nmr problem set
DESCRIPTION
NMR ProblemTRANSCRIPT
10 9 8 7 6 5 4 3 2 1ppm
2.4 2.33.70 3.65
1.20 1.15 1.10
Problem R-16A: C4H8O2
300 MHz 1H NMR Spectrum in CDCl3Source: Aldrich Spectral Viewer/Reich
Problem R-16B: C4H8O2
300 MHz 1H NMR Spectrum in CDCl3Source: Aldrich Spectral Viewer/Reich
Problem Set 1 - NMR Spectra
Two Isomers of C4H8O2
0Hz
102030
O
O
10 9 8 7 6 5 4 3 2 1 0ppm
4.2 4.1 2.05 2.00 1.30 1.25
0Hz
102030O
O
ReichChem 345
2
3
3
CH2 CH3
CH2 CH3
CH3
O
2
3
3
CH2 CH3
CH2 CH3
CH3O
IHD = 1
IHD = 1
X
quartet
Observed protons are circledprotons causing splitting are underlined
X
Problem R-16C: C4H8O2
300 MHz 1H NMR Spectrum in CDCl3Source: Aldrich Spectral Viewer/Reich
Two Isomers of C4H8O2
10 9 8 7 6 5 4 3 2 1 0ppm
3.00
2.06
1.01
2.11
4.15 4.10
1.8 1.7 1.6
1.00 0.95
8.1 8.0
0Hz
102030
H O
O
Problem R-16D: C4H8O2
100 MHz 1H NMR SpectrumSource: Reich
12 11 10 9 8 7 6 5 4 3 2 1 0ppm
0.32
1.000.94
1.17
2.4 2.3 1.7 1.61.0
OH
O
CH2 CH3CH2CH2 CH3
CH2 CH2 O
CH2 CH3CH2 CH3CH2CH2 CH2
OH
IHD = 1
IHD = 1
H
O
Problem R-18L C4H9Br300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich
0Hz
102030
10 9 8 7 6 5 4 3 2 1 0
1.00
2.20
1.42
3.45 3.40 3.35
1.00 0.95 0.90
1.95 1.90 1.85 1.80 1.75 1.70 1.65 1.60 1.55 1.50 1.45 1.40 1.35
BrCH2 CH3CH2
CH2 CH2CH2
CH2 CH3X
pentet sextet
CH2 CH3IHD = 0
200 180 160 140 120 100 80 60 40 20 0ppm
13.2
21.3
33.6
34.8Problem R-18L: C4H9Br
75 MHz 13C NMR spectrumSolv: CDCl3Source: ASV/Reich
CD
Cl 3
9 8 7 6 5 4 3 2 1 0ppm
1.4
4.0 3.9
7.3 7.2 7.1 7.0 6.9 6.8
Problem R-7N C8H10O300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer 33-09
Problem R-7N: C8H10O75 MHz 13C NMR spectrumSolv: CDCl3Source: ASV/Reich
210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0ppm
14.9
63.3
114.
6
120.
6
129.
5
Infrared Spectrum:
USED PS-1 2003
5
2
3
CH2 CH3
CH2 CH3O
R
O
IHD = 4
20
40
60
80
100
Problem R-7R: C5H9BrO2300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer 33-09
cm-1
10 9 8 7 6 5 4 3 2 1 0ppm
1.001.05
1.00
1.43
4.2
3.6 3.5 3.4 3.3 3.2 3.1 3.0 2.9
1.3
1800 1600 1400 1200 1000 800 600 400 200
Pe
rce
nt T
ran
smitt
an
ce
0Hz
102030
Change of scaleat 2000 cm-1
Infrared Spectrum:
Br O
O
4000 3800 3600 3400 3200 3000 2800 2600 2400 2200 2000
CH2 CH2X CH2 CH2CH2 CH3
23
22
CH2 CH3O
IHD = 1
X
Problem R-7R: C5H9BrO2
75 MHz 13C NMR spectrumSolv: CDCl3Source: ASV/Reich
210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0ppm
14.2
25.9
37.8
61.0
170.
4
CD
Cl 3
Problem R-7P C9H10O300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer 33-09
10 9 8 7 6 5 4 3 2 1 0ppm
2.32
1.00
1.54
1.25 1.203.1 3.0 3.0
8.0 7.9 7.8 7.7 7.6 7.5 7.4
0Hz
102030
Infrared Spectrum:
O
USED PS-1 2003IHD = 5
CH2 CH3
CH2 CH3
2
3
5
R
R = electron withdrawing group
X
Problem R-7P C9H10O75 MHz 13C NMR spectrumSolv: CDCl3Source: ASV/Reich
210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0ppm
8.2
31.8
127.
912
8.5
132.
8
136.
9
200.
6
CD
Cl 3O
4
1
1
4
8 7 6 5 4 3 2 1 0ppm
2.00
2.98
2.03
6.14
1.01
7.95 7.90ppm
7.60 7.55 7.50 7.45 7.40 7.35ppm
2.85 2.80ppm
1.00ppm
2.40 2.35 2.30 2.25 2.20ppm
Problem R-7M: C11H14O300 MHz 1H NMRSolv: CDCl3Source: A. Fiedler/Reich
Ph
O
Problem R-7M C11H14O75 MHz 13C NMR spectrumSolv: CDCl3Source: ASV/Reich
210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0ppm
22.8
25.1
47.4
128.
012
8.5
132.
7
137.
3
199.
9
0Hz
102030
CH (CH3)2
CH (CH3)2CH2
CHCH2
R
R = electron withdrawing group
IHD = 5
nonet
X
Problem R-7K C5H9ClO2250 MHz 1H NMR spectrum in CDCl3Source: Adam Fiedler/Reich
8 7 6 5 4 3 2 1 0ppm
1.00
2.16
3.263.13
4.45 4.40 4.35 4.30 4.25 4.20 4.15ppm 1.70 1.65
ppm
1.35 1.30 1.25ppm
O
O
Cl
220 210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0
ppm
14.0
21.5
52.6
62.0
170.
1Problem R-7K: C5H9ClO2
75 MHz 13C NMR spectrumSolv: CDCl3Source: ASV/Reich
0Hz
102030
20
40
60
80
100
cm-11800 1600 1400 1200 1000 800 600 400 200
Pe
rce
nt T
ran
smitt
an
ce
Change of scaleat 2000 cm-1
Infrared Spectrum:
4000 3800 3600 3400 3200 3000 2800 2600 2400 2200 2000
CH2 CH3
CH CH3
CH2 CH3X
CH CH3X
X = O, Cl
IHD = 1
Cl
O
O
This is a reasonable alternativestructure - also fits the 1H NMR
data
Correct structure
CD
Cl 3
Problem R-7O C5H8300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer 33-09
10 9 8 7 6 5 4 3 2 1 0ppm
2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1 1.0 0.9
Hz02040
Problem R-7O: C5H8
75 MHz 13C NMR spectrumSolv: CDCl3Source: ASV/Reich
210 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 0ppm
H
CH2 CH3
2
13
2
CH2 CH3CH2CH2CH2H
triplet of doublets (td)
CH2H
IHD = 2
There is a long-range coupling(over 4 bonds) between thealkyne proton and the CH2 group
H
C
HC
Problem R-18C C10H12O300 MHz 1H NMR spectrum in CDCl3Source: Aldrich Spectra Viewer/Reich
0Hz
102030O
10 9 8 7 6 5 4 3 2 1 0
1.00
1.52
1.02
2.51
7.35 7.30 7.25 7.20 7.153.70 3.65 3.60
2.50 2.45 2.40
1.05 1.00 0.95
Ph
220 200 180 160 140 120 100 80 60 40 20 0ppm
7.8
35.1
49.8
126.
812
8.6
129.
313
4.4
208.
8
Problem R-18C: C10H12O75 MHz 13C NMR spectrum CDCl3Source: ASV/Reich
CD
Cl 3
CH2 CH3
quartetCH2 CH3
CH2
5R
22
3
O
R
IHD = 5
ReichChem 345
O
Problem R-18R: C5H10O2
300 MHz 1H NMR Spectrum in CDCl3Source: Aldrich Spectral Viewer/Reich
0Hz
102030
10 9 8 7 6 5 4 3 2 1 0ppm
1.00
1.49
1.09
1.57
1.00 0.95 0.90
1.70 1.65 1.60
2.05 2.004.05 4.00
O
Propyl acetate
Problem R-18S: C5H10O2
300 MHz 1H NMR Spectrum in CDCl3Source: Aldrich Spectral Viewer/Reich
3.70 3.65
10 9 8 7 6 5 4 3 2 1 0ppm
1.00
0.64 0.66
0.96
2.35 2.30 2.25
1.7 1.6
0.95 0.90
O
O
0Hz
102030methyl butyrate
CH2 CH3CH2 CH3CH2
CH2 CH2
CH2 CH3
CH2 CH3CH2
CH2 CH2
O
22
3
2 2
3
3
O CH3
3
IHD = 1
IHD = 1
CH3