HSC Exam Practice

Sample response. An ester is an organic compound formed by the condensation reaction of a carboxylic acid with an alcohol, with water as the by-product. The ester functional group is –COO– (written in full as –C(=O)–O–), also described by the general formula RCOOR′.

Marking notes. 1 mark for a definition that identifies ester as a product of carboxylic acid + alcohol (or identifies the –COOH and –OH condensation). 1 mark for the correct functional group notation (–COO–, –C(=O)O–, or RCOOR′).

Sample response. The ester produced is propyl methanoate (methanoate from methanoic acid; propyl from propan-1-ol; alcohol-derived part named first). Functional group: ester linkage –COO– (or –C(=O)–O–).

Marking notes. 1 mark for “propyl methanoate” (accept “propyl formate” as a common name only if IUPAC name is clearly intended). 1 mark for correctly identifying the ester linkage (–COO–). Do not award the first mark for “methanoate propyl” — word order is part of the mark.

Sample response. CH₃CH₂COOH + C₂H₅OH ⇌ CH₃CH₂COOC₂H₅ + H₂O; catalyst: conc. H₂SO₄; conditions: heat under reflux.

Marking notes. 1 mark for correct balanced equation with all species present (propanoic acid + ethanol → ethyl propanoate + water). 1 mark for the reversible equilibrium arrow ⇌ (single arrow loses this mark). 1 mark for both conditions: concentrated H₂SO₄ AND heat under reflux (must have both for full mark; each alone is 0).

Sample response. Carboxylic acids have an O–H bond and form hydrogen-bonded dimers — each pair of carboxylic acid molecules forms two simultaneous H-bonds, requiring significantly more energy to break. Esters have no O–H bond and therefore cannot donate hydrogen bonds; ester–ester interactions are limited to dipole–dipole forces, which are weaker than H-bonds. Because more energy is required to separate carboxylic acid dimers, carboxylic acids have higher boiling points despite comparable molecular mass.

Marking notes. 1 mark for identifying that carboxylic acids form H-bonds (accept: “form dimers via two H-bonds”). 1 mark for correctly stating that esters cannot donate H-bonds due to the absence of an O–H bond. 1 mark for explicitly linking the stronger IMF in carboxylic acids to the higher boiling point.

Sample response. Acid hydrolysis: RCOOR′ + H₂O ⇌ RCOOH + R′OH; reversible equilibrium — the carboxylic acid and alcohol are produced but can re-esterify under the same conditions. Saponification (base hydrolysis): RCOOR′ + NaOH → RCOONa + R′OH; irreversible — the carboxylate salt produced cannot re-esterify under basic conditions, so the reaction goes to completion.

Marking notes. 1 mark for a correct general equation for acid hydrolysis with reversible arrow ⇌. 1 mark for stating acid hydrolysis is reversible and identifying why (carboxylic acid and alcohol are formed and can re-esterify). 1 mark for a correct general equation for saponification with single arrow → and NaOH. 1 mark for stating saponification is irreversible because the carboxylate salt cannot re-esterify under basic conditions.

Sample response. Butter contains saturated fatty acid chains. Saturated chains have no C=C bonds and are straight, allowing close parallel alignment. This close packing maximises the contact surface area between adjacent chains, producing strong cumulative dispersion forces and a relatively high melting point, so butter is solid at room temperature. Olive oil contains unsaturated fatty acid chains with one or more cis C=C bonds, which introduce rigid kinks. Kinked chains cannot pack closely — the gaps between adjacent chains reduce the contact surface area, weakening dispersion forces and lowering the melting point, so olive oil is liquid at room temperature.

Marking notes. 1 mark for identifying that butter has saturated (straight) chains and olive oil has unsaturated (kinked/cis C=C) chains. 1 mark for the packing argument: straight chains pack closely → large contact area → strong dispersion forces → solid. 1 mark for the contrast: kinked chains cannot pack closely → weaker dispersion forces → liquid. Must reference dispersion forces (not “H-bonds”) for the third mark.

Part (a) — Theoretical yield. Moles of ethanoic acid = 6.00 ÷ 60.05 = 0.0999 mol. From the balanced equation, 1 mol acid produces 1 mol ester. Theoretical moles of ethyl ethanoate = 0.0999 mol. Theoretical mass = 0.0999 × 88.11 = 8.80 g. [2 marks: 1 for correct moles of acid; 1 for correct theoretical mass of ester]

Part (b) — Percentage yield. % yield = (5.46 / 8.80) × 100 = 62.0%. [1 mark]

Part (c) — Reason for yield below 100%. Accept any one of: esterification is a reversible equilibrium — at equilibrium all four species (acid, alcohol, ester, water) remain present, so 100% conversion is never achieved; product is lost during isolation steps (separating funnel, washing, drying, distillation); some ester may hydrolyse back to acid and alcohol in the presence of water. [2 marks: 1 for identifying the reversible equilibrium as the fundamental reason; 1 for a second valid reason or fuller explanation]

Part (a) — Ethyl ethanoate vs butan-1-ol. Ethyl ethanoate (BP 77°C) has a lower boiling point than butan-1-ol (BP 118°C) despite having a higher molecular mass (88 vs 74 g mol&supmin;¹). Butan-1-ol has an O–H bond and forms hydrogen bonds (as both donor and acceptor), requiring more energy to vaporise. Ethyl ethanoate has no O–H bond and cannot donate hydrogen bonds; its intermolecular forces are limited to dipole–dipole and dispersion. H-bonds are stronger than dipole–dipole forces, so butan-1-ol has the higher boiling point despite lower molecular mass. [3 marks: 1 for comparing BPs with values; 1 for identifying H-bonding in butan-1-ol via O–H; 1 for explicitly stating ethyl ethanoate cannot donate H-bonds and is limited to dipole–dipole]

Part (b) — Butanoic acid vs ethyl ethanoate. Butanoic acid forms hydrogen-bonded dimers: each pair of butanoic acid molecules simultaneously forms two H-bonds (the O–H of one molecule donates to the C=O of the other). Breaking both H-bonds to vaporise the molecule requires more energy than breaking the single dipole–dipole interactions of ethyl ethanoate. Consequently, butanoic acid has a much higher boiling point (164°C vs 77°C) despite the same molecular mass. [2 marks: 1 for dimer / two simultaneous H-bonds explanation; 1 for linking to higher energy required to vaporise]

Sample Band 6 response.

Balanced equation: CH₃COOH + CH₃(CH₂)₄OH ⇌ CH₃COO(CH₂)₄CH₃ + H₂O; conc. H₂SO₄ (catalyst), heat under reflux. The reversible arrow ⇌ is essential; a single arrow loses marks.

Why yield is below 100%: Esterification is a reversible equilibrium. At equilibrium, the forward and reverse reactions occur at equal rates and all four species (pentyl ethanoate, water, ethanoic acid, pentan-1-ol) coexist. The equilibrium constant K_eq ≈ 1–4 for simple esterification reactions, meaning the equilibrium lies roughly in the middle and 100% conversion is never achieved. Under equimolar conditions, yield is approximately 50–65%.

Modification 1 — Excess pentan-1-ol (e.g. 3:1 molar ratio): Increasing the concentration of a reactant (pentan-1-ol) stresses the equilibrium. By Le Chatelier’s Principle, the system shifts right to oppose the increase by consuming more pentan-1-ol and producing more ester. Yield can rise from ~52% to ~78–80% with a 3:1 excess.

Modification 2 — Remove water as it forms (using conc. H₂SO₄ as a dehydrating agent, or adding anhydrous molecular sieves): Removing water (a product) reduces [H₂O]. By Le Chatelier’s Principle, the system shifts right to replace the removed water, producing more ester. This modification is complementary to Modification 1 and can push the yield above 80% when combined.

Isolation of pentyl ethanoate: (1) Pour reaction mixture into separating funnel with cold water — ester (less dense, insoluble in water) forms upper layer; acid, alcohol, H₂SO₄, and water form lower aqueous layer; drain and discard lower layer. (2) Wash with Na₂CO₃ solution to remove residual ethanoic acid; equation: 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂; drain and discard aqueous layer. (3) Wash with fresh water; drain. (4) Add anhydrous Na₂SO₄ to absorb residual water; filter. (5) Distil to collect pure pentyl ethanoate (BP 148°C).

Marking criteria (9 marks):

• 1 mark — Balanced esterification equation for pentyl ethanoate (CH₃COOH + C₅H₁₁OH → CH₃COOC₅H₁₁ + H₂O, all species balanced).
• 1 mark — Correct reversible arrow ⇌ AND conditions (conc. H₂SO₄ catalyst; heat under reflux) — both needed for this mark.
• 1 mark — Correctly identifies the reversible equilibrium and K_eq ≈ 1–4 as the reason yield is below 100% under equimolar conditions (not just “reaction incomplete”).
• 1 mark — Modification 1 (excess pentan-1-ol): states it increases a reactant concentration and by Le Chatelier shifts equilibrium right.
• 1 mark — Modification 1: gives a reasonable estimate of yield improvement (or states “increases toward 80%+”).
• 1 mark — Modification 2 (remove water): states product removal and by Le Chatelier shifts equilibrium right.
• 1 mark — Correct isolation Step 1: separating funnel — ester forms upper layer (non-polar, insoluble); aqueous lower layer drained and discarded.
• 1 mark — Isolation Step 2: Na₂CO₃ wash with correct ionic/equation (or words equivalent) showing it removes residual ethanoic acid as water-soluble carboxylate salt.
• 1 mark — Isolation Steps 3–5: water wash to remove Na₂CO₃; anhydrous drying agent (Na₂SO₄ or MgSO₄) to remove water; distillation to collect pure ester. (All three needed for this mark.)