Topic 19 · page 3 of 13
Salts and complex ions
The same lone pair that makes an amine a base lets it join H⁺ to make a salt, and join a copper(II) ion to make a complex. Both reactions use the lone pair as an electron pair donor.
Read first: Amines and water
Reaction with acids 19.2(ii)
An amine reacts with an acid by accepting H⁺ on the nitrogen lone pair. The product is a salt that contains an ammonium-type cation.
The product is butylammonium chloride, a white ionic solid. The ionic equation shows the change at the nitrogen:
Phenylamine behaves in the same way. It is an oily liquid that is only slightly soluble in water, but it dissolves in hydrochloric acid because it forms a soluble salt:
The product is phenylammonium chloride.
With a dibasic acid such as sulfuric acid, two amine molecules accept one H₂SO₄ between them:
The salts are ionic, so they are soluble in water and odourless (amines have a fishy smell).
Getting the amine back
Adding a strong base such as sodium hydroxide solution removes the proton again and releases the free amine:
You will see this step again on page 6: phenylamine is made in acid as its salt, and sodium hydroxide frees it.
Worked exampleSalt and sodium hydroxide
1.46 g of butylamine, M = 73.0 g mol⁻¹, reacts with excess hydrochloric acid. Find the mass of butylammonium chloride formed, and the volume of 1.00 mol dm⁻³ NaOH needed to free all the amine again.
- n(C₄H₉NH₂) = 1.46 ÷ 73.0 = 0.0200 mol
- Ratio 1 : 1, so n(C₄H₉NH₃⁺Cl⁻) = 0.0200 mol
- M(C₄H₉NH₃⁺Cl⁻) = 4 × 12.0 + 12 × 1.0 + 14.0 + 35.5 = 109.5 g mol⁻¹, so mass = 0.0200 × 109.5 = 2.19 g
- n(NaOH) = 0.0200 mol, so V = 0.0200 ÷ 1.00 = 0.0200 dm³ = 20.0 cm³
Reaction with copper(II) ions 19.2(v)
An amine can also donate its lone pair to a metal ion. A ligand is a species that donates a lone pair to a metal ion, forming a dative covalent bond. Butylamine behaves like ammonia with aqueous copper(II) ions, which are present as the pale blue hexaaquacopper(II) ion, [Cu(H₂O)₆]²⁺.
A little butylamine acts as a base. It takes H⁺ from water ligands, and a pale blue precipitate of copper(II) hydroxide forms:
Excess butylamine acts as a ligand. It replaces four of the water ligands, and the precipitate dissolves to give a deep blue solution of a complex ion:
The complex has six ligands round the copper ion: four butylamine molecules and two water molecules.
Phenylamine also has a lone pair to donate, so in principle it can do the same. It is a much weaker base and a much weaker ligand, because its lone pair is delocalised into the ring. The spec names butylamine as the main example, and mark schemes describe the colours for the aliphatic amine.
Check yourself
- Write the equation for the reaction of butylamine with hydrochloric acid and name the product.
- Phenylamine does not dissolve in water, but it dissolves in hydrochloric acid. Explain, with an equation.
- Write the equation for butylamine and sulfuric acid.
- How can the free amine be recovered from butylammonium chloride solution? Give an equation.
- 3.65 g of butylamine reacts with excess dilute hydrochloric acid. Calculate the mass of salt formed (M of the salt = 109.5 g mol⁻¹).
- Describe what you would see when butylamine is added slowly, to excess, to aqueous copper(II) sulfate, and write equations for the two changes.
Answers cover these until you have tried the questions
- . The product is butylammonium chloride.
- Phenylamine is only slightly soluble in water, but the lone pair on N accepts H⁺ from the acid, forming the ionic salt phenylammonium chloride, which is soluble: .
- , giving butylammonium sulfate. Two amine molecules per sulfuric acid.
- Add sodium hydroxide solution (a strong base) to remove H⁺ from the ammonium ion: .
- n(amine) = 3.65 ÷ 73.0 = 0.0500 mol. Ratio 1 : 1, so n(salt) = 0.0500 mol. Mass = 0.0500 × 109.5 = 5.475 g = 5.48 g.
- First a pale blue precipitate forms, then it dissolves to give a deep blue solution. (amine as a base), then (amine as a ligand).