What are addition polymers?

Addition polymerisation of four HSC alkene monomers A reaction scheme showing ethene, chloroethene, styrene and tetrafluoroethene monomers each with a carbon to carbon double bond opening under heat, pressure and catalyst to form the repeat unit of polyethylene, PVC, polystyrene and PTFE respectively, each drawn in brackets with subscript n. CH2=CH2 (ethene) -(CH2-CH2)n- polyethylene

What is chain length?

Longer chains give greater dispersion forces overall, higher melting point and stronger material. Industrial polymers are typically 1000 to 10,000 monomer units long.

What is functional groups in the chain?

Hydrogen-bond-capable groups (amide, hydroxyl) raise melting point and tensile strength considerably. Halogen substituents add dipole-dipole forces. Aromatic rings add rigidity.

What is crystallinity?

Regular, regularly-spaced chains can crystallise (form ordered regions); irregular chains stay amorphous. Crystalline regions are stronger and more dense. HDPE is about 90% crystalline; LDPE only about 50%.

§-Syllabus dot point

NSWChemistrySyllabus dot point

Inquiry Question 7: How are addition and condensation polymers made and how do their structures determine their uses?

Investigate the structural formulae, properties, formation and uses of addition polymers (polyethylene, polyvinyl chloride, polystyrene, polytetrafluoroethylene) and condensation polymers (nylon, polyester)

A focused answer to the HSC Chemistry Module 7 dot point on polymers. The addition polymerisation of alkenes to make polyethylene, PVC, polystyrene and PTFE, the condensation polymerisation of diacid plus diamine (nylon) and diacid plus diol (polyester), structure-property relationships, and worked HSC past exam questions.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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In plain English

Think of a monomer as a single LEGO brick. Addition polymerisation is like snapping identical bricks together in a long row: each brick (an alkene with a $C=C$ "clip") just clicks onto the next, nothing is left over, and you get one continuous chain, whether that is polyethylene bags, PVC pipe, polystyrene cups or Teflon coating. Condensation polymerisation is different: you need TWO types of brick, each with connectors on both ends (a diacid and a diamine for nylon, or a diacid and a diol for polyester), and every time two bricks click together a small "spacer" molecule, water, pops out and is thrown away. The polymer's real-world behaviour then comes down to how tidily the finished chains stack: straight, unbranched chains (HDPE, drawn nylon fibre) pack together like a neat stack of firewood and feel rigid and strong, while tangled, branched chains (LDPE) pack loosely like a pile of sticks and feel soft and flexible. Extra "stickiness" between chains, hydrogen bonding in nylon and polyester, aromatic rings in polystyrene and PET, makes the material tougher again.

What this dot point is asking

NESA wants you to draw the structures of common polymers, write polymerisation equations using either an addition (one monomer with $C=C$ , no byproduct) or condensation (two monomers with reactive groups at each end, releases water) mechanism, and explain how the chain architecture and intermolecular forces between chains determine real-world properties.

The answer

Two mechanisms, two polymer families

Feature	Addition polymerisation	Condensation polymerisation
Monomer	one type, contains $C=C$	two types, each with two reactive groups
Byproduct	none	small molecule (usually water)
Backbone	C-C only	C-C plus amide or ester linkages
Examples	polyethylene, PVC, polystyrene, PTFE	nylon, polyester (PET)

Spotting addition vs condensation on sight

One monomer with a $C=C$ , no byproduct written on the equation: addition polymerisation. The repeat unit is enclosed in brackets with subscript $n$ outside; every monomer atom appears in the product.
Two different monomers, each drawn with a reactive group at BOTH ends (diacid + diamine, or diacid + diol), plus $nH_2O$ on the product side: condensation polymerisation.
Linkage in the backbone tells you the family: an amide linkage $-CONH-$ means a polyamide (nylon); an ester linkage $-COO-$ means a polyester (PET).
Degree of polymerisation $n = \dfrac{\text{average molar mass of polymer}}{\text{molar mass of repeat unit (or monomer, for addition)}}$ .

Addition polymers

The $\pi$ bond of an alkene opens; the carbons join in a long chain. The general scheme:

nCH_2=CHX \rightarrow -(CH_2-CHX)_n-

The square-bracketed unit is the repeat unit. Every atom of the monomer ends up in the polymer; nothing is lost.

Polyethylene (PE), from ethene $CH_2=CH_2$ . The simplest polymer. Two grades:

LDPE (low-density): branched chains, free-radical catalysis at high pressure. Flexible, used for plastic bags, squeeze bottles, cling film.
HDPE (high-density): linear unbranched chains, Ziegler-Natta catalyst at low pressure. Rigid, used for milk bottles, piping, hard hats.

The same monomer produces dramatically different materials because chain architecture controls packing and intermolecular forces.

Polyvinyl chloride (PVC), from chloroethene $CH_2=CHCl$ :

nCH_2=CHCl \rightarrow -(CH_2-CHCl)_n-

The chlorine atoms make PVC denser and stronger than PE, and the C-Cl dipole adds dipole-dipole forces on top of dispersion. PVC is rigid in pure form (pipes, window frames) and softened to flexible form with plasticisers (cables, hoses, vinyl flooring).

Polystyrene (PS), from styrene $CH_2=CHC_6H_5$ (phenylethene):

nCH_2=CHC_6H_5 \rightarrow -(CH_2-CH(C_6H_5))_n-

The pendant phenyl rings make the polymer rigid and brittle. PS is used for plastic cutlery, CD cases, and (blown with $CO_2$ or pentane) as expanded polystyrene foam (cups, packaging).

Polytetrafluoroethylene (PTFE, Teflon), from tetrafluoroethene $CF_2=CF_2$ :

nCF_2=CF_2 \rightarrow -(CF_2-CF_2)_n-

The C-F bonds are very strong and the fluorine shield is chemically inert. PTFE is heat-resistant up to 260 degrees C, has a very low coefficient of friction (used as non-stick coating), and resists almost all chemicals.

Condensation polymers

Two monomers, each with two reactive groups, react head-to-tail-to-head-to-tail. A small molecule (water) is expelled at each linkage.

Nylon 6,6 (polyamide), from hexane-1,6-diamine and hexanedioic acid:

nH_2N(CH_2)_6NH_2 + nHOOC(CH_2)_4COOH \rightarrow -[NH(CH_2)_6NH-CO(CH_2)_4CO]_n- + 2nH_2O

The repeat unit contains two amide bonds. The "6,6" refers to the carbon count in each monomer (6 in the diamine, 6 in the diacid). The amide $N-H$ and $C=O$ groups hydrogen bond between adjacent chains, giving nylon high tensile strength, toughness and a high melting point (about 265 degrees C). Used for textiles (stockings, climbing ropes), engineering plastics (gears, bearings), and fishing line.

Polyester (PET, polyethylene terephthalate), from ethane-1,2-diol and benzene-1,4-dicarboxylic acid (terephthalic acid):

nHO-CH_2CH_2-OH + nHOOC-C_6H_4-COOH \rightarrow -[O-CH_2CH_2-O-CO-C_6H_4-CO]_n- + 2nH_2O

The repeat unit contains two ester linkages. The aromatic rings make PET rigid and dimensionally stable; the polymer can be drawn into strong fibres or blown into bottles. Used for soft-drink bottles, polyester clothing, packaging films.

Structure-property relationships

Chain length: Longer chains give greater dispersion forces overall, higher melting point and stronger material. Industrial polymers are typically 1000 to 10,000 monomer units long.
Branching: Linear chains pack closely (HDPE, drawn nylon fibre); branched chains pack loosely (LDPE). Closer packing means more dispersion force contact and higher density.
Functional groups in the chain: Hydrogen-bond-capable groups (amide, hydroxyl) raise melting point and tensile strength considerably. Halogen substituents add dipole-dipole forces. Aromatic rings add rigidity.
Crystallinity: Regular, regularly-spaced chains can crystallise (form ordered regions); irregular chains stay amorphous. Crystalline regions are stronger and more dense. HDPE is about 90% crystalline; LDPE only about 50%.
Crosslinking: Covalent bonds between adjacent chains turn a thermoplastic into a thermoset (vulcanised rubber, epoxy). Not usually examined at HSC but worth a mention.

Fully worked polymer questions

Worked example 1: writing the addition polymerisation equation for PVC.

Monomer: chloroethene, $CH_2=CHCl$ . Locate the $C=C$ and identify the substituent (Cl) that will hang off the backbone.
Open the $\pi$ bond and join monomers head to tail: $n\,CH_2=CHCl \rightarrow -(CH_2-CHCl)_n-$ .
Check: repeat unit in brackets, subscript $n$ outside, no byproduct on the product side.

Marker's note: full marks require the bracket-and-subscript- $n$ convention; drawing a short chain of only two or three repeat units without brackets loses the structural-formula mark.

Worked example 2: identifying the polymer family and monomers from a repeat unit.

Given repeat unit: $-[O-CH_2CH_2-O-CO-C_6H_4-CO]_n-$ .
Locate the recurring linkage: two $-CO-O-$ (ester) groups per repeat unit, so this is a condensation polyester.
Split the repeat unit at each ester linkage to recover the two monomers: $HO-CH_2CH_2-OH$ (ethane-1,2-diol) and $HOOC-C_6H_4-COOH$ (benzene-1,4-dicarboxylic acid).
Name the polymer: polyethylene terephthalate (PET).

Marker's note: markers reward correctly identifying the linkage type FIRST (this fixes addition vs condensation and amide vs ester), then correctly reverse-engineering both monomers; naming only one monomer or missing the diacid's aromatic ring loses marks.

Worked example 3: explaining a property from structure.

Claim to explain: "PTFE has an extremely low coefficient of friction and is chemically inert."
Structural reasoning: the carbon backbone is completely surrounded by fluorine atoms. C-F bonds are among the strongest single bonds in organic chemistry and fluorine's high electronegativity and small size let the fluorine atoms form a dense, unreactive shield around the backbone.
Link structure to property: this fluorine shield prevents other molecules (acids, bases, organic solvents, or a frying pan's food surface) from interacting with the carbon backbone, giving chemical inertness and a slippery, non-stick surface with very low intermolecular attraction to other materials.

Marker's note: a full-mark structure-to-property answer names the specific structural feature (the fluorine shield / C-F bond strength) and explicitly links it to the named property, rather than stating the property alone.

Examples in context

Example 1. Polyethene at Qenos Botany. The Qenos plant at Botany Industrial Park produces both low-density and high-density polyethene from ethene feed cracked at the adjacent ethylene plant. LDPE forms in high-pressure (around 2000 atm) free-radical polymerisation, giving branched chains used in NSW supermarket plastic bags. HDPE forms with a Ziegler-Natta catalyst at low pressure, giving linear chains with higher crystallinity used in milk bottles and pipes. The chemistry is identical at the monomer level: $n CH_2=CH_2 \rightarrow [-CH_2-CH_2-]_n$ , no byproduct. The HSC distinction between addition mechanism and structural outcome (branched vs linear) explains why one Qenos product is soft and the other rigid.

Example 2. Lycra and the NSW activewear supply chain. Activewear sold by Lorna Jane and 2XU outlets in NSW uses Lycra (spandex), a block copolymer containing both polyurea hard segments (amide-like) and polyether soft segments (ether). Each amide bond forms by condensation of a diisocyanate with a diamine with no net water loss but the same condensation principle. The HSC framework for condensation polymers (loss of small molecule, repeating amide linkage) generalises to this commercial example. Students predicting the monomer of a polymer from a fragment of its repeat unit are doing exactly the analysis a textile chemist does when reverse-engineering a competitor's fibre.

Try this

Q1. Distinguish between addition and condensation polymerisation, naming one example polymer of each. [3 marks]

Cue. Addition: alkene monomer, no byproduct (PE); condensation: difunctional monomers, water lost per bond (nylon).

Q2. Calculate the average degree of polymerisation $n$ of a polyethene chain with average molar mass $5.6 \times 10^4$ g mol $^{-1}$ . [2 marks]

Cue. Monomer mass 28.05 g mol $^{-1}$ ; $n = 5.6 \times 10^4 / 28.05 = 2000$ .

Q3. Polyester (PET) is made from ethane-1,2-diol and benzene-1,4-dicarboxylic acid. (a) Draw the repeat unit. (b) Explain why PET fibre has high tensile strength. (c) State one environmental consequence of PET use in NSW. [2+2+1 marks]

Cue. (a) $-OCH_2CH_2OOC{-}C_6H_4{-}CO-$ . (b) Linear chains with strong intermolecular forces (hydrogen bonding to ester carbonyl, dispersion forces between aromatic rings). (c) Non-biodegradable; landfill burden or recycling load.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC4 marksCompare the formation of polyethylene with the formation of nylon 6,6 in terms of monomer type, reaction mechanism, byproducts, and one structural feature of the resulting polymer.

Show worked answer →

A 4 mark answer needs both polymerisation equations and a comparison across four dimensions.

Polyethylene formation. Monomer is ethene $CH_2=CH_2$ (one type, contains a $C=C$ ). Mechanism is addition polymerisation: under pressure and heat with an initiator (or Ziegler-Natta catalyst), the $\pi$ bond opens and successive monomers add to a growing chain. No byproducts are formed; every atom of the monomer ends up in the polymer.

nCH_2=CH_2 \xrightarrow{\text{catalyst, heat, pressure}} -(CH_2-CH_2)_n-

Structural feature: a saturated hydrocarbon backbone, no functional groups, non-polar.

Nylon 6,6 formation. Monomers are two types: hexane-1,6-diamine $H_2N(CH_2)_6NH_2$ and hexanedioic acid $HOOC(CH_2)_4COOH$ . Each monomer has two reactive groups, so chains can grow at both ends. Mechanism is condensation polymerisation: the amine attacks the acid carbonyl and water is lost at each bond formed.

nH_2N(CH_2)_6NH_2 + nHOOC(CH_2)_4COOH \rightarrow -[NH(CH_2)_6NHCO(CH_2)_4CO]_n- + 2nH_2O

Structural feature: amide linkages along the chain, capable of hydrogen bonding between strands.

Summary table:

	Polyethylene	Nylon 6,6
Monomer	one ( $CH_2=CH_2$ )	two (diamine + diacid)
Mechanism	addition	condensation
Byproduct	none	water
Linkage	C-C	amide (CONH)

Markers reward (1) addition vs condensation, (2) one vs two monomer types, (3) no byproduct vs water, (4) the linkage difference with hydrogen bonding.

2019 HSC3 marksExplain why high-density polyethylene (HDPE) is stronger and more rigid than low-density polyethylene (LDPE), even though both are made of the same monomer.

Show worked answer →

Both HDPE and LDPE are polymers of ethene $CH_2=CH_2$ , but they differ in chain architecture, which controls packing and intermolecular forces.

HDPE is made with a Ziegler-Natta or metallocene catalyst at low pressure (about 5 atm) and low temperature (about 60 degrees C). The chains are linear and unbranched. Linear chains pack closely together, allowing strong dispersion forces to act over a large contact area. Result: high density (about 0.95 g/mL), high tensile strength, rigid, opaque, high softening temperature.

LDPE is made with a free-radical initiator at high pressure (about 1500 atm) and high temperature (about 200 degrees C). Random branching occurs through chain transfer reactions. The branched chains cannot pack closely, leaving more space between them. Dispersion forces are weaker because contact area is smaller. Result: lower density (about 0.92 g/mL), lower tensile strength, flexible, translucent, lower softening temperature.

Markers reward (1) linear vs branched chain architecture, (2) packing efficiency and contact area, (3) linking dispersion force strength to mechanical properties.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksFor the monomer

CH_2=CHCl

: (a) name the monomer and the resulting polymer; (b) write the balanced polymerisation equation using the bracket-and-subscript-n convention; (c) state whether a byproduct forms.

Show worked solution →

(a) Names. Monomer: chloroethene (vinyl chloride). Polymer: polyvinyl chloride (PVC).

(b) Equation.

nCH_2=CHCl \rightarrow -(CH_2-CHCl)_n-

(c) Byproduct. None. This is addition polymerisation; every atom of the monomer appears in the polymer.

Marking criteria: 1 mark for both correct names, 1 mark for the correctly bracketed equation with subscript $n$ , 1 mark for correctly stating no byproduct forms.

foundation3 marksState the two monomers used to make nylon 6,6, name the type of polymerisation, and identify the linkage found in the repeat unit.

Show worked solution →

Monomers: Hexane-1,6-diamine, $H_2N(CH_2)_6NH_2$ , and hexanedioic acid, $HOOC(CH_2)_4COOH$ .
Polymerisation type: Condensation polymerisation (two monomer types, each with two reactive end groups, water lost at each bond).
Linkage: Amide linkage, $-CONH-$ , formed between the amine and the carboxylic acid group.

Marking criteria: 1 mark for both correct monomer names, 1 mark for correctly naming condensation polymerisation, 1 mark for correctly identifying the amide linkage.

core5 marksA sample of nylon 6,6 has an average molar mass of

2.26 \times 10^4\ \text{g mol}^{-1}

. The repeat unit

-[NH(CH_2)_6NHCO(CH_2)_4CO]-

has molar mass

226.32\ \text{g mol}^{-1}

. Calculate the average degree of polymerisation,

n

, to 3 significant figures, and state what this number represents.

Show worked solution →

Step 1: write the degree-of-polymerisation formula.

n = \frac{\text{average molar mass of polymer}}{\text{molar mass of repeat unit}}

Step 2: substitute the given values.

n = \frac{2.26 \times 10^4\ \text{g mol}^{-1}}{226.32\ \text{g mol}^{-1}}

Step 3: compute.

n = 99.858...

Step 4: round to 3 significant figures (matching the 3 s.f. of $2.26 \times 10^4$ ).

n = 99.9

Interpretation. On average, each nylon 6,6 chain in the sample contains about 99.9 (approximately 100) repeat units joined end to end; because polymer chains vary in length, this is an AVERAGE degree of polymerisation, not an exact count for every chain.

Marking criteria: 1 mark for the correct formula, 1 mark for correct substitution, 1 mark for the correct unrounded value, 1 mark for rounding to 3 significant figures with no units (n is a pure ratio), 1 mark for correctly interpreting n as an average, not an exact, chain length.

core4 marksA polymer sample has the repeat unit

-[NH(CH_2)_6NHCO(CH_2)_4CO]-

. (a) Identify the polymer family and name the polymer. (b) Draw or describe the two monomers that combine to form it. (c) State the byproduct and its stoichiometric coefficient relative to n repeat units.

Show worked solution →

(a) Family and name: The repeat unit contains two amide linkages ( $-CONH-$ ), so this is a condensation polyamide: nylon 6,6 (the "6,6" reflects 6 carbons in each monomer).
(b) Monomers: Splitting the repeat unit at each amide bond recovers hexane-1,6-diamine, $H_2N(CH_2)_6NH_2$ , and hexanedioic acid, $HOOC(CH_2)_4COOH$ .
(c) Byproduct: Water, $H_2O$ ; for $n$ repeat units formed from $n$ molecules of each monomer, $2n$ molecules of water are released (one water molecule per amide bond, two amide bonds per repeat unit).

Marking criteria: 1 mark for correctly identifying nylon 6,6 as a condensation polyamide, 1 mark for both correct monomer names/structures, 1 mark for correctly identifying water as the byproduct, 1 mark for the correct stoichiometric coefficient $2n$ .

core4 marksThe rate vs time graph below is an owned illustrative plot of the mass of water released per minute during the melt condensation of hexanedioic acid with hexane-1,6-diamine, tracked over 60 minutes. The rate is highest in the first 10 minutes, falls steadily, and approaches zero by about 50 minutes. Explain this shape in terms of the condensation mechanism and monomer concentration, and predict the effect on the reaction rate of removing water vapour continuously instead of letting it accumulate.

Show worked solution →

Explaining the shape. At the start, the diamine and diacid monomers are at their highest concentration, so collisions between the two reactive end groups (amine and carboxylic acid) are most frequent, giving the fastest rate of water release. As the reaction proceeds, monomers are consumed and converted into longer chains; the concentration of unreacted end groups available to collide falls, so the rate of new amide-bond formation (and hence water release) steadily decreases. By about 50 minutes, most accessible end groups have reacted, so the rate approaches zero even though the condensation is reversible and never completely stops.

Effect of continuously removing water vapour. Condensation polymerisation to form nylon is an equilibrium process at the level of each individual bond (amide bond formation is reversible with hydrolysis). By Le Chatelier's principle, continuously removing the water product as it forms shifts the equilibrium at every linkage further towards bond formation, increasing both the overall rate of net polymer formation and the final average chain length (degree of polymerisation) compared with allowing water to accumulate in the reaction vessel.

Marking criteria: 1 mark for linking the initial high rate to high monomer/end-group concentration, 1 mark for linking the falling rate to falling end-group concentration as monomers are consumed, 1 mark for correctly applying Le Chatelier's principle to removing a product, 1 mark for correctly predicting an increase in both rate and average chain length.

exam6 marksCompare polyethylene and nylon 6,6 in terms of monomer type, polymerisation mechanism, and the dominant intermolecular force between chains, and use this comparison to explain why nylon fibre is used for climbing rope while polyethylene is used for plastic bags.

Show worked solution →

Band 6 plan. State the three-point structural comparison (monomer/mechanism/IMF) explicitly, then link EACH difference forward to the named application, ending with an explicit judgement tying strength requirements to intermolecular force.

Model answer.

Polyethylene is formed by addition polymerisation of a single monomer, ethene ( $CH_2=CH_2$ ), producing a purely hydrocarbon backbone with no polar functional groups; the only intermolecular force between chains is dispersion forces, which are relatively weak and depend heavily on chain packing (branched LDPE chains, used for bags, pack loosely and rely on weak dispersion forces alone, giving a soft, flexible, low-strength material suited to a lightweight, disposable application).

Nylon 6,6 is formed by condensation polymerisation of two monomers, hexane-1,6-diamine and hexanedioic acid, each with reactive end groups that link with loss of water to form amide bonds along the backbone. The amide group's $N-H$ and $C=O$ can hydrogen bond to the corresponding groups on an adjacent chain. Hydrogen bonding is substantially stronger than dispersion forces alone, and when nylon chains are drawn (aligned) into a fibre, these hydrogen bonds form efficiently along the whole chain length, giving very high tensile strength.

This IMF difference explains the application choice directly: climbing rope must resist a large tensile (pulling) force without snapping, which requires the strong, directional hydrogen bonding available only to nylon's amide backbone; a plastic bag only needs to hold its shape and resist tearing under light load, a demand fully met by polyethylene's cheaper dispersion-force-only structure. Using nylon for bags would be needlessly expensive, and using polyethylene for climbing rope would be dangerous, as its intermolecular forces cannot provide the tensile strength required for life-safety equipment.

Marker's note: top-band answers (1) name both monomers and both mechanisms correctly, (2) explicitly name the dominant IMF for EACH polymer (dispersion vs hydrogen bonding), not just "stronger forces", and (3) connect the IMF difference to the SPECIFIC mechanical demand of each named application, rather than restating that one polymer is "stronger" without saying why that matters for that use.