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
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
7
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
8
34.2 Isolation and Purification of Organic Compounds (SB p.78)
Filtration
The laboratory set-up of filtration
9
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
10
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
11
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
12
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
13
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
14
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
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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
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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
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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
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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
19
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
20
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
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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
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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
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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
24
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
25
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
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34.2 Isolation and Purification of Organic Compounds (SB p.81)
Distillation
The laboratory set-up of distillation
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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
28
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
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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
30
34.2 Isolation and Purification of Organic Compounds (SB p.82)
Fractional Distillation
The laboratory set-up of
fractional distillation
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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
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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
33
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
•
34
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
35
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
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34.2 Isolation and Purification of Organic Compounds (SB p.83)
Sublimation
A mixture of two compounds can be
separated by sublimation
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34.2 Isolation and Purification of Organic Compounds (SB p.83)
Chromatography
38
•
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
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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
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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
41
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
42
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
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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
44
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
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34.4
Qualitative
Analysis of
Elements in an
Organic Compound
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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
47
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
48
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
49
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
50
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
51
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
52
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).
A creamy yellow
precipitate is formed. It
is insoluble in excess
NH3(aq).
iodide ion (I-)
53
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
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34.5
Determination of
Empirical Formula
and Molecular
Formula from
Analytical Data
55
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
56
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
57
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
58
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
59
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
60
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
61
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
62
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
63
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
64
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
65
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
66
34.6
Structural
Information from
Physical Properties
67
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
68
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
69
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
70
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
71
34.6 Structural Information from Physical Properties (SB p.90)
Physical properties of some common organic compounds
Organic
compound
Hydrocarbo
ns
(saturated
and
unsaturated)
72
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
73
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
74
• 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)
75
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
76
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)
77
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)
78
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
79
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
80
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
81
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
82
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
83
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
84
Check Point 34-6
34.7
Structural
Information from
Chemical Properties
85
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
86
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:
87
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
88
34.7 Structural Information from Chemical Properties (SB p.93)
Chemical tests for different groups of organic compounds
Organic
compound
Saturated
hydrocarbons
89
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
90
• 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°)
91
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
92
34.7 Structural Information from Chemical Properties (SB p.94)
Chemical tests for different groups of organic compounds
Organic
compound
Alcohols
(  OH)
93
Test
• Add a small piece of
sodium metal
• Esterification: Add
ethanoyl chloride
Observation
• A colourless gas is
evolved
• 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)
94
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
95
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)
96
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  )
97
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)
98
• 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:
99
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
100
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
101
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
102
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
(
)
103
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
104
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
105
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
106
Example 34-7B
Check Point 34-7
The END
107
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
108
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
109
(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
110
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
111
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
112
∴ 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
113
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
114
∴ 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
115
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
116
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
117
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.
118
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
119
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
120break 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.
121
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
122
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
123
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)
124
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.
125
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.
126
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)
127
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.
128
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
129
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:
130
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
131
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.
132
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
133
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
134
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.
135
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
136
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.

137
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.
138
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
139
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.
140
2
34.7 Structural Information from Chemical Properties (SB p.99)
(a) (i)
(ii)
141
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
142
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
143
34.7 Structural Information from Chemical Properties (SB p.99)
(b) (i)
(ii)
144
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
145
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.
146
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.
147
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
148