Year 12 Chemistry Module 7 · Organic Chemistry ⏱ ~45 min 5 MC · 3 Short Answer Lesson 22 of 23

Condensation Polymers — Polyesters & Polyamides

Discover how PET bottles, polyester fabrics, and nylon ropes are built from bifunctional monomers — and why the same chemistry that makes plastic also makes the proteins in your body.

Today's hook: A PET bottle and a polyester shirt are made from the same polymer — so why does one hold water and the other absorb it?

0/5TASKS

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Before You Read

A PET plastic bottle and a polyester shirt are made from the same polymer. A nylon rope and a silk thread have the same type of chemical linkage — both are polyamides. The proteins in your muscles are also polyamides.

Before reading: what do you think the structural difference is between a polyester and a polyamide? What connecting group holds the monomers together in each case? And why do you think water is lost when these polymers form?

→

Key Condensation Polymerisation Equations

General pattern: Monomer A (group X) + Monomer B (group Y) → polymer + small molecule (H₂O or HCl)

        Polyester (diol + diacid):

        n HO-R-OH + n HOOC-R'-COOH → [-O-R-O-CO-R'-CO-]ₙ + 2n H₂O

        PET specifically:

        n HOCH₂CH₂OH + n HOOC-C₆H₄-COOH → [-OCH₂CH₂O-CO-C₆H₄-CO-]ₙ + 2n H₂O

        Polyester (hydroxy acid):

        n HO-R-COOH → [-O-R-CO-]ₙ + n H₂O

        Polyamide (diamine + diacid):

        n H₂N-R-NH₂ + n HOOC-R'-COOH → [-NH-R-NHCO-R'-CO-]ₙ + 2n H₂O

        Nylon 6,6 specifically:

        n H₂N(CH₂)₆NH₂ + n HOOC(CH₂)₄COOH → [-NH(CH₂)₆NHCO(CH₂)₄CO-]ₙ + 2n H₂O

Linkages: ester = -COO- (from -OH + -COOH) | amide = -CO-NH- (from -NH₂ + -COOH)

Learning Intentions

By the end of this lesson you will:

Know

What condensation polymerisation is and that H₂O is always released
PET monomers (ethylene glycol + terephthalic acid) and Nylon 6,6 monomers (hexane-1,6-diamine + hexanedioic acid)
Ester linkage (-COO-) vs amide linkage (-CO-NH-)

Understand

Why condensation polymers are hydrolysable (ester/amide bonds) but addition polymers are not (C-C bonds)
Why two functional groups per monomer are required
How to identify the linkage and break it to recover monomers

Can Do

Draw the repeat unit of PET and Nylon 6,6 from their monomers
Identify monomers from a given polymer repeat unit (insert H₂O at each linkage)
Compare addition and condensation polymerisation across three criteria

Scan these before reading

Key Terms

Condensation polymerisationMonomers react repeatedly with loss of a small molecule (water or HCl) at each step; requires two functional groups per monomer.

PolyesterA condensation polymer formed by reaction between a diol (–OH at both ends) and a dicarboxylic acid (–COOH at both ends); e.g., PET.

Polyamide (nylon)A condensation polymer formed by reaction between a diamine and a dicarboxylic acid; linked by amide (peptide) bonds; e.g., nylon-6,6.

DiolA compound with two hydroxyl groups (–OH); one of the two monomers required to make a polyester.

Dicarboxylic acidA compound with two carboxylic acid groups (–COOH); reacts with a diol (for polyester) or diamine (for polyamide).

Degree of polymerisationThe number of repeat units in a polymer chain; higher degree = higher molar mass = generally higher melting point and strength.

📚 Core Content

Condensation Polymerisation: What Makes It Different

In condensation polymerisation, every bond that forms between two monomers also releases a small molecule — usually water — so the polymer has fewer atoms than the sum of its monomers, unlike addition polymers where every atom is retained.

Requirements: Monomers must be bifunctional — each monomer has two reactive groups (one at each end), allowing the chain to grow from both ends simultaneously. When two groups react to form a linkage, a small molecule is eliminated:

Ester linkage
-OH + HOOC- → -OOC- + H₂O
diol -OH + diacid -COOH

Amide linkage
-NH₂ + HOOC- → -NH-CO- + H₂O
diamine -NH₂ + diacid -COOH

Industrial (acid chloride route)
-OH + ClOC- → -OOC- + HCl
by-product = HCl, not H₂O

Addition vs Condensation — Complete Comparison

Feature	Addition polymerisation	Condensation polymerisation
Monomer requirement	ONE functional group — C=C per monomer	TWO functional groups per monomer (or two bifunctional monomer types)
By-product	NONE — all atoms retained in polymer	H₂O (or HCl) — released at every bond formed
Polymer formula	Same empirical formula as monomer	Different — atoms lost in by-product
Linkage type	C-C (from C=C pi bond opening)	-COO- ester or -CO-NH- amide
Monomer type	Alkene (vinyl compound)	Diol + diacid; diamine + diacid; hydroxy acid; amino acid
Examples	PE, PP, PVC, PTFE, PS	PET, Nylon 6,6, Dacron, silk, proteins, collagen
Hydrolysable?	No — C-C bonds not susceptible to water	Yes — ester and amide bonds cleave with acid, base, or enzymes
Environmental persistence	Very high — no hydrolysable linkage; degrades only by UV/fragmentation	High, but ester/amide bonds can hydrolyse slowly; proteins rapidly digested

Mark-earning distinction: When a question asks "what type of polymerisation produces water as a by-product?" → condensation. When a question asks "what is the by-product of polymerising styrene?" → NONE (addition). This distinction earns a mark on nearly every HSC polymer question that asks about reaction type.

Common error — "addition produces water": ADDITION polymerisation: monomers add together WITH NO by-product. CONDENSATION polymerisation: monomers condense WITH a small molecule released. The name "condensation" refers to the elimination (condensation) of a small molecule — never attribute water production to addition polymerisation.

Exam Tip: For organic chemistry questions, draw full structural formulas showing all atoms and bonds — condensed or skeletal formulas alone may lose marks in HSC extended-response questions.

Quick check: Which correctly describes condensation polymerisation?

Polyesters: PET and Dacron

Polyesters form when hydroxyl groups (-OH) and carboxyl groups (-COOH) react — the same esterification from Module 7, but repeated n times using bifunctional monomers to produce a polymer chain instead of a small ester molecule.

PET Monomers (memorise these)

Diol: ethylene glycol (ethane-1,2-diol) — HOCH₂CH₂OH — two -OH groups

Diacid: terephthalic acid (benzene-1,4-dicarboxylic acid) — HOOC-C₆H₄-COOH — two -COOH groups

PET Properties & Uses

High MP (~260°C) — benzene ring rigidity. High tensile strength. Gas barrier properties. Recyclable (RIC code 1).

Uses: fizzy drink bottles, water bottles, polyester fabric (Dacron/Terylene — spun PET fibres), food packaging, photographic film.

Memorise PET monomers by formula: ethylene glycol (HOCH₂CH₂OH) and terephthalic acid (HOOC-C₆H₄-COOH). HSC questions ask for both by name AND formula. Writing "ethanoic acid" instead of terephthalic acid costs marks — the benzene ring in terephthalic acid is what gives PET its rigidity and high melting point.

Common error — ONE ester linkage per PET repeat unit: PET has TWO ester linkages per repeat unit — one at each end of the diacid component. [-O-CH₂CH₂-O-CO-C₆H₄-CO-]ₙ shows two -COO- groups. Drawing only one ester linkage earns partial marks.

True or False: The PET repeat unit [-O-CH₂CH₂-O-CO-C₆H₄-CO-]ₙ contains two ester linkages per repeat unit.

Polyamides: Nylon 6,6 and Proteins

Polyamides form by the same condensation mechanism as polyesters, but using amino groups (-NH₂) instead of hydroxyl groups — and the amide linkage that results is the same bond that holds every protein in your body together.

Nylon 6,6 Monomers

Diamine: hexane-1,6-diamine — H₂N(CH₂)₆NH₂ — 6 carbons, -NH₂ at each end

Diacid: hexanedioic acid (adipic acid) — HOOC(CH₂)₄COOH — 6 carbons total, -COOH at each end

"6,6" = 6 carbons in each monomer

Nylon 6,6 Properties & Uses

High MP (~265°C). N-H···O=C H-bonds between chains. High tensile strength + flexibility. Absorbs water (hygroscopic).

Uses: toothbrush bristles, clothing, ropes, parachutes, fishing line, gears, bearings, tyre cords.

Insight — the Nylon rope trick: Nylon was invented by Wallace Carothers at DuPont in 1935. In the famous "Nylon rope trick," hexane-1,6-diamine in water is layered over hexanedioyl dichloride (the acid chloride) in organic solvent. Nylon 6,6 forms instantly at the interface and can be pulled out as a continuous rope. The interfacial polymerisation still works in school labs today.

Common error — "hexane" is not a monomer: Hexane is an alkane with no reactive functional groups — it cannot undergo condensation polymerisation. The correct monomer is hexane-1,6-diamine (H₂N-(CH₂)₆-NH₂). The "di-" prefix and "-amine" suffix are both critical. An alkane cannot form a polymer by any Module 7 mechanism.

A polymer has repeat unit [-CO-(CH₂)₄-CO-NH-(CH₂)₆-NH-]ₙ. What are the monomers and polymer type?

Drawing Polymers and Identifying Monomers

The two-way conversion skill — polymer from monomers, and monomers from polymer — is the core exam skill of IQ6 and applies identically to polyesters and polyamides.

Monomers → Polymer (5 steps)

Write both monomers side by side.

Identify the reacting groups: -OH (or -NH₂) and -COOH.

Remove H from -OH (or -NH₂) and OH from -COOH → H₂O released.

Bond the remaining O (or N) to the carbonyl C → ester -COO- or amide -CO-NH- forms.

Enclose one diol + one diacid (minus water) in square brackets with subscript n and open bonds.

Polymer → Monomers (3 steps)

Identify the linkage type: -COO- = polyester; -CO-NH- = polyamide.

Break each linkage by inserting H₂O: add H to the O (or N) end and OH to the C=O end.

The two fragments produced are the monomers — name each by IUPAC rules.

Example — break Nylon 6,6 at amide bonds:
[-NH-(CH₂)₆-NH-CO-(CH₂)₄-CO-]ₙ
Break -CO-NH-: add H to N → H₂N-; add OH to C=O → -COOH.
Fragment 1: H₂N(CH₂)₆NH₂ = hexane-1,6-diamine ✓
Fragment 2: HOOC(CH₂)₄COOH = hexanedioic acid ✓

Identifying Linkage Type at a Glance:

-C(=O)-O-
Ester linkage
→ polyester
formed from -OH + -COOH

-C(=O)-NH-
Amide linkage
→ polyamide
formed from -NH₂ + -COOH

-C-C-
C-C backbone only
→ addition polymer
formed from C=C opening

Hydrolysis — Why Condensation Polymers Can Be Broken Down

Because ester and amide bonds form by condensation of functional groups, they can be broken by the reverse — hydrolysis (adding H₂O across the bond). This explains:

PET chemical recycling: controlled hydrolysis or methanolysis regenerates ethylene glycol and terephthalic acid — can be repolymerised.
Protein digestion: stomach acid + proteases hydrolyse amide bonds → amino acids absorbed.
PET vs polyethylene persistence: PET's ester bonds are slowly hydrolysable (and some PETase-producing bacteria exist); polyethylene's C-C bonds are not hydrolysed by water or enzymes.

Three criteria for comparing addition vs condensation in an extended response:
(1) Monomer functional groups: C=C (addition) vs two reactive groups per monomer (condensation).
(2) By-product: none (addition) vs H₂O or HCl (condensation).
(3) Linkage type: C-C (addition) vs ester (-COO-) or amide (-CO-NH-) (condensation).
Covering all three with structural justification is required for Band 5–6 marks.

Common error — water retained inside the chain: Students draw condensation polymer structures with -OH groups still present at what should be the ester or amide linkage. The water is RELEASED — only the linkage remains in the chain. The ester oxygen comes from the diol -OH (minus H); the C=O comes from the diacid (minus -OH). Nothing is retained inside the chain between the two monomer fragments.

Which of these is the odd one out? (Does NOT belong with the others as a condensation polymer monomer pairing)

🔬 Worked Examples

Worked Example 1 — Drawing PET and Nylon 6,6 Repeat Units

Problem: (a) Draw the repeat unit of PET from ethylene glycol and terephthalic acid. (b) Draw the repeat unit of Nylon 6,6 from hexane-1,6-diamine and hexanedioic acid. Label the linkage in each.

PET — diol + diacid → polyester:
HOCH₂CH₂OH + HOOC-C₆H₄-COOH → react at each -OH with each -COOH.
Remove: H from -OH, OH from -COOH → H₂O per linkage. Bond: diol O to diacid C=O.
Repeat unit: [-O-CH₂CH₂-O-CO-C₆H₄-CO-]ₙ
Linkage: ester (-COO-) — appears TWICE per repeat unit. By-product: 2H₂O per repeat unit.

Nylon 6,6 — diamine + diacid → polyamide:
H₂N(CH₂)₆NH₂ + HOOC(CH₂)₄COOH → react at each -NH₂ with each -COOH.
Remove: H from -NH₂, OH from -COOH → H₂O per linkage. Bond: N to carbonyl C.
Repeat unit: [-NH-(CH₂)₆-NH-CO-(CH₂)₄-CO-]ₙ
Linkage: amide (-CO-NH-) — appears TWICE per repeat unit. By-product: 2H₂O per repeat unit.

Worked Example 2 — Identifying Monomers from Polymer Structure

Problem: A condensation polymer has repeat unit [-O-(CH₂)₄-O-CO-(CH₂)₄-CO-]ₙ. Identify the monomers, state the linkage type, and compare this polymer with polystyrene.

Identify linkage: -CO-O- (reading from right to left: C=O then O) = ester linkage (-COO-) → polyester.

Break at each ester bond — insert H₂O:
Add H to each O end → HO-(CH₂)₄-OH = butane-1,4-diol (4C diol) ✓
Add OH to each C=O end → HOOC-(CH₂)₄-COOH = hexanedioic acid (6C total: 2 carbonyl C + 4 CH₂) ✓

Comparison — polyester vs polystyrene:

Feature	This polyester	Polystyrene
Monomers	Butane-1,4-diol + hexanedioic acid (bifunctional, -OH and -COOH)	Styrene — one C=C only
By-product	H₂O (2 per repeat unit)	None
Linkage	-COO- (ester)	-C-C-
Hydrolysable?	Yes — ester bonds cleave under acid/base	No — C-C bonds not hydrolysed

Worked Example 3 — Extended Response: Nylon Formation and Properties (5 marks)

Problem (5 marks): (a) Write the equation for Nylon 6,6 formation, including conditions. (b) Explain why Nylon 6,6 has such a high melting point (~265°C) using bond and IMF reasoning. (c) Compare the environmental persistence of PET with polyethylene, explaining the chemical basis for any difference.

n H₂N(CH₂)₆NH₂ + n HOOC(CH₂)₄COOH → [-NH(CH₂)₆NHCO(CH₂)₄CO-]ₙ + 2n H₂O Conditions: heat (~250–270°C), removal of water (drives equilibrium toward polymer). Type: condensation polymerisation.

High MP — two reasons:
(1) Amide bonds (-CO-NH-) are strong covalent bonds within each chain.
(2) The N-H of one chain forms hydrogen bonds (N-H···O=C) with the C=O of an adjacent chain. This inter-chain H-bond network significantly increases the energy required to separate chains during melting. Compare with addition polymers (only dispersion forces between chains) — H-bonding in Nylon contributes substantially to its high melting point despite having a more flexible aliphatic backbone.

PET vs polyethylene environmental persistence:
Both are persistent over human timescales. However, PET contains ester linkages (-COO-) which are susceptible to slow hydrolysis under acidic, alkaline, or enzymatic conditions. Some bacteria (e.g. Ideonella sakaiensis, discovered 2016) produce PETase enzymes that hydrolyse PET's ester bonds.
Polyethylene has only C-C and C-H bonds — no hydrolysable linkage. Water and microbial enzymes cannot attack C-C bonds directly. Polyethylene degrades only by UV photodegradation and mechanical fragmentation into microplastics.
Conclusion: both are environmentally persistent plastics over human-relevant timescales, but PET has a small additional degradation mechanism (ester hydrolysis) that polyethylene completely lacks.

Interactive Tool — Polymers Builder Open fullscreen ↗

True or false?

🔀 Sort the Steps +7 XP

Arrange these steps for forming a polyester from a diol and a dicarboxylic acid (e.g. PET from ethane-1,2-diol and benzene-1,4-dicarboxylic acid):

Mix the diol monomer (HO–R–OH) and the dicarboxylic acid monomer (HOOC–R'–COOH)

Heat with acid catalyst — the OH of the diol reacts with the COOH of the acid

An ester linkage (–COO–) forms and one molecule of water is eliminated

The growing chain still has reactive –OH and –COOH end groups that continue reacting

Long polymer chains form with repeating –OOC–R'–COO–R– ester linkage units

Complete the Learn phase to unlock Practice.

✏️ Activities

Activity 1 — Drawing Repeat Units from Monomers

Draw the repeat unit in correct square bracket notation for each of the following. State the linkage type and the by-product.

(a) Butane-1,4-diol (HOCH₂CH₂CH₂CH₂OH) + hexanedioic acid (HOOC(CH₂)₄COOH)
(b) Hydroxy acid: lactic acid (HO-CH(CH₃)-COOH) — polymerises with itself
(c) Hexane-1,6-diamine (H₂N(CH₂)₆NH₂) + hexanedioic acid (HOOC(CH₂)₄COOH)

Activity 2 — Identifying Monomers from Polymer Structures

For each repeat unit below, identify the linkage type, name and draw the monomers, and classify the polymer.

(i) [-O-(CH₂)₄-O-CO-(CH₂)₄-CO-]ₙ
(ii) [-NH-(CH₂)₅-CO-]ₙ (Nylon 6 — single amino acid monomer)
(iii) [-O-CH₂CH₂-O-CO-C₆H₄-CO-]ₙ (PET)

📝 Check Your Understanding

Q1. Which correctly describes condensation polymerisation?

Q2. A polymer has repeat unit [-CO-(CH₂)₄-CO-NH-(CH₂)₆-NH-]ₙ. What are the monomers and polymer type?

Q3. PET is susceptible to hydrolysis, but polystyrene is not. Which statement best explains this difference?

Q4. A student draws the repeat unit of Nylon 6,6 as [-NH-(CH₂)₆-NH-CO-(CH₂)₄-CO-]ₙ and states it contains one amide linkage per repeat unit. What is wrong?

Q5. Which monomer pairing CANNOT undergo condensation polymerisation to form a polyamide?

Q6. (3 marks)

(a) Draw the repeat unit of PET using the correct square bracket notation. (b) Name both monomers by IUPAC name and state the linkage type. (c) State the by-product and how many moles are produced per mole of repeat unit.

Q7. (4 marks)

A polymer X has repeat unit [-O-(CH₂)₂-O-CO-C₆H₄-CO-]ₙ. (a) Identify the linkage type and classify the polymer. (b) Identify the two monomers by name and structural formula. (c) Explain why polymer X is hydrolysable while polyethylene is not, using bond chemistry. (d) State one application of polymer X and explain how the benzene ring in one monomer influences the polymer's melting point.

Q8. (5 marks)

Compare addition polymerisation (using polypropylene as an example) with condensation polymerisation (using Nylon 6,6 as an example) across the following criteria: (i) monomer functional group requirements; (ii) by-product of polymerisation; (iii) linkage type and how it forms; (iv) hydrolysability; (v) environmental persistence in ocean water. Give structural justification for each point.

Show All Answers

Q1 — Answer: B

Condensation polymerisation requires bifunctional monomers — two reactive groups per monomer. At each bond formed, a small molecule (H₂O, HCl) is released. Option A describes addition polymerisation. Option C is partially correct about the single bifunctional monomer but wrong about no by-product — condensation ALWAYS produces a small molecule. Option D is completely wrong — the two types are fundamentally different.

Q2 — Answer: A

Linkage: -CO-NH- = amide bond → polyamide. Break each amide bond: add H to N → H₂N-; add OH to C=O → -COOH. Fragment 1: HOOC-(CH₂)₄-COOH = hexanedioic acid (6 carbons total). Fragment 2: H₂N-(CH₂)₆-NH₂ = hexane-1,6-diamine (6 carbons). This is Nylon 6,6. Option B gives diol → polyester (wrong linkage identified). Option D: the diacid fragment has 6 carbons total (hexanedioic acid), not 4 (butanedioic acid).

Q3 — Answer: C

PET's ester (-COO-) linkages can be cleaved by water under acidic or alkaline conditions, or by esterase/PETase enzymes — the C-O bond of the ester is susceptible to nucleophilic attack by water. Polystyrene has only C-C and C-H bonds — no ester or amide groups — and is not attacked by water or hydrolase enzymes. Option D is wrong — neither hydrolyses rapidly in ocean water at ambient conditions.

Q4 — Answer: D

The repeat unit [-NH-(CH₂)₆-NH-CO-(CH₂)₄-CO-]ₙ contains TWO amide linkages per repeat unit — one -CO-NH- on the left side (between (CH₂)₄ and (CH₂)₆) and one -CO-NH- on the right side. The diacid contributes two carbonyl groups, each forming one amide bond with the two ends of the diamine. One amide linkage per repeat unit would require only one -COOH end reacting — which would not produce a linear chain.

Q5 — Answer: B

Hexane (CH₃(CH₂)₄CH₃) is an alkane — it has no reactive functional groups (-NH₂, -OH, -COOH). Without functional groups at each end, hexane cannot form covalent bonds with hexanedioic acid by any condensation mechanism. It cannot form a polyamide (or any polymer). Option A: correct diamine + diacid pairing → Nylon 6,6. Option C: amino acid with -NH₂ + -COOH can self-condense → Nylon 6 (or protein). Option D: diamine + acid chloride → polyamide via HCl by-product (industrial route).

Q6 — Sample Answer (3 marks)

(a) PET repeat unit: [-O-CH₂CH₂-O-CO-C₆H₄-CO-]ₙ — square brackets, subscript n, open bonds at each end. [1 mark]
(b) Monomer 1: ethylene glycol (ethane-1,2-diol, HOCH₂CH₂OH). Monomer 2: terephthalic acid (benzene-1,4-dicarboxylic acid, HOOC-C₆H₄-COOH). Linkage type: ester (-COO-). [1 mark]
(c) By-product: H₂O. Amount: 2 moles of H₂O per mole of repeat unit (one H₂O per ester linkage formed; two ester linkages per repeat unit). [1 mark]

Q7 — Sample Answer (4 marks)

(a) Linkage: -COO- (ester bond). Polymer X is a polyester — specifically PET (polyethylene terephthalate). [0.5 mark]
(b) Break ester bonds (insert H₂O): Monomer 1: HO-CH₂CH₂-OH = ethylene glycol (ethane-1,2-diol). Monomer 2: HOOC-C₆H₄-COOH = terephthalic acid (benzene-1,4-dicarboxylic acid). [1 mark]
(c) PET's ester bonds (-COO-) contain a C-O single bond that can be attacked by water (hydrolysis) under acidic or alkaline conditions — the lone pairs on water's oxygen attack the electrophilic carbonyl carbon, regenerating the diol and diacid. Polyethylene has only C-C and C-H bonds — no electrophilic carbonyl carbon for water to attack. C-C bonds are not hydrolysed under any normal conditions. [1.5 marks]
(d) Application: plastic drink bottles (or polyester fabric/Dacron). Benzene ring contribution: the rigid planar benzene ring restricts free rotation of the polymer backbone, promoting crystalline ordering and increasing the energy required to disrupt the chain — resulting in a high melting point (~260°C). [1 mark]

Q8 — Sample Answer (5 marks)

(i) Monomer requirements: PP (addition) — propene has ONE functional group: C=C double bond per monomer. Nylon 6,6 (condensation) — requires TWO bifunctional monomers: hexane-1,6-diamine (two -NH₂ groups) and hexanedioic acid (two -COOH groups). [1 mark]
(ii) By-product: PP — no by-product. Nylon 6,6 — H₂O released at each bond formed; 2 moles H₂O per repeat unit. [1 mark]
(iii) Linkage formation: PP — C=C pi bond opens; two new C-C sigma bonds form between monomers. Nylon 6,6 — N from -NH₂ bonds to carbonyl C of -COOH; H₂O leaves. Linkage: amide (-CO-NH-). [1 mark]
(iv) Hydrolysability: PP — not hydrolysable; C-C bonds have no susceptibility to water. Nylon 6,6 — amide bonds (-CO-NH-) can be hydrolysed under acidic conditions or by proteases. [1 mark]
(v) Environmental persistence: PP — very high; no hydrolysable linkage; degrades only by UV photodegradation and mechanical fragmentation into microplastics. Nylon 6,6 — also high persistence in ocean water, but amide bonds offer a potential enzymatic degradation pathway not available to PP. In practice, both persist for decades to centuries in ocean environments. [1 mark]

How did your thinking change?

Back at the start, you were asked why a PET bottle holds water while a polyester shirt wicks sweat away from skin. Now you know: PET (polyethylene terephthalate) is a polyester — the ester linkage (–COO–) is moderately polar but the aromatic rings in the backbone make the bulk polymer relatively hydrophobic, so liquid water cannot penetrate the dense bottle wall. Sports polyester fabrics are made from the same class of polymer but are engineered as thin, textured fibres with capillary channels between them — the wicking is a physical fibre-structure effect, not a chemical difference in the polymer backbone. Both are polyesters; the difference is in how the material is processed and shaped, not in the repeat unit.

You also now know why water is lost during formation: polyesters have an ester linkage (–COO–) formed when –OH reacts with –COOH, losing H₂O. Polyamides have an amide linkage (–CO–NH–) formed when –NH₂ reacts with –COOH, also losing H₂O. The H from –OH or –NH₂ and the OH from –COOH combine and leave as water — the linkage is what remains between the two chains.

Look back at what you wrote before reading this lesson. How has your understanding changed?

I can draw PET and Nylon 6,6 repeat units, identify monomers from polymer structures, and compare addition with condensation polymerisation across three criteria

🏆 Review

What are the two monomers of PET? Write their structural formulas and state the linkage type.

State the key difference between addition and condensation polymerisation. Give one example of each type.

How do you identify the monomers from a given condensation polymer repeat unit?

Why does Nylon 6,6 have a higher melting point (~265°C) than polyethylene (~130°C for HDPE)?

Why is PET slightly more environmentally degradable than polyethylene, even though both are synthetic plastics?