SYNTHETIC AND NATURA L

CHAPTER 25

SYNTHETIC AND NATURA L

ORGANIC POLYMERS

Problem Categories Biological: 25.19, 25.20, 25.21, 25.35, 25.36, 25.39, 25.41, 25.43, 25.44, 25.51, 25.52, 25.53.Conceptual: 25.22, 25.27, 25.28, 25.29, 25.30, 25.32, 25.34, 25.37, 25.38, 25.40, 25.42, 25.45.

Descriptive: 25.7, 25.9, 25.10, 25.31, 25.33.

Industrial: 25.51.

25.7 T he reaction is initiated be a radical, R·

R·  CF

2 CF 2

® RCF

2 CF 2 · The product is also a radical, and the reaction continues.

RCF

2 CF 2

·  CF

2 CF 2

® RCF

2 CF 2 CF 2 CF 2 · etc...

25.8 The repeating structural unit of the polymer is:

Does each carbon atom still obey the octet rule?

25.9 The general reaction is a condensation to form an amide.

The polymer chain looks like:

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CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 954

Note that both reactants are disubstituted benzene derivatives with the substituents in the para or 1,4 positions.

25.10 Polystyrene is formed by an addition polymerization reaction with the monomer, styrene, which is a phenylsubstituted ethylene. The structures of styrene and polystyrene are shown in Table 25.1 of your text.

25.11 The structures are as shown.

25.12 The structures are shown.

25.19 alanylglycine and glycylalanine are shown in Figure 25.8.

25.20 The main backbone of a polypeptide chain is made up of the a carbon atoms and the amide group repeating alternately along the chain.

C H R 1 C O N H C R 2 H C O N H amide groups α carbon α carbon

For each R group shown above, substitute the distinctive side groups of the two amino acids. Their are two possible dipeptides depending on how the two amino acids are connected, either glycinelysine or

lysineglycine. The structures of the dipeptides are:

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CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 955

and

25.21 The structure of the polymer is:

25.22 The rate increases in an expected manner from 10°C to 30°C and then drops rapidly. The probable reason for this is the loss of catalytic activity of the enzyme because of denaturation at high temperature.

25.27 There are two common structures for protein molecules, an a helix and a b pleated sheet. The ahelical structure is stabilized by intramolecular hydrogen bonds between the NH and CO groups of the main chain, giving rise to an overall rodlike shape. The CO group of each amino acid is hydrogen-bonded to the NH group of the amino acid that is four residues away in the sequence. In this manner all the main-chain CO and NH groups take part in hydrogen bonding. The bpleated structure is like a sheet rather than a rod. The polypeptide chain is almost fully extended, and each chain forms many intermolecular hydrogen bonds with adjacent chains. In general, then, the hydrogen bonding is responsible for the three dimensional geometry of the protein molecules.

In nucleic acids, the key to the double-helical structure is the formation of hydrogen bonds between bases in the two strands. Although hydrogen bonds can form between any two bases, called base pairs, the most favorable couplings are between adenine and thymine and between cytosine and guanine.More information concerning the importance of hydrogen bonding in biological systems is in Sections 25.3 and 25.4 of the text.

25.28 Nucleic acids play an essential role in protein synthesis. Compared to proteins, which are made of up to 20 different amino acids, the composition of nucleic acids is considerably simpler. A DNA or RNA molecule Copyright © McGraw Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw Hill Education. 3 / 4

CHAPTER 25: SYNTHETIC AND NATURAL ORGANIC POLYMERS 956

contains only four types of building blocks: purines, pyrimidines, furanose sugars, and phosphate groups.Nucleic acids have simpler, uniform structures because they are primarily used for protein synthesis, whereas proteins have many uses.

25.29 When proteins are heated above body temperature they can lose some or all of their secondary and tertiary structure and become denatured. The denatured proteins no longer exhibit normal biological activity.

25.30 The sample that has the higher percentage of CG base pairs has a higher melting point because CG base pairs are held together by three hydrogen bonds. The AT base pair interaction is relatively weaker because it has only two hydrogen bonds. Hydrogen bonds are represented by dashed lines in the structures below.

25.31 As is described in Section 25.3 of the text, acids denature enzymes. The citric acid in lemon juice denatures the enzyme that catalyzes the oxidation so as to inhibit the oxidation (browning).

25.32 Leg muscles are active having a high metabolism, which requires a high concentration of myoglobin. The high iron content from myoglobin makes the meat look dark after decomposition due to heating. The breast meat is “white” because of a low myoglobin content.

25.33 The cleavage reaction is:

25.34 Insects have blood that contains no hemoglobin. Thus, they rely on simple diffusion to supply oxygen. It is unlikely that a human-sized insect could obtain sufficient oxygen by diffusion alone to sustain its metabolic requirements.

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