Polyesters and polyamides

In a nutshell

Polyesters and polyamides are condensation polymers: monomers join through ester or amide linkages, and a small molecule (H₂O or HCl) is lost for every link that forms.

Because those linkages can be broken by water, these polymers can be hydrolysed by hot aqueous acid or alkali back to their monomers. You also need to work in both directions: monomers to repeat unit, and polymer section back to monomers.

Assumed knowledge: Carboxylic acids and esters, Amino acids, amides and chirality, Alkenes (addition polymerisation).

Core content

Condensation polymers: what condensation polymerisation is

In condensation polymerisation, monomers join together and a small molecule, such as H₂O or HCl, is lost each time a link forms.

Compare addition polymerisation (4.1.3): alkene monomers join by opening the C=C, and the polymer is the only product.

For a chain to keep growing, each monomer needs a reactive group at both ends. So the usual monomers are:

• dicarboxylic acids (HOOC-R-COOH) or diacyl dichlorides (ClOC-R-COCl)
• diols (HO-R'-OH)
• diamines (H₂N-R'-NH₂)

These are the same reactions you met in 6.1.3, just done at both ends of the molecule at once:

You will not be asked to recall the structures of particular polyesters or polyamides or their monomers. The examples below show the method; in the exam the monomers or the polymer are given to you.

Condensation polymers: forming polyesters

A polyester forms from a dicarboxylic acid (or a diacyl dichloride) and a diol. Each link is an ester group, -COO-.

The equation for this example:

n HOOC-C₆H₄-COOH + n HOCH₂CH₂OH → -[OC-C₆H₄-COO-CH₂CH₂-O]_n- + 2n H₂O

Using the diacyl dichloride instead gives the same polymer, but the small molecule lost is HCl:

n ClOC-C₆H₄-COCl + n HOCH₂CH₂OH → -[OC-C₆H₄-COO-CH₂CH₂-O]_n- + 2n HCl

Why chemists use the acyl chloride: it is more reactive than the carboxylic acid, so it reacts more readily with the -OH (6.1.3).

Counting the small molecules: with n of each monomer, the bracketed repeat unit contains two new links, so write 2n H₂O (or 2n HCl). If you draw one complete chain with its end groups (H- at one end, -OH at the other) the exact number is 2n - 1, because the two ends have not reacted.

One monomer can also do it. A molecule with a -COOH (or -COCl) at one end and an -OH at the other can form a polyester on its own. The repeat unit then comes from one monomer, and n H₂O is lost.

n CH₃CH(OH)COOH → -[O-CH(CH₃)-CO]_n- + n H₂O

(2-hydroxypropanoic acid, lactic acid, gives poly(lactic acid).)

Condensation polymers: forming polyamides

A polyamide forms from a dicarboxylic acid (or a diacyl dichloride) and a diamine. Each link is an amide group, -CONH-.

The amide link in the chain has one H on the nitrogen and the N bonded to a carbon on each side: it is a secondary amide linkage (6.2.2).

With the dicarboxylic acid:

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

With the diacyl dichloride (hexanedioyl dichloride):

n ClOC(CH₂)₄COCl + n H₂N(CH₂)₆NH₂ → -[OC(CH₂)₄CO-NH(CH₂)₆NH]_n- + 2n HCl

A single monomer with an -NH₂ at one end and a -COOH at the other also forms a polyamide on its own:

n H₂N(CH₂)₅COOH → -[NH(CH₂)₅CO]_n- + n H₂O

(6-aminohexanoic acid gives nylon 6.)

Condensation polymers: acid and base hydrolysis of polyesters

Hydrolysis breaks the ester link by reaction with water, giving back the monomers. It is the reverse of the condensation.

The ester link breaks between the C=O carbon and the -O-. The -OH goes back onto the carbonyl carbon, and the H goes back onto the oxygen.

The conditions and products are the same as for simple esters (6.1.3):

For the polyester in Figure 2:

• Acid: -[OC-C₆H₄-COO-CH₂CH₂-O]_n- + 2n H₂O → n HOOC-C₆H₄-COOH + n HOCH₂CH₂OH
• Alkali: -[OC-C₆H₄-COO-CH₂CH₂-O]_n- + 2n NaOH → n NaOOC-C₆H₄-COONa + n HOCH₂CH₂OH

The diol is the same in both conditions. Only the acid end changes: free -COOH in acid, -COO^- salt in alkali.

Condensation polymers: acid and base hydrolysis of polyamides

The amide link breaks between the C=O carbon and the N. The -OH goes back onto the carbonyl carbon, and the H goes back onto the nitrogen.

This time both ends are affected by the conditions, because an amine is a base:

For nylon 6,6:

• Acid: -[OC(CH₂)₄CO-NH(CH₂)₆NH]_n- + 2n H₂O + 2n HCl → n HOOC(CH₂)₄COOH + n ClH₃N(CH₂)₆NH₃Cl
• Alkali: -[OC(CH₂)₄CO-NH(CH₂)₆NH]_n- + 2n NaOH → n NaOOC(CH₂)₄COONa + n H₂N(CH₂)₆NH₂

The ammonium salt in the acid equation can also be written as the ion, ⁺H₃N(CH₂)₆NH₃⁺.

Still don't get it? · which product is a salt, and why

Imagine the hydrolysis happens in two moments. First, water cuts every link and you get back the plain monomers. Second, those monomers find themselves sitting in a beaker full of acid or alkali, and they react with it like any acid or base would.

Step by step:

1. Cutting the links always gives a -COOH end and either an -OH end (polyester) or an -NH₂ end (polyamide).
2. In alkali, the OH^- ions neutralise every -COOH, so it ends up as -COO^- (a carboxylate salt with Na⁺).
3. In acid, the H⁺ ions find any base present. An -NH₂ group is a base (it accepts a proton), so it ends up as -NH₃⁺.
4. An alcohol -OH is neither acidic enough nor basic enough to change. So a diol comes out as the diol in both conditions.

The exam version: acid hydrolysis gives the carboxylic acid and, for a polyamide, the ammonium salt of the amine; alkaline hydrolysis gives the carboxylate salt and the free amine (or alcohol).

Condensation polymers: predicting the repeat unit from given monomers

To draw the repeat unit of a condensation polymer:

1. Identify the functional groups at each end of each monomer (-COOH, -COCl, -OH, -NH₂).
2. Remove the small molecule at each joining point: -OH from a -COOH (or -Cl from a -COCl), and H from an -OH or -NH₂.
3. Join one of each monomer through the new ester or amide link.
4. Leave one open link at each end, with a trailing bond through each bracket, and add n outside.

Check your answer: the repeat unit contains one of each monomer (one monomer if it is self-condensing), and no -OH, -H or -Cl at the ends.

Condensation polymers: predicting the monomers from a section of polymer

Work backwards by hydrolysing on paper:

1. Find every linkage in the section: -COO- (ester) or -CONH- (amide).
2. Cut each linkage between the C=O carbon and the O or N.
3. Add back -OH to each carbonyl carbon (giving -COOH) and H to each O or N (giving -OH or -NH₂).
4. Read off the distinct monomers. If you get the same molecule each time, the polymer came from one monomer.

The acid monomer can be given as the dicarboxylic acid or the diacyl dichloride; both make the same polymer.

Still don't get it? · finding the monomers from a polymer section

Picture a paper chain where every join was glued with a little drop of water. To get the original strips back, you find each glue point, cut through it, and give each cut edge back the bit of glue it lost.

Step by step:

1. Scan along the chain until you see C=O. Look at what is right next to it: an -O- means an ester link; an -NH- means an amide link.
2. Draw a cut between the C=O carbon and that O or N. Do this at every link, not just one.
3. On the carbonyl side, add -OH: the C=O becomes -COOH.
4. On the other side, add -H: the O becomes -OH, the NH becomes -NH₂.
5. Now count the different pieces. Two different pieces repeating means two monomers; one piece repeating means one monomer.

The exam version: the polymer section is a polyester (or polyamide) because it contains ester (or amide) linkages; the monomers are the dicarboxylic acid (or diacyl dichloride) and the diol (or diamine) obtained by hydrolysing each linkage.

Condensation polymers: predicting the type of polymerisation

You may be given a monomer (or monomers) or a polymer section and asked which type of polymerisation made it.

Keep two questions separate:

• Type of polymerisation: addition or condensation.
• Type of polymer: e.g. poly(alkene), polyester or polyamide.

A monomer with a C=C and one other group, such as propenoic acid CH₂=CHCOOH, still polymerises by addition: the C=C opens and the -COOH just hangs off the chain as a side group.

Worked examples

Example 1: repeat unit from two given monomers

Task: butanedioyl dichloride, ClOC(CH₂)₂COCl, reacts with ethane-1,2-diamine, H₂NCH₂CH₂NH₂. Give the repeat unit, the type of polymer, and the equation.

1. Groups: -COCl at both ends of one monomer, -NH₂ at both ends of the other. Acyl chloride + amine gives an amide link.
2. Remove Cl from each -COCl and one H from each -NH₂: these leave as HCl.
3. Join one of each: -OC(CH₂)₂CO-NHCH₂CH₂NH-
4. Repeat unit: -[OC(CH₂)₂CO-NHCH₂CH₂NH]- with a trailing bond through each bracket.
5. Type of polymer: a polyamide, made by condensation polymerisation.
6. Equation: n ClOC(CH₂)₂COCl + n H₂NCH₂CH₂NH₂ → -[OC(CH₂)₂CO-NHCH₂CH₂NH]_n- + 2n HCl

Where marks go: the small molecule is HCl, not H₂O, because an acyl chloride was used. The repeat unit ends in -CO on one side and -NH on the other, with no Cl or H left on the ends.

Atom check for one repeat unit: C₄H₄O₂Cl₂ + C₂H₈N₂ → C₆H₁₀N₂O₂ + 2 HCl. Both sides: C 6, H 12, N 2, O 2, Cl 2.

Example 2: monomers from a polymer section

Task: a section of a polymer is

-O-CH₂CH₂CH₂-O-CO-CH₂-CO-O-CH₂CH₂CH₂-O-CO-CH₂-CO-

Name the type of polymer and give the monomers.

1. Linkages: each C=O is bonded to an -O-, so these are ester links. It is a polyester.
2. Cut between each C=O carbon and the neighbouring O.
3. Add back: -OH on each carbonyl carbon gives HOOC-CH₂-COOH; H on each O gives HO-CH₂CH₂CH₂-OH.
4. Monomers: propanedioic acid, HOOCCH₂COOH (or propanedioyl dichloride, ClOCCH₂COCl), and propane-1,3-diol, HOCH₂CH₂CH₂OH.
5. Repeat unit: -[O-CH₂CH₂CH₂-O-CO-CH₂-CO]-. The section shown is two repeat units.

Where marks go: students often give "-CO-CH₂-CO-" as a monomer. That is a fragment, not a molecule: the monomer needs its -OH groups restored (HOOC-CH₂-COOH).

Example 3: hydrolysis of a polymer with both linkages

Task: a polymer has the repeat unit -[OC(CH₂)₂CO-NH-CH₂CH₂-O]-. Give the organic products of hydrolysis with (a) hot HCl(aq) and (b) hot NaOH(aq).

1. Linkages: the second C=O is bonded to -NH- (an amide link). The -O- at the right-hand end bonds, across the repeat-unit boundary, to the first C=O of the next unit (an ester link).
2. Cut both links and add back water: this gives HOOC(CH₂)₂COOH (butanedioic acid) and H₂NCH₂CH₂OH (2-aminoethanol).
3. (a) Acid: -COOH stays as it is; the basic -NH₂ is protonated. Products: HOOC(CH₂)₂COOH and HOCH₂CH₂NH₃⁺ (as its chloride salt).
4. (b) Alkali: both -COOH groups are neutralised; the amine and alcohol are unchanged. Products: ^-OOC(CH₂)₂COO^- (as sodium butanedioate) and H₂NCH₂CH₂OH.

Equations for one repeat unit:

• Acid: C₆H₉NO₃ + 2H₂O + H⁺ → HOOC(CH₂)₂COOH + HOCH₂CH₂NH₃⁺
• Alkali: C₆H₉NO₃ + 2OH^- → ^-OOC(CH₂)₂COO^- + H₂NCH₂CH₂OH

Where marks go: the -OH of 2-aminoethanol is not changed by acid or alkali. Only the amine (in acid) and the carboxylic acid (in alkali) change form.

Example 4: naming the type of polymerisation

Task: for each monomer (or pair), state the type of polymerisation.

Where marks go: "polyamide" is the type of polymer; the type of polymerisation is "condensation". Answer the question that was asked.

Common exam mistakes

• Writing n H₂O instead of 2n H₂O for two monomers. A repeat unit made from a diacid and a diol (or diamine) contains two new links, so 2n small molecules are lost.
• Saying the small molecule is always water. With a diacyl dichloride the by-product is HCl.
• End groups inside the brackets. The repeat unit must not carry the -OH, -H or -Cl that were lost; it ends in open bonds with trailing bonds through the brackets.
• Repeat unit too long or too short. One repeat unit has one of each monomer (or one monomer if self-condensing). Drawing two of each, or only half, loses the mark.
• Ester link drawn wrongly, e.g. -CO-O-O-CO- or a C-O-C with no C=O. The ester link is -C(=O)-O-, with the O of the diol joined directly to the carbonyl carbon.
• Amide link drawn as -CONH₂ in the middle of a chain. Inside a polyamide the N has one H: -CO-NH-.
• Monomer given as a fragment (e.g. -OC(CH₂)₄CO-). A monomer is a molecule: restore -COOH, -OH or -NH₂.
• Cutting in the wrong place when finding monomers or hydrolysis products. The ester or amide link breaks between the C=O carbon and the O or N, never at a C-C bond.
• Acid hydrolysis of a polyamide giving the free diamine. In acid the amine is protonated: the product is the ammonium salt (-NH₃⁺).
• Alkaline hydrolysis giving the free dicarboxylic acid. In alkali the acid is neutralised: the product is the carboxylate salt (-COO^-Na⁺).
• Changing the diol in alkali (writing an alkoxide, -O^-). The diol is unchanged in both acid and alkali.
• Vague hydrolysis conditions such as "add water". Hydrolysis needs hot aqueous acid or hot aqueous alkali (heat under reflux).
• Mixing up type of polymer and type of polymerisation. "Polyester" answers the first; "condensation" answers the second.
• Calling a substituted alkene a condensation monomer because it has a -COOH or -OH. If the monomer polymerises through its C=C, it is addition.

Key definitions

• Condensation polymerisation: the joining of monomers, each with two reactive functional groups, to form a polymer, with the loss of a small molecule (such as H₂O or HCl) for each link formed.
• Addition polymerisation: the joining of many unsaturated monomers (containing C=C) to form a polymer as the only product.
• Repeat unit: the smallest section of a polymer chain that repeats to make the whole polymer (drawn in brackets with trailing bonds and n).
• Polyester: a condensation polymer in which the monomers are joined by ester linkages, -COO-, formed from a dicarboxylic acid (or diacyl dichloride) and a diol.
• Polyamide: a condensation polymer in which the monomers are joined by amide linkages, -CONH-, formed from a dicarboxylic acid (or diacyl dichloride) and a diamine.
• Hydrolysis: the breaking of a bond by reaction with water (here catalysed by acid or using alkali), giving back the monomers.
• Acid hydrolysis (of a polyester or polyamide): heating with hot aqueous acid gives the dicarboxylic acid, plus the diol or the ammonium salt of the diamine.
• Base (alkaline) hydrolysis (of a polyester or polyamide): heating with hot aqueous alkali gives the carboxylate salt, plus the diol or the diamine.

Specification

• I can describe condensation polymerisation to form a polyester from a dicarboxylic acid (or diacyl dichloride) and a diol, or from one monomer with both groups, and state the small molecule lost. (6.2.3 a i)
• I can describe condensation polymerisation to form a polyamide from a dicarboxylic acid (or diacyl dichloride) and a diamine, or from one monomer with both groups, and state the small molecule lost. (6.2.3 a ii)
• I can give the products of the acid and base hydrolysis of the ester groups in a polyester. (6.2.3 b i)
• I can give the products of the acid and base hydrolysis of the amide groups in a polyamide, including the ammonium salt in acid and the carboxylate salt in alkali. (6.2.3 b ii)
• I can draw the repeat unit of an addition or condensation polymer from given monomer(s). (6.2.3 c i, with 4.1.3 j)
• I can deduce the monomer(s) needed for a given section of an addition or condensation polymer. (6.2.3 c ii, with 4.1.3 j)
• I can state the type of polymerisation (addition or condensation) from a given monomer or polymer. (6.2.3 c iii)

Related notes

• Alkenes: addition polymerisation, repeat units of poly(alkenes) and processing waste polymers
• Basic concepts of organic chemistry: functional groups, naming and condensed formulae
• Carboxylic acids and esters: esterification, acyl chlorides and the acid and alkaline hydrolysis of esters
• Amines: amines as bases, forming ammonium salts
• Amino acids, amides and chirality: primary and secondary amides
• Organic synthesis: heating under reflux, and using these reactions in multi-stage routes