Electrophilic addition to alkenes

The big idea: An alkene contains a carbon–carbon double bond, C=C (e.g. ethene, C₂H₄). A double bond is two shared pairs of electrons, so the C=C region is electron-rich.

That electron-rich double bond is an easy target for an electrophile — an electron-pair acceptor (an electron-poor or positively charged species). The C=C π electrons attack the electrophile, the double bond opens up, and a new group adds across the two carbons. Because something is added (nothing leaves), this is an addition reaction.

Key terms: - Alkene — a hydrocarbon with a C=C double bond (general formula C_nH_2n); it is unsaturated. - Saturated vs unsaturated — saturated = only single bonds (an alkane); unsaturated = has a C=C (or C≡C) so more atoms can be added. - Electrophile — an electron-pair acceptor, attracted to the electron-rich C=C (e.g. Br₂, HBr, H⁺). - Addition — two molecules join to make one product; the C=C becomes a C–C single bond, and no atom leaves.

A curly arrow shows the movement of a pair of electrons: the tail starts at the electrons that move (here the C=C π bond) and the head points to where the pair ends up (a new bond). In electrophilic addition the double bond is the electron-pair donor — it brings the electrons and attacks the electrophile.

Why the alkene is so reactive: The C=C double bond is electron-rich and the π electrons are exposed (held loosely above and below the carbons). This makes the alkene a good electron-pair donor, so it readily attacks electrophiles — far more readily than a saturated alkane, where the electrons are locked in strong, unreactive single bonds.

Common arrow mistakes: - Drawing the arrow from the electrophile to the C=C (backwards) — the C=C π electrons are the ones that move, so arrow 1 must start at the double bond. - Forgetting the second arrow — the Br–Br (or H–Br) bond must break, so that bonding pair has to go onto the leaving atom. - Calling it substitution — nothing is replaced; the reactant adds across the double bond.

Alkenes add a range of small molecules across the C=C. The double bond opens to a single bond, and one part of the reactant goes onto each carbon.

Bromine adds across the double bond: Ethene reacting with bromine gives 1,2-dibromoethane:

C₂H₄ + Br₂ → CH₂BrCH₂Br

One bromine atom adds to each carbon and the C=C becomes a C–C single bond. Nothing leaves the molecule — it is an addition.

The bromine-water test for unsaturation: Bromine water is orange/yellow-brown. Shake it with an alkene and it is decolourised (turns colourless) as the bromine adds across the C=C.

- Alkene (unsaturated) → bromine water decolourises (orange → colourless). - Alkane (saturated) → no change (stays orange); there is no C=C to add to.

This colour change is the standard test to tell an alkene (unsaturated) from an alkane (saturated).

Markovnikov — the major product: When an unsymmetrical reagent like HBr adds to an unsymmetrical alkene (e.g. propene, CH₃CH=CH₂), two products are possible. The major product follows Markovnikov's rule: the hydrogen adds to the carbon that already has more hydrogens.

For propene + HBr, H goes to the end CH₂ and Br goes to the middle carbon, giving 2-bromopropane, CH_{3}CHBrCH_{3}, as the major product.

How this is tested: R3.4 electrophilic addition shows up several ways:

- Paper 2: predict the product of an alkene + bromine (the dibromo compound), and outline why unsaturated molecules react readily. - Paper 1A (MCQ): identify which species can act as an electrophile (an electron-pair acceptor / electron-poor species).

The markers want the correct addition product (added across the double bond), the reason the C=C is electron-rich, and a clear definition of an electrophile.

Predict-the-product checklist: Open the C=C to a single bond, then add one part of the reactant to each carbon — and remember nothing leaves (it is addition, not substitution). For HBr on an unsymmetrical alkene, apply Markovnikov for the major product.