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Chapter 17: Problem 44

We have studied Fischer esterification, in which a carboxylic acid is reacted with an alcohol in the presence of an acid catalyst to form an ester. Suppose that you start instead with a dicarboxylic acid such as terephthalic acid and a diol such as ethylene glycol. Show how Fischer esterification in this case can lead to a macromolecule with a molecular weight several thousands of times that of the starting materials. O=C(O)c1ccc(C(=O)O)cc1 1,4-Benzenedicarboxylic acid \(\quad 1,2\)-Ethanediol (Terephthalic acid) (Ethylene glycol) As we shall see in Section \(29.5 \mathrm{~B}\), the material produced in this reaction is a highmolecular-weight polymer, which can be fabricated into Mylar films, and into the textile fiber known as Dacron polyester.

Short Answer

Expert verified
In this exercise, terephthalic acid (a dicarboxylic acid) and ethylene glycol (a diol) undergo Fischer esterification to form an ester linkage. Due to the presence of multiple functional groups on both molecules, these compounds can react further, leading to a high-molecular-weight polyester copolymer. This type of polymer can be used in various applications, such as Mylar films and Dacron polyester fibers.

Step by step solution

01

Understand the reagents and the Fischer esterification reaction

Fischer esterification is a reaction in which a carboxylic acid reacts with an alcohol to form an ester and water in the presence of an acid catalyst. In this exercise, we are given terephthalic acid, a dicarboxylic acid, and ethylene glycol, a diol. These two compounds can react to form a copolymer due to the presence of multiple functional groups. Terephthalic acid: 1,4-Benzenedicarboxylic acid Structure: O=C(O)c1ccc(C(=O)O)cc1 Ethylene glycol: 1,2-Ethanediol Structure: C(O)C(O)
02

Perform the Fischer esterification reaction

The Fischer esterification reaction occurs between the carboxylic acid group of terephthalic acid and the hydroxyl group of ethylene glycol. The acid catalyst will protonate the carboxyl group, making it more electrophilic and allowing the nucleophilic attack by the hydroxyl group of the ethylene glycol. This reaction will form an ester linkage with the release of water. The reaction can be represented as follows: Terephthalic acid + Ethylene glycol -> ester linkage + H2O O=C(O)c1ccc(C(=O)O)cc1 + C(O)C(O) -> O=C(O)C1=CC=C(C(=O)OC(O)C(O))C=C1 + H2O
03

Formation of the polymer

As both terephthalic acid and ethylene glycol each contain two functional groups, further esterification reactions can take place between the remaining carboxylic acid groups and hydroxyl groups, leading to the formation of a long chain copolymer. This polymer will have a repeating unit derived from both terephthalic acid and ethylene glycol. The overall polymerization can be described as: Terephthalic acid + n Ethylene glycol -> polyester copolymer + n H2O O=C(O)c1ccc(C(=O)O)cc1 + n C(O)C(O) -> [-O=C(O)C1=CC=C(C(=O)OCC(O)C)C=C1-]_n + n H2O
04

Conclusion

Fischer esterification between terephthalic acid and ethylene glycol, both having two functional groups, can lead to the formation of a polyester copolymer with a molecular weight several thousands of times greater than the starting materials. This high-molecular-weight polymer can be used to create Mylar films and Dacron polyester fibers.

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Most popular questions from this chapter

Select the stronger acid in each set. (a) Phenol \(\left(\mathrm{p} K_{\mathrm{a}} 9.95\right)\) and benzoic acid \(\left(\mathrm{p} K_{\mathrm{a}} 4.19\right)\) (b) Lactic acid \(\left(K_{a} 8.4 \times 10^{-4}\right)\) and ascorbic acid \(\left(K_{a} 7.9 \times 10^{-5}\right)\)

Given here are \({ }^{1} \mathrm{H}-\mathrm{NMR}\) and \({ }^{19} \mathrm{C}-\mathrm{NMR}\) spectral data for nine compounds. Each compound shows strong absorption between 1720 and \(1700 \mathrm{~cm}^{-1}\), and strong, broad absorption over the region \(2500-3300 \mathrm{~cm}^{-1}\). Propose a structural formula for each compound. Refer to Appendices 4,5, and 6 for spectral correlation tables. $$ \begin{aligned} &\text { (a) } \mathrm{C}_{5} \mathrm{H}_{10} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 0.94(\mathrm{t}, 3 \mathrm{H}) & 180.71 \\ 1.39(\mathrm{~m}, 2 \mathrm{H}) & 33.89 \\ 1.62(\mathrm{~m}, 2 \mathrm{H}) & 26.76 \\ 2.35(\mathrm{t}, 2 \mathrm{H}) & 22.21 \\ 12.0(\mathrm{~s}, 1 \mathrm{H}) & 13.69 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (b) } \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{19} \text { C-NMR } \\ \hline 1.08(\mathrm{~s}, 9 \mathrm{H}) & 179.29 \\ 2.23(\mathrm{~s}, 2 \mathrm{H}) & 47.82 \\ 12.1(\mathrm{~s}, 1 \mathrm{H}) & 30.62 \\ & 29.57 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (c) } \mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}_{4}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 0.93(\mathrm{t}, 3 \mathrm{H}) & 170.94 \\ 1.80(\mathrm{~m}, 2 \mathrm{H}) & 53.28 \\ 3.10(\mathrm{t}, 1 \mathrm{H}) & 21.90 \\ 12.7(\mathrm{~s}, 2 \mathrm{H}) & 11.81 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (d) } \mathrm{C}_{5} \mathrm{H}_{8} \mathrm{O}_{4}\\\ &\begin{array}{cr} \hline{ }^{1} \text { H-NMR } & { }^{19} \text { C-NMR } \\ \hline 1.29(\mathrm{~s}, 6 \mathrm{H}) & 174.01 \\ 12.8(\mathrm{~s}, 2 \mathrm{H}) & 48.77 \\ & 22.56 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (e) } \mathrm{C}_{4} \mathrm{H}_{6} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 1.91(\mathrm{~d}, 3 \mathrm{H}) & 172.26 \\ 5.86(\mathrm{~d}, 1 \mathrm{H}) & 147.53 \\ 7.10(\mathrm{~m}, 1 \mathrm{H}) & 122.24 \\ 12.4(\mathrm{~s}, 1 \mathrm{H}) & 18.11 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (f) } \mathrm{C}_{3} \mathrm{H}_{4} \mathrm{Cl}_{2} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{19} \text { C-NMR } \\ \hline 2.34(\mathrm{~s}, 3 \mathrm{H}) & 171.82 \\ 11.3(\mathrm{~s}, 1 \mathrm{H}) & 79.36 \\ & 34.02 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (g) } \mathrm{C}_{5} \mathrm{H}_{8} \mathrm{Cl}_{2} \mathrm{O}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 1.42(\mathrm{~s}, 6 \mathrm{H}) & 180.15 \\ 6.10(\mathrm{~s}, 1 \mathrm{H}) & 77.78 \\ 12.4(\mathrm{~s}, 1 \mathrm{H}) & 51.88 \\ & 20.71 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (h) } \mathrm{C}_{5} \mathrm{H}_{9} \mathrm{BrO}_{2}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 0.97(\mathrm{t}, 3 \mathrm{H}) & 176.36 \\ 1.50(\mathrm{~m}, 2 \mathrm{H}) & 45.08 \\ 2.05(\mathrm{~m}, 2 \mathrm{H}) & 36.49 \\ 4.25(\mathrm{t}, 1 \mathrm{H}) & 20.48 \\ 12.1(\mathrm{~s}, 1 \mathrm{H}) & 13.24 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (i) } \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}_{3}\\\ &\begin{array}{cc} \hline{ }^{1} \text { H-NMR } & { }^{13} \text { C-NMR } \\ \hline 2.62(\mathrm{t}, 2 \mathrm{H}) & 177.33 \\ 3.38(\mathrm{~s}, 3 \mathrm{H}) & 67.55 \\ 3.68(\mathrm{~s}, 2 \mathrm{H}) & 58.72 \\ 11.5(\mathrm{~s}, 1 \mathrm{H}) & 34.75 \\ \hline \end{array} \end{aligned} $$

On a cyclohexane ring, an axial carboxyl group has a conformational energy of \(5.9 \mathrm{~kJ}\) (1.4 kcal)/mol relative to an equatorial carboxyl group. Consider the equilibrium for the alternative chair conformations of trans-1,4-cyclohexanedicarboxylic acid. Draw the less stable chair conformation on the left of the equilibrium arrows and the more stable chair on the right. Calculate \(\Delta G^{0}\) for the equilibrium as written, and calculate the ratio of the more stable chair to the less stable chair at \(25^{\circ} \mathrm{C}\).

Acetic acid has a boiling point of \(118^{\circ} \mathrm{C}\), whereas its methyl ester has a boiling point of \(57^{\circ} \mathrm{C}\). Account for the fact that the boiling point of acetic acid is higher than that of its methyl ester, even though acetic acid has a lower molecular weight.

Low-molecular-weight dicarboxylic acids normally exhibit two different \(\mathrm{p} K_{\mathrm{a}}\) values. Ionization of the first carboxyl group is easier than the second. This effect diminishes with molecular size, and, for adipic acid and longer chain dicarboxylic acids, the two acid ionization constants differ by about one \(\mathrm{p} K\) unit. $$ \begin{array}{llcc} \hline \begin{array}{l} \text { Dicarboxylic } \end{array} & \text { Structural } & & \\ \hline \text { Oxalic } & \text { Formula } & \mathrm{p} K_{\mathrm{a} 1} & \mathrm{p} K_{\mathrm{a} 2} \\ \text { Malonic } & \mathrm{HOOCCOOH} & 1.23 & 4.19 \\ \text { Succinic } & \mathrm{HOOCCH}{ }_{2} \mathrm{COOH} & 2.83 & 5.69 \\ \text { Glutaric } & \mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{2} \mathrm{COOH} & 4.16 & 5.61 \\ \text { Adipic } & \mathrm{HOOC}\left(\mathrm{CH}_{2}\right)_{3} \mathrm{COOH} & 4.31 & 5.41 \\ \hline \end{array} $$ Why do the two \(\mathrm{p} K_{\mathrm{a}}\) values differ more for the shorter chain dicarboxylic acids than for the longer chain dicarboxylic acids?
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