R3.4 Electron-pair sharing reactionsIB Chemistry SL: Revision notes

A nucleophile forms a new bond to its reaction partner (the electrophile) by donating both bonding electrons. It must have a lone pair available. Nucleophiles can be negatively charged (OH⁻, CN⁻, Cl⁻, Br⁻) or neutral (H₂O, NH₃). Charged nucleophiles are usually more effective, because they are more strongly attracted to an electron-deficient (δ+) atom.

An electrophile forms a new bond by accepting both bonding electrons from a nucleophile. Electrophiles are attracted to regions of high electron density. They can be positively charged (H⁺) or neutral (Br₂, which becomes polarised near a C=C bond, and the δ+ hydrogen of HBr).

Heterolytic fission is the breaking of a covalent bond in which both bonding electrons stay with one fragment, so ions form. The pair goes to the more electronegative atom: HBr → H⁺ + Br⁻ and CH₃CH₂Br → CH₃CH₂⁺ + Br⁻. Compare homolytic fission, where each fragment keeps one electron and radicals form.

In nucleophilic substitution, a nucleophile donates an electron pair to form a new bond while another bond breaks heterolytically, releasing a leaving group. In a halogenoalkane the C–X bond is polar, so the carbon is δ+. Example: CH₃CH₂CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂CH₂OH + Br⁻. The lone pair on O forms the C–O bond; the C–Br bonding pair moves onto Br, which leaves as Br⁻. In a curly-arrow mechanism each arrow starts at an electron pair and ends where that pair forms a new bond or becomes a lone pair.

The C=C double bond (σ + π) is a region of high electron density, so alkenes are attacked by electrophiles and undergo electrophilic addition: the π bond breaks and two atoms or groups add, giving one saturated product.

• Halogens: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (1,2-dibromoethane); bromine water is decolourised — the test for unsaturation.
• Hydrogen halides: CH₂=CH₂ + HBr → CH₃CH₂Br.
• Water: CH₂=CH₂ + H₂O → CH₃CH₂OH, using steam with an acid catalyst (H₃PO₄ or H₂SO₄) and heat.