Halogenoalkanes and nucleophilic substitutionEdexcel International A Level Chemistry: Revision notes

A halogenoalkane is an alkane in which one or more hydrogen atoms have been replaced by a halogen (F, Cl, Br, I). The halogen is named as a prefix (chloro-, bromo-, iodo-) with a number for its position on the longest carbon chain, numbered from the end that gives the lowest number, e.g. 2-bromobutane, 1-chloro-2-methylpropane. Alphabetical order is used for several prefixes.

You must be able to draw structural, displayed and skeletal formulae. Halogenoalkanes are classified by the carbon atom bonded to the halogen:

• Primary: that carbon is attached to one other carbon (e.g. 1-bromopropane)
• Secondary: attached to two other carbons (e.g. 2-bromopropane)
• Tertiary: attached to three other carbons (e.g. 2-bromo-2-methylpropane)

Halogens are more electronegative than carbon, so the C–X bond is polar: carbon is δ+ and the halogen δ−. The electron-deficient carbon is attacked by a nucleophile, a species that donates a lone pair of electrons to form a new covalent bond. Examples are OH⁻, :NH₃, CN⁻ and H₂O.

In nucleophilic substitution the nucleophile replaces the halogen, which leaves as a halide ion. In a mechanism, a curly arrow shows the movement of an electron pair: from a lone pair (or bond) to the atom it attacks.

Aqueous KOH (heated under reflux): OH⁻ is the nucleophile and an alcohol forms. CH₃CH₂CH₂Br + OH⁻ → CH₃CH₂CH₂OH + Br⁻

Alcoholic KCN (ethanol solvent, heated under reflux): CN⁻ is the nucleophile and a nitrile forms. This lengthens the carbon chain by one carbon. CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻

Ammonia (excess, in ethanol, sealed tube under pressure): :NH₃ is the nucleophile and a primary amine forms. CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br

A large excess of ammonia is used because the amine product is also a nucleophile and would otherwise react with more halogenoalkane to give secondary and tertiary amines.

For a primary halogenoalkane with OH⁻:

1. The lone pair on OH⁻ attacks the δ+ carbon (curly arrow from the lone pair to carbon).
2. As the C–O bond forms, the C–Br bond breaks heterolytically (curly arrow from the bond to Br), and Br⁻ leaves.

With ammonia, the lone pair on nitrogen attacks the δ+ carbon and Br⁻ leaves, giving R–NH₃⁺. A second NH₃ molecule then removes H⁺ from the nitrogen, giving the amine R–NH₂ and NH₄⁺. (The names SN1 and SN2 are not needed at this stage.)

Ethanolic KOH (hot, ethanolic): OH⁻ acts as a base, removing H⁺ from a carbon next to C–X. The halogen leaves and a C=C forms: an alkene, plus H₂O and a halide ion. This is elimination. CH₃CH₂CH₂Br + KOH → CH₃CH=CH₂ + KBr + H₂O

The difference is the solvent: aqueous KOH gives substitution (alcohol), ethanolic KOH gives elimination (alkene).

Aqueous silver nitrate in ethanol: water is the nucleophile and hydrolyses the halogenoalkane. The halide ion released precipitates: AgCl white, AgBr cream, AgI yellow. Ethanol is the solvent in which both reagents mix.