OxfordAQA A-Level Chemistry: Polymers (9620) – 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.

Section 3.3.12, Polymers, is the focus of this set, taken from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020: 3.3.12.1 Condensation polymers and 3.3.12.2 Biodegradability and disposal of polymers. The whole section is International A-level only. It belongs to the International A2 organic content assessed in Unit 4, and Unit 5 may test anything in the course.

If the ideas are new, start with the polymers study guide and the polymers revision notes. Addition polymer questions are in the alkenes practice questions. The course hub, the printable checklist and the free 10-minute diagnostics help you plan. Where a question needs relative atomic masses, use Cl = 35.5, C = 12.0, H = 1.0, N = 14.0 and O = 16.0. C₆H₄ in a formula means a benzene ring bonded at positions 1 and 4.

Questions

1. Which of these monomer pairs forms a polyester?

• A benzene-1,4-diamine and benzene-1,4-dicarboxylic acid
• B propene and but-1-ene
• C butane-1,4-diol and benzene-1,4-dicarboxylic acid
• D hexane-1,6-diamine and propene

[1]

2. Benzene-1,4-dicarboxylic acid reacts with pentane-1,5-diol, HO–(CH₂)₅–OH, to form a condensation polymer.

(a) Draw the repeating unit of the polymer. [2] (b) Complete the equation for n molecules of each monomer, showing the inorganic product and its amount. [1]

3. A section of a polymer chain is

–CO–(CH₂)₄–CO–NH–(CH₂)₈–NH–CO–(CH₂)₄–CO–NH–(CH₂)₈–NH–

(a) Draw the repeating unit. [1] (b) Draw the structures of the two monomers and name them. [2]

4. A section of a polyamide made from two amino acids is

–NH–CH(CH₂CH₂CH₃)–CO–NH–(CH₂)₃–CO–NH–CH(CH₂CH₂CH₃)–CO–NH–(CH₂)₃–CO–

(a) Draw the structures of the two amino acids. [2] (b) Explain why an amino acid can form a polymer without a second type of monomer. [1]

5. State one typical use of each of Terylene, nylon 6,6 and Kevlar. [3]

6. Kevlar fibres are much stronger than Terylene fibres. Explain this difference in terms of the intermolecular forces between the polymer chains. [4]

7. This question is about the hydrolysis of polymers.

(a) The polyester in question 2 is refluxed with dilute hydrochloric acid. Give the names of both organic products. [2] (b) The polyamide in question 3 is boiled with aqueous sodium hydroxide. What are the two organic products? [2] (c) Poly(phenylethene) is not affected by either treatment. Explain why. [2]

8. This question is about the polyamide in question 3.

(a) Calculate the Mr of its repeating unit. Show your working. [2] (b) One sample has an average chain Mr of 6.35 × 10⁴. How many repeating units does an average chain contain? [1] (c) Calculate the percentage by mass of nitrogen in the polymer. Give your answer to 3 significant figures. [1]

9. Polymer X is heated with dilute hydrochloric acid. The only products are benzene-1,4-dicarboxylic acid and a salt Y, with formula C₁₀H₂₆N₂Cl₂, made from an unbranched diamine with an –NH₂ group at each end.

(a) Give the structure and the name of the diamine that Y is made from. [2] (b) Draw the repeating unit of X. [1] (c) State the type of condensation polymer that X is. [1] (d) Name the two organic products formed if X is heated with aqueous sodium hydroxide instead. [2] (e) Calculate the Mr of the repeating unit of X. [1]

10. A sportswear firm collects worn-out Terylene shirts and used poly(ethene) packaging film from its shops. Evaluate landfill, incineration and recycling as ways of dealing with this waste. [6]

11. 11-aminoundecanoic acid, H₂N–(CH₂)₁₀–COOH, forms a polyamide.

(a) Write an equation for the polymerisation, showing the repeating unit. [2] (b) Chains in a sample have a mean Mr of 3.66 × 10⁴. Work out the mean number of repeating units in one chain. [2] (c) Calculate the mass of monomer needed to make 50.0 kg of the polymer, assuming complete conversion. [2] (d) Explain why this polyamide is biodegradable but poly(ethene) is not. [3]

Answers

1. C [1] [1] Examiner insight: A diol with a dicarboxylic acid is the only pair that gives ester links; A gives a polyamide, and B and D cannot form ester links at all.

2. (a) One residue of each monomer joined by an ester link, with a second link split across the brackets [1]; bonds through both brackets with n outside: –[CO–C₆H₄–CO–O–(CH₂)₅–O]ₙ– [1] (b) n HOOC–C₆H₄–COOH + n HO–(CH₂)₅–OH → –[CO–C₆H₄–CO–O–(CH₂)₅–O]ₙ– + 2n H₂O [1] [3] Examiner insight: Check the count of CH₂ groups in the diol residue against the name; a unit with four or six CH₂ groups is drawn from the wrong monomer and is unlikely to score.

3. (a) –[CO–(CH₂)₄–CO–NH–(CH₂)₈–NH]– with bonds through the brackets [1] (b) HOOC–(CH₂)₄–COOH, hexanedioic acid [1]; H₂N–(CH₂)₈–NH₂, octane-1,8-diamine [1] [3] Examiner insight: Give each structure with its correct name, and count the carbons in each residue before naming, including the two carbonyl carbons of the acid.

4. (a) H₂N–CH(CH₂CH₂CH₃)–COOH [1]; H₂N–(CH₂)₃–COOH [1] (b) Each molecule has both an –NH₂ group and a –COOH group, so the amine end of one molecule reacts with the acid end of another, and the chain can grow at both ends [1]. [3] Examiner insight: Drawing the residues (–NH–…–CO–) rather than complete amino acids does not answer the question; add H to the N and OH to the C.

5. Terylene: clothing fibres or drinks bottles [1]. Nylon 6,6: clothing, carpets or ropes [1]. Kevlar: body armour or bulletproof vests [1]. [3] Examiner insight: Give a specific use for each polymer; “making plastics” or “materials” is too vague to credit.

6. Terylene chains are held by van der Waals forces and permanent dipole–dipole forces between polar C=O groups [1]. No hydrogen in the ester link is attached to an O or N atom, so Terylene chains cannot hydrogen bond to each other [1]. Kevlar chains form hydrogen bonds: the δ+ hydrogen of an N–H group is drawn to a lone pair on a carbonyl oxygen in the next chain [1]. Kevlar’s flat, rigid chains line up in parallel, so there are very many of these stronger hydrogen bonds, and more energy is needed to separate the chains [1]. [4] Examiner insight: An answer that talks about breaking covalent bonds in the chain contradicts itself; the comparison must be about forces between chains.

7. (a) Benzene-1,4-dicarboxylic acid [1]; pentane-1,5-diol [1] (b) Sodium hexanedioate [1]; octane-1,8-diamine [1] (c) Its chain holds only C–C and C–H bonds, which are essentially non-polar [1]; with no δ+ carbon present, neither water nor OH⁻ has a site to attack, so there is no link to hydrolyse [1]. [6] Examiner insight: In (b) the free acid is not the product in alkali; it is the sodium salt. In (c), saying the bonds are “too strong” without mentioning polarity misses the key idea.

8. (a) Repeating unit formula C₁₄H₂₆N₂O₂ [1]; Mr = 14(12.0) + 26(1.0) + 2(14.0) + 2(16.0) = 254.0 [1] (check: 146.0 + 144.0 − 2 × 18.0 = 254.0) (b) 6.35 × 10⁴ ÷ 254.0 = 250 [1] (c) % N = 28.0 ÷ 254.0 × 100 = 11.0% [1] [4] Examiner insight: Write the formula of the repeating unit before the Mr, so your method is clear and any slip in the arithmetic is easy to spot.

9. (a) H₂N–(CH₂)₁₀–NH₂ [1]; decane-1,10-diamine [1] (b) –[CO–C₆H₄–CO–NH–(CH₂)₁₀–NH]– with bonds through the brackets [1] (c) Polyamide [1] (d) Sodium benzene-1,4-dicarboxylate [1]; decane-1,10-diamine [1] (e) C₁₈H₂₆N₂O₂: 18(12.0) + 26(1.0) + 2(14.0) + 2(16.0) = 302.0 [1] [7] Examiner insight: For (a), Y is the diammonium chloride, ⁺H₃N–(CH₂)₁₀–NH₃⁺ 2Cl⁻; work back by removing the two HCl units (C₁₀H₂₆N₂Cl₂ − 2HCl = C₁₀H₂₄N₂) before drawing the diamine.

10. Landfill: low cost, and the two materials need not be separated [1]; against this, it occupies land, and poly(ethene) film is non-biodegradable, so it persists almost unchanged [1]. Incineration: shrinks the waste and the heat given out can be used to generate electricity [1]; but it produces CO₂, a greenhouse gas, and the raw materials are lost [1]. Recycling: conserves crude oil, uses less landfill and recovers the material [1]; however, the Terylene and poly(ethene) must be collected, separated and cleaned first, which costs energy and money, and quality can fall; Terylene could instead be hydrolysed back to its monomers, which needs energy and specialist plant [1]. [6] Examiner insight: “Evaluate” needs both sides for each method, and each point should be tied to these polymers; generic statements such as “bad for the environment” earn little.

11. (a) n H₂N–(CH₂)₁₀–COOH → –[NH–(CH₂)₁₀–CO]ₙ– [1] + n H₂O [1] (b) Repeating unit C₁₁H₂₁NO, Mr = 11(12.0) + 21(1.0) + 14.0 + 16.0 = 183.0 [1]; 3.66 × 10⁴ ÷ 183.0 = 200 [1] (c) Each repeating unit (183.0 g mol⁻¹) comes from one monomer molecule (Mr 201.0) [1]; mass = 50.0 × 201.0 ÷ 183.0 = 54.9 kg [1] (d) The amide links contain a polar C=O bond, so the carbonyl carbon is δ+ [1]; water (or OH⁻ or H⁺/H₂O) can attack it and hydrolyse the link, breaking the chain into monomers [1]. Poly(ethene) contains just C–C and C–H bonds, which are essentially non-polar, so a nucleophile finds no target and the polymer is inert [1]. [9] Examiner insight: In (a), one monomer gives n H₂O, not 2n; in (c), using the polymer Mr for both masses gives 50.0 kg, which ignores the water lost.

Where marks are usually lost

• Drawing a two-monomer repeating unit with a single residue, or with the same monomer’s residue twice.
• Unit ends left with –H or –OH instead of bonds through the brackets.
• 2n H₂O written for a single amino acid, or n H₂O for a diacid with a diol.
• Monomer names that miscount carbons, especially forgetting the two carbonyl carbons of a diacid.
• Hydrogen bonds claimed between polyester chains.
• Hydrogen bonds drawn from N–H to N instead of to the C=O oxygen.
• Free acid given as an alkaline-hydrolysis product, or free diamine as an acid-hydrolysis product.
• Polyalkene inertness explained by “strong bonds” with no mention of non-polar bonds or the lack of a δ+ carbon.
• Disposal answers that give only advantages, or only problems.

Next steps

• Polymers revision notes
• Polymers study guide
• Alkenes practice questions
• Shapes, polarity and intermolecular forces
• OxfordAQA A-Level Chemistry hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.12 Polymers (3.3.12.1 Condensation polymers; 3.3.12.2 Biodegradability and disposal of polymers), International A-level only.