The portion of polystyrene molecule with head to tail arrangement of styrene (monomer unit) needs to be sketched. 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 .

Question

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

Answer 1

Question

Chapter 23, Problem 4QAP

Interpretation Introduction

Interpretation:

The portion of polystyrene molecule with head to tail arrangement of styrene (monomer unit) needs to be sketched.

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.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Students have asked these similar questions

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

Part IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained. Draw the structures of these fragments.

For each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.