Halogen Compounds

Classification — count the alkyl groups on the carbon bearing the halogen:

• Primary (1°): one alkyl group (e.g. CH₃CH₂Br).
• Secondary (2°): two.
• Tertiary (3°): three.

Nucleophilic substitution — a nucleophile replaces the halogen (which leaves as X⁻):

Reagent / conditions	Nucleophile	Product
NaOH(aq), heat	OH⁻	alcohol
KCN in ethanol, heat	CN⁻	nitrile (adds one carbon)
NH₃ in ethanol, heat under pressure	NH₃	amine

Examples: $\text{CH}_3\text{CH}_2\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{Br}^-$ $\text{CH}_3\text{CH}_2\text{Br} + \text{CN}^- \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{Br}^-$ The CN⁻ reaction is useful in synthesis because it lengthens the carbon chain.

SN2 (primary halogenoalkanes) — one step. The nucleophile attacks the δ+ carbon from the opposite side to the halogen, as the C–X bond breaks, passing through a single transition state. $\text{Nu}^- + \text{R–X} \rightarrow [\text{Nu}\cdots\text{R}\cdots\text{X}]^{\ddagger} \rightarrow \text{Nu–R} + \text{X}^-$ Rate depends on both the halogenoalkane and the nucleophile (bimolecular).

SN1 (tertiary halogenoalkanes) — two steps. First the C–X bond breaks heterolytically to form a carbocation (slow, rate-determining); then the nucleophile attacks. $\text{R–X} \rightarrow \text{R}^+ + \text{X}^- \quad(\text{slow})$ $\text{R}^+ + \text{Nu}^- \rightarrow \text{R–Nu} \quad(\text{fast})$ Rate depends only on the halogenoalkane (unimolecular).

Why this split? Tertiary carbocations are stabilised by the electron-donating (+I) inductive effect of three alkyl groups, so SN1 is favoured. Primary cations are unstable, so primary halogenoalkanes react by SN2 instead. Secondary halogenoalkanes react by a mixture of both.

Elimination. With NaOH dissolved in ethanol and heat, a halogenoalkane undergoes elimination of HX to form an alkene (compare with substitution, which uses aqueous NaOH): $\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \xrightarrow{\text{ethanol, heat}} \text{CH}_2\text{=CH}_2 + \text{NaBr} + \text{H}_2\text{O}$

Identifying the halogen (and comparing reactivity). Warm the halogenoalkane with aqueous silver nitrate in ethanol; the halide ion released forms a silver halide precipitate:

• Cl → white AgCl (slowest to form)
• Br → cream AgBr
• I → yellow AgI (fastest to form)

The rate of precipitate formation shows the reactivity: C–I reacts fastest, C–Cl slowest, because the C–I bond is the weakest (longest bond, lowest bond energy) and breaks most easily. So reactivity is C–I > C–Br > C–Cl — the opposite order to bond strength.