Lesson 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)

I can interpret halogenoalkane reactivity, especially hydrolysis.Syllabus
Syllabus 9476, 11.3(f). Interpret halogenoalkane reactivity, especially hydrolysis.
I can describe the specified halogenoalkane reactions.Syllabus
Syllabus 9476, 11.5(a). Describe the specified halogenoalkane reactions.

An aqueous nucleophile substitutes; hot ethanolic base favours elimination

Solvent, reagent and temperature select different products from the same carbon skeleton.

The specified halogenoalkane transformations
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 next to the carbon that carries the leaving group. If there are different neighbouring carbons with hydrogens, elimination can give more than one alkene. Draw the carbon chain and mark each beta carbon before you predict the products. Do not just remove Br and any H you choose.

The primary amine formed still has a lone pair, so it is still a nucleophile. It can react with more halogenoalkane to give secondary and tertiary amines, and then a quaternary ammonium salt. Using excess ammonia means the halogenoalkane is more likely to meet NH₃ than the amine, so more primary amine forms. Further substitution is still possible.

Check your understanding

Bromoethane is heated with KCN in ethanol, and the nitrile is then reduced. Which amine forms?

Ethylamine, CH₃CH₂NH₂, because the chain stays two carbons
Propylamine, CH₃CH₂CH₂NH₂, because CN^- adds one carbon
Methylamine, CH₃NH₂, because reduction removes a carbon

Show the answer

Propylamine, CH₃CH₂CH₂NH₂, because CN^- adds one carbon

Substitution gives propanenitrile, CH₃CH₂CN. Reduction turns CN into CH₂NH₂, so the amine has three carbons.