Condensation polymersEdexcel A-Level Chemistry: Revision notes

In condensation polymerisation, monomers join in a chain and a small molecule (usually water, or HCl if acyl chlorides are used) is eliminated each time a link forms. Each monomer needs two functional groups, so the chain can grow at both ends. The monomers are either two different molecules (each with two identical groups) or one molecule with two different groups.

Compare addition polymerisation: unsaturated monomers (alkenes) open their C=C bonds, and the polymer is the only product.

A polyester forms from a dicarboxylic acid and a diol. The –COOH of one monomer reacts with the –OH of another to form an ester link, –COO–, and water.

PET: benzene-1,4-dicarboxylic acid + ethane-1,2-diol → –[OC–C₆H₄–COO–CH₂CH₂–O]ₙ– + water. A hydroxy acid, such as 2-hydroxypropanoic acid (lactic acid), can polymerise on its own to make PLA, because it has both an –OH and a –COOH group.

A polyamide forms from a dicarboxylic acid and a diamine. The –COOH of one monomer reacts with the –NH₂ of another to form an amide link, –CONH–, and water.

Nylon-6,6: hexanedioic acid, HOOC(CH₂)₄COOH, + 1,6-diaminohexane, H₂N(CH₂)₆NH₂ → –[CO(CH₂)₄CONH(CH₂)₆NH]ₙ– + water. The numbers show the carbons in each monomer. A single monomer with an amine and a carboxylic acid group, such as 6-aminohexanoic acid, makes nylon-6.

Amino acids, H₂N–CHR–COOH, have an amine and a carboxylic acid group, so they condense with one another to form polypeptides (proteins), with amide (peptide) links. Glycine, H₂NCH₂COOH, gives the repeat unit –NH–CH₂–CO–, and a general amino acid gives –NH–CHR–CO–. Water is eliminated at each link.

To draw a repeat unit:

1. Identify the functional groups on the monomers.
2. Remove the –OH from the acid group and the –H from the –OH or –NH₂ group (together they form the water).
3. Join the remaining parts, then draw one repeat unit in brackets with open bonds at both ends.

Worked example: ethane-1,2-diol and hexanedioic acid give –O–CH₂CH₂–O–CO–(CH₂)₄–CO–. Diacid + diamine: –CO–R–CO–NH–R′–NH–. Amino acid: –NH–CHR–CO–.

To find the monomers, locate the ester or amide link, break it, and add –OH back to the carbonyl side and –H to the O or N side.