OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2023: Question 20

✅ Correct Flowchart Products

• Top-Left Box (Br₂ reaction): Salicylic acid with a bromine atom ( Br ) substituted onto the benzene ring (typically ortho/para to the activating -OH group).
• Top-Right Box (Na₂CO₃ reaction): The sodium carboxylate salt: COO⁻ Na⁺ (or COONa ) attached to the benzene ring, leaving the phenol -OH group unreacted.
• Bottom Boxes (Propanoic Anhydride reaction): Forms an ester on the phenolic oxygen (acyl group -COCH₂CH₃ ) plus propanoic acid ( CH₃CH₂COOH ) as the by-product. Bottom two boxes can be in either order.

💡 Key Knowledge

• Phenol vs. Carboxylic Acid: Carboxylic acids react with weak bases like Na₂CO₃(aq) to form soluble carboxylate salts and release CO₂ . Phenols are too weakly acidic to react with Na₂CO₃ .
• Acylation: Phenols react with acid anhydrides (like propanoic anhydride) to form esters.
• Directing Groups: The -OH group on a benzene ring is strongly activating and directs incoming electrophiles (like Br₂ ) to the 2-, 4-, and 6-positions.

🧠 Exam Technique

• Check functional group compatibilities carefully: ensure you don't accidentally react groups that remain inert under the given reagents.
• Make sure structural formulas or skeletal/displayed formats clearly show the points of attachment for substituents.

❌ Common Errors

• Attempting to react the phenol group with sodium carbonate.
• Failing to include the by-product ( CH₃CH₂COOH ) in the bottom reaction box alongside the main ester product.

✅ Correct Answer Structure

• Must feature a displayed amide linkage ( -CONH- ) between two benzene rings.
• Must feature a displayed ester linkage ( -COO- ) between two benzene rings.
• Must include at least one terminal "end bond" on each end of the section, with correct positioning of all three functional groups ( -OH , -NH₂ , -COOH derivatives) across the repeating units.

💡 Key Knowledge

• Bifunctional Monomers: PAS contains an amine ( -NH₂ ), a phenol ( -OH ), and a carboxylic acid ( -COOH ). It can self-condense.
• Linkages: Amines react with carboxylic acids to form amides; phenols react with carboxylic acids to form esters.

🧠 Exam Technique

• Draw out the linkages fully displayed ( -C(=O)-NH- and -C(=O)-O- ) so examiners can explicitly award connectivity marks.
• Verify that the chain sequence correctly alternates reactions between the amine/phenol groups of one molecule and the carboxylic acid of another.

❌ Common Errors

• Reversing atom connectivity in the amide link (writing -NH-C(=O)- incorrectly relative to the chain orientation).
• Omitting the terminal bonds or failing to show the exact required number of amide and ester bridges.

📐 Step-by-Step Calculation

Step 1: Determine the Molar Mass of PAS

Molecular formula from structure: C₇H₇NO₃ → Mr = (7×12.0) + (7×1.0) + 14.0 + (3×16.0) = 153 g mol⁻¹

Step 2: Calculate Total Mass for the Child's Dosage

Dosage rate = 300 mg per kg

Child mass = 20.0 kg

Total daily dose in mg = 300 × 20.0 = 6000 mg = 6.0 g

Step 3: Convert Mass to Moles

Moles = Mass / Mr = 6.0 / 153 = 0.0392... mol (or calculate via moles per kg first: 300×10⁻³ / 153 = 1.96×10⁻³ mol kg⁻¹, then multiply by 20.0 kg to get 0.0392 mol )

Step 4: Use Avogadro's Constant to Find Molecules

Number of molecules = Moles × L = 0.039215... × 6.02 × 10²³ = 2.36 × 10²²

💡 Key Rules & Formatting

• Significant Figures: Give your final answer to 3 significant figures (matching the precision of the input data: 300 mg, 20.0 kg).
• Units Check: Remember to convert milligrams (mg) to grams (g) or moles (mol) using the ×10⁻³ factor.

❌ Calculation Traps

• Forgetting to multiply the unit dosage (300 mg/kg) by the child's entire body mass (20.0 kg).
• Failing to convert mg into g before dividing by the molar mass in g mol⁻¹.