Primary amines

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br. Ammonia attacks the δ+ carbon with its lone pair (nucleophilic substitution). It is used in large excess, in ethanol, and heated in a sealed tube so the volatile ammonia stays in the mixture at a useful concentration. A limited amount of ammonia at room temperature is too slow and lets the amine react further, giving secondary and tertiary amines. KCN in ethanol gives a nitrile, propanenitrile, not an amine. NaOH(aq) gives ethanol by hydrolysis.

The nitrogen in ethylamine still has a lone pair, so it can attack another bromoethane molecule to give diethylamine, then triethylamine and finally tetraethylammonium bromide. With a large excess of NH₃, a bromoethane molecule almost always collides with NH₃ rather than with the amine, so the primary amine is the main product. Ammonia is a reactant, not a catalyst: 2 mol are used per mol of bromoethane. The reaction is done in ethanol and the issue is further substitution, not hydrolysis. Ammonia is a reasonable nucleophile; the excess is about controlling which nucleophile the halogenoalkane meets.

Alkyl groups are electron-donating (a positive inductive effect). In ethylamine the ethyl group increases the electron density on N, so its lone pair forms a dative bond to H⁺ more readily: ethylamine is a slightly stronger base than ammonia. Bases accept protons; donating H atoms is what acids do, so the number of N-H hydrogens is not the reason. Alkyl groups donate, not withdraw, electron density. Both have one lone pair, but how available that lone pair is differs, so they are not equally strong.

CH₃CH₂NH₂ + HCl → CH₃CH₂NH₃⁺Cl⁻. The lone pair on N accepts H⁺, exactly as NH₃ forms NH₄⁺. The salt is a white ionic solid, soluble in water; adding NaOH(aq) releases the amine again. No substitution happens on the carbon chain, so chloroethane does not form. Dilute acid does not hydrolyse the C-N bond to give ethanol. Amines are bases (their solutions are alkaline), so they certainly react with acids.

The nitrogen is bonded to two ethyl groups and one H, so it is a secondary amine called diethylamine. It forms when ethylamine attacks a second bromoethane molecule. A primary amine has N bonded to only one carbon, such as ethylamine, CH₃CH₂NH₂. The prefix di- shows two ethyl groups, so it is not plain ethylamine. A quaternary ammonium ion has four carbons on N and a positive charge, such as (CH₃CH₂)₄N⁺; this molecule is neutral.

n(bromoethane) = 5.45 / 108.9 = 0.0500 mol. The ratio is 1:1, so the theoretical mass of ethylamine = 0.0500 × 45.0 = 2.25 g, and 60.0% of that is 1.35 g. 2.25 g is the theoretical mass, forgetting the yield. 3.27 g is 60.0% of the mass of bromoethane, which ignores the change in M_r. 0.675 g halves the answer, as if 2 mol of bromoethane gave 1 mol of amine; it is the ammonia, not the bromoethane, that is used in a 2:1 ratio.

CH₃CH₂Br + 2NH₃ → CH₃CH₂NH₂ + NH₄Br. n(bromoethane) = 10.9 / 108.9 = 0.100 mol, so n(NH₃) = 0.200 mol and the volume = 0.200 × 24.0 = 4.80 dm³. The first NH₃ substitutes for Br; the second removes H⁺ from CH₃CH₂NH₃⁺. 2.40 dm³ uses a 1:1 ratio, forgetting the second ammonia. 1.20 dm³ halves instead of doubling. 9.60 dm³ uses a 1:4 ratio. In practice far more than the minimum is used, to limit further substitution.

Each step is nucleophilic substitution by the N lone pair: NH₃ gives CH₃CH₂NH₂, then (CH₃CH₂)₂NH, then (CH₃CH₂)₃N. Triethylamine still has a lone pair, so it attacks once more, forming (CH₃CH₂)₄N⁺Br⁻, a quaternary ammonium salt. Now N has four bonds and no lone pair, so substitution stops. Triethylamine is not the end: a nucleophile needs a lone pair, not an N-H hydrogen. Ethylammonium bromide is only the first intermediate, and with excess halogenoalkane the substitution continues. Electron-rich N would react more readily, not less; the reaction stops because the lone pair has been used.