Answer for CBC311 – Chemical Spectroscopy and Applications (2012-2013)
Provided by Vu Minh Duy (CBC/04)
Question 1
(a) (i) By EI-MS
(1)
m/z = 97.10
(2)
m/z = 83.09
m/z = 69.07
(ii) Differentiate cis/trans isomers by C-H deformations which give IR absorption at
970 cm-1 (trans) and 690 cm-1 (cis)
(b) C7H10 degree of unsaturation = 3
UV absorbance at 197 nm is a clue for non-conjugated system, normally it is signal of C=C
MS shows fragment ion peak at m/z = 66  [C5H6](+) fragment, the other corresponding
fragment should be ethylene C2H4. Not so many fragmenting mechanisms can give alkene
fragment from C=C system, ex. retro-Diels Alder and Mc.Lafferty. In this case, it is retro-D.A.
mechanism, since Mc.Lafferty is normally observed for C=O type bond.
And based on 13C-NMR, the compound has 7C but gives only 4 signal, one of those is for
alkene (135.4 ppm). It can be concluded that it has a symmetrical structure.
retro-Diels Alder
fragmentation
m/z = 66
(c) Please refer to IR Lecture Notes for more details.
(d) M = 66  C5H6 (degree of unsaturation = 3)
Normally 13C-NMR peak at 194.1 ppm belongs to the region of aldehyde or ketone.
However, the compound is Hydrocarbon (no Oxygen or Nitrogen)  should think of a weird
structure, in this case, allene structure (C=C=C) where anisotropic effect at Csp can help
increasing chemical shift to downfield area (≈ 200ppm). Additionally, 1H-NMR shows 2 sets
of proton: 4 protons connecting to Csp3 (1.40ppm) and 2 protons connecting to Csp2
(4.65ppm). Therefore, the following structure can be deduced:
Additionally, UV-vis peak at 256 nm is characteristic for allene structure.
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Answer for CBC311 – Chemical Spectroscopy and Applications (2012-2013)
Question 2
(a) Based on given spectra, structures of (6), (7) and (8) are deduced as below:
O
H
N
OH
O
HO
O
O
OH
O
Chemical Formula: C9H8O4 Chemical Formula: C8H9NO2
Chemical Formula: C13H18O2
(6) – Commercial name: Aspirin
1
H-NMR: -Me (singlet, 3H, 2.5 ppm); COO-H (broad, 1H, 11 ppm); 4 different aromatic
protons (this indicates ortho position between the two substituents).
13
C-NMR: -Me (20 ppm); 2 almost similar C from ester and carboxylic acid group give
overlap signal (170 ppm); total 9 different Carbon nuclei.
(7) – Commercial name: Paracetamol (Acetaminophen)
H-NMR: -Me (singlet, 3H, 2 ppm); O-H (phenol) and N-H (amide) ranging from 9 ppm to 10
ppm (two peaks); 2 sets of aromatic protons, each set got 2 similar protons (this indicates
para position between the two substituents).
13
C-NMR: in total, there are only 6 different types of Carbon nuclei, since 4 aromatic Carbon
(except 2 ipso-Carbon nuclei) are divided into 2 set of similar (chemical equivalent) Carbon.
1
(8) – Commercial name: iBuprofen
H-NMR: From signal of 6H doublet and 1H multiplet  presence of iso-propyl group. One
proton is not observed due to fast exchange with solvent (acidic proton). 2 sets of aromatic
protons indicating para position between the two substituents.
13
C-NMR: -COOH (180 ppm); 4 aromatic Carbon (except 2 ipso-Carbon nuclei) are divided
into 2 set of similar (chemical equivalent) Carbon  6 aromatic Carbon shows only 4 peaks.
2 signal overlapped is from 2-Me of iso-propyl group.
1
(b) Characteristic IR absorption:
For (6), λC=O stretching for ester; λC=O stretching for carboxylic acid; λO-H stretching (broad).
For (7), λO-H stretching (broad); λC=O stretching for amide.
For (8), λO-H stretching (broad); λC=O stretching for carboxylic acid.
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Answer for CBC311 – Chemical Spectroscopy and Applications (2012-2013)
(c) MS spectra explanation:
COOH
COOH
O
O
O
m/z = 43
O
O
(6)
H
N
H
N
O
HO
HO
NH
O
HO
H
O
OH
(8)
C
O
m/z = 109
Mc.Lafferty
(7)
H 2C
O
OH
- OH
O
- CO
m/z = 161
(d) DEPT135 spectrum
Copy the normal 13C-NMR spectrum, and do the following modification:
Remove the peak at 180 ppm (it is –COO- group) and the two peaks near 140 ppm (they are
2C at ipso position on benzene ring).
Inverse the peak at 30 ppm (it is CH2 group)
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Answer for CBC311 – Chemical Spectroscopy and Applications (2012-2013)
Question 3
(a) 1H-NMR
JPt-H = 3000Hz or 30 ppm
JP-H trans = 100Hz or 1 ppm
JP-H cis = 10Hz or 0.1 ppm
1. Consider both Pt are 195Pt (NMR active) – 11.11% probability.
H coupling with P to give triplet of nonet
H coupling with Pt to give triplet
2. Consider both Pt are not NMR active – 44.44% probability.
H coupling with P to give triplet of nonet
No H-Pt couling
3. Consider 1 Pt is NMR active, the other one is not – 44.44% probability.
H coupling with P to give triplet of nonet
H coupling with Pt to give doublet
Important Note: Each line below represents a nonet
2 NMR inactive Pt
1 NMR active + 1 NMR inactive
2 NMR active Pt
The whole spectrum
21ppm
20ppm
19ppm
5ppm
-10ppm
-25ppm
-40ppm
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Answer for CBC311 – Chemical Spectroscopy and Applications (2012-2013)
(b)
31
P{1H}-NMR
P
P
Pt
P
P
P
P
H
Pt
P
P
P
P
There will be two signals for 2 different P nuclei (red and blue).
Consider 3 different situations as above (NMR activity of Pt) to get 3 different set of spectra,
and then merge them together to obtain the final spectra.
-
-
When both Pt are NMR active, Pred is doublet of quintet, Pblue is doublet of doublet.
When either of Pt is NMR active, Pred is doublet of quintet, Pred’ is quintet; Pblue is
doublet of doublet and Pblue’ is doublet (Pred’ and Pblue’ connecting to the NMRinactive-Pt nucleus).
When both Pt are NMR inactive, Pred is quintet; Pblue is doublet.
Note: the chemical shift of all Pred (in those situations) are the same. Assume that different
Pt nuclei does not change much the chemical environment of the P nuclei, then Pred and Pred’
share the same chemical shift also. Similarly applied for Pblue.
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Answer for CBC311 – Chemical Spectroscopy and Applications (2012-2013)
Question 4
PMe3
Cl
Pd
Pd
Me3P
PMe3
Cl
Cl
Cl
PMe3
1
Using H-NMR to determine the ratio between cis and trans form in equilibrium.
For cis complex, 1H-NMR gives doublet (P-H coupling)
For trans complex, 1H-NMR gives abnormal triplet (H coupling with both P, since the two P
bonding with Pd by the same d orbital, therefore, H can see them as 2 similar P  triplet
signal instead of doublet one like previous case). (Prof. Leung did discuss details about this
phenomenon in lecture).
Then, you can base on the ratio between the integration of two signals to estimate the
quantity of each form.
Illustration (example only):
Trans form
Cis form
End of paper
All the best for your exams!
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