2.10.2c: carry out the preparation of an halogenoalkane from an alcohol and explain why a metal halide and concentrated sulfuric acid should not be used when making a bromoalkane or an iodoalkane

Nucleophilic substitution

• INSERT MECHANISM HERE

To prepare:

• a chloroalkane = alcohol + concentrated hydrochloric acid
• a bromoalkane = alcohol + sodium/potassium bromide and concentrated sulphuric acid
• an iodoalkane = alcohol + sodium/potassium iodide and concentrated phosphoric (V) acid/red phosphorus and iodine
• PI3 + 3C4H9OH → 3C4H9I + H3PO3

Substitution and elimination:

• primary = mainly substitution
• secondary = both substitution and elimination
• tertiary = mainly elimination
• INSERT MECHANISM HERE
• Inducing elimination and substitution:
• weaker C-X bond = substitution (C-I undergoes substitution more readily)
• base strength: the stronger the base used, the more elimination is favoured
• eg. ethanolic NaOH = elimination, aqueous NaOH = substitution
• temperature: the higher the temperature, the more elimination is favoured

Why should a metal halide and concentrated sulphuric acid not be used when making a bromo- or iodoalkane?:

• if concentrated sulphuric acid is used, the halide is oxidised back into itself and hardly any hydrogen halide is made, and the halide gets in the way of the main reaction

MAKING A BROMOALKANE

MAKING A CHLOROALKANE