CHEM1002 Answers to Problem Sheet 4
1.
CH3
C
O
CH2OH
O
OCH2CH3
A
B
C
C4H 8O2 m/z = 88
C7H 8O m/z = 108
C4H 8O m/z = 72
OH
OH
O
D
E
F
C4H 10O2 m/z = 74
C6H 10O2 m/z = 114
C5H 8 m/z = 68
(a)
D will give a molecular ion at m/z = 74 in the mass spectrum.
(b)
B and F both have conjugated (i.e. adjacent) π bonds and will show strong
absorption in the UV-Visible spectrum.
(c)
A, C and E all have carbonyl groups and will show absorption around
1700 cm–1 in the infrared region.
(d)
B, D and E all have OH groups and will show absorption around 3500 cm–1
in the infrared region.
(e)
F does not possess a carbonyl or OH group and will not show absorption
either around 1700 or 3500 cm–1 in the infrared region.
(f)
The number of signals shown by each compound in its 13C NMR spectrum
is shown in the diagram below.
1
CH3
O
2
C
1
CH2OH
3
3 4
OCH2CH3
4
5
A - 4 signals
2
3
4
B - 5 signals
O
1
2
4
3
C - 4 signals
4
2
1
2
OH
3
OH
2
2
3
1
6
5
1
5
2
4
O
D - 2 signals
E - 6 signals
F - 5 signals
There are a number of possibilities and more information (e.g. 13C NMR) would
be required to decide between them. The IR absorption indicates that the O is
present as C=O. Strong absorption in the UV indicates the presence of
conjugation (i.e. adjacent double bonds). Some possible structures include:
2.
O
O
H
H
O
3.
(a)
3 ether groups, an arene ring and a
primary amine are present (see
opposite)
(b)
There are 8 types of C in the molecule
so 8 signals will be observed in the 13C
NMR.
8
ether
OCH3
7
H3CO
ether
6
5
5
arene
4
(c)
The molecular formula of mescaline is
C11H17O3N. The molar mass is therefore
211
(d)
7
OCH3
ether
4
3
2
1
The arene is conjugated so strong
absorption in the UV-visible spectrum is
expected.
NH2
primary amine
Mescaline
The NH2 groups will give rise to IR absorption at ~3400 cm–1.
No absorption at ~1700 cm-1 is expected as no C=O groups are present.
(e)
(e)
4.
1
1
6
5
1
2
3
4
6 signals
2
2
3
2 1
3
3 signals
3
1
6
1
4
5
6 signals
2
3
2
5 signals
4
5
2
3
3
4
4 signals
5.
The number of signals in the 13C NMR reflects the number of different carbon
environments.
The position of each signal reflects the type of environment for each carbon:
Aspirin contains 9 different carbon environments (see below) and so the
spectrum shows 9 signals:
O
9
OH
6
5
1
4
•
O
7
2
3
•
•
O
8
Carbon atom s 1-6 are aromatic so will show signals in the region 120-150 ppm.
Carbon atoms 7 and 9 are in C=O groups so show signals in the region
180-200 ppm
Carbon 8 is in an alkane group so its signal is in the region 0-20 ppm.
A sketch of the spectrum is show below with each signal labelled according to the
number scheme above. Note the group of signals in the aromatic region.
7
180
9
1
160
3
140
5
2,4,6
8
120
100
80
PPM
60
40
Paracetamol contains 6 different carbon environments (see below) and so the
spectrum shows 6 signals:
5
HO
6
4
5
3
4
O
N
H
2
1
20
0
•
•
•
Carbon atoms 3-6 are aromatic so will show signals in the region 120-150
ppm.
Carbon 2 is in a C=O group so its signal is in the region 180-200 ppm
Carbon 1 is in an alkane group so its signal is in the region 0-20 ppm.
A sketch of the spectrum is show below with each signal labelled according to the
number scheme above.
4
2
180
6.
1
3
6
160
5
140
(a)
120
100
PPM
(b)
80
60
40
(c)
OH
OH
phenol
OH
secondary alcohol
(d)
20
(e)
tertiary alcohol
(f)
CH2OH
O CH3
ether
O
primary alcohol
(h)
(g)
OH
C
ether
(i)
OH
OH
secondary alcohol
O
CH3
H
secondary alcohol
tertiary alcohol
O
two ethers
0