Chemistry • Year 12 • Module 7 • Lesson 15
Esters: Structure, Naming & Esterification
Apply ester naming, Le Chatelier yield reasoning, boiling-point comparison, and Australian industrial context to real data and scenarios.
1. Ester smells and applications in Australian industry
The table below lists seven esters used in the Australian food, flavour, and solvent industries. Use the data to answer the questions. 8 marks
| Ester (IUPAC name) | Structural formula | Aroma / use | BP (°C) | Water solubility |
|---|---|---|---|---|
| Methyl methanoate | HCOOCH₃ | Raspberry / pungent solvent | 32 | Miscible |
| Ethyl ethanoate | CH₃COOC₂H₅ | Nail polish remover / pear notes in wine | 77 | 8.3 g/100 mL |
| Methyl butanoate | C₃H₇COOCH₃ | Apple / strawberry | 102 | Slightly soluble |
| Butyl ethanoate | CH₃COOC₄H₉ | Pineapple; used in FSANZ-approved confectionery | 126 | Slightly soluble |
| Pentyl ethanoate | CH₃COOC₅H₁₁ | Banana; used in Australian lolly manufacturing | 148 | Insoluble |
| Pentyl propanoate | C₂H₅COOC₅H₁₁ | Pear; fragrance in personal care products | 168 | Insoluble |
| Ethyl hexanoate | C₅H₁₁COOC₂H₅ | Pineapple / tropical; used in Australian wine aroma additives | 167 | Insoluble |
Data compiled from FSANZ food additive register and published boiling point data. IUPAC names follow the alkyl alkanoate convention.
1.1 State the general trend in boiling point as the total carbon chain length of the ester increases. Explain why this trend occurs using intermolecular force (IMF) reasoning. 3 marks
1.2 Identify the ester from the table that would be most easily separated from an aqueous reaction mixture in a separating funnel. Justify your answer using the solubility data. 2 marks
1.3 Identify the acid and alcohol that would be reacted together to produce pentyl propanoate. Write the balanced esterification equation with all conditions. 3 marks
2. Interpret graph — effect of alcohol excess on ester yield
A food chemist investigating the synthesis of ethyl ethanoate for the Australian confectionery industry varied the molar ratio of ethanol to ethanoic acid while keeping all other conditions constant (conc. H₂SO₄, reflux, 1 hour reaction time). The graph below shows the percentage yield of ethyl ethanoate at different molar ratios of ethanol : ethanoic acid. 8 marks
Figure 2.1. Percentage yield of ethyl ethanoate vs molar ratio of ethanol to ethanoic acid. Conc. H₂SO₄ catalyst, reflux, 1 hour reaction time. Hypothetical data modelled on esterification equilibrium Keq data.
2.1 Describe the trend shown in the graph as the molar ratio of ethanol increases from 1:1 to 6:1. Include figures from the graph. 2 marks
2.2 Using Le Chatelier’s Principle, explain why increasing the ethanol : ethanoic acid ratio increases the ester yield. 3 marks
2.3 The graph shows that yield never reaches 100%, even with a 6:1 molar ratio. Explain why 100% yield is theoretically impossible for esterification. 2 marks
2.4 Suggest one additional modification (other than adding more ethanol) that the food chemist could use to further increase the yield beyond 84%. Explain how it works using Le Chatelier’s Principle. 1 mark
3. Cause-and-effect chain — ester isolation using a separating funnel
Complete the cause-and-effect chain for the five-step isolation of ethyl ethanoate after esterification. Fill in each “effect” box with what happens as a result of the step described. 5 marks
| Cause (step performed) | Effect (what this achieves) |
|---|---|
| Step 1: Reaction mixture is poured into separating funnel with cold water. | |
| Step 2: Lower aqueous layer is drained and discarded; Na₂CO₃ solution is added to the ester layer. | |
| Step 3: Aqueous layer is again drained; fresh water is added and shaken. | |
| Step 4: Anhydrous Na₂SO₄ is added to the drained ester layer. | |
| Step 5: Dried ester is filtered then distilled. |
4. Apply to a real context — Palm oil hydrolysis and Australian biodiesel
Palm oil is a triglyceride (triester of glycerol with three fatty acid chains). Australian researchers at the University of Queensland have investigated replacing petroleum-derived diesel with biodiesel produced from triglycerides. In the process, palm oil is hydrolysed using a base catalyst (NaOH) and then converted. 5 marks
4.1 Write the general equation for the saponification of a triglyceride (use R to represent a fatty acid chain). State the reagent and conditions. 2 marks
4.2 Explain why saponification is irreversible whereas acid hydrolysis of the same ester is reversible. 2 marks
4.3 The palm oil used in biodiesel production is mainly saturated (high proportion of palmitic acid, C₁₅H₁₃COOH). Would you expect palm oil to be liquid or solid at room temperature? Justify using an IMF argument. 1 mark
Q1.1 — Boiling point trend (3 marks)
As total carbon chain length increases, boiling point increases (e.g. from 32°C for methyl methanoate to 168°C for pentyl propanoate) [1]. This occurs because longer chains increase molecular mass and surface area [1], which strengthens London (dispersion) forces between molecules. More energy is required to overcome stronger dispersion forces, raising the boiling point [1].
Q1.2 — Easiest separation (2 marks)
Pentyl ethanoate (or pentyl propanoate / ethyl hexanoate — any ester listed as “insoluble” with a high BP) [1]. These esters are insoluble in water, so they form a completely distinct upper organic layer in the separating funnel, making the two-phase separation straightforward without requiring additional washing steps [1]. Accept any insoluble ester with valid justification.
Q1.3 — Pentyl propanoate equation (3 marks)
Acid: propanoic acid (CH₃CH₂COOH); Alcohol: pentan-1-ol (CH₃(CH₂)₄OH) [1].
Balanced equation: CH₃CH₂COOH + CH₃(CH₂)₄OH ⇌ CH₃CH₂COO(CH₂)₄CH₃ + H₂O [1, must use reversible arrow ⇌].
Conditions: concentrated H₂SO₄ (catalyst); heat under reflux [1].
Q2.1 — Graph trend (2 marks)
As the molar ratio of ethanol to ethanoic acid increases from 1:1 to 6:1, the percentage yield of ethyl ethanoate increases [1], rising steeply from ~50% at 1:1 to ~78% at 3:1, then levelling off (diminishing returns) to approximately 86% at 6:1 [1].
Q2.2 — Le Chatelier explanation (3 marks)
The esterification is a reversible equilibrium: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O [1]. Increasing the molar ratio of ethanol to ethanoic acid increases the concentration of a reactant (ethanol) [1]. By Le Chatelier’s Principle, the system opposes this increase by shifting the equilibrium to the right — producing more ester and water — until a new equilibrium position is reached with a higher ester yield [1].
Q2.3 — Why 100% is impossible (2 marks)
Esterification is a reversible equilibrium (Keq ≈ 1–4). At equilibrium, both the forward reaction (esterification) and the reverse reaction (acid hydrolysis) occur simultaneously [1]. Even with excess ethanol, the water produced shifts the equilibrium slightly back to the left, so all four species remain present at equilibrium and 100% conversion is never achieved [1].
Q2.4 — Additional modification (1 mark)
Remove water as it forms, e.g. using concentrated H₂SO₄ as a dehydrating agent, adding molecular sieves, or distilling off water continuously. Removing water (a product) reduces [H₂O], which shifts equilibrium to the right by Le Chatelier’s Principle to replace the removed product, increasing ester yield [1].
Q3 — Cause-and-effect chain (5 marks, 1 each)
Step 1 effect: The ester (less dense, low water solubility) separates into an upper organic layer; water, unreacted acid, alcohol, and H₂SO₄ dissolve in the lower aqueous layer — the two layers can be separated.
Step 2 effect: Na₂CO₃ reacts with residual carboxylic acid (forming water-soluble carboxylate salt + CO₂) and neutralises residual H₂SO₄; both are removed in the lower aqueous layer, leaving a purer ester in the organic layer.
Step 3 effect: Residual Na₂CO₃ and sodium carboxylate salt are washed out of the organic layer, further purifying the ester.
Step 4 effect: Anhydrous Na₂SO₄ absorbs any residual water from the organic layer; the agent clumps as it hydrates. Filtering removes the drying agent and leaves a dry ester.
Step 5 effect: Distillation separates the pure ester from any remaining impurities by collecting the fraction at the ester’s boiling point, giving a high-purity product.
Q4.1 — Saponification equation (2 marks)
General equation: C₃H₅(OOCR)₃ + 3NaOH → C₃H₅(OH)₃ + 3RCOONa [1, must use single arrow → not ⇌].
Reagent: concentrated NaOH(aq); conditions: heat under reflux [1].
Q4.2 — Irreversible vs reversible (2 marks)
In acid hydrolysis, the products are a carboxylic acid and an alcohol — both can re-esterify under the same conditions, so the reverse reaction occurs and equilibrium is established [1]. In saponification, the product is a carboxylate salt (RCOO⁻Na⁺) — the negatively charged carboxylate anion is too weakly reactive as an acylating agent to re-esterify with the alcohol under basic conditions, so the reaction proceeds to completion (irreversible) [1].
Q4.3 — Palm oil state at room temperature (1 mark)
Palm oil would be solid or semi-solid at room temperature (it is commonly described as a semi-solid fat). Its saturated fatty acid chains (palmitic acid) are straight and pack closely together, giving strong cumulative dispersion forces and a high enough melting point to be solid or semi-solid at ambient temperatures [1].