Swap a halogen atom for an NH2 group and you have a primary amine. Making one cleanly needs a big excess of ammonia, and the lone pair on nitrogen that does the attacking also makes amines bases, a little stronger than ammonia itself.
A primary amine has an NH2 group joined to one carbon atom, for example ethylamine (ethanamine), CH3CH2NH2. It is made by heating a halogenoalkane with excess ammonia in ethanol under pressure (in a sealed tube): CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br. This is nucleophilic substitution: the lone pair on N attacks the δ+ carbon. The amine product still has a lone pair, so it can attack more halogenoalkane, giving secondary and tertiary amines and finally a quaternary ammonium salt; a large excess of ammonia makes that less likely. The same lone pair accepts a proton, so amines are bases: CH3CH2NH2 + HCl → CH3CH2NH3+Cl−. The electron-donating ethyl group makes ethylamine a slightly stronger base than ammonia.
Choose a halogenoalkane in the simulator and step through the attack by ammonia. Watch the lone pair arrive, the carbon turn inside out, and a second ammonia molecule remove the proton.
| Halogenoalkane | Reagent and conditions | Organic product |
|---|---|---|
| bromoethane | excess NH3 in ethanol, heat in a sealed tube | ethylamine, CH3CH2NH2 |
| 1-chloropropane | excess NH3 in ethanol, heat in a sealed tube | propylamine, CH3CH2CH2NH2 |
| bromoethane | excess bromoethane with NH3 | mixture: (CH3CH2)2NH, (CH3CH2)3N, (CH3CH2)4N+Br− |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Which reagent and conditions convert bromoethane into ethylamine?
What is the name of CH3CH2NH2?
Which feature of an amine makes it a base?
What is formed when ethylamine reacts with hydrochloric acid?
Compare ammonia, methylamine and ethylamine in the simulator and read off the pH of each solution, then apply the Brønsted–Lowry idea to the lone pair on an amine nitrogen.
| Reaction | Equation | Note |
|---|---|---|
| with water | CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH− | weak base: solution is alkaline, pH about 11 to 12 |
| with HCl | CH3CH2NH2 + HCl → CH3CH2NH3+Cl− | ethylammonium chloride, a white ionic solid |
| salt with NaOH | CH3CH2NH3+Cl− + NaOH → CH3CH2NH2 + NaCl + H2O | a stronger base frees the amine (fishy smell) |
Use excess ammonia, not excess halogenoalkane, to get the primary amine; excess halogenoalkane pushes the substitution on to secondary and tertiary amines and the quaternary salt. Two moles of NH3 are used per mole of halogenoalkane, because the second one takes the H+. Amines are stronger bases than ammonia, not weaker: the alkyl group pushes electron density onto the nitrogen. The salt with HCl is ionic, CH3CH2NH3+Cl−; nothing is substituted on the carbon chain.
Unlocks once the four checks above are done. Worth more XP, written to AS Paper 1 and Paper 2 standard.
Why is a large excess of ammonia used to make ethylamine from bromoethane?
Which is the correct order of base strength, strongest first?
Bromoethane is heated with ammonia that is NOT in excess. Which compound could NOT be formed?
5.45 g of bromoethane (Mr 108.9) gives ethylamine (Mr 45.0) in 60.0% yield. What mass of ethylamine forms?
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