Edexcel A-Level Chemistry Paper 3, November 2020: Question 10

✅ Correct Answer / Marking Points (4 Marks)

• Mark 1: Lone pair on carbonyl oxygen with a curly arrow pointing to the H⁺ catalyst.
• Mark 2: Curly arrow from ethanol oxygen's lone pair to the carbonyl carbon atom.
• Mark 3: Curly arrow from the C–O single bond of the intermediate pointing directly to the oxygen of the water molecule.
• Mark 4: Curly arrow from the O–H bond back to the positively charged oxygen atom, regenerating the H⁺ catalyst.

💡 Key Knowledge

• Protonation of the carbonyl oxygen increases the $\delta^+$ charge on the carbon, making it much more susceptible to nucleophilic attack.
• Always display lone pairs explicitly on oxygen atoms when they participate in forming dative covalent bonds with curly arrows.

🧠 Exam Technique

Start your curly arrows precisely from bonds or lone pairs, and point arrowheads exactly to where new bonds form or electron pairs land. Penalties are enforced for sloppy arrow origins.

❌ Common Errors

• Drawing curly arrows starting inside empty space rather than from the electron source (lone pair or bond).
• Forgetting to show lone pairs on reacting oxygen atoms. (Penalised only once across M1 and M2).

✅ Correct Answer (2 Marks)

• Label: The 18-O atom is located in the single-bonded oxygen atom of the ethyl ethanoate ester product (bridging the carbonyl carbon and the ethyl group).
• Justification: The single bond C–O in the carboxylic acid breaks during the reaction, meaning the oxygen in ethanol acts as the nucleophile and ends up preserved in the ester.

💡 Key Knowledge

Isotopic labelling proves the origin of atoms in condensation reactions. In Fischer esterification, the water by-product always forms from the –OH group of the carboxylic acid and the –H from the alcohol.

📐 Step-by-Step Calculation

1. Step 1: Calculate Gibbs Free Energy ($\Delta G$)
   $\Delta G = -RT \ln K$
   $\Delta G = -8.31 \times 298 \times \ln(4.0) = -3433\text{ J mol⁻¹}$ (-3.433 kJ mol⁻¹)
2. Step 2: State the Entropy-Free Energy Equation
   $\Delta G = \Delta H - T\Delta S_{\text{system}}$
3. Step 3: Rearrange for $\Delta S_{\text{system}}$
   $\Delta S_{\text{system}} = (\Delta H - \Delta G) \div T$
4. Step 4: Substitute Values (Watch Units!)
   Convert $\Delta H$ to J mol⁻¹: $-6.0\text{ kJ mol⁻¹} = -6000\text{ J mol⁻¹}$
   $\Delta S_{\text{system}} = (-6000 - (-3433)) \div 298 = -2567 \div 298 = -8.614\text{ J K⁻¹ mol⁻¹}$
5. Step 5: Relate System Entropy to Standard Molar Entropies ($S^{\theta}$)
   $\Delta S_{\text{system}} = \Sigma S^{\theta}(\text{products}) - \Sigma S^{\theta}(\text{reactants})$
   $\Delta S_{\text{system}} = [S^{\theta}(\text{ester}) + S^{\theta}(\text{water})] - [S^{\theta}(\text{acid}) + S^{\theta}(\text{ethanol})]$
6. Step 6: Isolate and Calculate $S^{\theta}(\text{ethyl ethanoate})$
   $S^{\theta}(\text{ester}) = \Delta S_{\text{system}} + S^{\theta}(\text{acid}) + S^{\theta}(\text{ethanol}) - S^{\theta}(\text{water})$
   $S^{\theta}(\text{ester}) = -8.614 + 159.8 + 160.7 - 69.9 = \mathbf{+242\text{ J K⁻¹ mol⁻¹}}$ (or $240$ depending on rounding)

❌ Common Calculation Traps

• Unit Mismatch: Forgetting to multiply $\Delta H$ by $1000$ to convert from kJ mol⁻¹ to J mol⁻¹ while $\Delta G$ is calculated in J mol⁻¹.
• Sign Errors: Missing negative signs when substituting negative enthalpy or Gibbs values into subtraction arrangements.

✅ Correct Answer / Marking Points (3 Marks)

Any three valid comparisons demonstrating superior reactivity of acyl chlorides:

• The reaction with ethanoyl chloride is irreversible (compared to the reversible equilibrium with ethanoic acid).
• Hydrogen chloride ($\text{HCl}$) gas is produced as a by-product instead of water ($\text{H₂O}$).
• The reaction is very fast / occurs spontaneously at room temperature without requiring an acid catalyst or heating.

💡 Key Knowledge

Acyl chlorides have a much better leaving group ($\text{Cl⁻}$) compared to carboxylic acids ($\text{OH⁻}$ derivative), making their carbonyl carbon significantly more electrophilic and reactions with nucleophiles completely non-reversible.