Question 16.3: Condensation Polymerization of 6,6-Nylon Nylon was first rep......
Condensation Polymerization of 6,6-Nylon
Nylon was first reported by Wallace Hume Carothers of Du Pont^{TM} in about 1934. In 1939, Charles Stine, also from Du Pont^{TM}, reported the discovery of this first synthetic fiber to a group of 3000 women gathered for the New York World’s Fair. The first application was nylon stockings. Today nylon is used in hundreds of applications. Prior to nylon, Carothers had discovered neoprene (an elastomer).
The linear polymer 6,6-nylon is to be produced by combining 1000 g of hexamethylene diamine with adipic acid. A condensation reaction then produces the polymer. The molecular structures of the monomers are shown below. The linear nylon chain is produced when a hydrogen atom from the hexamethylene diamine combines with an OH group from adipic acid to form a water molecule.
\underset{\text{Hexamethylene diamine}}{\begin{matrix}\ \ \ \ \ \ \ \ H \ _{_{\diagdown }} \ \ \ \ \ \ \ \ \ \ \underset{|}{H} \quad \underset{|}{H} \quad \underset{|}{H} \quad \underset{|}{H} \quad \underset{|}{H} \quad \ \underset{|}{H} \quad \ \ \ \ \ _{_{\diagup }} \ H \ \ \ \ \ \\ \ \ \ \ \ N-C-C-C-C-C-C-N \ \\ \ \ \ H^{^{^{\ \diagup} }} \ \ \ \ \ \quad \overset{|}{H} \quad \overset{|}{H} \ \quad \overset{|}{H} \ \quad \overset{|}{H} \quad \overset{|}{H} \quad \overset{|}{H} \ \ \ \ \ \ \ \ \ ^{^ {^\diagdown} } H \end{matrix}} + \underset{\text{Adipic acid}}{\begin{matrix} \ \ \underset{||}{O} \quad \underset{|}{H} \quad \ \underset{|}{H} \quad \underset{|}{H} \quad \underset{|}{H} \quad \ \underset{||}{O} \ \ \ \ \\ H-O- C-C-C-C-C-C-O-H \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \quad \quad \ \ \ \ \quad \ \overset{|}{H} \quad \overset{|}{H} \quad \overset{|}{H} \quad \ \overset{|}{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix}}
\begin{matrix} \ H \ _{_{\diagdown }} \ \ \ \ \ \ \ \ \ \ \underset{|}{H} \quad \underset{|}{H} \quad \underset{|}{H} \quad \underset{|}{H} \quad \underset{|}{H} \quad \ \underset{|}{H} \quad \ \underset{|}{\overset{\downarrow }{H} } \ \quad \underset{||}{O} \quad \underset{|}{H} \quad \ \underset{|}{H} \quad \underset{|}{H} \quad \underset{|}{H} \quad \ \underset{||}{O} \ \ \ \ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ N-C-C-C-C-C-C-N-C-C-C-C-C-C-O-H \ \\ \ \ \ \ \ \ \ \ H^{^{^{\ \diagup} }} \ \ \ \ \ \quad \overset{|}{H} \quad \overset{|}{H} \ \quad \overset{|}{H} \ \quad \overset{|}{H} \quad \overset{|}{H} \quad \overset{|}{H} \ \ \ \ \quad \quad \ \ \ \ \quad \ \overset{|}{H} \quad \overset{|}{H} \quad \overset{|}{H} \quad \ \overset{|}{H}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{matrix} + H_{2}O
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6,6-\text{Nylon} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{ Water}
Note that the reaction can continue at both ends of the new molecule; consequently, long chains may form. This polymer is called 6,6-nylon because both monomers contain six carbon atoms.
How many grams of adipic acid are needed, and how much 6,6-nylon is produced, assuming 100% efficiency?
The molecular weights of hexamethylene diamine, adipic acid, and water are 116, 146, and 18 g/mol, respectively. The number of moles of hexamethylene diamine is equal to the number of moles of adipic acid:
\frac{1000 \ g}{116 \ g/mol} = 8.62 \ moles = \frac{x \ g}{146 \ g/mol}
x = 1259 g of adipic acid required
One water molecule is lost when hexamethylene diamine reacts with adipic acid. Each time a monomer is added to the chain, one molecule of water is lost. Thus, when a long chain forms, there are (on average) two water molecules released for each repeat unit of the chain (each repeat unit being formed from two monomers). Thus, the number of moles of water lost is 2 (8.62 moles) = 17.24 moles or
17.24 moles H_{2}O (18 g/mol) = 310 g H_{2}O
The total amount of nylon produced is
1000 g + 1259 g – 310 g = 1949 g = 1.95 kg