R Spectrum (KBr disc) 100 % of base peak 3000 40 80 13C NMR Spectrum (50 0 MHz. CDCI, solution) 91 2000 V (cm¹) 1600 120 160 m/e DEPT CH₂ CH₁ CH proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 160 M** 182 1200 800 Mass Spectrum C14H14 280 two signals at higher field two signals at higher field solvent 80 200 240 -Resolves into -Resolves into 120 0.0 0.5 1.0 1.5 absorbance 200 40 250 UV spectrum 185 3 mg/100 mis path length 0.2 cm solvent: ethanol 300 λ (nm) 350 08 (ppm) TMS 0

Organic Chemistry
8th Edition
ISBN:9781305580350
Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote
Chapter14: Mass Spectrometry
Section: Chapter Questions
Problem 14.37P
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Question
What is the IR and HNMRAndCNMR and mass
IR Spectrum
(KBr disc)
4000
100
80
60
408
20
% of base peak
3000
80
40
13C NMR Spectrum
(50 0 MHz. CDCI, solution)
91
2000
120
V (cm¹)
DEPT CH₂ CH CH
proton decoupled
200
¹H NMR Spectrum
(200 MHz, CDCI, solution)
10
9
8
m
1600
1200
160
m/e
160
M**
182
800
0.0
Mass Spectrum 0.5
1.0
C14H14
240
280
Resolves into
two signals at higher field
two signals at higher field
solvent
80
5 4
200
<-Resolves into
120
7 6
1.5
Problem 36
UV spectrum
1853 mg/100 mis
path length 02 cm
solvent: ethanol
300
200
40
3
350
08 (ppm)
TMS
1
250
λ (nm)
2
1
0
8 (ppm
Transcribed Image Text:IR Spectrum (KBr disc) 4000 100 80 60 408 20 % of base peak 3000 80 40 13C NMR Spectrum (50 0 MHz. CDCI, solution) 91 2000 120 V (cm¹) DEPT CH₂ CH CH proton decoupled 200 ¹H NMR Spectrum (200 MHz, CDCI, solution) 10 9 8 m 1600 1200 160 m/e 160 M** 182 800 0.0 Mass Spectrum 0.5 1.0 C14H14 240 280 Resolves into two signals at higher field two signals at higher field solvent 80 5 4 200 <-Resolves into 120 7 6 1.5 Problem 36 UV spectrum 1853 mg/100 mis path length 02 cm solvent: ethanol 300 200 40 3 350 08 (ppm) TMS 1 250 λ (nm) 2 1 0 8 (ppm
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