Reactivity of the Alkene Double Bond
Alkenes undergo electrophilic addition because the \(\text{C}=\text{C}\) double bond consists of an exposed \(\pi\) bond with high electron density that readily attracts electron-deficient electrophiles.
Why Do Alkenes React with Electrophiles?
The C=C double bond consists of a sigma (σ) bond and a pi (π) bond. The electron density in the π bond sits above and below the plane of the molecule, making it exposed and available for attack by electrophiles.
Definition and Atom Economy of Addition
In an addition reaction, two reactant molecules combine to form a single product. The \(\text{C}=\text{C}\;\pi\) bond breaks, and two new \(\sigma\) bonds form to the added atoms (100% atom economy).
Reactions of Alkenes
At SL you need to be able to deduce equations for the reactions of alkenes with the following reagents.
1. Reaction with Hydrogen Halides (HBr, HCl, HI)
Ethene + Hydrogen bromide → Bromoethane
CH₂=CH₂ + HBr → CH₃CH₂Br
The H and Br atoms add across the double bond. The π bond breaks and two new single bonds form.
2. Reaction with Halogens (Br₂, Cl₂)
Ethene + Bromine → 1,2-dibromoethane
CH₂=CH₂ + Br₂ → CH₂BrCH₂Br
This is the classic bromine water test for unsaturation. The orange bromine is decolourised as it adds across the double bond.
3. Reaction with Water (H₂O)
Ethene + Water → Ethanol
CH₂=CH₂ + H₂O → CH₃CH₂OH
This is the hydration of an alkene. In practice, requires an acid catalyst (e.g. concentrated H₂SO₄) or high temperature and pressure with steam.
Summary Table
| Reagent | Product from ethene | Type of product |
|---|---|---|
| HBr | CH₃CH₂Br | Halogenoalkane |
| HCl | CH₃CH₂Cl | Halogenoalkane |
| Br₂ | CH₂BrCH₂Br | Dihalogenoalkane |
| Cl₂ | CH₂ClCH₂Cl | Dihalogenoalkane |
| H₂O | CH₃CH₂OH | Alcohol |
HL Mechanistic Scope
The detailed step-by-step curly arrow mechanisms (carbocation intermediates, Markovnikov's rule, and asymmetric addition products) are covered at Higher Level in R3.4.11 and R3.4.12.
Thermodynamics of Alkene Addition
Alkenes undergo addition because breaking the weaker \(\pi\) bond (\(\approx 270\text{ kJ mol}^{-1}\)) to form two stronger \(\sigma\) bonds is energetically favourable (\(\Delta H < 0\)). Alkanes have only strong \(\sigma\) bonds, so they undergo substitution instead.
Addition vs Substitution Distinction
- Addition vs Substitution: Never write "substitution" for alkene reactions with \(\text{Br}_2\) or \(\text{HBr}\).
- Single Product: Addition yields one single product molecule. If an equation shows two products (like \(\text{HCl}\)), it is substitution, not addition!
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