Markovnikov's Rule & Anti-Markovnikov (Kharasch / Peroxide Effect)

Why Regiochemistry Matters

A double bond is electron-rich: the loosely held $\pi$ electrons sit above and below the molecular plane and behave as a source of electrons towards incoming electron-seeking species. Addition of a hydrogen halide $\ce{H-X}$ across an alkene is therefore an electrophilic addition reaction, and the order of reactivity of the halides follows their acid strength, $\ce{HI > HBr > HCl}$.

When the alkene is symmetrical — identical groups on each carbon of the double bond — orientation does not matter. Whichever way the reagent adds, the product is the same. Ethene and but-2-ene illustrate this clean, single-answer case:

$$\ce{CH2=CH2 + HBr -> CH3CH2Br}$$ $$\ce{CH3CH=CHCH3 + HBr -> CH3CH2CHBrCH3}$$

The interesting chemistry begins with unsymmetrical alkenes such as propene, where the two carbons of the double bond carry different numbers of hydrogen atoms. Now $\ce{H-Br}$ can add in two ways, giving 1-bromopropane or 2-bromopropane. Predicting which one dominates is the entire subject of this article.

Markovnikov's Rule

In 1869 the Russian chemist Vladimir Markovnikov, after studying many such additions in detail, framed the generalisation that now carries his name. Markovnikov's rule states that the negative part of the addendum (the adding molecule) gets attached to the carbon atom that possesses the lesser number of hydrogen atoms. The corollary, easier to recall in the exam hall, is that the hydrogen goes to the carbon already richer in hydrogen — colloquially, "the rich get richer."

Applied to propene plus hydrogen bromide, the bromine (the negative part) attaches to the central, more substituted carbon (C-2), and the product is 2-bromopropane:

$$\ce{CH3-CH=CH2 + HBr -> CH3-\underset{\underset{\Large Br}{|}}{C}H-CH3}$$

2-Bromopropane is observed as the principal product in actual practice. The reason behind the rule, however, is not a superstition about hydrogen distribution — it is carbocation stability, which is best seen in the mechanism.

Mechanism: Carbocation Pathway

Hydrogen bromide first supplies an electrophile, the proton $\ce{H+}$, which attacks the $\pi$ bond. The proton can add to either carbon, and each choice generates a different carbocation:

$$\ce{CH3-CH=CH2 ->[H+] CH3-\overset{+}{C}H-CH3} \quad (\text{secondary, b})$$ $$\ce{CH3-CH=CH2 ->[H+] CH3-CH2-\overset{+}{C}H2} \quad (\text{primary, a})$$

The secondary carbocation (b) is more stable than the primary carbocation (a), because alkyl groups release electron density and disperse the positive charge through hyperconjugation and the inductive effect. The more stable cation is formed at a faster rate, so it predominates. The bromide ion then attacks this secondary carbocation to give 2-bromopropane:

$$\ce{CH3-\overset{+}{C}H-CH3 + Br^- -> CH3-CHBr-CH3}$$

This is the deep meaning of Markovnikov's rule: the proton adds to the alkene so as to generate the more stable carbocation, and the halide simply follows. Stating the rule in terms of carbocation stability — rather than merely counting hydrogens — lets you handle alkenes where the "more hydrogens" shortcut is ambiguous, and it links directly to the carbocation-stability order you learn in Organic Chemistry — Basic Principles.

Anti-Markovnikov / Peroxide Effect

In 1933, at the University of Chicago, M. S. Kharasch and F. R. Mayo discovered that when the addition of HBr to an unsymmetrical alkene such as propene is carried out in the presence of a peroxide (for example benzoyl peroxide), the reaction runs contrary to Markovnikov's rule. The product is now 1-bromopropane, with the bromine on the terminal carbon:

$$\ce{CH3-CH=CH2 + HBr ->[(C6H5CO)2O2] CH3-CH2-CH2Br}$$

This reversal is called the peroxide effect, the Kharasch effect, or simply anti-Markovnikov addition. The single most examined fact about it is its scope: it happens only with HBr — never with HCl or HI. The reason lies entirely in the mechanism, which is no longer ionic but a free-radical chain.

Mechanism: Free-Radical Pathway

The peroxide effect proceeds via a free-radical chain mechanism. The peroxide undergoes homolysis to give radicals, which abstract hydrogen from HBr to liberate a bromine radical; this $\ce{Br.}$ is the chain carrier. It is the bromine radical — not a proton — that adds first to the double bond:

$$\ce{(C6H5CO)2O2 ->[\Delta] 2\,C6H5CO.} \quad\quad \ce{C6H5CO. + H-Br -> C6H5COOH + Br.}$$ $$\ce{CH3-CH=CH2 + Br. -> CH3-\overset{.}{C}H-CH2Br} \quad (\text{2° radical, more stable})$$ $$\ce{CH3-\overset{.}{C}H-CH2Br + H-Br -> CH3-CH2-CH2Br + Br.}$$

The bromine radical adds to the terminal carbon because doing so leaves the unpaired electron on the more substituted carbon, generating the more stable secondary free radical (just as alkyl substitution stabilises a carbocation, it stabilises a radical). That radical then abstracts a hydrogen from another HBr molecule, delivering 1-bromopropane and regenerating $\ce{Br.}$ to continue the chain. Because bromine attaches first — and to the opposite carbon from where the cation mechanism would place it — the regiochemistry inverts.

Why Only HBr

The peroxide effect is observed in the addition of HBr but not of HCl or HI. NCERT attributes this to bond energies and radical behaviour:

In short, the chain needs a halogen radical that both forms readily and adds efficiently to the alkene. Chlorine fails the first test, iodine fails the second; only bromine clears both, which is why the peroxide effect is the exclusive property of HBr.

The Two Pathways Side by Side

Holding the contrast in one frame is the fastest way to avoid mistakes under time pressure. Everything that distinguishes the two routes is summarised below.

Worked Product Predictions

The skill NEET tests is fast, confident product assignment. Work through these the way the rule demands — identify the more stable intermediate, then place the bromine.

The same Markovnikov logic governs the closely related additions of water and cold concentrated sulphuric acid to alkenes — both add in accordance with Markovnikov's rule to give the more substituted product — and it carries over to the hydration of alkynes, where Markovnikov addition of water gives a ketone via an enol. That alkyne case is exactly what NEET 2017 examined.