Condensation Polymerisation: Monomers and Repeating Units Revision Notes | A Level AQA Chemistry

1. What Makes Condensation Polymerisation Different?

You already know that addition polymerisation joins alkene monomers together without losing any atoms — the monomer and the repeating unit have the same molecular formula. Condensation polymerisation works differently: every time two monomers join, a small molecule is expelled. That expelled molecule is almost always water, though hydrogen chloride is lost in some industrial processes.

The consequence of losing atoms at each step is that the repeating unit is smaller than the combined monomers. This is the first thing examiners test — students who confuse addition and condensation polymers often draw repeating units that are too large. The key diagnostic question to ask yourself is: has a small molecule been lost? If yes, you are dealing with condensation polymerisation.

For condensation polymerisation to produce a long chain, each monomer must carry two reactive functional groups — one at each end. This bifunctionality is what allows the chain to grow in both directions. A monomer with only one functional group would terminate the chain; a monomer with three or more would cause branching or cross-linking.

2. The Three Monomer Combinations You Must Know

AQA specifies three distinct monomer pairings that produce condensation polymers. Two of them produce polyesters or polyamides from separate monomer types; the third uses a single bifunctional monomer. Getting these combinations right is the foundation for every drawing question on this topic.

Notice that both polyamide routes produce the same linkage group. The difference is whether you need two different monomers (dicarboxylic acid + diamine) or one monomer that already carries both functional groups (an amino acid with –NH$_2$ at one end and –COOH at the other).

A diol carries two hydroxyl groups (–OH), a diamine carries two amine groups (–NH$_2$), and an amino acid carries one of each (–NH$_2$ and –COOH). Each of these functional groups is capable of reacting with its complementary partner to form the linkage and expel water.

3. Polyesters: Terylene and the Ester Linkage

Terylene (also known as PET, poly(ethylene terephthalate)) is the classic AQA polyester example. It is made from two monomers: benzene-1,4-dicarboxylic acid (terephthalic acid) and ethane-1,2-diol. The carboxylic acid group of one monomer reacts with the hydroxyl group of the other, releasing water and forming an ester linkage –COO–.

The reaction at each bond-forming step can be summarised as: –COOH + HO– → –COO– + H$_2$O. The ester linkage is the bridge between the two monomer fragments in the repeating unit. When you draw the repeating unit, you must show this –COO– group explicitly — it is the mark-winning feature that distinguishes a polyester from any other polymer type.

The repeating unit of Terylene contains the benzene ring from terephthalic acid, the two –COO– ester linkages (one at each end of the dicarboxylic acid fragment), and the –CH$_2$CH$_2$– fragment from ethane-1,2-diol. The brackets of the repeating unit are drawn through the bonds that connect one unit to the next, with the letter n outside the bracket to indicate repetition.

Worked example: Drawing the Repeating Unit of Terylene from Monomers

Benzene-1,4-dicarboxylic acid and ethane-1,2-diol react to form Terylene. Draw the repeating unit and identify the linkage group.

1. Monomer 1: HOOC–C$_6$H$_4$–COOH (benzene-1,4-dicarboxylic acid) Monomer 2: HO–CH$_2$CH$_2$–OH (ethane-1,2-diol) — Identify the functional groups: two –COOH on monomer 1, two –OH on monomer 2. Both are bifunctional, so the chain can grow in both directions.
2. –COOH + HO– → –COO– + H$_2$O — One –COOH reacts with one –OH. Water is the small molecule eliminated. The ester linkage –COO– is formed.
3. –OC–C$_6$H$_4$–CO–O–CH$_2$CH$_2$–O– — Remove one H from each –OH and one OH from each –COOH (these combine to form H$_2$O). What remains is the repeating unit fragment.
4. $\left[–OC–C_6H_4–CO–O–CH_2CH_2–O–\right]_n$ — The brackets are drawn through the bonds at each end of the repeating unit. The subscript n indicates the unit repeats along the chain. Two ester linkages (–COO–) are visible per repeat unit.
5. Ester linkage: –COO– (or written as –OC(=O)–) — This is the defining feature of a polyester. Examiners expect you to name this linkage explicitly in 'describe' or 'identify' questions.

Answer: Repeating unit: $\left[–OC–C_6H_4–CO–O–CH_2CH_2–O–\right]_n$. The linkage is the ester group –COO–. Two ester linkages are present per repeating unit.

4. Polyamides: Nylon 6,6, Kevlar, and the Amide Linkage

Polyamides are formed when a carboxylic acid group reacts with an amine group: –COOH + H$_2$N– → –CONH– + H$_2$O. The amide linkage –CONH– is the defining feature. You will encounter two named polyamides at A-level: nylon 6,6 (made from two separate monomers) and Kevlar (also made from two separate monomers, but with a benzene ring in each).

Nylon 6,6 is synthesised from hexanedioic acid (a six-carbon dicarboxylic acid) and hexane-1,6-diamine (a six-carbon diamine). The name '6,6' refers to the six carbon atoms in each monomer. Each bond-forming step joins one –COOH to one –NH$_2$, releasing water and creating an amide linkage. The repeating unit therefore contains one fragment from each monomer, joined by an amide group at each end.

Worked example: Drawing the Repeating Unit of Nylon 6,6

Hexanedioic acid (HOOC–(CH$_2$)$_4$–COOH) and hexane-1,6-diamine (H$_2$N–(CH$_2$)$_6$–NH$_2$) react to form nylon 6,6. Draw the repeating unit.

1. Hexanedioic acid: –COOH at each end (dicarboxylic acid) Hexane-1,6-diamine: –NH$_2$ at each end (diamine) — Both monomers are bifunctional. The –COOH and –NH$_2$ groups are the reactive partners.
2. –COOH + H$_2$N– → –CO–NH– + H$_2$O — Water is eliminated. The amide linkage –CONH– is formed. This happens at both ends of each monomer as the chain grows.
3. From –COOH: remove –OH From –NH$_2$: remove –H These combine to give H$_2$O — What remains of the dicarboxylic acid fragment is –CO–(CH$_2$)$_4$–CO–. What remains of the diamine fragment is –NH–(CH$_2$)$_6$–NH–.
4. $\left[–CO–(CH_2)_4–CO–NH–(CH_2)_6–NH–\right]_n$ — The repeating unit contains two amide linkages (–CONH–), one from each end of the dicarboxylic acid reacting with the diamine. Brackets are drawn through the bonds at each end.

Answer: Repeating unit of nylon 6,6: $\left[–CO–(CH_2)_4–CO–NH–(CH_2)_6–NH–\right]_n$. The linkage is the amide group –CONH–.

Kevlar is structurally similar to nylon 6,6 in that it is also a polyamide, but its monomers both contain benzene rings: benzene-1,4-dicarboxylic acid (terephthalic acid) and benzene-1,4-diamine (1,4-diaminobenzene). The rigid benzene rings in every repeating unit, combined with the hydrogen bonding between amide groups on adjacent chains, give Kevlar its exceptional tensile strength — it is used in bullet-resistant vests and high-performance composites.

5. Polyamides from Amino Acids

The third monomer combination is the most elegant: a single amino acid monomer that already carries both a carboxylic acid group (–COOH) and an amine group (–NH$_2$). Because both reactive groups are present on the same molecule, only one type of monomer is needed. The –COOH of one amino acid reacts with the –NH$_2$ of the next, releasing water and forming an amide linkage — exactly the same chemistry as nylon 6,6, but with a single monomer species.

The repeating unit from an amino acid monomer H$_2$N–CHR–COOH (where R is the side chain) is written as $\left[–NH–CHR–CO–\right]_n$. Notice that the –NH– comes from the amine end and the –CO– comes from the carboxylic acid end, with water (one H from –NH$_2$ and the –OH from –COOH) eliminated at each step. This is the same pattern as nylon 6,6 — the only difference is that both functional groups originate from the same molecule.

Worked example: Drawing the Repeating Unit from an Amino Acid Monomer

An amino acid has the structure H$_2$N–CH(CH$_3$)–COOH. Draw the repeating unit of the polyamide it forms.

1. –NH$_2$ at one end; –COOH at the other end Side chain R = –CH$_3$ — The molecule is bifunctional. The –NH$_2$ of one monomer will react with the –COOH of the next.
2. –COOH + H$_2$N– → –CO–NH– + H$_2$O — Identical chemistry to nylon 6,6. Water is eliminated; the amide linkage forms.
3. From –COOH: remove –OH From –NH$_2$: remove –H Remaining fragment: –NH–CH(CH$_3$)–CO– — The H from –NH$_2$ and the OH from –COOH leave together as H$_2$O. The central carbon and its side chain are retained in full.
4. $\left[–NH–CH(CH_3)–CO–\right]_n$ — Brackets are placed through the bonds at each end of the repeating unit. The amide linkage –NHCO– (or –CONH– depending on direction of writing) is clearly visible.

Answer: Repeating unit: $\left[–NH–CH(CH_3)–CO–\right]_n$. The linkage is the amide group –CONH–.

6. Working Backwards: Identifying Monomers from a Polymer Section

AQA exam questions frequently show you a section of a polymer chain and ask you to identify the monomers. This is the reverse of drawing the repeating unit, and it requires you to locate the linkage group, break the bond at that linkage, and restore the atoms that were lost as water. Think of it as 'adding water back' across the linkage.

For an ester linkage –COO–: break the C–O bond and add –OH to the acid fragment (restoring –COOH) and –H to the alcohol fragment (restoring –OH). For an amide linkage –CONH–: break the C–N bond and add –OH to the acid fragment (restoring –COOH) and –H to the amine fragment (restoring –NH$_2$). The atoms you add back are always H and OH — the components of water.

Worked example: Identifying Monomers from a Polymer Chain Section

A section of a polymer chain contains the fragment: –OC–C$_6$H$_4$–CO–NH–(CH$_2$)$_6$–NH–. Identify the monomers and state the polymer type.

1. The linkage is –CO–NH– (amide linkage) — The presence of –CONH– confirms this is a polyamide. If the linkage were –COO–, it would be a polyester.
2. Break the C–N bond in –CO–NH– Fragment 1: –OC–C$_6$H$_4$–CO– Fragment 2: –NH–(CH$_2$)$_6$–NH– — Cut at the amide bond. You now have two fragments, each missing the atoms that were lost as water during polymerisation.
3. Fragment 1: add –OH to each –CO– end → HOOC–C$_6$H$_4$–COOH Fragment 2: add –H to each –NH– end → H$_2$N–(CH$_2$)$_6$–NH$_2$ — Adding –OH restores the carboxylic acid group; adding –H restores the amine group. These are the atoms that left as water (H + OH = H$_2$O) during polymerisation.
4. Monomer 1: benzene-1,4-dicarboxylic acid Monomer 2: hexane-1,6-diamine Polymer type: polyamide (specifically Kevlar if the diamine also contained a benzene ring, but here it is an aliphatic diamine) — The two monomers are a dicarboxylic acid and a diamine — the second AQA monomer combination. The polymer is a polyamide.

Answer: Monomers: benzene-1,4-dicarboxylic acid and hexane-1,6-diamine. Polymer type: polyamide. The linkage is the amide group –CONH–.

Drawing questions are the most common format for this topic. The mark scheme rewards three things: the correct atoms in the repeating unit, the correct linkage group explicitly shown, and the brackets with subscript n in the right position. Missing any one of these typically costs a mark. Always draw the linkage group in full — do not abbreviate –CONH– as a line or label.

7. Exam Technique: Drawing, Naming, and Evaluating Condensation Polymers

When a question asks you to 'identify the linkage' in a polymer, name it precisely: 'ester linkage' or 'amide linkage', and write the structural formula –COO– or –CONH– respectively. Writing 'ester bond' without the structural formula may not score the mark if the question asks you to 'draw' or 'show'. Read the command word carefully.