ChemistryYear 11 · Module 1 · IQ2⏱ ~30 min

Polymers: Structure and Properties

Think of polymer chains like trains. A monomer is a single train carriage. Polymerisation is coupling thousands of carriages end-to-end. The type of carriage (monomer structure) and how they connect determines everything — from the flexibility of a shopping bag to the strength of a bulletproof vest. Understanding polymer structure is a direct application of the bonding and IMF principles you've already built up.

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📝 Choose how you work: type answers below, or work in your book.

📚 Know

  • The difference between monomers and polymers
  • Addition and condensation polymerisation
  • Common polymers and their uses (polyethylene, PVC, nylon, polyester)

🔗 Understand

  • How monomer structure determines polymer properties
  • How chain length, branching, and cross-linking affect properties
  • Why IMFs between polymer chains matter for physical properties

✅ Can Do

  • Identify monomers from polymer structures and vice versa
  • Compare addition and condensation polymerisation mechanisms
  • Relate polymer structural features to observed physical properties
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Key Definitions

monomerA small molecule that can react repeatedly to form a polymer. Must have a reactive site (double bond or functional groups).
polymerA large molecule made of many repeating monomer units joined by covalent bonds. Molecular weights typically range from thousands to millions of g mol⁻¹.
addition polymerisationMonomers with C=C double bonds join together; the double bond opens to form new C–C single bonds. No atoms are lost — 100% atom economy. Product: only the polymer.
condensation polymerisationMonomers with two functional groups (e.g. –OH and –COOH) react repeatedly, releasing a small molecule (usually water) at each step. Product: polymer + small molecule byproduct.
cross-linkingCovalent bonds between adjacent polymer chains, making the material rigid, insoluble, and resistant to deformation. Example: vulcanised rubber (S–S cross-links).
degree of polymerisationThe number of monomer units in a polymer chain (n). Higher n → longer chain → generally higher strength and viscosity.

Core Content

⛓️

Addition Polymerisation

In addition polymerisation, alkene monomers (containing C=C) react through a chain reaction mechanism. Each C=C bond opens, and the carbons form new single bonds to adjacent monomers. No atoms are lost — every atom in the monomers appears in the polymer.

General equation: n(CH₂=CHR) → –[CH₂–CHR]ₙ– where R is the substituent group that varies between polymers.
MonomerPolymerSubstituent RKey property / use
Ethylene (CH₂=CH₂)Polyethylene (PE)–HFlexible, chemical resistant; bags, bottles, piping
Propylene (CH₂=CHCH₃)Polypropylene (PP)–CH₃Stiffer than PE; food containers, carpet fibre
Vinyl chloride (CH₂=CHCl)PVC–ClRigid or flexible (with plasticiser); pipes, electrical insulation
Styrene (CH₂=CHC₆H₅)Polystyrene (PS)–C₆H₅Rigid or foamed; packaging, insulation
Tetrafluoroethylene (CF₂=CF₂)PTFE (Teflon)–FNon-stick, high MP; cookware, bearings
Why does –Cl make PVC stiffer than PE? The C–Cl bond is polar (Cl is electronegative) → PVC chains have permanent dipoles → dipole-dipole forces between chains → stronger IMFs than the dispersion-only PE → higher resistance to deformation. Adding plasticisers (small molecules that wedge between chains) disrupts these IMFs, making PVC flexible.
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Condensation Polymerisation

Condensation polymerisation requires monomers with two reactive functional groups (bifunctional monomers). Each reaction between functional groups releases a small molecule — usually water (H₂O) or HCl. This reaction is called a condensation reaction.

Polyester (e.g. PET — polyethylene terephthalate)

Formed from a diol (2 × –OH groups) + a dicarboxylic acid (2 × –COOH groups). At each junction, an ester linkage (–COO–) forms and water is released.

Monomers → PET: Ethylene glycol (HO–CH₂CH₂–OH) + Terephthalic acid (HOOC–C₆H₄–COOH) → –[–OC–C₆H₄–CO–O–CH₂CH₂–O–]ₙ– + n H₂O
Uses: Plastic drink bottles, polyester clothing fibre.

Polyamide (e.g. Nylon-6,6)

Formed from a diamine (2 × –NH₂ groups) + a dicarboxylic acid. At each junction, an amide linkage (–CO–NH–) forms and water is released.

Monomers → Nylon-6,6: Hexamethylenediamine (H₂N–(CH₂)₆–NH₂) + Adipic acid (HOOC–(CH₂)₄–COOH) → –[–NH–(CH₂)₆–NH–CO–(CH₂)₄–CO–]ₙ– + n H₂O
Uses: Clothing fibre, rope, toothbrush bristles, engineering components.
FeatureAdditionCondensation
Monomer requirementC=C double bond2 functional groups (bifunctional)
ByproductNone (100% atom economy)Small molecule released (H₂O, HCl)
Linkage formedC–C bond (chain)Ester (–COO–), amide (–CONH–), etc.
ExamplesPE, PVC, polystyrene, PTFENylon, polyester (PET), polycarbonate
⚙️

Structural Features and Physical Properties

As you learned in L10, IMF strength determines physical properties. For polymers, the same rules apply — but the sheer size of polymer chains and how they interact also matters enormously.

Structural featureEffect on propertiesExample
Chain length (n)↑ chain length → ↑ total IMF surface → ↑ MP/viscosity/tensile strengthWax (short PE) vs. HDPE plastic (long PE)
Branching↑ branching → chains can't pack closely → ↓ density, ↓ MP, ↑ flexibilityLDPE (branched, flexible bags) vs. HDPE (linear, rigid pipes)
Cross-linkingCovalent bonds between chains → rigid, insoluble, thermosettingVulcanised rubber, bakelite, epoxy resin
Polar substituents↑ polarity → ↑ dipole-dipole or H-bonding between chains → ↑ MP, ↑ stiffnessNylon (N–H···O H-bonds) vs. polyethylene (dispersion only)
Thermoplastic vs. thermosetting: Thermoplastics (PE, PVC, nylon) soften on heating — chains can slide past each other because only IMFs hold them together. Thermosetting polymers (bakelite, epoxy) are cross-linked with covalent bonds — heating cannot soften them; they char or decompose. This determines recyclability: thermoplastics can be remelted and remoulded; thermosets cannot.

Worked Examples

1

Worked Example 1 — Stepwise: Identify monomer and classify polymer type

The following repeat unit is found in a common polymer: –[CH₂–CHCl]ₙ–. (a) Identify the monomer. (b) Classify the polymerisation type (addition or condensation). (c) Explain why this polymer is stiffer than polyethylene.
Step 1 — Identify the repeat unit Repeat unit: –CH₂–CHCl– Each repeat unit contains 2 carbons. Single bonds throughout in the chain. No ester or amide linkages present in the repeat unit. Step 2 — Work backwards to find monomer (addition) For addition polymerisation: monomer = repeat unit with one C=C bond replacing the two C–C bonds that form during polymerisation. Monomer: CH₂=CHCl (vinyl chloride) Step 3 — Classify polymerisation type No small molecule byproduct, no ester/amide linkages, C=C double bond in monomer → Addition polymerisation Step 4 — Explain relative stiffness PVC has –Cl substituent. Chlorine is electronegative (χ = 3.0) → C–Cl bond is polar → each PVC chain has permanent dipoles (δ+ and δ−) along its length. Dipole-dipole forces act between adjacent PVC chains. Polyethylene has only –H substituents → non-polar → only dispersion forces between chains. Dipole-dipole forces > dispersion forces → PVC chains are held more tightly → greater resistance to chain sliding → PVC is stiffer than PE.
Answer
(a) Monomer: CH₂=CHCl (vinyl chloride / chloroethene). (b) Addition polymerisation — no byproduct, monomer had C=C double bond that opened during polymerisation. (c) PVC is stiffer because the polar C–Cl bonds create dipole-dipole forces between chains; polyethylene has only dispersion forces. Dipole-dipole forces are stronger, requiring greater energy to slide chains past each other, hence greater stiffness.
2

Worked Example 2 — Stepwise: Compare two polymers from data

Two polymers A and B are tested: A has MP 130°C, is rigid at room temperature, and does not dissolve in any solvent. B has MP 65°C, is flexible at room temperature, and dissolves in some organic solvents. Suggest the structural feature responsible for each difference.
Step 1 — Analyse A's properties MP 130°C (relatively high for a polymer) AND insoluble in all solvents. Does not dissolve = cannot be separated by any solvent → suggests covalent cross-linking between chains (solvents can't penetrate to break apart IMFs because there are covalent bonds between chains). High MP + rigid = strong resistance to chain movement. Conclusion: A is likely a thermosetting/cross-linked polymer. Step 2 — Analyse B's properties MP 65°C (low) AND dissolves in organic solvents. Dissolves = only IMFs between chains (no covalent cross-links) → solvent molecules can displace chain-chain interactions. Flexible + low MP = weak IMFs, chains can slide easily. Conclusion: B is likely a thermoplastic polymer (short chain or non-polar substituents). Step 3 — Attribute differences Rigidity/non-solubility of A: covalent cross-links between chains lock structure. No solvent disrupts covalent bonds. Flexibility/solubility of B: only dispersion/dipole-dipole IMFs hold chains together; solvent molecules can compete for these interactions and cause dissolution; heating above 65°C provides energy to overcome IMFs.
Answer
A is rigid and insoluble due to covalent cross-links between polymer chains — these cannot be broken by heating or solvent. B is flexible and soluble because chains are held only by IMFs (dispersion or dipole-dipole); these are disrupted by organic solvent molecules or gentle heating.
⚠️

Common Mistakes

Writing the polymer formula without the repeat unit brackets. Always write –[repeat unit]ₙ– with brackets and subscript n. Without brackets, the structure is ambiguous. The n indicates the chain is made of thousands of these units.
Claiming condensation polymers always release water. The byproduct depends on the functional groups involved. Polyurethanes release no small molecule (the –OH reacts with –NCO directly). Always specify the actual byproduct for the reaction.
Confusing cross-linking with chain length. Both increase strength, but through different mechanisms: longer chains increase IMF contact area (more dispersion surface); cross-linking adds covalent bonds between chains. Cross-linked polymers are insoluble; long-chain thermoplastics can still dissolve.

📓 Copy Into Your Books

⛓️ Addition Polymerisation

  • Monomer: alkene with C=C double bond
  • Mechanism: C=C opens → new C–C bonds form
  • No byproduct (100% atom economy)
  • Examples: PE, PP, PVC, polystyrene, PTFE
  • Monomer = repeat unit + C=C bond

💧 Condensation Polymerisation

  • Monomer: bifunctional (2 reactive groups)
  • Byproduct: small molecule released (usually H₂O)
  • Linkages: ester (–COO–) or amide (–CONH–)
  • Examples: Nylon (amide links), PET (ester links)

⚙️ Structure → Properties

  • ↑ chain length → ↑ strength, ↑ MP
  • Branching → ↓ density, ↑ flexibility
  • Cross-linking → rigid, insoluble, thermosetting
  • Polar groups → stronger IMFs → stiffer

⚠️ Exam Tips

  • Monomer identification: reverse the polymerisation
  • Addition: add C=C back to repeat unit
  • Condensation: split at ester/amide linkage, add –OH/–H back
  • Thermoplastic: IMFs only → recyclable
  • Thermosetting: cross-linked → not recyclable

Activities

🔬 Activity 1 — Identification Drill

Monomer and Polymer Classification

1 The repeat unit of polypropylene is –[CH₂–CH(CH₃)]ₙ–. (a) Write the structural formula of the monomer. (b) Name the type of polymerisation. (c) Predict whether polypropylene or polyethylene would have a higher melting point. Justify with reference to IMFs.

✏️ Answer in your book

2 Nylon-6,6 has the repeat unit –[NH–(CH₂)₆–NH–CO–(CH₂)₄–CO]ₙ–. (a) What type of polymerisation produced nylon? (b) Identify the linkage that connects the monomers. (c) What small molecule was released at each step?

✏️ Answer in your book

3 Explain why nylon has a higher melting point than polyethylene, even though both are thermoplastic polymers.

✏️ Answer in your book
📊 Activity 2 — Data Analysis

Comparing Polyethylene Variants

Polymer typeStructureDensity (g/cm³)Tensile strength (MPa)Melting point (°C)
LDPE (low density)Highly branched chains0.91–0.948–20105–115
HDPE (high density)Linear chains, minimal branching0.94–0.9720–37120–140
UHMWPE (ultra high MW)Very long linear chains0.93–0.94~150130–135

A Explain why HDPE has a higher density and tensile strength than LDPE, using molecular structural reasoning.

✏️ Answer in your book

B UHMWPE has similar density to LDPE but much higher tensile strength. What structural feature explains this? Why would UHMWPE be difficult to process by injection moulding?

✏️ Answer in your book

Multiple Choice

Multiple Choice Questions

1. Which feature of a monomer is required for addition polymerisation?

A
Two carboxylic acid groups (–COOH)
B
A carbon–carbon double bond (C=C)
C
Two hydroxyl groups (–OH)
D
An amide linkage (–CONH–)

2. Which correctly identifies the small molecule released during polyester formation from a diol and a dicarboxylic acid?

A
CO₂
B
HCl
C
H₂O
D
NH₃

3. An unknown polymer is rigid, does not soften on heating, and is insoluble in all solvents. The most likely structural explanation is:

A
Extensive covalent cross-linking between polymer chains (thermosetting)
B
Very long chain length with strong dispersion forces
C
Strong hydrogen bonding between polar chain segments
D
The polymer is made from a single monomer type

4. PVC can be made either rigid (for pipes) or flexible (for electrical cable insulation) by adding plasticisers. How do plasticisers increase flexibility?

A
They break the C–Cl covalent bonds, removing the polar groups
B
They cross-link PVC chains, allowing them to slide more freely
C
They shorten the PVC chain length, reducing tensile strength
D
Small plasticiser molecules insert between PVC chains, disrupting dipole-dipole interactions and increasing chain mobility

5. The repeat unit of a polymer is –[NH–(CH₂)₅–CO]ₙ–. This polymer was formed by:

A
Addition polymerisation of an alkene monomer
B
Condensation polymerisation — the amide (–CO–NH–) linkage indicates a reaction between –NH₂ and –COOH groups with loss of H₂O
C
Addition polymerisation — no small molecule byproduct was released
D
Condensation polymerisation releasing HCl

Short Answer

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Short Answer Questions

6. Compare addition and condensation polymerisation. In your answer, describe the monomer requirements, the mechanism, and the products (including any byproducts) for each. Provide one named example of each type. 5 MARKS

✏️ Answer in your book

7. LDPE (low-density polyethylene) is used for flexible shopping bags while HDPE (high-density polyethylene) is used for rigid pipes. Both are made from the same monomer (ethylene). Explain how the difference in chain structure leads to these different applications. 4 MARKS

✏️ Answer in your book

8. Nylon-6,6 has a higher melting point (265°C) than polyethylene (~130°C for HDPE), despite both being addition-type synthetic polymers. Explain the molecular basis for this difference, referencing IMF types. 3 MARKS

✏️ Answer in your book

✅ Comprehensive Answers

🔬 Activity 1

1. (a) Monomer: CH₂=CHCH₃ (propylene/propene). (b) Addition polymerisation — the repeat unit has no ester or amide linkages; the monomer had a C=C double bond. (c) PP and PE would have similar MPs (both ~130°C) because both have only dispersion forces between chains. The methyl group in PP adds slightly more electrons per repeat unit → marginally stronger dispersion forces → PP is generally slightly stiffer and has a slightly higher MP than LDPE, but similar to HDPE. The key difference is physical: PP is stiffer due to greater chain rigidity from the methyl substituent.

2. (a) Condensation polymerisation. (b) Amide linkage: –CO–NH– (connecting a carbonyl C to a nitrogen). (c) Water (H₂O) — released when –COOH reacts with –NH₂: –COOH + H₂N– → –CO–NH– + H₂O.

3. Nylon has N–H bonds in its amide linkages (–CO–NH–). The N–H group can form hydrogen bonds (N–H···O=C) with adjacent nylon chains because N is electronegative enough to create H-bonding. These hydrogen bonds are far stronger than the dispersion-only forces between PE chains. Overcoming nylon's H-bonds requires more energy → higher MP (265°C vs ~130°C).

📊 Activity 2

A: HDPE's linear chains can pack closely together in a regular arrangement → high packing density → higher physical density. Close packing also maximises the surface area contact between chains → stronger total dispersion forces → chains are harder to separate → higher tensile strength. LDPE's branches prevent close packing: side chains physically block neighbouring chains from approaching → larger gaps between chains → lower density and weaker total dispersion forces → lower tensile strength.

B: UHMWPE has the same monomer and negligible branching as HDPE, but the chains are far longer (molecular weight ~3–6 million g/mol vs ~50,000 for HDPE). Longer chains → more total IMF contact area along each chain → much stronger total adhesion between chains even though the force per unit area is the same (dispersion only) → exceptional tensile strength. Injection moulding difficulty: very long entangled chains have extremely high viscosity even when melted — the polymer barely flows under typical moulding pressures, making processing very difficult.

❓ Multiple Choice

1. B — Addition polymerisation requires C=C. Bifunctional monomers with –COOH, –OH, or –NH₂ are for condensation polymerisation.

2. C — Ester formation (diol + dicarboxylic acid): –OH + HOOC– → –COO– + H₂O. Water is always the byproduct for diol/diacid condensation.

3. A — Insolubility + resistance to heating = covalent cross-links. Long chains alone still dissolve (HDPE does dissolve in hot organic solvents). H-bonds and dispersion forces are broken by solvents.

4. D — Plasticisers are small molecules that wedge between PVC chains, disrupting dipole-dipole interactions between chain segments → chains can slide past each other more easily → more flexible. They don't break covalent bonds.

5. B — The –CO–NH– linkage is an amide bond = condensation polymerisation product from –COOH + –NH₂ groups → releases H₂O. This is Nylon-6 (one monomer type: 6-aminohexanoic acid or caprolactam ring-opening).

📝 Short Answer Model Answers

Q6 (5 marks): Addition polymerisation: monomer requires a C=C double bond (alkene); the double bond opens and adjacent monomers form new C–C single bonds in a chain reaction; products are only the polymer — no byproduct (100% atom economy); example: polyethylene from ethylene (CH₂=CH₂) (2 marks). Condensation polymerisation: monomer must be bifunctional (two reactive functional groups, e.g. –COOH and –OH, or –COOH and –NH₂); functional groups react to form ester or amide linkages; byproduct (usually H₂O) is released at each junction; products are polymer + small molecule; example: nylon-6,6 from hexamethylenediamine + adipic acid, releasing H₂O (3 marks).

Q7 (4 marks): LDPE has highly branched chains (1 mark). Branches prevent adjacent chains from packing closely → less dense, fewer chain–chain contacts → weaker total dispersion forces between chains (1 mark) → flexible and easy to stretch → suitable for thin flexible bags. HDPE has linear chains with minimal branching (1 mark). Linear chains pack closely in an ordered arrangement → high density → greater chain–chain contact area → stronger total dispersion forces → rigid and strong (1 mark) → suitable for pressure pipes.

Q8 (3 marks): Nylon contains amide linkages (–CO–NH–); the N–H groups can form hydrogen bonds (N–H···O=C) with the carbonyl oxygen on adjacent nylon chains (1 mark). Hydrogen bonding is significantly stronger than dispersion forces (the only IMF in polyethylene, which has only C–H groups) (1 mark). More energy must be supplied to overcome nylon's hydrogen bonds during melting → higher melting point (265°C vs ~130°C) (1 mark).

Mark lesson as complete

Tick when you've finished all activities and checked your answers.

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