AQA A-Level Chemistry Paper 3, June 2018: Question 21
1 mark · Easy difficulty · Multiple Choice
Identify the monomer or pair of monomers used to form the specified condensation polymer repeating unit.
Practise this questionQuestion
Question text
21 The repeating unit of a polymer is shown.
Which monomer or pair of monomers could be used to make this polymer?
[1 mark]
A ClOC(CH2)4NH2 only
B ClOC(CH2)4COCl only
C ClOC(CH2)4COCl and H2N(CH2)6NH2
D ClOC(CH2)6COCl and H2N(CH2)4NH2
Mark scheme
Show the mark scheme
21 C
How to answer it
Condensation Polymers: Identifying Monomers of Nylon-6,6
Core syllabus focus: Condensation polymers, specifically polyamides (Nylon) and nucleophilic addition-elimination reactions of acyl chlorides.
- Recognising the repeating unit of a polyamide and identifying the amide link: -CO-NH- .
- Deducing monomer structures by "breaking" condensation links back to their functional group precursors (acyl chloride vs amine).
- Careful counting of carbon atoms in aliphatic chains, particularly inside brackets like -(CH₂)₄- versus -(CH₂)₆- .
Question 21 Analysis
Multiple Choice: Monomers for Polyamide Repeat Unit [1 Mark]
✅ Correct Answer
C: ClOC(CH₂)₄COCl and H₂N(CH₂)₆NH₂
💡 Key Knowledge
- Polyamide linkage: Formed when a carbonyl carbon reacts with a nitrogen, eliminating a small molecule (here, HCl ).
- Acyl chlorides vs Carboxylic acids: Diacyl chlorides react much faster with diamines at room temperature than dicarboxylic acids.
- Nylon-6,6 structure: Hexanedioyl dichloride (6 carbons total: 2 carbonyl carbons + 4 in chain) + Hexane-1,6-diamine (6 carbons total).
📐 Step-by-Step Structural Breakdown ("Unzipping" the Repeat Unit)
- Locate the amide bond: Find the -CO-NH- connection in the repeating unit:
-CO-(CH₂)₄-CO- attached to -NH-(CH₂)₆-NH- . - Cleave the amide bond: Cut between the carbonyl carbon ( C=O ) and the amine nitrogen ( N-H ).
- Restore monomer functional groups:
- Add -Cl back to both carbonyl groups: gives ClOC(CH₂)₄COCl (hexanedioyl dichloride).
- Add -H back to both amine groups: gives H₂N(CH₂)₆NH₂ (hexane-1,6-diamine).
- Verify total carbon counts:
- Diacyl chloride: 1 + 4 + 1 = 6 carbons.
- Diamine: 6 carbons.
🧠 Exam Technique: Elimination Strategy
- Rule out B immediately: A diacyl chloride cannot polymerise alone; it lacks nucleophilic groups to attack the acyl carbon.
- Rule out A: Option A has an amine and an acyl chloride on the same molecule with only 4 CH₂ groups. Its polymer would repeat every 5 carbons, not alternate between 4 and 6 methylene units.
- Differentiate C and D: Look closely at which block has 4 methylenes and which has 6. The C=O groups flank the (CH₂)₄ group, so the diacyl chloride must contain (CH₂)₄ , not (CH₂)₆ . This eliminates D.
❌ Common Student Pitfalls
- Swapping chain lengths (Picking D): Glancing too quickly and matching the 6-carbon chain to the acyl chloride and 4-carbon chain to the diamine. Always follow the connectivity of the C=O group.
- Miscounting chain length in the monomer: Forgetting that ClOC(CH₂)₄COCl has 6 carbons in total (hexanedioic acid derivative), not 4.
- Overlooking eliminated molecule: Forgetting that reaction of an acyl chloride with an amine produces HCl (condensation), not water.
Topics
Organic Chemistry · 3.3.12 Polymers · 3.3.9 Carboxylic Acids and Derivatives
Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.