HookOne chlorine atom, a hundred thousand dead ozone molecules
In 1985 scientists from the British Antarctic Survey at Halley Bay reported something almost no one wanted to believe: the ozone above Antarctica had thinned so severely each spring that there was, in effect, a hole. The culprit had been named a decade earlier by Molina and Rowland — chlorofluorocarbons, the seemingly perfect, unreactive gases used in fridges and aerosols. Their very stability was the danger: they survive long enough to drift up into the stratosphere, where ultraviolet light finally breaks them and frees chlorine radicals. A single chlorine radical then destroys on the order of a hundred thousand ozone molecules before it is removed, because it works as a catalyst — regenerated on every cycle.
Halogenoalkanes are defined by one feature, the carbon–halogen bond, and Section 3.3.3 turns on that bond twice over. In the lab the \(\text{C}-\text{X}\) bond is polar, so halogenoalkanes are attacked by nucleophiles (substitution) and, under different conditions, undergo elimination to make alkenes. Their reactivity, though, is set not by polarity but by bond enthalpy — and getting that distinction right is the single most examined idea in the topic. In the atmosphere, the \(\text{C}-\text{Cl}\) bond in a CFC is the weak link UV light cleaves to start the ozone-destroying chain. Master the two mechanisms, the rate trend and the radical chemistry, and you have the whole section.
ModelThe C–X bond — polar enough to attack, but enthalpy sets the pace
Every halogen is more electronegative than carbon, so the carbon–halogen bond is polar: the carbon carries a partial positive charge (\(\delta+\)) and the halogen a partial negative charge (\(\delta-\)). That \(\delta+\) carbon is the target a nucleophile attacks — the origin of every substitution reaction in this topic. But here is the trap AQA sets almost every year: the rate at which different halogenoalkanes react is governed by the \(\text{C}-\text{X}\) bond enthalpy, not by the polarity.
The \(\text{C}-\text{F}\) bond is the most polar of the four, yet it is also the strongest (mean bond enthalpy 484 kJ mol⁻¹) and the least reactive. The \(\text{C}-\text{I}\) bond is the least polar, yet the weakest (238 kJ mol⁻¹) and the most reactive, because a weak bond breaks most easily. So reactivity rises in the order \(\text{C}-\text{F} < \text{C}-\text{Cl} < \text{C}-\text{Br} < \text{C}-\text{I}\) — the exact reverse of the polarity order. Whenever a question asks you to compare or explain reactivity, reach for bond enthalpy, never for polarity.
MechanismNucleophilic substitution — three reagents, one move
A nucleophile is an electron-pair donor — it carries a lone pair and is drawn to the \(\delta+\) carbon. The mechanism is always the same choreography: a curly arrow from the nucleophile’s lone pair to the \(\delta+\) carbon forms the new bond, while a second curly arrow from the \(\text{C}-\text{X}\) bond to the halogen breaks it heterolytically, so the halogen leaves as \(\text{X}^{-}\). Only the nucleophile changes between the three AQA reactions.
With warm aqueous hydroxide (NaOH or KOH), the product is an alcohol — this is hydrolysis. With ethanolic potassium cyanide heated under reflux, the product is a nitrile, and crucially the cyanide adds a carbon to the chain, making it the reagent of choice when a synthesis needs to lengthen a carbon skeleton. With excess ammonia dissolved in ethanol in a sealed tube, the product is a primary amine, the excess ammonia there to limit further substitution of the product. Same arrows every time; only the attacking species differs.
Hydrolysis of bromoethane by warm aqueous sodium hydroxide: \(\text{CH}_3\text{CH}_2\text{Br} + \text{OH}^{-} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^{-}\). Mark \(\delta+\) on the carbon bonded to bromine and \(\delta-\) on the bromine. Draw a curly arrow from a lone pair on the hydroxide oxygen (\(:\text{OH}^{-}\)) to that carbon, and a second from the middle of the \(\text{C}-\text{Br}\) bond onto the bromine; the products are ethanol and a bromide ion. Swap in cyanide, \(\text{CH}_3\text{CH}_2\text{Br} + \text{CN}^{-} \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{Br}^{-}\), and the identical mechanism builds propanenitrile — a three-carbon product from a two-carbon start, which is exactly why cyanide is the standard trick for growing a chain by one carbon.
DataThe silver-nitrate rate test — watching bond enthalpy decide
The reactivity trend can be measured directly in a school laboratory, and the experiment is a clean lesson in controlling variables. Set up separate tubes of 1-chlorobutane, 1-bromobutane and 1-iodobutane; add ethanol as a common solvent so the organic halogenoalkane and the aqueous reagent mix into one layer; stand the tubes in a water bath at a fixed temperature; then add aqueous silver nitrate and start a stopwatch. As each \(\text{C}-\text{X}\) bond hydrolyses, the halide ion released reacts with the silver ion to give a silver-halide precipitate — white \(\text{AgCl}\), cream \(\text{AgBr}\), yellow \(\text{AgI}\) — and you time how long the precipitate takes to appear.
The independent variable is the halogen; the dependent variable is the time to precipitate; the control variables are temperature (the water bath), the concentration and volume of silver nitrate, the alkyl chain (all four-carbon, all primary) and the total volume. The iodo compound reacts fastest and the chloro slowest — the weakest \(\text{C}-\text{I}\) bond breaks first, precisely as bond enthalpy predicts and precisely opposite to what polarity would predict.
Typical class results at 50 °C: 1-iodobutane gives a yellow precipitate in about 5 s, 1-bromobutane a cream one in about 40 s, and 1-chlorobutane often shows only a faint cloudiness after several minutes. Line those times up against the mean \(\text{C}-\text{X}\) bond enthalpies — \(\text{C}-\text{I}\) 238, \(\text{C}-\text{Br}\) 276, \(\text{C}-\text{Cl}\) 338 kJ mol⁻¹ — and the fastest reaction has the weakest bond. The main errors are subjective: ‘first appearance of a precipitate’ is a judgement call, so onset times scatter between observers; and unless every tube sits in the same water bath, a few degrees’ difference changes the rate more than the halogen does, which is why temperature is the control you guard most carefully. The whole test works precisely because it isolates bond enthalpy — polarity would give the reverse order, and the reverse is not what you see.
MechanismElimination — the same molecule, a different fork in the road
Give a halogenoalkane hydroxide ions in hot, ethanolic conditions rather than warm aqueous ones, and the hydroxide stops behaving as a nucleophile and behaves as a base. Instead of attacking the \(\delta+\) carbon, it pulls a hydrogen off the carbon adjacent to the one bearing the halogen. A curly arrow from that \(\text{C}-\text{H}\) bond forms a \(\text{C}=\text{C}\) double bond, a second arrow from the \(\text{C}-\text{X}\) bond expels \(\text{X}^{-}\), and the hydroxide leaves with the hydrogen as water. The product is an alkene.
Substitution and elimination therefore compete for the same starting material, and the conditions tip the balance. Warm aqueous hydroxide favours substitution (an alcohol); hot ethanolic hydroxide favours elimination (an alkene). Structure matters too: primary halogenoalkanes lean towards substitution, tertiary ones towards elimination. The same reagent, potassium hydroxide, gives opposite products depending on the solvent and the structure — which is why reading the conditions in the question is not optional.
2-bromopropane, \((\text{CH}_3)_2\text{CHBr}\), with potassium hydroxide can go two ways. Warm aqueous KOH gives substitution: \((\text{CH}_3)_2\text{CHBr} + \text{OH}^{-} \rightarrow (\text{CH}_3)_2\text{CHOH} + \text{Br}^{-}\), producing propan-2-ol, the hydroxide acting as a nucleophile. Hot ethanolic KOH gives elimination: \((\text{CH}_3)_2\text{CHBr} + \text{OH}^{-} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O} + \text{Br}^{-}\), producing propene, the hydroxide acting as a base that removes a hydrogen from a neighbouring carbon. The exam hands you the conditions and expects you to name the product and the role of the hydroxide; miss the word ‘ethanolic’ or ‘aqueous’ and you take the wrong fork entirely.
CaseCFCs and ozone — a catalyst you never wanted
High in the stratosphere, ozone (\(\text{O}_3\)) absorbs harmful ultraviolet-B radiation and shields life below. CFCs such as dichlorodifluoromethane, \(\text{CCl}_2\text{F}_2\), were engineered to be inert — non-flammable, non-toxic, unreactive — which made them superb refrigerants and aerosol propellants, but also means nothing destroys them at ground level. They diffuse up into the stratosphere, where high-energy UV is at last strong enough to break their weakest bond, \(\text{C}-\text{Cl}\) (weaker than \(\text{C}-\text{F}\)), by homolytic fission, releasing a chlorine radical.
That radical then catalyses the breakdown of ozone in a chain: \(\text{Cl}\bullet + \text{O}_3 \rightarrow \text{ClO}\bullet + \text{O}_2\), then \(\text{ClO}\bullet + \text{O}_3 \rightarrow \text{Cl}\bullet + 2\text{O}_2\). The chlorine radical is regenerated at the end, so one atom cycles through thousands of ozone molecules — the very definition of a catalyst. Because the science was decisive, the 1987 Montreal Protocol phased CFCs out, replacing them with compounds that carry no \(\text{C}-\text{Cl}\) bond; the ozone layer is now slowly recovering, one of the few global pollution problems chemistry has largely solved.
Trace the catalytic cycle for one chlorine radical. Initiation, under stratospheric UV: \(\text{CCl}_2\text{F}_2 \rightarrow \bullet\text{CClF}_2 + \text{Cl}\bullet\), homolytic fission of the weakest bond. Propagation: \(\text{Cl}\bullet + \text{O}_3 \rightarrow \text{ClO}\bullet + \text{O}_2\), then \(\text{ClO}\bullet + \text{O}_3 \rightarrow \text{Cl}\bullet + 2\text{O}_2\). Add the two propagation steps and the chlorine species cancel, leaving the overall change \(2\text{O}_3 \rightarrow 3\text{O}_2\) and a \(\text{Cl}\bullet\) free to begin again. That regeneration is why a single chlorine radical is estimated to destroy around 100,000 ozone molecules before it is finally removed — it is a homogeneous catalyst, never used up.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
Mechanism marks (AO2) are won and lost on the curly arrows. Draw the lone pair on the nucleophile explicitly and start the arrow FROM that lone pair — an arrow starting on the negative charge symbol rather than the electrons can be penalised. Show the second arrow running from the \(\text{C}-\text{X}\) bond onto the halogen, mark the \(\delta+\) and \(\delta-\), and let the halogen leave as \(\text{X}^{-}\). For elimination, the give-away arrow is the one from a \(\text{C}-\text{H}\) bond forming the new \(\text{C}=\text{C}\).
The single most examined idea is the rate trend: always attribute it to bond enthalpy, quoting the order \(\text{C}-\text{I} < \text{C}-\text{Br} < \text{C}-\text{Cl}\) for bond strength, and never to polarity — examiner reports flag ‘the C–F bond is most polar so reacts fastest’ as a top dropped mark. State conditions with full precision, because they choose the product: aqueous versus ethanolic, warm versus hot, reflux, excess ammonia.
For the ozone chemistry, identify the initiation as homolytic fission of the C–Cl bond, show the chlorine radical acting as a catalyst that is regenerated, and give the overall \(2\text{O}_3 \rightarrow 3\text{O}_2\); be ready to explain why CFCs reach the stratosphere at all — their inertness. This section is examined on Paper 2 and returns synoptically on Paper 3, where the silver-nitrate practical and the radical mechanism both recur.