Transcript Structural determination of organic compounds

34
34.1
34.2
34.3
34.4
34.5
34.6
34.7
34.8
34.9
1
Structural Determination
of Organic Compounds
Introduction
Isolation and Purification of Organic Compounds
Tests for Purity
Qualitative Analysis of Elements in an Organic Compound
Determination of Empirical Formula and Molecular
Formula from Analytical Data
Structural Information from Physical Properties
Structural Information from Chemical Properties
Use of Infra-red Spectrocopy in the Identification of
Functional Groups
Use of Mass Spectra to Obtain Structural Information
The general steps to determine the structure
of an organic compound
2
Isolation and
Purification of
Organic
Compounds
3
Technique
1. Filtration
To separate an insoluble solid
from a liquid (slow)
2. Centrifugation
To separate an insoluble solid
from a liquid (fast)
To separate a solid from other
solids based on their different
solubilities in suitable solvent(s)
To separate a component from a
mixture with a suitable solvent
3. Recrystallization
4. Solvent extraction
5. Distillation
4
Aim
To separate a liquid from a
solution containing non-volatile
solutes
Technique
Aim
6. Fractional distillation To separate miscible liquids
with widely different boiling
points
7. Steam distillation
To separate liquids which are
immiscible with water and
The mixture boils
decompose easily below their
below 100C
b.p.
8. Vacuum distillation
ditto
9. Sublimation
To separate a mixture of solids
in which only one can sublime
10. Chromatography
5
To separate a complex mixture
of substances (large/small scale)
Tests for Purity
•
If the substance is a solid,
 its purity can be checked by
determining its melting point
•
If it is a liquid,
 its purity can be checked by
determining its boiling point
6
Use of Infra-red
Spectrocopy in the
Identification of
Functional Groups
7
Infra-red Spectroscopy
Arises from absorption of IR radiation by
organic compounds
Causes atoms and groups of atoms of organic
compounds to vibrate with increased
amplitude about the covalent bonds that
connect them
8
Modes of vibrations
Two basic modes : stretching and bending
Two atoms joined by a covalent bond can
undergo a stretching vibration where the
atoms move back and forth as if they were
joined by a spring
A stretching vibration
of two atoms
9
Modes of vibrations
Symmetrical stretching : -
Asymmetrical stretching : -
10
Modes of vibrations
Bending (scissoring) : - two modes
Behind paper
Out of paper
11
Modes of vibrations
Symmetric stretching
12
Rocking
Wagging
Asymmetric stretching
An in-plane bending
An out-of-plane bending
(scissoring)
(twisting)
The frequency() of a given stretching
vibration of a covalent bond depends on
k
ν 
μ
Bond strength
Reduced mass of
the system
For a diatomic molecule X – Y
μ
13
mx my
mx  my
Bond strength 
k

k
ν 
μ
C – C
<
C = C
C – O
<
C = O
<
C  C
Increasing frequency of vibration
14
Masses of bonding atoms 
k
ν 
μ


C – H
O - H
N - H
Chemical bonds containing H atoms have
high frequency of vibration due to the
small mass of H
15
Infra-red Spectroscopy
• Molecular vibrations are quantized
the molecules absorb IR radiation of a
particular amount of energy only.
E = h
• Only IR radiation with the same frequency
as the vibrational frequency can be
absorbed by the molecules.
16
Infra-red Spectroscopy
E = h
After the absorption of a quantum of energy
(h), the amplitude of vibration  but the
frequency of vibration () remains unchanged.
17
The electromagnetic spectrum
Wavelength (m) 
 Frequency (Hz / s1)
UV / visible / Near IR  electronic transition
E = hL = 400 – 1000 kJ mol1
18
The electromagnetic spectrum
Wavelength (m) 
 Frequency (Hz / s1)
Mid IR  vibrational transition
E = hL = 5 – 40 kJ mol1
19
The electromagnetic spectrum
Wavelength (m) 
 Frequency (Hz / s1)
Far IR / microwave  rotational transition
E = hL  0.02 kJ mol1
20
The electromagnetic spectrum
Wavelength (m) 
 Frequency (Hz / s1)
translational transition
E = hL  0 kJ mol1
21
Infra-red Spectroscopy
• Different functional groups (e.g. C=O, O-H)
have different vibration frequencies
 They have characteristic absorption
frequencies
• Functional groups can be identified from
their characteristic absorption frequencies
22
Infra-red Spectroscopy
• The stretching vibrations of single bonds
involving hydrogen (C  H, O  H and
N  H) occur at relatively high frequencies
23
Bond
Range of wavenumber (cm-1)
CH
2840 – 3095
OH
3230 – 3670
NH
3350 – 3500
Characteristic absorption wavenumbers of some
single bonds in infra-red spectra
Infra-red Spectroscopy
• Triple bonds are stronger and vibrate at
higher frequencies than double bonds
Bond
Range of wavenumber (cm-1)
CC
2070 – 2250
CN
2200 – 2280
C=C
1610 – 1680
C=O
1680 – 1750
Characteristic absorption wavenumbers of some
double bonds and triple bonds in infra-red spectra
24
Infra-red Spectroscopy
• An IR spectrum is obtained by scanning
the sample with IR radiations from
1.21013 Hz to 1.21014 Hz Or,
Wavenumber : 400 cm1 to 4000 cm-1
1 
wavenumber (cm )  
 c
1
• % transmittance rather than absorbance is
displayed
25
Dips(peaks) show absorptions by functional
groups
C=C
H
C
H
C
26
Fingerprint
region
27
Infra-red Spectroscopy
• Infra-red spectrometer is used to
measure the amount of energy absorbed
at each wavelength of the IR region
An infra-red
spectrometer
28
29
Infra-red Spectroscopy
• A beam of IR radiation is passed through
the sample
 the intensity of the emergent radiation
is carefully measured
• The spectrometer plots the results as a
graph called infra-red spectrum
 shows the absorption of IR radiation by
a sample at different frequencies
30
Use of IR Spectrum in the
Identification of Functional Groups
• When a compound absorbs IR radiation,
 the intensity of transmitted radiation
decreases
 results in a decrease in percentage of
transmittance
 a dip in the spectrum
 often called an absorption peak or
absorption band
31
Use of IR Spectrum in the
Identification of Functional Groups
• In general, an IR spectrum can be split
into four regions for interpretation
purpose
32
Range of
wavenumber (cm-1)
Interpretation
400 – 1500
• Often consists of many complicated
bands (stretching and bending)
• Unique to each compound
• Often called the fingerprint region
• Not used for identification of
particular functional groups
1500 – 2000
Absorption of double bonds,
e.g. C = C, C = O
2000 – 2500
Absorption of triple bonds,
e.g. C  C, C  N
2500 – 4000
Absorption of single bonds involving
hydrogen, e.g. C  H, O  H, N  H
33
Use of IR Spectrum in the
Identification of Functional Groups
• The region between 4 000 cm-1 and
1 500 cm-1 is often used for
 identification of functional groups
from their characteristic
absorption wavenumbers
34
What is the characteristic range of wavenumber of C=N bond?
Bond strength/wavenumber : C=C < C=N < C=O
Compound
Bond
Alkenes
Aldehydes, ketones,
acids, esters
Alkynes
Nitriles
Acids (hydrogen-bonded)
C=C
C=O
Characteristic range of
wavenumber (cm-1)
1610 – 1680
1680 – 1750
CC
CN
OH
2070 – 2250
2200 – 2280
2500 – 3300
Alkanes, alkenes, arenes
Alcohols, phenols
(hydrogen-bonded)
CH
OH
2840 – 3095
3230 – 3670
Primary amines
NH
3350 – 3500
35
Strategies for the Use of IR Spectra in
the Identification of Functional Groups
1. Focus at the IR absorption peak at or
above 1500 cm–1
 Concentrate initially on the major
absorption peaks
36
Strategies for the Use of IR Spectra in
the Identification of Functional Groups
2. The absence and presence of absorption
peaks at some characteristic ranges of
wavenumbers are equally important
 the absence of particular absorption
peaks can be used to eliminate the
presence of certain functional groups
or bonds in the molecule
37
Limitation of the Use of IR Spectroscopy in
the Identification of Organic Compounds
1. Some IR absorption peaks have very
close wavenumbers and the peaks always
coalesce
2. Not all vibrations give rise to strong
absorption peaks
38
Limitation of the Use of IR Spectroscopy in
the Identification of Organic Compounds
3. Not all absorption peaks in a spectrum can
be associated with a particular bond or
part of the molecule
4. Intermolecular interactions in molecules
can result in complicated infra-red spectra
39
1680-1750 cm1
N
1610-1680 cm1
Alkene
Arene
Y
Y
Absorption
Slightly above
3000 cm1
C=C bonds in benzene are weaker  ~1610 cm1
40
Contain
oxygen
Carboxylic
acid
Y
Y
1680-1750 cm1
Y
Broad
2500-3300 cm1
N
Aldehyde, Ketone,
Ester
41
N
2070-2280 cm1
Aldehyde : lower C-H
absorption at 2720-2820 cm-1
Nitrile
2200-2280 cm1
Y
2070-2280
cm1
Y
Contain
nitrogen
N
N
3200-3700 cm1
Alkyne
2070-2250 cm1
42
Additional sharp peak
~3250 cm1 for termnal
alkyne, -CC-H
Amine
3350-3500 cm1
Single peak for 2 amine;
double peaks for 1 amine
Y
Contain
nitrogen
3200-3700 cm1
N
Alcohol, phenol
3230-3700 cm1
N
Alkane
2840-3000 cm1
Y
Contain
oxygen
43
Y
weak
Several peaks
(2840-3000 cm1)
due to
different modes
of Symmetrical
& asymmetrical
stretching
Different
modes of C-H
bending
(~1400 cm1)
44
3300
3100
medium
H
C
C
C
45
Q.82
46
Q.82
47
Q.82
48
Q.82
H
C
C
C
49
Q.82
50
Q.83
51
Q.83
52
Q.83
28403000
28403000
53
27202820
Q.83
28403000
28403000
54
27202820
Q.84
C3H5N
IOU = 3 + 1 -5/2 + ½ = 2
CC, CN, diene
CC, CN
55

H
C
C
H
H
C
N
H
H
H
H
C
H
H
C
C
N
H
H
H
H
C
C
H
H
1 amine
C-H
56
CC, CN
C
N
Q.85
C6H10O
IOU = 6 + 1 -10/2 = 2
Ozonolysis gives CO2  terminal C=CH2
57
Q.85
C6H10O
IOU = 6 + 1 -10/2 = 2
+ve iodoform test 
OH
H3C
C
O
H3C
C
H
58
Q.85
C6H10O
IOU = 6 + 1 -10/2 = 2
O
No broadband within 3200-3600  H3C
=C-H
-C-H
C=C
C=O
59
C
O
C
H3C
O
C
H3C
O
C
H3C
60
*

H
O
C
C
61
H
C
O
H
C
62
H
C
C
C
H
C
63
No C-H stretching
above 3000 cm1
1 amine
N-H
bending
H
C
64
 No C=C-H
Broad band with peak
at about 2900 cm1
C
65
O
H
C
C
66
O
1 amine
H
C
67
H
C
68
Broad band with peak
at about 2900 cm1
C
69
O
Broad band with peak
at about 3300 cm1
H
C
70
H
C
C
71
O
H
C
72
C
O
H
C
73
No aliphatic C-H
stretching
O
C
H
C
74
O
No aliphatic C-H
stretching
75
Broad band with peak
at about 3300 cm1
H
C
76
Broad band with peak
at about 3300 cm1
H
C
77
Broad band with peak
at about 2900 cm1
H
C
C
78
O
No C-H stretching
above 3000 cm1
 No C=C-H
1 amine
N-H
bending
H
C
79
Broad band with peak
at about 3300 cm1
H
C
80
H
C
81
C
O
H
C
82
Broad band with peak
at about 3300 cm1
H
C
83
H
C
C
84
O
Broad band with peak
at about 3300 cm1
H
C
85
H
C
86
Broad band with peak
at about 3400 cm1
H
C
C
87
O
H
C
C
88
O
Broad band with peak
at about 2900 cm1
H
C
C
89
O
H
C
90
C
O
O
C
H
H
C
91
C
O
Broad band with peak
at about 3300 cm1
No aliphatic C-H
stretching
92
Broad band with peak
at about 3300 cm1
H
C
93
H
C
94
C
O
H
C
95
H
C
2 amine
96
C
O
4. Butanone
The IR spectrum of butanone
97
4. Butanone
Wavenumber (cm-1)
Intensity
2983
Strong
2925
Strong
1720
1416
Indication
C  H stretching
Very strong C = O stretching
Medium
C  H bending
(shifted as adjacent
to C = O)
Interpretation of the IR spectrum of butanone
98
5. Butan-1-ol
 a broad band is observed
99
 the vibration of the O  H group is
complicated by the hydrogen bonding
formed between the molecules
5. Butan-1-ol
Wavenumber
(cm-1)
3330
Intensity
Broad band
2960
Medium
2935
Medium
2875
Medium
Indication
OH
stretching
CH
stretching
Interpretation of the IR spectrum of butan-1-ol
100
6. Butanoic Acid
 a broad band is observed
 the vibration of the O  H group is
complicated by the hydrogen bonding
formed between the molecules
101
6. Butanoic Acid
Wavenumber (cm-1)
3100
1708
Intensity
Indication
Broad band O  H
stretching
Strong
C=O
stretching
Interpretation of the IR spectrum of
butanoic acid
102
7. Butylamine
The IR spectrum of butylamine
103
7. Butylamine
Wavenumber (cm-1)
Intensity
Indication
3371
Strong
N  H stretching
3280
Strong
2960 – 2875
Weak
1610
Medium
N  H bending
1475
Medium
C  H bending
C  H stretching
Interpretation of the IR spectrum of butylamine
104
8. Butanenitrile
The IR spectrum of butanenitrile
105
8. Butanenitrile
Wavenumber
(cm-1)
2990 – 2895
Intensity
2246
Very strong
1420
1480
Strong
Strong
Strong
Indication
CH
stretching
CN
stretching
CH
bending
Interpretation of the IR spectrum of
butanenitrile
106
Example 34-8
107
Check Point 34-8
34.9
Use of Mass Spectra
to Obtain Structural
Information
108
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Spectrometry
• One of the most sensitive and versatile
analytical tools
• More sensitive than other spectroscopic
methods (e.g. IR spectroscopy)
• Only a microgram or less of materials is
required for the analysis
109
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Spectrometry
In a mass spectrometric analysis, it involves:
1. the conversion of molecules to ions
2. separation of the ions formed according to
their mass-to-charge (m/e) ratio
 m is the mass of the ion in atomic
mass units and e is its charge
110
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Spectrometry
• Finally, the number of ions of each type
(i.e. the relative abundance of ions of
each type) is determined
• The analysis is carried out using a mass
spectrometer
111
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
Components of a mass spectrometer
112
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
In the vaporization chamber,
• the sample is heated until it vaporizes
 changes to the gaseous state
113
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
• The molecules in the gaseous state are
bombarded with a beam of fast-moving
electrons
 Positively-charged ions called the
molecular ions are formed
 One of the electrons of the molecule
is knocked off
114
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
• Molecular ions are sometimes referred to
as the parent ion
115
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
•
 one of the electrons is removed from
the molecules during the ionization
process
 the molecular ion contains a single
unpaired electron
 the molecular ion is not only a cation,
it is also a free radical
116
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
• e.g.
if a molecule of methanol (CH3OH) is
bombarded with a beam of fast-moving
electrons
 the following reaction will take place:
117
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
• The molecular ions formed in the ionization
chamber are energetically unstable
 undergo fragmentation
• Fragmentation can take place in a variety
of ways
 depend on the nature of the
particular molecular ion
118
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
• The way that a molecular ion fragments
 give us highly useful information
about the structure of a complex
molecule
119
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
•
The positively charged ions formed are
then accelerated by electric field and
deflected by magnetic field
 causes the ions to arrive the ion
detector
•
120
The lighter the ions, the greater the
deflection
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Spectrometry
121
•
Positively charged ions of higher
charge have greater deflection
•
Ions with a high m/e ratio are deflected
to smaller extent than ions with a low
m/e ratio
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Mass Spectrometry
• In the ion detector,
 the number of ions collected is measured
electronically
• The intensity of the signal is
 a measure of the relative abundance of
the ions with a particular m/e ratio
122
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Mass Spectrometry
• The spectrometer shows the results by
 plotting a series of peaks of varying
intensity
 each peak corresponds to ions of a
particular m/e ratio
• The graph obtained is known as a mass
spectrum
123
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Mass Spectrum
•
124
Generally published as bar graphs
.
Mass spectrum
of methanol
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Mass Spectrum
Corresponding ion
H3C+
m/e ratio
15
H  CO+
29
H2C = OH+
31
CH3OH
32
Interpretation of the mass spectrum of methanol
125
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Formation of Fragments
• The molecular ions formed in the ionization
chamber are energetically unstable
 Some of them may break up into
smaller fragments
 Called the daughter ions
126
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Formation of Fragments
• These ionized fragments are accelerated and
deflected by the electric field and magnetic
field
• Finally, they are detected by the ion detector
and
 their m/e ratios are measured
 explains why there are so many peaks
appeared in mass spectra
127
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
• The peak at m/e 31
 the most intense
peak
• Arbitrarily assigned
an intensity of 100%
 Called the base
peak
Mass spectrum of methanol
128
 Corresponds to
the most common
ion formed
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
• The peak at m/e 31
 corresponds to the ion H2C = OH+
 formed by losing one hydrogen atom
from the molecular ion
129
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
• The ion H2C = OH+ is a relatively stable ion
  the positive charge is not localized on
a particular atom
 it spreads around the carbon and the
oxygen atoms to form a delocalized
system
130
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
• The peak at m/e 29 corresponds to the ion
HC  O+
 formed by losing two hydrogen atoms
from the ion H2C = OH+
131
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
• The ion HC  O+ has two resonance
structures:
132
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
• The peak at m/e 15 corresponds to the
ion H3C+
 formed by the breaking of the C  O
bond in the molecular ion
133
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
Mass spectrum of pentan-3-one
134
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
Corresponding ion
m/e ratio
CH3CH2+
29
CH3CH2CO+
57
CH3CH2COCH2CH3
86
Interpretation of the mass
spectrum of pentan-3-one
135
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)
Formation of Fragments
• The fragmentation pattern of pentan-3-one
is summarized below:
136
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
Example 34-9A
Example 34-9B
Example 34-9C
137
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)
Fragmentation Pattern
1. Straight-chain Alkanes
•
Simple alkanes tend to undergo
fragmentation by
 the initial loss of a • CH3 to give a peak
at M – 15
 This carbocation can then undergo
stepwise cleavage down the alkyl chain
138
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)
1. Straight-chain Alkanes
•
139
Take hexane as an example:
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)
2. Branched-chain Alkanes
•
Tend to cleave at the “branch point”
 more stable carbocations are formed
140
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)
2. Branched-chain Alkanes
•
141
e.g.
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
3. Alkyl-substituted Aromatic Hydrocarbons
• Undergo loss of a hydrogen atom or alkyl
group
 yield the relatively stable tropylium ion
• Gives a prominent peak at m/e 91
142
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
3. Alkyl-substituted Aromatic Hydrocarbons
•
143
e.g.
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
4. Aldehydes and Ketones
• Frequently undergo fragmentation by losing
one of the side chains
 generate the substituted oxonium ion
 often represents the base peak in the
mass spectra
144
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
4. Aldehydes and Ketones
145
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
5. Esters, Carboxylic Acids and Amides
•
Often undergo cleavage that involves the
breaking of the C X bond
 form substituted oxonium ions as
shown below:
(where X = OH, OR, NH2, NHR, NR2)
146
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
5. Esters, Carboxylic Acids and Amides
• For carboxylic acids and unsubstituted
amides,
 characteristic peaks at m/e 45 and 44
are observed respectively
147
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
6. Alcohols
• In addition to the loss of a proton and the
hydroxyl radical,
 alcohols tend to lose one of the  alkyl
groups (or  hydrogen atoms)
 form oxonium ions
148
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
6. Alcohols
• For primary alcohols,
 the peak at m/e 31, 45, 59 or 73 often
appears
 depends on what the R1 group is
149
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
7. Haloalkanes
• Haloalkanes simply break at the C  X
bond
150
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
7. Haloalkanes
• In the mass spectra of chloroalkanes,
 two peaks, separated by two mass
units, in the ratio 3 : 1 will be
appeared
151
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
7. Haloalkanes
• In the mass spectra of bromoalkanes,
 two peaks, separated by two mass
units, having approximately equal
intensities will be appeared
152
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
Check Point 34-9
153
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Isolation and Purification of
Organic Compounds
•
The selection of a proper technique
 depends on the particular differences
in physical properties of the
substances present in the mixture
154
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Filtration
• To separate an insoluble solid from a
liquid particularly when the solid is
suspended throughout the liquid
• The solid/liquid mixture is called a
suspension
155
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Filtration
The laboratory set-up of filtration
156
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Filtration
• There are many small holes in the filter
paper
 allow very small particles of solvent
and dissolved solutes to pass through
as filtrate
• Larger insoluble particles are retained on
the filter paper as residue
157
34.2 Isolation and Purification of Organic Compounds (SB p.79)
Centrifugation
•
When there is only a small amount of
suspension, or when much faster
separation is required
 Centrifugation is often used
instead of filtration
158
34.2 Isolation and Purification of Organic Compounds (SB p.79)
Centrifugation
• The liquid containing
undissolved solids is put
in a centrifuge tube
• The tubes are then put
into the tube holders in a
centrifuge
A centrifuge
159
34.2 Isolation and Purification of Organic Compounds (SB p.79)
Centrifugation
• The holders and tubes are spun around at a
very high rate and are thrown outwards
• The denser solid is collected as a lump at
the bottom of the tube with the clear liquid
above
160
34.2 Isolation and Purification of Organic Compounds (SB p.79)
Crystallization
• Crystals are solids that have
 a definite regular shape
 smooth flat faces and straight edges
• Crystallization is the process of forming
crystals
161
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot Concentrated
Solution
• To obtain crystals from an unsaturated
aqueous solution
 the solution is gently heated to make it
more concentrated
• After, the solution is allowed to cool at
room conditions
162
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot Concentrated
Solution
• The solubilities of most solids increase
with temperature
• When a hot concentrated solution is
cooled
 the solution cannot hold all of the
dissolved solutes
• The “excess” solute separates out as
crystals
163
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot Concentrated
Solution
Crystallization by cooling a hot concentrated solution
164
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a Cold Solution
at Room Temperature
• As the solvent in a solution evaporates,
 the remaining solution becomes
more and more concentrated
 eventually the solution becomes
saturated
 further evaporation causes
crystallization to occur
165
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a Cold Solution
at Room Temperature
• If a solution is allowed to stand at room
temperature,
 evaporation will be slow
• It may take days or even weeks for crystals
to form
166
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a Cold Solution
at Room Temperature
Crystallization by slow evaporation of a solution
(preferably saturated) at room temperature
167
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Extraction
• Involves extracting a component from a
mixture with a suitable solvent
• Water is the solvent used to extract salts
from a mixture containing salts and sand
• Non-aqueous solvents (e.g. 1,1,1trichloroethane and diethyl ether) can be
used to extract organic products
168
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Extraction
• Often involves the use of a separating
funnel
• When an aqueous solution containing
the organic product is shaken with
diethyl ether in a separating funnel,
 the organic product dissolves into
the ether layer
169
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Extraction
The organic product in an aqueous solution can be
extracted by solvent extraction using diethyl ether
170
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Extraction
• The ether layer can be run off from the
separating funnel and saved
• Another fresh portion of ether is shaken
with the aqueous solution to extract any
organic products remaining
• Repeated extraction will extract most of
the organic product into the several
portions of ether
171
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Extraction
• Conducting the extraction with several
small portions of ether is more efficient
than extracting in a single batch with the
whole volume of ether
• These several ether portions are
combined and dried
 the ether is distilled off
 leaving behind the organic product
172
34.2 Isolation and Purification of Organic Compounds (SB p.81)
Distillation
• A method used to separate a solvent
from a solution containing non-volatile
solutes
• When a solution is boiled,
 only the solvent vaporizes
 the hot vapour formed condenses to
liquid again on a cold surface
• The liquid collected is the distillate
173
34.2 Isolation and Purification of Organic Compounds (SB p.81)
Distillation
The laboratory set-up of distillation
174
34.2 Isolation and Purification of Organic Compounds (SB p.81)
Distillation
• Before the solution is heated,
 several pieces of anti-bumping
granules are added into the flask
 prevent vigorous movement of the
liquid called bumping to occur during
heating
 make boiling smooth
175
34.2 Isolation and Purification of Organic Compounds (SB p.81)
Distillation
• If bumping occurs during distillation,
 some solution (not yet vaporized)
may spurt out into the collecting
vessel
176
34.2 Isolation and Purification of Organic Compounds (SB p.81)
Fractional Distillation
• A method used to separate a mixture of
two or more miscible liquids
177
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Distillation
The laboratory set-up of
fractional distillation
178
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Distillation
• A fractionating column is attached vertically
between the flask and the condenser
 a column packed with glass beads
 provide a large surface area for the
repeated condensation and vaporization
of the mixture to occur
179
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Distillation
• The temperature of the escaping vapour
is measured using a thermometer
• When the temperature reading becomes
steady,
 the vapour with the lowest boiling
point firstly comes out from the top of
the column
180
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Distillation
•
When all of that liquid has distilled off,
 the temperature reading rises and
becomes steady later on
 another liquid with a higher boiling
point distils out
•
181
Fractions with different boiling points can
be collected separately
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Sublimation
• Sublimation is the direct change of
 a solid to vapour on heating, or
 a vapour to solid on cooling
 without going through the liquid
state
182
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Sublimation
• A mixture of two compounds is heated in an
evaporating dish
• One compound changes from solid to
vapour directly
 The vapour changes back to solid on a
cold surface
• The other compound is not affected by
heating and remains in the evaporating dish
183
34.2 Isolation and Purification of Organic Compounds (SB p.83)
Sublimation
A mixture of two compounds can be
separated by sublimation
184
34.2 Isolation and Purification of Organic Compounds (SB p.83)
Chromatography
185
•
An effective method of separating a
complex mixture of substances
•
Paper chromatography is a common
type of chromatography
34.2 Isolation and Purification of Organic Compounds (SB p.83)
Chromatography
The laboratory set-up of paper chromatography
186
34.2 Isolation and Purification of Organic Compounds (SB p.83)
Chromatography
• A solution of the mixture is dropped at
one end of the filter paper
187
34.2 Isolation and Purification of Organic Compounds (SB p.83)
Chromatography
•
The thin film of water adhered onto the surface
of the filter paper forms the stationary phase
•
The solvent is called the mobile phase or eluent
188
34.2 Isolation and Purification of Organic Compounds (SB p.83)
Chromatography
•
When the solvent moves across the sample spot
of the mixture,
 partition of the components between the
stationary phase and the mobile phase occurs
189
34.2 Isolation and Purification of Organic Compounds (SB p.83)
Chromatography
• As the various components are being
adsorbed or partitioned at different rates,
 they move upwards at different rates
• The ratio of the distance travelled by the
substance to the distance travelled by the
solvent
 known as the Rf value
 a characteristic of the substance
190
34.2 Isolation and Purification of Organic Compounds (SB p.84)
A summary of different techniques of
isolation and purification
Technique
(a) Filtration
Aim
To separate an insoluble solid from a
liquid (slow)
(b) Centrifugation To separate an insoluble solid from a
liquid (fast)
(c) Crystallization To separate a dissolved solute from
its solution
(d) Solvent
To separate a component from a
extraction
mixture with a suitable solvent
(e) Distillation
To separate a liquid from a solution
containing non-volatile solutes
191
34.2 Isolation and Purification of Organic Compounds (SB p.84)
A summary of different techniques of
isolation and purification
Technique
(f) Fractional
distillation
(g) Sublimation
Aim
To separate miscible liquids with
widely different boiling points
To separate a mixture of solids in
which only one can sublime
(h)
To separate a complex mixture of
Chromatography substances
Check Point 34-2
192
34.4
Qualitative
Analysis of
Elements in an
Organic Compound
193
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Qualitative Analysis of an
Organic Compound
• Qualitative analysis of an organic
compound is
 to determine what elements are
present in the compound
194
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Carbon and Hydrogen
• Tests for carbon and hydrogen in an
organic compound are usually
unnecessary
 an organic compound must
contain carbon and hydrogen
195
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Carbon and Hydrogen
• Carbon and hydrogen can be detected by
heating a small amount of the substance
with copper(II) oxide
• Carbon and hydrogen would be oxidized
to carbon dioxide and water respectively
• Carbon dioxide turns lime water milky
• Water turns anhydrous cobalt(II) chloride
paper pink
196
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Sulphur
• Halogens, nitrogen and sulphur in organic
compounds can be detected
 by performing the sodium fusion test
197
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Sulphur
• The compound under test is
 fused with a small piece of sodium
metal in a small combustion tube
 heated strongly
• The products of the test are extracted with
water and then analyzed
198
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Sulphur
• During sodium fusion,
 halogens in the organic compound is
converted to sodium halides
 nitrogen in the organic compound is
converted to sodium cyanide
 sulphur in the organic compound is
converted to sodium sulphide
199
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Results for halogens, nitrogen and sulphur in the
sodium fusion test
Element
Material used
Observation
Halogens, as
Acidified silver
nitrate solution
chloride ion (Cl-)
A white precipitate is
formed. It is soluble in
excess NH3(aq).
bromide ion (Br-)
A pale yellow
precipitate is formed. It
is sparingly soluble in
excess NH3(aq).
iodide ion (I-)
A creamy yellow
precipitate is formed. It
is insoluble in excess
NH3(aq).
200
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Results for halogens, nitrogen and sulphur in the
sodium fusion test
Element
Material used
Observation
Nitrogen,as
cyanide ion (CN-)
A mixture of
A blue-green colour is
iron(II) sulphate observed.
and iron(III)
sulphate
solutions
Sulphur, as
sulphide ion (S2-)
Sodium
pentacyanonitr
osylferrate(II)
solution
A black precipitate is
formed
Check Point 34-4
201
34.5
Determination of
Empirical Formula
and Molecular
Formula from
Analytical Data
202
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
Quantitative Analysis of
an Organic Compound
• After determining the constituent elements
of a particular organic compound
 perform quantitative analysis to find the
percentage composition by mass of the
compound
 the masses of different elements in an
organic compound are determined
203
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
1. Carbon and Hydrogen
• The organic compound is burnt in excess
oxygen
• The carbon dioxide and water vapour
formed are respectively absorbed by
 potassium hydroxide solution and
anhydrous calcium chloride
204
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
1. Carbon and Hydrogen
• The increases in mass in potassium
hydroxide solution and calcium chloride
represent
 the masses of carbon dioxide and
water vapour formed respectively
205
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
2. Nitrogen
• The organic compound is heated with
excess copper(II) oxide
• The nitrogen monoxide and nitrogen
dioxide formed are passed over hot
copper
 the volume of nitrogen formed is
measured
206
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
3. Halogens
• The organic compound is heated with
fuming nitric(V) acid and excess silver
nitrate solution
• The mixture is allowed to cool
 then water is added
 the dry silver halide formed is weighed
207
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
4. Sulphur
• The organic compound is heated with
fuming nitric(V) acid
• After cooling,
 barium nitrate solution is added
 the dry barium sulphate formed is
weighed
208
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
Quantitative Analysis of
an Organic Compound
• After determining the percentage
composition by mass of a compound,
 the empirical formula of the compound
can be calculated
209
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
Quantitative Analysis of
an Organic Compound
The empirical formula of a compound is
the formula which shows the simplest
whole number ratio of the atoms present
in the compound
210
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.87)
Quantitative Analysis of
an Organic Compound
• When the relative molecular mass and
the empirical formula of the compound
are known,
 the molecular formula of the
compound can be calculated
211
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.88)
Quantitative Analysis of
an Organic Compound
The molecular formula of a compound
is the formula which shows the actual
number of each kind of atoms present
in a molecule of the compound
212
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.88)
Example 34-5A
Example 34-5B
Check Point 34-5
213
34.6
Structural
Information from
Physical Properties
214
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from
Physical Properties
• The physical properties of a compound
include its colour, odour, density, solubility,
melting point and boiling point
• The physical properties of a compound
depend on its molecular structure
215
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from
Physical Properties
• From the physical properties of a
compound,
 obtain preliminary information about
the structure of the compound
216
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from
Physical Properties
•
e.g.
 Hydrocarbons have low densities,
often about 0.8 g cm–3
 Compounds with functional groups
have higher densities
217
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from
Physical Properties
• The densities of most organic compounds
are < 1.2 g cm–3
• Compounds having densities > 1.2 g cm–3
must contain multiple halogen atoms
218
34.6 Structural Information from Physical Properties (SB p.90)
Physical properties of some common organic compounds
Organic
compound
Hydrocarbo
ns
(saturated
and
unsaturated)
219
Density
at 20oC
Melting point and
boiling point
Solubility
In water
In nonor highly
polar
polar
organic
solvents solvents
All
• Generally low but
Insoluble
increases with
have
number of carbon
densities
atoms in the molecule
< 0.8 g
cm–3
• Branched-chain
hydrocarbons have
lower boiling points
but higher melting
points than the
corresponding
straight-chain isomers
Soluble
34.6 Structural Information from Physical Properties (SB p.90)
Physical properties of some common organic compounds
Organic
compound
Density at
20oC
Aromatic
Between
hydrocarbons 0.8 and 1.0
g cm–3
220
Melting point and
boiling point
Generally low
Solubility
In water
In nonor highly
polar
polar
organic
solvents solvents
Insoluble
Soluble
34.6 Structural Information from Physical Properties (SB p.90)
Physical properties of some common organic compounds
Organic Density at
compound
20oC
Haloalkanes
221
• 0.9 - 1.1
g cm–3
for
chloroalkanes
• >1.0 g
cm–3 for
bromoalkanes
and
iodoalkanes
Melting point and
boiling point
• Higher than alkanes of
similar relative
molecular masses
( haloalkane
molecules are polar)
• All haloalkanes are
liquids except
halomethanes
• Both the m.p. and b.p.
increase in the order:
RCH2F < RCH2Cl <
RCH2Br < RCH2I
Solubility
In water
In nonor highly
polar
polar
organic
solvents solvents
Insoluble
Soluble
34.6 Structural Information from Physical Properties (SB p.90)
Physical properties of some common organic compounds
Organic
compound
Density at
20oC
Melting point and
boiling point
Alcohols • Simple
• Much higher than
alcohols are
hydrocarbons of
liquids and
similar relative
alcohols with molecular masses
( formation of
> 12
carbons are
hydrogen bonds
waxy solids
between alcohol
molecules)
222
Solubility
In water or
highly polar
solvents
In nonpolar
organic
solvents
• Lower
members:
Completely
miscible with
water (
formation of
hydrogen
bonds
between
alcohol
molecules
and water
molecules)
Soluble
34.6 Structural Information from Physical Properties (SB p.90)
Physical properties of some common organic compounds
Organic
compound
Density at
20oC
Melting point and
boiling point
Alcohols • All simple
• Straight-chain
alcohols
alcohols have
have
higher b.p. than the
densities
corresponding
< 1.0 g cm–3
branched-chain
alcohols
223
Solubility
In water or
highly polar
solvents
In nonpolar
organic
solvents
• Solubility
Soluble
decreases
gradually as
the
hydrocarbon
chain
lengthens
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compounds
Organic
compound
Density at
20oC
Carbonyl • <1.0 g cm–3
for aliphatic
compcarbonyl
ounds
compounds
(aldehydes
and
ketones)
224
Melting point and
boiling point
Higher than alkanes
but lower than
alcohols of similar
relative molecular
masses (Molecules
of aldehydes or
ketones are held
together by strong
dipole-dipole
interactions but not
hydrogen bonds)
Solubility
In water or
highly polar
solvents
In nonpolar
organic
solvents
• Lower
Soluble
members:
Soluble in
water ( the
formation of
hydrogen
bonds between
molecules of
aldehydes or
ketones and
water
molecules)
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compounds
Organic
compound
Density at
20oC
Carbonyl • > 1.0 g cm–3
for aromatic
compcarbonyl
ounds
compounds
(aldehydes
and
ketones)
225
Melting point and
boiling point
Solubility
In water or
highly polar
solvents
In nonpolar
organic
solvents
• Solubility
decreases
gradually as
the
hydrocarbon
chain
lengthens
Soluble
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compounds
Organic
compound
Carboxylic
acids
226
Density at
20oC
Melting point and
boiling point
• Lower
members
have
densities
similar to
water
• Methanoic
acid has a
density of
1.22 g cm–3
Higher than alcohols
of similar relative
molecular masses
( the formation of
more extensive
intermolecular
hydrogen bonds)
Solubility
In water or
highly polar
solvents
• First four
members are
miscible with
water in all
proportions
• Solubility
decreases
gradually as
the
hydrocarbon
chain
lengthens
In nonpolar
organic
solvents
Soluble
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compounds
Organic
compound
Esters
227
Density at
20oC
Lower
members
have
densities
less than
water
Melting point and
boiling point
Slightly higher than
hydrocarbons but
lower than carbonyl
compounds and
alcohols of similar
relative molecular
masses
Solubility
In water or
highly polar
solvents
Insoluble
In nonpolar
organic
solvents
Soluble
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compounds
Organic
compound
Amines
228
Density at
20oC
Melting point and
boiling point
Most amines • Higher than
have
alkanes but lower
densities
than alcohols of
less than
similar relative
water
molecular masses
Solubility
In water or
highly polar
solvents
• Generally
soluble
• Solubility
decreases in
the order:
1o amines >
2o amines >
3o amines
In nonpolar
organic
solvents
Soluble
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compounds
Organic
compound
Amines
229
Density at
20oC
Melting point and
boiling point
• 1o and 2o amines
are able to form
hydrogen bonds
with each other but
the strength is less
than that between
alcohol molecules
(NH bond is less
polar than O  H
bond)
Solubility
In water or
highly polar
solvents
In nonpolar
organic
solvents
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compounds
Organic
compound
Amines
230
Density at
20oC
Melting point and
boiling point
• 3o amines have
lower m.p. and b.p.
than the isomers
of 1o and 2o
amines
( molecules of 3o
amines cannot
form
intermolecular
hydrogen bonds)
Solubility
In water or
highly polar
solvents
In nonpolar
organic
solvents
34.6 Structural Information from Physical Properties (SB p.92)
Example 34-6
231
Check Point 34-6
34.7
Structural
Information from
Chemical Properties
232
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from
Chemical Properties
• The molecular formula of a compound
 does not give enough clue to the
structure of the compound
• Compounds having the same molecular
formula
 may have different arrangements of
atoms and even different functional groups
233
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from
Chemical Properties
• e.g.
The molecular formula of C2H4O2 may
represent a carboxylic acid or an ester:
234
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from
Chemical Properties
• The next stage is
 to find out the functional group(s)
present
 to deduce the actual arrangement
of atoms in the molecule
235
34.7 Structural Information from Chemical Properties (SB p.93)
Chemical tests for different groups of organic compounds
Organic
compound
Saturated
hydrocarbons
236
Test
Observation
• Burn the
• A blue or clear
saturated
yellow flame is
hydrocarbon
observed
in a nonluminous
Bunsen flame
34.7 Structural Information from Chemical Properties (SB p.93)
Chemical tests for different groups of organic compounds
Organic
compound
Test
Observation
Unsaturated
• Burn the unsaturated • A smoky flame is
hydrocarbons
hydrocarbon in a
observed
(C = C,
non-luminous Bunsen
C  C)
flame
• Add bromine in 1,1,1- • Bromine
trichloroethane at
decolourizes rapidly
room temperature
and in the absence of
light
237
• Add 1% (dilute)
acidified potassium
manganate(VII)
solution
• Potassium
manganate(VII)
solution decolourizes
rapidly
34.7 Structural Information from Chemical Properties (SB p.93)
Chemical tests for different groups of organic compounds
Organic
compound
Haloalkanes
(1°, 2° or 3°)
238
Test
Observation
• Boil with ethanolic
• For chloroalkanes, a
potassium hydroxide
white precipitate is
solution, then acidify
formed
with excess dilute
• For bromoalkanes, a
nitric(V) acid and add
pale yellow
silver nitrate(V)
precipitate is formed
solution
• For iodoalkanes, a
creamy yellow
precipitate is formed
34.7 Structural Information from Chemical Properties (SB p.93)
Chemical tests for different groups of organic compounds
Organic
Test
Observation
compound
Halobenzenes • Boil with ethanolic
• No precipitate is
potassium hydroxide
formed
solution, then acidify
with excess dilute
nitric(V) acid and add
silver nitrate(V)
solution
239
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compounds
Organic
compound
Alcohols
(  OH)
240
Test
Observation
• Add a small piece of
sodium metal
• A colourless gas is
evolved
• Esterification: Add
ethanoyl chloride
• The temperature of
the reaction mixture
rises
• A colourless gas is
evolved
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compounds
Organic
compound
Alcohols
(  OH)
241
Test
• Add acidified
potassium
dichromate(VI)
solution
Observation
• For 1° and 2°
alcohols, the clear
orange solution
becomes opaque
and turns green
almost immediately
• For 3° alcohols, there
are no observable
changes
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compounds
Organic
compound
Alcohols
(  OH)
Test
• Iodoform test for:
Add iodine in sodium
hydroxide solution
242
Observation
• A yellow precipitate
is formed
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compounds
Organic
compound
Alcohols
(  OH)
243
Test
• Lucas test: add a
solution of zinc
chloride in
concentrated
hydrochloric acid
Observation
• For 1° alcohols, the
aqueous phase
remains clear
• For 2° alcohols, the
clear solution
becomes cloudy
within 5 minutes
• For 3° alcohols, the
aqueous phase
appears cloudy
immediately
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compounds
Organic
compound
Ethers
( O  )
244
Test
• No specific test for
ethers but they are
soluble in
concentrated
sulphuric(VI) acid
Observation

34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compounds
Organic
Test
Observation
compound
Aldehydes
• Add aqueous sodium
• Crystalline salts are
hydrogensulphate(IV)
formed
(
)
• Add 2,4• A yellow, orange or
dinitrophenylhydrazine red precipitate is
formed
• Silver mirror test: add
Tollens’ reagent (a
solution of aqueous
silver nitrate in
aqueous ammonia)
245
• A silver mirror is
deposited on the
inner wall of the test
tube
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compounds
Organic
Test
compound
Ketones
• Add aqueous sodium
hydrogensulphate(IV)
Observation
• Crystalline salts are
formed (for
unhindered ketones
)
only)
• Add 2,4• A yellow, orange or
dinitrophenylhydrazine red precipitate is
formed
(
• Iodoform test for:
246
Add iodine in sodium
hydroxide solution
• A yellow precipitate
is formed
34.7 Structural Information from Chemical Properties (SB p.95)
Chemical tests for different groups of organic compounds
Organic
Test
compound
Carboxylic
• Esterification: warm
acids
the carboxylic acid
with an alcohol in the
(
) presence of
concentrated
sulphuric(VI) acid,
followed by adding
sodium carbonate
solution
• Add sodium
hydrogencarbonate
247
Observation
• A sweet and fruity
smell is detected
• The colourless gas
produced turns lime
water milky
34.7 Structural Information from Chemical Properties (SB p.95)
Chemical tests for different groups of organic compounds
Organic
Test
compound
Esters
• No specific test for
esters but they can be
(
) distinguished by its
characteristic smell
248
Observation
• A sweet and fruity
smell is detected
34.7 Structural Information from Chemical Properties (SB p.95)
Chemical tests for different groups of organic compounds
Organic
Test
compound
Acyl halides • Boil with ethanolic
potassium hydroxide
(
) solution, then acidify
with excess dilute
nitric(V) acid and add
silver nitrate(V)
solution
249
Observation
• For acyl chlorides, a
white precipitate is
formed
• For acyl bromides, a
pale yellow
precipitate is formed
• For acyl iodides, a
creamy yellow
precipitate is formed
34.7 Structural Information from Chemical Properties (SB p.95)
Chemical tests for different groups of organic compounds
Organic
Test
compound
Amides
• Boil with sodium
hydroxide solution
(
)
250
Observation
• The colourless gas
produced turns moist
red litmus paper or
pH paper blue
34.7 Structural Information from Chemical Properties (SB p.95)
Chemical tests for different groups of organic compounds
Organic
Test
Observation
compound
Amines
• 1o aliphatic amines:
• Steady evolution of
dissolve the amine in
N2(g) is observed
(NH2)
dilute hydrochloric acid
at 0 – 5 oC, then add
cold sodium nitrate(III)
solution slowly
• 1o aromatic amines:
• An orange or red
add naphthalen-2-ol in
precipitate is formed
dilute sodium
hydroxide solution
251
34.7 Structural Information from Chemical Properties (SB p.95)
Chemical tests for different groups of organic compounds
Organic
Test
compound
Aromatic
• Burn the aromatic
compounds
compound in a nonluminous Bunsen
(
) flame
• Add fuming
sulphuric(VI) acid
252
Observation
• A smoky yellow
flame with black soot
is produced
• The aromatic
compound dissolves
• The temperature of
the reaction mixture
rises
34.7 Structural Information from Chemical Properties (SB p.96)
Example 34-7A
Example 34-7C
253
Example 34-7B
Check Point 34-7
The END
254
34.1 Introduction (SB p.77)
What are the necessary information to determine the
structure of an organic compound?
Answer
Molecular formula from analytical data,
functional group present from physical and
chemical properties, structural information from
infra-red spectroscopy and mass spectrometry
Back
255
34.2 Isolation and Purification of Organic Compounds (SB p.84)
For each of the following, suggest a separation technique.
(a) To obtain blood cells from blood
(b) To separate different pigments in black ink
(c) To obtain ethanol from beer
(d) To separate a mixture of two solids, but only one
sublimes
(e) To separate an insoluble solid from a liquid
256
(a)
(b)
(c)
(d)
(e)
Centrifugation
Chromatography
Fractional distillation
Sublimation
Filtration
Back
Answer
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.87)
(a) Why is detection of carbon and hydrogen in organic
compounds not necessary?
(b) What elements can be detected by sodium fusion test?
Answer
(a) All organic compounds contain carbon
and hydrogen.
(b) Halogens, nitrogen and sulphur
Back
257
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.88)
An organic compound was found to contain 40.0% carbon,
6.7% hydrogen and 53.3% oxygen by mass. Calculate the
empirical formula of the compound.
Answer
258
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.88)
Back
Let the mass of the compound be 100 g. Then,
mass of carbon in the compound = 40.0 g
mass of hydrogen in the compound = 6.7 g
mass of oxygen in the compound = 53.3 g
Mass (g)
Carbon
Hydrogen
Oxygen
40.0
6.7
6.7
 6 .7
1.0
53.3
53.3
 3.33
16.0
6.7
2
3.33
3.33
1
3.33
Number of 40.0
 3.33
moles (mol) 12.0
Relative
3.33
1
number of
3.33
moles
Simplest
1
2
1
mole ratio
259
∴ The empirical formula of the organic compound is CH2O.
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.88)
An organic compound Z has the following composition by
mass:
Element
Percentage
by mass (%)
Carbon
Hydrogen
Oxygen
60.00
13.33
26.67
(a) Calculate the empirical formula of compound Z.
(b) If the relative molecular mass of compound Z is 60.0,
determine the molecular formula of compound Z.
Answer
260
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.89)
(a) Let the mass of the compound be 100 g. Then,
mass of carbon in the compound = 60.00 g
mass of hydrogen in the compound = 13.33 g
mass of oxygen in the compound = 26.67 g
Carbon
Hydrogen
60.00
13.33
Number of
moles (mol)
26.67
26.67
60.00
13.33
 1.67
5
 13.33
16.0
12.0
1.0
Relative
number of
moles
5
3
1.67
13.33
8
1.67
Mass (g)
Oxygen
1.67
1
1.67
Simplest
3
8
1
mole ratio
261
∴ The empirical formula of the organic compound is C3H8O.
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.89)
(b) The molecular formula of the compound is (C3H8O)n.
Relative molecular mass of (C3H8O)n = 60.0
n × (12.0 × 3 + 1.0 × 8 + 16.0) = 60.0
n =1
∴ The molecular formula of compound Z is C3H8O.
Back
262
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.89)
An organic compound was found to contain carbon,
hydrogen and oxygen only. On complete combustion, 0.15
g of this compound gave 0.22 g of carbon dioxide and 0.09 g
of water. If the relative molecular mass of this compound is
60.0, determine the molecular formula of this compound.
Answer
263
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.89)
Relative molecular mass of CO2 = 12.0 + 16.0 × 2 = 44.0
12.0
Mass of carbon in 0.22 g of CO2 = 0.22 g ×
44.0
= 0.06 g
Relative molecular mass of H2O = 1.0 × 2 + 16.0
= 18.0
Mass of hydrogen in 0.09 g of H2O = 0.09 g × 2.0
18.0
= 0.01 g
Mass of oxygen in the compound = (0.15 – 0.06 – 0.01) g
= 0.08 g
264
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.89)
Mass (g)
Number of
moles (mol)
Relative
number of
moles
Simplest
mole ratio
Carbon
Hydrogen
Oxygen
0.06
0.06
 0.005
12.0
0.01
0.01
 0.01
1.0
0.08
0.08
 0.005
16.0
0.005
1
0.005
0.01
2
0.005
1
0.005
1
0.005
2
∴ The empirical formula of the organic compound is CH2O.
265
1
34.5 Determination of Empirical Formula and Molecular Formula from
Analytical Data (SB p.89)
Let the molecular formula of the compound be (CH2O)n.
Relative molecular mass of (CH2O)n = 60.0
n × (12.0 + 1.0 × 2 + 16.0) = 60.0
n =2
∴ The molecular formula of the compound is C2H4O2.
Back
266
34.6 Structural Information from Physical Properties (SB p.92)
Back
Why do branched-chain hydrocarbons have lower
boiling points but higher melting points than the
corresponding straight-chain isomers?
Answer
Branched-chain hydrocarbons have lower boiling points than the
corresponding straight-chain isomers because the straight-chain
isomers are being flattened in shape. They have greater surface
area in contact with each other. Hence, molecules of the straightchain isomer are held together by greater attractive forces. On the
other hand, branched-chain hydrocarbons have higher melting
points than the corresponding straight-chain isomers because
branched-chain isomers are more spherical in shape and are
packed more efficiently in solid state. Extra energy is required to
267break down the efficient packing in the process of melting.
34.6 Structural Information from Physical Properties (SB p.92)
Back
Why does the solubility of amines in water decrease in
the order:
1o amines > 2o amines > 3o amines?
Answer
The solubility of primary and secondary amines is
higher than that of tertiary amines because tertiary
amines cannot form hydrogen bonds between
water molecules. On the other hand, the solubility
of primary amines is higher than that of secondary
amines because primary amines form a greater
number of hydrogen bonds with water molecules
than secondary amines.
268
34.6 Structural Information from Physical Properties (SB p.92)
Match the boiling points 65oC, –6oC and –88oC with the
compounds CH3CH3, CH3NH2 and CH3OH. Explain your
answer briefly.
Answer
269
34.6 Structural Information from Physical Properties (SB p.92)
Back
Compounds Boiling point (°C)
CH3CH3
–88
CH3NH2
–6
CH3OH
65
Ethane (CH3CH3) is a non-polar compound. In pure liquid form, ethane
molecules are held together by weak van der Waals’ forces. However,
both methylamine (CH3NH2) and methanol (CH3OH) are polar
substances. In pure liquid form, their molecules are held together by
intermolecular hydrogen bonds. As van der Waals’ forces are much
weaker than hydrogen bonds, ethane has the lowest boiling point
among the three. Besides, as the O  H bond in alcohols is more
polar than the N  H bond in amines, the hydrogen bonds formed
between methylamine molecules are weaker than those formed
between methanol molecules. Thus, methylamine has a lower boiling
270
point than methanol.
34.6 Structural Information from Physical Properties (SB p.92)
(a) Butan-1-ol boils at 118°C and butanal boils at 76°C.
(i) What are the relative molecular masses of butan-1ol and butanal?
(ii) Account for the higher boiling point of butan-1-ol.
(a) (i)
(ii)
271
The relative molecular masses of butan-1-ol
and butanal are 74.0 and 72.0 respectively.
Butan-1-ol has a higher boiling point because
it is able to form extensive hydrogen bonds
with each other, but the forces holding the
butanal molecules together are dipole-dipole
interactions only.
Answer
34.6 Structural Information from Physical Properties (SB p.92)
(b) Arrange the following compounds in order of
increasing solubility in water. Explain your answer.
Ethanol, chloroethane, hexan-1-ol
Answer
(b) The solubility increases in the order: chloroethane <
hexan-1-ol < ethanol. Both hexan-1-ol and ethanol are
more soluble in water than chloroethane because
molecules of the alcohols are able to form extensive
hydrogen bonds with water molecules. Molecules of
chloroethane are not able to form hydrogen bonds with
water molecules and that is why it is insoluble in water.
Hexan-1-ol has a longer carbon chain than ethanol and
this explains why it is less soluble in water than ethanol.
272
34.6 Structural Information from Physical Properties (SB p.92)
(c) Explain why (CH3)3N (b.p.: 2.9°C) boils so much lower
than CH3CH2CH2NH2 (b.p.: 48.7°C) despite they have
the same molecular mass.
Answer
(c) They are isomers. The primary amine is able to form
hydrogen bonds with the oxygen atom of water
molecules, but there is no hydrogen atoms directly
attached to the nitrogen atom in the tertiary amine.
273
34.6 Structural Information from Physical Properties (SB p.92)
Back
(d) Match the boiling points with the isomeric carbonyl
compounds.
Compounds:
Heptanal, heptan-4-one,
2,4-dimethylpentan-3-one
Boiling points: 124°C, 144°C, 155°C
(d)
274
Compound
Boiling point (oC)
Heptanal
155
Heptan-4-one
144
2,4-Dimethylpentan-3-one
125
Answer
34.7 Structural Information from Chemical Properties (SB p.96)
The empirical formula of an organic compound is CH2O
and its relative molecular mass is 60.0. It reacts with
sodium hydrogencarbonate to give a colourless gas which
turns lime water milky.
(a) Calculate the molecular formula of the compound.
Answer
(a) Let the molecular formula of the compound be (CH2O)n.
Relative molecular mass of (CH2O)n= 60.0
n  (12.0 + 1.0  2 + 16.0) = 60.0
n =2
∴ The molecular formula of the compound is C2H4O2.
275
34.7 Structural Information from Chemical Properties (SB p.96)
The empirical formula of an organic compound is CH2O
and its relative molecular mass is 60.0. It reacts with
sodium hydrogencarbonate to give a colourless gas which
turns lime water milky.
(b) Deduce the structural formula of the compound.
Answer
276
34.7 Structural Information from Chemical Properties (SB p.96)
(b) The compound reacts with sodium hydrogencarbonate to give a
colourless gas which turns lime water milky. This indicates that the
compound contains a carboxyl group ( COOH). Eliminating the
 COOH group from the molecular formula of C2H4O2, the atoms
left are one carbon and three hydrogen atoms. This obviously
shows that a methyl group ( CH3) is present. Therefore, the
structural formula of the compound is:
277
34.7 Structural Information from Chemical Properties (SB p.96)
The empirical formula of an organic compound is CH2O
and its relative molecular mass is 60.0. It reacts with
sodium hydrogencarbonate to give a colourless gas which
turns lime water milky.
(c) Give the IUPAC name for the compound.
(c) The IUPAC name for the compound is
ethanoic acid.
Back
278
Answer
34.7 Structural Information from Chemical Properties (SB p.96)
15 cm3 of a gaseous hydrocarbon were mixed with 120 cm3
of oxygen which was in excess. The mixture was exploded.
After cooling, the residual volume was 105 cm3. On
adding concentrated potassium hydroxide solution, the
volume decreased to 75 cm3.
(a) Calculate the molecular formula of the compound,
assuming all the volumes were measured under room
temperature and pressure.
(b) To which homologous series does the hydrocarbon
belong?
(c) Give the structural formula of the hydrocarbon.
279
Answer
34.7 Structural Information from Chemical Properties (SB p.97)
(a) Let the molecular formula of the compound be CxHy.
Volume of CxHy reacted = 15 cm3
Volume of unreacted oxygen = 75 cm3
Volume of oxygen reacted = (120 - 75) cm3 = 45 cm3
Volume of carbon dioxide formed = (105 - 75) cm3 = 30 cm3
CxHy + (x + y )O2  xCO2 + y H2O
2
4
Volume of CxHy reacted : Volume of CO2 formed = 1 : x = 15 : 30

1 15

x 30

x=2
280
34.7 Structural Information from Chemical Properties (SB p.97)
(a) Volume of CxHy reacted : Volume of O2 reacted = 1 : ( x  y )
4
= 15 : 45

1

15
45
y
(2  )
4
y
2  3
4

y=4

The molecular formula of the compound is C2H4.
(b) C2H4 belongs to alkenes.
(c) The structural formula of the hydrocarbon is:
Back
281
34.7 Structural Information from Chemical Properties (SB p.97)
Answer
20 cm3 of a gaseous organic compound containing only
carbon, hydrogen and oxygen were mixed with 110 cm3 of
oxygen which was in excess. The mixture was exploded at
105oC and the volume of the gaseous mixture was 150 cm3.
After cooling to room temperature, the residual volume was
reduced to 90 cm3. On adding concentrated potassium
hydroxide solution, the volume further decreased to 50 cm3.
(a) Calculate the molecular formula of the compound,
assuming that all the volumes were measured under
room temperature and pressure.
(b) The compound is found to contain a hydroxyl group
( OH) in its structure. Deduce its structural formula.
(c) Is the compound optically active? Explain your answer.
282
34.7 Structural Information from Chemical Properties (SB p.97)
(a) Let the molecular formula of the compound be CxHyOz.
Volume of CxHyOz reacted = 20 cm3
Volume of unreacted oxygen = 50 cm3
Volume of oxygen reacted = (110 - 50) cm3 = 60 cm3
Volume of carbon dioxide formed = (90 - 50) cm3 = 40 cm3
Volume of water (in the form of steam) formed
= (90 - 50) cm3 = 40 cm3
y
CxHyOz + (x + - z )O2  xCO2 + y H2O
4 2
2
Volume of CxHyOz reacted : Volume of CO2 formed = 1 : x = 20 : 40
 1 20

x 40

x=2
283
34.7 Structural Information from Chemical Properties (SB p.98)
y
(a) Volume of CxHyOz reacted : Volume of H2O formed = 1 : = 20 : 60
2
2 20


y 60

y=6
y z
Volume of CxHyOz reacted : Volume of O2 reacted = 1 : ( x   )
4 2
= 20 : 60
1
20

y z
60
(x  - )
4 2
6 z
2   3
4 2

z=1

The molecular formula of the compound is C2H6O.

284
34.7 Structural Information from Chemical Properties (SB p.98)
Back
(b) As the compound contains a OH group, the hydrocarbon
skeleton of the compound becomes C2H5 after eliminating the
 OH group from the molecular formula of C2H6O. The structural
formula of the compound is:
(c) The compound is optically inactive as both carbon atoms in the
compound are not asymmetric, i.e. both of them do not attach to
four different atoms or groups of atoms.
285
34.7 Structural Information from Chemical Properties (SB p.99)
(a) A substance contains 42.8% carbon, 2.38% hydrogen,
16.67% nitrogen by mass and the remainder consists of
oxygen.
(i) Given that the relative molecular mass of the
substance is 168.0, deduce the molecular formula
of the substance.
(ii) The substance is proved to be an aromatic
compound with only one type of functional group.
Give the names and structural formulae for all
isomers of the substance.
Answer
286
34.7 Structural Information from Chemical Properties (SB p.99)
(a) (i)
Let the mass of the compound be 100 g.
Mass (g)
Carbon
Hydrogen
Nitrogen
Oxygen
42.8
2.38
16.67
38.15
Number of
38.15
42.8
16.67
2.38
 2.38
 3.57
 1.19
 2.38
moles (mol) 12.0
16
.
0
14
.
0
1.0
Relative
2.38
3.57
2.38
1.19
2
3
number of
2
1
1.19
1.19
1.19
1.19
moles
Simplest
mole ratio
3
2
1
∴ The empirical formula of the compound is C3H2NO2.
287
2
34.7 Structural Information from Chemical Properties (SB p.99)
(a) (i)
(ii)
288
Let the molecular formula of the compound be (C3H2NO2)n.
Molecular mass of (C3H2NO2)n = 168.0
n × (12.0 × 3 + 1.0 × 2 + 14.0 + 16.0 × 2) = 168.0
∴
n =2
∴ The molecular formula of the compound is C6H4N2O4.
34.7 Structural Information from Chemical Properties (SB p.99)
(b) 30 cm3 of a gaseous hydrocarbon were mixed with 140
cm3 of oxygen which was in excess, and the mixture
was then exploded. After cooling to room temperature,
the residual gases occupied 95 cm3 by volume. By
adding potassium hydroxide solution, the volume was
reduced by 60 cm3. The remaining gas was proved to be
oxygen.
(i) Determine the molecular formula of the
hydrocarbon.
(ii) Is the hydrocarbon a saturated, an unsaturated or
an aromatic hydrocarbon?
Answer
289
34.7 Structural Information from Chemical Properties (SB p.99)
(b) (i)
Volume of hydrocarbon reacted = 30 cm3
Volume of unreacted oxygen = (95 – 60) cm3 = 35 cm3
Volume of oxygen reacted = (140 - 35) cm3 = 105 cm3
Volume of carbon dioxide formed = 60 cm3
CxHy + (x + y )O2  xCO2 + y H2O
2
4
Volume of CxHy reacted : Volume of CO2 formed
= 1 : x = 30 : 60
 1 30

x 60
 x=2
290
34.7 Structural Information from Chemical Properties (SB p.99)
(b) (i)
(ii)
291
Volume of CxHy reacted : Volume of O2 reacted
= 1 : ( 2  y ) = 30 : 105
4
1
30


y
105
(x  )
4
y
30  (2  )  105
4

y=6
 The molecular formula of the compound is C2H6.
From the molecular formula of the hydrocarbon, it can be
deduced that the hydrocarbon is saturated because it fulfils
the general formula of alkanes CnH2n+2.
34.7 Structural Information from Chemical Properties (SB p.99)
(c) A hydrocarbon having a relative molecular mass of 56.0
contains 85.5% carbon and 14.5% hydrogen by mass.
Detailed analysis shows that it has two geometrical
isomers.
(i) Deduce the molecular formula of the hydrocarbon.
(ii) Name the two geometrical isomers of the
hydrocarbon.
(iii) Explain the existence of geometrical isomerism in
the hydrocarbon.
Answer
292
34.7 Structural Information from Chemical Properties (SB p.99)
(c) (i)
Let the mass of the compound be 100 g.
Mass (g)
Number of
moles (mol)
Relative number
of moles
Simplest mole
ratio
Carbon
Hydrogen
85.5
14.5
85.5
 7.125
12.0
7.125
1
7.125
1
14.5
 14.5
1. 0
14.5
2
7.125
2
∴ The empirical formula of the compound is CH2.
293
34.7 Structural Information from Chemical Properties (SB p.99)
Back
(c) (i)
Let the molecular formula of the hydrocarbon be (CH2)n.
Molecular mass of (CH2)n = 56.0
n × (12.0 + 1.0 × 2) = 56.0
n =4
∴ The molecular formula of the hydrocarbon is C4H8.
(ii)
(iii) Since but-2-ene is unsymmetrical and free rotation of but-2-ene
is restricted by the presence of the carbon-carbon double bond,
geometrical isomerism exists.
294
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.102)
What is the relationship between frequency and
wavenumber?
Answer
The higher the frequency, the higher the
wavenumber.
Back
295
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.110)
An organic compound with a relative molecular mass of
72.0 was found to contain 66.66% carbon, 22.23% oxygen
and 11.11% hydrogen by mass. A portion of its infra-red
spectrum is shown below.
296
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.110)
(a) Determine the molecular formula of the compound.
(b) Deduce two possible structures of the compound, each
of which belongs to a different homologous series.
Answer
297
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.110)
(a) Let the mass of the compound be 100 g. Then,
mass of carbon in the compound = 66.66 g
mass of hydrogen in the compound = 11.11 g
mass of oxygen in the compound = 22.23 g
Mass (g)
Number of moles
(mol)
Relative number
of moles
Simplest mole
ratio
298
Carbon
Hydrogen
Oxygen
66.66
11.11
22.23
11.11
66.66
22.23
 11.11
 5.56
 1.39
1 .0
12.0
16.0
1.39
5.56
11.11
1
4
8
1.39
1.39
1.39
4
8
1
∴ The empirical formula of the compound is C4H8O.
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.110)
Let the molecular formula of the compound be (C4H8O)n.
Relative molecular mass of (C4H8O)n = 72.0
n × (12.0 × 4 + 1.0 × 8 + 16.0) = 72.0
∴
n =1
∴ The molecular formula of the compound is C4H8O.
299
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.110)
(b) From the IR spectrum, it can be observed that there are absorption
peaks at 2 950 cm–1 and 1 700 cm–1. The absorption peak at 2 950
cm–1 corresponds to the stretching vibration of the C  H bond, and
the absorption peak at 1 700 cm–1 corresponds to the stretching
vibration of the C = O bond. Since there is only one oxygen atom in
the molecule of the compound, we can deduce that the compound is
either an aldehyde or a ketone.
If it is an aldehyde, its possible structure will be:
300
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.110)
(b) If it is a ketone, its possible structure will be:
Back
301
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.111)
(a) An organic compound X forms a silver mirror with
ammoniacal silver nitrate solution. Another organic
compound Y reacts with ethanoic acid to give a product
with a fruity smell. The portions of infra-red spectra of
X and Y are shown below.
302
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.111)
Sketch the infra-red spectrum of a carboxylic acid based on
the IR spectra of X and Y.
Answer
303
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.111)
(a) From the information given, X would be an aldehyde and Y would be
an alcohol. Comparing the structures of an aldehyde and an alcohol
with that of a carboxylic acid, some common features are found
between the two. In the IR spectrum of a carboxylic acid, it is
expected that it contains the characteristic O — H (similar to the
alcohol) and C = O (similar to the aldehyde) absorption peaks. Thus,
peak values at around 3300 cm–1 and 1720 cm–1 are expected. A
broad band at around 3300 cm–1 is observed due to the complication
of the stretching vibration of the O — H group by hydrogen bonding
and it overlaps with the absorption of the C — H bond in the 2950 –
2875 cm–1 region.
304
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.111)
(a) The infrared spectrum of a carboxylic acid is as follows:
305
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
Answer
(b) The infra-red spectra of two organic compounds A and B
are shown below.
Decide which compound could be an alcohol. Explain your
306 answer briefly.
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
(b) Compound B could be an alcohol. From the two spectra given,
compound B shows a broad band at 3300 cm–1 and several peaks at
2960 – 2875 cm–1. This broad band corresponds to the complication
of the stretching vibration of the O — H bond by hydrogen bonding
occurring among alcohol molecules.
307
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
(c) The table below shows the characteristic absorption
wavenumbers of some covalent bonds in infra-red
spectra.
Bond
C=O
OH
Range of wavenumber (cm-1)
1680 – 1750
2500 – 3300
CH
NH
2840 – 3095
3350 – 3500
Answer
308
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
Sketch the expected infra-red spectrum for an amino acid
with the following structure:
309
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
(c) The infra-red spectrum of the amino acid is shown as follows:
310
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
(d) A portion of the infra-red spectrum of an organic
compound X is shown below. To which homologous series
does it belong?
Answer
311
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
(d) In the IR spectrum of compound X, the wide absorption band at
3500 – 3000 cm–1 corresponds to the stretching vibration of the
O — H bond. Besides, the absorption peak at 1760 – 1720 cm–1
corresponds to the stretching vibration of the C = O bond. Therefore,
compound X is a carboxylic acid.
312
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
(e) A portion of the infra-red spectrum of an organic
compound Y is shown on the right. Identify the functional
groups that it contains.
Answer
313
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.112)
(e) From the IR spectrum of compound Y, the two peaks in the 3300 –
3180 cm–1 region show that the compound contains the –NH2 group.
Besides, the sharp peak at 1680 cm–1 implies that the compound
also contains the C = O bond.
314
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.113)
(f) An organic compound Z with a relative molecular mass of
88.0 was found to contain 54.54% carbon, 36.36% oxygen
and 9.10% hydrogen by mass. A portion of its infra-red
spectrum is shown below:
315
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.113)
(i) Determine the molecular formula of compound Z.
(ii) Based on the result from (i), draw two possible structures
of the compound, each of which belongs to a different
homologous series.
(iii) Using the information from the IR spectrum, name the
homologous series that compound Z belongs to. Explain
your answer.
Answer
316
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.113)
(f) (i)
Let the mass of the compound be 100 g.
Mass (g)
Number of moles
(mol)
Relative number
of moles
Simplest mole
ratio
Carbon
Hydrogen
Oxygen
54.54
9.10
36.36
9.10
54.54
36.36

9
.
10
 4.55
 2.27
1
.
0
12.0
16.0
9.10
4.55
2.27
4
2
1
2.27
2.27
2.27
2
4
1
∴ The empirical formula of the compound is C2H4O.
317
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.113)
(f) (i)
(ii)
318
Let the molecular formula of the compound be (C2H4O)n.
Relative molecular mass of (C2H4O)n = 88.0
n × (12.0 × 2 + 1.0 × 4 + 16.0) = 88.0
n =2
∴ The molecular formula of the compound is C4H8O2.
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups
(SB p.113)
(f) (iii) From the IR spectrum of compound Z, the absorption peak at
3200 – 2800 cm–1 corresponds to the stretching vibration of
the C — H bond. Besides, the absorption peak at
1800 – 1600 cm–1 corresponds to the stretching vibration of
the C = O bond. The absence of the characteristic peak of the
O — H bond in the 3230 – 3670 cm–1 region indicates that
compound Z is an ester.
Back
319
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
Answer
The mass spectrum of pentan-2-one (CH3COCH2CH2CH3)
is shown below:
What ions do the peaks at m/e 86, 71 and 43 represent?
320
Explain your answer.
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
The relative molecular mass of pentan-2-one is 86. Therefore, the peak
at m/e 86 corresponds to the molecular ion of pentan-2-one. When the
C1  C2 bond is broken, the ion CH3CH2CH2CO+ (m/e = 71) is formed.
When the C2  C3 bond is broken, the ion CH3CO+ (m/e = 43) is formed.
Back
321
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
The mass spectrum of hydrocarbon X is shown below:
322
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
(a) What is the relative molecular mass of hydrocarbon
X?
(b) Which peak is the base peak?
(c) How many mass units is the base peak less than the
peak for the molecular ion?
(d) Deduce the structures of hydrocarbon X.
(e) Explain the peak at m/e 43.
(f) Propose the fragmentation pattern of the molecular
ion which gives rise to the peaks at m/e 58, 43, 29
Answer
and 15.
323
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
(a)
(b)
(c)
(d)
324
The relative molecular mass of hydrocarbon X is 58.0.
The base peak is at m/e 43.
15 mass units
Since the compound is a hydrocarbon, the molecular formula of the
compound must be CxHy. From the relative molecular mass of the
compound (i.e. 58.0), we can deduce that the compound contains 4
carbon atoms only. (If the compound contains 5 carbon atoms, the
relative molecular mass would be more than 12.0 × 5 = 60.0). The
number of hydrogen atoms in the compound is (58.0 - 12.0 × 4 = 10)
10. Therefore, the hydrocarbon is butane.
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
Back
(e) The peak at m/e 43 is 15 mass units less than the molecular ion.
This suggests that a methyl group is lost during the fragmentation of
the molecular ion. The peak at m/e 43 corresponds to CH3CH2CH2+.
(f)
325
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120)
An organic compound is investigated. The structural
formula of this compound is shown below:
326
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120)
The mass spectrum of the compound is shown below:
Interpret the peaks at m/e 134, 119, 91 and 43.
327
Answer
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120)
Back
The peak at m/e 134 corresponds to the molecular ion. The peak at m/e
119 corresponds to the ion that is 15 mass units less than the molecular
ion. This suggests that a methyl group is lost from the molecular ion. The
peak at m/e 91 is the base peak, which corresponds to the ion
C6H5CH2+ . The peak at m/e 43 corresponds to the ion CH3CO+.
328
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
Why would the molecular ion compound have two peaks,
separated by two mass units, in the ratio 3 : 1?
Answer
Chlorine has two isotopes, chlorine-35
and chlorine-37. Their relative
abundances are in the ratio of 3 : 1.
Back
329
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
Why would the molecular ion of a bromine-containing
compound have two peaks, separated by two mass units,
having approximately equal intensities?
Answer
Bromine has two isotopes, bromine-79
and bromine-81. Their relative
abundances are in the ratio of 1 : 1.
Back
330
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(a) What is base peak in a mass spectrum? Why is the m/e
of the base peak not the molecular mass of the
compound?
Answer
(a) The base peak is the most intense peak in a mass spectrum. It
represents the most stable ion formed during fragmentation or the
ion that can be formed in various ways during fragmentation of
the molecular ion. As molecular ions are usually unstable and will
undergo fragmentation, they do not normally show up as base
peaks in mass spectra.
331
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(b) The following is the mass spectrum of
bromomethylbenzene (benzyl bromide).
Interpret the peaks at m/e = 172, 170 and 91.
Answer
332
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(b) The relative molecular mass of bromomethylbenzene (benzyl
bromide) is 171.0. However, as bromine contains equal
abundances of the 79Br and 81Br isotopes, the spectrum shows
two small peaks of equal intensity at m/e = 172 and 170. The
base peak at m/e = 91 is due to the formation of the ion
C6H5CH2+.
333
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
Answer
(c) Study the following spectrum carefully and deduce
what group of organic compound it is. The compound
has a relative molecular mass of 114.
334
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(c) The base peak is at m/e = 57 which may be an oxonium ion or a
carbocation. This is a mass spectrum of a ketone, an aldehyde or
a hydrocarbon.
Back
335