WJEC A-Level Chemistry AS Unit 2, June 2025: Question 9

💡 Spectral Evidence Analysis

• IR Spectrum: Broad absorption peak at ~3350 cm⁻¹ shows the presence of an O–H alcohol group. (No C=O absorption at ~1700 cm⁻¹).
• Mass Spec Fragmentation:
  • Peak at m/z = 71 : Loss of •OH ( 88 − 17 = 71 ) → [C₅H₁₁]⁺
  • Peak at m/z = 15 : Methyl cation [CH₃]⁺
  • Peak at m/z = 17 : [OH]⁺
• ¹H NMR Spectrum: 3 distinct peaks → 3 proton environments.
  • Peak at δ 4.5 ppm (area 1): proton on O–H
  • Peak at δ 3.3 ppm (area 2): protons on –CH₂–O–
  • Peak at δ 0.9 ppm (area 9): 9 equivalent protons of three identical methyl groups –C(CH₃)₃
• ¹³C NMR Spectrum: 3 distinct peaks → 3 carbon environments:
  • δ 73 ppm : C–O carbon
  • δ 33 ppm : central quaternary carbon ( C–C )
  • δ 26 ppm : three equivalent –CH₃ carbons

✅ Final Identification & Structure

Compound A is 2,2-dimethylpropan-1-ol (commonly called neopentyl alcohol).

All 3 methyl groups are attached to a central quaternary carbon. They are chemically identical, giving a single 9H singlet in ¹H NMR and a single signal in ¹³C NMR.

🧠 Exam Technique (10-Mark Question)

• Systematic Approach: Structure your response under 5 clear headings: Elemental Analysis, Mass Spectrum, IR Spectrum, ¹H NMR, and ¹³C NMR.
• The mark scheme awards up to 9 marks for extracting data points (11 possible points) and 1 distinct mark for correctly identifying the final compound.
• Always state what the number of peaks means (e.g. "3 peaks = 3 chemical environments") before interpreting specific shifts.

❌ Common Errors to Avoid

• Forgetting % Hydrogen: The question gives %C and %O only. You must subtract from 100% to find %H.
• Misidentifying the 9H peak: Suggesting a straight-chain pentanol (like pentan-1-ol). Pentan-1-ol would give multiple complex splitting patterns, not 3 clean sharp singlets.
• Omitting fragment charges: In mass spectrometry, fragments must carry a positive charge (e.g. C₅H₁₁⁺ , not just C₅H₁₁).

✅ Correct Responses

(b)(i) Oxidising Agent:
Potassium dichromate(VI) / K₂Cr₂O₇ (or Potassium manganate(VII) / KMnO₄ )

(b)(ii) Balanced Oxidation Equation:
CH₃CH₂CH₂OH + 2 [O] → CH₃CH₂COOH + H₂O

(b)(iii) Chloride Ion Oxidation Half-Equation:
2Cl⁻ → Cl₂ + 2e⁻  (or 2Cl⁻ − 2e⁻ → Cl₂ )

💡 Key Concepts

• Primary Alcohol Oxidation: Propan-1-ol oxidises in two stages:
  Propan-1-ol → Propanal (+ 1 [O]) → Propanoic acid (+ 2 [O] total).
  To obtain the carboxylic acid, reflux with excess oxidising agent is required.
• Why Not HCl? Hydrochloric acid cannot acidify the reaction because Cr₂O₇²⁻ is a strong enough oxidising agent to oxidise Cl⁻ ions to toxic chlorine gas ( Cl₂ ). Hence, H₂SO₄ must be used.

🧠 Exam Technique

• In (b)(ii), note the question specifies propanoic acid, NOT propanal. Therefore, the stoichiometric coefficient in front of [O] must be 2.
• State oxidation numbers if writing names: write "potassium dichromate(VI)", not just "potassium dichromate".

❌ Common Errors

• Writing 1 [O]: Giving CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O scores 0 because the question explicitly states propanoic acid is formed.
• Balancing charges in half equations: Writing Cl⁻ → Cl + e⁻ loses both marks. Diatomic chlorine is Cl₂ , needing 2Cl⁻ and 2e⁻ .

✅ Correct Answers

(iv) I. Reaction Products:

Must show clearly: ethyl propanoate structure AND H₂O.

(iv) II. Name of Sweet-Smelling Compound:
ethyl propanoate

(iv) III. Purpose of Adding Acid:
Any one of:

• Acts as a catalyst / increases the rate of reaction
• Acts as a dehydrating agent / removes water to shift the equilibrium position to the right

💡 Ester Chemistry Essentials

• Naming Rule: Alkyl group attached to oxygen comes from the alcohol (→ ethyl). Carboxylate chain containing C=O comes from the carboxylic acid (→ propanoate).
• Reversibility: Esterification is an equilibrium reaction:
  RCOOH + R'OH ⇌ RCOOR' + H₂O
• Concentrated H₂SO₄ provides H⁺ ions to protonate the carbonyl oxygen, speeding up the reaction (catalytic role).

🧠 Exam Technique

• "Clearly show the structure": WJEC mark schemes penalise ambiguous or grouped formulas when displayed structure is asked. Draw out the ester linkage –C(=O)–O– fully showing all bonds.
• Don't forget the second product: condensation reactions always eliminate a small molecule ( H₂O ). 1 mark is specifically allocated for H₂O .

❌ Common Errors

• Backwards naming: Calling it "propyl ethanoate" instead of ethyl propanoate. Check your oxygen bridge carefully!
• Missing H₂O: Only drawing the organic product and forgetting to include water as the inorganic co-product.