Track where electrons begin and end so that charges, bonds and radical dots all tell the same chemical story.
Two ways to break a covalent bond
A covalent bond contains a shared electron pair. Homolytic fission gives one electron to each atom, producing radicals. Heterolytic fission gives both bonding electrons to one atom; for a neutral starting molecule this produces oppositely charged ions.
A radical is a species with an unpaired electron, represented by a dot. The chlorine and alkyl radicals used in alkane substitution are neutral. Do not define a radical simply as an uncharged species: many neutral molecules have all their electrons paired.
A curly arrow starts at electrons
A full-headed curly arrow represents movement of an electron pair. Its tail begins at a lone pair or a bond; its head shows where that pair goes. To break C–Br heterolytically, start at the C–Br bond and end at Br. To form a bond using oxygen’s lone pair, start at that lone pair and end at the atom receiving it.
A fishhook arrow, with one half of an arrowhead, represents movement of one electron. Balanced radical equations with dots are sufficient for the AQA radical mechanisms here; curly arrows are not required. A straight reaction arrow separates reactants and products and is not an electron-movement arrow.
A formal + or − is a whole charge. A δ+ or δ− label shows partial charge within a polar bond. In bromoethane the bonded C and Br are partially charged, not an isolated carbon cation and bromide ion.
Recognise the electron-pair donor and acceptor
A nucleophile donates an electron pair to form a covalent bond. OH⁻ and neutral NH₃ can both act as nucleophiles because they possess available lone pairs. Negative charge is not part of the definition. An electrophile accepts an electron pair; H⁺ and carbocations are useful examples.
For a preview of nucleophilic substitution, hydroxide attacks the carbon bonded to Br in bromoethane. At the same time the C–Br bonding pair moves to Br. Ethanol and Br⁻ form. The full halogenoalkane lesson develops the reagents and conditions; here the purpose is to learn arrow placement.
Diagram placeholder
Two electron-pair arrows in a substitution step
Labels to include:
- lone pair on O of OH⁻
- δ+ carbon bonded to Br
- C–Br bond
- δ− on bonded Br
- ethanol plus Br⁻
Draw bromoethane with the C–Br bond explicit. Start one full-headed curly arrow at a lone pair on hydroxide oxygen and point it to the carbon attached to Br. Start the second at the C–Br bond and point it to Br. Keep OH⁻ negative before reaction and show a separate Br⁻ product. Both sides have a total charge of −1.
Reaction names describe the overall change
Substitution replaces an atom or group with another. Addition combines reactants across a multiple bond without eliminating a small molecule. Elimination removes atoms or groups from a molecule, commonly creating C=C and a small molecule. These descriptions do not replace a mechanism: a mechanism explains the individual steps.
In radical substitution a chlorine atom replaces a hydrogen in an alkane, and HCl is also produced. The next topic applies the dot convention to initiation, propagation and termination. For any mechanism, finish by checking atom counts, total charge and the valency at every carbon.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official AQA mark allocations.
Q1. What distinguishes homolytic from heterolytic bond fission?Show answer
Homolysis divides the shared pair equally, one electron to each atom. Heterolysis transfers both bonding electrons to one atom.
Q2. Why is NH₃ a possible nucleophile even though it has no negative charge?Show answer
Nitrogen has a lone pair that it can donate to form a covalent bond. Nucleophilicity is defined by electron-pair donation, not charge alone.
Q3. Where must the arrow start and end when a C–Cl bond breaks to form Cl⁻?Show answer
Start the full-headed curly arrow at the C–Cl bond and end it at Cl. It represents both electrons from that bond moving to chlorine.
Q4. Write the homolytic fission equation for Br₂ and explain the dot.Show answer
Br₂ → 2Br•. Each dot represents an unpaired electron on a bromine radical; it is not a positive or negative charge.
Q5. A student draws OH⁻ attacking bromoethane with an arrow beginning at H. Correct the mechanism in words and check its overall charge.Show answer
The first arrow must begin at a lone pair on oxygen and end at the carbon bonded to Br. A second arrow runs from the C–Br bond to Br. Products are ethanol and Br⁻, so the overall charge is −1 on both sides.
Sources
Sources and examiner guidance (reviewed 2 October 2026)
- Chemrevise: Introduction to organic chemistry — p6: bond fission and mechanism conventions.
- AQA 7405 specification — 3.3.1.2: electron-pair arrows and radical equations.
- AQA June 2023 AS Paper 2 mark scheme — Q03.1 p16 and general mechanism guidance p9.
- AQA June 2023 AS Paper 2 examiner report — Q03.1 p3: substitution mechanism.
- AQA June 2022 AS Paper 2 examiner report — Q05.1 p4: ammonia charge error.
Finesse Tuition is not endorsed by AQA or Chemrevise. All explanations and examples here are our own.