Discover how the same chemical reaction that produces fruity fragrances also builds the fats stored in food — and learn to name, write, and manipulate ester reactions at HSC exam level.
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
A food chemist wants to recreate the smell of pineapple for a new range of confectionery. She reacts ethanoic acid with butan-1-ol in the presence of concentrated sulfuric acid, heats the mixture under reflux, then separates the oily product layer using a separating funnel, washes it with sodium carbonate solution, dries it, and distils it. The result is a clear liquid with a powerful fruity pineapple odour.
The reaction she used is the same one that originally formed the fats in the avocado she had for breakfast.
Before you read on: Write down what you think the structural connection is between pineapple flavouring and avocado fat. What chemical feature do they share? Why might the same type of reaction produce both a volatile fragrant liquid and a solid fat?
An ester is built from two halves — an acid half and an alcohol half — and its name reflects exactly this two-part origin, always listing the alcohol-derived part first and the acid-derived part second.
The ester functional group is –COO– (written as –C(=O)–O– in full structural form). The carbonyl carbon has a double bond — trigonal planar geometry, ~120° bond angles. The two oxygens in the ester group are NOT equivalent:
No H is attached to either oxygen in an ester — this is why esters cannot donate hydrogen bonds. This single structural fact explains nearly everything unusual about ester properties.
Ethyl ethanoate — ester linkage showing the cut at the single-bond O. Left of cut: ethanoate (from ethanoic acid). Right of cut (including the O): ethyl (from ethanol). Name = ethyl ethanoate.
Rule: Cut the ester at the single-bond O. Left = alkanoate (acid half, contains C=O). Right O + carbon chain = alkyl (alcohol half). Name: [alkyl] [alkanoate] — alcohol part always first.
MICROTASK · +5 XP
What is the IUPAC name of the ester formed from propanoic acid and methanol?
Esters are less polar than carboxylic acids and cannot H-bond with each other — which makes them more volatile, which is exactly why they are nature's choice for dispersing scent signals across distance.
Esters have a polar –COO– group (net permanent dipole) → dipole-dipole forces between ester molecules, plus London dispersion forces from the carbon chains. Crucially: esters have no O–H bond — they cannot donate H-bonds. The two oxygens can accept H-bonds from water, but ester-to-ester interactions are limited to dipole-dipole and dispersion only.
Esters sit between aldehydes/ketones and alcohols in boiling point — consistent with dipole-dipole forces but no H-bond donation.
Short-chain esters are partially soluble in water — the lone pairs on the ester oxygens can accept H-bonds from water (one-directional interaction). Ethyl ethanoate: ~8.3 g/100 mL. Longer-chain esters: essentially insoluble. All esters are readily miscible with non-polar organic solvents.
MICROTASK · +5 XP
True or False: Ethyl ethanoate (BP 77°C) has a lower boiling point than propan-1-ol (BP 97°C) because ethyl ethanoate has no O–H bond and therefore cannot donate hydrogen bonds, while propan-1-ol can.
Esterification is a reversible reaction that never goes to completion — understanding the equilibrium position and applying Le Chatelier's Principle to maximise yield is exactly the same skill as the industrial processes studied in Module 5, now applied to a functional group transformation.
R–COOH + R'OH ⇌ R–COOR' + H₂O
Both forward (esterification) and reverse (acid-catalysed hydrolysis) reactions occur simultaneously. For simple acids + primary alcohols, Keq ≈ 1–4 — equilibrium lies roughly in the middle, giving only ~50–70% ester without modification.
Excess alcohol — increases [alcohol], system shifts right to oppose the increase
Remove water as it forms — conc. H₂SO₄ as dehydrating agent absorbs water (product), shifts right
Remove ester as it forms — distil off continuously; permanently shifts equilibrium right
Add excess water — increases [H₂O] (product), equilibrium shifts left → acid hydrolysis
Acid-catalysed: R–COOR' + H₂O ⇌ R–COOH + R'OH (reversible — equilibrium)
Saponification (base): R–COOR' + NaOH → R–COONa + R'OH (irreversible — carboxylate salt does not re-esterify under basic conditions → drives to completion)
Both the ester (e.g. ethyl ethanoate BP 77°C) and the alcohol (e.g. ethanol BP 78°C) are volatile — they would escape from an open flask at the reaction temperature. Reflux uses a condenser to cool rising vapours back to liquid, which drips back — all volatile components are retained, allowing equilibrium to be established and maximum yield to be achieved.
MICROTASK · +5 XP
Which modification would shift the esterification equilibrium to the RIGHT and produce more ester?
Making the ester in the flask is only half the job — isolating it from the acidic reaction mixture requires a sequence of physical separation steps that rely directly on the solubility differences between esters and aqueous solutions.
Fats (solid) and oils (liquid) are both triesters of glycerol (propane-1,2,3-triol — three –OH groups) with three long-chain fatty acids (C12–C22 carboxylic acids). The product is a triglyceride — three ester linkages formed by three esterification reactions.
Triglyceride structure. Three ester linkages (–O–CO–, pink) connect the glycerol backbone (green) to three fatty acid chains. Saturated chains (straight, orange) pack closely → fat (solid). Unsaturated chains with C=C kinks (red) pack poorly → oil (liquid).
MICROTASK · +5 XP
Fill in the blank: Sunflower oil is liquid at room temperature because its unsaturated fatty acid chains have C=C bonds that introduce rigid cis _______, preventing close packing and reducing dispersion forces compared to a saturated fat.
Given: Name (i) CH₃COOC₃H₇ (ii) C₃H₇COOCH₃ (iii) HCOOC₂H₅; identify carboxylic acid and alcohol for each.
(i) CH₃COO– (from ethanoic acid = ethanoate) + –C₃H₇ (propyl from propan-1-ol) → propyl ethanoate; ethanoic acid + propan-1-ol.
(ii) C₃H₇COO– (from butanoic acid = butanoate) + –CH₃ (methyl from methanol) → methyl butanoate; butanoic acid + methanol.
(iii) HCOO– (from methanoic acid = methanoate) + –C₂H₅ (ethyl from ethanol) → ethyl methanoate; methanoic acid + ethanol.
Equation: CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄, heat under reflux).
Increase yield (1): Excess ethanol → increases [reactant] → Le Chatelier shifts right → more ester.
Increase yield (2): Remove water with conc. H₂SO₄ (dehydrating agent) → reduces [product] → Le Chatelier shifts right → more ester.
Why reflux: Both ester (BP 77°C) and alcohol (BP 78°C) volatile — would escape open flask. Reflux condenser returns vapours as liquid → all components retained → equilibrium reached → maximum yield.
Formation: C₃H₅(OH)₃ + 3C₁₇H₃₅COOH ⇌ C₃H₅(OOCC₁₇H₃₅)₃ + 3H₂O. Esterification (condensation). By-product: water (3 mol).
Solid fat vs liquid oil: Saturated chains (tristearate): straight → close packing → strong dispersion → solid. Unsaturated (triolein): cis C=C kinks → poor packing → weaker dispersion → liquid.
Saponification to soap: C₃H₅(OOCC₁₇H₃₅)₃ + 3NaOH → C₃H₅(OH)₃ + 3C₁₇H₃₅COONa. Conc. NaOH(aq), reflux. Products: glycerol + sodium stearate (soap). Irreversible — carboxylate salt cannot re-esterify under basic conditions.
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Esterification: CH₃COOH + CH₃CH₂CH₂OH ⇌ CH₃COOCH₂CH₂CH₃ + H₂O. The ester is propyl ethanoate (C₅H₁₀O₂). Adding NaOH undergoes saponification (hydrolysis under basic conditions): CH₃COOCH₂CH₂CH₃ + NaOH → CH₃COONa + CH₃CH₂CH₂OH. The ester is cleaved back into the sodium salt of the acid (sodium ethanoate) and propan-1-ol. This is the reverse of esterification but is non-reversible because the carboxylate salt doesn't re-react under these conditions.
For each of the following, (i) name the ester and (ii) identify the acid and alcohol needed to make it by esterification.
| Structural formula | (i) IUPAC name | (ii) Acid + alcohol |
|---|---|---|
| CH₃COOC₅H₁₁ | ? | ? |
| C₄H₉COOCH₂CH₃ | ? | ? |
| HCOOCH₃ | ? | ? |
Rank and explain the boiling points of: methyl propanoate (ester, MW 88 g/mol, BP 79°C), butan-1-ol (alcohol, MW 74 g/mol, BP 118°C), and butanoic acid (carboxylic acid, MW 88 g/mol, BP 164°C). Use intermolecular force reasoning specifically.
Q1. A student esterifies propanoic acid with methanol using conc. H₂SO₄ at reflux. What is the IUPAC name of the ester produced, and what is the role of the conc. H₂SO₄?
Q2. At equilibrium in an esterification, the flask contains ethyl ethanoate, water, ethanoic acid, and ethanol. A student adds anhydrous Na₂SO₄ to the flask. What effect does this have on the equilibrium position?
Q3. A student compares ethyl ethanoate (ester, MW 88 g/mol, BP 77°C) and propan-1-ol (alcohol, MW 60 g/mol, BP 97°C). Which explanation correctly accounts for propan-1-ol having a higher boiling point despite lower molecular mass?
Q4. A saponification reaction is performed by heating glyceryl trioleate (an unsaturated oil) with excess concentrated NaOH solution under reflux. Which statement correctly describes this reaction?
Q5. Which of the following correctly explains why butter (a fat) is solid at room temperature while sunflower oil is liquid, given that both are triglycerides?
Question 6 (4 marks) — A student has a sample of pentyl ethanoate (pear aroma, BP 148°C). They are told it was made from ethanoic acid and pentan-1-ol. (a) Write the IUPAC name of this ester and draw its structural formula showing the ester linkage –COO– explicitly. (b) Write the balanced esterification equation, including all conditions.
Question 7 (5 marks) — Compare the boiling points of the following three compounds: ethyl ethanoate (MW 88 g/mol, BP 77°C); propan-1-ol (MW 60 g/mol, BP 97°C); propanoic acid (MW 74 g/mol, BP 141°C). Explain the trend using intermolecular force reasoning.
Question 8 (6 marks) — A food scientist wants to produce methyl butanoate (strawberry/apple fragrance) by reacting butanoic acid with methanol. She runs the reaction at reflux with equal moles of both reactants and a few drops of conc. H₂SO₄, obtaining a 55% yield. (a) Write the balanced esterification equation with correct conditions. (b) Suggest and explain TWO modifications that would increase the yield, applying Le Chatelier's Principle in each case. (c) The scientist proposes to isolate the product using a separating funnel. Outline the separation steps, naming the reagent used to remove residual butanoic acid from the ester layer and writing the equation for that removal.
Q1 — C. Propanoic acid + methanol → methyl propanoate (methyl from methanol, propanoate from propanoic acid). Option B reverses the naming. Conc. H₂SO₄ is a catalyst — not consumed; also acts as a dehydrating agent.
Q2 — C. Anhydrous Na₂SO₄ is a drying agent — it absorbs water. Removing water (a product) reduces [H₂O] → Le Chatelier shifts equilibrium right → more ester produced. Option A is wrong because removing a product does affect equilibrium even though Na₂SO₄ itself is not in the equilibrium expression.
Q3 — C. Propan-1-ol (3C) has fewer carbons than ethyl ethanoate (4C), so dispersion forces are actually weaker for propan-1-ol (A is wrong). The key is H-bonding: propan-1-ol has O–H → forms H-bonds (donor and acceptor). Ethyl ethanoate has no O–H → only dipole-dipole forces. H-bonds are significantly stronger → propan-1-ol has higher BP despite lower mass.
Q4 — D. Saponification is irreversible — the carboxylate salt cannot re-esterify under basic conditions. Products are glycerol + sodium oleate (soap). No H₂SO₄ needed. NaOH is a reagent, not a catalyst — the actual product is the sodium salt of the acid, not the free acid (C wrong).
Q5 — A. Both are triglycerides. Saturated chains (butter) are straight → close packing → strong dispersion → solid. Unsaturated chains (sunflower oil) have cis C=C kinks → poor packing → weaker dispersion → liquid. B is wrong (same number of ester linkages — both triglycerides). C is completely false. D: chain length similar for both; key is shape, not mass.
Q6 (4 marks): (a) Pentyl ethanoate. Structural formula: CH₃–C(=O)–O–CH₂CH₂CH₂CH₂CH₃ [ester linkage –C(=O)–O– must be drawn explicitly] (2 marks: 1 for name, 1 for correct structure). (b) CH₃COOH + CH₃(CH₂)₄OH ⇌ CH₃COO(CH₂)₄CH₃ + H₂O; conc. H₂SO₄ (catalyst); heat under reflux. (2 marks: 1 for ⇌ and balanced, 1 for conditions)
Q7 (5 marks): Order: ethyl ethanoate (77°C) < propan-1-ol (97°C) < propanoic acid (141°C) (1 mark). Ethyl ethanoate lowest: no O–H bond → cannot donate H-bonds → only dipole-dipole forces, weaker than H-bonds (1 mark). Propan-1-ol intermediate: has O–H → forms H-bonds (donor and acceptor) → stronger than dipole-dipole → higher BP despite lower mass (2 marks). Propanoic acid highest: forms H-bonded dimers — two simultaneous H-bonds per pair, effective unit is dimer — requires the most energy to vaporise (1 mark).
Q8 (6 marks): (a) CH₃CH₂CH₂COOH + CH₃OH ⇌ CH₃CH₂CH₂COOCH₃ + H₂O; conc. H₂SO₄; heat under reflux (2 marks). (b) Mod 1: Excess methanol — increases [methanol] (reactant) → Le Chatelier shifts right to oppose increase → more methyl butanoate formed (1 mark). Mod 2: Remove water via conc. H₂SO₄ as dehydrating agent — reduces [H₂O] (product) → Le Chatelier shifts right to replace removed product → more ester forms (1 mark). (c) Step 1: separating funnel + cold water; ester layer upper, aqueous layer lower (contains butanoic acid, methanol, H₂SO₄); drain aqueous layer. Step 2: Na₂CO₃ wash; removes residual butanoic acid: 2CH₃CH₂CH₂COOH + Na₂CO₃ → 2CH₃CH₂CH₂COONa + H₂O + CO₂; also neutralises H₂SO₄; drain aqueous layer. Step 3: water wash; drain. Step 4: anhydrous Na₂SO₄ (or MgSO₄); filter. (Distil for purity.) (2 marks)
The structural connection between pineapple flavouring (butyl ethanoate) and avocado fat (a triglyceride) is the ester linkage — both contain one or more –COO– functional groups formed by the same type of reaction: esterification of a carboxylic acid with an alcohol, with water as the by-product. The pineapple ester has one ester linkage and is a small, volatile molecule (low BP → fragrant). The triglyceride in avocado has three ester linkages — three fatty acids bonded to glycerol — and is a much larger molecule with longer non-polar chains that make it non-volatile and waxy. Same functional group, same reaction, vastly different physical properties due to chain length and degree of saturation.
Now review your Think First prediction below:
1. Why do esters have lower boiling points than alcohols of similar molecular mass?
Esters have no O–H bond and therefore cannot donate hydrogen bonds. Ester-to-ester interactions are limited to dipole-dipole forces and dispersion forces, which are weaker than the hydrogen bonds formed between alcohol molecules. Alcohols have one O–H that both donates and accepts H-bonds — more energy required to vaporise.
2. Name the ester formed from ethanoic acid and propan-1-ol, and write the esterification equation.
Propyl ethanoate. CH₃COOH + CH₃CH₂CH₂OH ⇌ CH₃COOCH₂CH₂CH₃ + H₂O; conc. H₂SO₄ catalyst; heat under reflux. Reversible arrow (⇌) is essential.
3. Explain the role of conc. H₂SO₄ in esterification. Is it a reactant or a catalyst?
Conc. H₂SO₄ is the catalyst — it is not consumed in the overall reaction and does not appear in the products. It also acts as a dehydrating agent, absorbing water (a product) as it forms, which shifts the equilibrium toward more ester production (Le Chatelier's Principle). Write it above or below the equilibrium arrow, not as a reactant.
4. What are triglycerides, and how do fats differ from oils in terms of structure?
Triglycerides are triesters of glycerol (propane-1,2,3-triol) with three fatty acids (long-chain carboxylic acids), connected by three ester linkages. Fats (solid): saturated fatty acid chains — straight, pack closely → strong dispersion forces → high melting point. Oils (liquid): unsaturated chains with C=C kinks — cannot pack closely → weaker dispersion forces → lower melting point.
5. Write the equation for saponification of a triglyceride and explain why this reaction is irreversible.
R–COOR' + NaOH → R–COONa + R'OH (single arrow — irreversible). Saponification is irreversible because the carboxylate salt (R–COONa) formed is a resonance-stabilised anion that is not reactive enough to undergo re-esterification under basic conditions. The reaction proceeds to completion, producing soap (sodium carboxylate) and glycerol.
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