Halogenoalkanes and nucleophilic substitutionEdexcel A-Level Chemistry: Revision notes

A halogenoalkane has a halogen atom in place of a hydrogen atom of an alkane. They are classified by the number of carbon atoms attached to the carbon bearing the halogen:

• primary: one other carbon, e.g. 1-bromobutane
• secondary: two other carbons, e.g. 2-bromobutane
• tertiary: three other carbons, e.g. 2-bromo-2-methylpropane

The C–X bond is polar: the carbon is δ+ and the halogen is δ−. The δ+ carbon is attacked by nucleophiles.

A nucleophile is an electron pair donor that attacks a δ+ carbon, e.g. OH⁻, CN⁻, NH₃ and H₂O. In nucleophilic substitution the halogen is replaced by the nucleophile.

• Aqueous KOH, warm: alcohol. CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻ (OH⁻ is a nucleophile)
• Aqueous silver nitrate in ethanol: water is the nucleophile, forming an alcohol and halide ions, which precipitate AgX: AgCl white, AgBr cream, AgI yellow
• KCN in ethanol, heated under reflux: nitrile, lengthening the carbon chain. CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻
• Excess ammonia in ethanol, sealed tube, heated: primary amine. CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br
• Ethanolic KOH, heated: elimination to an alkene, with OH⁻ acting as a base.

For a primary halogenoalkane with aqueous KOH the mechanism has one step:

1. The lone pair on OH⁻ attacks the δ+ carbon (curly arrow from the lone pair to the carbon).
2. At the same time the C–Br bond breaks by heterolytic fission (curly arrow from the bond to the Br), forming Br⁻.

Products: an alcohol and Br⁻. The same pattern applies to CN⁻.

Ammonia reacts in two stages:

1. The lone pair on N attacks the δ+ carbon and the C–Br bond breaks, forming Br⁻ and an intermediate R–NH₃⁺.
2. A second NH₃ removes H⁺ from the intermediate, giving the primary amine R–NH₂ and NH₄⁺.

A large excess of ammonia is used so that the amine, which is also a nucleophile, does not react with more halogenoalkane.

Effect of the halogen. Rate: iodoalkane > bromoalkane > chloroalkane. The C–X bond breaks in the slow step. Bond enthalpy falls from C–Cl (346) to C–Br (290) to C–I (228 kJ mol⁻¹), so C–I is easiest to break. Polarity does not explain it, as C–Cl is the most polar.

Effect of the structure. Rate: tertiary > secondary > primary. Tertiary halogenoalkanes form the most stable carbocation, since three alkyl groups release electron density towards the positive carbon.

Core practical 4. Add equal volumes of each halogenoalkane to ethanol and aqueous silver nitrate in a water bath at the same temperature, and time how long a precipitate takes to appear. Keep volumes, concentration and temperature constant.