# Electrophilic addition to alkenes

The pi bond of an alkene is electron-rich and attacks electrophiles such as H–Br. H⁺ adds first, forming a carbocation, and then Br⁻ adds. Markovnikov's rule: H goes to the carbon that already has more hydrogens, so the more stable cation forms.

Chemistry (university) · Part 10: Organic reactions and mechanisms · Card 6 of 10 · Last updated 2026-10-19

## Analogy

The rich get richer: the hydrogen goes where there are already more hydrogens.

## Example

**Propene + HBr**

H⁺ adds to CH₂, giving the secondary cation
Br⁻ attacks that carbon
Product: 2-bromopropane (major)
Not 1-bromopropane

## In practice

Alkene additions turn cheap petrochemicals, like propene, into useful alcohols and halides.

## Practice

### Flashcards

- **State Markovnikov's rule.** In HX addition, H adds to the alkene carbon with more hydrogens.

### Interview questions

**Predict the major product of propene with HBr and explain it using carbocation stability.**

- Alkene pi bond attacks H⁺
- Secondary cation forms in preference
- Br⁻ attacks the cation
- Product: 2-bromopropane (Markovnikov)

## Questions people ask

### Why does Markovnikov addition happen?

The first step forms the more stable carbocation. Adding H⁺ to the carbon with more hydrogens gives the positive charge on the more substituted carbon, which is better stabilised. The bromide then attacks that carbon. So the intermediate decides which product is major.

### What changes when peroxides are present?

With peroxides, HBr adds by a radical path instead, and bromine ends up on the less substituted carbon. This anti-Markovnikov result applies to HBr only, not HCl or HI. Radical chain reactions have different intermediates, so the rule changes and the product is 1-bromopropane for propene.

## Related concepts

- [Carbocation stability](https://studyvisually.com/learn/science/chemistry-university/chemistry-university-part-10/chemistry-university-part-10-concept-4)

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