Edexcel A-Level Chemistry Paper 2, November 2020: Question 5

✅ Correct Answer

D ( C₆H₁₄ )

💡 Key Knowledge

A non-cyclic, saturated hydrocarbon is an alkane, which follows the general formula CₙH₂ₙ₊₂ . For n = 6 , the number of hydrogen atoms is (2 × 6) + 2 = 14 .

❌ Common Errors

Options A, B, and C correspond to unsaturated or cyclic structures ( C₆H₆ is benzene, C₆H₁₀ and C₆H₁₂ contain rings or multiple bonds).

✅ Correct Answer

B (3)

💡 Key Knowledge

Pentane has 3 structural isomers: pentane (straight chain), 2-methylbutane (branched), and 2,2-dimethylpropane (double branched).

🧠 Exam Technique

Systematically draw the longest carbon chain from 5 carbons down to 3 carbons to ensure you miss none and avoid counting duplicate rotations.

✅ Correct Answer

D (16 sigma bonds and 1 pi bond)

💡 Key Knowledge

Cyclohexene has a 6-carbon ring with one double bond. Total bonds = 6 C-C single bonds, 10 C-H single bonds (total 16 sigma bonds), and 1 double bond containing 1 pi bond.

❌ Common Errors

Forgetting to count the individual C-H sigma bonds across the ring structure is the primary reason students pick incorrect low totals.

✅ Correct Answer

A ( 50 cm³ )

📐 Calculation Breakdown

• Equation: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g)
• Reactant moles/volume ratio: 2 moles of gas react with 7 moles of gas to produce 4 moles of gas (water is liquid, but note the question states "when water is a gas").
• Total initial gas = 2 + 7 = 9 volumes. Total final gas = 4 + 6 = 10 volumes.
• Net change per 100 cm³ of ethane reacted: 100 × (10 - 9) / 2 = +50 cm³ increase.

✅ Correct Answer

Formula: C₅H₈ (Cyclopentene skeletal structure)

📐 Step-by-Step Calculation

1. Find x (carbons): KOH absorbs CO₂ . The decrease in volume is 125 cm³ . Since 25 cm³ of CₓHᵧ forms 125 cm³ of CO₂ , x = 125 / 25 = 5 .
2. Find y (hydrogens): Total volume decrease = 75 cm³ . Using the stoichiometric relation: 25 + 25(5 + y/4) - 75 = 125 , solving yields y = 8 .
3. Deduce Structure: C₅H₈ cyclic hydrocarbon with one double bond is cyclopentene.

🧠 Exam Technique

Make sure to draw a clear skeletal formula of a pentagon containing one double bond for the final mark point.

💡 Key Mechanism Features

• Polarity: ICl is polar with δ+ on I and δ- on Cl because iodine is less electronegative than chlorine.
• Curly Arrow 1: From the C=C double bond to the δ+ iodine atom.
• Curly Arrow 2: From the I-Cl bond to the Cl atom, forming a chloride ion ( Cl⁻ ).
• Intermediate: A secondary carbocation (CH₃-CH⁺-CH₂I) is formed as the major product pathway because it is more stable than the primary carbocation.
• Final step: A curly arrow from the lone pair on Cl⁻ to the positive carbon atom.

❌ Common Errors

Placing the partial positive charge ( δ+ ) directly onto the carbon intermediate instead of the I-Cl reagent, or reversing the iodine/chlorine positions on the major product, resulting in loss of marks.

📐 Step-by-Step Calculation

1. Moles of Br₂: (30.6 × 0.200) / 1000 = 0.00612 mol (or 6.12 × 10⁻³ mol )
2. Moles of Limonene: Since limonene has two double bonds, 1 mole reacts with 1 mole of Br₂? Wait, looking at the structure, limonene has two double bonds so it reacts in a 1:2 ratio with Br₂.
   Moles of limonene = 0.00612 / 2 = 0.00306 mol .
3. Mass of Limonene: Molar mass of limonene ( C₁₀H₁₆ ) = (10×12) + (16×1) = 136 g mol⁻¹ .
   Mass = 0.00306 × 136 = 0.41616 g .
4. Percentage by Mass: (0.41616 / 0.500) × 100 = 83.232% .
   Expressed to 3 significant figures: 83.2% (or 2 SF: 83%).

🧠 Exam Technique & Traps

Always inspect structural drawings carefully to count active sites (alkene double bonds). Limonene contains two double bonds, meaning it reacts with a 1:2 stoichiometric ratio of Br₂. Missing this factor halves or doubles your final answer incorrectly!