15. Halogen compounds

Halogenoalkanes can be made by radical substitution of alkanes with Cl₂/Br₂ under UV, electrophilic addition of X₂ or HX to an alkene, or substitution of an alcohol using a hydrogen halide or halogenating reagent.

Choose the route from the starting material: alkane needs UV radicals, alkene consumes π bonds, and alcohol replaces –OH. Conditions are part of the reaction, not optional decoration.

Ethene + HBr → bromoethane; ethanol + PCl₅ → chloroethane + POCl₃ + HCl; ethane + Br₂ requires UV light and gives bromoethane plus HBr.

Do not swap aqueous and ethanolic conditions or call every halogenoalkane preparation an addition reaction.

A primary halogenoalkane has the C–X carbon attached to one other carbon, secondary to two, and tertiary to three. The classification does not count the halogen itself.

Locate the carbon directly bonded to X, then count its carbon neighbours. This classification helps predict substitution mechanism and relative reaction rate.

CH₃CH₂Br is primary; CH₃CHBrCH₃ is secondary; (CH₃)₃CBr is tertiary. The carbon skeleton, not the name ending alone, determines the label.

Do not classify by the total number of carbons in the molecule, and do not call a carbon attached to two hydrogens “secondary”.

In nucleophilic substitution, a nucleophile donates an electron pair to the carbon attached to X while the C–X bond breaks and X⁻ leaves. Different nucleophiles give different products.

Aqueous OH⁻ gives an alcohol, CN⁻ in ethanol gives a nitrile with one extra carbon, and NH₃ in ethanol gives an amine. AgNO₃ in ethanol can compare halide hydrolysis by precipitating AgX.

CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻; CH₃CH₂Br + CN⁻ → CH₃CH₂CN + Br⁻. The solvent and nucleophile explain the product.

Do not confuse CN⁻ substitution with elimination, and do not forget that the nitrile route increases the carbon count by one.

In elimination, a base removes H from a carbon next to the C–X carbon while X⁻ leaves, forming a C=C. Hot ethanolic NaOH favours elimination of a halogenoalkane.

The solvent and temperature matter: aqueous hydroxide more often gives substitution, while ethanolic hydroxide and heat favour elimination. Balance atoms and identify the small molecules removed.

CH₃CH₂Br + OH⁻(ethanol) → CH₂=CH₂ + H₂O + Br⁻. The product is an alkene, not an alcohol.

Do not use the aqueous substitution equation under ethanolic conditions, and do not call elimination an addition reaction.

SN2 substitution occurs in one step as the nucleophile attacks while the leaving group leaves. SN1 occurs in two main steps: C–X ionisation, then nucleophile attack on the carbocation.

SN2 is favoured by less hindered primary centres; SN1 is favoured when a more substituted carbocation is stabilised by alkyl inductive effects. Solvent and leaving-group ability also matter.

OH⁻ attacking bromoethane illustrates SN2. A tertiary halogenoalkane can ionise first and then react with water by SN1.

SN1 does not mean “one molecule reacts”; it names the rate-determining molecularity of the ionisation step.

Primary halogenoalkanes usually react by SN2, tertiary by SN1, and secondary can use either pathway depending on steric hindrance, carbocation stability and conditions.

Use structure as a starting prediction, not a rigid law. A primary centre is accessible to a nucleophile; a tertiary centre is crowded but can form a stabilised carbocation.

Bromoethane favours direct OH⁻ attack, whereas tert-butyl bromide favours ionisation in a polar protic solvent. A secondary substrate needs the surrounding conditions to decide.

Primary/secondary/tertiary describes the carbon bearing X, not the total carbon count, and the trend is not independent of solvent.

In hydrolysis, water or hydroxide replaces X in a halogenoalkane. A weaker C–X bond generally hydrolyses more readily, and aqueous silver nitrate detects the halide ion formed as AgX.

Keep the test fair: same concentration, temperature, solvent and volume. The precipitate time is evidence about relative rate, not a direct measurement of bond energy alone.

Iodoalkanes usually form AgI more quickly than bromo- or chloroalkanes because C–I cleavage is easier. The precipitate colour helps identify the halide.

A faster precipitate does not prove the reaction is purely SN1; substrate structure and nucleophile conditions also affect the rate.