HookThe plastic age began with a leaky flask
On 27 March 1933, two ICI chemists at Winnington in Cheshire set out to react ethene with benzaldehyde at about 170 °C and nearly 2,000 atmospheres. The experiment failed — the pressure vessel leaked overnight — but when Eric Fawcett and Reginald Gibson opened it they found the walls coated with a waxy white solid: roughly 0.4 g of poly(ethene). Nobody could repeat the result reliably, because the true initiator had been a trace of oxygen sneaking in through the leak, and it took until December 1935 for ICI to make the material on purpose. The first production plant came on stream on 1 September 1939 — the day the Second World War began — and the new plastic promptly vanished into military secrecy, insulating the cables of airborne radar sets, because it was light, waterproof and a superb electrical insulator.
Everything in that story runs on one structural feature: the carbon–carbon double bond. Section 3.3.4 asks three things of you. Explain why C=C makes alkenes reactive when alkanes sit inert — the double bond as a centre of high electron density. Predict and draw what happens when electrophiles attack it — electrophilic addition, carbocation intermediates and Markownikoff's rule. And scale it up: when the double bond opens once per molecule across tens of thousands of molecules, you get addition polymers, whose usefulness and whose disposal problem both trace straight back to the bonding.
ModelThe double bond — a fat, exposed centre of electron density
An alkene is an unsaturated hydrocarbon: it contains a carbon–carbon double bond, so it carries fewer hydrogens than the alkane with the same skeleton (general formula CₙH₂ₙ for one double bond and no rings). Four electrons sit in that one small region between the two carbons, and the phrase AQA pays for is exactly that: the double bond is a centre of high electron density. The fuller textbook picture is that one electron pair forms a normal σ bond on the line between the nuclei, while the second pair — the π component — sits in lobes above and below the plane of the molecule, held less tightly. The bond-enthalpy data agree: C=C is about 612 kJ mol⁻¹ against 347 kJ mol⁻¹ for C–C, so the second bond adds only around 265 kJ mol⁻¹ — much less than doubling the strength. The weaker half is the one reactions spend.
The geometry does the rest. The two double-bond carbons and the four atoms attached to them lie flat in one plane, bond angles about 120°. Rotation about the double bond is restricted — twisting would break the sideways overlap — which is why E and Z stereoisomers of alkenes such as but-2-ene are separate, isolable compounds (that idea belongs to 3.3.1.3, but it is the same bonding fact wearing different clothes).
Put the two halves together and the chemistry writes itself: an exposed, electron-rich region attracts electrophiles — species that can accept an electron pair — and the energetically cheap move is to break the weaker second bond while keeping the σ framework intact. That is why alkenes do addition where alkanes can only manage substitution, and why alkenes react in seconds at room temperature with reagents that leave alkanes untouched.
MechanismElectrophilic addition — three reagents, one mechanism
Three electrophiles cover the whole of 3.3.4.2: hydrogen bromide, bromine and concentrated sulfuric acid. HBr arrives ready-polarised — bromine is the more electronegative atom, so the molecule is permanently Hδ+–Brδ−. Br₂ has no permanent dipole, and AQA expects the extra sentence: as the alkene approaches, its electron-rich double bond repels the bonding pair in Br–Br, inducing a dipole so the nearer bromine becomes δ+. From there every reaction is the same three-move dance, drawn with curly arrows. Move one: an arrow from the double bond to the δ+ atom of the electrophile. Move two: the electrophile's own bond breaks heterolytically, an arrow carrying its pair onto the leaving atom. What remains is a carbocation intermediate — one carbon short of an octet, positively charged — plus a negative ion. Move three: an arrow from a lone pair on that anion (Br⁻ or hydrogensulfate) into the positive carbon.
Cold concentrated sulfuric acid adds exactly this way: ethene gives ethyl hydrogensulfate, CH₃CH₂OSO₂OH, and warming the product with water hydrolyses it to ethanol and regenerates the H₂SO₄ — the old industrial route to ethanol, and a preview of the hydration chemistry in 3.3.5.
The bromine reaction doubles as the standard test for unsaturation: shake the unknown with orange bromine water. A C=C bond consumes the bromine by addition, so the orange colour is decolourised; an alkane leaves the flask stubbornly orange. Two observations, stated before and after — that is the whole test.
Propene plus bromine, narrated the way the exam wants it drawn. Arrow one starts unambiguously at the double bond and lands on the δ+ bromine of the induced dipole. Arrow two takes the Br–Br bonding pair onto the far bromine, releasing Br⁻. The intermediate is the secondary carbocation CH₃CH⁺CH₂Br — draw the + on the middle carbon, not floating in space. Arrow three runs from a lone pair on Br⁻ to that carbon, giving 1,2-dibromopropane: \(\text{CH}_3\text{CH}=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3\text{CHBrCH}_2\text{Br}\). Observation: the orange solution becomes colourless. AQA pays per arrow, and every arrow must start at an electron pair — a bond or a lone pair — never at an atom's symbol or at a charge.
ModelMarkownikoff's rule — the carbocation decides the major product
Add HBr to propene and two products are possible: 1-bromopropane if the hydrogen bonds to the middle carbon, 2-bromopropane if it bonds to the end. In 1869 the Russian chemist Vladimir Markownikoff codified what experiment shows: hydrogen adds to the carbon that already carries more hydrogens, so 2-bromopropane dominates. But the rule is only the shadow of the cause, and AQA marks the cause: carbocation stability.
The two routes pass through different intermediates. Hydrogen adding to the end carbon parks the positive charge on the middle carbon — a secondary carbocation, two alkyl groups attached. Hydrogen adding to the middle carbon leaves a primary carbocation with just one. Alkyl groups release electron density towards the positive centre through the σ framework — the positive inductive effect — spreading the charge and lowering the intermediate's energy. Stability therefore climbs primary < secondary < tertiary. The secondary route runs through the lower-energy intermediate, so it happens faster and supplies the major product.
Mind the honest wording. Major, not only: some of the less stable carbocation still forms, so the minor product genuinely exists in the flask. And when the alkene is symmetrical the whole question dissolves — but-2-ene plus HBr gives 2-bromobutane whichever carbon takes the hydrogen, a favourite AQA trap for students who reach for the rule before looking at the molecule.
But-1-ene reacts with HBr. Route one: H adds to carbon-1, giving CH₃CH₂CH⁺CH₃ — a secondary carbocation stabilised by electron release from two alkyl groups. Br⁻ attacks it to give 2-bromobutane, the major product. Route two: H adds to carbon-2, giving the primary carbocation CH₃CH₂CH₂CH₂⁺, stabilised by only one alkyl group; its product, 1-bromobutane, is the minor one. The sentence that earns the explanation marks: 'the secondary carbocation is more stable than the primary because two electron-releasing alkyl groups spread the positive charge, so it forms preferentially and leads to the major product.' Additions like this typically run around nine parts major to one part minor — so write major and minor, never 'the product'.
ModelAddition polymers — one molecule, repeated ten thousand times
Persuade the double bonds of many alkene molecules to open — high pressure and a trace of oxygen in ICI's original process, or modern catalysts — and each monomer forms two new σ bonds to its neighbours. Tens of thousands of units link into one long saturated chain: addition polymerisation. Nothing else is produced, so the atom economy is 100% and the polymer's empirical formula is the monomer's.
AQA's marks here are mostly notation, and the conventions are strict. The repeating unit comes from the monomer by converting the double bond to a single bond and extending continuation bonds out through square brackets, subscript n outside: poly(ethene) is [–CH₂–CH₂–]ₙ. Substituents stay exactly where they started — chloroethene, CH₂=CHCl, gives poly(chloroethene), PVC, with a chlorine on every second backbone carbon. Naming is mechanical: poly(monomer name), brackets included. You must also read the notation backwards: given a drawn stretch of chain, find the shortest block that regenerates it, then restore the double bond to identify and name the monomer.
Chemically, the backbone you have built is a very long alkane: saturated, non-polar, σ-bonded. That is why addition polymers are so unreactive — no electron-rich double bond remains for electrophiles to attack, and the strong C–C and C–H framework shrugs off water, acids and bases. Useful in a food wrapper; a genuine problem in landfill, where the same inertness means the material persists for centuries (the disposal chemistry is developed at A2 in 3.3.12).
Propene, CH₃CH=CH₂, polymerises. Convert C=C to C–C, run the continuation bonds through the brackets, and the repeating unit is [–CH₂–CH(CH₃)–]ₙ: a two-carbon backbone with a methyl branch. The methyl group hangs off the chain — drawing a three-carbon backbone is the classic dropped mark, because polymerisation joins monomers only through the two double-bond carbons. Name: poly(propene). In reverse: a chain drawn …–CH₂–CHCl–CH₂–CHCl–… has repeating unit [–CH₂–CHCl–]ₙ, so the monomer is chloroethene. The bookkeeping scales, too: a poly(ethene) molecule with \(M_r \approx 280{,}000\) contains \(280{,}000 \div 28 = 10{,}000\) monomer units.
CaseTuning a polymer without touching a covalent bond
Between the chains of an addition polymer there are only weak intermolecular forces — but multiply weak by a chain thousands of atoms long and packing starts to dictate properties. ICI's high-pressure process makes low-density poly(ethene): the chains grow branched, cannot pack closely, and the product is soft and flexible — density about 0.92 g cm⁻³, softening around 105 °C, the stuff of carrier bags and squeezy bottles. Catalyst processes developed in the 1950s make high-density poly(ethene) instead: almost unbranched chains that pack tightly, giving more surface contact and stronger total van der Waals attraction — density about 0.96 g cm⁻³, softening nearer 130 °C, stiff enough for water pipes and wheelie bins. Same monomer, identical repeating unit; the difference between a sandwich bag and a bin is entirely intermolecular.
PVC pushes the idea further. Its polar C–Cl bonds add permanent dipole–dipole attractions between chains, so unmodified PVC is rigid — uPVC window frames and guttering. To make cable sheathing and flooring, manufacturers blend in a plasticiser: small molecules that sit between the polymer chains, prising them apart and letting them slide over one another. Weaker forces between chains, flexible material — a complete change of mechanical character achieved without altering a single covalent bond. When AQA hands you an unfamiliar polymer and asks why one grade is flexible and another rigid, this is always the shape of the answer: name the forces between chains, then explain what branching, polar groups or plasticiser molecules do to them.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Mechanism questions are marked arrow by arrow, and AQA's conventions are non-negotiable: every curly arrow starts at an electron pair — the double bond, a bonding pair, or a lone pair — and ends where the pair goes. Arrows that start at atoms, charges or hydrogens score nothing. For bromine, an early mark is often the induced dipole: label Brδ+–Brδ− and say the alkene's electron-rich double bond repels the Br–Br bonding pair. Draw the carbocation intermediate explicitly with the + on the correct carbon, and finish with an arrow from a lone pair on the anion. A mechanism that reaches the right product with arrows from the wrong places is a wrong mechanism.
'Explain why X is the major product' is a three-step AO2 answer: name both possible carbocations (secondary versus primary), state that alkyl groups release electron density so the more substituted ion is more stable, and conclude that the more stable intermediate forms faster and leads to the major product. Quoting the rule alone caps you at the bottom of the marks — the rule is AO1 recall, the carbocation argument is the application AQA is actually testing, and application carries more of the paper than recall does.
Observations are before-and-after statements: 'orange solution becomes colourless', never 'it decolourises' on its own and never 'clear'. Polymer drawing is pure accuracy — square brackets, continuation bonds passing through them, no double bond inside, subscript n, side groups on a two-carbon backbone — and examiners accept no partial credit for a repeat unit with the substituent absorbed into the chain. Finally, always check whether the alkene is symmetrical before writing about major and minor products; with but-2-ene there is only one addition product, and spotting that is worth the same marks as the full inductive-effect argument would have been.