Edexcel A-Level Chemistry AS Paper 2, June 2025: Question 6

✅ Correct Answer

A: C₄H₉OH + PCl₅ → C₄H₉Cl + POCl₃ + HCl

💡 Key Knowledge

Reaction with solid PCl₅ at room temperature is the standard chemical test for the -OH group:

• Observations: Vigorous reaction, steamy/misty acidic fumes of HCl(g) that turn damp blue litmus red.
• Inorganic by-products are phosphorus oxychloride ( POCl₃ , liquid) and hydrogen chloride ( HCl , gas).

Step 1: Use the stoichiometric ratio from the balanced equation
Equation: Cr₂O₇²⁻ + 8H⁺ + 3CH₃CH(OH)CH₂CH₃ → 3CH₃COCH₂CH₃ + 2Cr³⁺ + 7H₂O
Molar ratio of Cr₂O₇²⁻ : butan-2-ol = 1 : 3
Moles of Na₂Cr₂O₇ required = 0.150 ÷ 3 = 0.050 mol

Step 2: Calculate molar mass (Mᵣ) of anhydrous Na₂Cr₂O₇
Mᵣ = (23.0 × 2) + (52.0 × 2) + (16.0 × 7) = 46.0 + 104.0 + 112.0 = 262 g mol⁻¹

Step 3: Calculate required mass
Mass = moles × Mᵣ = 0.050 mol × 262 g mol⁻¹ = 13.1 g

❌ Common Traps

• Ignoring the 1:3 ratio: Multiplying by 3 or assuming a 1:1 ratio leads to calculating mass from 0.150 mol or 0.450 mol.
• Wrong Mᵣ formula: Leaving out one Na atom or miscalculating oxygen mass. Mᵣ must be based on Na₂Cr₂O₇ (262), not K₂Cr₂O₇ or Cr₂O₇²⁻ alone.

🧠 Exam Technique

Always double check if the question asks for moles or mass, and look carefully at the reacting ratio in any given ionic equation. Transfer of error (TE) is allowed if you carry through an incorrect molar mass.

🧠 How to Score 6/6 (Logical Reasoning)

• Structure Matters: Marks are split into Indicative Content (max 4 marks) + Logical Linkages (max 2 marks).
• Correct Sequence: Reflux → Distil crude product → Dry → Filter/Decant → Final Redistillation. If reflux is omitted or placed after distillation, top marks cannot be awarded.
• Equipment Names: Always explicitly name the condenser orientation (vertical for reflux, sloping/downward for distillation).

❌ Common Errors from Examiner Reports

• Sealing the apparatus: Placing a stopper on top of the reflux condenser creates a sealed system under heat — an explosion hazard that loses marks.
• Unsuitable drying agents: Using reactive substances like concentrated H₂SO₄ or solid NaOH instead of neutral anhydrous salts (CaCl₂, MgSO₄).
• Vague temperatures: Forgetting to mention the exact boiling temperature range (around 353 K / 80 °C) for the final redistillation.

✅ Correct Answer

Compound: 2-bromobutane

Skeletal formula: A 4-carbon zig-zag chain with a single bond to a -Br atom on carbon 2.

❌ Common Mistakes

• Drawing structural or displayed formulas (e.g., CH₃CH(Br)CH₂CH₃ ) — the question strictly specifies skeletal formula.
• Drawing the bromine atom on carbon 1 (1-bromobutane) instead of carbon 2. The substitution occurs specifically at the secondary carbon where the -OH group was attached.

Step 1: Calculate moles of PI₃ produced
Moles of PI₃ = mass ÷ Mᵣ = 22.1 ÷ 411.7 = 0.053680 mol

Step 2: Relate moles of PI₃ to moles of I₂
From the equation: 6 mol I₂ → 4 mol PI₃ (ratio 6 : 4 = 1.5 : 1)
Moles of I₂ needed = 0.053680 × (6 ÷ 4) = 0.080520 mol

Step 3: Calculate number of I₂ molecules
Number of molecules = moles × Avogadro constant (L)
Number of I₂ molecules = 0.080520 × (6.02 × 10²³) = 4.85 × 10²² molecules
(Full value: 4.8473 × 10²²)

❌ Common Traps

• Inverting the mole ratio: Multiplying by 4/6 instead of 6/4.
• Confusing molecules with atoms: The question asks for iodine molecules (I₂), not iodine atoms. Multiplying further by 2 is incorrect.

🧠 Exam Technique

Always state unrounded values in working steps to avoid rounding errors, and quote your final answer to 3 significant figures (4.85 × 10²²), matching the data given in the question.

Step 1: Calculate moles of limiting reactant (butan-2-ol)
Mᵣ of butan-2-ol = 74.0 g mol⁻¹
Moles of butan-2-ol = 11.9 ÷ 74 = 0.16081 mol

Step 2: Determine maximum theoretical mass of 2-iodobutane
Stoichiometry: 3 CH₃CH(OH)CH₂CH₃ → 3 CH₃CH(I)CH₂CH₃ (1 : 1 molar ratio)
Theoretical moles of 2-iodobutane = 0.16081 mol
Theoretical mass = 0.16081 mol × 183.9 g mol⁻¹ = 29.573 g

Step 3: Calculate percentage yield
Percentage yield = (actual mass ÷ theoretical mass) × 100
% Yield = (22.8 ÷ 29.573) × 100 = 77.1% (or 77.097%)

✅ Correct Answer

1. Structural Formulae:

• Product of partial oxidation of butan-1-ol (aldehyde):
  CH₃CH₂CH₂CHO
• Product of oxidation of butan-2-ol (ketone):
  CH₃COCH₂CH₃ (or CH₃COCH₂CH₃ / C₂H₅COCH₃ )

2. Molecular Formula & Isomerism:

• Both share the identical molecular formula: C₄H₈O
• They are functional group isomers (or structural isomers) because they have the same molecular formula but different functional groups / arrangements of atoms.

❌ Common Formula Pitfalls

• Writing -COH instead of -CHO: Writing the aldehyde as CH₃CH₂CH₂COH is penalised as it incorrectly implies an alcohol. Aldehydes must always end in -CHO.
• Displaying bonds when asked for structural: Drawing full C-H bonds when structural formula is explicitly required.
• Giving incomplete explanations of isomerism: Stating only that "they have different shapes" without explicitly mentioning that both have the same molecular formula ( C₄H₈O ).