A2 Chemistry 9701: Polyester Repeat Units from Diols, Diacids and a Hydroxy Acid

Propane-1,3-diol, $\text{HOCH}_2\text{CH}_2\text{CH}_2\text{OH}$ , and pentanedioic acid, $\text{HOOC}(\text{CH}_2)_3\text{COOH}$ , react together by condensation polymerisation to form a polyester.

(a) Deduce the repeat unit of this polyester, and state the number of water molecules released per repeat unit. [3]

(b) A different polyester has the repeat unit $-\text{O}(\text{CH}_2)_4\text{OCO}(\text{CH}_2)_4\text{CO}-$ . Identify, by name and structural formula, the two monomers used to form this polyester. [2]

(c) 6-Hydroxyhexanoic acid, $\text{HO}(\text{CH}_2)_5\text{COOH}$ , can also undergo condensation polymerisation on its own to form a polyester. Deduce the repeat unit formed, and state the number of water molecules eliminated per repeat unit. [2]

Part (a): Repeat unit of the diol/diacid polyester

Propane-1,3-diol has two $-\text{OH}$ groups and pentanedioic acid has two $-\text{COOH}$ groups. Each $-\text{OH}$ reacts with a $-\text{COOH}$ to form an ester link ( $-\text{O-CO}-$ ), releasing one molecule of water per link:

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

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

$\left(-\text{O}(\text{CH}_2)_3\text{OCO}(\text{CH}_2)_3\text{CO}-\right)_n$

Part (b): Identifying the monomers from a repeat unit

Splitting the given repeat unit $-\text{O}(\text{CH}_2)_4\text{OCO}(\text{CH}_2)_4\text{CO}-$ at the two ester oxygens:

The $-\text{O}(\text{CH}_2)_4\text{O}-$ portion, with an $-\text{OH}$ added back at each end, gives $\text{HO}(\text{CH}_2)_4\text{OH}$ , butane-1,4-diol.
The $-\text{CO}(\text{CH}_2)_4\text{CO}-$ portion, with an $-\text{OH}$ added back at each carbonyl carbon, gives $\text{HOOC}(\text{CH}_2)_4\text{COOH}$ , hexanedioic acid.

Part (c): Self-condensation of a single hydroxycarboxylic acid

6-Hydroxyhexanoic acid, $\text{HO}(\text{CH}_2)_5\text{COOH}$ , carries both functional groups needed for ester formation in the same molecule: one $-\text{OH}$ and one $-\text{COOH}$ . Each molecule reacts with the next by forming a single ester link between its $-\text{OH}$ and the neighbouring molecule’s $-\text{COOH}$ :

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

Since only one $-\text{OH}$ and one $-\text{COOH}$ are contributed per monomer, only one ester link (and therefore one water molecule) forms per repeat unit added to the chain. The repeat unit is:

$\left(-\text{O}(\text{CH}_2)_5\text{CO}-\right)_n$

Final answers

(a) Repeat unit $\left(-\text{O}(\text{CH}_2)_3\text{OCO}(\text{CH}_2)_3\text{CO}-\right)_n$ ; 2 water molecules released per repeat unit.
(b) Butane-1,4-diol, $\text{HO}(\text{CH}_2)_4\text{OH}$ , and hexanedioic acid, $\text{HOOC}(\text{CH}_2)_4\text{COOH}$ .
(c) Repeat unit $\left(-\text{O}(\text{CH}_2)_5\text{CO}-\right)_n$ ; 1 water molecule eliminated per repeat unit.