OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2021: Question 20

✅ Correct Answer & Mark Breakdown

• Stage 1: Curly arrow starting from the O–H bond (or its lone pair) going to the oxygen atom; curly arrow starting from the O–H bond going to the H atom. Co-product is H₂O (1 mark per curly arrow).
• Stage 2: First curly arrow starts from the delocalised ᵵ-ring and goes to the electrophilic carbon atom in CO₂ . Second curly arrow starts from the C=O bond and goes to the oxygen atom.
• Intermediate: Must show a horseshoe-shaped partial delocalisation with a positive charge inside, retaining the O⁻ group and attaching the new COO⁻ group.
• Stage 2 Final Arrow: Curly arrow starting from the C–H bond (where COO⁻ entered) going back into the ring to restore aromaticity, generating H⁺ .

💡 Key Knowledge

• Phenol is more acidic than aliphatic alcohols due to the partial delocalisation of the p-type lone pair from oxygen into the benzene ring, weakening the O–H bond.
• Sodium hydroxide ( NaOH ) removes a proton from phenol, turning it into a stable phenoxide ion ( C₆H₅O⁻ ).
• Carbon dioxide ( CO₂ ) acts as a weak electrophile attacked by the nucleophilic phenoxide ring at carbon-2 (ortho position).

🧠 Exam Technique

• Curly arrows: Must start precisely at a bond or a lone pair and end precisely where electrons go. Arrowheads must point at the destination atom/space.
• Horseshoe intermediate: The positive charge must be enclosed inside an incomplete ring that covers less than half of the ring, opening towards the incoming COO⁻ group.

❌ Common Errors

• Drawing the intermediate horseshoe so it covers more than 50% of the ring circumference.
• Omitting the O⁻ group from the intermediate structure.
• Starting the electrophilic attack arrow from outside the ring instead of directly from the ᵵ-electron cloud.

✅ Correct Answers

• -OH (from NaOH): Base (Accept: alkali).
• CO₂: Electrophile (Accept: electron pair acceptor).

❌ Common Errors

• Calling OH⁻ a "nucleophile" in this specific acid-base context (it acts strictly to deprotonate the phenol).

✅ Correct Answer & Equation

• Two molecules of salicylic acid react, condensing with loss of water to form a cyclic/dimeric structure containing two ester linkages.
• Balanced Equation: 2 C₇H₆O₃ ➔ C₁₄H₈O₄ + 2 H₂O (or showing structural formulas).

💡 Key Knowledge

Salicylic acid contains both a phenol/alcohol-derived group and a carboxylic acid group, allowing intermolecular condensation reactions (esterification) between separate molecules when catalyzed by an acid.

✅ Marking Points

• Dissolve impure crystals in a minimum volume of hot solvent/water.
• Filter hot (if necessary for insoluble impurities - though note: initial filtering/drying agents are ignored).
• Allow to cool so crystals form, then filter cold/vacuum filter and dry.

❌ Common Errors

• Using too much solvent (prevents crystallisation upon cooling, destroying yield).
• Forgetting to state that the solvent must be hot during dissolution.

📐 Step-by-Step Calculation: Empirical Formula

1. Divide percentage by relative atomic mass (Ar):
   C: 31.44 / 12.0 = 2.62
   H: 1.31 / 1.0 = 1.31
   N: 18.34 / 14.0 = 1.31
   O: 48.91 / 16.0 = 3.06
2. Find simplest whole number ratio (divide by smallest value, 1.31):
   C: 2.00 | H: 1.00 | N: 1.00 | O: 2.33
3. Scale up by factor of 3 to clear fractions:
   Empirical Formula = C₆H₃N₃O₇

💡 Molecular Formula & NMR Deduction

• Molar Mass check: Empirical formula mass of C₆H₃N₃O₇ = (6×12) + (3×1) + (3×14) + (7×16) = 229.0 g mol⁻¹.
• The mass spectrum gives a molecular ion peak at m/z = 229.0 , meaning the molecular formula is identical to the empirical formula ( C₆H₃N₃O₇ ).
• Functional groups & directing effects: Phenol reacts with nitric/sulfuric acid at 100°C to undergo exhaustive nitration (forming picric acid / 2,4,6-trinitrophenol).
• The –OH group is 2,4-directing, while incoming nitro groups direct 3-relative. This accounts for the 2,4,6-trisubstituted symmetry, explaining the low number of carbon environments observed in the ¹³C NMR spectrum.