OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2023: Question 16

✅ Correct Answer

Trend: Boiling point increases as branching decreases (or as the hydrocarbon chain becomes more straight-chained, moving from 2,2-dimethylbutane to 2-methylpentane to hexane).

Explanation: Hexane has fewer branches, leading to a greater molecular surface area contact between molecules. This allows for stronger London forces (induced dipole-dipole interactions), requiring more energy to overcome.

💡 Key Knowledge

• All three isomers have the molecular formula C₆H₁₄ and contain the same number of electrons. Do not talk about electron count!
• Branching creates more spherical molecules, reducing the surface contact area between adjacent molecules.

🧠 Exam Technique

• Always make a clear comparison across all three molecules (use words like "fewer branches", "higher surface area").
• Explicitly link surface contact to the strength of London forces and the energy required to break them.

❌ Common Errors

• Mentioning chain length (these are isomers, chain length is identical).
• Stating that molecules with more branches have stronger forces (getting the trend backwards).
• Crediting vague statements like "harder to break bonds" without specifying intermolecular forces.

✅ Correct Answer

Initiation: Br₂ → 2Br• (with ultraviolet / UV )

Propagation:
1) C₆H₁₄ + Br• → C₆H₁₃• + HBr
2) C₆H₁₃• + Br₂ → C₆H₁₃Br + Br•

Termination (Any three equations combining two radicals):
1) 2Br• → Br₂
2) C₆H₁₃• + Br• → C₆H₁₃Br
3) 2C₆H₁₃• → C₁₂H₂₆ (dimerization showing two 2-methylpentyl radicals combining)

💡 Key Knowledge

• Radicals are reactive species with an unpaired electron, represented by a clear dot ( • ).
• Propagation cycles regenerate a radical, allowing the chain reaction to continue.
• Termination removes radicals by combining two of them into a stable molecule.

🧠 Exam Technique

• Make sure the radical dot ( • ) is clearly placed on the correct carbon atom (e.g., carbon 3 of the hexyl chain) for substituted structures.
• Skeletal formulas can be used cleanly to represent the carbon chains, but ensure the radical dot is clearly visible next to the correct vertex.

❌ Common Errors

• Omitting the radical dot or placing it incorrectly on non-carbon atoms where it makes no chemical sense.
• Forgetting to include UV light in the initiation step.
• Writing propagation steps that consume radicals without generating new ones, or creating termination steps that produce radicals.

✅ Correct Answer

Formula of A: C₆Br₁₄

Balanced Equation:
C₆H₁₄ + 14Br₂ → C₆Br₁₄ + 14HBr

🧠 Exam Technique

• Read the question carefully: "all 14 H atoms" are substituted by bromine atoms, meaning every hydrogen is replaced by a Br atom, and an equal moles of HBr is formed as a byproduct.
• Double check atom counts on both sides to ensure mass balance.

❌ Common Errors

• Failing to balance the halogen molecules ( Br₂ ) and hydrogen bromide ( HBr ) coefficients.
• Writing an incomplete substitution formula.

📐 Step-by-Step Calculation

Step 1: Find moles of compound B using molar gas volume

n(B) = Volume / Molar gas volume = 72.0 cm³ / 40.0 dm³ mol⁻¹

Convert units: 72.0 cm³ = 72.0 × 10⁻³ dm³

n(B) = (72.0 × 10⁻³) / 40.0 = 1.80 × 10⁻³ mol

Step 2: Calculate molar mass M(B)

M(B) = mass / moles = 0.8649 g / (1.80 × 10⁻³ mol) = 480.5 g mol⁻¹

Step 3: Determine formula by subtracting carbon mass and finding bromine count

Mass of 6 carbons = 6 × 12.0 = 72.0 g mol⁻¹

Remaining mass for (H + Br) = 480.5 - 72.0 = 408.5 g mol⁻¹

Testing combinations of H and Br in C₆H₍₁₄₋ₓ₎Brₓ :

If formula is C₆H₉Br₅ : (6×12.0) + (9×1.0) + (5×79.9) = 72 + 9 + 399.5 = 480.5 g mol⁻¹.

Molecular Formula: C₆H₉Br₅

❌ Common Calculation Traps

• Unit mismatch trap: Using standard molar volume 24.0 dm³ mol⁻¹ instead of the given 40.0 dm³ mol⁻¹ . Watch out for exam-specific data values!
• Forgetting to convert cm³ to dm³ (dividing by 1000).