4000 100 IR Spectrum (quid film) 88888 80 Få 60 40% 8 3000 CO 80 40 13C NMR Spectrum (50.0 MHz. CDCI, solution) proton coupled proton decoupled 200 2000 v (cm'¹) 1600 127 120 160 m/e 160 M 156 prapon 1 12 50 800 0.0 Mass Spectrum 0.5 Mt M 1.0 1.5 120 1200 contains a halogen 200 240 1=2 280 solvent 80 Problem 12 بما انه خفي لا يوجد اي تفاعل كمان بين لاكي له absorbance 200 40 UV spectrum 5.00 mg/10 mis path length :0.50 cm solvent cyclohexane 300 350 Calz IF No TMS in this sample nea= 12ntent/t = 156 42 250 λ (nm) ele 0 8 (ppm)

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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Problem 12
IR Spectrum
(quid fim)
--
12 s0
4000
3000
2000
1600
1200
600
0.0P
v (cm")
100
Mass Spectrum 0.5
80
M 156
UW spectrum
5.00 mg/10 mis
path length 050 cm
solvent cyclohexane
127
1.0
29
contains a halogen
1.5
40
80
120
160
200
240
280
200
250
300
350
m/e
2 (nm)
13C NMR Spectrum
No TMS in
this sample
(50.0 MHz CDCI, solution)
solvent
proton coupled
= 156
proton decoupled
200
160
120
80
40
8 (ppm)
H NMR Spectrum
(200 MHz, CDCI, solution)
10
8.
7
4
3
8 (ppm)
% of base peak
absorbance
Transcribed Image Text:Problem 12 IR Spectrum (quid fim) -- 12 s0 4000 3000 2000 1600 1200 600 0.0P v (cm") 100 Mass Spectrum 0.5 80 M 156 UW spectrum 5.00 mg/10 mis path length 050 cm solvent cyclohexane 127 1.0 29 contains a halogen 1.5 40 80 120 160 200 240 280 200 250 300 350 m/e 2 (nm) 13C NMR Spectrum No TMS in this sample (50.0 MHz CDCI, solution) solvent proton coupled = 156 proton decoupled 200 160 120 80 40 8 (ppm) H NMR Spectrum (200 MHz, CDCI, solution) 10 8. 7 4 3 8 (ppm) % of base peak absorbance
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