WJEC A-Level Chemistry Unit 4, June 2025: Question 6

✅ Correct Answers

(i) Initiation Step [1 mark]:

Cl₂ → 2Cl•

(ii) Termination forming 1,2-dichloroethane [1 mark]:

•CH₂Cl + •CH₂Cl → ClH₂C—CH₂Cl

(Accept: 2 •CH₂Cl → CH₂ClCH₂Cl or C₂H₄Cl₂ )

🧠 Exam Technique: Radicals

• Always display the unpaired electron radical dot clearly (e.g., Cl• or •CH₂Cl ). Place the dot near the atom carrying the odd electron.
• In (a)(ii), notice that chloromethane undergoes further substitution. Chloromethyl radicals ( •CH₂Cl ) form during propagation. When two collide, they undergo a termination step to form a 2-carbon chain.

❌ Common Errors

• Forgetting the radical dot entirely, writing Cl₂ → 2Cl (0 marks).
• Suggesting homolysis of methane in initiation: CH₄ → •CH₃ + •H (UV light initiates Cl—Cl cleavage, not C—H).
• Trying to form 1,2-dichloroethane via a propagation step with Cl₂ rather than a radical-radical termination.

Step 1: Convert all given quantities to base SI units

• Pressure, p = 9.86 × 10⁴ Pa (already in Pa)
• Volume, V = 42.9 cm³ = 42.9 × 10⁻⁶ m³
• Temperature, T = 363 K (already in K)
• Gas constant, R = 8.31 J mol⁻¹ K⁻¹

Step 2: Calculate moles (n) of vapour [1 mark for formula/sub, 1 mark for answer]

n = pV / RT

n = (9.86 × 10⁴ Pa × 42.9 × 10⁻⁶ m³) / (8.31 × 363 K)

n = 4.22994 / 3016.53 = 1.402 × 10⁻³ mol (or 1.40 × 10⁻³ mol)

Step 3: Calculate Relative Molecular Mass (M_r) [1 mark]

M_r = mass / moles = 0.168 g / (1.402 × 10⁻³ mol) = 119.8 (or 120)

(Allow error carried forward from step 2 if calculation is clearly shown).

✅ Correct Answers for (b)(ii)

(ii) I. Why a water bath is unsuitable [1 mark]:

• Water boils at 100 °C / temperature of boiling water cannot exceed 100 °C.
• Therefore, it cannot reach 112 °C and will not vaporise the liquid sample completely.

(ii) II. Suggested modification [1 mark]:

Use an oil bath, an electrical heating mantle, or an electrical heater.

❌ Common Errors in (b)

• Volume conversion: Multiplying by 10⁻³ instead of 10⁻⁶ when converting cm³ to m³.
• Vague temperature answers: Writing "the water bath is not hot enough" scores 0. You must state that water boils at 100 °C / cannot reach 112 °C.
• Unsafe alternatives: Suggesting a "Bunsen burner directly heating the flask" — volatile organic vapours are flammable!

✅ Correct Answers

(c)(i) Reaction with steam [1 mark]:

CCl₄ + H₂O → COCl₂ + 2HCl

(c)(ii) Formula of Alkyl Group R [2 marks]:

Formula: C₃H₇ (or —CH₂CH₂CH₃ / —CH(CH₃)₂ )

📐 Working for (c)(ii)

Reaction Equation:
C₆H₅CHCl₂ + CaCO₃ → C₆H₅CHO + CaCl₂ + CO₂

Step 1: Calculate M_r of reactants & desired product [1 mark]

• Desired product (benzaldehyde, C₆H₅CHO ): (7 × 12.01) + (6 × 1.008) + 16.00 = 106.06 (accept 106)
• Reactant 1 ((dichloromethyl)benzene, C₆H₅CHCl₂ ): (7 × 12.01) + (6 × 1.008) + (2 × 35.45) = 161.06 (accept 161)
• Reactant 2 ( CaCO₃ ): 100 (given in question)
• Total mass of all reactants = 161.06 + 100 = 261.06 (accept 261)

Step 2: Calculate % Atom Economy [1 mark]

% Atom Economy = (M_r of desired product / Total M_r of reactants) × 100

% Atom Economy = (106.06 / 261.06) × 100 = 40.6%

(Accept 40.6% or 40.61% based on rounding).

✅ (d)(ii) Methods to Study Rate [2 Marks]

Any two of the following methods:

• Measure the volume of CO₂ gas evolved at timed intervals (using a gas syringe).
• Measure the loss in mass of the reaction vessel at timed intervals (as CO₂ escapes).
• Take samples at timed intervals and measure the intensity of the C=O infrared absorption peak (at 1650–1750 cm⁻¹).
• Use turbidimetry / light sensor at timed intervals as the insoluble CaCO₃ suspension reacts away and clears.

🧠 Crucial Examiner Rule: "At Timed Intervals"

To obtain both marks in kinetics questions, at least one method must explicitly mention time (e.g., "at regular time intervals", "record time taken", "continuous logging over time").

Simply stating "measure the volume of gas" will lose marks because rate is defined as change per unit time!

Given Equation:
C₆H₅CCl₃ + 2H₂O → C₆H₅COOH + 3HCl

Step 1: Calculate total moles of H⁺ (acid) produced [1 mark]

1 mole of C₆H₅CCl₃ produces:
• 1 mole of benzoic acid ( C₆H₅COOH )
• 3 moles of hydrochloric acid ( HCl )

Total moles of monoprotic acid per mole reactant = 1 + 3 = 4 moles of H⁺.

Total moles of H⁺ = 4 × 0.050 mol = 0.200 mol

Step 2: Calculate volume of 2.50 mol dm⁻³ NaOH needed [1 mark]

Neutralisation ratio is 1:1 ( H⁺ + OH⁻ → H₂O ), so moles of NaOH needed = 0.200 mol .

Volume (dm³) = moles / concentration = 0.200 / 2.50 = 0.080 dm³

Volume (cm³) = 0.080 × 1000 = 80.0 cm³

❌ The Classic 4x Stoichiometry Trap

The most common blunder in this question is forgetting that both products are acids:

• Only counting HCl: 3 × 0.050 = 0.150 mol → gives 60.0 cm³ (incorrect!).
• Only counting benzoic acid: 1 × 0.050 = 0.050 mol → gives 20.0 cm³ (incorrect!).

The question specifies: "needed to completely neutralise the two acids produced". Always sum all acidic species!