The smell of pineapple, banana, pear, and strawberry in sweets and soft drinks is almost entirely synthetic — each aroma is a single ester molecule, made by the same reaction type, under the same conditions, just with different acid and alcohol starting materials.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
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 is sp²-hybridised (planar geometry). 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.
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 (hexane, diethyl ether).
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 you studied in Module 5, now applied to a functional group transformation.
R–COOH + R'OH ⇌ R–COOR' + H₂O
Both the forward reaction (esterification) and the reverse reaction (acid-catalysed hydrolysis) occur simultaneously. At equilibrium, all four species are present. For simple acids + primary alcohols, Keq ≈ 1–4 — equilibrium lies roughly in the middle, giving only ~50–70% ester without modification.
Use excess alcohol — increases [alcohol], system shifts right to oppose the increase. Yield may rise from ~65% to ~75–80%.
Remove water as it forms — adds a drying agent, molecular sieves, or relies on conc. H₂SO₄ as dehydrating agent. Removing a product shifts equilibrium right (Le Chatelier).
Remove ester as it forms — if ester BP is accessible, distil it off continuously. Permanently shifts equilibrium right. (Practical only if BP differences allow.)
Add excess water — increases [H₂O] (product), equilibrium shifts left. This is acid-catalysed hydrolysis — the reverse of esterification.
Acid-catalysed: R–COOR' + H₂O ⇌ R–COOH + R'OH (reversible — equilibrium, excess H₂O shifts left)
Base-catalysed (saponification): 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, reducing reactant concentrations and preventing equilibrium from being reached. 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.
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.
After reflux, the mixture contains: ester, unreacted carboxylic acid, unreacted alcohol, conc. H₂SO₄, and water. Procedure:
The ester (non-polar, low water solubility) forms an upper layer. Water, unreacted acid, alcohol, and H₂SO₄ dissolve in the lower aqueous layer. Drain and discard the lower aqueous layer.
Na₂CO₃ reacts with residual carboxylic acid (2RCOOH + Na₂CO₃ → 2RCOONa + H₂O + CO₂) and neutralises residual H₂SO₄. The carboxylate salt is water-soluble → stays in aqueous layer. Drain and discard. Caution: CO₂ is produced — release pressure regularly by opening the stopcock.
Removes residual Na₂CO₃ and sodium carboxylate from the ester layer. Separate and discard aqueous layer.
Add anhydrous Na₂SO₄ or anhydrous MgSO₄ to absorb residual water. The agent clumps as it hydrates. Filter to remove.
If high purity required, distil the dried ester. Collect the fraction at the known boiling point of the target ester.
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 at room temperature).
Fats vs Oils — IMF Argument:
This connects to Lesson 6: vegetable oil (unsaturated, liquid) + H₂ (Ni catalyst, ~200°C) → partially hydrogenated fat (more saturated, solid or semi-solid = margarine).
"Esterification goes to completion." It is a reversible equilibrium (⇌). Even with excess alcohol or a drying agent, 100% yield is never achieved under standard conditions. All four species remain present at equilibrium.
"Esters can form hydrogen bonds with each other." Esters have no O–H bond and therefore cannot donate H-bonds. They can only accept H-bonds via the lone pairs on their oxygens (e.g. when dissolved in water). Ester-to-ester IMF is dipole-dipole only.
"Conc. H₂SO₄ is a reactant in esterification." Conc. H₂SO₄ is the catalyst — it is not consumed and does not appear in the products. It also acts as a dehydrating agent (removes water), which helps shift equilibrium toward the ester, but it is still classified as a catalyst.
(a) Name the following esters: (i) CH₃COOC₃H₇ (ii) C₃H₇COOCH₃ (iii) HCOOC₂H₅. (b) For each, identify the carboxylic acid and alcohol used in esterification.
Cut at the single-bond O. Left (including C=O) = alkanoate. Right O + chain = alkyl. Write: alkyl first, alkanoate second.
Left: CH₃COO– → from ethanoic acid → ethanoate. Right: –C₃H₇ → propyl (from propan-1-ol). Name: propyl ethanoate.
Left: C₃H₇COO– → from butanoic acid (CH₃CH₂CH₂COOH) → butanoate. Right: –CH₃ → methyl (from methanol). Name: methyl butanoate.
Left: HCOO– → from methanoic acid (HCOOH) → methanoate. Right: –C₂H₅ → ethyl (from ethanol). Name: ethyl methanoate.
(a)(i) Propyl ethanoate. (a)(ii) Methyl butanoate. (a)(iii) Ethyl methanoate.
(b)(i) Ethanoic acid + propan-1-ol. (b)(ii) Butanoic acid + methanol. (b)(iii) Methanoic acid + ethanol.
(a) Write the balanced equation for the esterification of ethanoic acid with ethanol to produce ethyl ethanoate. Include all conditions. (b) The student obtains a 65% yield with equal moles of acid and alcohol. Suggest TWO modifications to increase yield. Explain each using Le Chatelier's Principle. (c) Why is reflux used rather than heating in an open flask?
CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O
Conditions: conc. H₂SO₄ (catalyst, written above/below arrow), heat under reflux. Reversible arrow (⇌) — compulsory.
Modification 1 — Use excess ethanol: Increasing [ethanol] (a reactant) shifts equilibrium to the right by Le Chatelier's Principle, opposing the increase and producing more ester. Yield can rise from ~65% toward ~75–80%.
Modification 2 — Remove water as it forms: Conc. H₂SO₄ acts as a dehydrating agent, absorbing water (a product) as it is produced. Removing a product reduces [H₂O], shifting the equilibrium right to replace the removed product — more ester is produced.
Both ethyl ethanoate (BP 77°C) and ethanol (BP 78°C) are volatile — they would evaporate and escape from an open flask at the reaction temperature, reducing reactant concentrations and preventing equilibrium from being reached. Reflux uses a condenser to return rising vapours as liquid dripping back into the flask — all volatile components are retained, allowing equilibrium to be established and maximum yield to be achieved.
(a) CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O (conc. H₂SO₄, heat under reflux).
(b) (1) Excess ethanol — shifts equilibrium right (Le Chatelier: increase reactant concentration). (2) Remove water via conc. H₂SO₄ as dehydrating agent — shifts equilibrium right (Le Chatelier: decrease product concentration).
(c) Volatile reactants and product would escape from an open flask. Reflux condenser returns vapours to the flask, retaining all components for equilibrium to be established.
Glyceryl tristearate is a fat formed from glycerol and stearic acid (octadecanoic acid, C₁₇H₃₅COOH — saturated C18). (a) Write the formation equation, state the reaction type and the small molecule produced. (b) Explain why glyceryl tristearate is solid while triolein (corresponding unsaturated fat from oleic acid — one C=C per chain) is liquid. (c) Explain how glyceryl tristearate could be hydrolysed to produce soap — state reagent, conditions, and products.
Glycerol (3 –OH groups) + 3 stearic acid molecules → triester + 3 water:
C₃H₅(OH)₃ + 3C₁₇H₃₅COOH ⇌ C₃H₅(OOCC₁₇H₃₅)₃ + 3H₂O
Reaction type: esterification (condensation reaction). Small molecule produced: water (H₂O), 3 moles per mole of triglyceride.
Glyceryl tristearate (saturated, solid): Stearic acid chains (C₁₇H₃₅–) are fully saturated — no C=C bonds. Saturated chains are straight and can align parallel, packing closely together. Close packing → large surface area of contact → strong cumulative dispersion forces → high melting point → solid at room temperature.
Triolein (unsaturated, liquid): Each oleic acid chain contains one C=C bond (cis-configuration), introducing a rigid kink. Kinked chains cannot pack closely → gaps between adjacent chains → reduced contact surface area → weaker dispersion forces → lower melting point → liquid at room temperature.
Base-catalysed hydrolysis (saponification) with NaOH. Irreversible because carboxylate salt does not re-esterify under basic conditions → drives to completion.
C₃H₅(OOCC₁₇H₃₅)₃ + 3NaOH → C₃H₅(OH)₃ + 3C₁₇H₃₅COONa
Reagent: Concentrated NaOH(aq). Conditions: Heat under reflux. Products: (1) Glycerol (propane-1,2,3-triol) — the alcohol component recovered; (2) Sodium stearate (C₁₇H₃₅COONa) — the soap (sodium salt of fatty acid, with hydrophobic C₁₇ tail and hydrophilic –COONa head).
(a) C₃H₅(OH)₃ + 3C₁₇H₃₅COOH ⇌ C₃H₅(OOCC₁₇H₃₅)₃ + 3H₂O. Esterification (condensation). By-product: water.
(b) Saturated chains (tristearate): straight → close packing → strong dispersion forces → solid. Unsaturated chains (triolein): cis C=C kinks → poor packing → weaker dispersion → liquid.
(c) C₃H₅(OOCC₁₇H₃₅)₃ + 3NaOH → C₃H₅(OH)₃ + 3C₁₇H₃₅COONa. Conc. NaOH(aq), heat under reflux. Products: glycerol + sodium stearate (soap). Irreversible — carboxylate salt cannot re-esterify under basic conditions.
Structure: Functional group –COO– (–C(=O)–O–). No O–H bond → cannot donate H-bonds → dipole-dipole forces only → lower BP than alcohols and carboxylic acids of same chain length.
Naming: Alkyl alkanoate. Cut at single-bond O. Left (with C=O) = alkanoate. Right O + chain = alkyl. Write alkyl first.
Aromas: Butyl ethanoate = pineapple; pentyl ethanoate = banana; pentyl propanoate = pear.
Esterification: R–COOH + R'OH ⇌ R–COOR' + H₂O. Conc. H₂SO₄ catalyst; heat under reflux. Arrow is ⇌ (reversible). Yield <100%.
Increasing yield: (1) Excess alcohol shifts right; (2) Remove water (conc. H₂SO₄ as dehydrating agent) shifts right.
Hydrolysis: Acid-catalysed: R–COOR' + H₂O ⇌ R–COOH + R'OH (reversible). Base-catalysed (saponification): R–COOR' + NaOH → R–COONa + R'OH (irreversible → complete).
Fats and oils: Both are triglycerides (triesters of glycerol + 3 fatty acids). Fats: saturated chains (straight) → close packing → strong dispersion → solid. Oils: unsaturated chains (C=C kinks) → poor packing → weaker dispersion → liquid.
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 | ? | ? |
A student attempts the esterification of propanoic acid with butan-1-ol under reflux with equal moles of each reactant and a few drops of conc. H₂SO₄. They obtain a 58% yield of butyl propanoate after 45 minutes.
(a) Write the balanced equation for this reaction, including the correct arrow type and conditions. (b) Suggest three specific modifications the student could make to improve yield. For each, identify the Le Chatelier principle being applied. (c) Explain why the yield cannot reach 100% even with all three modifications applied simultaneously.
1. 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?
2. 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?
3. 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?
4. 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?
5. Which of the following correctly explains why butter (a fat) is solid at room temperature while sunflower oil (an 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 smells like pear and that 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, all with similar molecular masses. Explain the boiling point trend using intermolecular force reasoning. 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).
Question 8 (6 marks) — A food scientist wants to produce methyl butanoate (an ester with a 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. She obtains 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 also 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 (propyl comes from propan-1-ol, not propanoic acid). 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 (base-catalysed ester hydrolysis with NaOH) is irreversible — the carboxylate salt (soap) cannot re-esterify under basic conditions because carboxylate anions are not reactive enough as acylating agents. Products are glycerol + sodium oleate (soap). No H₂SO₄ needed (A, B wrong). NaOH is a reagent, not a catalyst — the actual product is the sodium salt of the acid, not the free acid (C is wrong).
Q5 — A. Both are triglycerides. The difference is chain saturation. 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 is similar for both; the key is not mass but shape.
Q6 (4 marks): (a) Pentyl ethanoate. Structural formula: CH₃–C(=O)–O–CH₂CH₂CH₂CH₂CH₃ [the ester linkage –C(=O)–O– must be drawn explicitly, showing C=O and single-bond O to the pentyl chain] (2 marks: 1 for name, 1 for correct structure with ester linkage visible). (b) CH₃COOH + CH₃(CH₂)₄OH ⇌ CH₃COO(CH₂)₄CH₃ + H₂O; conc. H₂SO₄ (catalyst); heat under reflux. (2 marks: 1 for ⇌ and balanced equation, 1 for conditions)
Q7 (5 marks) — sample: Order (lowest to highest BP): ethyl ethanoate (77°C) < propan-1-ol (97°C) < propanoic acid (141°C) (1 mark). Ethyl ethanoate has the lowest BP despite the highest molecular mass: it has no O–H bond and cannot donate H-bonds. Ester-ester interactions are limited to dipole-dipole forces, which are weaker than H-bonds (1 mark). Propan-1-ol is intermediate: it has one O–H bond and forms H-bonds (donor and acceptor) with neighbouring molecules. H-bonds are stronger than dipole-dipole forces, so more energy is required to vaporise propan-1-ol than ethyl ethanoate, despite propan-1-ol having a lower molecular mass (2 marks). Propanoic acid has the highest BP: it forms hydrogen-bonded dimers — two molecules simultaneously form two H-bonds (O–H···O=C from each molecule to the other), creating a cyclic pair. Breaking both H-bonds simultaneously requires more energy than breaking ethanol's single H-bond, giving propanoic acid the highest BP of the three (1 mark).
Q8 (6 marks) — sample: (a) CH₃CH₂CH₂COOH + CH₃OH ⇌ CH₃CH₂CH₂COOCH₃ + H₂O; conc. H₂SO₄ (catalyst); heat under reflux (2 marks). (b) Modification 1: Use excess methanol. By increasing [methanol] (reactant), Le Chatelier's Principle predicts the equilibrium shifts to the right to oppose the increase, producing more methyl butanoate at equilibrium (1 mark). Modification 2: Remove water as it forms using conc. H₂SO₄ as a dehydrating agent (or molecular sieves). Removing water (product) reduces [H₂O], Le Chatelier shifts equilibrium right to replace the removed product — more ester forms (1 mark). (c) Step 1 — pour into separating funnel with cold water; ester layer (upper) separates from aqueous layer (lower, containing butanoic acid, methanol, H₂SO₄); drain and discard aqueous layer. Step 2 — add dilute Na₂CO₃ solution; reacts with residual butanoic acid: 2CH₃CH₂CH₂COOH + Na₂CO₃ → 2CH₃CH₂CH₂COONa + H₂O + CO₂; also neutralises H₂SO₄; drain and discard aqueous layer. Step 3 — wash with fresh water; drain. Step 4 — add anhydrous Na₂SO₄ (or MgSO₄); filter. (Distil for highest 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 at room temperature. Same functional group, same reaction, vastly different physical properties due to chain length and degree of saturation.