A compound of mass 149 has the following elemental analysis: C = 80.48%, H=10.13%, N=9,39%. The IR, 1H and 13C NMR spectra are given below. In the 13C spectrum decoupled from the 1H the nature of the carbon was determined using the 13C DEPT spectra (not shown here). Assign the spectra and identify the compound. You can: 1) Calculate the molecular formula of compound A; 2) Calculate the unsaturation number; 3) Analysethe characteristic bands of the IR spectrum; 4) Assign and interpret all the data from the 1H and 13C NMR spectra).
A compound of mass 149 has the following elemental analysis: C = 80.48%, H=10.13%, N=9,39%. The IR, 1H and 13C NMR spectra are given below. In the 13C spectrum decoupled from the 1H the nature of the carbon was determined using the 13C DEPT spectra (not shown here). Assign the spectra and identify the compound. You can: 1) Calculate the molecular formula of compound A; 2) Calculate the unsaturation number; 3) Analysethe characteristic bands of the IR spectrum; 4) Assign and interpret all the data from the 1H and 13C NMR spectra).
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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![A compound of mass 149 has the following elemental analysis : C = 80.48%, H=10.13%,
N=9,39%. The IR, 1H and 13C NMR spectra are given below. In the 13C spectrum decoupled from the 1H the nature
of the carbon was determined using the 13C DEPT spectra (not shown here). Assign the spectra and identify
the compound. You can: 1) Calculate the molecular formula of compound A; 2) Calculate the unsaturation number; 3)
Analysethe characteristic bands of the IR spectrum; 4) Assign and interpret all the data from the 1H and 13C NMR
spectra).
3500
1800 1600 1400 1200 1000 800 600 cm¹
го протру
AU
I=3
I=1
I=1
4.0
8.0
I=2
I=2
71
150
3000
с
7.0
CH
C
125
2500
6.0
CH
2000
2,8; 40 ppm
5.0
100
Explain why the answer Is this one :
4,5 (6,8; 115)
C
CH3-NH THÀ
ppm
ppm 75
3.0
7,2; 128)
132
50
CH₂
2.0
CH
CH₂
25
CH 3
-CH
3,2; 27 CH3
I=6
1.0
1,2; 22](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F67ac0f1e-6d5e-4078-a32a-bf478f74a856%2F9b96f9b6-3c63-4d35-a2ff-edd9f8c57de7%2Fyhzqg57_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A compound of mass 149 has the following elemental analysis : C = 80.48%, H=10.13%,
N=9,39%. The IR, 1H and 13C NMR spectra are given below. In the 13C spectrum decoupled from the 1H the nature
of the carbon was determined using the 13C DEPT spectra (not shown here). Assign the spectra and identify
the compound. You can: 1) Calculate the molecular formula of compound A; 2) Calculate the unsaturation number; 3)
Analysethe characteristic bands of the IR spectrum; 4) Assign and interpret all the data from the 1H and 13C NMR
spectra).
3500
1800 1600 1400 1200 1000 800 600 cm¹
го протру
AU
I=3
I=1
I=1
4.0
8.0
I=2
I=2
71
150
3000
с
7.0
CH
C
125
2500
6.0
CH
2000
2,8; 40 ppm
5.0
100
Explain why the answer Is this one :
4,5 (6,8; 115)
C
CH3-NH THÀ
ppm
ppm 75
3.0
7,2; 128)
132
50
CH₂
2.0
CH
CH₂
25
CH 3
-CH
3,2; 27 CH3
I=6
1.0
1,2; 22
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