Wavelengh m) 10 12 13 14 15 16 26 27 28 29 ! 300 3600 3400 3200 3000 2800 2600 2400 23200 2000 1S00 1600 1400 1200 1000 S00 600 Wavenumber icm 100 |44 72 57 20 40 60 80 100 mlz Relative abundance
Wavelengh m) 10 12 13 14 15 16 26 27 28 29 ! 300 3600 3400 3200 3000 2800 2600 2400 23200 2000 1S00 1600 1400 1200 1000 S00 600 Wavenumber icm 100 |44 72 57 20 40 60 80 100 mlz Relative abundance
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...
Related questions
Question
How do I identify the compounds with IR and Mass Spectra, These are three compounds
Transcribed Image Text:Wavelength (im)
5,5
10
12
13
15
16
14
26 27 28 29 3
3600
3400 3200 3000 2800 2600 2400 2200
2000
1800
1600
1400
1200
1000
800
600
3800
Wavenumber iem
100
| 44
72
50
57
20
40
60
80
100
mlz
Relative abundance
Transcribed Image Text:b.
Wavelength (um)
5.5
IS 16
12
13
14
9. The IR and mass spectra of three different compounds are shown below. Identify each
compound.
10
2.5 26 2.7 28 2.9 3
а.
Wivelength (um
35
12
15 16
3.5
4.5
10
11
13
14
25 26 2.7 28 29 3
1600
1400
1000
S00
600
2000
I800
1200
4000 3800 3600 3400 3200 3000 2800 2600 2400 2200
Wavenunber (em
BR00 W0 1400 3200 MO00 2500 2600 2400 2200
2000
1400
1200
100
S00
600
1800
1600
a00
180
Wavenumber iem
100
83
100
43
55
58
50
50
41
29
100
71
85
20
40
60
80
100
120
140
160
20
40
60
80
100
120
mlz
mlz
Relative abundance
Relative abundance
Expert Solution
Step 1
Spectra 1 : 2800 cm-1 and 2700 cm-1 aldehyde C-H stretch. 1730 cm-1 carbonyl stretch. Therefore the compound is n-butanal.
Spectra 2 ; 1720 cm-1 carbonyl stretch. Therefore the compound is 2-hexanone.
Spectra 3: 1630 cm-1 C=C stretch. M:M+2 ratio is 1:1. Therefore the compound is having bromine. Therefore the compound is 6-bromo-1-hexene.
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