A2 Chemistry 9701: Polyamide Repeat Unit and Alkaline vs Acidic Hydrolysis

Hexane-1,6-diamine, $\text{H}_2\text{N}(\text{CH}_2)_6\text{NH}_2$ , and heptanedioic acid, $\text{HOOC}(\text{CH}_2)_5\text{COOH}$ , react together by condensation polymerisation to form a polyamide used in synthetic fibres.

(a) Deduce the repeat unit of this polyamide, showing the amide link clearly, and state the number of water molecules eliminated per repeat unit. [3]

(b) The polyamide is hydrolysed by heating under reflux with an excess of aqueous sodium hydroxide. Name the type of bond broken in the polymer backbone, and give the structural formula(e) of the organic product(s) formed per repeat unit under these alkaline conditions. [3]

(c) State and explain how the form of the amine-containing product in (b) would differ if the same polyamide were instead hydrolysed by heating under reflux with an excess of dilute hydrochloric acid. [2]

Part (a): Repeat unit of the diamine/diacid polyamide

Hexane-1,6-diamine has two $-\text{NH}_2$ groups and heptanedioic acid has two $-\text{COOH}$ groups. Each $-\text{NH}_2$ reacts with a $-\text{COOH}$ to form an amide link ( $-\text{NH-CO}-$ ), releasing one molecule of water per link:

$-\text{NH}_2 + \text{HOOC}- \;\longrightarrow\; -\text{NH-CO}- + \text{H}_2\text{O}$

Because a repeat unit is built from one diamine molecule joined to one diacid molecule, two such amide links form per repeat unit (one at each end), so two molecules of water are eliminated per repeat unit. The repeat unit is:

$\left(-\text{NH}(\text{CH}_2)_6\text{NHCO}(\text{CH}_2)_5\text{CO}-\right)_n$

Part (b): Alkaline hydrolysis of the polyamide

Refluxing with excess aqueous sodium hydroxide hydrolyses the amide bonds ( $-\text{NH-CO}-$ ) in the polymer backbone. Hydroxide ions attack the polarised carbonyl carbon of each amide link, breaking the chain and regenerating the diamine and the diacid:

$-\text{NH-CO}- + \text{OH}^- \;\longrightarrow\; -\text{NH}_2 + {}^-\text{OOC}-$

The diamine product, hexane-1,6-diamine, $\text{H}_2\text{N}(\text{CH}_2)_6\text{NH}_2$ , is a base and is not protonated by the hydroxide ions present, so it is recovered in its free, neutral form. The carboxylic acid released from the diacid portion, however, is deprotonated by the excess $\text{NaOH}$ , so it is recovered as its sodium salt, disodium heptanedioate:

$\text{NaOOC}(\text{CH}_2)_5\text{COONa}$

Part (c): Contrast with acidic hydrolysis

If the same polyamide were instead hydrolysed under reflux with excess dilute hydrochloric acid, the amide bonds would still be broken, again regenerating the diamine and the diacid, but the ionisation state of the amine-containing product would be different. With an excess of acid present, the basic $-\text{NH}_2$ groups of the diamine are readily protonated:

$-\text{NH}_2 + \text{H}^+ \;\longrightarrow\; -\text{NH}_3^+$

so the diamine is recovered not in its free form but as the diammonium salt, $^+\text{H}_3\text{N}(\text{CH}_2)_6\text{NH}_3^+$ (isolated as its dichloride salt, since $\text{HCl}$ is in excess). This mirrors the diacid’s behaviour in reverse: under alkaline conditions the acidic product is converted to a salt while the amine stays neutral, whereas under acidic conditions the basic amine product is converted to a salt while the diacid stays neutral.

Final answers

(a) Repeat unit $\left(-\text{NH}(\text{CH}_2)_6\text{NHCO}(\text{CH}_2)_5\text{CO}-\right)_n$ ; 2 water molecules eliminated per repeat unit.
(b) Amide bond broken; products: hexane-1,6-diamine, $\text{H}_2\text{N}(\text{CH}_2)_6\text{NH}_2$ (free amine), and disodium heptanedioate, $\text{NaOOC}(\text{CH}_2)_5\text{COONa}$ .
(c) Under acidic conditions the diamine is instead recovered as the diammonium salt, $^+\text{H}_3\text{N}(\text{CH}_2)_6\text{NH}_3^+$ , because the excess HCl protonates its basic $-\text{NH}_2$ groups.