OCR A-Level Chemistry AS Breadth in chemistry (01), June 2019: Question 25

✅ Correct Answers

• Alcohol A: 2-methylpropanoic acid (Carboxylic acid structure drawn correctly).
• Alcohol B: NONE (Tertiary alcohol cannot be oxidized).
• Alcohol C: Butan-2-one (Ketone structure drawn correctly).

💡 Key Knowledge

• Primary alcohols (A): Oxidized fully to carboxylic acids under reflux with acidified dichromate(VI).
• Tertiary alcohols (B): Resistant to oxidation because the carbon bonded to the -OH has no adjacent hydrogens.
• Secondary alcohols (C): Oxidized to ketones under reflux.

❌ Common Errors

Students often incorrectly attempt to oxidize tertiary alcohols or draw incorrect skeletal structures with erroneous carbon chain lengths. Ensure you count your vertices carefully!

✅ Correct Answer

butan-2-ol (Hyphens and commas are flexible, but spelling must be chemically sound).

❌ Common Errors

Writing butan-3-ol . Remember to number from the end that gives the lowest possible locant for the functional group.

✅ Correct Answer

C₄H₁₀O + 6O₂ → 4CO₂ + 5H₂O

🧠 Exam Technique

Balance elements in alphabetical order for organic combustion: Carbon first, then Hydrogen, and balance Oxygen last using fractions if necessary, though whole numbers apply here.

✅ Correct Answers

• Initiation Equation: Cl₂ → 2Cl•
• Initiation Conditions: UV (or sunlight)
• Propagation Equation 1: C₄H₁₀ + Cl• → C₄H₉• + HCl
• Propagation Equation 2: C₄H₉• + Cl₂ → C₄H₉Cl + Cl•

🧠 Exam Technique

Radical dots ( • ) are strictly required in propagation steps and must not be omitted. However, dots are not strictly penalized in initiation equations by this specific mark scheme, though it is best practice to include them. Never write ionic charges like Cl⁻ instead of radicals!

✅ Correct Answer

C₄H₁₀ + 10Cl₂ → C₄Cl₁₀ + 10HCl

💡 Key Knowledge

Substitution of all hydrogen atoms in butane ( C₄H₁₀ ) means every single one of the 10 hydrogens is replaced by a chlorine atom, producing decachlorobutane and 10 moles of hydrogen chloride.

📐 Step-by-Step Calculation

1. Calculate moles of compound E:
   n(E) = Volume / Molar Gas Volume = 78.0 / 32500 = 2.40 × 10⁻³ mol
2. Calculate molar mass of compound E:
   M(E) = Mass / Moles = 0.636 / (2.40 × 10⁻³) = 265 g mol⁻¹
3. Determine the molecular formula:
   Butane skeleton contributes: (4 × 12.0) + (H remaining) = 48.0 + mass of remaining atoms.
   By subtracting carbon and accounting for substituted halogens (Cl = 35.5), we deduce the formula: C₄H₄Cl₆ .

❌ Common Calculation Traps

Watch out for unit conversions! If a student mistakenly uses standard molar gas volume at RTP ( 24000 cm³ mol⁻¹ ) instead of the non-standard conditions given in the stem ( 32500 dm³ mol⁻¹ ), they get n = 3.25 × 10⁻³ mol and M = 196 g mol⁻¹ . Fortunately, the mark scheme allows ECF (Error Carried Forward) so full marks can still be secured if subsequent steps are mathematically sound.