Module 6: Organic Chemistry and Analysis · Year 13
6.2.1 Amines
Start with nitrogen’s lone pair: use it to explain basicity and salt formation, then follow ammonia and amines through haloalkane substitution and make aromatic amines from nitroarenes.
What you need to know
Open a line for a quick recap. If it feels obvious, move straight to the linked practice.
6.2.1(a) Explain why amines are Brønsted–Lowry bases. Quick revision
The nitrogen atom in an amine has a lone pair of electrons. It can use that pair to accept H⁺, so an amine is a Brønsted–Lowry base.
Write the protonation as RNH₂ + H⁺ → RNH₃⁺. For OCR this explanation is enough: you are not expected to compare the basicity of different amines in this section.
6.2.1(a) Write equations and draw salts formed when amines react with dilute acids. Quick revision
With a dilute inorganic acid such as HCl, the amine accepts H⁺ and forms an ammonium salt. For ethylamine: CH₃CH₂NH₂ + HCl → CH₃CH₂NH₃⁺ Cl⁻.
When you draw the salt, put the positive charge on nitrogen and include the counter-ion. The carbon skeleton and the C–N bond stay unchanged.
6.2.1(b)(i) Know the reagents, conditions and substitution equation for preparing a primary aliphatic amine from a haloalkane. Quick revision
Heat the haloalkane with excess ethanolic ammonia. NH₃ acts as a nucleophile and replaces the halogen by –NH₂, so this is nucleophilic substitution.
An overall equation such as CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br is useful. Use excess ammonia because the primary amine product can also attack haloalkanes; a large NH₃ concentration improves the proportion of primary amine.
6.2.1(b)(i) Predict secondary amines and further substituted products formed when amines react with haloalkanes. Quick revision
A primary amine still has a nitrogen lone pair, so it can attack another haloalkane. Each further substitution replaces another N–H by an alkyl group: primary amine → secondary amine → tertiary amine.
Further reaction of a tertiary amine can give a quaternary ammonium salt, R₄N⁺X⁻. If the haloalkane is in excess, expect more substituted nitrogen products as well as the primary amine.
6.2.1(b)(ii) Know the reagents and conditions for reducing a nitroarene using tin and concentrated hydrochloric acid. Quick revision
Heat the nitroarene with tin and concentrated hydrochloric acid. The –NO₂ group is reduced to –NH₂; nitrobenzene therefore gives phenylamine.
The strongly acidic mixture protonates the amine as it forms. If a question asks for the free amine after the reduction, an alkaline work-up releases it. For the reagent recall, I’d learn Sn/concentrated HCl as a pair.