Chem 460, Problem Set #3 (Answer)
(1) Identify protons with unique chemical shifts by labeling them with a, b, c etc
in the following molecules. Identify all the enantiotopic protons and diastereotopic
protons.
(a)
(b)
diastereotopic protons
He
Hd
Hd
O
Hg
O
Hf
Hb
O
Ha
Hg
Hg
Ha
Hm
Hg
Hg
Hc
Ha
Hh
He
Hg
Hc
Ha
enantiotopic
protons
Hi
Cl
Hg
Hk Hj
Hf
Hb
Hl
axial and equatorial geminal protons are diastreotopic protons
(2) Download both 1H- and 13C-NMR data from the SDBS database for the
compounds shown below. Calculate the expected chemical shift for all the
protons and carbons. Do they match to the observed shifts? Make a comparison
table.
(a)
(b)
O
CH3
CH3
HO
Cl
Cl
(a)
SDBS#51723
1
H-NMR
Assign.
A
B
C
D
E
F
Shift(ppm)
7.430
7.409
7.302
2.897
1.731
0.981
CH3
HA: 7.27 + 0.02 + 0.02 + 0.09 = 7.40
HB: 7.27 + 0.64 – 0.06 – 0.06 = 7.79
HC: 7.27 + 0.02 – 0.04 + 0.09 = 7.34
HD: 1.2 + 1.7 = 2.9
HE: 1.2 + 0.3 = 1.5
HF: 0.9 + 0.1 = 1.0
13
C-NMR
ppm Int. Assign.
202.33 231 1
137.93 292 2
137.08 297 3
131.95 231 4
130.39 917 5
129.93 877 6
127.30 872 7
44.81 743 8
17.67 1000 9
13.70 823 10
C1: ~ 200 ppm
C2: 128.5 + 7.8 + 0.2 – 2.0 =134.5
C3: 128.5 + 6.4 + 1.0 + 2.8 = 138.7
C4: 128.5 + 6.4 + 1.0 – 0.4 = 135.5
C5: 128.5 + 0.2 + 0.2 - 0.4 = 128.5
C6: 128.5 + 1.0 + 1.0 – 0.4 = 130.1
C7: 128.5 + 0.2 – 2.0 - 0.4 = 126.3
For C8, C9 and C10, calculate the shifts in propane, and then attach a COR group at the
terminal.
C8: -2.5 + 9.1 + 9.4 + 31 = 47
C9: - 2.5 + 9.1 x 2 + 1 = 16.7
C10: -2.5 + 9.1 + 9.4 – 2 =14
(b)
SDBS#1808
1
H-NMR
Assign.
A
B
C
D
E
F
G
Shift(ppm)
7.095
6.656
6.624
4.8
3.055
2.280
1.187
HA: 7.27 – 0.15 – 0.09 – 0.14 = 6.89
HB: 7.27 – 0.5 – 0.06 – 0.18 = 6.53
HC: 7.27 - 0.5 – 0.17 – 0.06 = 6.54
HD: 1 ~ 10 ppm
HE:1.5 +1.4 = 2.9
HF: 0.9 + 1.4 = 2.3
HG: 0.9 + 0.4 = 1.3
13
C-NMR
ppm Int. Assign.
152.82 677 1
139.38 542 2
136.60 577 3
125.90 488 4
117.15 463 5
113.05 458 6
28.65 517 7
23.43 1000 8
19.24 373 9
C1: 128.5 + 26.6 – 0.1 - 2.5 = 152.5
C2: 128.5 + 20.1 – 7.3 – 0.1 =141.2
C3: 128.5 + 9.3 + 1.6 – 0.5 = 138.9
C4: 128.5 + 1.6 – 0.1 – 2.0 = 128
C5: 128.5 – 12.7 + 0.7 + 0 = 116.5
C6: 128.5 – 12.7 – 2.9 + 0 = 112.9
For C7 and C9, calculate the shifts in propane, and add an internal Phenyl substituent
C7: - 2.5 + 9.1 x 2 + 17 = 32.7
C8: -2.5 + 9.1 + 9.4 + 7 = 23
C9: - 2.5 + 23 = 20.5
(3) Go to the "Webspectra" page, and do the suggested problems. For all the
problems suggested, molecular formula, 1H-NMR and 13C-NMR are provided.
Writing the structure of the compound is not enough for full credit. Clearly show
how you interpreted the spectra, what functional groups/partial structures could
be deduced from the given data, how you eliminated the wrong structures, etc.
http://www.chem.ucla.edu/~webspectra/
Note that the answer is provided in the website, and you can use it as a hint, if
needed.
Intermediate (1); 11(Determine the benzene substitution pattern first based on
spin-spin coupling, and then place substituents to match the chemical shifts of
aromatic protons or carbons) and 14.
Compound 11
C8H8O4: IHD = 5
13
C-NMR
A peak at 167 ppm is possibly due to a C=O of carboxylic acid derivative. Six
aromatic carbons are seen at 151, 147, 123, 121, 115 and 112 ppm. Aliphatic
carbon attached to an electronegative element is seen at 55.5 ppm.
1
H-NMR
A broad peak ~ 11pm suggest –COOH group. A methyl proton at 3.8 ppm
suggests – OCH3 group. There are only 3 aromatic protons. Thus, this is a
trisubstituted benzene with –COOH, -OCH3 and OH groups. The coupling pattern
of aromatic protons suggest 1,2,4 substitution (a doublet with a large J(~6.83
ppm) and overlapping doublet of doublets and a doublet of a small J (~7.42
ppm).
C6
C5
HO
C1
C4
CO2H
C2
C3
OCH3
Calculated 13C-NMR chemical shifts
C1: 128.5 + 2.9 + 1.0 – 7.3 = 125.1
C2: 128.5 + 1.3 -14.4 + 1.6 = 117
C3: 128.5 + 31.4 – 12.7 + 0.4 = 147.6
C4: 128.5 + 26.6 -14.4 + 4.3 = 145
C5: 128.5 – 12.7 + 1.0 + 0.4 = 117.2
C6: 128.5 + 1.3 -7.7 + 1.6 = 123.7
Compound 14
C4H10O2: IHD = 0
13
C-NMR
There are only three carbon peaks, suggesting a symmetry. Aliphatic carbon(s)
attached to an electronegative element at 66 ppm.
1
H-NMR
Methyl protons at 0.77 ppm are a doublet, suggesting CH-CH3. The coupling CH
proton shows a multiplet, suggesting more neighbor protons in addition to the
methyl group. The peaks at 3.5 ppm are complex, but it looks like overlapping
doublet of doublets. The broad peak at 4.5 ppm is probably –OH (2 Hs).