AQA A-Level Chemistry Paper 2, 2019: Question 3

📐 Step-by-Step Calculation

1. Find % Hydrogen:
   % H = 100% − 88.2% = 11.8%
2. Calculate molar amounts (M1):
   Moles of C = 88.2 / 12.0 = 7.35 mol
   Moles of H = 11.8 / 1.0 = 11.8 mol
3. Divide by smallest amount (M2):
   C = 7.35 / 7.35 = 1.00
   H = 11.8 / 7.35 = 1.605 (or 1.61)
4. Convert to whole-number ratio (M3):
   1.60 is 8/5, so multiply both by 5:
   C = 1 × 5 = 5, H = 1.605 × 5 = 8
   Formula = C₅H₈

✅ Correct Deductions & Structure (M4)

Molecular Formula: C₅H₈

Structure: Must be branched and non-cyclic.

Acceptable structures include:

• 2-methylbuta-1,3-diene (Isoprene):
  CH₂=C(CH₃)−CH=CH₂
• 3-methylbut-1-yne:
  HC≡C−CH(CH₃)₂
• 3-methylbuta-1,2-diene:
  CH₂=C=C(CH₃)₂

❌ Common Errors

• Rounding 1.61 to 2: Gives an incorrect formula of C₁H₂ or C₄H₈, completely failing to realise that .60/.61 represents the fraction 3/5 (requiring multiplication by 5).
• Ignoring question constraints: Drawing straight-chain pentadienes (e.g. penta-1,3-diene) or cyclic isomers (cyclopentene). The question explicitly demands a non-cyclic branched hydrocarbon.
• Drawing 5-valent carbons: Adding too many hydrogens to the branched carbon.

🧠 Exam Technique

Notice the ratio: decimals like .5 require ×2; .33/.67 require ×3; .25/.75 require ×4; and .2, .4, .6, .8 require ×5. Never round a decimal like 1.6 to 2.

✅ Correct Answer

Skeletal formula of monomer:

Draw a 4-carbon conjugated diene skeleton: a zigzag of 4 carbon atoms with double bonds at positions 1 and 3.

//=//  →  CH₂=CH−CH=CH₂

IUPAC Name: Buta-1,3-diene

💡 Key Knowledge

The repeat unit shows a 2-carbon backbone with a −CH=CH₂ (vinyl) side group: −CH₂−CH(CH=CH₂)− .

• This is a 1,2-addition polymer of buta-1,3-diene, where only one of the double bonds opened during polymerisation.
• Monomer deduction: remove the continuation bonds from the two backbone carbons and place a double bond between them, yielding CH₂=CH−CH=CH₂ .

❌ Common Errors

• Not drawing skeletal formula: Drawing full structural or displayed formulae when the prompt specifically requested skeletal. This forfeits M1.
• Missing the 'a' in 'Buta': Writing "But-1,3-diene" or omitting locants ("Butadiene"). AQA IUPAC nomenclature rules strictly require Buta-1,3-diene.

🧠 Exam Technique

The two marks are independent: even if you misdraw the skeletal formula, you can still get the second mark for the correct IUPAC name Buta-1,3-diene .

✅ Correct Answer

Repeating unit of stereoisomer (E / trans):

The original diagram showed the Z (cis) isomer (both −CH₂− groups on the same side of the C=C). The third polymer is the E (trans) isomer:

—CH₂            H
    \          /
     C == C
    /          \
   H            CH₂—

(Trailing extension bonds must extend out from the −CH₂− groups on opposite sides).

Reason: Restricted rotation about the C=C (double) bond

💡 Key Knowledge

• Requirements for E/Z isomerism:
  1. Restricted rotation around a bond (here, the π-bond of the C=C double bond).
  2. Two different groups attached to each carbon atom of the C=C double bond (here, −H and −CH₂−).
• When asked why this type of stereoisomerism arises, examiners strictly look for the physical barrier: restricted rotation about the double bond.

❌ Common Errors

• Forgetting trailing bonds: Leaving out the continuation/trailing bonds on the −CH₂− groups makes it a molecule rather than a repeating unit.
• Incomplete explanation: Writing only that "there are two different groups on each carbon". While true, that is a condition for stereoisomers to exist, not the fundamental physical reason why the two forms cannot interconvert.

🧠 Exam Technique

Marks are awarded independently. You can still score the reason mark even if your drawing has errors, provided you clearly mention restricted rotation about the C=C bond.

✅ Correct Answer

Give any one of the following points:

• Carbon–carbon (C−C) bonds are non-polar (so are unreactive / not attacked by nucleophiles / water).
• Carbon–carbon (C−C) bonds are too strong to break under environmental conditions.
• Carbon–carbon (C−C) bonds cannot be hydrolysed (unlike condensation polymers such as polyesters/polyamides).

💡 Addition vs Condensation Polymers

• Polyalkenes (Addition polymers): Saturated/unsaturated non-polar hydrocarbon backbones made entirely of C−C bonds. Resistant to chemical attack, enzyme action, and hydrolysis.
• Condensation polymers (Polyesters/Polyamides): Contain polar ester ( −COO− ) or amide ( −CONH− ) linkages with δ+ carbonyl carbons, rendering them vulnerable to nucleophilic attack and biodegradation via hydrolysis.

❌ Common Errors

• Vague answers: Stating simply that "it is an addition polymer" or "it is inert" without mentioning the C−C bonds or their non-polarity.
• Mentioning C−H bonds: The mark scheme specifically notes: "Ignore C−H bonds". The stability of the polymer chain lies in its C−C backbone.

🧠 Exam Technique

Always name the specific bond in question: "C−C bonds are non-polar" is the safest, most concise response for why addition polymers do not biodegrade.