Halogenoalkanes react with hydroxide ions from sodium hydroxide or potassium hydroxide, but the product depends on the solvent. Two reactions compete.Aqueous alkali: substitution
Warmed under reflux with KOH(aq), the OH− acts as a nucleophile and replaces the halogen, giving an alcohol. (Details in the next lesson.)Ethanolic alkali: elimination
Heated under reflux with KOH dissolved in ethanol, the OH− acts as a base. It removes an H+ from a carbon next to the C-X carbon, and the halide ion leaves. A C=C double bond forms, so the product is an alkene. A molecule of HX is eliminated overall:CH3CHBrCH3 + KOH → CH3CH=CH2 + KBr + H2O Here 2-bromopropane gives propene.More than one alkene: if the C-X carbon has different neighbouring carbons bearing H atoms, elimination can give several alkenes. For example, 2-bromobutane with ethanolic KOH gives but-1-ene and but-2-ene (which exists as E and Z isomers).Which reaction wins?
• aqueous solvent and warm conditions favour substitution
• ethanolic solvent and higher temperatures favour elimination
• primary halogenoalkanes tend to undergo substitution; tertiary halogenoalkanes tend to undergo eliminationIn practice a mixture of products often forms, with one reaction dominating. The same reagent, OH−, can therefore make either an alcohol or an alkene, so it is important to state the solvent.
Warmed under reflux with KOH(aq), the OH− acts as a nucleophile and replaces the halogen, giving an alcohol. (Details in the next lesson.)Ethanolic alkali: elimination
Heated under reflux with KOH dissolved in ethanol, the OH− acts as a base. It removes an H+ from a carbon next to the C-X carbon, and the halide ion leaves. A C=C double bond forms, so the product is an alkene. A molecule of HX is eliminated overall:
• aqueous solvent and warm conditions favour substitution
• ethanolic solvent and higher temperatures favour elimination
• primary halogenoalkanes tend to undergo substitution; tertiary halogenoalkanes tend to undergo eliminationIn practice a mixture of products often forms, with one reaction dominating. The same reagent, OH−, can therefore make either an alcohol or an alkene, so it is important to state the solvent.