Part 1: Condensation polymers and repeat-unit construction

Build repeat units from functional-group reactions and reverse the process to recover the monomers.

Two reactive ends allow a chain to keep growing

Condensation polymerisation joins monomers while eliminating small molecules. To produce a long linear chain, each monomer must supply two appropriately reactive functional groups overall. A diol can react with a dicarboxylic acid to form a polyester; a diamine can react with a dicarboxylic acid to form a polyamide.

The functional groups react by the same chemistry as small-molecule ester or amide formation. Each new link leaves another reactive end available, allowing further joining. A molecule with only one suitable group would tend to cap a chain rather than extend it indefinitely in this simple linear model.

A single monomer type can also polymerise if it contains both required groups: a hydroxycarboxylic acid can form a polyester and an amino acid can form a polyamide. Do not assume that condensation always requires two different monomers.

Worked construction: keep every atom in the backbone

Take HO–CH₂–CH₂–OH and HOOC–CH₂–CH₂–COOH. Join an OH of the diol to an acid’s carbonyl carbon by ester formation. A convenient repeat unit is [–O–CH₂–CH₂–O–C(=O)–CH₂–CH₂–C(=O)–]ₙ. Read across a bracket boundary to verify another ester link is formed.

The ester link is –C(=O)–O–. Retain the single-bonded oxygen supplied by the diol; removing it would create a different backbone. Both open bonds must cross the brackets to show continuation. The n belongs outside, and terminal H/OH groups are not inserted into every repeat unit.

To recover the monomers, break each acyl–oxygen connection conceptually and restore H to the alcohol oxygen and OH to the acid carbonyl carbon. Then check that each reconstructed monomer has two functional groups and the correct carbon skeleton.

Replace the oxygen linkage with the correct nitrogen linkage

With H₂N–(CH₂)₄–NH₂ and HOOC–(CH₂)₂–COOH, a repeat unit is [–NH–(CH₂)₄–NH–C(=O)–(CH₂)₂–C(=O)–]ₙ. The linkage is –C(=O)–NH– and contains no extra oxygen between carbonyl carbon and nitrogen.

For the single monomer H₂N–CH₂–COOH, condensation gives [–NH–CH₂–C(=O)–]ₙ. Each amino-acid residue contributes one nitrogen and one carbonyl group. Draw two adjacent repeats to check that the boundary is an amide link.

Dicarboxylic acid chlorides can replace the dicarboxylic acids in the corresponding polymer-forming reactions. With diols or diamines they form the same types of links but eliminate HCl rather than water. Identify the actual leaving groups before naming the small molecule.

Distinguish a repeat-unit equation from an exact finite chain

Each ester or amide link made from a carboxylic acid eliminates one H₂O. A finite unbranched chain assembled from m monomer molecules contains m − 1 links, provided it forms one chain without rings. For example five bifunctional monomer molecules joined into one chain form four links and eliminate four waters.

An idealised polymer equation often uses repeat-unit notation and omits end-group detail. Do not use that shorthand to claim an exact small-molecule count for a finite chain when a question explicitly asks about its ends. Count the connections shown.

To distinguish addition from condensation, inspect both monomers and linkage formation. Addition of an alkene opens C=C and retains all monomer atoms in the backbone without a small-molecule by-product. Condensation uses functional groups and eliminates a small molecule in the reactions studied here.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

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