OxfordAQA A-Level Chemistry: Polymers (9620) – Revision Notes

These notes condense section 3.3.12, Polymers, of OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020. They cover 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 sits in the International A2 organic content, which Unit 4 assesses, and Unit 5 may test anything in the course.

Worked examples and fuller explanations are in the polymers study guide, and the polymers practice questions let you check yourself. Addition polymers are in the alkenes revision notes. Use the course hub, the printable checklist and the free 10-minute diagnostics to plan the rest of your revision.

Key words

Term	Meaning
Condensation polymer	Polymer made when monomers with two reactive groups join, giving off a small molecule (usually water) at each new link
Polyester	Chain joined by ester groups, –COO–, built from a diol and a diacid
Polyamide	Chain joined by amide groups, –CONH–, built from a diamine and a diacid, or from amino acids
Repeating unit	The shortest part of the chain that repeats; with two monomers, it contains a residue from each
Hydrolysis	Splitting a bond by reaction with water; for these polymers it reverses the condensation
Biodegradable	Broken down in the environment, here by hydrolysis of the links
Chemically inert	Unreactive; describes polyalkenes

3.3.12.1 The three named polymers

C₆H₄ means a benzene ring bonded at positions 1 and 4.

Polymer	Diol or diamine	Diacid	Repeating unit	Used for
Terylene (polyester)	HO–CH₂CH₂–OH	HOOC–C₆H₄–COOH	–[CO–C₆H₄–CO–O–CH₂CH₂–O]–	polyester fabric; bottles
Nylon 6,6 (polyamide)	H₂N–(CH₂)₆–NH₂	HOOC–(CH₂)₄–COOH	–[CO–(CH₂)₄–CO–NH–(CH₂)₆–NH]–	rope, carpet, clothing
Kevlar (polyamide)	H₂N–C₆H₄–NH₂	HOOC–C₆H₄–COOH	–[CO–C₆H₄–CO–NH–C₆H₄–NH]–	bulletproof vests, armour

Names: ethane-1,2-diol, hexane-1,6-diamine and benzene-1,4-diamine; the diacids are benzene-1,4-dicarboxylic acid (Terylene and Kevlar) and hexanedioic acid (nylon 6,6).

Memory hook: the “6,6” in nylon 6,6 counts the carbon atoms in each monomer, six and six.

Which monomers give which linkage

Second monomer reacting with the diacid	Linkage	Polymer
diol, HO–R–OH	ester –COO–	polyester
diamine, H₂N–R–NH₂	amide –CONH–	polyamide
(amino acid reacts with itself)	amide –CONH–	polyamide, one monomer only

3.3.12.1 Drawing skills

Method: monomers → repeating unit

1. Remove OH from each –COOH and H from each –OH or –NH₂.
2. Join carbonyl C to O (ester) or to N (amide).
3. Bracket one residue of each monomer (one residue only for an amino acid).
4. Bonds through both brackets; n outside.
5. In the equation: 2n H₂O for two monomers, n H₂O for one amino acid.

Method: chain section → repeating unit

1. Mark every –CO–O– or –CO–NH–.
2. Find the shortest repeating stretch.
3. Start and end the unit by cutting through linkages.

Method: chain section → monomers

1. Split each ester link at its C–O single bond, or each amide link at its C–N bond.
2. Add OH to each carbonyl carbon → –COOH.
3. Add H to each O or N → –OH or –NH₂.
4. Count carbons in each residue before naming.

Small reminder. Section –O–(CH₂)₆–O–CO–(CH₂)₇–CO–O–(CH₂)₆–O–CO–(CH₂)₇–CO–. Cutting the ester links gives HO–(CH₂)₆–OH, hexane-1,6-diol, and HOOC–(CH₂)₇–COOH, nonanedioic acid (seven CH₂ carbons plus two carbonyl carbons = nine).

Small reminder. 5-aminopentanoic acid, H₂N–(CH₂)₄–COOH, is a single monomer with both groups. Its polymer has the repeating unit –[NH–(CH₂)₄–CO]–, and one water forms per unit.

Mr of a repeating unit

• Two monomers: Mr(unit) = Mr(diacid) + Mr(diol or diamine) − 2 × 18.0
• One amino acid: Mr(unit) = Mr(amino acid) − 18.0
• Units per chain = Mr(chain) ÷ Mr(unit)

3.3.12.1 Forces between chains

Polymer	Forces between chains	Reason
Polyalkene	van der Waals only	only non-polar C–C and C–H bonds
Polyester	dipole–dipole between C=O groups, plus van der Waals	the ester link has no N–H or O–H
Polyamide	hydrogen bonds, plus van der Waals	the N–H hydrogen of one chain bonds to an O lone pair on a carbonyl of the next
Kevlar	hydrogen bonds in very large numbers	rigid, flat chains align side by side

Chain of reasoning for a melting-point answer: name the forces in each polymer → say which is stronger → more energy needed to separate chains → higher melting point. Melting separates chains; covalent bonds in the chain stay intact.

3.3.12.2 Hydrolysis and biodegradability

Why polyesters and polyamides hydrolyse: the C=O in each link makes the carbonyl carbon δ+. Water or OH⁻ attacks it as a nucleophile and the link splits. So these polymers are biodegradable.

Why polyalkenes do not: the chain is only C–C and C–H bonds, which are non-polar. There is no δ+ carbon to attack, so polyalkenes are chemically inert and non-biodegradable.

Conditions	Polyester →	Polyamide →
Dilute acid, heat	diacid + diol	diacid + diammonium salt
Aqueous alkali, heat	dicarboxylate salt + diol	dicarboxylate salt + diamine

Rule of thumb: in acid, the nitrogen ends up protonated; in alkali, the acid ends up as its salt.

3.3.12.2 Disposal methods

Method	For	Against
Landfill	cheap; no sorting	uses land; polyalkenes last a very long time
Incineration	reduces volume; heat can generate electricity	CO₂ (greenhouse gas); toxic gases such as HCl from chlorinated polymers must be removed
Mechanical recycling	conserves crude oil and energy; cuts landfill	collection, sorting by type and cleaning cost money and energy; quality falls
Chemical recycling	recovers monomers or useful chemicals	energy and specialist plant needed

Method: a disposal evaluation

1. Name each method you discuss.
2. For each, give one advantage and one disadvantage.
3. Link each point to chemistry: biodegradability, combustion products, sorting different polymer types.
4. Finish with a judgement if the question says “evaluate” or “suggest the best”.

Must-know distinctions

• Addition vs condensation: addition polymers come from C=C monomers and form no other product; condensation polymers come from monomers with two reactive groups and release water.
• Polyester vs polyamide: –COO– vs –CONH–; permanent dipole–dipole forces vs hydrogen bonds between chains.
• Two monomers vs one: a diacid with a diol or diamine gives two residues per unit; an amino acid gives one.
• Acid vs alkaline hydrolysis: ammonium salt in acid; carboxylate salt in alkali.
• Biodegradable vs inert: polar links can be hydrolysed; a non-polar C–C backbone cannot.

Quick self-test

1. Name the small molecule released when two amino acid molecules join.
2. Name both monomers of Kevlar.
3. Name the linkage in Terylene.
4. Give one typical use of nylon 6,6.
5. Draw the repeating unit of the polymer made from 5-aminopentanoic acid, H₂N–(CH₂)₄–COOH.
6. Hexane-1,6-diol and nonanedioic acid form a polyester. Calculate the Mr of its repeating unit, then the number of units in a chain of Mr 5.40 × 10⁴. (Ar: H = 1.0, C = 12.0, O = 16.0)
7. What is the strongest intermolecular force between Terylene chains?
8. Why do Kevlar chains form so many hydrogen bonds with each other?
9. Name the organic products when Terylene is heated with dilute hydrochloric acid.
10. Which of nylon 6,6 and poly(propene) is biodegradable? Explain why.
11. Give one reason why mixed plastic waste must be sorted before mechanical recycling.
12. Name the monomers of the polymer with the section –NH–(CH₂)₆–NH–CO–(CH₂)₇–CO–.

Answers

1. Water.
2. Benzene-1,4-diamine and benzene-1,4-dicarboxylic acid.
3. Ester, –COO–.
4. Clothing, carpets or ropes.
5. –[NH–(CH₂)₄–CO]ₙ–, with bonds through both brackets.
6. Unit C₁₅H₂₆O₄: Mr = 15(12.0) + 26(1.0) + 4(16.0) = 270.0 (check: 118.0 + 188.0 − 36.0). Units = 5.40 × 10⁴ ÷ 270.0 = 200.
7. Permanent dipole–dipole forces between C=O groups (van der Waals forces are also present).
8. The chains are flat and rigid, so they pack side by side and many N–H and C=O groups line up.
9. Ethane-1,2-diol and benzene-1,4-dicarboxylic acid.
10. Nylon 6,6. Its amide links have a δ+ carbonyl carbon that water can attack; poly(propene) has only non-polar C–C and C–H bonds.
11. Different polymers have different properties, so a melted mixture gives a poor-quality product.
12. Hexane-1,6-diamine and nonanedioic acid.

Where marks are usually lost

• Putting one residue, or two of the same residue, inside a two-monomer repeating unit.
• Leaving –H or –OH on the ends of a repeating unit.
• Writing n H₂O instead of 2n H₂O for a diacid with a diol or diamine.
• Joining the ring of Terylene or Kevlar at the wrong positions; it is 1,4.
• Saying polyesters hydrogen bond between chains.
• Drawing a hydrogen bond from N–H to another N, rather than to the C=O oxygen.
• Saying polyalkenes are inert because their bonds are weak; they are inert because the bonds are non-polar.
• Giving the free diamine as an acid-hydrolysis product, or the free acid as an alkaline-hydrolysis product.
• Disposal answers that list advantages only, or give no chemistry.
• Miscounting carbons when naming a monomer from a chain section.

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.