Post on 24-Dec-2015
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Structural Determination Structural Determination of Organic Compoundsof Organic Compounds
34.134.1 IntroductionIntroduction
34.234.2 Isolation and Purification of Organic CompoundsIsolation and Purification of Organic Compounds
34.334.3 Tests for PurityTests for Purity
34.434.4 Qualitative Analysis of Elements in an Organic CompoundQualitative Analysis of Elements in an Organic Compound
34.534.5 Determination of Empirical Formula and Molecular Determination of Empirical Formula and Molecular
Formula from Analytical DataFormula from Analytical Data
34.634.6 Structural Information from Physical PropertiesStructural Information from Physical Properties
34.734.7 Structural Information from Chemical PropertiesStructural Information from Chemical Properties
34.834.8 Use of Infra-red Spectrocopy in the Identification of Use of Infra-red Spectrocopy in the Identification of
Functional GroupsFunctional Groups
34.934.9 Use of Mass Spectra to Obtain Structural InformationUse of Mass Spectra to Obtain Structural Information
3344
2
The general steps to determine the structure of an organic compound
3
Isolation and Isolation and Purification Purification of Organic of Organic
CompoundsCompounds
4
Technique Aim
1. Filtration To separate an insoluble solid from a liquid (slow)
2. Centrifugation To separate an insoluble solid from a liquid (fast)
3. Recrystallization To separate a solid from other solids based on their different solubilities in suitable solvent(s)
4. Solvent extraction
To separate a component from a mixture with a suitable solvent
5. Distillation To separate a liquid from a solution containing non-volatile solutes
5
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 decompose easily below their b.p.
8. Vacuum distillation ditto
9. Sublimation To separate a mixture of solids in which only one can sublime
10. Chromatography To separate a complex mixture of substances (large/small scale)
The mixture boils below 100C
6
• 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
Tests for PurityTests for Purity
7
Use of Infra-red Use of Infra-red Spectrocopy in Spectrocopy in
the the Identification of Identification of
Functional Functional GroupsGroups
8
Infra-red Infra-red SpectroscopySpectroscopyArises 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
9
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
Modes of vibrationsModes of vibrations
10
Symmetrical stretching : -
Asymmetrical stretching : -
Modes of vibrationsModes of vibrations
11
Bending (scissoring) : - two modes
Behind paperOut of paper
Modes of vibrationsModes of vibrations
12
Modes of vibrationsModes of vibrations
Symmetric stretching
Asymmetric stretching
An in-plane bending
(scissoring)
An out-of-plane bending
(twisting)
Rocking
Wagging
13
The frequency() of a given stretching vibration of a covalent bond depends on
μk
νBond strength
Reduced mass of the system
yx
yx
mm
mmμ
For a diatomic molecule X – Y
14
Bond strength
k
C – C < C = C < C C
C – O < C = O
Increasing frequency of vibration
μk
ν
15
Masses of bonding atoms
C – H O - H N - H
μk
ν
Chemical bonds containing H atoms have high frequency of vibration due to the small mass of H
16
• Molecular vibrations are quantized
the molecules absorb IR radiation of a particular amount of energy only.
Infra-red Infra-red SpectroscopySpectroscopy
• Only IR radiation with the same frequency as the vibrational frequency can be absorbed by the molecules.
E = h
17
Infra-red Infra-red SpectroscopySpectroscopy
E = h
After the absorption of a quantum of energy (h), the amplitude of vibration but the frequency of vibration () remains unchanged.
18
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
UV / visible / Near IR electronic transition
E = hL = 400 – 1000 kJ mol1
19
Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
Mid IR vibrational transition
E = hL = 5 – 40 kJ mol1
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Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
Far IR / microwave rotational transition
E = hL 0.02 kJ mol1
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Frequency (Hz / s1)
Wavelength (m)
The electromagnetic spectrumThe electromagnetic spectrum
translational transitionE = hL 0 kJ mol1
22
• Different functional groups (e.g. C=O, O-H) have different vibration frequencies
They have characteristic absorption frequencies
Infra-red Infra-red SpectroscopySpectroscopy
• Functional groups can be identified from their characteristic absorption frequencies
23
• The stretching vibrations of single bonds involving hydrogen (C H, O H and N H) occur at relatively high frequencies
3350 – 3500N H
3230 – 3670O H
2840 – 3095C H
Range of wavenumber (cm-1)
Bond
Characteristic absorption wavenumbers of some single bonds in infra-red spectra
Infra-red Infra-red SpectroscopySpectroscopy
24
• Triple bonds are stronger and vibrate at higher frequencies than double bonds
1680 – 1750C = O
1610 – 1680C = C
2200 – 2280C N
2070 – 2250C C
Range of wavenumber (cm-1)
Bond
Characteristic absorption wavenumbers of some double bonds and triple bonds in infra-
red spectra
Infra-red Infra-red SpectroscopySpectroscopy
25
• An IR spectrum is obtained by scanning the sample with IR radiations from
1.21013 Hz to 1.21014 Hz
Infra-red Infra-red SpectroscopySpectroscopy
Or,
• % transmittance rather than absorbance is displayed
Wavenumber : 400 cm1 to 4000 cm-1
c
1)(cmwavenumber 1
26
Dips(peaks) show absorptions by functional groups
C=C
C
HC
H
27
Fingerprintregion
Band region
28
• Infra-red spectrometer is used to
measure the amount of energy absorbed at each wavelength of the IR region
An infra-red spectrometer
Infra-red Infra-red SpectroscopySpectroscopy
29
30
• 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
Infra-red Infra-red SpectroscopySpectroscopy
31
• 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
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional GroupsIdentification of Functional Groups
32
• In general, an IR spectrum can be split into four regions for interpretation purpose
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional GroupsIdentification of Functional Groups
33
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
34
• 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
Use of IR Spectrum in the Use of IR Spectrum in the Identification of Functional GroupsIdentification of Functional Groups
35
Compound Bond Characteristic range of wavenumber (cm-1)
Alkenes C = C 1610 – 1680
Aldehydes, ketones, acids, esters
C = O 1680 – 1750
Alkynes C C 2070 – 2250
Nitriles C N 2200 – 2280
Acids (hydrogen-bonded) O H 2500 – 3300
Alkanes, alkenes, arenes C H 2840 – 3095
Alcohols, phenols (hydrogen-bonded)
O H 3230 – 3670
Primary amines N H 3350 – 3500
What is the characteristic range of wavenumber of C=N bond?Bond strength/wavenumber : C=C < C=N < C=O
36
Strategies for the Use of IR Spectra in Strategies for the Use of IR Spectra in the Identification of Functional the Identification of Functional GroupsGroups
1. Focus at the IR absorption peak at or above 1500 cm–1
Concentrate initially on the major absorption peaks
37
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
Strategies for the Use of IR Spectra in Strategies for the Use of IR Spectra in the Identification of Functional the Identification of Functional GroupsGroups
38
Limitation of the Use of IR Spectroscopy Limitation of the Use of IR Spectroscopy in the Identification of Organic in the Identification of Organic CompoundsCompounds
1. Some IR absorption peaks have very close wavenumbers and the peaks always coalesce
2. Not all vibrations give rise to strong absorption peaks
39
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
Limitation of the Use of IR Spectroscopy Limitation of the Use of IR Spectroscopy in the Identification of Organic in the Identification of Organic CompoundsCompounds
40
1610-1680 cm1
1680-1750 cm1
Absorption Slightly above
3000 cm1
AlkeneArene
Y
N
Y
C
H
C
H
C=C bonds in benzene are weaker ~1610 cm1
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1680-1750 cm1
Broad2500-3300 cm1
YY
2070-2280 cm1
N
Aldehyde, Ketone, Ester
N
Y
Carboxylic acid
Containoxygen
C
O
OH
C
O
H
C
O
C
O
O
Aldehyde : lower C-H absorption at 2720-2820 cm-1
42
2070-2280 cm1 Containnitrogen
Y
Nitrile2200-2280 cm1
Y
Alkyne2070-2250 cm1
N
N
3200-3700 cm1
Additional sharp peak ~3250 cm1 for termnal alkyne, -CC-H
43
Containnitrogen
Y
Amine3350-3500 cm1
Y
NAlkane
2840-3000 cm1
N
3200-3700 cm1
Containoxygen
Y
Alcohol, phenol3230-3700 cm1
Single peak for 2 amine; double peaks for 1 amine
44
3300
3100
Several peaks (2840-3000 cm1) due todifferent modes of Symmetrical & asymmetrical stretching
Different modes of C-H bending (~1400 cm1)
weak
medium
45
C
H
C
C
46
Q.82
47
Q.82
48
Q.82
49
Q.82
C
C
C
H
50
Q.82
51
Q.83
52
Q.83
53
Q.83
2720- 28202840-
3000
2840-3000
54
Q.83
2720- 28202840-
3000
2840-3000
55
Q.84 C3H5N IOU = 3 + 1 -5/2 + ½ = 2
CC, CN, diene
CC, CN
56
C
H
H
H
C
H
H
C NC CH C N H
HH
H
H C
H
H
C C N
H
H
CC, CN
C-H1 amine
57
Q.85 C6H10O IOU = 6 + 1 -10/2 = 2
Ozonolysis gives CO2 terminal C=CH2
58
Q.85 C6H10O IOU = 6 + 1 -10/2 = 2
+ve iodoform test C O
H3C
H3C C
OH
H
59
Q.85 C6H10O IOU = 6 + 1 -10/2 = 2
No broadband within 3200-3600 H3C C
O
=C-H
C=C
C=O
-C-H
60
C
O
H3C *
C
O
H3C
C
O
H3C
61
H
C
C H
O
C O
62
C
C
C
H
C
H
63
H
C
C C
C C H
64
H
C
1 amine
No C-H stretching above 3000 cm1
No C=C-H
N-H bendin
g
65
O H
C
H
C O
C
CBroad band with peak at about 2900 cm1
C O
66
H
C
C O
C O
67
H
C
C
C
1 amine
C
H
68
H
C
C N
69
O H
C
C
C OBroad band with peak at about 2900 cm1
C O
70
H
C
O H
C C
C C H
C O
Broad band with peak at about 3300 cm1
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H
C
C C
N C H
C N
C O
72
H
C
C
H
C
C
C O
C O
73
H
CC
H
C
C
74
C
H
C
C
No aliphatic C-H stretching
C H
O
C O
75
C
HC
C
C N
No aliphatic C-H stretching
76
H
C
O H
C C
C C H
C O
Broad band with peak at about 3300 cm1
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H
C
O HC O
Broad band with peak at about 3300 cm1
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H
C
O HC O
Broad band with peak at about 2900 cm1
C O
79
H
C
1 amine
No C-H stretching above 3000 cm1
No C=C-H
N-H bendin
g
80
H
C
O HC O
Broad band with peak at about 3300 cm1
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H
CC O
82
H
C
C N
83
H
C
O H
C O
Broad band with peak at about 3300 cm1
84
H
C
C
H
C
C
C O
C O
85
H
CO H
C C
C C H
C O
Broad band with peak at about 3300 cm1
86
H
C
C C
C C H
87
H
C
O HC O
Broad band with peak at about 3400 cm1
C O
88
H
C
C O
C O
89
H
C
O H C O
Broad band with peak at about 2900 cm1
C O
90
H
C
C O
91
H
C
C H
O
C O
92
O H
C
H
C
C
C O
Broad band with peak at about 3300 cm1
No aliphatic C-H stretching
93
H
C
O H
C O
Broad band with peak at about 3300 cm1
94
H
CC O
95
H
C
C
C
C
H
96
C O
H
C
C
C
2 amine
C
H
97
4. 4. ButanoneButanone
The IR spectrum of butanone
98
4. 4. ButanoneButanoneWavenumber (cm-1) Intensity Indication
2983 Strong C H stretching
2925 Strong
1720 Very strong C = O stretching
1416 Medium C H bending (shifted as adjacent to C = O)
Interpretation of the IR spectrum of butanone
99
5. Butan-1-5. Butan-1-olol
a broad band is observed
the vibration of the O H group is complicated by the hydrogen
bonding formed between the molecules
100
5. Butan-1-5. Butan-1-ololWavenumber
(cm-1)Intensity Indication
3330 Broad band O H stretching
2960 Medium C H stretching
2935 Medium
2875 Medium
Interpretation of the IR spectrum of butan-1-ol
101
6. Butanoic 6. Butanoic AcidAcid
a broad band is observed
the vibration of the O H group is complicated by the hydrogen
bonding formed between the molecules
102
6. Butanoic 6. Butanoic AcidAcidWavenumber (cm-
1)Intensity Indication
3100 Broad band O H stretching
1708 Strong C = O stretching
Interpretation of the IR spectrum of butanoic acid
103
7. 7. ButylaminButylaminee
The IR spectrum of butylamine
104
7. 7. ButylaminButylamineeWavenumber (cm-
1)Intensity Indication
3371 Strong N H stretching
3280 Strong
2960 – 2875 Weak C H stretching
1610 Medium N H bending
1475 Medium C H bending
Interpretation of the IR spectrum of butylamine
105
8. 8. ButanenitrileButanenitrile
The IR spectrum of butanenitrile
106
8. 8. ButanenitrileButanenitrile
Wavenumber (cm-1)
Intensity Indication
2990 – 2895 Strong C H stretching
2246 Very strong C N stretching
1420 Strong C H bending
1480 Strong
Interpretation of the IR spectrum of butanenitrile
107
Example 34-8Example 34-8 Check Point 34-8Check Point 34-8
108
34.34.99 Use of Mass Use of Mass
Spectra to Spectra to Obtain Obtain
Structural Structural InformationInformation
109
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Mass SpectrometrySpectrometry
• 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
110
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Mass SpectrometrySpectrometryIn 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
111
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.113)
Mass Mass SpectrometrySpectrometry
• 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
112
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
Components of a mass spectrometer
113
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
In the vaporization chamber,
• the sample is heated until it vaporizes
changes to the gaseous state
114
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)Mass Mass
SpectrometrySpectrometry• 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
115
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)Mass Mass
SpectrometrySpectrometry
• Molecular ions are sometimes referred to as the parent ion
116
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)Mass Mass
SpectrometrySpectrometry 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
117
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)Mass Mass
SpectrometrySpectrometry• e.g.
if a molecule of methanol (CH3OH) is bombarded with a beam of fast-moving electrons
the following reaction will take place:
118
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
• 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
119
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
• The way that a molecular ion fragments
give us highly useful information about the structure of a complex molecule
120
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry• The positively charged ions formed are
then accelerated by electric field and deflected by magnetic field
causes the ions to arrive the ion detector
• The lighter the ions, the greater the deflection
121
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.114)
Mass Mass SpectrometrySpectrometry
• 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
122
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Mass Mass SpectrometrySpectrometry• 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
123
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)
Mass Mass SpectrometrySpectrometry
• 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
124
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)Mass Mass
SpectrumSpectrum• Generally published as bar graphs.
Mass spectrum of methanol
125
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)Mass Mass
SpectrumSpectrum
Corresponding ion m/e ratio
H3C+ 15
H CO+ 29
H2C = OH+ 31
CH3OH 32
Interpretation of the mass spectrum of methanol
126
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)Formation of Formation of
FragmentsFragments
• The molecular ions formed in the ionization chamber are energetically unstable
Some of them may break up into smaller fragments
Called the daughter ions
127
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.115)Formation of Formation of
FragmentsFragments• 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
128
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments
Mass spectrum of methanol
• The peak at m/e 31
the most intense peak
• Arbitrarily assigned an intensity of 100%
Called the base peak
Corresponds to the most
common ion formed
129
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments• The peak at m/e 31
corresponds to the ion H2C = OH+
formed by losing one hydrogen atom from the molecular ion
130
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments• 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
131
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments• The peak at m/e 29 corresponds to the ion
HC O+
formed by losing two hydrogen atoms from the ion H2C = OH+
132
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments• The ion HC O+ has two resonance
structures:
133
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments• The peak at m/e 15 corresponds to the
ion H3C+
formed by the breaking of the C O bond in the molecular ion
134
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments
Mass spectrum of pentan-3-one
135
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments
Corresponding ion m/e ratio
CH3CH2+ 29
CH3CH2CO+ 57
CH3CH2COCH2CH3 86
Interpretation of the mass spectrum of pentan-3-one
136
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.116)Formation of Formation of
FragmentsFragments• The fragmentation pattern of pentan-3-one
is summarized below:
137
Example 34-9AExample 34-9A Example 34-9BExample 34-9B
Example 34-9CExample 34-9C
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
138
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)Fragmentation Fragmentation
PatternPattern1. Straight-chain 1. Straight-chain AlkanesAlkanes• Simple alkanes tend to undergo
fragmentation by
This carbocation can then undergo stepwise cleavage down the alkyl chain
the initial loss of a • CH3 to give a peak at M – 15
139
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)1. Straight-chain 1. Straight-chain
AlkanesAlkanes• Take hexane as an example:
140
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)
2. Branched-chain 2. Branched-chain AlkanesAlkanes
• Tend to cleave at the “branch point”
more stable carbocations are formed
141
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.121)
2. Branched-chain 2. Branched-chain AlkanesAlkanes• e.g.
142
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
3. Alkyl-substituted Aromatic 3. Alkyl-substituted Aromatic HydrocarbonsHydrocarbons• Undergo loss of a hydrogen atom or alkyl
group
yield the relatively stable tropylium ion
• Gives a prominent peak at m/e 91
143
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
3. Alkyl-substituted Aromatic 3. Alkyl-substituted Aromatic HydrocarbonsHydrocarbons
• e.g.
144
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
4. Aldehydes and 4. Aldehydes and KetonesKetones• Frequently undergo fragmentation by losing
one of the side chains
generate the substituted oxonium ion
often represents the base peak in the mass spectra
145
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)
4. Aldehydes and 4. Aldehydes and KetonesKetones
146
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)5. Esters, Carboxylic Acids and 5. Esters, Carboxylic Acids and
AmidesAmides• 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)
147
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.122)5. Esters, Carboxylic Acids and 5. Esters, Carboxylic Acids and
AmidesAmides• For carboxylic acids and unsubstituted
amides,
characteristic peaks at m/e 45 and 44 are observed respectively
148
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)6. 6.
AlcoholsAlcohols• 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
149
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)6. 6.
AlcoholsAlcohols• For primary alcohols,
the peak at m/e 31, 45, 59 or 73 often appears
depends on what the R1 group is
150
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)7. 7.
HaloalkaneHaloalkaness • Haloalkanes simply break at the C X
bond
151
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)7. 7.
HaloalkaneHaloalkaness
• In the mass spectra of chloroalkanes,
two peaks, separated by two mass units, in the ratio 3 : 1 will be appeared
152
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
7. 7. HaloalkaneHaloalkaness
• In the mass spectra of bromoalkanes,
two peaks, separated by two mass units, having approximately equal intensities will be appeared
153
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.123)
Check Point 34-9Check Point 34-9
154
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Isolation and Purification Isolation and Purification of Organic Compoundsof Organic Compounds
• The selection of a proper technique
depends on the particular differences in physical properties of the substances present in the mixture
155
34.2 Isolation and Purification of Organic Compounds (SB p.78)
FiltrationFiltration
• 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
156
34.2 Isolation and Purification of Organic Compounds (SB p.78)
FiltrationFiltration
The laboratory set-up of filtration
157
34.2 Isolation and Purification of Organic Compounds (SB p.78)
FiltrationFiltration
• 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
158
34.2 Isolation and Purification of Organic Compounds (SB p.79)
CentrifugatioCentrifugationn• When there is only a small amount of
suspension, or when much faster separation is required
Centrifugation is often used instead of filtration
159
34.2 Isolation and Purification of Organic Compounds (SB p.79)
CentrifugatioCentrifugationn
• 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
160
34.2 Isolation and Purification of Organic Compounds (SB p.79)CentrifugatioCentrifugationn• 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
161
34.2 Isolation and Purification of Organic Compounds (SB p.79)
CrystallizatioCrystallizationn
• Crystals are solids that have
a definite regular shape
smooth flat faces and straight edges
• Crystallization is the process of forming crystals
162
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot 1. Crystallization by Cooling a Hot Concentrated SolutionConcentrated 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
163
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot 1. Crystallization by Cooling a Hot Concentrated SolutionConcentrated 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
164
34.2 Isolation and Purification of Organic Compounds (SB p.79)
1. Crystallization by Cooling a Hot 1. Crystallization by Cooling a Hot Concentrated SolutionConcentrated Solution
Crystallization by cooling a hot concentrated solution
165
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a 2. Crystallization by Evaporating a Cold Solution at Room TemperatureCold 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
166
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a 2. Crystallization by Evaporating a Cold Solution at Room TemperatureCold 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
167
34.2 Isolation and Purification of Organic Compounds (SB p.80)
2. Crystallization by Evaporating a 2. Crystallization by Evaporating a Cold Solution at Room TemperatureCold Solution at Room Temperature
Crystallization by slow evaporation of a solution (preferably saturated) at room
temperature
168
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
• 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. 1,1,1-trichloroethane and diethyl ether) can be used to extract organic products
169
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
• 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
170
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
The organic product in an aqueous solution can be extracted by solvent extraction using diethyl
ether
171
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
• 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
172
34.2 Isolation and Purification of Organic Compounds (SB p.80)
Solvent Solvent ExtractionExtraction
• 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
173
34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
• 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
174
34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
The laboratory set-up of distillation
175
34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
• 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
176
34.2 Isolation and Purification of Organic Compounds (SB p.81)
DistillatioDistillationn
• If bumping occurs during distillation,
some solution (not yet vaporized) may spurt out into the collecting vessel
177
34.2 Isolation and Purification of Organic Compounds (SB p.81)
Fractional DistillationFractional Distillation
• A method used to separate a mixture of two or more miscible liquids
178
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional DistillationFractional Distillation
The laboratory set-up of fractional distillation
179
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional DistillationFractional 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
180
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Fractional DistillationDistillation• 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
181
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Fractional DistillationDistillation• 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
• Fractions with different boiling points can be collected separately
182
34.2 Isolation and Purification of Organic Compounds (SB p.82)
SublimatioSublimationn
• 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
183
34.2 Isolation and Purification of Organic Compounds (SB p.82)
SublimatioSublimationn• 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
184
34.2 Isolation and Purification of Organic Compounds (SB p.83)
SublimatioSublimationn
A mixture of two compounds can be separated by sublimation
185
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
• An effective method of separating a complex mixture of substances
• Paper chromatography is a common type of chromatography
186
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
The laboratory set-up of paper chromatography
187
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
• A solution of the mixture is dropped at one end of the filter paper
188
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
• 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
189
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy
• When the solvent moves across the sample spot of the mixture,
partition of the components between the stationary phase and the mobile phase
occurs
190
34.2 Isolation and Purification of Organic Compounds (SB p.83)
ChromatographChromatographyy• 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
191
34.2 Isolation and Purification of Organic Compounds (SB p.84)
Technique Aim
(a) Filtration 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 extraction
To separate a component from a mixture with a suitable solvent
(e) Distillation To separate a liquid from a solution containing non-volatile solutes
A summary of different techniques of isolation and purification
192
34.2 Isolation and Purification of Organic Compounds (SB p.84)
Technique Aim
(f) Fractional distillation
To separate miscible liquids with widely different boiling points
(g) Sublimation To separate a mixture of solids in which only one can sublime
(h) Chromatography
To separate a complex mixture of substances
A summary of different techniques of isolation and purification
Check Point 34-2Check Point 34-2
193
34.34.44 Qualitative Qualitative
Analysis of Analysis of Elements in an Elements in an
Organic Organic CompoundCompound
194
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
• Qualitative analysis of an organic compound is
to determine what elements are present in the compound
Qualitative Analysis of Qualitative Analysis of an Organic Compoundan Organic Compound
195
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Carbon and Carbon and HydrogenHydrogen• Tests for carbon and hydrogen in an
organic compound are usually unnecessary
an organic compound must contain carbon and hydrogen
196
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Carbon and Carbon and HydrogenHydrogen• 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
197
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur• Halogens, nitrogen and sulphur in organic
compounds can be detected
by performing the sodium fusion test
198
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur• 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
199
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
Halogens, Nitrogen and Halogens, Nitrogen and SulphurSulphur
• 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
200
34.4 Qualitative Analysis of Elements in an Organic Compound (SB p.86)
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).
Results for halogens, nitrogen and sulphur in the sodium fusion test
201
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 iron(II) sulphate and iron(III) sulphate solutions
A blue-green colour is observed.
Sulphur, assulphide ion (S2-)
Sodium pentacyanonitrosylferrate(II) solution
A black precipitate is formed
Check Point 34-4Check Point 34-4
202
34.34.55Determination of Determination of
Empirical Empirical Formula and Formula and
Molecular Molecular Formula from Formula from
Analytical DataAnalytical Data
203
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
• 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
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan Organic Compound
204
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
1. Carbon and 1. Carbon and
HydrogenHydrogen• 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
205
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
1. Carbon and 1. Carbon and
HydrogenHydrogen• The increases in mass in potassium hydroxide solution and calcium chloride represent
the masses of carbon dioxide and water vapour formed respectively
206
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
2. 2.
NitrogenNitrogen• 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
207
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
3. 3.
HalogensHalogens• 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
208
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
4. 4.
SulphurSulphur• The organic compound is heated with
fuming nitric(V) acid
• After cooling,
barium nitrate solution is added
the dry barium sulphate formed is weighed
209
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan Organic Compound
• After determining the percentage composition by mass of a compound,
the empirical formula of the compound can be calculated
210
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan 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
211
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.87)
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan 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
212
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
Quantitative Analysis of Quantitative Analysis of an Organic Compoundan 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
213
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
Example 34-5AExample 34-5A Example 34-5BExample 34-5B
Check Point 34-5Check Point 34-5
214
34.34.66 Structural Structural
Information Information from Physical from Physical
PropertiesProperties
215
34.6 Structural Information from Physical Properties (SB p.89)
• 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
Structural Information from Structural Information from Physical PropertiesPhysical Properties
216
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from Structural Information from Physical PropertiesPhysical Properties
• From the physical properties of a compound,
obtain preliminary information about the structure of the compound
217
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from Structural Information from Physical PropertiesPhysical Properties• e.g.
Hydrocarbons have low densities, often about 0.8 g cm–3
Compounds with functional groups have higher densities
218
34.6 Structural Information from Physical Properties (SB p.89)
Structural Information from Structural Information from Physical PropertiesPhysical 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
219
34.6 Structural Information from Physical Properties (SB p.90)
Organic compound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly
polar solvents
In non-polar
organic solvents
Hydrocarbons (saturated and unsaturated)
All
have densities < 0.8 g cm–3
• Generally low but increases with number of carbon atoms in the molecule
• Branched-chain hydrocarbons have lower boiling points but higher melting points than the corresponding straight-chain isomers
Insoluble Soluble
Physical properties of some common organic compounds
220
34.6 Structural Information from Physical Properties (SB p.90)
Organic compound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly
polar solvents
In non-polar
organic solvents
Aromatic hydrocarbons
Between 0.8 and 1.0 g cm–3
Generally low Insoluble Soluble
Physical properties of some common organic compounds
221
34.6 Structural Information from Physical Properties (SB p.90)
Organic compound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly
polar solvents
In non-polar
organic solvents
Halo-alkanes
• 0.9 - 1.1 g cm–3 for chloro-alkanes
• >1.0 g cm–3 for bromo-alkanes and iodo-alkanes
• 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
Insoluble Soluble
Physical properties of some common organic compounds
222
34.6 Structural Information from Physical Properties (SB p.90)
Organic comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Alcohols • Simple alcohols are liquids and alcohols with > 12 carbons are waxy solids
• Much higher than hydrocarbons of similar relative molecular masses ( formation of hydrogen bonds between alcohol molecules)
• Lower members: Completely miscible with water ( formation of hydrogen bonds between alcohol molecules and water molecules)
Soluble
Physical properties of some common organic compounds
223
34.6 Structural Information from Physical Properties (SB p.90)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Alcohols • All simple alcohols have densities < 1.0 g cm–3
• Straight-chain alcohols have higher b.p. than the corresponding branched-chain alcohols
• Solubility decreases gradually as the hydrocarbon chain lengthens
Soluble
224
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Carbonyl
comp-ounds
(alde-hydes
and ketones)
• <1.0 g cm–3 for aliphatic carbonyl compounds
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)
• Lower members:Soluble in water ( the formation of hydrogen bonds between molecules of aldehydes or ketones and water molecules)
Soluble
225
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Carbonyl
comp-ounds
(alde-hydes
and ketones)
• > 1.0 g cm–3 for aromatic carbonyl compounds
• Solubility decreases gradually as the hydrocarbon chain lengthens
Soluble
226
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Carbo-xylic
acids
• 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)
• First four members are
miscible with water in all proportions
• Solubility decreases gradually as the hydrocarbon chain lengthens
Soluble
227
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Esters Lower members
have densities less than water
Slightly higher than hydrocarbons but lower than carbonyl compounds and
alcohols of similar relative molecular masses
Insoluble Soluble
228
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Amines Most amines have densities less than water
• Higher than alkanes but lower than alcohols of similar relative molecular masses
• Generally soluble
• Solubility decreases in the order:
1o amines > 2o amines > 3o amines
Soluble
229
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Amines • 1o and 2o amines are able to form hydrogen bonds with each other but the strength is less than that between alcohol molecules (NH bond is less polar than O H bond)
230
34.6 Structural Information from Physical Properties (SB p.91)
Physical properties of some common organic compoundsOrganic
comp-ound
Density at 20oC
Melting point and boiling point
Solubility
In water or highly polar
solvents
In non-polar
organic solvents
Amines • 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)
231
34.6 Structural Information from Physical Properties (SB p.92)
Example 34-6Example 34-6 Check Point 34-6Check Point 34-6
232
34.34.77 Structural Structural
Information Information from Chemical from Chemical
PropertiesProperties
233
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from Structural Information from Chemical PropertiesChemical 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
234
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from Structural Information from Chemical PropertiesChemical Properties
• e.g.The molecular formula of C2H4O2 may represent a carboxylic acid or an ester:
235
34.7 Structural Information from Chemical Properties (SB p.93)
Structural Information from Structural Information from Chemical PropertiesChemical Properties
• The next stage is
to find out the functional group(s) present
to deduce the actual arrangement of atoms in the molecule
236
34.7 Structural Information from Chemical Properties (SB p.93)
Organic compound
Test Observation
Saturated
hydrocarbons
• Burn the saturated hydrocarbon in a non-luminous Bunsen flame
• A blue or clear yellow flame is observed
Chemical tests for different groups of organic compounds
237
34.7 Structural Information from Chemical Properties (SB p.93)
Organic compound
Test Observation
Unsaturated hydrocarbons (C = C, C C)
• Burn the unsaturated hydrocarbon in a non-luminous Bunsen flame
• A smoky flame is observed
• Add bromine in 1,1,1-trichloroethane at room temperature and in the absence of light
• Bromine decolourizes rapidly
• Add 1% (dilute) acidified potassium manganate(VII) solution
• Potassium manganate(VII) solution decolourizes rapidly
Chemical tests for different groups of organic compounds
238
34.7 Structural Information from Chemical Properties (SB p.93)
Organic compound
Test Observation
Haloalkanes
(1°, 2° or 3°)
• Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution
• For chloroalkanes, a white precipitate is formed
• For bromoalkanes, a pale yellow precipitate is formed
• For iodoalkanes, a creamy yellow precipitate is formed
Chemical tests for different groups of organic compounds
239
34.7 Structural Information from Chemical Properties (SB p.93)
Organic compound
Test Observation
Halobenzenes • Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution
• No precipitate is formed
Chemical tests for different groups of organic compounds
240
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Alcohols ( OH)
• 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
241
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Alcohols ( OH)
• Add acidified potassium dichromate(VI) solution
• For 1° and 2° alcohols, the clear orange solution becomes opaque and turns green almost immediately
• For 3° alcohols, there are no observable changes
242
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Alcohols ( OH)
• Iodoform test for:
Add iodine in sodium hydroxide solution
• A yellow precipitate is formed
243
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Alcohols ( OH)
• Lucas test: add a solution of zinc chloride in concentrated hydrochloric acid
• 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)
244
Chemical tests for different groups of organic compoundsOrganic
compoundTest Observation
Ethers ( O )
• No specific test for ethers but they are soluble in concentrated sulphuric(VI) acid
34.7 Structural Information from Chemical Properties (SB p.94)
245
34.7 Structural Information from Chemical Properties (SB p.94)
Organic compound
Test Observation
Aldehydes
( )
• Add aqueous sodium hydrogensulphate(IV)
• Crystalline salts are formed
• Add 2,4-dinitrophenylhydrazine
• A yellow, orange or red precipitate is formed
• Silver mirror test: add Tollens’ reagent (a solution of aqueous silver nitrate in aqueous ammonia)
• A silver mirror is deposited on the inner wall of the test tube
Chemical tests for different groups of organic compounds
246
34.7 Structural Information from Chemical Properties (SB p.94)
Organic compound
Test Observation
Ketones
( )
• Add aqueous sodium hydrogensulphate(IV)
• Crystalline salts are formed (for unhindered ketones only)
• Add 2,4-dinitrophenylhydrazine
• A yellow, orange or red precipitate is formed
• Iodoform test for:
Add iodine in sodium hydroxide solution
• A yellow precipitate is formed
Chemical tests for different groups of organic compounds
247
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Carboxylic acids
( )
• Esterification: warm the carboxylic acid with an alcohol in the presence of concentrated sulphuric(VI) acid, followed by adding sodium carbonate solution
• A sweet and fruity smell is detected
• Add sodium hydrogencarbonate
• The colourless gas produced turns lime water milky
Chemical tests for different groups of organic compounds
248
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Esters
( )
• No specific test for esters but they can be distinguished by its characteristic smell
• A sweet and fruity smell is detected
Chemical tests for different groups of organic compounds
249
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Acyl halides
( )
• Boil with ethanolic potassium hydroxide solution, then acidify with excess dilute nitric(V) acid and add silver nitrate(V) solution
• 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
Chemical tests for different groups of organic compounds
250
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Amides
( )
• Boil with sodium hydroxide solution
• The colourless gas produced turns moist red litmus paper or pH paper blue
Chemical tests for different groups of organic compounds
251
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Amines
(NH2)
• 1o aliphatic amines: dissolve the amine in dilute hydrochloric acid at 0 – 5 oC, then add cold sodium nitrate(III) solution slowly
• Steady evolution of N2(g) is observed
• 1o aromatic amines: add naphthalen-2-ol in dilute sodium hydroxide solution
• An orange or red precipitate is formed
Chemical tests for different groups of organic compounds
252
34.7 Structural Information from Chemical Properties (SB p.95)
Organic compound
Test Observation
Aromatic compounds
( )
• Burn the aromatic compound in a non-luminous Bunsen flame
• A smoky yellow flame with black soot is produced
• Add fuming sulphuric(VI) acid
• The aromatic compound dissolves
• The temperature of the reaction mixture rises
Chemical tests for different groups of organic compounds
253
34.7 Structural Information from Chemical Properties (SB p.96)
Example 34-7AExample 34-7A Example 34-7BExample 34-7B
Example 34-7CExample 34-7C Check Point 34-7Check Point 34-7
254
The END
255
34.1 Introduction (SB p.77)
What are the necessary information to determine the structure of an organic compound? AnswerMolecular formula from analytical data,
functional group present from physical and
chemical properties, structural information from
infra-red spectroscopy and mass spectrometry
Back
256
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 liquidAnswer(a) Centrifugation
(b) Chromatography
(c) Fractional distillation
(d) Sublimation
(e) Filtration
Back
257
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
258
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
259
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.88)
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
∴ The empirical formula of the organic compound is CH2O.
Carbon Hydrogen Oxygen
Mass (g) 40.0 6.7 53.3
Number of moles (mol)
Relative number of moles
Simplest mole ratio
1 2 1
33.312.040.0 7.6
1.06.7 33.3
16.053.3
13.333.33 2
3.336.7 1
3.333.33
Back
260
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:
(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
Element Carbon Hydrogen
Oxygen
Percentage by mass (%)
60.00 13.33 26.67
261
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
∴ The empirical formula of the organic compound is C3H8O.
Carbon Hydrogen Oxygen
Mass (g) 60.00 13.33 26.67
Number of moles (mol)
Relative number of moles
Simplest mole ratio
3 8 1
512.060.00 33.13
1.013.33 67.1
16.026.67
81.67
13.33 11.671.67 3
1.675
262
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
263
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
264
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
Mass of carbon in 0.22 g of CO2 = 0.22 g ×
= 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 ×
= 0.01 g
Mass of oxygen in the compound = (0.15 – 0.06 – 0.01) g
= 0.08 g
44.012.0
18.02.0
265
34.5 Determination of Empirical Formula and Molecular Formula from Analytical Data (SB p.89)
∴ The empirical formula of the organic compound is CH2O.
Carbon Hydrogen Oxygen
Mass (g) 0.06 0.01 0.08
Number of moles (mol)
Relative number of moles
Simplest mole ratio
1 2 1
005.012.00.06 01.0
1.00.01 005.0
16.00.08
20.0050.01 1
0.0050.005 1
0.0050.005
266
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
267
34.6 Structural Information from Physical Properties (SB p.92)
Why do branched-chain hydrocarbons have lower boiling points but higher melting points than the
corresponding straight-chain isomers?
AnswerBranched-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 straight-
chain 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
break down the efficient packing in the process of melting.
Back
268
34.6 Structural Information from Physical Properties (SB p.92)
Why does the solubility of amines in water decrease in the order:
1o amines > 2o amines > 3o amines?
AnswerThe 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.
Back
269
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
270
34.6 Structural Information from Physical Properties (SB p.92)
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
point than methanol.
Back
271
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-1- ol and butanal?
(ii) Account for the higher boiling point of butan-1-ol.
Answer(a) (i) The relative molecular masses of butan-
1-ol and butanal are 74.0 and 72.0
respectively.
(ii) 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.
272
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-olAnswer(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.
273
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.
274
34.6 Structural Information from Physical Properties (SB p.92)
(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°CAnswer(d)
1252,4-Dimethylpentan-3-one
144Heptan-4-one
155Heptanal
Boiling point (oC)Compound
Back
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.
(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.
276
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
277
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:
278
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.Answer(c) The IUPAC name for the compound
is ethanoic acid.
Back
279
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.
Answer
280
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 + )O2 xCO2 + H2O
Volume of CxHy reacted : Volume of CO2 formed = 1 : x = 15 : 30
x = 2
2y
4y
3015
x1
281
34.7 Structural Information from Chemical Properties (SB p.97)
(a) Volume of CxHy reacted : Volume of O2 reacted = 1 : ( )
= 15 : 45
y = 4
The molecular formula of the compound is C2H4.
(b) C2H4 belongs to alkenes.
(c) The structural formula of the hydrocarbon is:
4y
x
4515
)4y
2(
1
34y
2
Back
282
34.7 Structural Information from Chemical Properties (SB p.97)
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.
Answer
283
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
CxHyOz + (x + - )O2 xCO2 + H2O
Volume of CxHyOz reacted : Volume of CO2 formed = 1 : x = 20 : 40
x = 2
2y
4y
2z
4020
x1
284
34.7 Structural Information from Chemical Properties (SB p.98)
(a) Volume of CxHyOz reacted : Volume of H2O formed = 1 : = 20 : 60
y = 6
Volume of CxHyOz reacted : Volume of O2 reacted = 1 :
= 20 : 60
z = 1
The molecular formula of the compound is C2H6O.
2y
6020
y2
)2z
4y
x(
6020
)2z
-4y
(x
1
32z
46
2
285
34.7 Structural Information from Chemical Properties (SB p.98)
(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.
Back
286
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
287
34.7 Structural Information from Chemical Properties (SB p.99)
(a) (i) Let the mass of the compound be 100 g.
∴ The empirical formula of the compound is C3H2NO2.
Carbon Hydrogen Nitrogen Oxygen
Mass (g) 42.8 2.38 16.67 38.15
Number of moles (mol)
Relative number of moles
Simplest mole ratio
3 2 1 2
57.30.128.42 38.2
0.138.2 19.1
0.1467.16 38.2
0.1615.38
319.157.3 2
19.138.2 1
19.119.1 2
19.138.2
288
34.7 Structural Information from Chemical Properties (SB p.99)
(a) (i) 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.
(ii)
289
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
290
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 + )O2 xCO2 + H2O
Volume of CxHy reacted : Volume of CO2 formed
= 1 : x = 30 : 60
x = 2
2y
4y
6030
x1
291
34.7 Structural Information from Chemical Properties (SB p.99)
(b) (i) Volume of CxHy reacted : Volume of O2 reacted
= 1 : ( ) = 30 : 105
y = 6
The molecular formula of the compound is C2H6.
(ii) 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.
4y
2
10530
)4y
x(
1
105)4y
2(30
292
(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
34.7 Structural Information from Chemical Properties (SB p.99)
293
34.7 Structural Information from Chemical Properties (SB p.99)
(c) (i) Let the mass of the compound be 100 g.
∴ The empirical formula of the compound is CH2.
Carbon Hydrogen
Mass (g) 85.5 14.5
Number of moles (mol)
Relative number of moles
Simplest mole ratio
1 2
125.70.125.85 5.14
0.15.14
1125.7125.7 2
125.75.14
294
34.7 Structural Information from Chemical Properties (SB p.99)
(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.
Back
295
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.102)
What is the relationship between frequency andwavenumber?
AnswerThe higher the frequency, the higher the
wavenumber.
Back
296
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.
297
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
298
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
∴ The empirical formula of the compound is C4H8O.
Carbon Hydrogen Oxygen
Mass (g) 66.66 11.11 22.23
Number of moles (mol)
Relative number of moles
Simplest mole ratio
4 8 1
56.50.1266.66 11.11
0.111.11 39.1
0.1623.22
439.156.5 8
39.111.11 1
39.139.1
299
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.
300
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:
301
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
302
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.
303
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
304
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.
305
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:
306
34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.112)
(b) The infra-red spectra of two organic compounds A and B are shown below.
Decide which compound could be an alcohol. Explain your answer briefly.
Answer
307
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.
308
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 Range of wavenumber (cm-1)
C = O 1680 – 1750
O H 2500 – 3300
C H 2840 – 3095
N H 3350 – 3500Answer
309
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:
310
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:
311
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
312
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.
313
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
314
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.
315
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:
316
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
317
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.
∴ The empirical formula of the compound is C2H4O.
Carbon Hydrogen Oxygen
Mass (g) 54.54 9.10 36.36
Number of moles (mol)
Relative number of moles
Simplest mole ratio
2 4 1
55.40.1254.54 10.9
0.110.9 27.2
0.1636.36
227.255.4 4
27.210.9 1
27.227.2
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34.8 Use of Infra-red Spectroscopy in the Identification of Functional Groups (SB p.113)
(f) (i) 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.
(ii)
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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.
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320
34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
The mass spectrum of pentan-2-one (CH3COCH2CH2CH3) is shown below:
What ions do the peaks at m/e 86, 71 and 43 represent? Explain your answer.
Answer
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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.
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34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
The mass spectrum of hydrocarbon X is shown below:
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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 and 15. Answer
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34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
(a) The relative molecular mass of hydrocarbon X is 58.0.
(b) The base peak is at m/e 43.
(c) 15 mass units
(d) 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.
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34.9 Use of Mass Spectra to Obtain Structural Information (SB p.118)
(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)
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326
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:
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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. Answer
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34.9 Use of Mass Spectra to Obtain Structural Information (SB p.120)
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+.
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329
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?
AnswerChlorine has two isotopes, chlorine-35
and chlorine-37. Their relative
abundances are in the ratio of 3 : 1.
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330
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?
AnswerBromine has two isotopes, bromine-79
and bromine-81. Their relative
abundances are in the ratio of 1 : 1.
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331
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.
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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
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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+.
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34.9 Use of Mass Spectra to Obtain Structural Information (SB p.124)
(c) Study the following spectrum carefully and deduce what group of organic compound it is. The compound has a relative molecular mass of 114.
Answer
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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.
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