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3.3.2Physical Chemistry

Reactions of Halogenoalkanes — AQA AS Chemistry Revision Notes & Topical Questions

5 min read 4 parts Lesson 74 of 102

📚 AQA AS Chemistry Hub / Topical Revision
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Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
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Part 1 of 4Nucleophilic Substitution Reactions

With hydroxide
  • With hydroxide CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻ converts the halogenoalkane into an alcohol.
With cyanide
  • With cyanide CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻ forms a nitrile and lengthens the carbon chain by one carbon.
With ammonia
  • With ammonia CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br gives an amine; excess ammonia reduces further substitution.
Nucleophile role
  • Nucleophile role In each case the nucleophile donates a lone pair to the δ+ carbon, and the halide ion leaves.
ReagentConditionsProduct (from bromoethane)
Aqueous NaOH or KOHWarm, aqueousEthanol
KCN in ethanol/waterHeat under refluxPropanenitrile
Excess NH₃ in ethanolHeat in a sealed tubeEthylamine

Part 2 of 4The Substitution Mechanism

The mechanism shows curved arrows that represent the movement of electron pairs.

1
The nucleophile approaches the δ+ carbon atom, using its lone pair of electrons.
2
A curly arrow is drawn from the lone pair of the nucleophile to the carbon atom.
3
A second curly arrow is drawn from the C–X bond to the halogen atom, showing heterolytic fission.
4
The halogen leaves as a halide ion, and a new bond forms between the nucleophile and the carbon.
5
For OH⁻ the product is an alcohol, and the halide is released as Br⁻ or Cl⁻.

Part 3 of 4Elimination Reactions

The same reagents can behave as bases under different conditions.

Elimination Reactions
Conditions
  • Conditions Hot ethanolic potassium hydroxide, heated under reflux, favours elimination over substitution.
Equation
  • Equation CH₃CH₂Br + KOH → CH₂=CH₂ + KBr + H₂O shows bromoethane eliminating HBr to form ethene.
Role of OH⁻
  • Role of OH⁻ Here hydroxide acts as a base, removing a hydrogen atom from the carbon next to the carbon bearing the halogen.
Solvent matters
  • Solvent matters Water favours substitution, while ethanol as the solvent favours elimination, so conditions must be stated precisely.

Part 4 of 4Exam Checklist & Tips

3.3.2 – Key Points for the Exam

  • Nucleophilic substitution Replacement of a halogen by a nucleophile through attack on the δ+ carbon.
  • Elimination A reaction in which a small molecule such as HBr is removed from a molecule to form a double bond.
  • Nitrile An organic compound containing the –C≡N group.
  • Reflux Heating with a condenser fitted so that vapours return to the flask.

Common Mistakes

  • Mixing up the conditionsaqueous hydroxide gives substitution, ethanolic hydroxide gives elimination.
  • Forgetting that cyanide adds an extra carbon atom to the chain.
  • Starting the curly arrow at the hydrogen of OH⁻ instead of the lone pair on oxygen.
  • Drawing the arrow for C–X bond breaking pointing to carbon instead of the halogen.

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