ORGANIC CHEMISTRY-W/S.G+SOLN.MANUAL
ORGANIC CHEMISTRY-W/S.G+SOLN.MANUAL
8th Edition
ISBN: 9780134595450
Author: Bruice
Publisher: PEARSON
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Chapter 20.3, Problem 4P

a)

Interpretation Introduction

Interpretation:

Whether D-erythrose and L-erythose are enantiomers or diastereomers has to be stated.

Concept Introduction:

Enantiomers are the compounds with same chemical formula.  These are known as optical isomers.  Enantiomers are basically stereo-isomers of the same compound.  Two compounds are called enantiomers of each other if these are formed a non-superimposable mirror images.

Diastereomers are also the compounds with the same chemical formula.  These are also the types of stereo-isomers.  Two compounds are called diastereomers of each other if both the compounds have different configuration at the one or more (not all) equivalent stereo centers.  These are not mirror images of each other.

b)

Interpretation Introduction

Interpretation:

Whether L-erythrose and L-threose, enantiomers or diastereomers are to be stated.

Concept Introduction:

Enantiomers are the compounds with same chemical formula.  These are known as optical isomers.  Enantiomers are basically stereo-isomers of the same compound. Two compounds are called enantiomers of each other if these are formed a non-superimposable mirror images.

Diastereomers are also the compounds with the same chemical formula.  These are also the types of stereo-isomers.  Two compounds are called diastereomers of each other if both the compounds have different configuration at the one or more (not all) equivalent stereo centers.  These are not mirror images of each other.

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Relative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 100 HS-NJ-0547 80 60 31 20 S1 84 M+ absent 10 30 40 50 60 70 80 90 100 100- MS2016-05353CM 80- 60 40 20 135 137 S2 164 166 0-m 25 50 75 100 125 150 m/z 60 100 MS-NJ-09-43 40 20 20 80 45 S3 25 50 75 100 125 150 175 m/z
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

Chapter 20 Solutions

ORGANIC CHEMISTRY-W/S.G+SOLN.MANUAL

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