WJEC A-Level Chemistry Unit 4, June 2025: Question 7

15 marks · Medium difficulty · Structured Questions

Devise a method to obtain pure benzoic acid crystals from contaminated sodium benzoate, determine the identity of an aromatic amine via gas calculations and NMR spectroscopy, name an azo dye component, and identify isomers based on characteristic chemical tests.

Practise this question

Question

Question 7 consists of multiple parts across five pages. Part (a) is a 6-mark QER question asking to devise a method to separate and recrystallise pure benzoic acid from a mixture of sodium benzoate and sand using solubility and reaction data. Part (b)(i) provides an equation of a primary aromatic amine V reacting with nitrous acid to release nitrogen gas, requiring a calculation of its molar mass using gas volume data. Part (b)(ii) shows a high-resolution proton NMR spectrum of compound V with singlets and doublets to deduce the structure of alkyl group R. Part (b)(iii) shows the coupling reaction between diazotised V and compound W to form an azo dye, asking for the name of W. Part (c) displays a table of four isomeric compounds R, S, T, and U containing various functional groups and asks to identify which fits descriptions involving reactions with NaHCO3, 2,4-DNP, Tollens' reagent, and NaOH.
Question text

7. (a) You are provided with a 5g sample of sodium benzoate contaminated with a small

quantity of sand.

Use the information below to devise a method that others could follow, to obtain crystals

of pure benzoic acid from this sample.

• The solubility of sodium benzoate in water is 66g/100g of water at 20°C

• Sand is insoluble in water and in hydrochloric acid

• Sodium benzoate reacts with hydrochloric acid to give benzoic acid and sodium

chloride

• The solubility of benzoic acid in water is 0.3g/100g of water at 20°C rising to

5.6g/100g of water at 100°C

In your answer you should explain the reasons for each step and suggest the volume of

water used at each step. [6 QER]

E

(b) (i) A primary aromatic amine, compound V, reacts with nitric(III) acid to give the

corresponding phenol and nitrogen gas.

R NH2 + HNO2 R OH + N2 + H2O

14 compound V © WJEC CBAC Ltd. (1410U40-1)

5.00 g of compound V gave 823.5 cm3 of nitrogen gas measured at 298 K and

1atm pressure.

Use this information to show that the molar mass of compound V is 149 g mol–1.

16 [2]

E

(ii) The high resolution 1H NMR spectrum of compound V is shown below.

The signal due to the —NH2 protons is at δ 3.5ppm.

22 2

87 6 5 4 3 2 1 0

δ/ppm

Use the molar mass from part (i) and this spectrum to deduce the structure of the

alkyl group R. Give your reasoning. [3]

15 © WJEC CBAC Ltd. (1410U40-1)

R has the structure 17

E

(iii) At 5°C compound V reacts with nitric(III) acid to give a diazonium compound

which then reacts with compound W to give the azo dye below.

CH3

16 © WJEC CBAC Ltd. (1410U40-1)

R N N OH

CH3

State the name of compound W. [1]

E

(c) Each of the four compounds whose formulae are shown below are isomers.

Compound Formula

O

R C

OCH3

O

S H3C C

OH

O

T CH3O C

H

O

U HO C

CH3

Select the compound that fits the descriptions below.

Give your reasoning in each case. [3]

Produces carbon dioxide with sodium hydrogencarbonate.

Gives an orange-red solid with 2,4-dinitrophenylhydrazine but does not react with Tollens’

17 reagent.

Produces methanol when heated with sodium hydroxide.

Mark scheme

Show the mark scheme Mark scheme for Question 7. Part (a) outlines indicative content for dissolving, filtration of sand, precipitation with HCl, filtration, recrystallisation from hot water, and drying, evaluated on a 3-tier marking band up to 6 marks. Part (b)(i) awards 2 marks for calculating moles of N2 (0.0336 mol) and determining molar mass as 149 g/mol. Part (b)(ii) awards 3 marks for deducing R contains 9 protons, has formula C4H9, and identifying it as a tert-butyl group. Part (b)(iii) awards 1 mark for 2,6-dimethylphenol. Part (c) awards 3 marks for identifying S (carboxylic acid), U (ketone not aldehyde), and R (methyl ester hydrolysed to methanol).

Marks available

Question Marking details

AO1 AO2 AO3 Total Maths Prac

7 (a) Indicative content

• add 5g of sodium benzoate to 10-100 cm3 of cold water

• stir to dissolve the sodium benzoate

o as sodium benzoate is very soluble in cold water

• filter the solution

o to remove insoluble sand

• add hydrochloric acid to the filtrate

• until no more benzoic acid is precipitated

o as benzoic acid is largely insoluble in cold water

22 2 6 6

• filter off the precipitated benzoic acid

• wash the benzoic acid with (cold) water

o to remove soluble impurities

• add the benzoic acid to 100-200 cm3 of cold water and heat to

boiling

o so that solid dissolves

• allow to cool to room temperature

o so that benzoic acid recrystallises

• filter off the benzoic acid and dry

5-6 marks

Essential stages of the preparation included with sensible volumes of water; clear reasoning for steps

There is a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. The information

included in the response is relevant to the argument.

3-4 marks

Most stages of the preparation included; good attempt to explain some steps

There is a line of reasoning which is partially coherent, largely relevant, supported by some evidence and with some structure.

Mainly relevant information is included in the response but there may be some minor errors or the inclusion of some

information not relevant to the argument.

1-2 marks

Basic description of one or two steps

There is a basic line of reasoning which is not coherent, supported by limited evidence and with very little structure. There may

be significant errors or the inclusion of information not relevant to the argument.

0 marks

No attempt made or no response worthy of credit.

9 Marks available

AO1 AO2 AO3 Total Maths Prac

(b) (i) 823.5

moles of nitrogen = 1000 × 24.5 = 0.0336 (1)

1:1 mole ratio therefore 0.0336 mol of V

5.00 2 2 2

molar mass of V = 0.0336 = 149 (1)

must show method

no ecf possible

(ii) alkyl group R contains 9 protons (1)

award (1) for either of following

• R has 4 carbon atoms (CnH2n+1) / molecular formula C4H9

• R has mass of 57

signal is a singlet therefore all alkyl protons are equivalent

so R is (1)

(iii) 2,6-dimethylphenol 1 1

(c) compound S it is a carboxylic acid (1) 1

compound U it has a carbonyl group but is not an aldehyde /

(cannot be oxidised so) it is a ketone (1) 1 3 3

compound R it is a methyl ester / it is an ester which is 1

hydrolysed to give an alcohol (1)

Question 7 total 5 5 5 15 2 9

How to answer it

Organic Preparation, Analysis & Functional Group Identification

WHAT THIS QUESTION TESTS

This synoptic question evaluates core practical and theoretical organic chemistry across several key units:

  • Practical Purification & Solubility (QER): Designing a multi-step laboratory method to separate a salt from an insoluble impurity, precipitating an organic acid via acidification, and recrystallising the product using precise water volumes.
  • Gas Molar Volume Stoichiometry: Using experimental gas volume at room temperature and pressure (24.5 dm³ mol⁻¹ at 298 K / 1 atm) to determine molar mass.
  • High-Resolution ¹H NMR Interpretation: Deduce the structure of an alkyl group using chemical shift, integration values, and splitting patterns (singlet vs multiplet).
  • Aromatic & Azo Chemistry: Identifying the coupling partner used in electrophilic substitution to synthesise an azo dye.
  • Qualitative Organic Analysis: Distinguishing structural isomers using sodium hydrogencarbonate (NaHCO₃), 2,4-DNP, Tollens' reagent, and alkaline ester hydrolysis.

Part (a) — Purification of Benzoic Acid from a Contaminated Salt

Devising a multi-step purification procedure using solubility data [6 Marks - QER]

✅ Model Method (Full 6-Mark Flow)

  1. Dissolve: Add the 5 g sample to 10–100 cm³ of cold water. Stir thoroughly until the sodium benzoate dissolves entirely (solubility is high: 66 g / 100 g cold water).
  2. Filter out sand: Filter the mixture (gravity filtration). Sand is insoluble and remains on the filter paper; collect the filtrate (aqueous sodium benzoate).
  3. Precipitate acid: Add excess dilute hydrochloric acid (HCl) to the filtrate with stirring until no more white precipitate forms. Benzoic acid forms and precipitates because it is sparingly soluble in cold water (0.3 g / 100 g).
  4. Collect crude product: Filter off the precipitated benzoic acid under reduced pressure (Buchner funnel) and wash with a small amount of cold water to remove soluble NaCl and excess acid.
  5. Recrystallise: Dissolve the crude benzoic acid in the minimum volume of boiling water (approx. 100–200 cm³, as solubility is 5.6 g / 100 g at 100 °C).
  6. Cool & Dry: Allow the hot solution to cool slowly to room temperature, then chill in ice to maximise yield. Filter off the pure crystals and dry in a desiccator or warm oven.

📐 Justification of Water Volumes

1. Cold dissolution of sodium benzoate:
At 20 °C, solubility = 66 g / 100 g water.
For ~5 g: minimum water needed = (5 / 66) × 100 ≈ 7.6 cm³.
Suggested volume: 10 to 100 cm³ (ensures complete dissolution while keeping concentration high).
2. Hot recrystallisation of benzoic acid:
Theoretical maximum benzoic acid from 5 g sodium benzoate is ~4.2 g.
At 100 °C, solubility = 5.6 g / 100 g water.
For ~4.2 g: minimum boiling water = (4.2 / 5.6) × 100 ≈ 75 cm³.
Suggested volume: 100 to 200 cm³ to fully dissolve the acid at boiling point.

🧠 Exam Technique: Structuring QER Responses

  • Logical Stage Headings: Group your points into Separation of Sand, Precipitation, and Recrystallisation.
  • State the "Why": High marks require reasoning alongside each action: e.g., "wash with cold water to remove NaCl without redissolving the acid".
  • Include Realistic Volumes: Examiners penalised answers quoting vague terms ("add some water") or impossible volumes (e.g. 1000 cm³, which would prevent recrystallisation).

❌ Common Errors to Avoid

  • Adding HCl directly to the solid mixture first: this precipitates benzoic acid mixed with sand, failing to separate the sand!
  • Forgetting to state minimum volume of hot water during recrystallisation, leading to poor crystal yield on cooling.
  • Washing crystals with warm or hot water instead of ice-cold water.
Mark scheme bands: 5–6 marks require all essential stages, sensible calculated water volumes, and clear reasons embedded throughout a logically structured account.

Part (b)(i) — Molar Mass Calculation from Gas Volume

Determining Mᵣ from reaction of primary aromatic amine with HNO₂ [2 Marks]

📐 Step-by-Step Calculation

Step 1: Calculate moles of N₂ gas produced
Molar volume at 298 K and 1 atm = 24.5 dm³ mol⁻¹ = 24 500 cm³ mol⁻¹
Moles of N₂ = Volume / Molar Volume
Moles of N₂ = 823.5 / (1000 × 24.5) = 823.5 / 24 500 = 0.03361 mol
Step 2: Use stoichiometry to find moles of Compound V
Equation: R-C₆H₄-NH₂ + HNO₂ → R-C₆H₄-OH + N₂ + H₂O
Molar ratio Compound V : N₂ = 1 : 1
Moles of Compound V = 0.03361 mol
Step 3: Calculate molar mass of Compound V
Molar mass (Mᵣ) = mass / moles = 5.00 g / 0.03361 mol = 148.8 g mol⁻¹ ≈ 149 g mol⁻¹

❌ Calculation Traps & Examiner Tips

  • Wrong molar volume: Watch out! At 298 K / 1 atm, molar gas volume is 24.5 dm³ mol⁻¹ (not 24.0 or 22.4). Check the WJEC data sheet.
  • Unit conversion: Convert 823.5 cm³ to dm³ by dividing by 1000 (0.8235 dm³).
  • "Show that" questions: You must write down the exact intermediate steps and numerical values; you cannot just write 149 without supporting working.
Mark allocation: 1 mark for correct moles of N₂ (0.0336 mol); 1 mark for calculating molar mass = 149 g mol⁻¹. No error carried forward (ECF).

Part (b)(ii) — High-Resolution ¹H NMR Analysis

Deducing the identity and structure of alkyl group R [3 Marks]

✅ Structure & Evidence

Identity of group R: tert-butyl group, -C(CH₃)₃

Structural formula of R:

     CH₃
     |
— C — CH₃
     |
     CH₃

💡 NMR Spectral Deduction

  • Integration = 9: The tall singlet at δ 1.3 ppm has an integration of 9, meaning R contains 9 equivalent protons.
  • Mass balance:
    Mᵣ of V = 149.
    Mass of benzene ring (-C₆H₄-) = 76.
    Mass of -NH₂ group = 16.
    Mass of R = 149 - (76 + 16) = 57.
    Formula: C₄H₉ (since 4×12 + 9×1 = 57).
  • Splitting: The peak at δ 1.3 ppm is a singlet, meaning these 9 protons are on carbon atoms with no adjacent protons (bonded to a quaternary carbon).
Mark allocation: 1 mark for stating R contains 9 protons; 1 mark for deducing R has 4 carbons / formula C₄H₉ / mass = 57; 1 mark for singlet reasoning leading to the tert-butyl structure.

Part (b)(iii) — Azo Coupling Component

Naming the aromatic coupling partner [1 Mark]

✅ Correct IUPAC Name

2,6-dimethylphenol

💡 Nomenclature Rules

  • The parent functional group takes position 1: the -OH group on the benzene ring makes it a phenol (C-1).
  • The two methyl groups (-CH₃) are located symmetrically on carbons 2 and 6.
  • Therefore, the systematic name is 2,6-dimethylphenol.
Mark allocation: 1 mark for correct full name.

Part (c) — Functional Group Tests on Isomers

Matching isomers R, S, T, and U to chemical descriptions [3 Marks]

✅ Identification & Explanations

1. Produces CO₂ with NaHCO₃:
Compound S
Reasoning: It contains a carboxylic acid group (-COOH). Carboxylic acids are acidic enough to react with hydrogencarbonates to liberate carbon dioxide gas. (Phenols are not acidic enough).
2. Orange-red solid with 2,4-DNP but no reaction with Tollens' reagent:
Compound U
Reasoning: Forms a precipitate with 2,4-DNP, so it contains a carbonyl group (C=O). It does not react with Tollens' reagent, meaning it is not an aldehyde; therefore, it is a ketone. (Compound T is an aldehyde and would form a silver mirror).
3. Produces methanol when heated with sodium hydroxide:
Compound R
Reasoning: It is a methyl ester (-COOCH₃). Alkaline hydrolysis of this ester cleaves the ester bond to release the sodium carboxylate and methanol (CH₃OH).

❌ Common Errors in Identification

  • Confusing T and U: Both give an orange precipitate with 2,4-DNP. But compound T has a formyl group (-CHO, aldehyde) which does reduce Tollens' reagent. Only ketone U fits the negative Tollens' result.
  • Phenol vs Carboxylic Acid with NaHCO₃: Phenols do not effervesce with carbonates or hydrogencarbonates. Only the carboxylic acid group in S will react.
Mark allocation: 1 mark for each correct compound matched with valid functional group reasoning (3 marks total).

Topics

Organic Chemistry · Physical Chemistry · Practical · 1.3 Chemical calculations · 4.4 Aldehydes and ketones · 4.5 Carboxylic acids and their derivatives · 4.6 Amines · 4.8 Organic synthesis and analysis · A2 Unit 4 practical work

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.