The problems are on the esterification worksheet. Try them on your own first, then compare with the answers below.
Quick answer: Carboxylic acids react with alcohols, in the presence of concentrated sulfuric acid, to form esters and water in a reversible equilibrium. These worked answers correct the 20 problems of the Chemistry 11-12 worksheet on esterification, from names of acids and esters to making aspirin: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the method and then the final result in bold. Atomic masses used: H 1.0, C 12.0, N 14.0, O 16.0, Na 23.0, Cl 35.5. Always check that equations balance and that equilibria are written with the reversible arrow.
Level: Chemistry grades 11-12 (A Level · IB · Class 11-12) · Curriculum: A Level 3.3.9
Problem 1 Names of acids and esters
1. Propanoic acid.2. The alkyl group on the oxygen is CH₃ (methyl) and the acid stem is propan: methyl propanoate.
3. Ethyl group on oxygen, one carbon in the acid: ethyl methanoate.
Problem 2 Acid reactions
1. 2CH₃COOH + Mg → (CH₃COO)₂Mg + H₂2. 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂
3. CH₃COOH + NaOH → CH₃COONa + H₂O
Names of the salts: magnesium ethanoate and sodium ethanoate.
Problem 3 Making methyl propanoate
1. CH₃CH₂COOH + CH₃OH ⇌ CH₃CH₂COOCH₃ + H₂O.2. Concentrated sulfuric acid; heat under reflux.
3. C₄H₈O₂ (4 C, 8 H, 2 O).
Problem 4 Hydrolysis products
1. Methanoic acid and ethanol.2. Sodium methanoate and ethanol.
3. The alkaline hydrolysis.
Problem 5 Making an acyl chloride
1. CH₃CH₂COOH + SOCl₂ → CH₃CH₂COCl + SO₂ + HCl.2. Propanoyl chloride.
3. Both by-products are gases, so they leave the mixture.
Problem 6 Reading an esterification scheme
1. The –COO– group between the C=O carbon and the propyl group.2. “Propyl” comes from propan-1-ol.
3. C₆H₁₂O₂ (3 + 3 carbons, 12 hydrogens, 2 oxygens).
Problem 7 Carbon dioxide from vinegar
1. CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂.2. n = 6.0 / 60.0 = 0.100 mol.
3. V = 0.100 × 24.0 = 2.40 dm³.
Problem 8 Calculating Kc
1. Acid = 0.500 − 0.333 = 0.167 mol (alcohol also 0.167 mol); water = 0.333 mol.2. Kc = [ester][water] / ([acid][alcohol]).
3. Kc = (0.333)² / (0.167)² = 4.0 (volume cancels).
Problem 9 Raising the yield
1. Use an excess of one reactant, and remove water or the ester as it forms.2. A catalyst only increases the rate, so equilibrium is reached sooner but the yield is the same.
3. Kc is unchanged, because it depends only on temperature.
Problem 10 Making soap
1. n = 8.90 / 890 = 0.0100 mol.2. n(NaOH) = 0.0300 mol; mass = 0.0300 × 40.0 = 1.20 g.
3. Mass of soap = 0.0300 × 306 = 9.18 g.
Problem 11 Banana flavour
1. CH₃COOH + C₅H₁₁OH ⇌ CH₃COOC₅H₁₁ + H₂O.2. n(acid) = 6.0 / 60.0 = 0.100 mol; theoretical mass = 0.100 × 130 = 13.0 g.
3. Actual = 0.70 × 13.0 = 9.1 g.
Problem 12 Why alkaline hydrolysis finishes
1. The acid-catalysed hydrolysis.2. In alkali, the acid is converted to its carboxylate ion, which is negatively charged and cannot be attacked by the alcohol, so the reverse reaction cannot happen.
3. Sodium ethanoate.
Problem 13 Acylation equations
1. CH₃COCl + H₂O → CH₃COOH + HCl.2. CH₃COCl + C₂H₅OH → CH₃COOC₂H₅ + HCl.
3. CH₃COCl + 2NH₃ → CH₃CONH₂ + NH₄Cl; the second ammonia molecule absorbs the HCl.
Problem 14 Why acyl chlorides are reactive
1. It is bonded to two electronegative atoms, O and Cl, which both pull electron density away.2. Cl⁻ is a stable ion that departs easily.
3. They are corrosive, react violently with water and give off HCl fumes, so they must be handled in a fume cupboard.
Problem 15 Atom economy of the acyl chloride route
1. CH₃CH₂COOH + SOCl₂ → CH₃CH₂COCl + SO₂ + HCl.2. 92.5 / (74.0 + 119.0) × 100 = 92.5 / 193.0 = 47.9 %.
3. Sulfur dioxide and hydrogen chloride.
Problem 16 Rise to equilibrium
1. About 0.67 mol/dm³ (plateau).2. Acid = alcohol = 0.33 mol/dm³; Kc = 0.67² / 0.33² = 4.0 (4.1 with rounded values).
3. Rate = 0.20 / 20 = 0.010 mol dm⁻³ min⁻¹.
Problem 17 Excess alcohol
1. Kc = x² / [(1.00 − x)(3.00 − x)] = 4.0.2. 3x² − 16x + 12 = 0; x = (16 − √112) / 6 = 0.903 mol (the other root is above 1.00).
3. Conversion = 0.903 / 1.00 × 100 = 90.3 %.
Problem 18 Find the errors
1. Chloride leaves first, forming an acylium ion; in reality the nucleophile adds first.2. The ethanol oxygen lone pair attacks the δ+ carbon, the C=O π electrons move onto O, giving a tetrahedral intermediate.
3. The intermediate loses Cl⁻ and an H⁺ from the oxygen, and these combine to make HCl.
Problem 19 Back titration of an ester
1. n = 0.400 × 0.0500 = 0.0200 mol.2. Excess = 0.500 × 0.0200 = 0.0100 mol, so reacted = 0.0200 − 0.0100 = 0.0100 mol.
3. M = 0.880 / 0.0100 = 88.0 g/mol, formula C₄H₈O₂ (4(12) + 8 + 32 = 88).
Problem 20 Making aspirin
1. C: 7 + 4 = 11 = 9 + 2; H: 6 + 6 = 12 = 8 + 4; O: 3 + 3 = 6 = 4 + 2. Balanced.2. n = 2.76 / 138.0 = 0.0200 mol; mass = 0.0200 × 180.0 = 3.60 g.
3. Yield = 2.52 / 3.60 = 70.0 %; atom economy = 180.0 / (138.0 + 102.0) = 75.0 %.
