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SAChemistrySyllabus dot point

How do monomers join to form addition and condensation polymers?

Distinguish addition and condensation polymerisation, identify monomers and repeating units, and relate polymer structure to properties.

Addition versus condensation polymerisation, identifying monomers and repeating units, the chemistry of polyesters and polyamides, structure-property links, and worked SACE-style repeating-unit and degree-of-polymerisation problems.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-02

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What this dot point is asking
Lead worked calculation
Addition polymerisation
Condensation polymerisation
Comparing the two
Structure determines properties
Recycling and degradation
Why it matters

What this dot point is asking

SACE expects you to distinguish the two mechanisms, identify monomers and repeating units, and explain how chain structure determines polymer properties.

Lead worked calculation

Finding the repeating unit and degree of polymerisation

A sample of poly(ethene) has an average molar mass of $2.80 \times 10^{4}\ \text{g mol}^{-1}$ . Identify the monomer, draw the repeating unit, and find the average number of monomers per chain. ( $M(\text{C}_2\text{H}_4) = 28.0\ \text{g mol}^{-1}$ .)

Identify the monomer and repeating unit

The monomer is ethene, $\text{CH}_2=\text{CH}_2$ . In addition polymerisation the double bond opens, so the repeating unit is $-[\text{CH}_2-\text{CH}_2]-$ (drawn with bonds extending out on each side and an $n$ subscript).

Find the degree of polymerisation

Each repeating unit has the same mass as the monomer ( $28.0\ \text{g mol}^{-1}$ , since nothing is lost), so:

\text{degree of polymerisation} = \frac{M(\text{polymer})}{M(\text{repeating unit})} = \frac{2.80 \times 10^{4}}{28.0} = 1.00 \times 10^{3}

Conclude

The chain contains about $1000$ ethene units. For an addition polymer the repeating-unit mass equals the monomer mass; for a condensation polymer you would subtract the mass of the small molecule lost.

Addition polymerisation

To draw the repeating unit, take the monomer, open the $\text{C}=\text{C}$ into a single bond, and show bonds extending from each end into the rest of the chain, all inside brackets with subscript $n$ .

Condensation polymerisation

The linkage chemistry is the same as esterification and amide formation: condensation polymers form ester or amide bonds exactly as a single ester or amide would, just repeated thousands of times.

Comparing the two

Structure determines properties

Polymer properties follow from chain structure and the forces between chains.

Chain length: longer chains have stronger total intermolecular forces, raising melting point and strength.
Branching: unbranched chains (HDPE) pack closely, giving a denser, stronger, more crystalline material; branched chains (LDPE) pack loosely, giving a softer, more flexible plastic.
Cross-linking: covalent bridges between chains give rigid, heat-resistant thermosets that do not melt.
Polar groups: ester and amide linkages allow hydrogen bonding between chains (nylon, polyester fibres), increasing strength and raising melting point compared with non-polar addition polymers.

Recycling and degradation

Condensation polymers contain ester or amide linkages that can be hydrolysed, so polyesters and nylons can be chemically broken back into monomers and are more biodegradable. Addition polymers have inert carbon-carbon backbones with no easily hydrolysed linkage, making them very durable but slow to break down, a key environmental drawback.

Why it matters

Polymers are everywhere, from packaging and fibres to biological macromolecules. Understanding the two mechanisms explains why some plastics are inert and persistent while others can be hydrolysed and recycled, and it connects directly to the condensation chemistry of proteins, carbohydrates and lipids.

Exam-style practice questions

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

SACE 20225 marksPropene (

\text{CH}_2=\text{CHCH}_3

) polymerises to poly(propene). (a) Draw the repeating unit and write the type of polymerisation. (b) Explain why no small molecule is lost. (c) A sample of poly(propene) has an average molar mass of

1.26 \times 10^{5}\ \text{g mol}^{-1}

; calculate the average number of monomer units per chain. (

M(\text{C}_3\text{H}_6) = 42.0\ \text{g mol}^{-1}

Show worked answer →

(a) The repeating unit is $-[\text{CH}_2-\text{CH}(\text{CH}_3)]-$ shown with continuation bonds; this is addition polymerisation. (2 marks)

(b) In addition polymerisation the $\text{C}=\text{C}$ double bonds simply open and the monomers add together; no atoms are removed, so no small molecule (such as water) is lost. (1 mark)

(c) Degree of polymerisation $= \dfrac{1.26 \times 10^{5}}{42.0} = 3.00 \times 10^{3}$ monomer units per chain. (2 marks)

SACE 20205 marksNylon-6,6 is formed from a diamine and a dicarboxylic acid. (a) State the type of polymerisation and name the small molecule released. (b) Name the type of linkage formed between monomers. (c) Compare condensation polymerisation with addition polymerisation in terms of the monomers required and whether a by-product forms.

Show worked answer →

(a) Condensation polymerisation; the small molecule released is water. (2 marks)

(b) The linkage formed is an amide (peptide) linkage, $\text{-CONH-}$ . (1 mark)

(c) Condensation polymerisation needs monomers with two reactive functional groups each (e.g. two $\text{-COOH}$ or two $\text{-NH}_2$ / $\text{-OH}$ ) and releases a small molecule (water) each time a bond forms. Addition polymerisation needs monomers with a $\text{C}=\text{C}$ double bond and releases no by-product, the double bond simply opens. (2 marks)