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Halogenoalkanes (RX) are saturated organic compounds containing a polar C–X bond, where X = F, Cl, Br, or I. The halogen is more electronegative than carbon, so the carbon carries a δ+ partial charge and the halogen a δ−. This polarisation makes the carbon electrophilic — vulnerable to nucleophiles, species that donate a lone pair. Halogenoalkane chemistry is dominated by nucleophilic substitution (replacement of X⁻ by a new group) and the competing elimination reaction (loss of HX to give an alkene). This lesson classifies halogenoalkanes as primary, secondary, or tertiary; develops SN1, SN2, E1, and E2 mechanisms; identifies the four exam-critical nucleophiles (H₂O, OH⁻, NH₃, CN⁻) and their products; analyses the substrate, nucleophile, and solvent factors that decide the pathway; and closes with the CFC ozone-depletion case study that drove a global treaty.
Spec mapping (AQA 7405): This lesson maps to §3.3.3 (halogenoalkanes — nucleophilic substitution, elimination, and the role of halogenoalkanes in the troposphere). It builds directly on lesson 4 of this course (alkenes — the source of halogenoalkanes via electrophilic addition of HX or X₂, and the elimination product of dehydrohalogenation) and lesson 7 (alcohols, which are formed by substitution with OH⁻ and which can be re-converted to halogenoalkanes by HX, PCl₃, or SOCl₂). The mechanism master class in lesson 8 of this course consolidates curly-arrow conventions across SN1, SN2, E1, and E2. The kinetics framework in §3.1.9 supplies the rate-equation language (rate-determining step, order with respect to substrate vs nucleophile) used to distinguish SN1 from SN2 experimentally. Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: AO1 recall items include classifying halogenoalkanes (primary/secondary/tertiary), defining SN1, SN2, E1, and E2, and naming the four standard nucleophiles. AO2 application appears in every Paper 2: draw mechanisms with correctly oriented curly arrows, identify intermediates and transition states, and predict the major product from given conditions. AO3 evaluation includes rationalising why a given substrate follows SN1 rather than SN2 (or vice versa) using substrate steric and electronic effects, nucleophile strength, and solvent polarity, and evaluating the environmental significance of the CFC-ozone catalytic cycle and the Montreal Protocol response.
On the Pauling scale, C = 2.55 against F = 3.98, Cl = 3.16, Br = 2.96, I = 2.66. The δ+ carbon is the site of nucleophilic attack. The dipole magnitude follows electronegativity difference: C–F is the most polar, C–I the least — yet reactivity is opposite. Reactivity is kinetically controlled by bond enthalpy, not polarity.
| Bond | Bond enthalpy / kJ mol⁻¹ | Bond length / pm | Dipole | Reactivity |
|---|---|---|---|---|
| C–F | 484 | 138 | Largest | Least reactive |
| C–Cl | 338 | 177 | Large | Moderate |
| C–Br | 276 | 194 | Moderate | Reactive |
| C–I | 238 | 214 | Smallest | Most reactive |
Bond enthalpy decreases down Group 17 because the halogen atom gets larger and orbital overlap with carbon's 2p orbital is poorer. Iodoalkanes are most reactive even though C–I is the least polar — kinetics (ease of bond breaking) dominates over thermodynamics (initial polarity). This is why the silver-nitrate test (below) gives the fastest precipitate for AgI.
Key Definition: A nucleophile is a species that donates a lone pair of electrons to form a new covalent bond with an electron-deficient atom. Nucleophiles are either negatively charged ions (OH⁻, CN⁻, RO⁻, X⁻) or neutral molecules with a lone pair (H₂O, NH₃, ROH). Strength depends on basicity, polarisability, and the solvent.
The mechanism a halogenoalkane follows depends critically on the number of carbon substituents on the δ+ carbon.
| Classification | Substituents on C–X carbon | Example | Preferred mechanism |
|---|---|---|---|
| Primary (1°) | 1 (or 0 for CH₃X) | CH₃CH₂Br (bromoethane) | SN2 / E2 |
| Secondary (2°) | 2 | (CH₃)₂CHBr (2-bromopropane) | mixture |
| Tertiary (3°) | 3 | (CH₃)₃CBr (2-bromo-2-methylpropane) | SN1 / E1 |
Two factors decide the mechanism. Steric crowding around the δ+ carbon disfavours SN2 (the nucleophile cannot reach the back face) and favours SN1. Electronic stabilisation of the carbocation intermediate also favours SN1 with more substitution: tertiary carbocations are stabilised by inductive donation of three alkyl groups and by hyperconjugation; methyl cations are too unstable to form. The effects reinforce: primary substrates have neither steric crowding nor cation stability so they cannot do SN1; tertiary substrates have both so they cannot do SN2.
AQA examines four nucleophiles. Each gives a distinct product class.
Aqueous NaOH (or KOH) at reflux. OH⁻ acts as a nucleophile, attacking the δ+ carbon and displacing halide. Bromoethane + aqueous NaOH:
CH₃CH₂Br + NaOH(aq) → CH₃CH₂OH + NaBr
Solvent matters: aqueous NaOH gives substitution; ethanolic NaOH gives elimination (below).
Water is a weaker nucleophile than OH⁻ — its lone pair on neutral O is less available than on O⁻. Slow but proceeds via an oxonium intermediate R–OH₂⁺ that loses a proton. Forms the basis of the silver-nitrate hydrolysis-rate test.
CH₃CH₂Br + H₂O → CH₃CH₂OH + HBr
NH₃'s lone pair on nitrogen attacks the δ+ carbon; after C–N bond formation, R–NH₃⁺ is deprotonated by a second NH₃ to give the neutral amine.
CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
Sealed tube keeps NH₃ in solution under pressure. Excess NH₃ is essential because the primary amine is itself nucleophilic and reacts further to give secondary amine (CH₃CH₂)₂NH, tertiary amine (CH₃CH₂)₃N, and ultimately the quaternary ammonium salt (CH₃CH₂)₄N⁺Br⁻.
Cyanide attacks via carbon (the more nucleophilic end). Product is a nitrile, R–C≡N. This extends the carbon chain by one — invaluable in synthesis.
CH₃CH₂Br + KCN → CH₃CH₂CN + KBr
Propanenitrile can be hydrolysed (aqueous acid reflux) to propanoic acid CH₃CH₂COOH, or reduced (LiAlH₄ or H₂/Ni) to propylamine CH₃CH₂CH₂NH₂. The CN⁻ → COOH route is a standard chain-extension strategy.
| Nucleophile | Conditions | Product class | Example product |
|---|---|---|---|
| OH⁻ (aq) | Aqueous NaOH, reflux | Alcohol | CH₃CH₂OH |
| H₂O | Water, heat (slow) | Alcohol | CH₃CH₂OH |
| NH₃ | Excess in ethanol, sealed tube, heat | Primary amine | CH₃CH₂NH₂ |
| CN⁻ | KCN in ethanol/water, reflux | Nitrile | CH₃CH₂CN |
SN2: Substitution, Nucleophilic, bi2molecular — two molecules in the rate-determining step.
A single concerted step: nucleophile approaches the δ+ carbon from the back side, opposite to the leaving group. As the new C–Nu bond forms, the old C–X bond breaks in the same motion. No intermediate; one transition state with the carbon approximately sp² hybridised, nucleophile and leaving group partially bonded on opposite faces.
Worked example: CH₃Br + OH⁻. OH⁻ approaches opposite to Br. At the transition state the carbon is sp²-planar with H atoms in the equatorial plane and OH⁻ and Br⁻ partially bonded on the two faces, each carrying δ−. The transition state collapses: OH–C becomes a full σ bond and Br departs with both electrons.
Curly arrows: From the OH⁻ lone pair to the δ+ C; from the C–Br bond to Br. Both on one diagram because the step is concerted.
Rate equation: rate = k[CH₃Br][OH⁻]. First order in each, second order overall.
Stereochemistry: inversion of configuration. The three substituents on the central carbon are forced through the planar transition state and emerge on the opposite face — like an umbrella inverting. (R)-2-bromobutane gives (S)-2-butanol (Walden inversion). For achiral CH₃Br the inversion is not observable, but the mechanism still operates.
Why primary? Steric: one alkyl group leaves the back face open. Tertiary substrates have three alkyl groups blocking the back face.
SN1: Substitution, Nucleophilic, uni1molecular — one molecule (the substrate) in the rate-determining step.
Two steps via a carbocation intermediate.
Step 1 (slow, rate-determining). The C–X bond breaks heterolytically; both electrons depart with the halide, leaving a carbocation with an empty p orbital. The tertiary cation is stabilised by three alkyl groups (positive inductive effect plus hyperconjugation from C–H σ bonds into the empty p orbital).
(CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻
Curly arrow: C–Br bond → Br.
Step 2 (fast). Nucleophile attacks the planar carbocation.
(CH₃)₃C⁺ + OH⁻ → (CH₃)₃COH
Curly arrow: OH⁻ lone pair → C⁺.
Rate equation: rate = k[RX]. The nucleophile appears in step 2 but not in the slow step, so doubling [OH⁻] does not double the rate — the diagnostic kinetic test distinguishing SN1 from SN2.
Stereochemistry: racemisation. The planar sp² carbocation is attacked from either face with equal probability. A single enantiomer gives a racemic 50:50 mixture — loss of optical activity is a hallmark of SN1.
Why tertiary? Tertiary carbocations are most stable (lowest activation energy for the slow step), and tertiary substrates are too crowded for SN2 to compete.
Real reactions span a continuum; secondary halogenoalkanes especially can react by either pathway depending on conditions. Four factors decide which mechanism dominates.
1. Substrate. Primary substrates can only do SN2; tertiary only SN1; secondary go either way. Methyl halides (CH₃X) react exclusively by SN2 — there is no carbocation stabilisation at all.
2. Nucleophile strength. Strong nucleophiles (OH⁻, CN⁻, RO⁻) favour SN2 because they participate in the rate-determining step and accelerate it. Weak nucleophiles (H₂O, ROH) cannot push out the leaving group concertedly and so allow SN1 to compete — the substrate ionises first and then the weak nucleophile attacks the carbocation.
3. Solvent. Polar protic solvents (water, alcohols) stabilise carbocations and halide ions by hydrogen bonding, favouring SN1. Polar aprotic solvents (acetone, DMSO, DMF) solvate cations but leave the nucleophile naked and reactive, favouring SN2.
4. Temperature. Higher temperature favours elimination over substitution (the elimination transition state typically has higher activation energy but more disorder — entropy-favoured), and also slightly favours SN1 over SN2 (the unimolecular ionisation has the higher activation energy).
| Feature | SN1 | SN2 |
|---|---|---|
| Steps | 2 (via carbocation) | 1 (concerted) |
| Rate-determining step | Ionisation of RX (slow) | Concerted attack and departure |
| Rate equation | rate = k[RX] | rate = k[RX][Nu] |
| Substrate preference | Tertiary > secondary >> primary | Primary > secondary >> tertiary |
| Stereochemistry | Racemisation | Inversion (Walden) |
| Nucleophile strength | Weak nucleophile tolerated | Strong nucleophile required |
| Solvent preference | Polar protic (H₂O, ROH) | Polar aprotic (acetone, DMSO) |
| Intermediate | Carbocation | None (transition state only) |
Exam Tip: Show curly arrows starting from the nucleophile's lone pair (or from a bond, never from a positive charge). For SN1, draw the carbocation intermediate with a clear + charge and an empty p orbital. For SN2, place both curly arrows on a single diagram showing the transition state with partial bonds; state explicitly that the mechanism is concerted.
Substitution is not the only fate of a halogenoalkane plus base. The base can also remove a β-hydrogen (a hydrogen on the carbon adjacent to the C–X carbon), leading to loss of HX and formation of an alkene.
Reaction: CH₃CH₂Br + NaOH → CH₂=CH₂ + NaBr + H₂O
Conditions: hot, concentrated, ethanolic NaOH. In ethanol (polar but less polar than water) the OH⁻ is a less effective nucleophile and a more effective base — it removes the β-proton rather than attacking the δ+ carbon.
E2 mechanism (bimolecular, for primary and secondary substrates). Concerted: the base removes the β-H as the C–X bond breaks and the C=C π bond forms. All three events happen in one step. Three curly arrows on one diagram: from the lone pair on OH⁻ to the β-H; from the β-C–H bond to form a new C=C π bond; from the C–X bond to X.
E1 mechanism (unimolecular, for tertiary substrates). Two steps: identical first step to SN1 (C–X bond breaks to form a carbocation), then a base removes a β-H to form the alkene. Rate = k[RX].
Effect of substrate. Primary halogenoalkanes overwhelmingly favour substitution (SN2) over elimination because the back face is open. Tertiary halogenoalkanes give substantial elimination because the crowded back face blocks substitution but the β-hydrogens are easy to reach. The competition between SN1 and E1 in tertiary substrates is mostly controlled by temperature — hot conditions favour elimination.
| Substrate | Aqueous NaOH | Ethanolic NaOH (hot) |
|---|---|---|
| Primary | Substitution (alcohol) | Mostly substitution; some elimination |
| Secondary | Mixture, substitution major | Mixture, elimination major |
| Tertiary | Substitution + elimination | Elimination (alkene) |
Common Misconception: Students write "NaOH gives substitution" without specifying solvent. The solvent is the deciding variable on the exam: aqueous OH⁻ = nucleophile = substitution; ethanolic OH⁻ = base = elimination. State both the role of OH⁻ and the solvent on every mechanism question.
Relative reactivity of C–Cl, C–Br, and C–I is shown by warming three halogenoalkanes (e.g. 1-chlorobutane, 1-bromobutane, 1-iodobutane) with aqueous silver nitrate dissolved in ethanol (ethanol co-solvent because halogenoalkanes are immiscible with water).
The C–X bond hydrolyses (slow nucleophilic substitution by water): RX + H₂O → ROH + H⁺ + X⁻. The released halide is immediately precipitated as silver halide:
X⁻(aq) + Ag⁺(aq) → AgX(s)
Fastest precipitate forms with the iodoalkane (C–I weakest); slowest with chloroalkane.
| Halogenoalkane | Precipitate | Colour | Time to appear |
|---|---|---|---|
| R–Cl | AgCl | White | Slowest |
| R–Br | AgBr | Cream | Moderate |
| R–I | AgI | Yellow | Fastest |
Confirmation by ammonia: AgCl dissolves in dilute NH₃; AgBr in concentrated NH₃; AgI insoluble in both. The experiment must use the same temperature, concentration, and solvent composition; the rate measurement confirms the bond-enthalpy order.
Chlorofluorocarbons (CFCs) are halogenoalkanes containing only Cl, F, and C, e.g. CCl₃F (CFC-11), CCl₂F₂ (CFC-12, freon). Their C–F bonds are very strong, they are non-flammable, non-toxic at ground level, and chemically inert — properties that made them ideal as refrigerants, aerosol propellants, and foam-blowing agents in the mid-twentieth century.
The inertness is the problem. CFCs released at ground level survive transit to the stratosphere (years), where short-wavelength UV-C radiation (λ < 240 nm) photolytically cleaves the weakest bond — the C–Cl bond, not the C–F bond:
CCl₂F₂ → ·CClF₂ + Cl·
The chlorine atom is a free radical that initiates a catalytic destruction cycle for ozone (O₃):
Cl· + O₃ → ClO· + O₂ ClO· + O → Cl· + O₂
Net: O₃ + O → 2 O₂
The Cl· radical is regenerated. A single chlorine atom can destroy tens of thousands of ozone molecules before it is removed (typically by reaction with methane or NO₂ to form HCl or ClONO₂ "reservoir species"). The dramatic seasonal collapse of stratospheric ozone over Antarctica — the "ozone hole" first reported in 1985 — was traced quantitatively to CFC photolysis.
The international response was the Montreal Protocol (1987), which phased out CFC production. CFCs were replaced first by HCFCs (hydrochlorofluorocarbons, e.g. CHClF₂) — which contain a C–H bond that is removed by hydroxyl radicals in the troposphere before reaching the stratosphere — and then by HFCs (hydrofluorocarbons, no chlorine at all) and increasingly by non-halogenated alternatives. The Montreal Protocol is widely regarded as the most successful environmental treaty: stratospheric chlorine peaked around 1997 and is now declining, and the ozone layer is projected to recover to 1980 levels by ~2070.
The case study illustrates a recurring theme of organic chemistry — the same property (kinetic inertness) that makes a molecule a useful industrial chemical can make it a long-term environmental liability. The catalytic nature of the Cl· cycle means that even tiny atmospheric concentrations of CFCs are damaging.
Question 1. [13 marks total]
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