5.7: 2D NMR Solutions
- Page ID
- 195117
Exercise 5.1.1:
ethyl butanoate
Exercise 5.1.2:
Exercise 5.1.3:
Exercise 5.1.4:
Exercise 5.1.5:
Exercise 5.1.6:
Exercise 5.1.7:
Exercise 5.1.8:
Exercise 5.1.9:
Exercise 5.1.10:
Exercise 5.1.11:
Exercise 5.1.12:
Exercise 5.2.1:
Exericse 5.2.2:
Exercise 5.2.3:
Exercise 5.3.1:
Exercise 5.3.2:
Exercise 5.3.3:
Exercise 5.3.4:
Exercise 5.3.5:
Exercise 5.3.6:
Exercise 5.3.7:
Exercise 5.3.8:
Exercise 5.4.1:
Exercise 5.4.2:
Exercise 5.4.3:
a) Due to resonance, there is substantial pi character to the amide bond which restricts free rotation around that bond.
b)
c)
Exercise 5.4.4:
Exercise 5.4.5:
Exercise 5.4.6:
a)
b)
c)
Exercise 5.4.7:
Exercise 5.5.1:
Exercise 5.5.2:
Exercise 5.5.3:
Exercise 5.5.4:
Exercise 5.5.5:
Exercise 5.6.1:
^{1}H NMR
Chemical shift (ppm) | Integration | Multiplicity | Partial Structure |
5.54 | 2H | multiplet | CH=C (x 2) |
4.17 | 2H | doublet | O-CH_{2}-CH |
2.08 | 2H | quintet | CH-CH_{2}-CH_{3} |
1.47 | 1H | boad singlet | OH |
0.95 | 3H | triplet | CH_{2}-CH_{3} |
* total # H: 10
^{13}C NMR
Chemical shift (ppm) |
Type of carbon |
134 | sp^{2} |
128 |
sp^{2} |
58 | sp^{3}-O |
21 | sp^{3} |
14 | sp^{3} |
*total # C: 5
COSY
Assignment | ^{1}H | COSY |
A | 0.95 | 2.08 |
B | 2.08 | 0.95, 5.54 |
C | 4.17 | 5.54 |
D | 5.54 | 2.08 |
E | 5.54 | 4.17 |
*HMQC indicates two hydrogens at 5.54 are in two different environments
HMQC
Assignment | ^{13}C | ^{1}H |
A | 14 | 0.95 |
B | 21 | 2.08 |
C | 58 | 4.17 |
D | 128 | 5.54 |
E | 134 | 5.54 |
Formula:
C_{5}H_{10}O (1 O indicated from shift in ^{13}C, ^{1}H NMR)
FW = \(5 \times 12) + (10 \times 1) + (1 \times 16) = 86\)
Compare C_{5}H_{10} ratio to C_{5}H_{12} in hydrocarbon
Degrees of unsaturation = \(\frac{(2 \times 5) + 2 - 10}{2} = 1 unit (1 double bond)
The data tables should be consitent with this structure:
pent-2-en-1-ol (could be cis or trans based on this analysis)
Exercise 5.6.2:
^{1}H NMR:
Chemical shift (ppm) | Integration | Multiplicity | Partial structure |
4.7 | 5H | singlet | solvent |
3.93 | 1H | triplet | CH_{2}-CH-N |
2.40 | 2H | multiplet | CH_{2}-CH_{2}? |
2.09 | 2H | multiplet | CH_{2}-CH_{2}? |
1.4 | 9H | singlet | C(CH_{3})_{3} |
*Total number of H: 19 H
^{13}C NMR:
Chemical shift (ppm) | Type of carbon |
170 | sp^{2} (C=O) |
80 | sp^{3} (C-O) |
52 | sp^{3 }(C-N) |
32 | sp^{3} |
28 | sp^{3} |
26 | sp^{3} |
*Total number of C: 6 apparent, but two more suggested by symmetry (3 methyl groups in ^{1}H NMR) for 8 C; a third extra suggested by MW fit for 9 C
COSY:
Assignment | ^{1}H | COSY |
Solvent | 4.7 | -- |
B | 3.93 | 2.40 |
D | 2.40 | 2.09 |
C | 2.09 | 2.40, 2.09 |
A | 1.4 | -- |
Formula:
C_{9}H_{18}O_{3}N_{2} (extra O indicated from shift in ^{13}C, ^{1}H NMR; second O suggested by C=O in ^{13}C NMR; additional CO needed to fit MW)
FW = \((9 \times 12) + (18 \times 1) + (3 \times 16) + (2 \times 14) = 202\)
FW = ((9 \times 12) + (18 \times 1) + (3 \times 16) + (2 \times 14) = 202\)
Compare C_{9}H_{18} to C_{9}H_{22} for the corresponding hydrocarbon corrected for two nitrogens (therefore two extra hydrogens)
Degrees of unsaturation = \(\frac{(2 \times 9) + 2 + 2 - 10}{2} = 2\) units (2 double bonds)
The data tables should be consistent with this structure:
Exercise 5.6.3:
The data should be consistent with this structure:
Exercise 5.6.4:
^{1}H NMR:
Chemical shift (ppm) |
Integration | Multiplicity | Partial Structure |
4.71 | -- | singlet | solvent |
4.17 | 1H | doublet? | CO-CH-N |
3.75 | 3H | singlet | O-CH_{3} |
2.25 | 1H | multiplet | CH-CH-(CH_{3})_{2} |
0.92 | 6H | triplet? | 2 x CH_{3} |
^{13}C NMR:
Chemical shift (ppm) | Type of carbon |
170 | sp^{2} C=O |
60 | sp^{3} C-N |
52 | sp^{3} C-O |
30 | sp^{3} C |
19 | sp^{3} C |
COSY:
Assignment | ^{1}H | COSY |
A | 4.17 | 2.25 |
B | 3.75 | -- |
C | 2.25 | 4.17 |
D | 0.92 | 2.25 |
Formula:
C_{6}H_{13}O_{2}N (1 O indicated from shift in ^{13}C, ^{1}H NMR)
FW = \((6 \times 120 + (13 \times 1) + (2 \times 16) + (1 \times 14) = 131\)
Degrees of unsaturation = \(\frac{(2 \times 6) + 2 + 1 - 13}{2} = 1\) unit (1 double bond)
The data should be consistent with this structure:
Exercise 5.6.5:
The data should be consistent with this structure:
Exercise 5.6.6:
The data should be consistent with this structure:
Exercise 5.6.7:
The data should be consistent with this structure:
Exercise 5.6.8:
The data should be consistent with this structure:
Exercise 5.6.9:
The data should be consistent with this structure:
Exercise 5.6.10:
The data should be consistent with this structure:
Exercise 5.6.11:
The data should be consistent with this structure:
Exercise 5.6.12:
The data should be consistent with this structure:
Exercise 5.6.13:
^{1}H NMR:
Chemical shift (ppm) | Integration | Multiplicity | Partial structure |
5.7 | 1H | singlet | C=CH-CO |
2.77 | 1H | multiplet | CH_{2}-CH-CO |
2.62 | 1H | multiplet | C-CH-C |
2.39 | 1H | multiplet | C-CH-C |
2.05 | 1H | doublet? | C-CH-CH? |
1.96 | 3H | singlet | C-CH_{3} |
1.48 | 3H | singlet | C-CH_{3} |
0.98 | 3H | singlet | C-CH_{3} |
^{13}C NMR:
Chemical shft (ppm) | Type of carbon |
204 | sp^{2} C=O |
170 | sp^{2} |
121 | sp^{2} |
59 | sp^{3} |
55 | sp^{3} |
50 | sp^{3} |
41 | sp^{3} |
28 | sp^{3} |
24 | sp^{3} |
22 | sp^{3} |
COSY:
Assignment | ^{1}H | COSY |
1 | 2.39 | 2.77, 2.62? |
3 | 5.7 | 2.05? |
5 | 2.62 | 2.62? |
7a | 2.77 | 2.05 |
7b | 2.05 | 2.77 |
8 | 1.48 | -- |
9 | 0.98 | -- |
10 | 1.96 | -- |
Formula:
C_{10}H_{14}O (1 O indicated from shift in ^{13}C, ^{1}H NMR)
FW = \((10 \times 12) + (14 \times 1) + (1 \times 16) = 150 \)
Degrees of unsaturation = \(\frac{(2 \times 10) + 2 - 14}{2} = 4 units (e.g. 2 rings, 2 double bonds)
The data should be consistent with this structure:
Exercise 5.6.14:
The data should be consistent with this structure:
Exercise 5.6.15:
The data should be consistent with this structure: