1. Carbon-Halogen Bond Polarity and Reactivity
Halogen atoms are more electronegative than carbon, creating a polar bond:
\[ \text{C}^{\delta+} - \text{X}^{\delta-} \]
The electron-deficient carbon atom (Cδ+) is vulnerable to attack by nucleophiles (electron-pair donors).
Bond Enthalpy Controls Reactivity
Although the C − F bond is the most polar, reactivity is dictated by bond enthalpy, not bond polarity. Typical values:
- C − F: 467 kJ mol⁻¹ (extremely strong, virtually unreactive).
- C − Cl: 338 kJ mol⁻¹.
- C − Br: 276 kJ mol⁻¹.
- C − I: 238 kJ mol⁻¹ (weakest bond, broken most easily).
Order of reactivity: iodoalkanes > bromoalkanes > chloroalkanes > fluoroalkanes. C − I is the weakest bond, so iodoalkanes react fastest. Bond enthalpy matters more than bond polarity.
2. Precise Curly Arrow Drawing Rules
- A curly arrow represents the movement of a pair of electrons.
- An arrow MUST start either directly on a lone pair of electrons (:) or at the centre of a covalent bond.
- The arrow head must terminate directly on the destination atom (or between atoms to form a new bond).
- Floating arrows or arrows originating from minus signs are penalised immediately.
3. Nucleophilic Substitution Reactions
A nucleophile replaces the halogen atom in three primary synthetic pathways:
| Nucleophile | Reaction Conditions | Organic Product | Stoichiometric Equation |
|---|---|---|---|
| Hydroxide ion (:OH⁻) | Warm aqueous NaOH or KOH, reflux | Alcohol | R − Br + OH⁻ → R − OH + Br⁻ |
| Cyanide ion (:CN⁻) | KCN in ethanol and water, reflux | Nitrile (extends carbon chain by +1) | R − Br + CN⁻ → R − CN + Br⁻ |
| Ammonia (:NH₃) | Excess NH₃ in ethanol, heated in a sealed tube | Primary Amine | R − Br + 2NH₃ → R − NH₂ + NH₄Br |
Mechanism with CN⁻: making a nitrile
The cyanide ion uses the lone pair on its carbon atom. The product is a nitrile, and the carbon chain gains one carbon. Bromoethane gives propanenitrile.
\[ \text{CH}_3\text{CH}_2\text{Br} + \text{CN}^- \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{Br}^- \]
Mechanism with NH₃: two steps
- The lone pair on NH₃ attacks the δ+ carbon, and Br⁻ leaves. This gives RNH₃⁺.
- A second NH₃ removes H⁺ from the nitrogen. This gives RNH₂ + NH₄⁺.
Why is EXCESS Ammonia Required?
The primary amine product (R − NH₂) still has a lone pair on its nitrogen atom and can act as a competing nucleophile. Using excess ammonia (in ethanol, in a sealed tube) ensures ammonia molecules outcompete the amine. This limits further substitution, which would form secondary and tertiary amines and quaternary ammonium salts.
4. Elimination Reactions to Form Alkenes
OH⁻ can react with a halogenoalkane in two ways. The spec example is 2-bromopropane with KOH.
In water, warm: substitution
OH⁻ acts as a nucleophile. It attacks the δ+ carbon.
\[ \text{CH}_3\text{CHBrCH}_3 + \text{OH}^- \rightarrow \text{CH}_3\text{CH(OH)CH}_3 + \text{Br}^- \]
Product: propan-2-ol.
In ethanol, hot: elimination
OH⁻ acts as a base. It removes an H⁺ from a carbon next to the C-Br carbon.
\[ \text{CH}_3\text{CHBrCH}_3 + \text{OH}^- \rightarrow \text{CH}_3\text{CH=CH}_2 + \text{H}_2\text{O} + \text{Br}^- \]
Product: propene.
Both reactions happen at the same time. The conditions decide which one gives the main product. Water and warming favour substitution. Ethanol and a high temperature favour elimination.
Elimination mechanism
- A lone pair on OH⁻ attacks an H atom on a carbon next to the C-Br carbon.
- The C-H bond pair moves in to form the C=C.
- The C-Br bond pair moves onto Br, and Br⁻ leaves.
Substitution Conditions
- Aqueous solvent (water).
- Warm / moderate temperature.
- Favoured by primary halogenoalkanes.
Elimination Conditions
- Ethanolic solvent (pure alcohol).
- High temperature / reflux.
- Favoured by tertiary halogenoalkanes.
5. Ozone Depletion by Chlorofluorocarbons (CFCs)
Beyond the OxfordAQA specification
9620 does not require ozone depletion by CFCs. This section is background only.
Chlorofluorocarbons (CFCs) were widely used as refrigerants and aerosol propellants due to their stability and non-toxicity. In the stratosphere, intense UV radiation breaks the weak C − Cl bond via homolytic fission:
\[ \text{CF}_2\text{Cl}_2 \xrightarrow{UV} \text{CF}_2\text{Cl}^\bullet + \text{Cl}^\bullet \]
Catalytic Ozone Destruction Cycle
The chlorine radical (\(\text{Cl}^\bullet\)) acts as a homogeneous catalyst in a two-step ozone destruction cycle:
\[ \text{Step 1: } \text{Cl}^\bullet + \text{O}_3 \rightarrow \text{ClO}^\bullet + \text{O}_2 \]
\[ \text{Step 2: } \text{ClO}^\bullet + \text{O}_3 \rightarrow \text{Cl}^\bullet + 2\text{O}_2 \]
\[ \text{Overall: } 2\text{O}_3 \rightarrow 3\text{O}_2 \]
Because the chlorine radical is regenerated in Step 2, a single \(\text{Cl}^\bullet\) can destroy over 100,000 ozone molecules before termination.
Go deeper
These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.
6. Practice Questions
Show Answer & Mark Scheme
Mechanism marks:
- Curly arrow from lone pair on OH⁻ to Cδ+ of bromoethane. (1 mark)
- Correct dipoles Cδ+−Brδ−. (1 mark)
- Curly arrow from C − Br bond to Br atom, forming Br⁻. (1 mark)
Show Answer & Mark Scheme
- KCN in ethanol and water, heated under reflux. (1 mark)
- Propanenitrile, CH₃CH₂CN. (1 mark)
- Curly arrow from the lone pair on the C of CN⁻ to the δ+ carbon, or from the C − Br bond to Br. (1 mark)
Show Answer & Mark Scheme
- A second NH₃ uses its lone pair to remove H⁺ from the CH₃CH₂NH₃⁺ ion. (1 mark)
- This gives CH₃CH₂NH₂ and NH₄⁺. (1 mark)
- Excess ammonia limits further substitution, because ethylamine also has a lone pair and could react again. (1 mark)
Show Answer & Mark Scheme
- (a) Propan-2-ol. (1 mark) OH⁻ acts as a nucleophile. (1 mark)
- (b) Propene. (1 mark) OH⁻ acts as a base. (1 mark)
Show Answer & Mark Scheme
- The C − I bond has a lower bond enthalpy than the C − Cl bond. (1 mark)
- So the C − I bond breaks more easily. Bond enthalpy matters more than bond polarity. (1 mark)