Swap a hydrogen of ammonia for an alkyl group and you have an amine. Learn the three routes to ethylamine, how a primary amine becomes a secondary one, and why ethylamine is a stronger base than ammonia.
An amine is ammonia with one or more H atoms replaced by alkyl or aryl groups: a primary amine is RNH2, a secondary amine is R2NH. Primary alkyl amines are made by heating a halogenoalkane with excess ammonia in ethanol in a sealed tube, by reducing an amide with LiAlH4 or by reducing a nitrile with LiAlH4 or H2/Ni. Because the amine product is itself a nucleophile, it can react with more halogenoalkane to give a secondary amine. The lone pair on nitrogen accepts H+, so amines are bases: ethylamine is stronger than ammonia because the electron-donating alkyl group makes the lone pair more available, while phenylamine is much weaker because its lone pair is delocalised into the benzene ring.
Making an amine from a halogenoalkane is covered in the AS nitrogen compounds lab. Here, reduce nitrobenzene in the simulator to make phenylamine, the route that halogenoarenes cannot take.
| Starting compound | Reagent and conditions | Equation |
|---|---|---|
| bromoethane | excess NH3 in ethanol, heat in a sealed tube | CH3CH2Br + 2NH3 → CH3CH2NH2 + NH4Br |
| ethanenitrile | LiAlH4 in dry ether, or H2 with Ni catalyst | CH3CN + 4[H] → CH3CH2NH2 |
| ethanamide | LiAlH4 in dry ether | CH3CONH2 + 4[H] → CH3CH2NH2 + H2O |
| ethylamine | heat with bromoethane | CH3CH2NH2 + CH3CH2Br → (CH3CH2)2NH + HBr |
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Which compound is a secondary amine?
What are the conditions for making ethylamine from bromoethane?
Which reagent reduces ethanenitrile to ethylamine?
Why is ethylamine a stronger base than ammonia?
Compare ethylamine, ammonia and phenylamine in the simulator and watch how available the lone pair is. Then use the table to rank the three nitrogen bases.
| Base | Kb / mol dm−3 | pKb | Lone pair on N |
|---|---|---|---|
| ethylamine, CH3CH2NH2 | 5.4 × 10−4 | 3.27 | made more available by the electron-donating ethyl group |
| ammonia, NH3 | 1.8 × 10−5 | 4.74 | reference: only H atoms on N |
| phenylamine, C6H5NH2 | 4.3 × 10−10 | 9.37 | partly delocalised into the benzene ring, so less available |
Basicity is about how available the lone pair on N is to accept H+, not about how many H atoms there are. Alkyl groups push electron density onto N (stronger base); a benzene ring pulls the lone pair into its delocalised system (weaker base). In the halogenoalkane route, forgetting the excess ammonia means a mixture of primary, secondary and tertiary amines and quaternary salts.
Unlocks once the four checks above are done. Worth more XP, written to A Level Paper 4 standard.
Which list puts the bases in order of DECREASING strength?
10.89 g of bromoethane (Mr 108.9) gives 3.15 g of ethylamine (Mr 45.0). What is the percentage yield?
What volume of hydrogen at room conditions is needed to reduce 0.0500 mol of ethanenitrile completely to ethylamine?
Bromoethane is heated with a LIMITED amount of ammonia. Why is the yield of ethylamine low?
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