Edexcel A-Level Chemistry AS Paper 2, June 2025: Question 1

✅ Correct Answer

D — C₁₀H₁₆

📐 Step-by-Step Counting

1. Carbon atoms: Count all vertices, line ends, and branches.
   • Left side: 2 methyl ends + 1 alkene carbon + 1 CH = 4 carbons
   • Central chain: 2 CH₂ groups = 2 carbons
   • Right side: 1 tertiary alkene carbon + 1 exocyclic =CH₂ + 1 -CH= + 1 =CH₂ end = 4 carbons
   Total carbons = 10.
2. Hydrogen atoms: Ensure each carbon forms 4 bonds:
   (3 + 3 + 1 + 2 + 2 + 0 + 2 + 1 + 2) = 16 hydrogens.
3. Quick check: An open-chain alkane with 10 carbons has formula C₁₀H₂₂. Each C=C double bond introduces 1 degree of unsaturation (removes 2 H atoms). With 3 double bonds: 22 − (3 × 2) = 16 H atoms → C₁₀H₁₆.

❌ Common Errors

• Choosing B (C₅H₈): This is the empirical formula (simplest whole-number ratio), not the molecular formula.
• Choosing C (C₁₀H₁₀): Miscounting hydrogens by forgetting that single-bonded saturated carbons in the chain have 2 hydrogens each and terminal methyls have 3.

🧠 Exam Technique

Use the Degrees of Unsaturation shortcut for acyclic alkenes: General formula = C_nH_2n+2−2d where d is the number of double bonds. For myrcene: n = 10, d = 3 → H = 2(10) + 2 − 2(3) = 16. This avoids miscounting hydrogen atoms under exam pressure!

✅ Correct Answer

B — 3

💡 Key Knowledge

• In the dark at room temperature, bromine ( Br₂ ) undergoes electrophilic addition across carbon–carbon double bonds ( C=C ).
• Each C=C double bond reacts with exactly 1 molecule of Br₂ to form a vicinal dibromo alkane.
• Myrcene contains 3 separate C=C double bonds, so 1 mole of myrcene reacts with a maximum of 3 moles of Br₂ molecules.

❌ Common Distractor Traps

• Option D (6): Confusing bromine atoms with bromine molecules. 6 bromine atoms are incorporated, but that equals 3 Br₂ molecules. Read the question stem carefully!
• "In the dark": Specifies that free-radical substitution of C–H bonds by UV light cannot occur; only electrophilic addition across the alkene bonds takes place.

✅ Correct Answer

B — 2-methylbuta-1,3-diene

💡 IUPAC Nomenclature Rules

1. Longest carbon chain containing both double bonds: 4 carbons → root name is buta- (the extra 'a' is added before a consonant suffix like 'diene').
2. Numbering the chain: Number from the end that gives lower locants to double bonds and substituents. Numbering from left: double bonds at carbons 1 and 3; methyl substituent at carbon 2.
3. Suffix: Two double bonds require -1,3-diene.
4. Combined Name: 2-methylbuta-1,3-diene.

❌ Why Other Options are Wrong

• A (2-methylbut-1,3-ene): Missing the prefix di- to indicate two double bonds.
• C (1,3-diene-2-methylbutane): Incorrect order; substituents precede the root chain, and the principal functional group suffix comes at the end.
• D (2-methylpent-1,3-ene): Incorrect chain length (pent = 5 carbons in main chain instead of 4).

✅ Correct Answer

C — acidified potassium manganate(VII)

💡 Key Reaction & Reagents

Alkenes react with cold, acidified potassium manganate(VII) ( KMnO₄ / H₂SO₄ ) to form diols (compounds with two −OH groups on adjacent carbon atoms):

R−CH=CH₂ + [O] + H₂O → R−CH(OH)−CH₂OH

This reaction also serves as a chemical test for unsaturation: the purple manganate(VII) solution is decolourised (turns colourless).

❌ Comparison with Alternative Reagents

• A (H₂ / Ni): Hydrogenation — reduces alkenes to alkanes, not diols.
• B (Acidified K₂Cr₂O₇): Potassium dichromate(VI) does not react with alkenes under standard conditions (it oxidises primary/secondary alcohols and aldehydes).
• D (Steam and acid catalyst): Hydration — adds one H and one OH across a double bond to form a monohydric alcohol, not a diol.