AS · Practice questions · Primary amines

Put the lone pair to work.

Six original Cambridge-style questions on primary amines.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Excess ammonia, sealed tube, lone pair.

Make: RX + excess NH₃ in ethanol, heat under pressure.
2 mol NH₃: one substitutes, one removes H⁺.
Further substitution: secondary, tertiary, quaternary salt.
Base: RNH₂ + HCl → RNH₃⁺Cl⁻; stronger than NH₃.

Recall

[3 marks]

Give the reagent and conditions, and write an equation, for making ethylamine from bromoethane.

Excess ammonia, dissolved in ethanol. ✓
Heated in a sealed tube (under pressure). ✓
CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br. ✓

Explain

[3 marks]

Explain why the reaction of bromoethane with ammonia can give a mixture of products, and how the yield of ethylamine is maximised.

Ethylamine has a lone pair on N, so it is also a nucleophile. ✓
It can attack more bromoethane, giving diethylamine, triethylamine and tetraethylammonium bromide. ✓
Using a large excess of ammonia makes collisions of bromoethane with NH₃ far more likely than with ethylamine. ✓

Explain

[3 marks]

Explain why ethylamine is a stronger base than ammonia.

Both act as bases by using the lone pair on N to accept H⁺. ✓
The ethyl group is electron-donating (positive inductive effect). ✓
This increases electron density on N, so the lone pair is more available to bond to H⁺. ✓

Calculation

[2 marks]

4.50 g of ethylamine reacts with excess hydrochloric acid. Calculate the mass of ethylammonium chloride formed. (M_r: CH₃CH₂NH₂ 45.0, CH₃CH₂NH₃Cl 81.5)

n(ethylamine) = 4.50 ÷ 45.0 = 0.100 mol; the ratio is 1 : 1. ✓
Mass of salt = 0.100 × 81.5 = 8.15 g. ✓

Calculation

[2 marks]

What is the minimum volume of ammonia gas, measured at room conditions, needed to convert 10.9 g of bromoethane (M_r 108.9) into ethylamine? (molar gas volume 24.0 dm³)

n(bromoethane) = 10.9 ÷ 108.9 = 0.100 mol, so n(NH₃) = 2 × 0.100 = 0.200 mol. ✓
Volume = 0.200 × 24.0 = 4.80 dm³ (in practice a large excess is used). ✓

Analysis

[2 marks]

Ethylammonium chloride is warmed with aqueous sodium hydroxide. Describe what is observed and write an equation.

A fishy-smelling gas (ethylamine) is released, which turns damp red litmus blue. ✓
CH₃CH₂NH₃⁺Cl⁻ + NaOH → CH₃CH₂NH₂ + NaCl + H₂O. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.