Unit 4 · Spec 3.3.12

Condensation Polymers (Polyesters & Polyamides)

Condensation polymerisation, Terylene (PET) polyester, Nylon 6,6 and Kevlar polyamides, repeating unit identification, hydrolysis, and polymer disposal for OxfordAQA A-Level Chemistry.

1. Condensation vs Addition Polymerisation

Polymers are giant macromolecules formed by linking thousands of small monomer units together:

Feature Addition Polymerisation (AS Unit 2) Condensation Polymerisation (A2 Unit 4)
Monomer Requirements Monomers contain a carbon-carbon double bond (C=C alkenes). Each monomer has two functional groups that react: either two monomers each with two identical groups (e.g. a diacid with a diol or a diamine), or one monomer with two different groups (e.g. an amino acid).
Byproducts Formed Zero byproducts: 100% atom economy. A small molecule is eliminated at each linkage (usually H₂O or HCl).
Polymer Backbone Non-polar, inert carbon-carbon chain (-C-C-C-C-). Polar chain containing ester (-COO-) or amide (-CONH-) linkages.
Biodegradability Non-biodegradable (chemically inert C-C bonds resist bacterial and enzymatic attack). Biodegradable (polar linkages can be slowly hydrolysed by water and environmental enzymes).

2. Polyesters: Terylene (PET)

Polyesters are formed when dicarboxylic acids react with diols (or from hydroxycarboxylic acids containing both groups):

Terylene (Poly(ethylene terephthalate) / PET) Synthesised from benzene-1,4-dicarboxylic acid and ethane-1,2-diol with the elimination of water:
n HOOC-C₆H₄-COOH + n HO-CH₂CH₂-OH → -[CO-C₆H₄-CO-O-CH₂CH₂-O]ₙ- + 2n H₂O
Applications: Synthetic textile clothing fibres (Terylene/Dacron), plastic beverage bottles, and food packaging trays.

3. Condensation Polymer Repeating Units Diagram

Each repeating unit has an open bond at each end. The ester or amide link is boxed. To find the monomers, break the link and add back -OH to the C=O and -H to the O or N.

Repeating units of Terylene, nylon 6,6, Kevlar and a polypeptide C O C O O CH₂ CH₂ O n ester link Terylene (PET), a polyester from benzene-1,4-dicarboxylic acid and ethane-1,2-diol N H (CH₂)₆ N H C O (CH₂)₄ C O n amide link Nylon 6,6, a polyamide from hexanedioic acid and hexane-1,6-diamine N H N H C O C O n amide link Kevlar, a polyamide from benzene-1,4-dicarboxylic acid and benzene-1,4-diamine N H C H R C O N H C H R′ C O peptide (amide) link Polypeptide (protein) chain from amino acids; R and R′ are the side chains of each amino acid

4. Polyamides: Nylon 6,6 and Kevlar

Polyamides are formed when dicarboxylic acids (or acyl dichlorides) react with diamines (or from amino acids):

1. Nylon 6,6

Formed from hexanedioic acid (6 carbons) and hexane-1,6-diamine (6 carbons):

n HOOC-(CH₂)₄-COOH + n H₂N-(CH₂)₆-NH₂ → -[CO-(CH₂)₄-CO-NH-(CH₂)₆-NH]ₙ- + 2n H₂O

The amide links (-CONH-) form extensive intermolecular hydrogen bonds between adjacent polymer chains, giving Nylon high tensile strength, elasticity, and abrasion resistance (ropes, textiles, carpets).

2. Kevlar (High-Performance Aromatic Polyamide)

Synthesised from benzene-1,4-dicarboxylic acid and benzene-1,4-diamine:

n HOOC-C₆H₄-COOH + n H₂N-C₆H₄-NH₂ → -[CO-C₆H₄-CO-NH-C₆H₄-NH]ₙ- + 2n H₂O

Why Kevlar is Exceptionally Strong

  • The planar benzene rings make the polymer chains extremely rigid and linear.
  • These straight chains pack together tightly in parallel sheets.
  • Immense numbers of regular intermolecular hydrogen bonds form between C=O and N-H groups on neighbouring chains.
  • This produces an extraordinarily high tensile strength-to-weight ratio: Kevlar is 5 times stronger than steel on an equal-weight basis (used in bulletproof vests, aerospace composites, and puncture-resistant tyres).

Forces Between Chains

  • Polyamides (nylon 6,6 and Kevlar): hydrogen bonds form between the N-H of one chain and the C=O of a neighbouring chain.
  • Polyesters (Terylene): there is no N-H or O-H in the chain, so there is no hydrogen bonding between chains. The polar C=O groups give permanent dipole-dipole forces between chains.

Uses

Polymer Uses
TeryleneClothing fibres, and bottles (as PET)
NylonRopes and clothing
KevlarBulletproof vests and helmets. Its strength comes from strong hydrogen bonds between the chains.

5. Polymer Hydrolysis & Disposal

Hydrolysis of Condensation Polymers

Unlike addition polymers, polyesters and polyamides can be chemically cleaved by hydrolysis under reflux:

  • Acid Hydrolysis (hot aqueous HCl):
    Polyester → original dicarboxylic acid + diol
    Polyamide → original dicarboxylic acid + diamine salt (e.g. Cl⁻ ⁺H₃N-(CH₂)₆-NH₃⁺ Cl⁻)
  • Alkaline Hydrolysis (hot aqueous NaOH):
    Polyester → dicarboxylate salt (e.g. Na⁺⁻OOC-R-COO⁻Na⁺) + diol
    Polyamide → dicarboxylate salt + diamine

Biodegradability

  • Polyesters and polyamides can be hydrolysed at the ester or amide link. So they are biodegradable.
  • Polyalkenes (addition polymers such as poly(ethene)) have only non-polar C-C and C-H bonds. There is nothing for water or nucleophiles to attack. So they are not biodegradable.

Disposal of Polymers

Method Advantages Disadvantages
Landfill Cheap. Uses land. Non-biodegradable polymers last for centuries.
Incineration Reduces the volume of waste. Energy from burning can be recovered. Makes CO₂. Toxic gases (for example HCl from PVC) must be removed.
Recycling Saves raw materials and energy. Sorting is costly. The recycled polymer can be lower quality.

6. Worked Repeating Unit Calculations

Worked Example 1: Deducing Monomers from a Polyamide Structure
Identify the two monomers required to synthesise the polymer with repeating unit:
-[CO-(CH₂)₈-CO-NH-(CH₂)₄-NH]ₙ-

Step 1: Locate the amide linkages:

Break the -CO-NH- bonds to restore the monomer functional groups.

Step 2: Add -OH to carbonyl groups to reform carboxylic acids:

HO-CO-(CH₂)₈-CO-OH = decanedioic acid (10 carbons).

Step 3: Add -H to nitrogen atoms to reform amines:

H-NH-(CH₂)₄-NH-H = butane-1,4-diamine (4 carbons).

Worked Example 2: Drawing a Biodegradable Polyester Repeating Unit
Polylactic acid (PLA) is a biodegradable polyester made from 2-hydroxypropanoic acid, CH₃-CH(OH)-COOH.
Draw the repeating unit of PLA.

Step 1: Identify reacting functional groups:

The single monomer contains both a carboxyl group (-COOH) and an alcohol group (-OH).

Step 2: Eliminate H₂O:

Loss of -OH from -COOH and -H from the alcohol group links monomers via ester bonds.

Step 3: Draw repeating unit with trailing bonds:

-[O-CH(CH₃)-CO]ₙ-

Go deeper

These UK AQA pages cover the same ideas in more detail. They may include content that is not in 9620.

7. Practice Questions

Question 1 (2 marks)
Draw the repeating unit of the polyester made from ethane-1,2-diol and hexanedioic acid.
Show Answer & Mark Scheme

Answer: -[O-CH₂CH₂-O-CO-(CH₂)₄-CO]ₙ-

  • Ester links correct, with the C=O drawn (1).
  • Correct repeating unit with an open bond at each end (1).
Question 2 (2 marks)
Name the strongest type of force between chains in (a) nylon 6,6 and (b) Terylene.
Show Answer & Mark Scheme

Answer:

  • (a) Hydrogen bonds, between N-H and C=O on neighbouring chains (1).
  • (b) Permanent dipole-dipole forces (1).
Question 3 (3 marks)
Explain why nylon 6,6 is biodegradable but poly(ethene) is not.
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Answer:

  • Nylon 6,6 has amide links, which can be hydrolysed (1).
  • Poly(ethene) has only non-polar C-C and C-H bonds (1).
  • There is nothing for water or nucleophiles to attack (1).
Question 4 (2 marks)
Give one advantage and one disadvantage of disposing of polymers by incineration.
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Answer:

  • Advantage: reduces the volume of waste, or energy can be recovered (1).
  • Disadvantage: makes CO₂, or toxic gases such as HCl (from PVC) must be removed (1).