Condensation Polymerisation

In condensation polymerisation, monomers join through their functional groups, and a small molecule is eliminated at every link (most often water, sometimes HCl when an acyl chloride is used).

Crucially, each monomer must carry two reactive groups — one at each end — so the chain can keep growing in both directions. Two families are examined:

• A diol (two –OH) + a dicarboxylic acid (two –COOH) → a polyester (ester linkages), losing H₂O each link.
• A diamine (two –NH₂) + a dicarboxylic acid → a polyamide (amide linkages), losing H₂O each link.

For example, a polyester forms as: $n\,\text{HO–R–OH} + n\,\text{HOOC–R'–COOH} \rightarrow [\text{–O–R–O–CO–R'–CO–}]_n + 2n\,\text{H}_2\text{O}$

This is the key contrast with addition polymerisation (from C=C), where the monomers simply add together with no atoms lost.

A repeat unit is the smallest section that repeats along the chain, drawn with a bond extending through each bracket to show it continues.

Polyester linkage — the ester group –CO–O– joins each pair of monomers: $\text{[–O–CH}_2\text{CH}_2\text{–O–CO–C}_6\text{H}_4\text{–CO–]}_n \quad (\text{e.g. Terylene/PET})$

Polyamide linkage — the amide group –CO–NH– joins each pair (this is the same linkage as the peptide bonds in proteins): $\text{[–NH–(CH}_2)_6\text{–NH–CO–(CH}_2)_4\text{–CO–]}_n \quad (\text{nylon-6,6})$

When you draw a repeat unit, make sure the continuation bonds pass through the brackets and that one small molecule's worth of atoms has been removed at the junction — not the whole –OH/–H from both ends.

Exams often show a section of a condensation polymer and ask for the monomers. The method:

1. Find the linkages — ester (–CO–O–) or amide (–CO–NH–).
2. Break the chain at each linkage (between the C of C=O and the O/N).
3. Add back the eliminated atoms: put –OH on the carbonyl carbon (→ –COOH) and –H on the O or N (→ –OH or –NH₂).

This recovers a dicarboxylic acid and a diol (polyester) or a diamine (polyamide). Because each monomer is bi-functional, you will get two different monomers (or one monomer that contains both groups, e.g. an amino acid).

The same logic in reverse confirms which small molecule is lost: every new linkage releases one H₂O (or HCl if an acyl-chloride monomer was used).