Condensation polymerisation: core idea

· Condensation polymerisation = many monomers join to form a polymer, with elimination of a small molecule.
· The small molecule is usually H₂O when using –COOH, –OH, –NH₂ groups.
· The small molecule is usually HCl when using acyl chlorides / dioyl chlorides.
· Monomers must have two reactive functional groups so the chain can keep growing.
· Main CIE examples: polyesters and polyamides.
· Key linkages to recognise: ester linkage, –COO– and amide linkage, –CONH–.

This diagram shows monomers joining by condensation to form a polyester chain. It highlights that a small molecule is eliminated as each new linkage forms. Use it to visualise how repeating ester links build up the polymer. Source

Polyesters

· Polyesters contain repeated ester linkages, –COO–.
· Route 1: diol + dicarboxylic acid → polyester + water.
· General idea: HO–R–OH + HOOC–R′–COOH → [–O–R–OOC–R′–CO–]ₙ + H₂O.
· Route 2: diol + dioyl chloride → polyester + hydrogen chloride.
· General idea: HO–R–OH + ClOC–R′–COCl → polyester + HCl.
· Route 3: hydroxycarboxylic acid → polyester + water.
· A hydroxycarboxylic acid has both –OH and –COOH in the same molecule, so it can self-condense.
· Exam focus: be able to draw the repeat unit from a given diol + dicarboxylic acid/dioyl chloride or from a hydroxycarboxylic acid.

This shows the standard CIE route for forming a polyester from a diacid and a diol. The repeat unit contains an ester linkage between the two monomer residues. It is useful for practising how to remove H and OH to form water. Source

This diagram shows a real polyester example, PET, formed by condensation polymerisation. It helps connect the general polyester repeat unit to a named polymer structure. Focus on the repeated –COO– ester linkages. Source

Polyamides

· Polyamides contain repeated amide linkages, –CONH–.
· Route 1: diamine + dicarboxylic acid → polyamide + water.
· General idea: H₂N–R–NH₂ + HOOC–R′–COOH → [–NH–R–NHCO–R′–CO–]ₙ + H₂O.
· Route 2: diamine + dioyl chloride → polyamide + hydrogen chloride.
· General idea: H₂N–R–NH₂ + ClOC–R′–COCl → polyamide + HCl.
· Route 3: aminocarboxylic acid → polyamide + water.
· Route 4: amino acids → polyamides/polypeptides + water.
· In amino acids, the –NH₂ group of one amino acid reacts with the –COOH group of another to form a peptide bond, which is an amide bond.
· Exam focus: identify –CONH– as the key linkage and use it to deduce the original monomer(s).

This is a clear polyamide example showing formation of nylon-6,6. The important feature is the repeated amide linkage, –CONH–. It is useful for seeing how a diamine and dicarboxylic acid build a polymer chain. Source

This diagram shows amino acids joining by condensation to form peptide bonds. Peptide bonds are amide linkages, so polypeptides count as polyamides. It is useful for recognising amino acid polymerisation in exam structures. Source

Deducing repeat units from monomers

· For a diol + dicarboxylic acid, remove H from each –OH group and OH from each –COOH group to form H₂O.
· For a diol + dioyl chloride, remove H from –OH and Cl from –COCl to form HCl.
· For a diamine + dicarboxylic acid, remove H from –NH₂ and OH from –COOH to form H₂O.
· For a diamine + dioyl chloride, remove H from –NH₂ and Cl from –COCl to form HCl.
· Put the remaining atoms into square brackets with n outside: [repeat unit]ₙ.
· The repeat unit must show the correct linkage atoms: –COO– for polyesters, –CONH– for polyamides.
· Do not include H₂O or HCl inside the repeat unit.

Identifying monomers from a condensation polymer

· First identify the linkage: –COO– = polyester, –CONH– = polyamide.
· For a polyester, split each ester linkage between C–O.
· Restore the monomers by adding –OH to the carbonyl side and H to the oxygen side.
· This gives either a diol + dicarboxylic acid or a hydroxycarboxylic acid.
· For a polyamide, split each amide linkage between C–N.
· Restore the monomers by adding –OH to the carbonyl side and H to the nitrogen side.
· This gives either a diamine + dicarboxylic acid, an aminocarboxylic acid, or amino acids.
· If the original monomer could be a dioyl chloride, replace –COOH with –COCl and the eliminated molecule is HCl instead of H₂O.

This shows how a diacid and diamine form a polyamide chain. The diagram is useful for practising how to reverse the process and identify monomers from a polymer section. Look for the repeated –CONH– link. Source

Common exam traps

· Do not confuse addition and condensation polymerisation: condensation forms a small molecule by-product.
· Do not draw monomers with only one functional group unless the molecule also has another reactive group elsewhere.
· Do not forget the second functional group on diols, diamines, dicarboxylic acids and dioyl chlorides.
· Do not reverse ester/amide linkages incorrectly: polyester = –COO–, polyamide = –CONH–.
· Do not include extra atoms from eliminated H₂O or HCl in the repeat unit.
· Do not miss amino acids: amino acids form polyamides because they form peptide/amide bonds.