WJEC A-Level Chemistry Unit 4, June 2025: Question 5

14 marks · Hard difficulty · Structured Questions

Determine the identity and structure of compounds derived from 4-hydroxycinnamic acid using titration data, percentage composition, polymerisation, and explain trends in boiling temperatures of aromatic compounds.

Practise this question

Question

Question 5 consists of several parts: (a) Titration calculation using 4.10 g of carboxylic acid E neutralised by 25.00 cm³ of 0.500 mol dm⁻³ Na2CO3 solution (2:1 ratio) to show Mr is 164. (b) Compound E is 3-(4-hydroxyphenyl)prop-2-enoic acid. Sub-parts ask for: (i) structure formed when E reacts with excess NaOH, (ii) observation with iron(III) chloride solution, (iii) deduction of structure of compound F formed by reaction with bromine using percentage carbon data (65.9% in E, 26.8% in F), and (iv) structures of addition and condensation polymer repeating units formed by E. (c) Boiling points table of ethylbenzene (136 °C), propylbenzene (159 °C), butylbenzene (183 °C). Requires explanation of the trend, prediction of the boiling point of hexylbenzene with reasoning, and comparison of boiling points between 4-ethylphenol and ethoxybenzene.
Question text

5. (a) Compound E is a carboxylic acid containing only one carboxylic acid group. Its relative

molecular mass was found by a titration with aqueous sodium carbonate solution.

4.10 g of compound E was just neutralised by 25.00 cm3 of 0.500 mol dm–3 sodium

carbonate. 2mol of the acid reacts with 1mol of sodium carbonate.

Use the information to show that the relative molecular mass of compound E is 164. [2]

(b) Compound E has the formula below.

H H

O

HO C C C

OH

(i) Draw the structure of the compound produced when compound E reacts with an

excess of aqueous sodium hydroxide. [1]

(ii) State what is seen when a solution of compound E is treated with aqueous iron(III)

chloride solution. [1]

E

(iii) Compound E reacts with bromine to give compound F.

Compound F does not contain an alkene functional group. Analysis of compounds

E and F for the percentage of carbon by mass in each compound gave the

following results.

06 © WJEC CBAC Ltd. (1410U40-1)

Compound Percentage of carbon

E 65.9

F 26.8

Use the information to suggest a structure for compound F. Show your reasoning.

8 [4]

E

(iv) The structure of compound E suggests that it can be polymerised.

I. Give the structure of the repeating section obtained when compound E

produces an addition polymer. [1]

II. Give the structure of the repeating section obtained when compound E

produces a condensation polymer.9 [1]

E

(c) The boiling temperatures of some alkylbenzenes are given in the table.

Compound Boiling temperature/°C

ethylbenzene 136

propylbenzene 159

butylbenzene 183

(i) Explain why the boiling temperatures of these compounds increase as shown. [1]

07 © WJEC CBAC Ltd. (1410U40-1)

(ii) Predict the boiling temperature of hexylbenzene. Give your reasoning. [2]

(iii) Explain why the boiling temperature of 4-ethylphenol is higher than that of

ethoxybenzene, even though they are both isomers of formula C8H10O. [1]

HO CH2CH3 CH3CH2O

08 © WJEC CBAC Ltd. (1410U40-1)

4-ethylphenol ethoxybenzene

Mark scheme

Show the mark scheme Mark scheme for Question 5 provides: (a) Moles of Na2CO3 = 0.0125, moles of E = 0.0250, Mr = 4.10 / 0.0250 = 164 (2 marks). (b)(i) Di-sodium salt structure showing phenolate and carboxylate ions with Na+ (1 mark). (ii) Purple coloration/solution (1 mark). (iii) Reasoning that both have 9 carbons (mass 108), Mr of F = 108 / 0.268 = 403, mass difference 403 - 164 = 239 corresponds to 3 Br atoms (addition across C=C plus electrophilic substitution on the aromatic ring), leading to 3-(3-bromo-4-hydroxyphenyl)-2,3-dibromopropanoic acid structure (4 marks). (iv) Addition polymer repeating unit showing saturated carbon backbone and intact functional groups (1 mark); Condensation polymer showing repeating polyester unit formed from phenolic -OH and carboxylic acid group (1 mark). (c)(i) Increased chain length increases London/van der Waals forces (1 mark). (ii) Predicted value between 222-234 °C based on ~23-25 °C increase per CH2 group (2 marks). (iii) 4-ethylphenol exhibits hydrogen bonding whereas ethoxybenzene does not (1 mark).

Marks available

Question Marking details

AO1 AO2 AO3 Total Maths Prac

5 (a) 0.500 × 25.00

moles of sodium carbonate = 1000 = 0.0125 (1)

moles of compound E = 2 × 0.0125 = 0.0250

mass 4.10 2 2 1

Mr = moles = 0.0250 = 164 (1)

must show method

no ecf possible

(b) (i)

(ii) purple coloration / solution 1 1 1

Marks available

AO1 AO2 AO3 Total Maths Prac

(iii) compounds E and F both have 9 carbon atoms representing 108

out of 164 (1) 1

108 represents 26.8% of the Mr of compound F

Mr = 108 × 26.8 = 403 (1) 1

403 – 164 = 239 suggests there are 3 bromine atoms (3 × 80)

(1)

(1)

credit possible for other structure if it follows from incorrect

calculations

(b) (iv) I

II

Marks available

AO1 AO2 AO3 Total Maths Prac

(c) (i) as the (alkyl) chain length increases, the van der Waals forces

between the molecules increase / there are more intermolecular

forces (therefore more energy needed to separate the molecules)

(ii) award (1) for any value in the range 222-234 C

the difference between consecutive alkyl groups increases by about

23 / 24 / 25 each time (as the homologous series is ascended) and

hexylbenzene contains 6 alkyl carbon atoms / ‘has two increases’ (1)

(iii) 4-ethylphenol can take part in hydrogen bonding but

ethoxybenzene does not show hydrogen bonding (therefore more

energy needed to separate molecules of 1 1

4-ethylphenol)

Question 5 total 1 7 6 14 2 2

How to answer it

Phenolic Carboxylic Acids, Dual Polymerisation & Arene Physical Trends

📌 What this question tests

This synoptic question covers key aspects of organic analysis, bifunctional reactivity, and physical properties across several core A-Level Chemistry modules:

  • Acid-Base Titrations: Reacting ratios involving diprotic carbonate ions (2:1 acid-to-carbonate stoichiometry) to deduce relative molecular mass ( Mr ).
  • Reactions of Phenols & Acids: Neutralisation forming diphenolate/dicarboxylate salts with excess alkali, and the qualitative diagnostic test for phenols with iron(III) chloride.
  • Structural Elucidation & Multi-Mechanistic Bromination: Combining alkene addition with electrophilic aromatic substitution on an activated benzene ring.
  • Addition vs Condensation Polymerisation: Recognising that a bifunctional molecule containing both a C=C alkene group and mutually reactive functional groups ( -OH and -COOH ) can undergo two completely different modes of polymerisation.
  • Intermolecular Forces: Explaining trends in boiling points of alkylbenzenes (London dispersion forces) and structural isomers (hydrogen bonding vs dipole interactions).

Part (a) — Titration Stoichiometry & Molecular Mass Determination

Calculation to show that relative molecular mass (Mr) is 164

📐 Step-by-Step Calculation

Step 1: Calculate moles of sodium carbonate reacted
Moles of Na₂CO₃ = concentration × volume (dm³)
Moles = 0.500 mol dm⁻³ × (25.00 / 1000 dm³) = 0.0125 mol
Step 2: Use reacting stoichiometry to find moles of Compound E
The question states: 2 mol of the acid reacts with 1 mol of sodium carbonate (2 : 1 ratio).
Moles of acid E = 2 × 0.0125 mol = 0.0250 mol
Step 3: Calculate the relative molecular mass (Mr)
Mr = mass / moles = 4.10 g / 0.0250 mol = 164
Award: [1 mark] for calculating 0.0125 mol of Na₂CO₃ and doubling to 0.0250 mol; [1 mark] for 4.10 / 0.0250 = 164. Full working must be shown.

🧠 Exam Technique

Always state the formula or method clearly before inserting numbers. Since the target answer (164) is given in the question, examiners will award zero marks if steps are omitted or fudge factors are used. You must explicitly show the multiplication by 2.

❌ Common Trap

Forgetting the 2:1 reacting ratio and calculating 4.10 / 0.0125 = 328. Be sure to check the acid:carbonate stoichiometry given in the prompt.

Part (b)(i) & (b)(ii) — Acidic Character and Functional Group Tests

Reactions with excess NaOH and aqueous FeCl₃

✅ (b)(i) Structure with Excess NaOH

Both the carboxylic acid group (–COOH) and the phenolic –OH group are acidic enough to react with strong base (NaOH).

Structure to draw:

Na⁺ O⁻—(C₆H₄)—CH=CH—COO⁻ Na⁺

Both –OH groups must be deprotonated to form their respective ionic sodium phenoxide and sodium carboxylate groups.

[1 mark] Awarded for correct sodium salt at both ends.

✅ (b)(ii) Observation with Aqueous FeCl₃

Phenols react with neutral iron(III) chloride solution to give an intensely coloured complex.

Observation: Purple colouration / purple solution

[1 mark] Awarded for specifying "purple" (or violet) coloration/solution.

💡 Acidity Hierarchy

Carboxylic acids (pKa ~4–5) react with both weak bases (Na₂CO₃) and strong bases (NaOH). Phenols (pKa ~10) are sufficiently acidic to react with strong bases (NaOH) forming phenoxide salts, but are not strong enough to react with carbonates. Alcohols (pKa ~16–18) react with neither.

❌ Common Student Errors

  • Only neutralising the –COOH group and leaving the phenolic –OH unchanged. The question specifies excess NaOH!
  • Drawing covalent O–Na bonds instead of ionic charges (O⁻ Na⁺).
  • Confusing the FeCl₃ test for phenols (purple) with precipitation of Fe(OH)₃ (brown/orange).

Part (b)(iii) — Deductive Structural Elucidation of Compound F

Reaction with Bromine: Calculating mass changes and deducing the structure

📐 Deducing the Number of Bromine Atoms

Step 1: Determine mass of carbon in the molecule
Compound E (C₉H₈O₃) contains 9 carbon atoms.
Mass contributed by carbon = 9 × 12.0 = 108 g mol⁻¹.
Step 2: Calculate Mr of Compound F
Compound F retains all 9 carbon atoms and has 26.8% carbon by mass:
Mr(F) = 108 × (100 / 26.8) = 403
Step 3: Calculate the total mass gained
Increase in molecular mass = 403 − 164 = 239
Step 4: Account for the chemical reactions with bromine
• Alkene double bond: Electrophilic addition of 1 mol Br₂ (+160).
• Phenol ring: Activating –OH directs electrophilic aromatic substitution to the ortho position: replacement of 1 × H with 1 × Br (gain of 80 − 1 = +79).
Total mass increase = 160 + 79 = 239 (indicates addition of 3 bromine atoms net).
[1 mark] 9 carbons = 108
[1 mark] Mr of F = 403
[1 mark] Deducting 239 corresponds to 3 bromine atoms (2 by addition, 1 by substitution)
[1 mark] Correct structure of Compound F

✅ Structure of Compound F

Draw 4-hydroxycinnamic acid modified as follows:

  • Ring: Bromine substituted on the ring ortho to the –OH group (at position 3 on the ring).
  • Side-chain: Single bond between carbons, with a bromine on each: –CH(Br)–CH(Br)–COOH .

🧠 Examiner Commentary

Top candidates recognised that bromine reacts with both the aliphatic alkene unit (electrophilic addition, consuming the C=C bond) and the highly activated phenol ring (electrophilic substitution). Candidates who only considered addition across the double bond missed the 3rd bromine atom.

Part (b)(iv) — Dual Polymerisation of Compound E

Addition vs Condensation Repeating Units

✅ I. Addition Polymer Repeating Unit

Polymerisation occurs across the C=C double bond, keeping all other groups intact as substituents.

Structure to draw:

–[ CH(C₆H₄OH) — CH(COOH) ]ₙ–

Draw two chain carbons linked by a single bond, open extension bonds passing through square brackets, with –C₆H₄OH attached to one carbon and –COOH attached to the other.

[1 mark] Addition repeating unit with open bonds.

✅ II. Condensation Polymer Repeating Unit

Condensation occurs between the phenolic –OH of one monomer and the –COOH of the next, eliminating H₂O to form a polyester link.

Structure to draw:

–[ O—(C₆H₄)—CH=CH—C(=O) ]ₙ–

The C=C double bond remains unchanged. Open single bonds must extend from the phenolic oxygen and the carbonyl carbon through the brackets.

[1 mark] Condensation repeating unit with ester linkages.

❌ Common Polymer Pitfalls

  • Retaining double bonds in addition polymers: The C=C must become a C–C single bond in the main backbone.
  • Breaking double bonds in condensation polymers: The alkene backbone is untouched during condensation esterification.
  • Missing extension bonds: Repeating units must clearly show open bonds extending through or outside brackets to denote continuity.

Part (c) — Physical Trends & Intermolecular Forces in Arene Derivatives

Boiling points, trend prediction, and structural isomer comparison

✅ (c)(i) Alkylbenzene Boiling Point Trend

As the alkyl chain length increases:

  • The molecules have more electrons and a larger surface area of contact.
  • Van der Waals / London dispersion forces increase in strength between molecules.
  • More thermal energy is required to overcome these stronger intermolecular forces.
[1 mark] van der Waals forces increase / more energy required.

✅ (c)(ii) Prediction for Hexylbenzene

Observe the incremental differences in boiling point:

  • Ethylbenzene (136°C) to Propylbenzene (159°C): +23°C
  • Propylbenzene (159°C) to Butylbenzene (183°C): +24°C

Each additional –CH₂– adds approximately 23–25°C. Hexylbenzene has two more carbons than butylbenzene (pentyl then hexyl):

Predicted b.p. = 183 + (2 × 24) = 231°C

[1 mark] Reasoning: two successive increases of ~24°C.
[1 mark] Predicted value in range 222°C – 234°C.

✅ (c)(iii) 4-Ethylphenol vs Ethoxybenzene Boiling Point

Both have identical molecular formula (C₈H₁₀O) and similar numbers of electrons, but very different intermolecular forces:

  • 4-ethylphenol has an –OH group directly attached to the aromatic ring, enabling molecules to form strong hydrogen bonds between each other.
  • Ethoxybenzene is an ether (contains no H atom bonded directly to O, N, or F); it exhibits only weaker dipole-dipole and van der Waals attractions and cannot form intermolecular hydrogen bonds.
  • Much more thermal energy is needed to break the stronger hydrogen bonds in 4-ethylphenol, giving it a higher boiling point.
[1 mark] Mentions hydrogen bonding in 4-ethylphenol and absence of hydrogen bonding in ethoxybenzene.

Topics

Organic Chemistry · Physical Chemistry · 1.3 Chemical calculations · 1.4 Bonding · 2.5 Hydrocarbons · 4.2 Aromaticity · 4.3 Alcohols and phenols · 4.5 Carboxylic acids and their derivatives · 4.8 Organic synthesis and analysis

Question and mark scheme from the WJEC A-Level Chemistry examination, Unit 4, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.