A3 2. Circle the compound that matches the given spectrum.For "Excellent", rank from strongest to weakest.Br 0H₂NOHOHpka = 3.44pka 4.26100-80-60-4020Relative Intensity11-4-435425#875m/zSource: SDBS100OHpka - 3.34mempen125150OHpka = 4.77

Presence of one chlorine atom gives M and M+2 peaks with 3:1 intensity ratio

2. Circle the compound that matches the given spectrum. For "Excellent", rank from strongest to weakest. Br O H₂N OH OH pka 3.44 pka 4.26 100- 80 Relative Intensity 20 0 1-4-4354 25 75 m/z Source: SDBS 100 OH pka 3.34 mythe 125 150 OH pka = 4.77

2. Circle the compound that matches the given spectrum. For "Excellent", rank from strongest to weakest. Br O H₂N OH OH pka 3.44 pka 4.26 100- 80 Relative Intensity 20 0 1-4-4354 25 75 m/z Source: SDBS 100 OH pka 3.34 mythe 125 150 OH pka = 4.77

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Concept explainers

Mass Spectrometry

This process is an instrumental method in which a sample is changed to fast moving positive ions by electron bombardment. The charged particles vary due to their different mass to charge ratio and are estimated with the help of this difference. It is also called mass spectroscopy.

Question

2. Circle the compound that matches the given spectrum.
For "Excellent", rank from strongest to weakest.
Br 0
H₂N
OH
OH
pka = 3.44
pka 4.26
100-
80-
60-
40
20
Relative Intensity
11-4-4354
25
#8
75
m/z
Source: SDBS
100
OH
pka - 3.34
mempen
125
150
OH
pka = 4.77

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