Nucleophilic substitution in halogenoalkanesAQA A-Level Chemistry: Revision notes
Section 1
Polar bonds and nucleophiles
Halogenoalkanes contain a polar C–X bond. The halogen is more electronegative than carbon, so the carbon atom has a partial positive charge (δ+) and the halogen is δ–.
The δ+ carbon is attacked by nucleophiles. A nucleophile is an electron pair donor: it uses a lone pair to form a new covalent bond with the carbon. The halogen leaves as a halide ion. This is a nucleophilic substitution: the nucleophile replaces the halogen.
The nucleophiles you need are OH⁻, CN⁻ and NH₃.
The carbon is δ+ because the halogen is more electronegative. Do not say the halogen 'attracts' the nucleophile. The nucleophile is attracted to the δ+ carbon.
Section 2
Substitution with OH⁻, CN⁻ and NH₃
With aqueous hydroxide (warm aqueous NaOH or KOH), the halogenoalkane forms an alcohol:
CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻
With cyanide (KCN in ethanol, heated under reflux), the product is a nitrile. This adds a carbon atom, so it lengthens the carbon chain:
CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻
With excess ammonia (in ethanol, heated in a sealed tube), the product is a primary amine:
CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
Name the reagent, the conditions and the product together, e.g. 'heat under reflux with KCN in ethanol gives a nitrile'.
Section 3
The mechanism
Curly arrows show the movement of electron pairs. In each mechanism:
- A curly arrow goes from the lone pair on the nucleophile to the δ+ carbon.
- A curly arrow goes from the C–X bond to the halogen atom, which leaves as X⁻ (heterolytic fission).
For OH⁻ and CN⁻ these two arrows give the organic product directly.
For ammonia there is an extra step. After the first attack, the nitrogen carries a positive charge in the intermediate R–NH₃⁺. A second NH₃ molecule uses its lone pair to remove an H⁺ from nitrogen, forming the amine R–NH₂ and NH₄⁺. This is why two moles of NH₃ are needed.
The curly arrow must start from the lone pair (or bond) and end where the electrons go. An arrow that starts from the nucleophile's charge sign loses the mark.
Section 4
Excess ammonia
The amine product also has a lone pair on nitrogen, so it is a nucleophile too. It can attack remaining halogenoalkane to form a secondary amine, then a tertiary amine and a quaternary ammonium salt. This produces a mixture of products.
Using a large excess of ammonia makes it more likely that a halogenoalkane molecule collides with NH₃ than with the amine, so mainly the primary amine forms.
Section 5
Rate and carbon-halogen bond enthalpy
The rate of substitution depends on how easily the C–X bond breaks. The weaker the bond (the lower the bond enthalpy), the faster the reaction.
- C–Cl: 338 kJ mol⁻¹
- C–Br: 276 kJ mol⁻¹
- C–I: 238 kJ mol⁻¹
So the rate of reaction increases in the order chloroalkane < bromoalkane < iodoalkane.
This is true even though C–Cl is the most polar of these bonds. The rate is controlled by bond enthalpy, not polarity.
In a test, the halogenoalkane is warmed with aqueous silver nitrate in ethanol. Water is the nucleophile and the halide ion released forms a precipitate with silver ions. The iodoalkane forms its precipitate fastest.
If asked why chloroalkanes react slowly, say 'C–Cl has a higher bond enthalpy, so more energy is needed to break it'. Do not use polarity.
Must know
- C–X bonds are polar: C is δ+, so nucleophiles attack it.
- Nucleophiles: OH⁻ (gives alcohol), CN⁻ (gives nitrile), NH₃ (gives amine).
- Mechanism: lone pair arrow to δ+ C, bond arrow to X.
- Excess NH₃ for a primary amine; two moles needed per mole of halogenoalkane.
- Weaker C–X bond = faster reaction; I > Br > Cl.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Nucleophilic substitution in halogenoalkanes
- 1-Bromobutane, CH₃CH₂CH₂CH₂Br, is warmed with aqueous sodium hydroxide, NaOH. Butan-1-ol is formed by a nucleophilic substitution reaction.Write an equation for the reaction of 1-bromobutane with hydroxide ions.2 marks
- A chemist converts 1-bromopropane, CH₃CH₂CH₂Br, into butanenitrile, CH₃CH₂CH₂CN, by heating it under reflux with a solution of potassium cyanide, KCN, in ethanol.Suggest why this reaction is useful in organic synthesis.2 marks
- A student compares the rates of reaction of three halogenoalkanes with water. Each of 1-chlorobutane, 1-bromobutane and 1-iodobutane is warmed with aqueous silver nitrate in ethanol and the time taken for a precipitate of the silver halide to form is measured. The bond enthalpies are: C–Cl 338 kJ mol⁻¹, C–Br 276 kJ mol⁻¹ and C–I 238 kJ mol⁻¹.Predict which halogenoalkane forms a precipitate first and explain your answer.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).