Summary — What This Chapter Covers
- Polyesters form from a diol and a dicarboxylic acid (or dioyl chloride), or from a hydroxycarboxylic acid.
- Polyamides form from a diamine and a dicarboxylic acid (or dioyl chloride), from an aminocarboxylic acid, or from amino acids.
- You can draw the repeat unit from given monomers, and work back from a polymer section to its monomers.
- Addition polymers come from C=C monomers with no by-product; condensation polymers lose H2O or HCl.
- Poly(alkenes) are inert and hard to biodegrade; some polymers break down in light; polyesters and polyamides are hydrolysed by acid or alkali.
1. Condensation Polymerisation
35.1
Polyesters
In condensation polymerisation, monomers join and a small molecule (H2O or HCl) is eliminated at each link.
(a) Diol + dicarboxylic acid (or dioyl chloride)
Ethane-1,2-diol and benzene-1,4-dicarboxylic acid give the polyester PET (Terylene):
Repeat unit of PET
n HOCH2CH2OH + n HOOC–C6H4–COOH → [–OCH2CH2O–CO–C6H4–CO–]n + 2n H2O
Using the dioyl chloride, ClOC–C6H4–COCl, gives the same polymer but eliminates HCl instead of water, and reacts faster.
(b) A hydroxycarboxylic acid
One monomer carries both groups. 2-hydroxypropanoic acid (lactic acid), CH3CH(OH)COOH, gives poly(lactic acid), repeat unit [–OCH(CH3)CO–], with water eliminated.
Polyamides
(a) Diamine + dicarboxylic acid (or dioyl chloride)
Hexane-1,6-diamine and hexanedioic acid give nylon-6,6:
Repeat unit of nylon-6,6
n H2N(CH2)6NH2 + n HOOC(CH2)4COOH → [–NH(CH2)6NH–CO(CH2)4CO–]n + 2n H2O
With hexanedioyl dichloride, ClOC(CH2)4COCl, HCl is eliminated instead.
(b) An aminocarboxylic acid
6-aminohexanoic acid, H2N(CH2)5COOH, gives nylon-6, repeat unit [–NH(CH2)5CO–].
(c) Amino acids
Amino acids join by peptide (amide) bonds to give polypeptides and proteins, e.g. [–NHCH2CO–NHCH(CH3)CO–] from glycine and alanine.
Repeat units and monomers
Monomers → repeat unit: remove –OH from each COOH (or –Cl from each COCl) and –H from each OH or NH2; join the ends with an ester (–COO–) or amide (–CONH–) link; show open bonds at both ends.
Polymer → monomers: find the ester or amide links, break them, and add back –OH to the C=O side and –H to the O or N side (this is the same as hydrolysis).
Worked Example
Draw the repeat unit of the polymer formed from butane-1,4-diol, HO(CH2)4OH, and propanedioyl dichloride, ClOCCH2COCl. What small molecule is eliminated?
Worked Example
A section of polymer is –NH–C6H4–NH–CO–C6H4–CO–NH–C6H4–NH–CO–C6H4–CO–. Identify the monomers.
Practice Question
What monomer forms the polymer [–OCH2CH2CH2CO–]n?
2. Predicting the Type of Polymerisation
35.2
Addition or condensation?
| Addition polymerisation | Condensation polymerisation |
|---|
| Monomer |
| By-product |
| Examples |