100 80 IR Spectrum (KBr disc) 4000 % of bane peak 3000 5 20 40 80 13C NMR Spectrum (100 0 MHz. CDCI, solution) DEPT CH, CH₂ CH proton decoupled 200 ¹H NMR Spectrum (400 MHz, CDCI, solution) 10 9 8 1693 v (cm³) 0 120 160 m/e 2000 M-112 160 7 1600 6 1200 120 800 Mass Spectrum C6H80₂ 280 200 240 5 solvent 80 4 Problem 10 M= 12+2 (0₂14-8 = C=0₂ No significant UV absorption above 220 nm 2*6 +2 (8-2) + 12+2 (6) 4 =16/6 0 8 (ppm) TMS L 0 8 (pp 3 40 2 11 J 1 L
100 80 IR Spectrum (KBr disc) 4000 % of bane peak 3000 5 20 40 80 13C NMR Spectrum (100 0 MHz. CDCI, solution) DEPT CH, CH₂ CH proton decoupled 200 ¹H NMR Spectrum (400 MHz, CDCI, solution) 10 9 8 1693 v (cm³) 0 120 160 m/e 2000 M-112 160 7 1600 6 1200 120 800 Mass Spectrum C6H80₂ 280 200 240 5 solvent 80 4 Problem 10 M= 12+2 (0₂14-8 = C=0₂ No significant UV absorption above 220 nm 2*6 +2 (8-2) + 12+2 (6) 4 =16/6 0 8 (ppm) TMS L 0 8 (pp 3 40 2 11 J 1 L
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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Question
What is the IR and HNMRAndCNMR and mass
![IR Spectrum
(KBr disc)
4000
100
80
% of base peak
3000
56
40
80
13C NMR Spectrum
(100 0 MHz. CDCI, solution)
DEPT CH₂ CH₂ CH
proton decoupled
200
¹H NMR Spectrum
(400 MHz, CDCI, solution)
10
9
8
7
1693
v (cm³)
O
120 160
m/e
2000
M-112
160
7
1600
6
1200
120
800
Mass Spectrum
C6H8O₂
280
200 240
5
solvent
80
4
Problem 10
M = 12+2
(0₂14-8 = 6
=3
C=0₂
No significant UV
absorption above 220 nm
2X6 + 2 (8-2) 1 (E
12+2-(6) +4
=16/2
0
3
40
2
116
1
8 (ppm)
TMS
L
0
8 (ppm)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0f909aea-afe8-4a7c-a10d-2d338ae92aaa%2F6455c3a8-e30b-422a-90f6-993c3363d1dd%2Fyilos48_processed.jpeg&w=3840&q=75)
Transcribed Image Text:IR Spectrum
(KBr disc)
4000
100
80
% of base peak
3000
56
40
80
13C NMR Spectrum
(100 0 MHz. CDCI, solution)
DEPT CH₂ CH₂ CH
proton decoupled
200
¹H NMR Spectrum
(400 MHz, CDCI, solution)
10
9
8
7
1693
v (cm³)
O
120 160
m/e
2000
M-112
160
7
1600
6
1200
120
800
Mass Spectrum
C6H8O₂
280
200 240
5
solvent
80
4
Problem 10
M = 12+2
(0₂14-8 = 6
=3
C=0₂
No significant UV
absorption above 220 nm
2X6 + 2 (8-2) 1 (E
12+2-(6) +4
=16/2
0
3
40
2
116
1
8 (ppm)
TMS
L
0
8 (ppm)
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