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Write the structural formula of two units of the polymer formed from (a) the reaction of terephthalic acid with 1,2 -diaminoethane, \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\); (b) the polymerization of 4-hydroxybenzoic acid.

Short Answer

Expert verified
For part (a), the structural formula for two units of the polymer is \( -C_6H_4(CO)NHCH_2CH_2NH(CO)C_6H_4- \). For part (b), it is \( -C_6H_4(O)CO -\).

Step by step solution

01

Understand the Reaction for Part (a)

The reaction of terephthalic acid with 1,2-diaminoethane is a condensation polymerization, which is a step-growth polymerization involving the loss of small molecules. In this case, water molecules are lost as terephthalic acid and 1,2-diaminoethane react to form amide bonds in a polyamide polymer. Terephthalic acid has the formula \(C_6H_4(CO_2H)_2\) and 1,2-diaminoethane is \(H_2NCH_2CH_2NH_2\). For each repeating unit, two hydrogen atoms from the amine groups of 1,2-diaminoethane react with the carboxyl groups of the terephthalic acid to release two water molecules and form amide bonds.
02

Write the Structural Formula for Part (a)

Considering the stoichiometry of the reaction, the repeating structural unit will consist of a single terephthalic acid unit connected to two 1,2-diaminoethane units. The acid (COOH) ends of terephthalic acid will form amide linkages with the amine (NH2) groups of the 1,2-diaminoethane by losing water molecules. Thus, the structural formula for two repeating units of the resultant polymer can be depicted with amide linkages as \( -C_6H_4(CO)NHCH_2CH_2NH(CO)C_6H_4- \) connected in a linear chain.
03

Understand the Polymerization of Part (b)

For part (b), the polymerization of 4-hydroxybenzoic acid is a type of step-growth polymerization as well. Here, the carboxylic acid group reacts with the hydroxy group of another 4-hydroxybenzoic acid molecule, resulting in the formation of an ester bond and the release of a water molecule. 4-hydroxybenzoic acid has the formula \(C_6H_4(OH)CO_2H\).
04

Write the Structural Formula for Part (b)

In the polymer, the ester linkages will be formed by the reaction of the carboxylic and hydroxy groups of adjacent monomer units. The resulting polymer will have repeated ester linkages and phenol rings from the continual reaction of monomer units. The structural formula for two repeating units of this polymer can be represented as \( -C_6H_4(O)CO -\) in a linear chain.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polymer Chemistry
The fascinating world of polymer chemistry is where small molecules called monomers transform into giant macromolecules known as polymers. Imagine linking many identical or similar building blocks, much like a child connects Lego pieces to create something more substantial. That's essentially how polymers form. Through specific types of chemical reactions, these monomers bond together, and as they do, they create durable materials that can have various properties based on their composition and structure.

These materials support our modern lifestyle in countless ways, from the plastic containers that keep our food fresh to the fibers in our clothing. Understanding how polymers form, their structures, and their unique properties equips us to engineer materials for specific needs, making the study of polymer chemistry an essential field in science and technology.
Amide Bond Formation
Amide bond formation is a central reaction in condensation polymerization, particularly in creating polyamides, proteins, and other essential biomolecules. These bonds form when a carboxylic acid group \( -COOH \) from one molecule reacts with an amine group \( -NH_2 \) from another. In this process, a molecule of water \( H_2O \) is released as a byproduct, typically referred to as condensation.

During the synthesis of a polyamide, such as nylon, the amide linkages (also known as peptide bonds) become the backbone of the polymer chain. This linkage is robust and contributes to the high tensile strength and resistance to harsh conditions seen in materials made from polyamides, which is why they're widely used in textiles and engineering plastics.
Ester Linkage
On the other hand, ester linkages are the name of the game when it comes to forming polyesters. Think of an ester bond as a bridge connecting a carboxylic acid group and an alcohol group. When these two groups react, they also partake in a condensation reaction, ejecting a water molecule and forming an ester linkage \( -COO- \).

The resulting ester bonds create a polymer structure that's flexible, durable, and has a range of applications like in drink bottles, clothing, and magnetic recording tapes. Polyesters can be easily modified to attain specific properties such as flame resistance or added strength, further increasing their versatility in the manufacturing world.
Step-Growth Polymerization
Step-growth polymerization is the elegant dance where small molecules join step by step, creating longer and longer chains until a polymer is born. This process differs from other polymerization methods because monomers, dimers, trimers, and higher oligomers combine at any stage of the reaction. Unlike chain-growth polymerization where the growth of polymer chains proceeds primarily at the ends, step-growth polymerization allows for the growth at any point along the reacting molecule.

Industrially and biologically important polymers, including polyamides and polyesters, are synthesized through this gradual and versatile process. Every bond formation is a step forward in creating a large macromolecule, giving polymers their remarkable properties. This method's central feature is that high molecular weight polymers are formed late in the reaction, requiring high purity and close control of the stoichiometry to ensure uniform polymer structures.

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Most popular questions from this chapter

Classify each of the following reactions as (1) an addition reaction, (2) a nucleophilic substitution reaction, (3) an clectrophilic substitution reaction, or (4) a condensation reaction: (a) the reaction of 1-butene with chlorine in the absence of light; (b) the polymerization of the amino acid glycine; (c) the hydrogenation of 1-butyne; (d) the polymerization of styrene, \(\mathrm{CH}_{2} \mathrm{CHC}_{6} \mathrm{H}_{5}\); (c) the reaction of methylamine with butanoic acid.

In the amide group, rotation is restricted about the \(C-N\) bond, so that the \(\mathrm{C}, \mathrm{N}\), and \(\mathrm{O}\) atoms in that group are normally in the same plane. This rigidity helps to form the primary structure of proteins. The amino acid glycine can form a dipeptide, composed of two glycine monomers. (a) Draw the Lewis structure of the glycine dipeptide. (b) Explain how resonance structures can account for the restricted rotation and draw a second resonance structure for the dipeptide.

Draw the structure of the principal product formed from cach of the following condensation reactions: (a) butanoic acid with 2-propanol; (b) ethanoic acid with 1-pentanol; (c) hexanoic acid with methylethyl amine; (d) ethanoic acid with propylamine.

Write the condensed structural formulas of the principal products of the reaction that takes place when (a) ethylene glycol, 1,2 -ethanediol, is heated with stearic acid, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{COOH}\); (b) ethanol is heated with oxalic acid, \(\mathrm{HOOCCOOH}\); (c) 1 -butanol is heated with propanoic acid.

You are given samples of 1-propanol, pentane, and ethanoic acid. Describe how you would use chemical tests, such as aqueous solubility and acid-base indicators, to distinguish among the three compounds.

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