6.2.3 Polyesters and Polyamides

A polyester needs monomers that can make ester links, –COO–, at both ends. A dicarboxylic acid plus a diol is the standard pair; a hydroxycarboxylic acid can also self-condense.

Each ester link formed from –COOH and –OH eliminates H₂O. If an acyl chloride is used in place of the carboxylic acid, the small molecule is HCl. Draw the repeat unit with bonds continuing through the brackets.

A polyamide contains repeated amide links, –CONH–. You can form one from a dicarboxylic acid and a diamine, or from a molecule such as an amino acid that carries both reacting groups.

Forming the amide from –COOH and –NH₂ eliminates H₂O; using a dioyl chloride instead eliminates HCl. Keep the amide link as –CO–NH–: an extra O gives the wrong connectivity.

Acid hydrolysis breaks every ester link and restores the functional groups that formed it. The acyl side becomes –COOH and the oxygen side becomes –OH.

So a polyester made from a dicarboxylic acid and a diol gives the dicarboxylic acid plus the diol. If the chain came from one hydroxycarboxylic acid, hydrolysis returns that hydroxy acid.

Hot aqueous alkali also breaks the ester links, but under alkaline conditions the acid products remain as carboxylate salts, –COO⁻ M⁺. The alcohol ends are still –OH.

With NaOH, a polyester from a dicarboxylic acid and a diol therefore gives the sodium dicarboxylate plus the diol. Do not convert the carboxylate to –COOH unless the mixture is acidified afterwards.

Acid hydrolysis breaks the amide links. The carbonyl side becomes a carboxylic acid, while the nitrogen side gives an amine that is protonated in the acidic mixture.

For a polyamide from a dicarboxylic acid and a diamine, expect the dicarboxylic acid plus a diammonium salt. When you draw the nitrogen product, use –NH₃⁺ where a primary amine group has been protonated.

Alkaline hydrolysis breaks the amide links and leaves the acid side as a carboxylate salt. The nitrogen-containing product is the amine, not an ammonium ion, because the conditions are alkaline.

For NaOH hydrolysis of a polyamide made from a dicarboxylic acid and a diamine, draw the sodium dicarboxylate and the neutral diamine. Use the conditions to decide the charges before you finish the structures.

For an addition polymer, open the monomer C=C and keep the substituents on the same two backbone carbons. For a condensation polymer, join the reacting functional groups and show the ester or amide link formed.

Whichever route you use, put one true repeat unit inside brackets and draw bonds passing through both bracket edges. I’d do that end-bond check before moving on; it catches a lot of polymer drawings that are otherwise nearly right.

To recover the monomer from an addition polymer, take one backbone repeat and restore the C=C between the two carbons that came from the alkene. The side groups stay attached to those carbons.

For a condensation polymer, cut through the ester or amide links and restore the functional groups: –COOH with –OH for a polyester, or –COOH with –NH₂ for a polyamide. If the question specifies an acyl-chloride route, restore –COCl on that monomer.

An addition polymer comes from opening C=C bonds and does not eliminate a small molecule. Its main chain comes from the alkene carbon skeleton.

A condensation polymer is built by reactions between functional groups, commonly giving ester or amide links in the chain and eliminating a small molecule such as H₂O or HCl. Name the process as addition or condensation polymerisation.