WJEC A-Level Chemistry Unit 4, June 2025: Question 9

✅ Correct Answers

(i) Reagent: Alkaline potassium manganate(VII) / MnO₄⁻ / OH⁻ [1 mark]

(ii) Type of reaction: Oxidation [1 mark]

(iii) Moisture explanation: Phosphorus(V) chloride (PCl₅) reacts with water OR benzoyl chloride reacts with water (hydrolyses back to benzoic acid) [1 mark]

💡 Key Knowledge

• Alkyl side chains attached directly to a benzene ring are vigorously oxidised to −COOH by alkaline KMnO₄ regardless of chain length.
• Acyl chlorides ( R−COCl ) and phosphorus halides ( PCl₅ ) hydrolyse readily in the presence of water/moisture, producing steamy fumes of HCl and reducing overall yield:
  C₆H₅COCl + H₂O → C₆H₅COOH + HCl

🧠 Exam Technique

In (i), WJEC expects you to specify the medium: state alkaline potassium manganate(VII). Simply writing “KMnO₄” without stating alkaline conditions or hydroxide ions ( OH⁻ ) often risks losing the mark on WJEC papers.

❌ Common Errors

• Acidified dichromate: Writing acidified potassium dichromate(VI) ( K₂Cr₂O₇ / H⁺ ) — this is not a strong enough oxidising agent to oxidise alkyl side chains on benzene rings!
• Vague moisture answers: Stating “water dilutes the mixture” rather than identifying a chemical hydrolysis reaction involving PCl₅ or the acyl chloride.

✅ Correct Answers

(iv) Structure of product:

N-methylbenzamide: A benzene ring bonded to a carbonyl group, which is bonded to a secondary amine nitrogen carrying a hydrogen atom and a methyl group:

C₆H₅−CO−NH−CH₃

Displaying: Carbonyl C=O bonded to −NH−CH₃ [1 mark]

(v) Missing compounds:

• Reactant: NH₃ (ammonia)

• Product: H₂O (water)

(Both required for 1 mark) [1 mark]

💡 Key Knowledge

• Nucleophilic addition-elimination: Primary amines ( CH₃NH₂ ) react like ammonia ( NH₃ ) with acyl chlorides, substituting −Cl with −NHCH₃ to yield secondary amides.
• Acid-Base Neutralisation & Thermal Dehydration:
  1. C₆H₅COOH + NH₃ → C₆H₅COO⁻ NH₄⁺ (salt formation)
  2. C₆H₅COO⁻ NH₄⁺ → C₆H₅CONH₂ + H₂O (dehydration upon heating)

✅ Full Mark Answer (3 Marks Breakdown)

• Absence of a characteristic peak at 1661 cm⁻¹ indicates the absence of a carbonyl ( C=O ) group, confirming the loss of the −C=O carbon from benzamide. [1 mark]
• Compound Z is phenylamine (aniline), formula C₆H₅NH₂ . [1 mark]
• The molecular ion peak at m/z = 93 corresponds to the molecular ion of phenylamine: [C₆H₅NH₂]⁺ • Mᵣ = (6 × 12) + (7 × 1) + 14 = 93 . [1 mark]

🧠 Exam Technique: Structuring the Explanation

To secure all 3 marks on this question type, explicitly link every piece of given data to your chemical deduction:

1. Spectroscopy deduction: Peak at 1661 cm⁻¹ = amide carbonyl C=O . Absence = loss of carbonyl group.
2. Name/structure: State the chemical formula or draw phenylamine.
3. Mass spec verification: Calculate Mᵣ(C₆H₅NH₂) = 93 and state it matches m/z = 93 .

❌ Common Calculation Traps

• Missing the 1:2 reacting ratio: Forgetting that 1 mole of diprotic H₂SO₄ neutralises 2 moles of NH₃ is the single most common reason candidates lose marks.
• Writing the whole molecule instead of R: The question asks for the formula of the R group, not the full amide. Writing CH₃CH₂CONH₂ on the final line may lose the final mark unless clearly identified.
• Rounding too early: Rounding intermediate values can skew the calculated Mᵣ away from 73. Always keep full calculator precision.

🧠 Error Carried Forward (ECF)

Full credit is possible for the final formula if it correctly matches an incorrect Mᵣ resulting from an arithmetic error earlier in your working, provided your method is clear and fully displayed.