OxfordAQA A-Level Chemistry: Polymers (9620)

This study guide teaches section 3.3.12, Polymers, from OxfordAQA’s International AS and A-level Chemistry specification (9620), Version 5.3, for International AS and A-level exams from May/June 2020, and spans 3.3.12.1 Condensation polymers and 3.3.12.2 Biodegradability and disposal of polymers. The specification places this section in the International A2 organic content, so all of it is International A-level only. You meet it in Unit 4 (Organic 2 and Physical 2), whose organic content runs from 3.3.7 to 3.3.16, while Unit 5 (practical and synoptic) may test anything in the course.

Pair this guide with the polymers revision notes and the polymers practice questions. The alkenes study guide deals with addition polymers (3.3.4.3), and the three kinds of intermolecular force are in shapes, polarity and intermolecular forces. See also the course hub, the printable checklist and the free diagnostics.

Section 3.3.12 at a glance

Section	Outcome
3.3.12.1	Know the three monomer combinations that give condensation polymers: diacid with diol, diacid with diamine, and amino acids
3.3.12.1	Repeating units and linkages of polyesters such as Terylene, and of polyamides such as nylon 6,6 and Kevlar
3.3.12.1	Typical uses of these polymers
3.3.12.1	Go from monomers to the repeating unit, from a chain section to the repeating unit, and from a chain section back to the monomers
3.3.12.1	Explain the nature of the intermolecular forces between condensation polymer molecules
3.3.12.2	Polyalkenes are chemically inert and non-biodegradable
3.3.12.2	Polyesters and polyamides are biodegradable because hydrolysis splits their links; explain why polyalkenes resist hydrolysis
3.3.12.2	Advantages and disadvantages of different disposal methods, including recycling

Every row is International A-level only. Section 3.3.12 lists no practical activity of its own.

3.3.12.1 How condensation polymers form

A condensation polymer forms when monomers that each carry two reactive groups join end to end. Every new bond releases a small molecule, usually water. Because each monomer reacts at both ends, the chain keeps growing in both directions.

Compare addition polymers (3.3.4.3), where a C=C bond opens and the polymer is the only product. A condensation polymer’s backbone contains repeated linkages with O or N in them.

The specification names three monomer combinations:

Linkage formed	Made from	Family
–COO– (ester)	a diol with a diacid	polyester
–CONH– (amide)	a diamine with a diacid	polyamide
–CONH– (amide)	amino acids alone	polyamide

The named polymers

In the formulas below, C₆H₄ stands for a benzene ring bonded at positions 1 and 4, so the chain runs straight through the ring.

• Terylene (a polyester) comes from benzene-1,4-dicarboxylic acid, HOOC–C₆H₄–COOH, and ethane-1,2-diol, HO–CH₂CH₂–OH. Repeating unit: –[CO–C₆H₄–CO–O–CH₂CH₂–O]–.
• Nylon 6,6 (a polyamide) comes from hexanedioic acid, HOOC–(CH₂)₄–COOH, and hexane-1,6-diamine, H₂N–(CH₂)₆–NH₂. Each monomer has six carbon atoms, which is where the name comes from. Repeating unit: –[CO–(CH₂)₄–CO–NH–(CH₂)₆–NH]–.
• Kevlar (a polyamide) pairs the same diacid as Terylene with benzene-1,4-diamine, H₂N–C₆H₄–NH₂. Repeating unit: –[CO–C₆H₄–CO–NH–C₆H₄–NH]–.

Typical uses. Terylene is spun into fibres for clothing and is moulded into drinks bottles. Nylon 6,6 makes clothing, carpets and ropes. Kevlar is used where great strength for low mass matters, such as body armour and bulletproof vests.

Drawing the repeating unit from monomers

1. Take OH from a –COOH group and H from an –OH or –NH₂ group; together they make water.
2. Join the carbonyl carbon to the O or N that lost its H.
3. Repeat at the other end, so one residue of each monomer sits inside the bracket.
4. Draw a bond through each bracket and write n after it.

Worked example 1. Heptanedioic acid, HOOC–(CH₂)₅–COOH, reacts with butane-1,4-diol, HO–(CH₂)₄–OH. Write an equation for the formation of the polymer and name the linkage.

• Each repeating unit contains one ester link inside the bracket and is joined to the next unit by another ester link, so two water molecules leave per repeating unit.
• Equation:

n HOOC–(CH₂)₅–COOH  +  n HO–(CH₂)₄–OH
      →  –[ CO–(CH₂)₅–CO–O–(CH₂)₄–O ]ₙ–  +  2n H₂O

• The linkage is an ester link, so the product is a polyester.

Polyamides from amino acids

An amino acid carries –NH₂ and –COOH on the same molecule, so a single kind of monomer is enough. The amine end of one molecule reacts with the acid end of the next. The repeating unit holds one residue, and one water molecule forms per unit. When several different amino acids join, the polyamide is a polypeptide; proteins are studied in section 3.3.13.

Worked example 2. Draw the repeating unit of the polymer formed from 2-aminobutanoic acid, H₂N–CH(C₂H₅)–COOH.

• Remove H from the N and OH from the carbonyl carbon.
• Repeating unit: –[NH–CH(C₂H₅)–CO]–, with bonds through both brackets and n outside.
• Equation: n H₂N–CH(C₂H₅)–COOH → –[NH–CH(C₂H₅)–CO]ₙ– + n H₂O

From a section of chain to the repeating unit and the monomers

1. Mark every –CO–O– or –CO–NH– group along the chain.
2. Find the shortest stretch that repeats. With two monomers it contains one residue from the diacid and one from the diol or diamine.
3. Start and finish the repeating unit by cutting through linkages, between C and O or between C and N.
4. To get the monomers, cut every linkage the same way. Add OH to each carbonyl carbon (giving –COOH) and H to each O or N (giving –OH or –NH₂).

Worked example 3. Part of a polymer chain is

–NH–CH₂CH₂–NH–CO–(CH₂)₆–CO–NH–CH₂CH₂–NH–CO–(CH₂)₆–CO–

Give the repeating unit and name both monomers.

• Every link is an amide group, –CO–NH–, so the polymer is a polyamide.
• Repeating unit: –[NH–CH₂CH₂–NH–CO–(CH₂)₆–CO]–.
• Cutting each C–N bond and adding H to N gives H₂N–CH₂CH₂–NH₂, ethane-1,2-diamine.
• Adding OH to each carbonyl carbon gives HOOC–(CH₂)₆–COOH, octanedioic acid. Count the carbons: six CH₂ groups plus two C=O carbons make eight.

Worked example 4. A polypeptide section is

–NH–C(CH₃)₂–CO–NH–CH₂CH₂–CO–NH–C(CH₃)₂–CO–NH–CH₂CH₂–CO–

Draw the two amino acids that formed it.

• Cut each C–N bond in the amide links. The residues are –NH–C(CH₃)₂–CO– and –NH–CH₂CH₂–CO–.
• Add H to each N and OH to each C: H₂N–C(CH₃)₂–COOH (2-amino-2-methylpropanoic acid) and H₂N–CH₂CH₂–COOH (3-aminopropanoic acid).

Using the repeating unit in calculations

The Mr of a two-monomer repeating unit equals the sum of the two monomer Mr values minus 2 × 18.0 for the water lost.

Worked example 5. The polyester from worked example 1 has an average chain Mr of 3.21 × 10⁴. Find the average number of repeating units per chain. (Ar: H = 1.0, C = 12.0, O = 16.0)

• Repeating unit formula: C₁₁H₁₈O₄. Mr = 11(12.0) + 18(1.0) + 4(16.0) = 214.0.
• Cross-check: Mr(heptanedioic acid) + Mr(butane-1,4-diol) − 2(18.0) = 160.0 + 90.0 − 36.0 = 214.0.
• Number of units = 3.21 × 10⁴ ÷ 214.0 = 150.

3.3.12.1 Forces between condensation polymer chains

Every polymer has van der Waals forces (induced dipole–dipole forces) between its chains. Condensation polymers also contain polar linkages, which add stronger forces.

• Polyamides form hydrogen bonds between chains. Nitrogen is far more electronegative than hydrogen, so the H of an N–H group carries a large δ+ charge. That hydrogen is drawn towards a lone pair on the δ− O atom of a carbonyl group in the next chain.
• Polyesters have permanent dipole–dipole forces between C=O groups on neighbouring chains. The ester link has no hydrogen bonded to oxygen or nitrogen, so these chains cannot hydrogen bond to each other.
• Kevlar chains are flat and rigid because of the benzene rings. They pack side by side in parallel, which lines up many N–H and C=O groups and gives a large number of hydrogen bonds. That explains the strength of Kevlar fibres.

Stronger forces between chains mean more energy is needed to separate them. So a polyamide usually melts higher than a polyester of similar chain size, because hydrogen bonds are stronger than dipole–dipole forces. Melting separates chains; it does not break covalent bonds inside a chain.

3.3.12.2 Biodegradability and hydrolysis

Why polyesters and polyamides break down

Ester and amide links both contain a polar C=O bond, which leaves the carbonyl carbon δ+. Water or hydroxide ions act as nucleophiles and attack that carbon, and the C–O or C–N bond of the link splits. The process is hydrolysis, and it undoes the condensation step. It is faster when the polymer is heated with acid or alkali. Because their links can be hydrolysed, polyesters and polyamides are biodegradable.

The products depend on the conditions, just as for simple esters in section 3.3.9.1:

Conditions	Polyester gives	Polyamide gives
Heat with dilute acid	diacid and diol	diacid and the diamine’s ammonium salt
Heat with aqueous alkali	the diacid’s salt and diol	the diacid’s salt and diamine

In acid, the amine groups are protonated, so the diamine leaves as an ammonium salt. In alkali, the –COOH groups are deprotonated, so the acid leaves as a carboxylate salt.

Worked example 6. The polyamide from worked example 3 is heated (a) with aqueous sodium hydroxide, and (b) with dilute hydrochloric acid. Name the organic products in each case.

• (a) Alkali: sodium octanedioate, NaOOC–(CH₂)₆–COONa, and ethane-1,2-diamine.
• (b) Acid: octanedioic acid and ethane-1,2-diammonium chloride, ⁺H₃N–CH₂CH₂–NH₃⁺ 2Cl⁻.

Why polyalkenes cannot be hydrolysed

A polyalkene backbone is made only of C–C and C–H bonds. These bonds are strong and essentially non-polar. No carbon in the chain carries a δ+ charge, so nucleophiles such as water or OH⁻ have nothing to attack. That is why polyalkenes are chemically inert and do not break down in the environment (they are non-biodegradable).

3.3.12.2 Disposing of polymers

Method	Advantages	Disadvantages
Landfill	Cheap; no sorting needed	Uses land; polyalkenes persist for a very long time
Incineration	Shrinks the volume of waste; the heat released can generate electricity	Produces CO₂, a greenhouse gas; chlorine-containing polymers such as PVC give toxic HCl, which must be removed from the waste gases
Mechanical recycling (melt and remould)	Saves crude oil and energy; less landfill	Must be collected, sorted by type and cleaned; quality often falls
Chemical recycling	Breaks polymers into monomers or other useful chemicals; condensation polymers can be hydrolysed back to their monomers	Needs energy and specialist plant

To evaluate disposal, pair each advantage with a disadvantage and link both to the chemistry.

Common errors

• Putting two residues of the same monomer, or only one residue, inside a two-monomer repeating unit.
• Leaving H or OH at the ends of a repeating unit instead of bonds through the brackets.
• Claiming hydrogen bonds between polyester chains.
• Saying polyalkenes resist hydrolysis because their bonds are weak; the reason is that they are non-polar.
• Giving the free diamine as a product of acid hydrolysis, or the free acid in alkali.

Where next

Next, use the revision notes, the practice questions and the diagnostics.

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.3.12 Polymers (3.3.12.1 Condensation polymers; 3.3.12.2 Biodegradability and disposal of polymers), International A-level only.