4.2 - Condensation Polymers
Condensation polymerisation
- A condensation polymer is a polymer formed when monomers join together with the loss of a small molecule, usually water.
- Each monomer carries two functional groups - One at each end - so it can react at both ends to continue the polymer chain.
- A functional group is the atom or group of atoms that decides how a molecule reacts.
How the monomers are arranged
- Two different monomers, each carrying two of the same functional group. The two monomers alternate along the chain.
- A single monomer carrying two different functional groups (one at each end) polymerising head to tail.
Water is formed as a product
- Every time two monomer ends join, a molecule of water is lost.
- Condensation polymerisation produces two products: the polymer and water.
- The repeat unit contains fewer atoms than the monomers, because atoms leave to form water.
Addition and condensation polymerisation compared
| Addition polymerisation | Condensation polymerisation | |
|---|---|---|
| Monomers | One kind, with a carbon-carbon double bond (C=C) | Usually two kinds, each with two of the same functional group; or one kind with two different groups |
| What reacts | The C=C double bond opens | A functional group at one end reacts with a functional group on the next monomer |
| Products | One product: the polymer | Two products: the polymer and water |
| Repeat unit | Contains all the atoms of the monomer | Contains fewer atoms than the monomers, because water is lost |
Ester and amide repeat units
Monomers for condensation polymerisation carry two functional groups:
- Carboxylic acid group (-COOH) - Two per molecule makes a dicarboxylic acid.
- Alcohol group (-OH) - Two per molecule makes a diol.
- Amine group (-NH2) - Two per molecule makes a diamine.
A dicarboxylic acid with a diol makes a polyester. A dicarboxylic acid with a diamine makes a polyamide.
Block equations
Because only the functional groups react, the carbon chains between them can be represented as blocks:
n HOOC-▭-COOH + n HO-▭-OH → [-CO-▭-CO-O-▭-O-]n + 2n H2O
n HOOC-▭-COOH + n H2N-▭-NH2 → [-CO-▭-CO-NH-▭-NH-]n + 2n H2O
Building the repeat unit
- Write the two monomers side by side - Acid on the left, the second monomer on the right, with reacting ends facing each other.
- Remove the -OH from the acid end and an H from the other monomer (the H from the diol's -OH, or one H from the diamine's -NH2). These leave as a molecule of water.
- Join what's left. The carbon from the -COOH bonds to the atom the hydrogen left:
- To the oxygen of a diol, giving -CO-O- (an ester link, forming a polyester).
- To the nitrogen of a diamine, giving -CO-NH- (an amide link, forming a polyamide).
- Repeat at the two outer ends - Another water molecule is lost, leaving open bonds pointing outwards to show the chain continues.
- Put brackets round the repeat unit with the open bonds passing through them, and write a subscript n outside the closing bracket.
- Add the balancing numbers - Write n in front of each monomer and 2n in front of H2O.
For a polyester:
n HOOC-CH2-CH2-COOH + n HO-CH2-CH2-OH → [-CO-CH2-CH2-CO-O-CH2-CH2-O-]n + 2n H2O
For a polyamide:
n HOOC-CH2-CH2-COOH + n H2N-CH2-CH2-NH2 → [-CO-CH2-CH2-CO-NH-CH2-CH2-NH-]n + 2n H2O
Working back to the monomers
- Find the links in the repeat unit: -CO-O- for a polyester, or -CO-NH- for a polyamide.
- Break each link between the carbon and the heteroatom (the single-bonded O or N).
- Add back the water atoms - Add an -OH to the C=O group (rebuilding -COOH), and an -H to the oxygen (rebuilding -OH) or nitrogen (rebuilding -NH2).
- Do the same to the outer ends, then remove the brackets and the n.
Worked example - A polyester's repeat unit
A dicarboxylic acid, HOOC-CH2-COOH, is polymerised with a diol, HO-CH2-CH2-CH2-OH. Write the repeat unit of the polyester and the equation for the reaction.
Step 1 - Write the monomers side by side
Put the acid on the left and the diol on the right, so the acid's right-hand -COOH faces the diol's left-hand -OH.
The coloured -OH and -H are the ones that leave as water.
Step 2 - Remove water at the facing ends
Take the -OH off the facing acid end and the H off the facing -OH of the diol to form water.
Step 3 - Join what's left
The carbon from the -COOH bonds to the oxygen from the diol to form the ester link, -CO-O-:
HOOC-CH2-CO-O-CH2-CH2-CH2-OH
Step 4 - Do the same at the two outer ends
Remove the -OH from the left-hand end of the acid and the H from the right-hand end of the diol as a second water molecule, leaving open bonds at both ends.
Step 5 - Bracket the repeat unit
Place brackets around the repeat unit with the open bonds passing through them, and add the subscript n.
Step 6 - Add the balancing numbers and write the equation
Write n before each monomer and 2n before H2O:
Repeat unit: [-CO-CH2-CO-O-CH2-CH2-CH2-O-]n
Equation: n HOOC-CH2-COOH + n HO-CH2-CH2-CH2-OH → [-CO-CH2-CO-O-CH2-CH2-CH2-O-]n + 2n H2O