Edexcel A-Level Chemistry Paper 2, June 2017: Question 7

✅ Correct Answer / Pathway

• Step 1 (Nitration): Concentrated HNO₃ and concentrated H₂SO₄, with heat/reflux (50-60 °C). Intermediate 1: Nitrobenzene ( C₆H₅NO₂ ).
• Step 2 (Reduction): Tin (Sn) and concentrated hydrochloric acid (HCl), heated under reflux. Intermediate 2: Phenylamine ( C₆H₅NH₂ ).
• Step 3 (Acylation): Ethanoyl chloride ( CH₃COCl ) reacted with phenylamine to form Antifebrin.

💡 Key Knowledge

• Direct conversion to phenylamine requires Sn/HCl reduction rather than catalytic hydrogenation in standard Edexcel synthetic routes.
• Acylation uses ethanoyl chloride rather than ethanoic acid to ensure a fast, high-yielding reaction without requiring severe conditions.

🧠 Exam Technique

Intermediate structures must be clearly drawn showing the benzene ring and functional groups. Always state both reagents and conditions (like heat or reflux) to secure individual mark points.

❌ Common Errors

• Using dilute acids instead of concentrated acids for nitration.
• Attempting to use AlCl₃ as a catalyst in Step 3 (acylation of phenylamine does not require a Lewis acid catalyst).
• Omitting "heat" or "reflux" conditions which loses marks.

✅ Correct Answer

B (6)

Antifebrin has a symmetrical aromatic ring structure, meaning carbons 2 & 6 and carbons 3 & 5 are chemically equivalent. Combined with the carbonyl carbon, methyl carbon, and ring carbons 1 and 4, this yields exactly 6 unique carbon environments.

❌ Common Errors & Distractor Analysis

• A (5): Incorrectly assumes only three unique peaks for the aromatic ring.
• C (7): Miscounts symmetric carbon environments.
• D (8): Counts total carbon atoms in the molecule instead of non-equivalent carbon environments.

✅ Correct Answer

• The oxygen atom of the hydroxy (-OH) group has a lone pair of electrons that interacts/delocalises with the pi-system of the benzene ring.
• This significantly increases the electron density of the benzene ring.
• As a result, incoming electrophiles (such as Br₂ ) are more strongly attracted, and the electrophile is more easily polarised, making substitution happen much more readily than in benzene.

💡 Key Knowledge

Groups with lone pairs directly attached to a benzene ring (like -OH or -NH₂) are activating, electron-releasing groups that direct incoming substituents to the 2-, 4-, and 6-positions.

❌ Common Errors

Mentioning "charge density" instead of "electron density", or incorrectly describing the mechanism as "electrophilic addition" instead of "electrophilic substitution".

📐 Step-by-Step Calculation

1. Find the molar mass of paracetamol ( C₈H₉NO₂ ):
   (8 × 12.0) + (9 × 1.0) + (1 × 14.0) + (2 × 16.0) = 151.0 g mol⁻¹
2. Calculate the mass of paracetamol in the tablet:
   Mass = moles × molar mass
   Mass = 3.10 × 10⁻³ mol × 151 g mol⁻¹ = 0.4681 g (or 468.1 mg)
3. Calculate the percentage by mass:
   Percentage = (0.4681 g / 0.500 g) × 100 = 93.62%
4. Apply appropriate significant figures:
   Since input data (3.10 × 10⁻³) is given to 3 significant figures, round the final answer to 93.6%. (Also accept 94% if 2 SF is justified).

❌ Calculation Traps

• Unit mismatch: Forgetting to convert mg to g or comparing moles directly to milligrams without converting units.
• Rounding intermediate numbers too early, which leads to final rounding discrepancies.