain reaction by generating the initial Br• radical. Once the chain starts, peroxide has done its job.
Alternative Method: Regioselectivity via Stability Arguments
Instead of drawing mechanisms, you can predict the product using a single thumb rule:
In both cases, the intermediate that forms is the more stable one. The difference is only what kind of intermediate — carbocation (ionic) or radical (free radical).
- More stable carbocation → more substituted carbon → Br ends up there (Markownikoff)
- More stable radical → also more substituted carbon → Br ends up at less substituted carbon because Br attacked first (anti-Markownikoff)
This “attack first” logic is the key to the peroxide effect. In free radical addition, Br• attacks the double bond. Since Br• goes to the less substituted carbon to give the more stable radical, the net result is Br ending up on the less substituted carbon — the opposite of Markownikoff.
Common Mistake
Applying Markownikoff’s rule to free radical addition. Many students write 2-bromopropane even when the question clearly states “in the presence of peroxide.” The peroxide effect only applies to HBr — not HCl or HI. HCl and HI do not show anti-Markownikoff addition with peroxides because the radical chain mechanism is not energetically feasible for them. If your question says HCl + peroxide, Markownikoff’s product still forms.
This question is directly from NCERT Class 11 Chapter 13 (Hydrocarbons) and appears regularly in CBSE board exams as a 3-mark question. In JEE Main, the peroxide effect appears in one-liners testing whether you know it’s exclusive to HBr. Getting the mechanism right — especially why Br• attacks first — is what separates a 3/3 answer from a 1/3 answer.