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Markovnikov's rule is usually the first organic "rule" students learn, and it is also one of the first they misapply. The rule tells you what usually happens. The mechanism tells you why, and when it will not. This post covers both, with worked examples, the peroxide effect, carbocation rearrangements and a short practice routine.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists for alkenes "mechanism of electrophilic addition, addition of hydrogen, halogens, water, hydrogen halides (Markownikoffs and peroxide effect)". The NEET (UG) 2026 syllabus uses nearly the same words, and the JEE (Advanced) 2026 brochure lists "Effect of peroxide on addition reactions" and "Rearrangement reactions of alkyl carbocation". The material is in Unit 9 of NCERT Class 11 Chemistry Part II. Check your own exam year's syllabus.
What does Markovnikov's rule say?
NCERT gives the rule as: the negative part of the adding molecule attaches to the carbon atom that has fewer hydrogen atoms. NCERT dates Markovnikov's generalisation to 1869. For propene and HBr, CH₃–CH=CH₂ + HBr, the rule predicts 2-bromopropane, CH₃–CHBr–CH₃, and in practice that is the main product.
Why does it work? Follow the carbocation
The mechanism has two steps.
H⁺ adds to the double bond. It can add to either carbon. Adding to the end CH₂ gives a secondary carbocation, CH₃–C⁺H–CH₃. Adding to the middle carbon gives a primary carbocation, CH₃–CH₂–C⁺H₂.
Br⁻ attacks the carbocation.
NCERT's explanation is that the secondary carbocation is more stable than the primary one, so it forms faster and dominates. The product comes from whichever carbocation forms more easily. For why alkyl groups stabilise carbocations, see our post on carbocations, carbanions and free radicals.
Worked example 1. Predict the main product of 2-methylpropene, (CH₃)₂C=CH₂, with HBr.
H⁺ on the CH₂ end gives (CH₃)₃C⁺, a tertiary carbocation.
H⁺ on the other carbon gives (CH₃)₂CH–CH₂⁺, a primary carbocation.
Tertiary is far more stable, so the product is 2-bromo-2-methylpropane, (CH₃)₃CBr.
What is the peroxide effect?
NCERT explains that with a peroxide such as benzoyl peroxide present, HBr adds to propene against the Markovnikov rule, giving 1-bromopropane. NCERT credits the observation to M. S. Kharasch and F. R. Mayo in 1933 at the University of Chicago, and it was reported in a 1933 series of papers on "the peroxide effect" in the addition of hydrogen bromide, including its addition to propylene (propene).
The mechanism is a free-radical chain, and the order of addition is reversed:
The peroxide breaks into radicals, which pull an H from HBr and release Br•.
Br• adds first, to the end CH₂, giving the more stable secondary radical CH₃–C•H–CH₂Br.
That radical takes H from another HBr, giving CH₃CH₂CH₂Br and a new Br•.
So the same idea applies: the more stable intermediate wins. Because bromine now adds first, it ends up on the end carbon.
Why only HBr?
NCERT gives bond enthalpies to explain this. The H–Cl bond (430.5 kJ/mol) is stronger than H–Br (363.7 kJ/mol), and it is not broken by the free radical. The H–I bond (296.8 kJ/mol) is weaker, but iodine radicals combine to form I₂ instead of adding to the double bond. So the peroxide effect is an HBr effect. With HCl or HI, use Markovnikov's rule whether or not peroxide is present.
What happens when the carbocation can rearrange?
Worked example 2. 3-Methylbut-1-ene, CH₂=CH–CH(CH₃)₂, with HBr and no peroxide.
H⁺ adds to the end CH₂ to give a secondary carbocation: CH₃–C⁺H–CH(CH₃)₂.
The neighbouring carbon carries an H. If that H moves over with its electron pair (a 1,2-hydride shift), the positive charge lands on a tertiary carbon: CH₃–CH₂–C⁺(CH₃)₂.
Br⁻ attacks the tertiary carbocation, so the main product is 2-bromo-2-methylbutane, not the 2-bromo-3-methylbutane that a rule-only answer predicts.
Rearrangement questions are JEE Advanced territory. If you are preparing only for boards or JEE Main, learn the idea and check your syllabus.
Common mistakes
Using the peroxide effect with HCl or HI. It applies to HBr only.
Applying the rule without drawing the intermediate. It fails on rearrangements and on alkenes with unusual substituents.
Symmetrical alkenes. But-2-ene with HBr gives 2-bromobutane either way, and the rule has nothing to decide.
Skipping the mechanism altogether. Over a year of organic chemistry, Grove, Cooper and Rush (2012) found a large proportion of second-year students chose not to use mechanisms in their work. Drawing the arrows is what lets you predict products you have not memorised.
When you can write the product but cannot explain the steps, the Fermi Problem Library has a step-by-step solver that explains the reasoning, so you can compare your mechanism with a full solution.
A practice routine
Write the ionic and radical mechanisms for propene + HBr from memory.
Do six predictions: propene, 2-methylpropene and but-1-ene, each with HBr with and without peroxide.
Add two with HCl and peroxide, to check you do not apply the effect.
If JEE Advanced is your target, add two rearrangement problems and draw the shift.
Then see how the reverse process works in our post on elimination reactions and Saytzeff's rule.
