OWLV2 FOR MASTERTON/HURLEY'S CHEMISTRY:
OWLV2 FOR MASTERTON/HURLEY'S CHEMISTRY:
8th Edition
ISBN: 9781305079304
Author: Hurley
Publisher: IACCENGAGE
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Chapter 23, Problem 14QAP
Interpretation Introduction

(a)

Interpretation:

The monomers from which the given condensation polymer is formed needs to be identified.

Concept introduction:

A polymer is a long chain consists of large number of monomer units. In a polymer, the monomers are linked to each other in a continuous or repetitive manner. These monomer units are linked to each other either through the formation of peptide linkage, glycosidic linkage or by removal of any moiety such as a water molecule. Polyvinyl chloride, Bakelite and polystyrene are some of the examples of the polymers.

Interpretation Introduction

(b)

Interpretation:

The monomers from which the given condensation polymer is formed needs to be identified.

Concept introduction:

A polymer is a long chain consists of large number of monomer units. In a polymer, the monomers are linked to each other in a continuous or repetitive manner. These monomer units are linked to each other either through the formation of peptide linkage, glycosidic linkage or by removal of any moiety such as a water molecule. Polyvinyl chloride, Bakelite and polystyrene are some of the examples of polymers.

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Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100
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Chapter 23 Solutions

OWLV2 FOR MASTERTON/HURLEY'S CHEMISTRY:

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