Nucleophilic Substitution vs Elimination in Halogenoalkanes & Ozone Depletion
Halogenoalkanes contain a polar carbon-halogen bond (C^δ+ - X^δ-) that makes the electron-deficient carbon atom susceptible to attack by lone-pair-donating nucleophiles (:OH⁻, :CN⁻, :NH₃), while simultaneously allowing the same hydroxide ion to act as a Brønsted-Lowry base by removing an adjacent β-hydrogen atom in an elimination reaction to form an alkene. Whether nucleophilic substitution or elimination predominates depends critically on both the solvent/temperature conditions (warm aqueous NaOH vs hot ethanolic KOH) and the structure of the halogenoalkane (1° vs 2° vs 3°). Also, experimental hydrolysis rates with aqueous silver nitrate prove that C-X bond enthalpy (not bond polarity) governs reactivity (R-I > R-Br > R-Cl). Examiners also test the free-radical chain propagation equations by which stratospheric chlorine radicals (Cl ) from CFCs catalytically destroy ozone (O₃).
- C-X Bond Polarity vs Bond Enthalpy (Rates of Hydrolysis)
- Halogens are more electronegative than carbon, giving a polar C^δ+ - X^δ- bond. Although C-Cl is the most polar bond, C-X bond enthalpy decreases sharply down Group 17 as atomic radius increases: C-F (484 kJ mol⁻¹) > C-Cl (338 kJ mol⁻¹) > C-Br (276 kJ mol⁻¹) > C-I (238 kJ mol⁻¹). Because the C-I bond is the weakest (lowest activation energy to break), iodoalkanes react fastest in nucleophilic substitution, proved by warming with aqueous ethanol and AgNO₃(aq): 1-iodobutane gives a yellow AgI precipitate immediately; 1-bromobutane gives a cream AgBr precipitate slowly; 1-chlorobutane gives a white AgCl precipitate very slowly.
- Nucleophilic Substitution with :OH⁻ and :CN⁻
- A nucleophile is an electron-pair donor. Curly-arrow mechanism: 1. Draw a curly arrow from the lone pair on :OH⁻ (or :CN⁻,lone pair on the C atom) to the δ⁺ carbon atom of C^δ+ - X^δ-. 2. Draw a curly arrow from the middle of the C-X covalent bond to the δ⁻ halogen atom (heterolytic fission), forming the alcohol (or nitrile) and :X⁻ halide ion. Note: Reaction with KCN in aqueous ethanol extends the carbon chain by one carbon atom (R-X + :CN⁻ → R-C≡N + X⁻), which can then be reduced to an amine or hydrolysed to a carboxylic acid.
- Nucleophilic Substitution with Excess Ammonia (:NH₃)
- Two-stage mechanism using excess concentrated ethanolic ammonia heated in a sealed tube under pressure: 1. Curly arrow from lone pair on :NH₃ to δ⁺ C of C^δ+ - X^δ-, and curly arrow from C-X bond to X, forming a positively charged alkylammonium intermediate (R-NH₃⁺) and :X⁻. 2. A second molecule of :NH₃ acts as a Brønsted-Lowry base: draw a curly arrow from the lone pair on the second :NH₃ to an H atom of the -NH₃⁺ group, and a curly arrow from that N-H bond to the N⁺ atom, producing the primary amine (R-NH₂) and NH₄⁺X⁻.
Curly-Arrow Organic Mechanisms (Nucleophilic Substitution & Addition)
3 Golden Rules for Curly Arrows:
1. Start at electrons: Every curly arrow MUST start at a lone pair of electrons or the middle of a covalent bond (never start an arrow on a δ+ or positive charge).
2. Show dipoles: Always label δ+ and δ− on polar bonds like C–Br or H–Br.
3. Balance charges: If a neutral molecule loses a leaving ion (:Br⁻), the intermediate carbocation must carry a formal + charge.
Key Concepts & Exam Definitions (6)
- C-X Bond Polarity vs Bond Enthalpy (Rates of Hydrolysis)
- Halogens are more electronegative than carbon, giving a polar C^δ+ - X^δ- bond. Although C-Cl is the most polar bond, C-X bond enthalpy decreases sharply down Group 17 as atomic radius increases: C-F (484 kJ mol⁻¹) > C-Cl (338 kJ mol⁻¹) > C-Br (276 kJ mol⁻¹) > C-I (238 kJ mol⁻¹). Because the C-I bond is the weakest (lowest activation energy to break), iodoalkanes react fastest in nucleophilic substitution, proved by warming with aqueous ethanol and AgNO₃(aq): 1-iodobutane gives a yellow AgI precipitate immediately; 1-bromobutane gives a cream AgBr precipitate slowly; 1-chlorobutane gives a white AgCl precipitate very slowly.
- Nucleophilic Substitution with :OH⁻ and :CN⁻
- A nucleophile is an electron-pair donor. Curly-arrow mechanism: 1. Draw a curly arrow from the lone pair on :OH⁻ (or :CN⁻,lone pair on the C atom) to the δ⁺ carbon atom of C^δ+ - X^δ-. 2. Draw a curly arrow from the middle of the C-X covalent bond to the δ⁻ halogen atom (heterolytic fission), forming the alcohol (or nitrile) and :X⁻ halide ion. Note: Reaction with KCN in aqueous ethanol extends the carbon chain by one carbon atom (R-X + :CN⁻ → R-C≡N + X⁻), which can then be reduced to an amine or hydrolysed to a carboxylic acid.
- Nucleophilic Substitution with Excess Ammonia (:NH₃)
- Two-stage mechanism using excess concentrated ethanolic ammonia heated in a sealed tube under pressure: 1. Curly arrow from lone pair on :NH₃ to δ⁺ C of C^δ+ - X^δ-, and curly arrow from C-X bond to X, forming a positively charged alkylammonium intermediate (R-NH₃⁺) and :X⁻. 2. A second molecule of :NH₃ acts as a Brønsted-Lowry base: draw a curly arrow from the lone pair on the second :NH₃ to an H atom of the -NH₃⁺ group, and a curly arrow from that N-H bond to the N⁺ atom, producing the primary amine (R-NH₂) and NH₄⁺X⁻.
- Concurrent Elimination with Hot Ethanolic OH⁻ (OH⁻ as a Base)
- In the presence of hot ethanolic KOH or NaOH (anhydrous conditions), :OH⁻ acts as a Brønsted-Lowry base (proton acceptor) rather than a nucleophile, converting a halogenoalkane into an alkene + H₂O + X⁻: 1. Curly arrow from lone pair on :OH⁻ to a hydrogen atom on the adjacent carbon atom (the β-carbon next to the C-X carbon). 2. Curly arrow from that C-H bond to the C-C single bond to form a C=C double bond. 3. Curly arrow from the C-X bond to the halogen atom X (releasing :X⁻). Unsymmetrical 2°/3° halogenoalkanes produce a mixture of structural and E/Z stereoisomeric alkenes.
- Controlling Substitution vs Elimination (Conditions & Structure)
- 1. Solvent & Temperature: Warm aqueous NaOH favours nucleophilic substitution (forming an alcohol); Hot ethanolic KOH favours elimination (forming an alkene). 2. Halogenoalkane Structure: Primary (1°) halogenoalkanes react predominantly by nucleophilic substitution (unhindered δ⁺ C); Tertiary (3°) halogenoalkanes react predominantly by elimination (steric hindrance from 3 alkyl groups blocks nucleophilic attack on C, while many exposed β-hydrogens favour base attack); Secondary (2°) halogenoalkanes give a mixture of both.
- CFCs & Catalytic Free-Radical Ozone Depletion
- In the stratosphere, ozone (O₃) absorbs harmful UV-B radiation (O₃ ⇌ O₂ + O ). Chlorofluorocarbons (CFCs, e.g. CF₂Cl₂) are unreactive in the troposphere due to strong C-F and C-Cl bonds, but UV radiation in the stratosphere causes homolytic fission of the weaker C-Cl bond: Initiation: CF₂Cl₂ → CF₂Cl + Cl . Propagation Step 1: Cl + O₃ → ClO + O₂. Propagation Step 2: ClO + O₃ → Cl + 2O₂ (or ClO + O → Cl + O₂). Overall Equation: 2O₃ → 3O₂. Because the Cl radical is regenerated in Propagation Step 2, it acts as a homogeneous catalyst (one Cl destroys ~100,000 O₃ molecules). Replaced by HFCs (hydrofluorocarbons, e.g. CH₂FCF₃), which contain no C-Cl bonds.
Quick-Fire Flashcards (6)
Click any card to check the answer, then mark it as Known to track your progress.
Why do iodoalkanes undergo nucleophilic substitution much faster than chloroalkanes, even though the C-Cl bond is much more polar than the C-I bond?
Compare the role of the OH⁻ ion and the reaction conditions in converting 2-bromopropane into propan-2-ol versus propene.
Why is an excess of ammonia used when reacting a halogenoalkane with ethanolic ammonia to make a primary amine?
In the nucleophilic substitution of bromoethane with KCN, why is this reaction so useful in organic synthesis?
Write the two propagation steps and the overall equation showing how chlorine free radicals (Cl ) catalyse the destruction of ozone (O₃).
Why is ethanol added as a co-solvent when comparing the rates of hydrolysis of 1-chlorobutane, 1-bromobutane, and 1-iodobutane with aqueous silver nitrate (AgNO₃)?
Exam-Style Questions & Mark Schemes (4)
Write or think through your answer first, then click Check Mark Scheme to see the exact points examiners award marks for.
- Q1
Describe the mechanism for the reaction of bromoethane (CH₃CH₂Br) with excess concentrated ethanolic ammonia (NH₃) to form ethylamine, naming the mechanism and clearly explaining every curly arrow and intermediate. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Name of mechanism: Nucleophilic substitution [1 mark]. 2. Stage 1 (Nucleophilic attack): Show the C^δ+ - Br^δ- dipole on bromoethane; draw a curly arrow starting from the lone pair on the nitrogen atom of :NH₃ to the δ⁺ carbon atom of bromoethane [1 mark]. 3. Draw a second curly arrow starting from the middle of the C-Br covalent bond to the δ⁻ bromine atom (heterolytic fission) [1 mark]. 4. Draw the positively charged ethylammonium intermediate, [CH₃CH₂N⁺H₃] (with the + charge on the nitrogen atom), plus the :Br⁻ ion [1 mark]. 5. Stage 2 (Deprotonation): Draw a second :NH₃ molecule acting as a base with a curly arrow from its lone pair to one of the H atoms attached to the N⁺, and a curly arrow from that N-H bond onto the N⁺ atom, producing ethylamine (CH₃CH₂NH₂) and NH₄⁺Br⁻ [1 mark].
- Q2
When 2-bromobutane is heated under reflux with potassium hydroxide dissolved in ethanol, three isomeric alkenes are formed. Name the mechanism, explain the role of the hydroxide ion in the mechanism, and name the three alkene isomers produced. (6 marks)
Click “Check Mark Scheme” to compare your answer, 1. Name of mechanism: Elimination (base elimination) [1 mark]. 2. Role of OH⁻: The hydroxide ion acts as a Brønsted-Lowry base (proton acceptor) [1 mark]. 3. Mechanism description: Draw a curly arrow from the lone pair on :OH⁻ to a hydrogen atom on an adjacent carbon (C1 or C3), a second curly arrow from that C-H bond into the C-C bond to form a C=C double bond, and a third curly arrow from the C-Br bond to the Br atom (forming H₂O and Br⁻) [2 marks]. 4. Removing H from C1 forms but-1-ene [1 mark]. 5. Removing H from C3 forms E-but-2-ene and Z-but-2-ene (stereoisomers due to restricted rotation around C=C with two different groups on each C=C carbon) [1 mark].
- Q3
A student investigated the relative rates of hydrolysis of 1-chloropropane, 1-bromopropane, and 1-iodopropane by placing three test tubes containing each halogenoalkane, ethanol, and aqueous silver nitrate in a water bath at 50°C. State and explain the order in which precipitates appear, including ionic equations for the precipitate formation. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. In all three tubes, water (not Ag⁺!) acts as the nucleophile in nucleophilic substitution to hydrolyse the halogenoalkane: R-X + H₂O → R-OH + H⁺ + X⁻(aq) [1 mark]. 2. The released halide ions (X⁻) immediately react with aqueous silver ions to form precipitates: Ag⁺(aq) + X⁻(aq) → AgX(s) [1 mark]. 3. 1-iodopropane forms a yellow precipitate (AgI) fastest; 1-bromopropane forms a cream precipitate (AgBr) second; 1-chloropropane forms a white precipitate (AgCl) slowest [1 mark]. 4. As halogen atomic radius increases down Group 17 from Cl to Br to I, the shared pair of electrons in the C-X bond is further from the halogen nucleus, making the C-X covalent bond progressively weaker (lower bond enthalpy: C-I < C-Br < C-Cl) [1 mark]. 5. Because the C-I bond has the lowest bond enthalpy, less activation energy is required to break it in the rate-determining step, giving the fastest rate of hydrolysis [1 mark].
- Q4
1,1,1-trichlorotrifluoroethane (CF₃CCl₃) was formerly used as a solvent before being banned under the Montreal Protocol. Write equations to show how CF₃CCl₃ initiates and propagates the destruction of ozone (O₃) in the upper atmosphere, explain why CF₃CH₂F is a safe replacement for the ozone layer, and state how nitrogen monoxide ( NO) from high-altitude aircraft also catalyses ozone breakdown. (5 marks)
Click “Check Mark Scheme” to compare your answer, 1. Initiation (UV radiation breaks the weaker C-Cl bond by homolytic fission, not the stronger C-F bond): CF₃CCl₃ → CF₃CCl₂ + Cl [1 mark]. 2. Propagation steps for Cl : Step 1: Cl + O₃ → ClO + O₂; Step 2: ClO + O₃ → Cl + 2O₂ (overall: 2O₃ → 3O₂) [1 mark]. 3. CF₃CH₂F is a hydrofluorocarbon (HFC) that contains no C-Cl bonds; the C-F and C-H bonds are too strong to be broken by stratospheric UV radiation, so no Cl radicals are generated [1 mark]. 4. Nitrogen monoxide ( NO) is a free radical (unpaired electron on N) that catalyses ozone destruction via two propagation steps: Step 1: NO + O₃ → NO₂ + O₂ [1 mark]. 5. Step 2: NO₂ + O₃ → NO + 2O₂ (or NO₂ + O → NO + O₂), regenerating the NO radical catalyst [1 mark].
Common Exam Questions (4)
When drawing the curly-arrow mechanism for nucleophilic substitution with KCN, where must the minus charge and lone pair be drawn on the cyanide ion?
They must be drawn on the carbon atom (:C⁻≡N), not on the nitrogen atom. If you draw the lone pair on nitrogen or start your curly arrow from the middle of the CN⁻ ion, examiners will deduct the mark because it is the carbon atom's lone pair that forms the new C-C covalent bond.
In the elimination mechanism of a halogenoalkane, which hydrogen atom does the :OH⁻ base attack?
Never attack the hydrogen attached to the same carbon as the halogen (the α-carbon). The :OH⁻ base must attack a hydrogen atom on an adjacent carbon atom (a β-carbon next to the C-X carbon) so that a C=C double bond can form between the two carbons.
Why don't UV rays break the C-F bond in CFCs in the stratosphere?
The C-F bond is extremely short and strong (bond enthalpy ~484 kJ mol⁻¹, compared to ~338 kJ mol⁻¹ for the C-Cl bond). Stratospheric UV radiation has enough photon energy (E = hν) to break the C-Cl bond homolytically, but not enough energy to break the C-F bond.
What happens if a primary amine keeps reacting with excess bromoethane?
Ethylamine (CH₃CH₂NH₂) attacks another bromoethane to form diethylamine ((CH₃CH₂)₂NH, secondary amine), which attacks a third bromoethane to form triethylamine ((CH₃CH₂)₃N, tertiary amine), which finally attacks a fourth bromoethane to form tetraethylammonium bromide ((CH₃CH₂)₄N⁺Br⁻, a quaternary ammonium salt used as a cationic surfactant in fabric conditioners because its nitrogen no longer has a lone pair).
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