Chapter 38
Structural Determination of
Organic Compounds
38.1
38.2
38.3
38.4
38.5
38.6
1
Introduction
Isolation and Purification of Organic Compounds
Qualitative Analysis of Elements in an Organic
Compound
Determination of Empirical and Molecular
Formulae from Analytical Data
Chemical Tests for Functional Groups
Use of Infra-red Spectroscopy in the Identification
of Functional Groups
New Way Chemistry for Hong Kong A-Level Book 3B
38.1 Introduction (SB p.137)
The determination of the structure of an organic compound
involves four steps:
1. Isolation and purification of the compound
2. Qualitative analysis of the elements present in the compound
3. Determination of the molecular formula of the compound
4. Determination of the functional group present in the compound
2
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.138)
Filtration
• The solid/ liquid mixture
to be filtered is poured
onto the folded filter
paper in the filter funnel
• Liquids are allowed to
pass through the filter
paper (known as filtrate)
while insoluble solids
(known as residue) are
retained on the filter
paper
3
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.139)
Centrifugation
• The tubes containing
undissolved solids are
spun around very rapidly
and are thrown outwards
in the centrifuge
• The denser solid collects
as a lump at the bottom
of the tube with the clear
liquid above it
• The liquid can be
removed by decantation
4
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.139)
Crystallization
Crystallization by Making and Cooling a
Hot Concentrated Solution
• The solubility of most solids increase with a rise of temperature
• As a hot concentrated solution cools, it cannot hold all of its
dissolve solute. The excess solute separates out as crystals
5
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.140)
Crystallization by Evaporating
a Solution at Room Temperature
• When the solution becomes saturated after evaporation,
further evaporation causes crystallization to occur
• The crystallization process will be slow if it occurs at
room temperature
6
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.141)
Solvent Extraction
• To dissolve out a component from a
mixture with a suitable solvent
• An aqueous solution containing the
organic product is usually shaken
with diethyl ether in a separating
funnel. The ether layer is then run off
• Repeated extraction with fresh
ether to extract any organic products
remaining
• The ether portions are combined and
dried, and distilled away to obtain
the organic products
7
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.141)
Simple Distillation
• To separate a liquid from a
solution of a liquid and a
non-volatile solid or liquid
• Only the liquid vapourizes
to form vapours which
condense to liquid on cold
surface which collected as
distillate
• Anti-bumping granules
are added to prevent
bumping
8
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.142)
Fractional Distillation
• To separate a liquid from
a mixture of two or more
miscible liquids which
have large difference in
boiling point
• Fractionating column
provides a large surface
area for condensation
and vapourization of the
mixture to occur
9
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.143)
Steam Distillation
• Many liquid or even solid compounds which are
immiscible with water can be purified by distillation
in a current of steam
10
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.143)
Principle of steam distillation:
• the total vapour pressure of an immiscible mixture is
equal to the sum of the vapour pressures of individual
components
 the water and the compound will distil at a lower
temperature than the boiling point of either one
of them
11
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.144)
Sublimation
• Direct change of a solid to vapour on heating or
vapour to solid on cooling
• A mixture of two compounds is
heated in an evaporating dish, one
compound sublimes. The vapour
changes back to solid on a cool
surface while the other compound
is not affected and remains in the
evaporating dish
12
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.144)
Chromatography
• To separate a complex mixture of substances
• As the various components are being absorbed or partitioned
at different rates, they move upwards to different extent
• The ratio of the distance travelled by the substance to the
distance travelled by the solvent is known as Rf value, which is
the characteristic of the substance
13
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.146)
A summary of different techniques of isolation and purification
Technique
(a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
14
Separation
Filtration
Centrifugation
Crystallization
Solvent extraction
An insoluble solid from a liquid (slow)
An insoluble solid from a liquid (fast)
A dissolved solute from its solution
Dissolving out of a component from a mixture with a
suitable solvent
Simple distillation
A liquid from a solution of non-volatile solutes
Fractional distillation Miscible liquids with widely different boiling points
Steam distillation
A liquid or a solid (that is immiscible with water and
is volatile in steam) from impurities
Sublimation
Two solids, only one of which can sublime
Chromatography
A complex mixture of substances
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.146)
Tests for Purity
Determination of Melting Point
• Some of the dry solid is placed in a thinwalled glass melting point tube. The
tube is attached to a thermometer
• The temperature at which the solid
melts is its melting point
• Pure solid has a sharp melting point
• Impure solid melts gradually over a
wide temperature range
15
New Way Chemistry for Hong Kong A-Level Book 3B
38.2 Isolation and Purification of Organic Compounds (SB p.147)
Determination of Boiling Point
• The boiling point of a liquid can be determined by using
the simple distillation apparatus
• The temperature at which the liquid boils steadily is
its boiling point
• The boiling point of a pure liquid is quite sharp, but
varies with changes in external pressure
• The presence of non-volatile solutes such as salts raises
the boiling point of a liquid
16
New Way Chemistry for Hong Kong A-Level Book 3B
38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.147)
Carbon and Hydrogen
• Carbon and hydrogen can be detected by heating a
small amount of the substance with Cu2O
• Carbon and hydrogen would be oxidized to CO2 and
H2O respectively
• CO2 turns lime water milky, and H2O turns
anhydrous CoCl2 paper pink
17
New Way Chemistry for Hong Kong A-Level Book 3B
38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.147)
Halogens, Nitrogen and Sulphur
• Sodium fusion test is performed to test for the presence
of halogens, nitrogen and sulphur
• The compound under test is fused with a small piece of
sodium metal in a small combustion tube and then heated
strongly
• The product of the test are extracted with water and
analyzed
• During sodium fusion, halogens, nitrogens and sulphur in
the organic compound are converted to sodium halide,
sodium cyanide and sodium sulphide respectively
18
New Way Chemistry for Hong Kong A-Level Book 3B
38.3 Qualitative Analysis of Elements in an Organic Compound (SB p.148)
Element
Halogens, as
Cl–
Br–
I–
19
Test
Positive result
Acidify and boil to remove any
CN– and S2– present, then add
A white ppt., soluble in
AgNO3(aq)
excess NH3(aq)
Ag+(aq) + X–(aq)  AgX(s) A pale yellow ppt., sparingly
soluble in excess NH3(aq)
A creamy yellow ppt.,
insoluble in excess NH3(aq)
Nitrogen, as
CN–
Add a mixture of iron(II)
sulphate and iron(III) sulphate
solutions
A blue-green colour is
observed
Sulphur, as
S2–
Add a solution of sodium
pentacyanonitrosylferrate(II)
A black ppt. is observed
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.148)
• Quantitative analysis is to find out the percentage
composition by mass of the compound
• Methods to determine the mass of the elements:
1. Carbon and hydrogen
• The organic compound is burnt in oxygen.
• The carbon dioxide and water vapour formed are
respectively absorbed by KOH solution and anhydrous
CaCl2.
• The increases in mass in KOH solution and CaCl2
represent the masses of carbon dioxide and water
vapour formed respectively
20
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.149)
2. Nitrogen
• The organic compound is heated with excess Cu2O
• The NO and NO2 formed are passed over hot Cu and
the volume of N2 formed is measured
21
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.149)
3. Halogens
• The organic compound is heated with fuming HNO3
and excess AgNO3 solution
• The mixture is allowed to cool and then water is added
• The dry AgX formed is weighed
22
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.149)
4. Sulphur
• The organic compound is heated with fuming HNO3
• After cooling, barium nitrate solution is added
• The dry barium sulphate formed is weighed
23
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.149)
• The empirical formula of the compound can be
calculated after the determination of percentage
composition by mass of a compound
• The molecular formula can be calculated after knowing
relative molecular mass and the empirical formula
24
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.149)
• The empirical formula of a compound is the formula
which shows the simplest whole number ratio of the
atoms present in the 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
25
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.149)
Example 38-1
On quantitative analysis, 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.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
Answer
26
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.149)
Solution:
Let the mass of the compound be 100 g. Then,
mass of C in the compound = 40.0 g
mass of H in the compound = 6.7 g
mass of O in the compound = 53.3 g
Mass (g)
C
H
O
40.0
6.7
53.3
6.7
 6.7
1.0
6.7
2
3.33
53.3
 3.33
16.0
3.33
1
3.33
2
1
Number of moles 40.0
 3.33
(mol)
12.0
3.33
Relative number
1
of moles
3.33
Simplest mole
ratio
1
∴ The empirical formula of the organic compound is CH2O.
27
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.150)
Example 38-2
An organic compound Z has the following composition by
mass:
Element
Carbon
Percentage
60.00
by mass (%)
Hydrogen
Oxygen
13.33
26.67
(a) Calculate the empirical formula of compound Z.
(b) If compound Z is found to have a relative molecular
mass of 60, determine the molecular formula of
compound Z.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
28
New Way Chemistry for Hong Kong A-Level Book 3B
Answer
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.150)
Solution:
Let the mass of the compound be 100 g. Then,
mass of C in the compound = 60.0 g
mass of H in the compound = 13.33 g
mass of O in the compound = 26.67 g
Mass (g)
Number of
moles (mol)
Relative number
of moles
Simplest mole
ratio
C
H
O
60.0
13.33
26.67
60.00
5
12.0
5
3
1.67
3
13.33
26.67
 13.33
 1.67
1.0
16.0
1.67
13.33
1
8
1.67
1.67
8
1
∴ The empirical formula of compound Z is C3H8O.
29
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.150)
Solution:
The molecular formula of the compound is (C3H8O)n.
Relative molecular mass of (C3H8O)n = 60
n  (12.0 3 + 1.0 8 + 16.0) = 60
n=1
∴ The molecular formula of compound Z is C3H8O.
30
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.150)
Example 38-3
An organic compound is found to contain carbon,
hydrogen and oxygen only. On complete combustion,
0.15 g of this compound gives 0.22 g of carbon dioxide
and 0.09 g of water. If the relative molecular mass of this
compound is found to be 60, determine the molecular
formula of this compound.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
Answer
31
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.151)
Solution:
Let the empirical formula of the compound be CxHyOz.
Mass of C in CxHyOz = Mass of C in CO2
Mass of H in CxHyOz = Mass of H in H2O
Mass of O in CxHyOz = Mass of CxHyOz – mass of C in CxHyOz
– mass of H in CxHyOz
Relative molecular mass of CO2 = 12.0 + 16.0  2
= 44.0
12 .0
Mass of C in 0.22 g of CO2 = 0.22 g 
44 .0
= 0.06 g
2 .0
Mass of H in 0.09 g of H2O = 0.09 g 
18 .0
= 0.01 g
32
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.151)
Solution:
Mass of O in the compound = (0.15 – 0.06 – 0.01) g
= 0.08 g
The simplest whole number ratio of x, y and z can be determined by
following the steps in the table below.
Mass (g)
Number of moles
(mol)
Relative number
of moles
Simplest mole
ratio
33
C
H
O
0.06
0.01
0.08
0.06
 0.005
12.0
0.005
1
0.005
0.01
 0.01
1.0
001
2
0005
0.08
 0.005
16.0
0.005
1
0.005
1
2
1
∴ The empirical
formula of the organic compound is CH2O.
New Way Chemistry for Hong Kong A-Level Book 3B
38.4 Determination of Empirical and Molecular Formulae
from Analytical Data (SB p.151)
Solution:
The molecular formula of the compound is (CH2O)n.
Relative molecular mass of (CH2O)n = 60
n  (12.0 + 1.0  2 +16.0) = 60
n =2
∴ The molecular formula of the compound is C2H4O2.
34
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.152)
• The molecular formula does not give enough
information on the structure of the compound
∵ compounds having the same molecular formula
may have widely different arrangement of atoms
and even different functional groups
e.g.
35
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.152)
• ∴ the following step is to find out the functional
group(s) present so as to deduce the actual
arrangement of atoms in the molecule
• The presence of certain functional groups in an organic
compound can be detected by simple chemical tests
36
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.152)
Group
Saturated
hydrocarbon
Test
• Combustion: burn a little
saturated hydrocarbon in a
non-luminous Bunsen flame
Positive result
• A blue or clear yellow
flame is observed
Unsaturated
• Combustion: burn a little
• A smoky flame is
hydrocarbon
unsaturated hydrocarbon in a observed
non-luminous Bunsen flame
(C = C, C  C)
37
• Addition of Br2 in CH3CCl3
at room temp. and in the
absence of light
• Rapid decolourization of
Br2
• Addition of 1% (dilute)
acidified KMnO4
• Immediate
decolourization of
KMnO4 solution and/ or
development of a brown
colour (MnO2)
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.152)
Group
38
Test
Positive result
Haloalkanes
(1°, 2° or 3°)
• Hydrolysis followed by
Ag+/HNO3(aq): boil with
ethanolic KOH, then
acidify with excess dilute
HNO3 and add AgNO3
solution
• Precipitation of AgX is
observed
(AgCl – white ppt.;
AgBr – pale yellow ppt.;
AgI – creamy yellow ppt.
Remarks: Halobenzenes
show no observable
changes
Halobenzene
• Hydrolysis followed by
Ag+/HNO3(aq): boil with
ethanolic KOH, then
acidify with excess dilute
HNO3 and add AgNO3
solution
• No precipitation of AgX is
observed
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.152)
Group
Test
Positive result
Alcohol (–OH)
• Oxidation by K2Cr2O7/H+(aq)
• 1° and 2° alcohols: the
clear orange solution
becomes opaque and turns
green almost immediately
• 3° alcohols: no observable
changes
• Test with sodium metal: add a
small piece of Na metal to the
alcohol
• Evolution of H2 is
observed
• Esterification with ethanoyl
chloride
• A rise in temp. together
with evolution of HCl is
observed
• Iodoform test for :
• A yellow ppt. of CHI3 is
produced
39
Addition
of I2 /forNaOH(aq)
New Way Chemistry
Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.152)
Group
40
Test
Positive result
Alcohol (–OH)
• Lucas test: treat the alcohol
with anhydrous ZnCl2 /
conc. HCl (Lucas reagent)
• 1° alcohols: do not
dissolve appreciably, the
aqueous phase remains
clear
• 2° alcohols: the clear
solution becomes cloudy
within 5 minutes
• 3° alcohols: a cloudy
phase separates almost
immediately
Ether (–O–)
• No specific test but it is
soluble in conc. H2SO4
Aldehyde
(
)
• Addition of NaHSO3
—
• Crystalline salts are
observed
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.153)
Group
Aldehyde
(
)
Test
Positive result
• Addition of
• A yellow, orange or red
2,4,-dinitrophenylhydrazine
ppt. is observed
• Iodoform test for:
• A yellow ppt. of CHI3 is
produced
Addition of I2 / NaOH(aq)
Carboxylic acid
(
)
41
• Ester formation: warm the
carboxylic acid with an
absolute alcohol in the
presence of conc.
H2SO4(aq), followed by
adding Na2CO3(aq)
• A sweet, fruity smell of
an ester is detected
• Addition of NaHCO3
• Evolution of CO2(g)
which turns lime water
milky
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.153)
Group
Test
Amine (–NH2)
• Test for 1° aliphatic amines: • Steady evolution of N2(g)
dissolve the amine in
is observed
dilute HCl at 0 – 5°C, then
add cold NaNO2(aq)
slowly
• Test for 1° aromatic
amines: addition of
naphthalen-2-ol in dilute
NaOH
Ester
(
42
)
Positive result
• An orange or red ppt. is
produced
• No specific test but can be • A sweet and fruity smell is
distinguished by its
detected
characteristic smell
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.154)
Group
Positive result
Acyl halide
(
)
• Hydrolysis followed by
Ag+/HNO3(aq): boil with
ethanolic KOH, then
acidify with excess dilute
HNO3 and add AgNO3
solution
• Precipitation of AgX is
observed
(AgCl – white ppt.;
AgBr – pale yellow ppt.;
AgI – creamy yellow ppt.
Remarks: Halobenzenes
show no observable
changes
Amide
(
• Boil the amide with
NaOH(aq)
• Evolution of NH3(g) is
detected
Phenol
(
43
Test
)
• Addition of neutral
) FeCl3(aq)
• Change in colour is
observed (may be the
transient appearance of a
red, violet or blue colour)
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.154)
Group
Test
Phenol
(
• Addition of Br2 in
) CH3CCl3
• A white ppt. of
2,4,6-tribromophenol is
formed
Aromatic
compound
(
• Combustion: burn a little
aromatic compound in a
) non-luminous Bunsen
flame
• A smoky yellow flame
with black soot is
produced
• Fuming H2SO4
44
Positive result
• Dissolution of compound
and evolution of heat
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.156)
Example 38-4
An organic compound is found to have an empirical
formula of CH2O and a relative molecular mass of 60. It
reacts with sodium hydrogencarbonate to give a
colourless gas which turns lime water milky.
Solution:
(a) Calculate the molecular formula of the compound.
(a) Let the molecular formula of the compound be (CH2O)n.
Answer
Relative molecular mass of (CH2O)n = 60
n  (12.0 + 1.0  2 + 16.0) = 60
n=2
∴ The molecular formula of the compound is C2H4O2.
45
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.156)
Example 38-4 (cont’d)
(b) Deduce the structural formula of the compound.
Solution:
(c) Give the IUPAC name for the compound.
(b) The compound reacts with sodium hydrogencarbonate to give
(R.a.m.:
= 1.0,gas
C =(which
12.0, turns
O = 16.0)
carbonHdioxide
lime water milky), indicating that
Answer
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. The
structural formula of the compound is therefore CH3COOH.
(c) The IUPAC name for the compound is ethanoic acid.
46
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.156)
Example 38-5
15 cm3 of a gaseous hydrocarbon were mixed with 120 cm3
of oxygen which was in excess. The mixture was exploded
and 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 volumes were measured under room
temperature and pressure.
Answer
47
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.156)
Solution:
(a) Let the molecular formula of the compound be CxHy.
Volume of hydrocarbon 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
y
y

C x H y   x  O2 
 xCO2  H 2 O
4
2

Volume of CxHy reacted : volume of CO2 formed = 1 : x
= 15 : 30
1 15

∵
x 30
∴x=2
48
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.156)
Solution:
Volume of CxHy reacted : volume of O2 reacted = 1 : ( x  y )
= 15 : 45 4
1
15

∵
y 45
(2  )
4
2
y
3
4
∴y=4
∴ The molecular formula of the compound is C2H4
49
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.156)
Example 38-5 (cont’d)
(b) To which homologous series does the hydrocarbon
belong?
(c) Give the structural formula of the hydrocarbon.
Answer
Solution:
(b) C2H4 belongs to the alkene series.
(c) The structural formula of the hydrocarbon is:
50
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.157)
Example 38-6
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
105°C 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 aqueous
potassium hydroxide, the volume further decreased to 50 cm3.
(a) Calculate the molecular formula of the compound,
assuming all volumes were measured under room
temperature and pressure.
Answer
51
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.157)
Solution:
(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 = (150 – 90) cm3
= 60 cm3
y z
y

C x H y Oz   x   O2 
 xCO2  H 2 O
4 2
2

Volume of CxHyOz reacted : volume of CO2 formed = 1 : x
= 20 : 40
1 20

∵
x 40
∴x=2
52
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.157)
Solution:
y
(a) Volume of CxHyOz reacted : volume of H2O formed = 1 :
2
= 20 : 60
2 20

∵
y 60
∴y=6
Volume of CxHyOz reacted : volume of O2 formed = 1 : ( x  y  z )
= 20 : 60 4 2
∵
1
20

y z
( x   ) 60
4 2
6 z
2  3
4 2
∴z=1
∴ The molecular formula of the compound is C2H6O.
53
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.157)
Example 38-6 (cont’d)
(b) The compound is found to contain an –OH functional
Solution:
group in its structure. Deduce its structural formula.
(b)
contains
a –OH
group
in its structure,
(c)AsIsthe
thecompound
compound
optically
active?
Explain
your
the
hydrocarbon skeleton of the compound becomes C2H5
answer.
after eliminating the –OH group from C2H6O. TheAnswer
structural
formula of the compound is CH3CH2OH.
(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 groups of atoms.
54
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
Check Point 38-1
(a) A substance contains 42.8% carbon, 2.38% hydrogen,
16.67% nitrogen by mass and the remainder consists of
oxygen.
(i) Given the fact that the relative molecular mass of the
substance is 168, deduce the molecular formula of the
substance.
(R.a.m.: H = 1.0, C = 12.0, N = 14.0, O = 16.0)
Answer
55
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
(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
16.67
38.15
42.8
No. of
2.38

1
.
19
 2.38
 3.57
 2.38
moles (mol) 12.0
14.0
16.0
1.0
2.38
1.19
2.38
Relative no. 3.57
2

1
2
3
of moles
1.19
1.19
1.19
1.19
Simplest
mole ratio
3
2
1
2
∴ The empirical formula of the compound is C3H2NO2.
Let the molecular formula of the compound be (C3H2NO2)n.
Molecular mass of (C3H2NO2)n = 168
n  (12.0  3 + 1.0  2 + 14.0 + 16.0  2) = 168
∴n =2
∴ The molecular formula of the compound is C6H4N2O4.
56
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
Check Point 38-1 (cont’d)
(ii) The substance
is proved to be an
aromatic compound
(a) (ii) 1,2-Dinitrobenzene
1,3-Dinitrobenzene
with only one type of functional group. Give the
structural formulae for all isomers of the substance
(the name of each isomer should also be included).
1,4-Dinitrobenzene
Answer
57
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
Check Point 38-1 (cont’d)
(b) 30cm3 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. Be 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.
Answer
58
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
(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
y
y

C x H y   x  O2 
 xCO2  H 2O
4
2

Volume of CxHy reacted : Volume of CO2 formed
=1:x
= 30 : 60
∵ 1  30
x 60
∴x=2
59
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
Volume of CxHy reacted : Volume of O2 reacted
= 1 :  x  y 
4

= 30 : 105
1
30

∵
y 105
x
4
y

30   2    105
4

∴y=6
∴ The molecular formula of the hydrocarbon is C2H6.
60
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
Check Point 38-1 (cont’d)
(ii) Is the hydrocarbon a saturated, unsaturated or
aromatic hydrocarbon?
Answer
(b) (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.
61
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
Check Point 38-1
(c) A hydrocarbon having a relative molecular mass of 56
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.
(R.a.m.: H = 1.0, C = 12.0)
62
New Way Chemistry for Hong Kong A-Level Book 3B
Answer
38.5 Chemical Tests for Functional Groups (SB p.159)
(c) (i) Let the mass of the compound be 100 g.
Mass (g)
No. of moles
(mol)
Relative no.
of moles
Simplest
mole ratio
Carbon
Hydrogen
85.5
14.5
85.5
 7.125
12.0
7.125
1
7.125
14.5
 14.5
1.0
14.5
2
7.125
1
2
∴ The empirical formula of the hydrocarbon is CH2.
Let the molecular formula of the hydrocarbon be (CH2)n.
Molecular mass of (CH2)n = 56
n  (12.0 + 1.0  2) = 56
n =4
∴ The molecular formula of the compound is C4H8.
63
New Way Chemistry for Hong Kong A-Level Book 3B
38.5 Chemical Tests for Functional Groups (SB p.159)
Check Point 38-1 (cont’d)
(ii) Name the two geometrical isomers of the
(c) (ii)
hydrocarbon.
(iii) Explain the existence of geometrical isomerism.
Answer
(iii) Since the but-2-ene molecule is unsymmetrical
and free rotation of the but-2-ene molecule is restricted
by the presence of the carbon-carbon double bond,
geometrical isomerism exists.
64
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.159)
The Electromagnetic Spectrum
• Electromagnetic radiation has dual property: as wave or
as photons
• The relationship of the frequency () of an electromagnetic
radiation, its wavelength () and velocity (c) is shown as
follows:
c


• The energy of a quantum of electromagnetic radiation is:
E = h where h is the Planck constant (i.e. 6.626  10–34 J s)
or
65
E
hc

New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.160)
• Visible light has wavelength between 400 nm and 800 nm
• Infra-red radiation has wavelength between 800 nm and 300 m
66
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.161)
Infra-red Spectroscopy
• Organic compounds absorb electromagnetic energy in the
infra-red (IR) region of the spectrum
• IR radiation causes atoms and groups of atoms of organic
compounds to vibrate with increased amplitude about the
covalent bonds that connect them
• The vibration is
quantized,
therefore,
the IR radiation
absorbed have
particular energy
67
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.161)
• Infra-red spectrometer:
Instrument used to measure the
amount of light absorbed at each
wavelength of the IR region
• Infra-red spectrum:
shows the absorption of IR radiation by a sample at different
frequencies
• The IR radiation is specified by its wavenumber (unit: cm–1)
which is the reciprocal of wavelength
68
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.161)
• Covalent bonds behave as if they were tiny springs
connecting the atoms in the vibrations
• The atoms only vibrate at certain frequencies, therefore
covalently bonded atoms have only particular
vibrational energy levels
• The excitation of a molecule from one vibrational energy
level to another occurs only when the compound absorbs
IR radiation of the exact energy required
69
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.161)
Molecules vibrate in a variety of ways:
Stretching vibration: two atoms joined by a covalent bond
move back and forth as if they were joined by a spring
70
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.162)
A variety of stretching and bending vibrations:
71
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.162)
• The frequency of a given stretching vibration of a
covalent bond depends on:
1.
The masses of the bonded atoms
2.
The strength of the bond
• Light atoms vibrate at higher frequencies than heavier ones
• The stretching frequencies of groups involving hydrogen such
as C – H, N – H and O – H occur at relatively high frequencies
Bond Range of wavenumber (cm–1)
C–H
2 840 – 3 095
O–H
3 230 – 3 670
N–H
3 350 – 3 500
72
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.162)
• Triple bonds are stronger (and vibrate at higher
frequencies) than double bonds
Bond Range of wavenumber (cm–1)
2 070 – 2 250
CC
2 200 – 2 280
CN
C=C
1 610 – 1 680
C=O
1 680 – 1 750
• The IR spectra of simple compounds contain many
absorption peaks
 possibility of two different compounds having the same
IR spectrum is very small
73
 IR spectrum has been called the ‘fingerprint’ of a
molecule
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.163)
Use of IR Spectrum in the Identification of Functional Groups
• A 100% transmittance in the spectrum implies no
absorption of IR radiation
• Absorption peak (or band): Due to the absorption of IR
radiation, a decrease in percent transmittance is resulted
and hence a dip in the spectrum
74
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.163)
The four regions of an IR spectrum
Range of
wavenumber (cm–1)
75
Interpretation
400 – 1 500
This region often consists of many different
complicated bands. This part of the spectrum is
unique to each compound and is often called
the fingerprint region. It is not used for
identification of particular functional groups.
1 500 – 2 000
Absorption of double bonds,
e.g. C = C, C = O
2 000 – 2 500
Absorption of triple bonds,
e.g. C  C, C  N
2 500 – 4 000
Absorption of single bonds formed by
hydrogen,
e.g. C – H, O – H and N – H
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.164)
• 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
Bond
Characteristic range
of wavenumber (cm–1)
Alkenes
C=C
1 610 – 1 680
Aldehydes, ketones, acids, esters
C=O
1 680 – 1 750
Alkynes
CC
2 070 – 2 250
Nitriles
CN
2 200 – 2 280
Acids (hydrogen-bonded)
O–H
2 500 – 3 300
Alkanes, alkenes, arenes
C–H
2 840 – 3 095
Alcohols, phenols (hydrogenbonded)
O–H
3 230 – 3 670
Primary amines
N–H
3 350 – 3 500
Compound
76
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.164)
Interpretation of IR Spectra
Interpretation of the IR spectrum of butane
77
Wavenumber (cm–1)
Intensity
2 968
2 890
vs
m
C – H stretching
1 468
s
C – H bending
Indication
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.166)
Interpretation of the IR spectrum of cis-but-2-ene
78
Wavenumber (cm–1)
Intensity
3 044
3 028
vs
vs
C – H stretching
2 952
vs
C – H stretching
1 677
1 657
m
m
C = C stretching
1 411
s
C – H bending
Indication
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.167)
The IR spectrum of hex-1-yne
79
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.167)
Interpretation of the IR spectrum of hex-1-yne
Wavenumber (cm–1) Intensity
3 313
vs
C – H stretching
(sp C – H)
2 963
2 938
2 874
vs
vs
s
C – H stretching
(sp3 C – H)
2 119
s
C  C stretching
s
C – H bending
(sp C – H)
m
C – H bending
(sp3 C – H)
1 468
1 445
80
Indication
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.168)
Interpretation of the IR spectrum of butanone
81
Wavenumber (cm–1)
Intensity
2 983
2 925
s
s
C – H stretching
1 720
vs
C = O stretching
1 416
m
C – H bending
Indication
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.169)
Interpretation of the IR spectrum of butan-1-ol
82
Wavenumber (cm–1)
Intensity
Indication
3 330
Broad band
O – H stretching
2 960
2 935
2 875
m
m
m
C – H stretching
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.170)
Interpretation of the IR spectrum of butanoic acid
83
Wavenumber (cm–1)
Intensity
Indication
3 100
Broad band
O – H stretching
1 708
s
C = O stretching
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.170)
•
The absorption of the O – H group in alcohols and
carboxylic acids usually appear as broad band instead
of sharp peak
∵
the vibration of the O – H group is complicated
by the hydrogen bonding formed between the
molecules
84
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.171)
Interpretation of the IR spectrum of butylamine
85
Wavenumber (cm–1)
Intensity
Indication
3 371
3 280
s
s
N – H stretching
2 960 – 2 875
w
C – H stretching
1 610
m
N – H bending
1 475
m
C – H bending
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.171)
Interpretation of the IR spectrum of butanenitrile
86
Wavenumber (cm–1)
Intensity
2 990 – 2 895
s
C – H stretching
2 246
vs
C  N stretching
1 420
s
1 480
s
Indication
C – H bending
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Uses of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.172)
Summary of strategies for the use of IR spectra in the
identification of functional groups in an organic compound:
1. Focus at the IR absorption peak at or above 1500 cm–1.
Concentrate initially on the major absorption peaks
2. For each absorption peak, try to list out all the possibilities
using a correlation table or chart. Note that not all
absorption peaks in the spectrum can be assigned
3. The absence and presence of absorption peaks at some
characteristic ranges of wavenumbers are equally
important. It is because the absence of particular absorption
peaks can be used to identify certain functional groups or
bonds do not exist in the molecule
87
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.173)
Example 38-7
An organic compound with a relative molecular mass of 72 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.
Answer
(a) Determine the molecular formula of the compound.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
88
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.173)
Solution:
(a) Let the mass of the compound be 100 g. Then,
mass of carbon in the compound = 66.66 g
mass of hydrogen in the compound = 11.11 g
mass of oxygen in the compound = 22.23 g
Mass (g)
Number of moles
(mol)
Relative number
of moles
Simplest mole
ratio
C
H
O
66.66
11.11
22.23
66.66
11.11
22.23
 5.56
 11.11
 1.39
12.0
1.0
16.0
5.56
1.39
11.11
4
1
8
1.39
1.39
1.39
4
8
1
∴ The empirical formula of the organic compound is C4H8O.
89
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.173)
Solution:
Let the molecular formula of the compound is (C4H8O)n.
Relative molecular mass of (C4H8O)n = 72
n  (12.0  4 + 1.0 8 + 16.0) = 72
∴
n=1
∴ The molecular formula of the compound is C4H8O.
90
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.173)
Example 38-7 (cont’d)
(b) Deduce two possible structures of the compound, each of
which belongs to a different homologous series.
Answer
91
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.173)
Solution:
(b) From the IR spectrum, it can be observed that there are
absorption peaks at 2950 cm–1 and 1700 cm–1. The absorption
peak at 2950 cm–1 corresponds to the stretching vibration of
the C – H bond, and the absorption peak at 1700 cm–1
corresponds to the stretching vibration of the C = O bond. As
there is only one oxygen atom in the molecule of the compound, it
is either an aldehyde or a ketone.
If it is an aldehyde, its possible structure will be:
92
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Solution:
(b) If it is a ketone, its possible structure will be:
93
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2
(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.
94
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(a)
Sketch the infra-red spectrum of a carboxylic acid based
on the IR spectra of X and Y.
Answer
95
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
(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 predicted. A wide 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. The infra-red spectrum of a carboxylic
acid is as follows:
96
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
(a)
97
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(b) The infra-red spectra of two organic compounds A and B
are shown below.
98
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(b)
(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.
State which compound could be an alcohol. Explain your
answer briefly.
Answer
99
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(c) The infra-red spectrum of the amino acid is
(c) Given the following
characteristic absorption
shown as follows:
wavenumbers of some covalent bonds in infra-red
spectra:
Bond
Range of wavenumber (cm–1)
C=O
O–H
C–H
N–H
1680 – 1750
2500 – 3300
2840 – 3095
3350 – 3500
Sketch the expected infra-red spectrum for an amino
acid with the following structure:
Answer
100
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(d) Below is a portion of the infra-red spectrum of an
organic compound X. To which homologous series
does it belong?
(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.
Answer
101
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(e) Below is a portion of the infra-red spectrum of an
organic compound Y. Identify the functional groups
that it contains.
(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.
Answer
102
New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(f) An organic compound Z with a relative molecular mass of
88 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:
Answer
(i) Determine the molecular formula of compound Z.
(R.a.m.: H = 1.0, C = 12.0, O = 16.0)
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New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
(f) (i) Let the mass of the compound be 100 g.
Carbon
Oxygen
Hydrogen
Mass (g)
54.54
36.36
9.10
36.36
9.10
No. of moles 54.54
 2.27
 4.55
 9.10
(mol)
16
.
0
12.0
1.0
Relative no.
4.55
2.27
9.10

2

1
4
of moles
2.27
2.27
2.27
Simplest
2
1
4
mole ratio
∴ The empirical formula of the compound is C2H4O.
Let the molecular formula of the compound be (C2H4O)n.
Molecular mass of (C2H4O)n = 88
n  (12.0  2 + 1.0  4 + 16.0) = 88
n =2
∴ The molecular formula of the compound is C4H8O2.
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New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(f) (ii) Based on the result from (i), draw two possible
structures of the compound, each of which belongs
to a different homologous series.
Answer
(f) (ii)
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New Way Chemistry for Hong Kong A-Level Book 3B
38.6 Use of Infra-red Spectroscopy in the
Identification of Functional Groups (SB p.174)
Check Point 38-2 (cont’d)
(f) (iii) Using the information from the IR spectrum, name
the homologous series that compound Z belongs to .
Answer
(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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New Way Chemistry for Hong Kong A-Level Book 3B
The END
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New Way Chemistry for Hong Kong A-Level Book 3B