EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
6th Edition
ISBN: 9781305687875
Author: Gilbert
Publisher: CENGAGE LEARNING - CONSIGNMENT
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Chapter 8.5, Problem 3E
(a)
Interpretation Introduction
Interpretation: The fragments responsible for the given m/z values needs to be determined.
Concept Introduction: A mass spectrum is formed between intensity and m/z ratio which represents a chemical analysis. For a sample, mass spectrum is a pattern that represents the distribution of ion by mass in sample. Here, m/z ratio is mass to charge ratio of the sample.
(b)
Interpretation Introduction
Interpretation: The reason for very low density of the peak at 113 needs to be explained.
Concept Introduction: A mass spectrum is formed between intensity and m/z ratio which represents a chemical analysis. For a sample, mass spectrum is a pattern that represents the distribution of ion by mass in sample. Here, m/z ratio is mass to charge ratio of the sample.
(c)
Interpretation Introduction
Interpretation: The value of m/z needs to be calculated for the parent peak and reason for it to not observed needs to be explained.
Concept Introduction: A mass spectrum is formed between intensity and m/z ratio which represents a chemical analysis. For a sample, mass spectrum is a pattern that represents the distribution of ion by mass in sample. Here, m/z ratio is mass to charge ratio of the sample.
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A compound gives a mass spectrum with essentially only three peaks at m/z = 77 (40%), 112 (100%), and 114 (33%). Identify the compound.
Q.1.
Predict the relative intensity of the M, M+1, and M+2 peaks for a low
resolution mass spectrum of a compound with molecular formula, C6H13Br. The
natural abundances of the select isotopes are given below.
35 Cl: 75.8%
79Br: 50.5%
12C: 98.89%
37Cl: 24.2% (The ratio is almost 3:1)
81Br: 49.5% (The ratio is almost 1:1)
13C: 1.11%
Mass spectroscopic.
What is the isotopic peak pattern of each :
Organic compound with 2 bromine atoms.
Organic compound with 2 chlorine atoms.
Chapter 8 Solutions
EBK EXPERIMENTAL ORGANIC CHEMISTRY: A M
Ch. 8.2 - Prob. 1ECh. 8.2 - Prob. 2ECh. 8.2 - Prob. 3ECh. 8.2 - Prob. 4ECh. 8.2 - Prob. 5ECh. 8.2 - Prob. 6ECh. 8.2 - Prob. 7ECh. 8.2 - Prob. 8ECh. 8.2 - Prob. 9ECh. 8.2 - Prob. 10E
Ch. 8.2 - Prob. 11ECh. 8.2 - Prob. 12ECh. 8.2 - Prob. 13ECh. 8.2 - Prob. 14ECh. 8.2 - Prob. 15ECh. 8.2 - Prob. 16ECh. 8.2 - Prob. 17ECh. 8.2 - Prob. 18ECh. 8.3 - Prob. 1ECh. 8.3 - Prob. 2ECh. 8.3 - Prob. 3ECh. 8.3 - Prob. 4ECh. 8.3 - Prob. 5ECh. 8.3 - Prob. 6ECh. 8.3 - Prob. 7ECh. 8.3 - Prob. 8ECh. 8.3 - Prob. 9ECh. 8.3 - Prob. 1.1ECh. 8.3 - Prob. 1.2ECh. 8.3 - Prob. 1.3ECh. 8.4 - Prob. 1ECh. 8.4 - Prob. 2ECh. 8.4 - Prob. 3ECh. 8.4 - Prob. 4ECh. 8.4 - Prob. 5ECh. 8.5 - Prob. 1ECh. 8.5 - Prob. 2ECh. 8.5 - Prob. 3ECh. 8.5 - Prob. 4E
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