IR Spectrum (KBr disc) ستا 4000 100 80 60 20 % of base peak 13C NMR Spectrum (100.0 MHz, D₂0 solution) 30 40 yu 3.0 3000 proton decoupled 200 ¹H NMR Spectrum (400 MHz. D₂0 solution) Note: there are 3 protons which exchange with the D₂0 solvent 3.2 2000 expansions ppm 1600 M+= 103 80 120 160 m/e DEPT CH₂ CH₂ CH 160 1600 V (cm¹) 2.2 200 120 2.0 ppm J 800 Mass Spectrum 1200 C4H₂NO₂ 280 240 80 H₂O and HOD in solvent Problem 21 No significant UV absorption above 220 nm 40 2 08 (ppm) 1 0 loom

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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80
um
ution)
ed
I
um
lution)
4000
100
80
IR Spectrum
(KBr disc)
60
40
20
% of base peak
10
30
110
40
13C NMR
Spectrum
(100.0 MHz,
D₂O solution)
proton decoupled
200
¹H NMR Spectrum
(400 MHz. D₂0 solution)
Note: there are 3 protons
which exchange with
the D₂O solvent
3.2
3000
3.0
9
2000
M+= 103
120
expansions
ppm
80
m/e
DEPT CH₂ CH₂ CH
160
8
1600
V (cm¹)
1600
160
1200
800
Mass Spectrum
C4H₂NO₂
200 240 280
120
Problem 21
No significant UV
absorption above 220 nm
40
08 (ppm)
0
8 (ppm
80
H₂O and HOD
in solvent
2.2 2.0
ppm
1
1
7 6 5 4 3 2 1
Transcribed Image Text:80 um ution) ed I um lution) 4000 100 80 IR Spectrum (KBr disc) 60 40 20 % of base peak 10 30 110 40 13C NMR Spectrum (100.0 MHz, D₂O solution) proton decoupled 200 ¹H NMR Spectrum (400 MHz. D₂0 solution) Note: there are 3 protons which exchange with the D₂O solvent 3.2 3000 3.0 9 2000 M+= 103 120 expansions ppm 80 m/e DEPT CH₂ CH₂ CH 160 8 1600 V (cm¹) 1600 160 1200 800 Mass Spectrum C4H₂NO₂ 200 240 280 120 Problem 21 No significant UV absorption above 220 nm 40 08 (ppm) 0 8 (ppm 80 H₂O and HOD in solvent 2.2 2.0 ppm 1 1 7 6 5 4 3 2 1
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