2.4.6 (HL)—Condensation polymers
- Syllabus
- First assessment 2025
- Objective
- 2.4.6
- Level
- HL
Condensation polymerization joins difunctional monomers and releases a small molecule. Polyamides and polyesters form through the corresponding functional-group linkages.
Use both functional groups to connect the monomers, show continuation bonds, and account for the eliminated small molecule. A diol plus dicarboxylic acid forms ester links; a diamine plus dicarboxylic acid forms amide links, –CO–NH–, with water eliminated. Check that the repeat unit contains residues from both monomers, that no unreacted end group is trapped inside the repeat, and that the linkage plus by-product conserves every atom.
For a diol and a dicarboxylic acid, join –OH and –COOH groups to make ester links and release water at each new link. The repeat unit must contain residues from both monomers with continuation bonds through the functional links; do not leave unreacted end groups inside the repeat.
Representative question
Nylon 6,6 is formed by condensation polymerisation.
Deduce the structures of the two monomers that form the polyamide nylon 6,6.
H2 N(CH2)6NH2HOOC(CH2)4COOH/ClOC(CH2)4COCl
Marking guidance:
Accept any type of structural or skeletal formula.
Penalize clear incorrect bond connections only once in the paper.
Retrieve the pathway: locate bonding contributions, connect them to material properties, distinguish alloy lattice disruption, and construct addition or condensation polymer repeating units from monomer evidence.
Check that the bonding description matches the material, the property explanation names the structural cause, and every polymer substituent, continuation bond, and released small molecule is represented.