AQA A-Level Chemistry: Polymers (7405) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.3.12 Polymers of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: 3.3.12.1 Condensation polymers and 3.3.12.2 Biodegradability and disposal of polymers. The whole section is A-level only. It is examined on A-level Paper 2, and A-level Paper 3 can test any content.

Learn the ideas first in the polymers study guide and the polymers revision notes. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show where your gaps are. Addition polymer questions are in the alkenes practice questions. Questions 7, 9, 10 and 12 need a calculator. Use Ar: H = 1.0, C = 12.0, N = 14.0, O = 16.0, Cl = 35.5. In formulas, C₆H₄ is a benzene ring joined at positions 1 and 4.

Questions

1. Which pair of compounds reacts to form a polyamide?

• A ethane-1,2-diol and butanedioic acid
• B propene and chloroethene
• C butane-1,4-diamine and hexanedioic acid
• D ethane-1,2-diol and ethene

[1]

2. Hexanedioic acid reacts with ethane-1,2-diol to form a condensation polymer.

(a) Name the type of linkage formed between the monomers. [1] (b) Name the other product of the reaction. [1]

3. Draw the repeating unit of the polymer formed from benzene-1,4-dicarboxylic acid and propane-1,3-diol. [2]

4. A section of a polymer chain is –CO–(CH₂)₂–CO–NH–C₆H₄–NH–CO–(CH₂)₂–CO–NH–C₆H₄–NH–

(a) Draw the repeating unit. [1] (b) Give the structures and names of the two monomers. [2] (c) State the type of condensation polymer. [1]

5. A section of a polypeptide chain is –NH–CH₂–CO–NH–CH(CH₃)–CO–NH–CH₂–CO–NH–CH(CH₃)–CO–

(a) Draw the structures of the two amino acids used to make it. [2] (b) State the other product formed as the chain is made. [1]

6. Explain, in terms of intermolecular forces, why nylon 6,6 has a higher melting point than poly(ethene) with chains of similar length. [3]

7. This question is about Kevlar.

(a) Describe how a hydrogen bond forms between two neighbouring Kevlar chains. [2] (b) The repeating unit of Kevlar is –[CO–C₆H₄–CO–NH–C₆H₄–NH]–. A sample of Kevlar has an average Mr of 1.19 × 10⁵. Calculate the average number of repeating units per chain. [2] (c) State one typical use of Kevlar. [1]

8. This question is about the breakdown of polymers.

(a) The polyester from Question 3 is heated with aqueous sodium hydroxide. Name the two organic products. [2] (b) Nylon 6,6 is heated with dilute hydrochloric acid. Name the two organic products. [2] (c) Explain why poly(ethene) is not broken down by heating with aqueous sodium hydroxide. [2] (d) State which of the polymers in (a), (b) and (c) is non-biodegradable. [1]

9. In a school laboratory, nylon 6,6 is made from hexanedioyl dichloride, ClCO(CH₂)₄COCl, and hexane-1,6-diamine.

(a) Write an equation for the formation of nylon 6,6 by this route. Show the repeating unit. [2] (b) Calculate the percentage atom economy of this route, taking two molecules of by-product per repeating unit. [2] (c) Suggest why the diacyl dichloride is used instead of hexanedioic acid. [1]

10. Terylene is made from benzene-1,4-dicarboxylic acid and ethane-1,2-diol.

(a) Calculate the Mr of the repeating unit of Terylene. [1] (b) Calculate the mass of water formed when 9.60 g of Terylene is made. Assume two water molecules form per repeating unit. [2]

11. A town collects mixed plastic waste containing poly(propene), PVC and nylon 6,6. Evaluate landfill, incineration and mechanical recycling as ways of disposing of this waste. [7]

12. A section of polymer Q is –CO–(CH₂)₃–CO–NH–(CH₂)₅–NH–CO–(CH₂)₃–CO–NH–(CH₂)₅–NH–

(a) Name the two monomers used to make Q. [2] (b) Calculate the Mr of the repeating unit of Q. Show your working. [2] (c) Name the strongest type of force between chains of Q and explain how it arises. [2] (d) Q is heated with excess aqueous sodium hydroxide. Name the two organic products. [2]

Answers

1. C [1] [1] Examiner insight: A diamine plus a dicarboxylic acid gives amide links; A gives a polyester and B and D are alkenes or a diol that cannot form amide links.

2. (a) Ester [1] (b) Water [1] [2] Examiner insight: “Hydrogen and oxygen” or “OH” is not accepted for (b); the product must be named as water or written as H₂O.

3. Ester links correct at both ends with one residue of each monomer [1]; extension bonds and n shown: –[CO–C₆H₄–CO–O–CH₂CH₂CH₂–O]ₙ– [1]. [2] Examiner insight: A unit that keeps an –OH or –H at either end, or contains two diol residues, loses the first mark.

4. (a) –[CO–(CH₂)₂–CO–NH–C₆H₄–NH]– with extension bonds [1] (b) HOOC–(CH₂)₂–COOH, butanedioic acid [1]; H₂N–C₆H₄–NH₂, benzene-1,4-diamine [1] (c) Polyamide [1] [4] Examiner insight: In (b) a correct structure with the wrong name, or the right name for the wrong ring positions, is not given the mark; check both.

5. (a) H₂N–CH₂–COOH [1] and H₂N–CH(CH₃)–COOH [1] (b) Water [1] [3] Examiner insight: Drawing the residues, –NH–CH₂–CO–, instead of the full amino acids does not score; add H to N and OH to C.

6. Both have van der Waals forces between chains [1]. Nylon 6,6 also forms hydrogen bonds between the N–H of one chain and a lone pair on the C=O oxygen of another [1]. Hydrogen bonds are stronger, so more energy is needed to separate the chains of nylon 6,6 [1]. [3] Examiner insight: Any mention of breaking covalent bonds in the chain on melting contradicts the answer and can cost the final mark.

7. (a) The N–H bond is polar, so the H is δ+ [1]; it is attracted to a lone pair on the δ− oxygen of a C=O group on the next chain [1]. (b) Mr(C₁₄H₁₀N₂O₂) = 14(12.0) + 10(1.0) + 2(14.0) + 2(16.0) = 238.0 [1]; 1.19 × 10⁵ / 238.0 = 500 [1]. (c) Bulletproof vests or body armour (or ropes) [1] [5] Examiner insight: In (a) a hydrogen bond drawn between two N–H groups, or between H and the N of another chain, does not score.

8. (a) Sodium benzene-1,4-dicarboxylate [1]; propane-1,3-diol [1] (b) Hexanedioic acid [1]; hexane-1,6-diammonium chloride (the salt of the diamine) [1] (c) Poly(ethene) has only C–C and C–H bonds, which are non-polar [1], so there is no δ+ carbon for the hydroxide ion (nucleophile) to attack [1]. (d) Poly(ethene) [1] [7] Examiner insight: In (b) the free diamine is not accepted, because in acid the amine groups are protonated; in (a) the free acid is not accepted in alkali.

9. (a) n ClCO(CH₂)₄COCl + n H₂N(CH₂)₆NH₂ → –[CO–(CH₂)₄–CO–NH–(CH₂)₆–NH]ₙ– [1] + 2n HCl [1] (b) Mr: ClCO(CH₂)₄COCl = 183.0, diamine = 116.0, repeating unit = 226.0 [1]; atom economy = 226.0 / (183.0 + 116.0) × 100 = 75.6% [1] (c) The acyl chloride is more reactive, so the reaction is fast at room temperature [1]. [5] Examiner insight: Allow error carried forward in (b) from a wrong Mr, but a by-product of water in (a) loses that mark because the reactant is an acyl chloride.

10. (a) C₁₀H₈O₄: 192.0 [1] (b) Repeating units = 9.60 / 192.0 = 0.0500 mol, so water = 0.100 mol [1]; mass of water = 0.100 × 18.0 = 1.80 g [1]. [3] Examiner insight: Using one water per repeating unit halves the answer; a wrong Mr from (a) can usually be carried forward, but a wrong mole ratio cannot.

11. Landfill: simple and needs no sorting of the mixed waste [1]; but uses land, and poly(propene) and PVC are non-biodegradable so persist for a very long time [1]. Incineration: reduces the volume and the heat can generate electricity [1]; but releases CO₂, a greenhouse gas [1], and burning PVC gives toxic HCl, which must be removed from the gases [1]. Mechanical recycling: saves crude oil and energy and reduces landfill [1]; but the three polymers must be collected, sorted by type and cleaned, which costs money and energy, and quality falls each cycle [1]. [7] Examiner insight: An evaluation needs an advantage and a disadvantage for each method; a list of only problems, or “it is bad for the environment” with no chemistry, limits the credit.

12. (a) Pentanedioic acid [1]; pentane-1,5-diamine [1] (b) Repeating unit C₁₀H₁₈N₂O₂ [1]; Mr = 10(12.0) + 18(1.0) + 2(14.0) + 2(16.0) = 198.0 [1] (check: 132.0 + 102.0 − 2 × 18.0 = 198.0) (c) Hydrogen bonds [1], between the δ+ H of N–H on one chain and a lone pair on the C=O oxygen of another chain [1]. (d) Sodium pentanedioate [1]; pentane-1,5-diamine [1] [8] Examiner insight: In (a) the number of carbons in each residue must be counted carefully: the acid residue –CO–(CH₂)₃–CO– has five carbons, not three.

Where marks are usually lost

• Repeating units with only one monomer residue when two monomers are used.
• Ends of a repeating unit drawn with –OH or –H instead of extension bonds.
• Writing –CO–NH₂– or –CO–OH– as a linkage.
• Showing water as the by-product when the reactant is an acyl chloride.
• Hydrogen bonds drawn between two N–H groups instead of N–H and C=O.
• Saying polyesters form hydrogen bonds between chains.
• Free amine given as a product of acid hydrolysis, or free acid from alkaline hydrolysis.
• Explaining polyalkene inertness with “weak bonds” rather than non-polar bonds and no δ+ carbon.
• Disposal answers without a disadvantage for each method.
• Miscounting carbons when naming monomers from a chain section.

Next steps

• Polymers revision notes
• Polymers study guide
• Alkenes practice questions
• Bonding
• AQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.12 Polymers (3.3.12.1 Condensation polymers and 3.3.12.2 Biodegradability and disposal of polymers), A-level only.