Electrophilic addition reactions

Why alkenes are targets. The $\sigma$-bonded carbon skeleton of an alkene is unremarkable, but the $\pi$ bond above and below it is a loosely held, polarisable electron cloud. Any species that is short of electrons — an electrophile $E^+$, or the electrophilic end of a polarised or polarisable reagent — is drawn to it.

The general two-step mechanism. Electrophilic addition always proceeds in two stages: the $\pi$ electrons attack the electrophile first, and a nucleophile completes the addition second.

where: $E^+$ is the electrophile (or its electrophilic end), the middle species is the carbocation intermediate formed on the carbon that did not bond to $E$, and $Nu^-$ is the nucleophile that attacks it to give the final addition product. Every one of the four named reactions below is this same two-step pattern wearing a different reagent.

The four reactions of the syllabus.

• Hydrogenation: $\text{H}_2$ adds across the double bond over a Ni, Pd or Pt catalyst surface to give the alkane. $\text{H}_2$ is not strongly polarised on its own, so full electrophilic-mechanism detail is not expected here — treat it mainly as alkene to alkane.
• Halogenation: $\text{X}_2$ (Br₂ or Cl₂) adds because the approaching alkene induces a temporary dipole in the halogen molecule, making one halogen atom electrophilic. Addition proceeds through a bridged halonium-ion-type intermediate to give a vicinal dihalide; decolourisation of bromine water or Br₂ in CCl₄ is the classic laboratory test for a C=C double bond.
• Hydrohalogenation: HX (HCl, HBr, HI) adds with $\text{H}^+$ as the electrophile, forming a carbocation that $\text{X}^-$ then attacks.
• Hydration: $\text{H}_2\text{O}$ adds under a trace of strong acid catalyst (H₂SO₄), again with $\text{H}^+$ as the electrophile, followed by attack of water and loss of a proton to give an alcohol.

Markovnikov's rule. When an unsymmetrical reagent H–X adds to an unsymmetrical alkene, H adds to the double-bond carbon that already carries more hydrogens, and X adds to the carbon with fewer — informally, "the rich get richer." The mechanistic root is carbocation stability: protonation happens wherever it generates the more stable carbocation (tertiary $>$ secondary $>$ primary), because more alkyl groups release electron density toward the positive carbon through the inductive and hyperconjugative effects, and X⁻ then attacks that more stable cation.

The peroxide (Kharasch) effect. In the presence of peroxides, and only with HBr, the mechanism switches from ionic to free-radical: peroxides generate a bromine radical, which adds first to whichever carbon gives the more stable free radical. The regiochemistry reverses — Br ends up on the carbon that originally had more hydrogens. HCl and HI do not show this effect because the corresponding radical-chain steps are energetically unfavourable for them.

There is no numeric formula to derive here — electrophilic addition is a mechanistic, not a quantitative, topic — but the reasoning behind Markovnikov's rule and the peroxide exception is exactly the kind of "why" that examiners expect you to be able to reconstruct on demand, so it is worth building carefully from first principles rather than memorising the two rules as slogans.

Step 1. Start from the fact that addition of H–X to an unsymmetrical alkene like propene can, in principle, happen two ways: H bonds to the terminal carbon or H bonds to the internal carbon. These are not equally likely, and the reason is what happens to the other carbon once H has bonded.

Step 2. Whichever carbon does not bond to H is left electron-deficient — it carries the positive charge, as a carbocation, in the ionic (no-peroxide) pathway. A carbocation's stability rises sharply with the number of alkyl groups attached to the charged carbon, because each adjacent C–H or C–C bond can donate electron density toward the empty orbital through hyperconjugation and induction. This gives the stability order tertiary $>$ secondary $>$ primary, and a difference of even one alkyl group changes which pathway is followed almost exclusively.

Step 3. Because the reaction always proceeds through whichever intermediate forms faster — and a more stable intermediate has a lower-energy, faster-forming transition state — the mechanism funnels through the more-substituted carbocation. For propene that means H adds to the terminal $\text{CH}_2$ carbon (giving a secondary cation) rather than to the internal carbon (which would leave only a primary cation). Markovnikov's rule is simply the observable consequence of always taking the lower-energy carbocation route.

Step 4. Under peroxides, the first species to add to the double bond is not $\text{H}^+$ but a bromine radical, generated when peroxides initiate a free-radical chain. Radical stability follows the same substitution trend as carbocation stability (tertiary $>$ secondary $>$ primary radical), but now it is the radical, not the cation, whose stability is being maximised — and the radical forms on whichever carbon does not receive the bromine atom. So Br adds to the terminal carbon (leaving the more stable secondary radical on the internal carbon), which is the opposite carbon from where Br ended up in the ionic pathway. The rule that decides the outcome — "form the more stable intermediate" — has not changed at all; only the identity and stability trend of the intermediate has, because the mechanism itself has switched from ionic to radical.

Problem: Propene reacts with HBr, once in the absence and once in the presence of peroxides. Predict the major product in each case and justify using intermediate stability.

Step 1. Without peroxide, the mechanism is ionic. Propene is $\text{CH}_3-\text{CH}=\text{CH}_2$; its $\pi$ electrons attack the electrophilic hydrogen of HBr, releasing $\text{Br}^-$ and forming a carbocation on whichever carbon did not bond to H.

Step 2. Compare the two possible carbocations: H adding to the terminal carbon leaves $\text{CH}_3-\overset{+}{C}H-\text{CH}_3$, a secondary cation; H adding to the internal carbon would leave a primary cation. The secondary cation is more stable, so the reaction proceeds through it.

Step 3. $\text{Br}^-$ attacks the secondary carbocation: $\text{CH}_3-\overset{+}{C}H-\text{CH}_3 + \text{Br}^- \longrightarrow \text{CH}_3-\text{CHBr}-\text{CH}_3$. The major product is 2-bromopropane, the Markovnikov product.

Step 4. With peroxide present, the mechanism switches to free-radical: peroxide homolysis ultimately generates $\text{Br}^{\bullet}$, which adds first to propene. Comparing the two possible radicals, addition to the terminal carbon leaves $\text{CH}_3-\overset{\bullet}{C}H-\text{CH}_2\text{Br}$ (secondary radical, more stable) rather than a primary radical from addition at the internal carbon, so Br adds to the terminal carbon this time. Chain propagation with another HBr molecule gives $\text{CH}_3-\text{CH}_2-\text{CH}_2\text{Br} + \text{Br}^{\bullet}$.

Final answer: without peroxide, the major product is 2-bromopropane (Markovnikov); with peroxide, the major product is 1-bromopropane (anti-Markovnikov) — the same reagents, opposite regiochemistry, because the intermediate being stabilised switches from a carbocation to a free radical.

Competitive stretch

If the alkene were 2-methylpropene, $(\text{CH}_3)_2\text{C}=\text{CH}_2$, instead of propene, the ionic pathway would funnel through a tertiary carbocation rather than a secondary one (even more stable), giving tert-butyl bromide as the exclusive Markovnikov product; the gap in stability between tertiary and primary is larger than between secondary and primary, so the regioselectivity would be sharper still.

• Applying the peroxide effect to HCl or HI — the Kharasch effect is observed only with HBr.
• Stating Markovnikov's rule backwards as "halogen adds to the carbon with more hydrogens" — it is hydrogen that adds to the carbon already richer in hydrogens.
• Quoting Markovnikov's rule without justifying it mechanistically — examiners expect the answer to reference carbocation stability (tertiary > secondary > primary), not the rule alone.
• Treating $\text{X}_2$ addition as going through a simple open carbocation exactly like HX addition — it actually proceeds through a bridged halonium-ion-type intermediate; full stereochemical detail is beyond this syllabus, but the intermediate itself is examinable.

Ask the tutor: if distinguishing the carbocation (ionic) mechanism from the free-radical (peroxide) mechanism is still confusing, ask the AI tutor to walk through both pathways side by side for the same alkene.