Topic 18 of 34
An aqueous nucleophile substitutes; hot ethanolic base favours elimination
Solvent, reagent and temperature select different products from the same carbon skeleton.
A-Level 9476 (2026-2027)
An aqueous nucleophile substitutes; hot ethanolic base favours elimination
Solvent, reagent and temperature select different products from the same carbon skeleton.
| Starting example and conditions | Balanced representative equation | Reasoning |
|---|---|---|
| Bromoethane, aqueous NaOH and heat | CH3CH2Br + OH- → CH3CH2OH + Br-. | OH- acts as a nucleophile; hydrolysis replaces Br by OH. |
| Bromoethane, KCN in ethanol and heat | CH3CH2Br + CN- → CH3CH2CN + Br-. | Cyanide attacks through carbon; propanenitrile has one more carbon than bromoethane. |
| Bromoethane, excess ammonia in ethanol, heat under pressure | CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br. | One ammonia replaces Br; another removes a proton from the initially formed alkylammonium ion. |
| 2-bromopropane, NaOH in ethanol and heat | CH3CHBrCH3 + OH- → CH3CH=CH2 + Br- + H2O. | Base removes a beta-H while Br leaves, forming C=C. |
A beta hydrogen is on a carbon adjacent to the carbon bearing the leaving group. Elimination may give more than one alkene if different adjacent positions are available. Draw the carbon skeleton and mark eligible beta carbons before predicting products; do not simply erase Br and one arbitrarily placed H.
The primary amine product still has a nucleophilic lone pair and can react further with halogenoalkane to give secondary and tertiary amines, then a quaternary ammonium salt. Excess ammonia favours encounters with NH3 and improves primary-amine formation. It does not make further substitution logically impossible.