Markovnikov's Rule, Anti-Markovnikov Addition and the Peroxide Effect

ChemistryClass 11, HydrocarbonsFor JEE and NEET aspirants

Markovnikov's rule predicts where the parts of an unsymmetrical reagent such as HBr go when it adds to an unsymmetrical alkene: the H goes to the double-bond carbon that already has more hydrogens, and the Br to the other one. Propene and HBr give 2-bromopropane, not 1-bromopropane. The reason is the intermediate: the reaction passes through the more stable carbocation. With HBr and a peroxide the orientation reverses (anti-Markovnikov addition, the peroxide effect). Markovnikov's rule is in the Class 11 NCERT chapter on hydrocarbons and is tested in JEE Main, JEE Advanced and NEET.

At a glance

Rule: in the addition of HX, the H goes to the double-bond carbon with more H atoms; X goes to the more substituted carbon.
Reason: the more stable carbocation forms: 3° $>$ 2° $>$ 1°; benzylic and allylic cations are stabilised too.
Example:
$C H_{3} CH = C H_{2} + HBr C H_{3} CHBrC H_{3}$
2-bromopropane (major).
Peroxide effect:
$C H_{3} CH = C H_{2} + HBr (C_{6} H_{5} CO)_{2} O_{2} C H_{3} C H_{2} C H_{2} Br$
1-bromopropane. Only HBr does this.
Hydration:
$(C H_{3})_{2} C = C H_{2} + H_{2} O dil. H_{2} S O_{4} (C H_{3})_{3} COH$
follows the rule; hydroboration-oxidation gives the opposite (anti-Markovnikov) alcohol.
Alkynes: $HgS O_{4}$ and dilute $H_{2} S O_{4}$ add water by the rule, so a terminal alkyne gives a methyl ketone (ethyne gives ethanal).

1. Mechanism

Addition of HX is an electrophilic addition in two steps. The first, slow step makes a carbocation, and the alkene can make one of two. The faster route, through the more stable cation, gives the major product.

Figure 1: HBr adds to propene by Markovnikov's rule. Step 1 decides the product: the H goes to the

C H_{2}

end because that leaves the more stable 2° carbocation.

Protonation (slow). The $π$ electrons bond to the H of HBr and $B r^{-}$ leaves. Putting the H on the $C H_{2}$ end leaves the + on the middle carbon, a 2° carbocation held up by two alkyl groups (inductive effect and hyperconjugation). The 1° cation from the other orientation is much less stable, so that route is much slower.
Capture (fast). $B r^{-}$ bonds to the positive carbon.

Because a carbocation is involved, it can rearrange first. A 1,2-shift of a hydride or a methyl group that turns a 2° cation into a 3° one happens before $X^{-}$ arrives (see Solved Example 2).

Exam Trick

Place the + first, then the X. Ask where the positive charge would be most stable (3°, benzylic, allylic, next to a lone pair). The H goes to the other carbon, and X goes where the + ends up (after any hydride or methyl shift). This works even when "the carbon with more H" gives the wrong answer.

2. The peroxide effect (anti-Markovnikov addition of HBr)

With an organic peroxide such as benzoyl peroxide (heat or light breaks the peroxide into radicals), or with UV light, HBr adds by a free-radical chain instead, and the Br ends up on the less substituted carbon. Heat alone, with no peroxide, does not do this: HBr then adds by Markovnikov's rule. This is the peroxide effect, also called the Kharasch effect.

2.1 Initiation

(C_{6} H_{5} COO)_{2} heat 2 C_{6} H_{5} CO O^{∙}

C_{6} H_{5} CO O^{∙} C_{6} H_{5}^{∙} + C O_{2}

C_{6} H_{5}^{∙} + HBr C_{6} H_{6} + B r^{∙}

The peroxide's job is to make bromine atoms. Benzene and carbon dioxide come out as by-products.

2.2 Propagation

B r^{∙} + C H_{3} CH = C H_{2} C H_{3} C^{∙} HC H_{2} Br

C H_{3} C^{∙} HC H_{2} Br + HBr C H_{3} C H_{2} C H_{2} Br + B r^{∙}

Here the bromine atom adds first, to the $C H_{2}$ end, because that leaves the more stable 2° radical. The H arrives second, from HBr. So the same idea (the more stable intermediate) gives the opposite orientation, because a different species adds first.

Only HBr. Both chain steps must release energy. With HCl the H–Cl bond is too strong for the radical to take its H; with HI the iodine atom does not add to the alkene (adding it releases no energy, and iodine atoms simply join up as $I_{2}$ , which is how NCERT puts it). HCl and HI therefore add by Markovnikov's rule even when a peroxide is present.

3. Where the rule applies

Reaction	Orientation	Note
HCl, HBr, HI (no peroxide)	Markovnikov	Through a carbocation; rearrangements possible
HBr with a peroxide (or UV light)	Anti-Markovnikov	Radical chain; only HBr. HBr with heat alone is Markovnikov
$H_{2} O$ with dilute $H_{2} S O_{4}$ (acid-catalysed hydration)	Markovnikov	Through a carbocation; rearrangements possible
Oxymercuration-demercuration ( $Hg (OAc)_{2}$ , $H_{2} O$ ; then $NaB H_{4}$ )	Markovnikov	No rearrangement
Hydroboration-oxidation ( $B_{2} H_{6}$ ; then $H_{2} O_{2}$ , $NaOH$ )	Anti-Markovnikov	H and OH add to the same face (syn); no rearrangement
Alkyne with $H_{2} O$ , $HgS O_{4}$ , dilute $H_{2} S O_{4}$	Markovnikov	Terminal alkynes give methyl ketones
Alkyne, hydroboration-oxidation	Anti-Markovnikov	Terminal alkynes give aldehydes (a bulky borane such as 9-BBN is used so it adds only once)
Symmetrical alkene (but-2-ene, cyclohexene)	No choice	Only one product

Styrene shows both sides. With HBr alone the cation forms next to the ring (benzylic), giving 1-bromo-1-phenylethane; with HBr and a peroxide the radical forms there instead, and the Br goes to the end carbon: 1-bromo-2-phenylethane.

JEE Advanced

The rule is about the cation, not the H count. In $C F_{3} CH = C H_{2}$ , a + next to the strongly electron-withdrawing $C F_{3}$ is very unstable, so HBr gives $C F_{3} C H_{2} C H_{2} Br$ , the opposite of what counting H atoms suggests. In $C H_{2} = CHCl$ , the lone pair of Cl stabilises a + on its own carbon, so HCl gives 1,1-dichloroethane, $C H_{3} CHC l_{2}$ .

Markovnikov without rearrangement. Oxymercuration goes through a bridged mercurinium ion, not a free carbocation, so 3,3-dimethylbut-1-ene gives 3,3-dimethylbutan-2-ol there, while acid-catalysed hydration gives the rearranged 2,3-dimethylbutan-2-ol.

4. Worked examples

Solved Example 1

Propene is treated with (a) HBr, (b) HBr and benzoyl peroxide, (c) HCl and benzoyl peroxide. Name the major product each time.

Solution:

(a) 2-Bromopropane, $C H_{3} CHBrC H_{3}$ : Markovnikov addition through the 2° carbocation.

(b) 1-Bromopropane, $C H_{3} C H_{2} C H_{2} Br$ : the peroxide effect; Br adds first and the 2° radical forms.

Solved Example 2

What is the major product when 3-methylbut-1-ene reacts with HBr (no peroxide)?

Solution:

H adds to C-1, giving the 2° cation at C-2: $C H_{3} C H^{+} CH (C H_{3})_{2}$ . A 1,2-hydride shift from C-3 turns it into the 3° cation $C H_{3} C H_{2} C^{+} (C H_{3})_{2}$ , and $B r^{-}$ bonds there. The major product is 2-bromo-2-methylbutane, $C H_{3} C H_{2} CBr (C H_{3})_{2}$ , not the unrearranged 2-bromo-3-methylbutane.

Solved Example 3

1-Methylcyclohexene is treated separately with (a) HBr, (b) HBr and a peroxide, (c) dilute

H_{2} S O_{4}

, (d)

B_{2} H_{6}

, then

H_{2} O_{2}

and NaOH. Give the major product of each.

Solution:

(a) 1-Bromo-1-methylcyclohexane: H goes to C-2, the 3° cation forms at C-1.

(b) 1-Bromo-2-methylcyclohexane: Br adds first, at C-2, leaving the 3° radical at C-1.

(d) trans-2-Methylcyclohexan-1-ol: anti-Markovnikov, and H and OH add to the same face, so the OH ends up trans to the methyl group.

5. Exam traps

Watch out

The peroxide effect works only with HBr. HCl and HI add by Markovnikov's rule even with a peroxide.
In the peroxide reaction the bromine atom adds first, to the $C H_{2}$ end. The peroxide makes Br atoms, not H atoms.
Benzoyl peroxide gives $C O_{2}$ and benzene as by-products: phenyl radicals take H from HBr.
Carbocation additions can rearrange (hydride or methyl shift). Hydroboration and oxymercuration do not.
A symmetrical alkene gives only one product, so the rule does not apply.
Hydration of a terminal alkyne with $H g^{2 +}$ gives a methyl ketone, not an aldehyde (ethyne alone gives ethanal).
Hydroboration-oxidation is anti-Markovnikov and syn: 1-methylcyclohexene gives trans-2-methylcyclohexan-1-ol.
Think "most stable intermediate", not "carbon with more H": $C F_{3} CH = C H_{2}$ and HBr give $C F_{3} C H_{2} C H_{2} Br$ .

Past-paper questions on Markovnikov's rule

Our past papers: JEE Main 2025-2026, JEE Advanced 2022-2026, NEET 2018-2019, 2022-2026. We found 20 questions in them that test this:

JEE Main 13 (2025-2026)
JEE Advanced 3 (2023-2024)
NEET 4 (2024-2026)

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