Module 4 is where organic chemistry becomes mechanistic. You stop memorising products and start explaining why a particular bond breaks and a particular product dominates.
Three mechanisms carry the module: free-radical substitution, electrophilic addition and nucleophilic substitution — plus alcohols, isomerism, synthesis routes, infrared and mass spectrometry.
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Structural isomers have the same molecular formula but a different structural formula: chain isomers, positional isomers and functional group isomers.
Stereoisomers have the same structural formula but a different arrangement in space. At this level that means E/Z isomerism about a C=C double bond, which cannot rotate.
Bond fission: homolytic fission gives each atom one electron — two radicals (shown with single-headed 'fish-hook' arrows). Heterolytic fission gives both electrons to one atom — an ion pair (double-headed curly arrows).
Ratios: C 4.35/2.175 = 2, H 13.0/2.175 = 5.98 ≈ 6, O 2.175/2.175 = 1 → empirical formula C₂H₆O, which has M = 46.0 — the same as Mr, so the molecular formula is also C₂H₆O, containing 6 hydrogen atoms (ethanol).
Alkanes are saturated and largely unreactive — the C–C and C–H bonds are strong and almost non-polar. With a halogen and UV light they undergo free-radical substitution, in three stages:
The limitation OCR want you to state: free-radical substitution gives a mixture of products. Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄, and termination steps give a range of by-products, so the yield of any one product is poor.
The C=C double bond has a σ bond plus a π bond formed by sideways overlap of p orbitals. The π electrons sit above and below the plane, are exposed and are attracted to electrophiles — electron-pair acceptors.
Electrophilic addition with HBr on propene:
Markovnikov's rule, explained properly: the major product forms via the more stable carbocation. Stability is tertiary > secondary > primary, because alkyl groups are electron-releasing (positive inductive effect) and spread the positive charge. Propene + HBr therefore gives mainly 2-bromopropane (via the secondary carbocation), not 1-bromopropane.
Alkenes also polymerise: n CH₂=CH₂ → –(CH₂–CH₂)–n. Addition polymerisation has 100% atom economy.
M(Br₂) = 159.8 g mol⁻¹. mass = 0.0500 × 159.8 = 7.99 g.
Alcohols hydrogen-bond, so they have far higher boiling points and much greater water solubility than alkanes of similar Mr.
The colour change to quote: acidified dichromate goes orange (Cr₂O₇²⁻) → green (Cr³⁺) when it oxidises an alcohol. No colour change means a tertiary alcohol.
Theoretical = 0.100 mol × 60.0 = 6.00 g. % yield = (4.20 ÷ 6.00) × 100 = 70.0%.
Tap a set of reagents and conditions on the left, then the product on the right.
The C–X bond is polar (Cδ+–Xδ−), so the carbon is open to attack by a nucleophile — an electron-pair donor with a lone pair.
Rate of hydrolysis is set by the C–X bond enthalpy, not the electronegativity. C–I is the weakest bond (about 228 kJ mol⁻¹) so iodoalkanes hydrolyse fastest; C–F is the strongest (about 467 kJ mol⁻¹) and is essentially inert. A silver-nitrate-in-ethanol test shows this beautifully: the yellow AgI precipitate appears first, the white AgCl last.
Ozone: CFCs undergo homolytic fission in the stratosphere, releasing Cl• radicals which catalyse ozone breakdown: Cl• + O₃ → ClO• + O₂, then ClO• + O → Cl• + O₂. The Cl• is regenerated, so one radical destroys many ozone molecules.
Infrared: bonds absorb IR at frequencies that make them vibrate. Key absorptions to know:
The fingerprint region (below 1500 cm⁻¹) is unique to each compound and is used for identification by comparison with a database.
Mass spectrometry: the peak at the highest m/z is the molecular ion peak, M⁺, and it gives the Mr directly. Fragment peaks arise when M⁺ breaks up (e.g. loss of CH₃ = 15, loss of OH = 17, loss of C₂H₅ = 29).
Chlorine gives itself away. Because ³⁵Cl and ³⁷Cl exist in a roughly 3 : 1 ratio, a chlorine-containing compound shows an M⁺ and an M+2 peak with a 3 : 1 height ratio.
Mr = 36 + 8 + 16 = 60. Loss of a CH₃ group (15) from this would give a fragment at m/z 45.
Tap a reaction, then tap the mechanism it follows.
OCR expect you to plan a two-stage route between any functional groups in the module. Learn the map:
The exam's favourite trap: the solvent decides the outcome for a haloalkane and KOH. Aqueous → OH⁻ acts as a nucleophile → substitution → alcohol. Hot and ethanolic → OH⁻ acts as a base → elimination → alkene.
Atom economy = (2 × 46.0) ÷ 180.0 × 100 = 92.0 ÷ 180.0 × 100 = 51.1%. Almost half the glucose mass ends up as waste CO₂ — which is why the hydration of ethene (100% atom economy) is used industrially.
Isomerism: structural (chain, position, functional group) and E/Z stereoisomerism about a C=C
Alkanes: free-radical substitution — initiation (UV, homolytic), propagation, termination; gives a mixture
Alkenes: electrophilic addition via a carbocation; the more stable carbocation gives the major product
Alcohols: primary → aldehyde (distil) → acid (reflux); secondary → ketone; tertiary not oxidised
Haloalkanes: nucleophilic substitution; rate follows C–X bond enthalpy — C–I fastest
Solvent decides: aqueous KOH → substitution; hot ethanolic KOH → elimination
Analysis: IR (broad 2500–3300 = COOH; 1700 = C=O) · MS (M⁺ gives Mr; M+2 at 3 : 1 = chlorine)
That is Module 4 — three mechanisms and a synthesis map you can navigate in both directions. Press Finish to see your score.
You've worked through Core Organic Chemistry for OCR A-level Chemistry A (H432). 🎉
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