A2 Chemistry 9701: Polyester from a Diacyl Dichloride, Polyamide Monomers and Acidic Hydrolysis

(a) Hexane-1,6-diol, $\text{HO}(\text{CH}_2)_6\text{OH}$ , is reacted with propanedioyl dichloride, $\text{ClOCCH}_2\text{COCl}$ , to form a polyester. Deduce the repeat unit formed, and state the number and identity of the small molecules eliminated per repeat unit. [3]

(b) A different polyamide has the repeat unit $-\text{NH}(\text{CH}_2)_6\text{NHCO}(\text{CH}_2)_2\text{CO}-$ . Identify, by name and structural formula, the two monomers used to form this polyamide. [2]

(c) The polyester formed in (a) is hydrolysed by heating under reflux with an excess of dilute hydrochloric acid. Give the structural formulae of the two organic products formed per repeat unit under these conditions. [2]

Part (a): Repeat unit and elimination product from a diacyl dichloride

Hexane-1,6-diol has two $-\text{OH}$ groups and propanedioyl dichloride, $\text{ClOCCH}_2\text{COCl}$ , has two acyl chloride groups, $-\text{COCl}$ . Unlike a plain dicarboxylic acid, an acyl chloride reacting with an alcohol releases hydrogen chloride, not water, as each ester link forms:

$-\text{OH} + \text{ClOC}- \;\longrightarrow\; -\text{O-CO}- + \text{HCl}$

Since a repeat unit is built from one diol molecule joined to one diacyl dichloride molecule, this reaction happens twice per repeat unit, once at each end, so two molecules of HCl are eliminated. The repeat unit is:

$\left(-\text{O}(\text{CH}_2)_6\text{OCOCH}_2\text{CO}-\right)_n$

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

Splitting the given repeat unit $-\text{NH}(\text{CH}_2)_6\text{NHCO}(\text{CH}_2)_2\text{CO}-$ at the two amide bonds:

The $-\text{NH}(\text{CH}_2)_6\text{NH}-$ portion, with a hydrogen added back at each nitrogen, gives $\text{H}_2\text{N}(\text{CH}_2)_6\text{NH}_2$ , hexane-1,6-diamine.
The $-\text{CO}(\text{CH}_2)_2\text{CO}-$ portion, with an $-\text{OH}$ added back at each carbonyl carbon, gives $\text{HOOC}(\text{CH}_2)_2\text{COOH}$ , butanedioic acid.

Part (c): Acidic hydrolysis of the polyester from (a)

Heating the polyester under reflux with excess dilute hydrochloric acid hydrolyses the ester bonds ( $-\text{O-CO}-$ ) in the backbone, regenerating the original diol and diacid. Unlike the amine group released when a polyamide is hydrolysed, neither an alcohol nor a carboxylic acid is significantly protonated or deprotonated by dilute HCl, so both products are recovered in their neutral forms:

$\text{HO}(\text{CH}_2)_6\text{OH} \quad \text{and} \quad \text{HOOCCH}_2\text{COOH}$

Final answers

(a) Repeat unit $\left(-\text{O}(\text{CH}_2)_6\text{OCOCH}_2\text{CO}-\right)_n$ ; 2 molecules of HCl eliminated per repeat unit.
(b) Hexane-1,6-diamine, $\text{H}_2\text{N}(\text{CH}_2)_6\text{NH}_2$ , and butanedioic acid, $\text{HOOC}(\text{CH}_2)_2\text{COOH}$ .
(c) Hexane-1,6-diol, $\text{HO}(\text{CH}_2)_6\text{OH}$ , and propanedioic acid, $\text{HOOCCH}_2\text{COOH}$ .