NORTHERN CAPE DEPARTMENT OF EDUCATION
PHYSICAL SCIENCES
GRADE 12 LESSONS
ORGANIC MOLECULES
TERM I
G. Izquierdo Rodríguez
&
G. Izquierdo Gómez
2018
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 1
ORGANIC MOLECULES
Lesson 1
Topic: Special properties of carbon. Representing organic molecules (structural-, condensed
structural-, and molecular formulae).
Objective:
Learners must be able to:
Explain the special properties of carbon atom that make it possible to form a variety
of bonds.
Define organic molecules, molecular formula, structural formula and condensed
structural formula.
Initial activities:
- The teacher must control the attendance of the learners.
- The teacher must moderate and discuss the homework.
Introduction:
The modern definition of Organic Chemistry is the chemistry of carbon compounds.
The misleading name “organic” is a relic of the days when chemical compounds were divided into
two from where they had come from. Organic compounds were those obtained from vegetable or
animal sources that are from material produced by living organisms.
These compounds from organic sources had in common: they all contained the element carbon.
Even after it had become clear that these compounds did not have to come from living sources
but could be made in the laboratory, it was convenient to keep the name ORGANIC to describe
them and compounds like them. The division between inorganic and organic compounds has been
kept up to now.
There are two large reservoirs of organic material from which simple organic compound can be
obtained: PETROLEUM and COAL. (Both of these are organic in the old serve, being products of
the decay of plants and animals. Organic molecules contain both carbon and hydrogen. Though
many organic chemicals also contain other elements, it is the carbon-hydrogen bond that defines
them as organic. There are millions of different organic molecules, each with different chemical
and physical properties.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 2
We use organic compounds everyday such as plastics, synthetics fibres such as polyester and
nylon, drugs, dyes, pesticides, crude oil and its derivatives such as petrol, paraffin and diesel, and
so many others.
- The teacher must orientate towards the objectives of the lesson and write the topic on the
blackboard.
Development:
The teacher must define organic molecules.
We know that Organic Chemistry is the branch of chemistry that deals with organic molecules.
An organic molecule is one which contains carbon, although not all compounds that contain
carbon are organic molecules. Noticeable exceptions are carbon monoxide (CO), carbon dioxide
(CO2), carbonates (e.g. calcium carbonate), carbides (e.g. calcium carbide) and cyanides (e.g.
sodium cyanide). Pure carbon compounds such as diamond and graphite are also not organic
compounds. Organic molecules can range in size from simple molecules to complex structures
containing thousands of atoms!
Although carbon is present in all organic compounds, other elements such as hydrogen (H),
oxygen (O), nitrogen (N), sulphur (S) and phosphorus (P) are also common in these molecules.
An organic molecule is a molecule that contains carbon atoms (generally bonded to other
carbon atoms as well as hydrogen atoms).
Organic compounds are very important in daily life and they range from simple to extremely
complex.
Organic molecules make up a big part of our own bodies, they are in the food we eat and in the
clothes we wear. Organic compounds are also used to make products such as medicines, plastics,
washing powders, dyes, along with a long list of other items. There are millions organic
compounds found in nature, as well as millions of synthetic (man-made) organic compounds.
- The teacher must discuss the special properties of carbon that makes it possible to form a
variety of bonds.
Carbon has a number of unique properties which influence how it behaves and how it bonds with
other atoms:
Carbon (C) has atomic number 6 and appears in the second row and Group IV of the
Periodic Table.
Carbon has four valence electrons which mean that each carbon atom can form a
maximum of four bonds with other atoms. Because of the number of bonds that carbon
can form with other atoms, organic compounds can be very complex.
The electron configuration of the carbon atom in the ground state can be illustrated as
follows: 1s2,2s2, 2px1, 2py1,2pz0
We can draw Aufbau diagram for carbon.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 3
Aufbau means “building up” in German, and the aufbau principle tells us how to build up the
electronic configuration of an atom’s ground (most stable) state. Starting with the lowest-energy
orbital, we fill the orbitals in order until we have added the proper number of electrons.
The orbital diagram for carbon can be shown as follows:
Energy
From this orbital diagram it is possible to observe that there are only two unpaired electrons
(valence electrons) available to form covalent bonds.
Why does carbon atom always form four bonds?
The addition of a small quantity of energy transfers one electron in the 2s-orbital to the empty 2pzorbital to give the exited state with four unpaired electrons, which are now available to form four
chemical bonds.
1s2,2s1, 2px1, 2py1,2pz1
2px
2py
2pz
2s
1s
Similar to other non-metals, carbon needs eight electrons to satisfy its valence shell. Carbon
therefore forms four bonds with other atoms (each bond consisting of one of carbon's electrons
and one of the bonding atom's electrons). Every valence electron participates in bonding, thus a
carbon atom's bonds will be distributed evenly over the atom's surface. These bonds form a
tetrahedron (a pyramid with a spike at the top), as illustrated below:
Carbon can form bonds with other carbon atoms to form single, double or triple covalent
bonds.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 4
Carbon can also form bonds with other atoms like hydrogen, oxygen, nitrogen and the
halogens.
Carbon can bond to form straight chain, branched, and cyclic molecules.
Because of this, long chain structures can form. This is known as catenation - the
bonding of atoms of the same element into longer chains.
o
These chains can either be:
Unbranched and can contain single carbon-carbon bonds only, or double and triple
carbon-carbon bonds as well.
o
Branched (have a branched group,) and can contain single carbon-carbon bonds only, or
double and triple carbon-carbon bonds as well.
Because of its position on the periodic table, most of the bonds that carbon forms with
other atoms are covalent. Think for example of a C - C bond. The difference in
electronegativity between the two atoms is zero, so this is a pure covalent bond. In the case
of a C - H bond, the difference in electronegativity between carbon (2,5) and hydrogen (2,2)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 5
is so small that C-H bonds are almost purely covalent. The result of this is that most organic
compounds are non-polar. This affects some of the properties of organic compounds.
The main source of the carbon in organic compounds is carbon dioxide in the atmosphere. Plants
use sunlight to convert carbon dioxide and water (inorganic compounds) into sugar (an organic
compound) through the process of photosynthesis.
6CO2(g) + 6H2O(ℓ) → C6H12O6(aq) + 6O2(g)
Plants are therefore able to make their own organic compounds through photosynthesis, while
animals feed on plants or plant products in order to gain the organic compounds that they need to
survive.
Other important sources of carbon are fossil fuels such as coal, petroleum and natural gas. This
is because fossil fuels are themselves formed from the decaying remains of dead organisms.
- The teacher must explain how to represent organic molecules.
There are a number of ways to represent organic compounds. It is useful to know all of these so
that you can recognise a molecule regardless of how it is shown. There are four main ways of
representing a compound in two dimensions (on your page). We will use the examples of two
molecules called 2-methylpropane and butane to help explain the difference between each.

Structural formula
Structural formula is a structural formula of a compound that shows which atoms are
attached to which within the molecule. Atoms are represented by their chemical symbols
and lines are used to represent ALL the bonds that hold the atoms together.
The structural formulae of 2-methylpropane and butane are shown below.

Semi-structural formula
It is possible to understand the structure of an organic molecule without writing out all the carbonhydrogen bonds. This way of writing a structure is called a semi-structural formula.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 6

Condensed structural formula
Condensed structural formula: This notation shows the way in which atoms are bonded
together in the molecule, but DOES NOT SHOW ALL bond lines.
As for a semi-structural representation, the carbon atoms are grouped with the hydrogen atoms
bonded directly to it. The bonds between these groups are not shown. Branched or substituent
groups are shown in brackets after the carbon atom to which they are bonded.
When a condensed structural formula is written for a compound containing double or triple bonds,
the multiple bonds are often drawn as they would be in the structural formula.
Example: 2- butene
CH3CHCHCH3
or CH3CH = CHCH3

Molecular formula
Is a chemical formula that indicates the type of atoms and the correct number of each in a
molecule.
Example: C4H8O
The molecular formula of a compound shows how many atoms of each type are in a molecule.
The number of each atom is written as a subscript after the atomic symbol. The molecular formula
of 2-methylpropane is: C4H10. This means that each molecule of 2-methylpropane consists of four
carbon atoms and ten hydrogen atoms. The molecular formula of butane is also C4H10. Molecular
formula gives no structural information about the compound.
Of course molecules are not two-dimensional so shown below are a few examples of different
ways to represent methane (CH4) and ethane (C2H6).
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 7
Different ways to represent methane
3 dimensional model
(Space-filling models)
3 dimensional model
(ball and stick model)
3 dimensional
structural formula
3 dimensional
tetrahedral
Different ways to represent ethane
3 dimensional model
(Space-filling models)
3 dimensional model
(ball and stick model)
3 dimensional
structural formula
3 dimensional structural
formula
Final activities:
- The teacher must summarise the lesson.
Summary:
o
o
Organic chemistry is the branch of chemistry that deals with organic molecules.
An organic molecule is a molecule that contains carbon atoms (generally bonded
to other carbon atoms as well as hydrogen atoms).
Carbon has a number of unique properties which influence how it behaves and how it
bonds with other atoms:
Carbon has four valence electrons which mean that each carbon atom can form a
maximum of four bonds with other atoms. Because of the number of bonds that carbon can
form with other atoms, organic compounds can be very complex.
Carbon can form bonds with other carbon atoms to form single, double or triple covalent
bonds.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 8
o
Carbon can bond to form straight chain, branched, and cyclic molecules.
o
Because of its position on the periodic table, most of the bonds that carbon forms with
other atoms are covalent.
o
There are a number of ways to represent organic compounds:
o Molecular formula: A chemical formula that indicates the type of atoms and the correct
number of each in a molecule.
o Structural formula: A structural formula of a compound shows which atoms are attached
to which within the molecule. Atoms are represented by their chemical symbols and lines
are used to represent ALL the bonds that hold the atoms together.
o Condensed structural formula: This notation shows the way in which atoms are bonded
together in the molecule, but DOES NOT SHOW ALL bond lines.
o Semi-structural formula: It is possible to understand the structure of an organic
molecule without writing out all the carbon-hydrogen bonds. This way of writing a structure
is called a semi-structural formula
- The teacher must orientate the homework.
Activity 1 (Exercise 4 - 1.1 Page 112, Siyavula, Grade 12):
For each of the following, give the structural formula and the molecular formula.
a)
b)
c)
CH3CH2CH3
CH3CH2CH(CH3)CH3
CH3CH3
Activity 2 (Exercise 4 - 1.2 Page 112, Siyavula, Grade 12):
For each of the following organic compounds, give the condensed structural formula and the
molecular formula.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 9
a)
b)
Activity 3 (Exercise 4 - 1.3 Page 112, Siyavula, Grade 12):
Give two possible structural formulae for the compound with a molecular formula of C 4H10.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 10
Solution of the activities:
Activity 1:
a)
Structural formula:
Molecular formula: C3H8
b) Structural formula:
or
Molecular formula: C5H12
c) Structural formula:
Molecular formula: C2H6
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 11
Activity 2:
a)
Condense structural formula: CH3CHCHCH3
Molecular formula: C4H8
b)
Condense structural formula: CH2CHCH(CH3)CH3
Molecular formula: C5H10
Activity 3:
There are only two possible options:
or
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 12
-ORGANIC MOLECULES
Lesson 2
Topic: Functional Groups and homologous series. Hydrocarbon. Classification of hydrocarbons.
Structural isomers (chain, positional and functional).
Objective:
Learners must be able to:
Define functional group, hydrocarbon and homologous series.
Define saturated and unsaturated compounds.
Define structural isomers.
Define chain isomers, positional isomers and functional isomers.
Identify compounds that are saturated, unsaturated and are isomers (up to 8 carbon
atoms).
Initial activities
- Control of the attendance of the learners.
- The teacher must check and discuss the homework.
Introduction
Because carbon compounds are so numerous, it is convenient to organize them into families that
have structural similarities. Organic molecules are divided into separate families according to their
chemical and physical properties, each with their own characteristic series reaction.
The way in which a compound will react is determined by a particular characteristic of a group of
atoms and the way they are bonded (e.g. double 𝐶 − 𝐶 bond, 𝐶 − 𝑂𝐻 group).
This is called the functional group. This group is important in determining how a compound will
react. The same functional group will undergo the same or similar chemical reaction(s) regardless
of the size of the molecule it is a part of. Molecules can have more than one functional group.
- The teacher must orientate the objectives of the lesson and write the topic on the blackboard.
Development:
- The teacher must explain the term hydrocarbon, saturated and unsaturated.
Certain organic compounds contain only two elements: hydrogen and carbon, and hence are
known as HYDROCARBONS. Hydrocarbons provide the backbone of all organic compounds.
Hydrocarbons are the simplest class of organic compounds.
Hydrocarbon: Organic compounds that consist of hydrogen and carbon only
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 13
Each carbon atom in a hydrocarbon forms a total of four bonds. These bonds are combinations of
single bonds with hydrogen atoms and single or multiple bonds with other carbon atoms.
HYDROCARBONS
Aromatic
Aliphatic
(arenes)
(Which have benzene
ring structures- with
double bonds)
Cyclic compound
Acyclic compound
Ring structures
Chain structure
Cycloalkanes
Saturated
Unsaturate
d
Compounds in which there are no
multiple bonds between C atoms in their
hydrocarbon chains
Alkanes
Compounds with one or more multiple
bonds between C atoms in their
hydrocarbon chains
Alkenes
Alkynes
- The teacher must explain the term functional group and homologous series.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 14
Organic molecules are divided into separate families according to their chemical and physical
properties, each with their own characteristic series reaction.
Functional group: A bond or an atom or a group of atoms that determine(s) the physical
and chemical properties of a group of organic compounds.
It is possible to say that a functional group is a reactive portion of a molecule that undergoes
predictable reactions.
A series of compounds of the same chemical function that have similar properties and structures
but differ in composition in one or more groups –CH2- form a homologous series.
Homologous series: A series of organic compounds that can be described by the same
general formula OR in which one member differs from the next with a CH2 group.
The following table gives the functional groups of some homologous series.
Homologous
series
General
formula
Alkanes
CnH2n+2
Functional group
Structure
No functional group
Example
Description
Only 𝐶 − 𝐻 Ethane
and 𝐶 − 𝐶
single
bonds
Alkenes
C2H2n
Ethene
𝐶=𝐶
double bond
Alkynes
CnH2n-2
𝐶 ≡ 𝐶 triple Ethane
bond
Halo
alkanes CnH2n+1X
(alkyl halides)
(X = F, Cℓ, Br, I)
Alcohols
CnH2n+1OH
Structural formula
Chloromethane
Halogen
atom
bonded to a
saturated 𝐶
atom.
Ethanol
Hydroxide
group
bonded to a
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 15
saturated 𝐶
atom
Aldehydes
CnH2nO
Formyl
group
Ketones
CnH2nO
Propanone
Carbon
(Acetone)
group
bonded to
two 𝐶 atoms
Carboxylic
CnH2nO2
Carboxyl
group
acids
Esters
CnH2nO2
-
Ethanol
Ethanoic acid
(acetic acid or
vinegar)
Ethyl
ethanoate
There are some important points to note as we discuss functional groups:
The beginning of a compound name (prefix) comes from the number of carbons in the
longest chain:
The end of a compound name (suffix) comes from the functional group, e.g. an alkane has
the suffix -ane.
- The teacher must explain the term isomers.
There are organic molecular which have the same number and types of atoms, but they are
arranged in a different ways. These molecules are called isomers.
The molecular formulae of isomers are the same, but have different structures. This phenomenon
is known as isomerism.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 16
Isomerism
No at this level
Structural isomers
Stereoisomers
Organic molecules with the same
molecular formula, but different
structural formulae
Chain Isomers
Optical Isomers
Positional Isomers
Geometrical isomers
Functional
isomers
Same
molecular
formula,
but
different types of
chains, e.g. butane
and
2methylpropane.
Same molecular formula, but
different positions of the side
chain, substituents or functional
groups on the parent chain, e.g. 1chloropropane and 2-chloropropane
or but-2-ene and but-1-ene
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Same
molecular
formula,
but
different functional
groups, e.g. methyl
methanoate
and
ethanoic acid
Page 17
Final activities:
- The teacher must do the conclusions of the lesson by means of activity 1.
Activity 1: One word answers:
Write only the word/term for each of the following descriptions next to the question number.
1.1
Organic compounds that contain only carbon and hydrogen.
1.2
Group of organic compounds all of which have the same functional group and whose
consecutive member differ by –CH2.
1.3
Organic compounds in which the carbon atoms only have single bonds.
1.4
Organic compounds in which the carbon atoms form double or triple bounds.
1.5
Organic molecules with the same molecular formula, but different structural formulae.
1.6
Organic molecules with the same molecular formula, but different types of chains.
1.7
Organic molecules with the same molecular formula, but different positions of the side
chain, substituents or functional groups on the parent chain.
1.8
Organic molecules with the same molecular formula, but different functional groups.
- The teacher must orientate the homework.
Homework:
Activity 2: Multiple-choice questions
Four options are provided as possible answers to the following questions. Each question has
only one correct answer. Choose the correct answer.
2.1
Which one of the following hydrocarbons is saturated?
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 18
Activity 3:
Rewrite the structural formula for each of the organic compound represented in ACTIVITY 2 as
condensed structural formulae.
Activity 4:
Write the molecular formula for each of the organic compounds represented in ACTIVITY 2.
Activity 5 (Exercise 4 - 7.1 page 131, Siyavula book grade 12):
Match the organic compound in Column A with its’ isomer in Column B:
Column A
Column B
1
A
2
B
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 19
3
C
Activity 6 (Exercise 4 - 7.2 page 131, Siyavula book grade 12):
a) Give the ketone isomer of butanol:
b) Give a carboxylic acid that is an isomer of:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 20
Solution of the activities:
Activity 1
1.1
Hydrocarbons
1.2
Homologous series
1.3
Saturated
1.4
Unsaturated
1.5
Structural isomers
1.6
Chain isomers
1.7
Positional isomers
1.8
Functional isomers
Activity 2:
A. Because there are no multiple bonds between C atoms in their hydrocarbon chains.
Activity 3:
A)
CH3CH3
B)
CH3C≡CH
C)
CH2=CH2
D)
𝐶𝐻3 = 𝐶𝐻𝐶𝐻2 𝐶𝐻 = 𝐶𝐻2
Activity 4:
A)
C2H6
B)
C3H4
C)
C2H4
D)
C5H9
Activity 5:
1-C
2-A
3-B
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 21
Activity 6:
a)
or
b)
Two isomers that have one branching methyl group:
or
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 22
One isomer that has two branching methyl groups:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 23
ORGANIC MOLECULES
Lesson 3
Topic: Naming alkanes, alkenes and alkynes.
Objective:
Learners must be able to:
Given the IUPAC name when given the formula for alkanes, alkenes and alkynes.
Write down the formula when given the IUPAC name for alkanes, alkenes and alkynes.
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
Introduction:
In order to give compounds a name, certain rules must be followed. When naming organic
compounds, the IUPAC (International Union of Pure and Applied Chemistry) nomenclature
(naming scheme) is used. This is to give consistency to the names. It also enables every
compound to have a unique name, which is not possible with the common names used (for
example in industry). We will first look at some of the steps that need to be followed when naming
a compound, and then try to apply these rules to some specific examples.
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must explain IUPAC steps to name organic compounds.
A good general rule to follow is to start at the end (the suffix) and work backwards (from right to
left) in the name.
1.
Recognise the functional group in the compound. This will determine the suffix of the
name
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 24
2.
Find the longest continuous carbon chain that contains the functional group (it won’t
always be a straight chain) and count the number of carbon atoms in this chain.
This number will determine the prefix (the beginning) of the compound’s name.
3.
Number the carbons in the longest carbon chain (Important: If the molecule is not an
alkane (i.e. has a functional group) you need to start numbering so that the functional
group is on the carbon with the lowest possible number). Start with the carbon at the end
closest to the functional group.
4.
Look for any branched groups: alkyl groups.
Group
- CH3
- CH2CH3
Name
Methyl
Ethyl
Name them by counting the number of carbon atoms in the branched group, these
groups will all end in -yl.
Note the position of the group on the main carbon chain. If there is more than one of the
same type of branched group then both numbers must be listed (e.g. 2,4 -) and one of
the prefixes must be used.
Important: If the molecule is an alkane the branched group must be on the carbon with
the lowest possible number.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 25
The branched groups must be listed before the name of the main chain in alphabetical
order (ignoring di/tri/tetra).
If there are no branched groups this step can be ignored.
5.
For the alkyl halides the halogen atom is treated in much the same way as branched
groups:
To name them take the name of the halogen atom (e.g. iodine) and replace the -ine with
-o (e.g. iodo).
Give the halogen atom a number to show its position on the carbon chain. If there is
more than one halogen atom the numbers should be listed and a prefix should be used
(e.g. 3,4-diiodo- or 1,2,2-trichloro-).
The halogen atoms must be listed before the name of the main chain in alphabetical
order (ignore di/tri/tetra).
If there are no halogen atoms this step can be ignored.
6.
Combine the elements of the name into a single word in the following order:
branched groups/halogen atoms in alphabetical order (ignoring prefixes)
prefix of main chain
name ending according to the functional group and its position on the longest carbon
chain.
The IUPAC (systematic) names of organic compound have three parts to them:
Prefix
What substituent?
Base
How many
carbons?
Suffix
What family?
- The teacher must explain how to give the name to alkanes with example 1.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 26
Example 1:
Give the IUPAC name for the following compound:
SOLUTION
Step 1: Identify the functional group
The compound is a hydrocarbon with single bonds between the carbon atoms. It is an
alkane and will have the suffix -ane.
Step 2: Find the longest carbon chain
or
There are four carbon atoms in the longest chain. The prefix for this compound is but-.
Step 3: Number the carbons in the carbon chain
Number the carbons in the parent chain beginning with the end of the chain nearest the
substituent.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 27
or
Step 4: Look for any branched groups, name them and give their position on the carbon chain
or
There is a branched group attached to the second carbon atom. In this case the methyl
group is on carbon 2. This group has the formula CH3, which is methane without a
hydrogen atom. However, because it is not part of the main chain, it is given the suffix -yl
(i.e. methyl). The position of the methyl group comes just before its name (see the next
step).
Step 5: Combine the elements of the compound’s name into a single word in the order of
branched group; prefix; name ending according to the functional group
The compound’s name is 2-methylbutane.
-
The teacher must explain with the following example how to write the formula from the
IUPAC name.
Example 2 (Example 3 page 135, Siyavula book, Grade 12):
Draw the semi-structural formula for the organic compound 2,2,4-trimethylhexane
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 28
SOLUTION:
Step 1: Identify the functional group
The name ends in -ane therefore the compound is an alkane.
Step 2: Determine the number of carbon atoms in the longest chain
The longest chain has the prefix hex-. There are therefore 6 carbon atoms in the longest
chain.
𝑪−𝑪−𝑪−𝑪−𝑪−𝑪
Step 3: Number the carbons in the carbon chain
𝑪𝟏 − 𝑪𝟐 − 𝑪𝟑 − 𝑪𝟒 − 𝑪𝟓 − 𝑪𝟔
Step 4: Look for any branched groups and place them on the structure
The compound is 2,2,4-trimethylhexane.
Therefore there are three branched groups - two on carbon 2 and one on carbon 4.
𝑪𝑯𝟑
𝑪𝑯𝟑
𝑪𝟏 − 𝑪𝟐 − 𝑪𝟑 − 𝑪𝟒 − 𝑪𝟓 − 𝑪𝟔
𝑪𝑯𝟑
Step 5: Combine this information and add the hydrogen atoms
Carbon atoms can have four single bonds.
Therefore wherever a carbon atom has less than four bonds draw in hydrogen atoms until
there are four bonds.
𝑪𝑯𝟑
𝑪𝑯𝟑
𝑪𝑯𝟑 − 𝑪 − 𝑪𝑯𝟐 − 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯𝟑
𝑪𝑯𝟑
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 29
- The teacher must explain how to name alkenes with example 3.
Example 3
Write down the IUPAC name for the following compound:
𝑪𝑯𝟑
𝑪𝑯𝟐 = 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯𝟑
𝑪𝑯𝟐 𝑪𝑯𝟑
SOLUTION
Step 1: Identify the functional group
The compound is a hydrocarbon with double bonds between the carbon atoms. It is an
alkene and will have the suffix -ene.
Step 2: Find the longest carbon chain
𝑪𝑯𝟑
𝑪𝑯𝟐 = 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯𝟑
𝑪𝑯𝟐 𝑪𝑯𝟑
There are six carbon atoms in the longest chain. The prefix for this compound is hex-.
Step 3: Number the carbons in the carbon chain
Remember that the carbon atoms must be numbered so that the functional group is at the
lowest numbered carbon atom possible (the end closer to the double bond).
𝑪𝑯𝟑
𝑪𝑯𝟐 = 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯𝟑
1
2
3
4
5
6
𝑪𝑯𝟐 𝑪𝑯𝟑
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 30
Step 4: Look for any branched groups, name them and give their position on the carbon chain.
𝑪𝑯𝟑 methyl
𝑪𝑯𝟐 = 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯𝟑
1
2
3
4
5
6
𝑪𝑯𝟐 𝑪𝑯𝟑
ethyl
There is a methyl in carbon 3 and an ethyl in carbon 4
Step 5: Combine the elements of the compound’s name into a single word in the order of
branched group; prefix; name ending according to the functional group
The branched groups must be listed before the name of the main chain in alphabetical
order (ignoring di/tri/tetra)
The compound’s name is 4-ehtyl-3-methyl-1-hexene.
-
The teacher must explain how to draw the structural and molecular formula for an alkene.
Example 4 (Example 5 page 136 Siyavula book Grade 12):
Draw the structural and molecular formula for the organic compound 3-methylbut-1-ene
SOLUTION
Step 1: Identify the functional group
The name ends in -ene therefore the compound is an alkene.
Step 2: Determine the number of carbon atoms in the longest chain containing the functional
Group.
The longest chain has the prefix but-. There are therefore 4 carbon atoms in the longest
chain. The double bond is in carbon 1.
𝑪=𝑪−𝑪−𝑪
Step 3: Number the carbons in the carbon chain
𝑪𝟏 = 𝑪𝟐 − 𝑪𝟑 − 𝑪𝟒
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 31
Step 4: Look for any branched groups and place them on the structure
The compound is 3-methylbut-1-ene. There is a methyl group at carbon 3.
𝑯
𝑯
𝑪
𝑯
𝑪𝟏 = 𝑪𝟐 − 𝑪𝟑 − 𝑪𝟒
Step 5: Combine this information and add the hydrogen atoms
Carbon atoms can have four single bonds. Therefore wherever a carbon atom has less
than four bonds draw in hydrogen atoms until there are four bonds.
𝑯
𝑯
𝑪
𝑯
𝑯
𝑯
𝑪=𝑪−𝑪−𝑪
𝑯
𝑯
𝑯
𝑯
𝑯
The molecular formula will be C5H10
-
The teacher must use example 5 to explain how to give the IUPAC name to an alkyne from
the structural formula.
Example 5 (Example 7 page 138 Siyavula book, Grade 12):
Give the IUPAC name for the following compound:
𝑯
𝑯
𝑯
𝑯
𝑯−𝑪−𝑪−𝑪−𝑪≡ 𝑪−𝑪−𝑯
𝑯
𝑯
𝑯
𝑯−𝑪−𝑯
𝑯
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 32
SOLUTION:
Step 1: Identify the functional group
There is a triple bond between two of the carbon atoms, so this compound is an alkyne.
The suffix will be -yne.
Step 2: Find the longest carbon chain containing the functional group
The functional group is a triple bond, so the longest chain must contain the triple bond.
𝑯
𝑯
𝑯
𝑯
𝑯
𝑯−𝑪−𝑪−𝑪−𝑪≡ 𝑪−𝑪−𝑯
𝑯
𝑯
𝑯
𝑯
𝑯
𝑯−𝑪−𝑪−𝑪−𝑪≡ 𝑪−𝑪−𝑯
𝑯
𝑯
𝑯−𝑪−𝑯
𝑯
𝑯
𝑯−𝑪−𝑯
𝑯
𝑯
or
There are six carbon atoms in the longest chain. The prefix of the compound’s name will
be hex-.
Step 3: Number the carbons in the longest chain.
In this example, you will need to number the carbons from right to left so that the triple
bond is between carbon atoms with the lowest numbers (the suffix for the compound will
therefore be -2-yne).
𝑯
𝑯
𝑯
𝑯
𝑯−𝑪−𝑪−𝑪−𝑪≡ 𝑪−𝑪−𝑯
6
5
4
3
2
𝑯
𝑯
1
𝑯
𝑯−𝑪−𝑯
𝑯
Step 4: Look for any branched groups, name them and assign the number of the carbon atom to
which the group is attached
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 33
There is a methyl (CH3) group attached to the fifth carbon (remember we have numbered
the carbon atoms from right to left).
𝑯
𝑯
𝑯
𝑯
𝑯−𝑪−𝑪−𝑪−𝑪≡ 𝑪−𝑪−𝑯
6
5
𝑯
4
3
2
𝑯
1
𝑯
𝑯−𝑪−𝑯
𝑯
Step 5: Combine the elements of the name into a single word in the following order: branched
groups; prefix; name ending according to the functional group and its position along the
longest carbon chain
If we follow this order, the name of the compound is 5-methylhex-2-yne.
Draw the structural formula for alkynes is in the same way as was done in example 4 changing
the double bond for triple bond.
Final activities:
- The teacher must summarise the steps to give the IUPAC name to hydrocarbons.
Summary:
Steps when given the IUPAC name of hydrocarbons.
1.
Recognise the functional group in the compound. This will determine the suffix of the
name:
 alkane (-ane)
 alkene (-ene)
 alkyne (-yne)
2.
Find the longest continuous carbon chain that contains the functional group (it won’t
always be a straight chain) and count the number of carbon atoms in this chain.
This number will determine the prefix (the beginning) of the compound’s name.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 34
3.
Number the carbons in the longest carbon chain (Important: If the molecule is not an
alkane (i.e. has a functional group) you need to start numbering so that the functional
group is on the carbon with the lowest possible number). Start with the carbon at the end
closest to the functional group.
4.
Look for any branched groups: functional group.
Group
Name
- CH3
Methyl
- CH2CH3 Ethyl
Note the position of the group on the main carbon chain. If there is more than one of the
same type of branched group then both numbers must be listed (e.g. 2,4 -) and one of
the prefixes must be used.
Important: If the molecule is an alkane the branched group must be on the carbon with
the lowest possible number.
The branched groups must be listed before the name of the main chain in alphabetical
order (ignoring di/tri/tetra).
5.
Combine the elements of the name into a single word in the following order:
branched groups atoms in alphabetical order (ignoring prefixes)
prefix of main chain
name ending according to the functional group and its position on the longest carbon
chain.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 35
- The teacher must orientate the homework:
Homework:
Activity 1
Give the complete IUPAC name for the following organic compounds.
1.1
Write the structural formula for each compound.
Activity 2
Write down the structural isomers for C5H12.
Activity 3
Give the complete IUPAC names for the following compounds.
a)
b)
Activity 4:
Write down the structural formula and the condensed structural formula for the following
compounds.
4.1
3- methylbut-1-ene.
4.2
pent-1-ene.
4.3
5- ethyl – 3- methylhept-1-ene
4.4
3,3 – dimethylbut-1-ene.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 36
4.5
2 – ethylpent-1-ene.
4.6
2,3 -dimethyl pentane.
4.7
methyl propane (2- methyl propane)
4.8
Methyl propene
4.9
4–ethyl–3 – methyl heptane
Activity 5:
Give the IUPAC names for the following structures.
5.1
CH3C≡CH
5.2
CH3CH2CH2C≡CH
5.3
Activity 6:
Write down the structural formula and the condensed structural formula for the following
compounds.
6.1
But-2-yne
6.2
3-methyl-1-butyne (3-methylbut-1-yne)
6.3
hex-3-yne
6.4
3,3-dimethylbut-1-yne
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 37
Solution of the activities:
Activity 1
2,3-dimethylheptane
3-methylpentane
2-methylhexane
1.1
Activity 2
Pentane
2-methylbutane
2,2- dimethyl propane
Activity 3:
a) 2,2-dimethylhept-3-ene
b) 1,4- Hexadiene
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 38
Activity 4:
Name
Structural formula
4.1
𝑯
3- methylbut-1-ene.
𝑯
Condensed structural formula
𝑯
𝑯
𝑯−𝑪−𝑪−𝑪=𝑪−𝑯
𝑪𝑯𝟑 − 𝑪𝑯 − 𝑪𝑯 = 𝑪𝑯𝟐
𝑪𝑯𝟑
𝑯
𝑯−𝑪−𝑯
𝑯
4.2
pent-1-ene.
𝑯
𝑯 𝑯
𝑯
𝑪𝑯𝟐 = 𝑪𝑯 − 𝑪𝑯𝟐 − 𝑪𝑯𝟐 − 𝑪𝑯𝟑
𝑯−𝑪= 𝑪−𝑪−𝑪−𝑪−𝑯
𝑯
𝑯
𝑯
𝑯
4.3
5-ethyl–3-methylhept1-ene
4.4
3,3–dimethylbut-1ene.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 39
4.5
2– ethylpent-1-ene.
4.6
2,3
pentane.
-dimethyl
4.7
Methyl propane
(2- methyl propane)
4.8
Methyl propene
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 40
4.9
4–ethyl–3 – methyl
heptane
Activity 5:
5.1
5.2
5.3
5.4
Propyne
1- pentyne (pent-1-yne)
4- methylhex-2-yne
2-methylhex-3-yne
If the triple bond occurs in the middle of the longest chain, the order of numbering will be
determined by the position of other substituents on the chain. Substituents must always
have the lowest possible numbers.
Activity 6:
6.1
Name
But-2-yne
Structural formula
Condensed structural formula
CH3−C≡C−CH3
(2-butyne)
6.2
(3-methylbut-1-yne)
CH3−CH(CH3)−C ≡C−CH3
3-methyl-1-butyne
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 41
6.3
hex-3-yne
CH3−CH2−C ≡C−CH2− CH3
3-hexyne
6.4
3,3-dimethylbut-1-yne
(CH3)3C−CH2−C ≡C−CH
- ORGANIC MOLECULES
Lesson 4
Topic: Naming carboxylic acids. Esters.
Objective:
Learners must be able to:
Write down the IUPAC name when given the formula for carboxylic acids and esters.
Write down the formula when given the IUPAC name for carboxylic acids and esters.
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and mark the homework.
Introduction:
The teacher must recall the steps to write down the IUPAC name of organic compounds.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 42
1.
Recognise the functional group in the compound. This will determine the suffix of the
name
2.
Find the longest continuous carbon chain that contains the functional group (it won’t
always be a straight chain) and count the number of carbon atoms in this chain.
This number will determine the prefix (the beginning) of the compound’s name.
3.
Number the carbons in the longest carbon chain (Important: If the molecule is not an
alkane (i.e. has a functional group) you need to start numbering so that the functional
group is on the carbon with the lowest possible number). Start with the carbon at the end
closest to the functional group.
4.
Look for any branched groups: Alkyl groups.
Group
Name
- CH3
Methyl
- CH2CH3
Ethyl
Name them by counting the number of carbon atoms in the branched group, these
groups will all end in -yl.
Note the position of the group on the main carbon chain. If there is more than one of the
same type of branched group then both numbers must be listed (e.g. 2,4 -) and one of
the prefixes must be used.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 43
Important: If the molecule is an alkane the branched group must be on the carbon with
the lowest possible number.
The branched groups must be listed before the name of the main chain in alphabetical
order (ignoring di/tri/tetra).
If there are no branched groups this step can be ignored.
5.
For the alkyl halides the halogen atom is treated in much the same way as branched
groups:
To name them take the name of the halogen atom (e.g. iodine) and replace the -ine with
-o (e.g. iodo).
Give the halogen atom a number to show its position on the carbon chain. If there is
more than one halogen atom the numbers should be listed and a prefix should be used
(e.g. 3,4-diiodo- or 1,2,2-trichloro-).
The halogen atoms must be listed before the name of the main chain in alphabetical
order (ignore di/tri/tetra).
If there are no halogen atoms this step can be ignored.
6.
Combine the elements of the name into a single word in the following order:
branched groups/halogen atoms in alphabetical order (ignoring prefixes)
prefix of main chain
name ending according to the functional group and its position on the longest carbon
chain.
The IUPAC systematic names of organic compound have three parts to them:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 44
prefix
What substituent?
base
How many
carbons?
suffix
What family?
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must explain how to write down the IUPAC name when given the formula for
carboxylic acids.
Carboxylic acids are the most important class of organic acids. In ancient times man used the
acetic acid in vinegar. Other known acids are tartaric to be found in grapes and citrus that is
contained in the lemon juice.
Organic acids can serve as food seasoning, laboratory reagent, to produce fabric, etc.
There are very important derivatives of carboxylic acid called esters which are found widely in
nature in fat and oils and are responsible for the aroma of many fruits.
The combination of carbonyl group (C=O) and a hydroxyl group (-OH) on the same carbon atom
is called a carboxyl group. Compound containing the carboxyl group are distinctly acidic and are
called carboxylic acids.
CARBOXYLIC ACIDS (General formula: CnH2nO2)
Functional group:
Carboxylic acids are classified according to the substituent bonded to the carboxyl group. An
aliphatic acid has an alkyl group bonded to the carboxyl group, and an aromatic acid has an
aryl group. The simplest acid is formic acid, with a hydrogen atom bonded to the carboxyl group.
Example 1
Write down the IUPAC name and molecular formula for the following compound:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 45
𝑯
𝑯
𝑯
𝑯
𝑪
𝑯
𝑯 𝑯
𝑶
𝑪−𝑪−𝑪−𝑪−𝑪−𝑶
𝑯
𝑯
𝑯
𝑯 𝑯
𝑪
𝑯
𝑯
SOLUTION
Step 1: Identify the functional group
The compound has a -COOH group and is therefore a carboxylic acid. The suffix will be
-oic acid.
Step 2: Find the longest carbon chain that contains the functional group.
𝑯
𝑯
𝑯
𝑯
𝑪
𝑯
𝑯 𝑯
𝑶
𝑪−𝑪−𝑪−𝑪−𝑪−𝑶
𝑯
𝑯 𝑯
𝑯
𝑪
𝑯
𝑯
𝑯
There are five carbon atoms in the longest chain that contains the functional group, and only single
bonds between carbon atoms. The prefix for this compound is pentan-.
Step 3: Number the carbon atoms in the carbon chain
The carbon atoms will be numbered from right to left so that the carboxylic acid functional
group has the lowest numbered carbon atom.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 46
𝑯
𝑯
𝑯
𝑯
𝑪
𝑯
𝑯 𝑯
𝑶
𝑪5− 𝑪4− 𝑪3− 𝑪2− 𝑪 1− 𝑶
𝑯
𝑯
𝑯
𝑯 𝑯
𝑪
𝑯
𝑯
Step 4: Look for any branched groups
𝑯
𝑯
𝑯
𝑯
𝑪
𝑯
𝑯 𝑯
𝑶
𝑪5− 𝑪4− 𝑪3− 𝑪2− 𝑪 1− 𝑶
𝑯
𝑯 𝑯
𝑯
𝑪
𝑯
𝑯
𝑯
There are two methyl groups, one in carbon 3 and the other in carbon 4. So it will be -3,4dimethylStep 5: Name the halogen atoms and assign the number for the carbon atom attached to it
There are no halogens in these compounds.
Step 6: Combine the elements of the name into a single word in the following order:
Branched groups; prefix, name ending according to the functional group
3,4-dimethylpentanoic acid
- The teacher must explain how to write down the IUPAC name when given the formula for
esters.
When an alcohol reacts with a carboxylic acid, an ester is formed. This process is called
esterification.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 47
The teacher can use the short video (Carboxylic Acids and Esters | The Chemistry Journey | The
Fuse School). https://www.youtube.com/watch?v=ecHKrd7hn4U,
Most esters have a characteristic smell. In the reaction a molecule of water is removed from the
two compounds and a new bond is formed between what remains of the alcohol and the carboxylic
acid. A catalyst is required in this reaction; in this case it must be an inorganic acid (e.g. H 2SO4).
Esters have the general formula: CnH2nO2. This general formula can also be applied to carboxylic
acids, but the more complex general formula for esters alone is not covered at this level.
Some common uses for esters are:
in cosmetics and beauty products because they typically have a fruity smell, making them
good as artificial flavourants and scents.
in nail varnish remover and model plane glue.
as solvents for non-water soluble compounds (e.g. oils, resins) because the ester of a
specific carboxylic acid will be less water soluble than the carboxylic acid.
as plasticisers because esters can make a compound less brittle, and more flexible.
Example 2
Write down the IUPAC name for the following compound:
𝑯 𝑯
𝑶 𝑯
𝑯−𝑪−𝑪−𝑶−𝑪−𝑪−𝑯
𝑯 𝑯
𝑯
SOLUTION
Step 1: Identify the functional group
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 48
There is a -C=O (carbonyl) group as well as an oxygen atom bonded to the carbon atom
of the carbonyl and another carbon atom. This is therefore an ester and the suffix is oate.
Step 2: Determine which part is from the alcohol and which is from the carboxylic acid
An ester is a carboxylic acid derivative. Divide the molecule in two with the carbonyl group
on one side and the oxygen bonded to two carbon atoms on the other.
𝑶 𝑯
𝑯 𝑯
𝑯−𝑪−𝑪−𝑶−𝑪−𝑪−𝑯
𝑯
𝑯 𝑯
The part containing the oxygen atom bonded to two different carbon atoms was formed
from the alcohol and is on the left here. The part containing the carbonyl group was
formed from the carboxylic acid and is on the right here.
𝑯 𝑯
𝑶 𝑯
𝑯−𝑪−𝑪−𝑶−𝑪−𝑪−𝑯
𝑯 𝑯
From alcohol
𝑯
From carboxylic acid
Step 3: Number the carbon atoms on the carbon chains
There are two carbon atoms in the left-hand chain (from the alcohol). Therefore this will be
ethyl. There are two carbon atoms in the right-hand chain (from the carboxylic acid)
therefore the prefix will be ethan-.
Step 4: Combine the elements of the compound’s name into a single word in the order of chain
from the alcohol; prefix (from chain containing carbonyl functional group), name ending
according to functional group
The compound’s name is ethyl ethanoate
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 49
Example 3 (Example 24 page 256, Siyavula book Grade 12):
Draw the structural representations for the organic compound ethyl hexanoate.
SOLUTION:
Step 1: Identify the functional group
The compound has the suffix -oate. It is therefore an ester and has a -C=O (carbonyl)
group as well as an oxygen atom bonded to the carbon atom of the carbonyl and another
carbon atom.
Step 2: Determine which part is from the alcohol and which is from the carboxylic acid
The ethyl tells us that there are two carbon atoms in the part of the chain from the alcohol.
The prefix hex- tells us that there are six carbon atoms from the part of the chain from the
carboxylic acid.
𝑪−𝑪
𝑪−𝑪−𝑪−𝑪−𝑪−𝑪
Step 3: Place the functional group as well as any branched groups.
The oxygen atom bonded to two different carbon atoms is located between the two
sections. The -C=O (carbonyl) group is located at the first carbon atom of the carboxylic
acid chain.
𝑶
𝑪−𝑪−𝑶−𝑪−𝑪−𝑪−𝑪−𝑪−𝑪
Step 4: Combine this information and add the hydrogen atoms.
𝑯
𝑯
𝑯
𝑶
𝑯
𝑯
𝑯
𝑯
𝑯
𝑪−𝑪−𝑶−𝑪−𝑪−𝑪−𝑪−𝑪−𝑪
𝑯
𝑯
𝑯
𝑯
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
𝑯
𝑯
𝑯
𝑯
Page 50
Final activities:
- The teacher must do the conclusion of the lesson.
Summary
 General formula of carboxylic acids: CnH2nO2)
 The carboxylic acid functional group has the formula –COOH
 The functional group of carboxylic acids is:
 The carboxylic functionality has the highest priority according to the
IUPAC-system of nomenclature.
 An ester molecule consists of an alcohol and a carboxylic acid condensed
together.
 Functional group of esters is:
 Esterification between a carboxylic acid and an alcohol is an equilibrium reaction .
The teacher must orientate the homework.
Homework:
Activity 1
Write down the IUPAC name for the following organic compounds.
1.1
1.2.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 51
Activity 2:
Give the structural formula of the following organic compounds
2.1
2.2
Pentanoic acid
Propyl ethanoate
Activity 3:
Write down the IUPAC names for the following compounds:
3.1
3.2
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 52
Solution of the activities:
Activity 1
1.1
Ethyl propanoate
1.2
Butanoic acid
Activity 2
2.1
2.2
Activity 3
2.1
Octyl butanoate
2.2
Hexanoic acid
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 53
ORGANIC MOLECULES
Lesson 5
Topic: Intermolecular forces. Physical properties (melting point, boiling point and vapour
pressure). Physical properties and types of functional groups.
Objective:
Learners must be able to:
Recognize and apply to given examples the relationship between:
 Physical properties and intermolecular forces (ethanol, dimethyl ether, ethanoic
acid, ethane).
 Physical properties and number and type of functional groups (ethanol, dimethyl
ether, ethanoic acid, ethane, chloro-ethane).
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
- Baseline assessment.
1. Define organic molecules.
Organic molecules are molecules containing carbon atoms.
2. Define the term functional group.
Functional group is a bond or an atom or a group of atoms that determine(s) the physical
and chemical properties of a group of organic compounds.
Introduction:
The physical properties of all organic compounds mainly depends on their structure, qualitative
and quantitative composition, type of chemical bonding, order of the joining of the atoms in the
molecule and special disposition of them.
What physical properties do you know?
Some physical properties of chemical compounds we are going to study include the following:
Phase
Boiling and melting points
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 54
Volatility
Vapour pressure
Viscosity
Solubility
Physical properties of organic compounds are mainly determined by the strength of
intermolecular forces.
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
-
The teacher must give the definition of the following physical properties of chemical
compounds:
Phase (whether the compound is a gas, liquid or solid at room temperature)
Boiling and melting points (at which temperature the liquid boils and the solid melts).
Volatility (how easy the liquid vaporises)
Vapour pressure (the pressure at which the vapour of a substance is in dynamic
equilibrium with its liquid or solid form. Substances with high vapour pressure are volatile,
and such substances have high volatility).
Viscosity (a measure of the resistance to flow of the liquid-the higher the viscosity the less
easy it flows).
Solubility (how easily it dissolves in non-polar solvents or in polar solvents)
- The teacher must recall intermolecular forces from grade 11.The short video “Major
intermolecular forces” must be used for this purpose.
https://www.youtube.com/watch?v=S8QsLUO_tgQ
Intermolecular forces are forces that act between molecules.
Intramolecular forces occur within a molecule and include covalent bonds between atoms.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 55
Remember from Grade 11 that a dipole molecule is a molecule that has its charge unevenly
distributed. One end of the molecule is slightly positive and the other is slightly negative. An
overview of the different types of intermolecular forces that are discussed previously are given
below:
1. Dipole-dipole forces
When one dipole molecule comes into contact with another dipole molecule, the positive pole of
the one molecule will be attracted to the negative pole of the other, and the molecules will be held
together in this way.
One special case of this is hydrogen bonding:
 Hydrogen bonds
As the name implies, this type of intermolecular bond involves a hydrogen atom. When a
molecule contains a hydrogen atom covalently bonded to a small, highly electronegative
atom (e.g. O, N or F) this type of intermolecular force can occur. The highly
electronegative atom on one molecule attracts the hydrogen atom on a nearby molecule.
2. Van der Waals forces
 Induced-dipole forces
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 56
In non-polar molecules the electronic charge is usually evenly distributed but it is possible
that at a particular moment in time, the electrons might not be evenly distributed
(remember that the electrons are always moving in their orbitals).
The molecule will have a temporary dipole. When this happens, molecules that are next to each
other attract each other very weakly.
𝒆−
+
𝜹
−
𝜹
𝒆−
+
𝜹
𝜹−
I2
I2
𝒆−
𝒆−
 Dipole-induced-dipole forces
These forces exist between dipoles and non-polar molecules. The dipole induces a dipole
in the non-polar molecule leading to a weak, short lived force which holds the compounds
together.
- The teacher must explain the relationship between physical properties and intermolecular
forces (ethanol, dimethyl ether, ethanoic acid, ethane).
We are going to study the relationship between the type of intermolecular forces and some of the
physical properties of the molecules.
The polarity of molecules determines the forces of attraction between the molecules in the liquid
state. Polar molecules are attracted by the opposite charge effect (the positive end of one
molecule is attracted to the negative end of another molecule). Molecules have different degrees
of polarity as determined by the functional group present.
For compounds of comparable molecular mass, the more polar the functional group, the
stronger the intermolecular forces.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 57
Polarity ranking of functional groups (studied):
𝐂𝐚𝐫𝐛𝐨𝐧𝐱𝐢𝐥𝐢𝐜 𝐚𝐜𝐢𝐝 > 𝐴𝑙𝑐𝑜ℎ𝑜𝑙 > 𝐾𝑒𝑡𝑜𝑛𝑒 > 𝐴𝑙𝑑𝑒ℎ𝑦𝑑𝑒 > 𝐸𝑠𝑡𝑒𝑟 > 𝐻𝑎𝑙𝑜𝑎𝑙𝑘𝑎𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑦𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑎𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑒𝑛𝑒
Functional group
Carboxylic acid
Alcohol
Explanation
Example
Hydrogen bonds can be Ethanoic Acid
formed and accepted on two
atoms. Carboxylic acids also
tend to form dimers due to
two
hydrogen
bonds
between
two
acid
molecules. This leads to
increased molecular size
resulting in stronger van der
Waals forces (dipole-dipole
& dispersion) between two
dimers
Alcohols have one site for
hydrogen bonding. In
addition, van der Waals
Ethanol
forces
(dipole-dipole
&
dispersion)
also
exist
between molecules.
Ketone
Since
ketones
and Propanoane
aldehydes lack hydroxyl
groups, they are incapable
of intermolecular hydrogen
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 58
bonds and are less polar
than alcohols.
Aldehyde
Boiling
points
are
considerable higher than
that of alkanes due to weak
intermolecular dipole-dipole
forces. The carbonyl group
is polar since oxygen is more
electronegative than carbon
Ethanal
and forms a partially
charged dipole. In Aldehyde
addition dispersion forces
also
exist
between
molecules.
Ester
The ester functional group
has a similar character to the
ketone
and
aldehyde
functional groups. Esters are
the least polar of the three.
Esters can form dipoledipole forces and dispersion
forces.
Haloalkanes
Somewhat polar molecules
due to the C-X bond
(X=halogen). Van der Waals
dipole-dipole forces exist
between the polar part of the
molecules and ordinary Van
der Waals forces between
non-polar alkyl parts.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 59
Alkane
Non-polar molecules - only
weak van der Waals forces
(dispersion forces)
alkene
Non-polar molecules - weak
van der Waals forces
(dispersion forces). For the
same
number of C atoms, boiling
points are slightly lower
than that of alkanes.
Dispersion
forces depend on the shape
of the molecule & the
number of electrons it
contains.
Each alkene has 2 fewer
electrons than the alkane
with the same number of C
atoms.
alkyne
Non-polar molecules - weak
van der Waals forces
(dispersion forces). Alkyne
molecules are more
polarisable due to the four
pi-electrons that are loosely
held. Therefore alkynes
have higher boiling points
than alkanes and alkenes
with the
same number of C atoms in
their chains .
For compounds of the same functional group the intermolecular forces depend on:
Numbers of functional group.
Surface area (chain length).
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 60
Symmetry (Branched chain).
Polarisability (size of the atoms)
BOILING AND MELTING POINTS
Intermolecular forces affect the boiling and melting points of substances. Substances with weak
intermolecular forces will have low melting and boiling points as less energy (heat) is needed to
overcome these forces. Those with strong intermolecular forces will have high melting and boiling
points as more energy (heat) is required to overcome these forces. When the temperature of a
substance is raised beyond its’ melting or boiling point the intermolecular forces are not weakened.
Rather, the molecules have enough energy to overcome those forces.
We can define the boiling point as the temperature at which the vapour pressure of a
substance equals the pressure above it surface (atmospheric pressure)
Melting point is the temperature at which a solid is converted to its liquid phase. Energy is
needed to overcome the attractive forces in the more ordered crystalline solid.
Boiling and melting points of organic compounds increase with increase in strength of
intermolecular forces - more energy is needed to break the intermolecular forces.
NUMBER OF FUNCTIONAL GROUPS
The effect of a functional group increases when the number of functional groups increases,
resulting in higher boiling and melting points.
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐎𝐇
ethanol
bp = 78°C
mp = -114 °C
One side for H bonding
𝐂𝐇𝟐 𝐎𝐇𝐂𝐇𝟐 𝐎𝐇
ethan-1,2-dinol
bp = 197°C
mp = -13 °C
Two side for H bonding
Increase number of the same functional group,
increasing boiling point
Note: Both molecules also
experience van der Waals
forces (dispersion; nonpolar C chain; dipoledipole: polar C-O and O-H
bonds), but the hydrogen
bonding (due to the O-H
bond) is most significant.
VAPOUR PRESSURE
When a substance is in the liquid or solid state there will be some molecules in the gas state.
These molecules have enough energy to overcome the intermolecular forces holding the majority
of the substance in the liquid or solid phase. These gas molecules exert a pressure on the liquid
or solid (and the container) and that pressure is the vapour pressure of that compound. The
weaker the intermolecular forces within a substance the higher the vapour pressure will be.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 61
Compounds with higher vapour pressures have lower flash points and are therefore more
flammable.
Vapour pressure is the pressure exerted (at a specific temperature) on a solid or liquid
compound by molecules of that compound that are in the gas phase.
Compounds with high boiling points have low vapour pressures. Vapour pressure decreases
with increase in strength of intermolecular forces.
POLARITY OF FUNCTIONAL GROUPS
For compounds of comparable molecular mass, the more polar the functional group, the stronger
the intermolecular forces, the lower the vapour pressure.
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
pentane
72 g·mol-1
vp = 441 mmHg (21°C)
Wan der Waals forces
(only dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝐎
butanal
72 g·mol-1
vp = 90 mmHg (21°C)
Wan der Waals forces
(dispersion + dipole-dipole)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐎𝐇
butan-1-ol
74 g·mol-1
vp = 4,5 mmHg (21°C)
Wan der Waals forces (dispersion +
dipole-dipole) + Hydrogen bonds
Increase polarity of functional group, increase strength of intermolecular forces,
decrease vapour pressure
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 62
NUMBER OF FUNCTIONAL GROUPS:
The effect of a functional group increases when the numbers of functional groups increases,
resulting in lower vapour pressures.
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐎𝐇
ethanol
vp = 59,02 mmHg
(25°C)
One side for H bonding
𝐂𝐇𝟐 𝐎𝐇𝐂𝐇𝟐 𝐎𝐇
ethan-1,2-dinol
vp = 0,07 mmHg
(25°C)
Two side for H bonding
Increase number of the same functional group,
decreasing vapour pressure
Note: Both molecules also
experience van der Waals
forces (dispersion; nonpolar C chain; dipoledipole: polar C-O and O-H
bonds), but the hydrogen
bonding (due to the O-H
bond) is most significant.
DENSITY
Density is a measure of the mass per unit of volume.
The stronger the intermolecular forces (between molecules), the shorter the distance between
those molecules and the denser it is. Think of a solid. It has a high density because it holds its
shape and volume. The particles are held together by ionic forces of attraction as opposed to
dipole-dipole forces or LDFs. Ionic forces are the strongest type of intermolecular force. Thus,
solids which use it, have the highest density. It is an inverse relationship, and density depends on
the strength of intermolecular forces.
Final activities
- The teacher must summarise the lesson.
Summary:
Intramolecular forces occur within a molecule and include covalent bonds between
atoms.
Intermolecular forces are forces that act between molecules.
 Dipole-dipole forces
o Hydrogen bond
 Van der Waals forces
o Induced-dipole forces
o Dipole-induced-dipole forces
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 63
Physical properties of organic compounds are mainly determined by the strength of
intermolecular forces.
The more polar the functional group, the stronger the intermolecular forces.
Polarity ranking of functional groups (studied):
𝐂𝐚𝐫𝐛𝐨𝐧𝐱𝐢𝐥𝐢𝐜 𝐚𝐜𝐢𝐝 > 𝐴𝑙𝑐𝑜ℎ𝑜𝑙 > 𝐾𝑒𝑡𝑜𝑛𝐞 > 𝐴𝑙𝑑𝑒ℎ𝑦𝑑𝑒 > 𝐸𝑠𝑡𝑒𝑟 > 𝐻𝑎𝑙𝑜𝑎𝑙𝑘𝑎𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑦𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑎𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑒𝑛𝑒
Boiling and melting points of organic compounds increase with increase in strength
of intermolecular forces - more energy is needed to break the intermolecular forces.
 If the number of functional groups increases, then it will be result in higher boiling and
melting points.
Vapour pressure is the pressure exerted (at a specific temperature) on a solid or
liquid compound by molecules of that compound that are in the gas phase. Vapor
pressure or equilibrium vapor pressure is the pressure of a vapor in thermodynamic
equilibrium with its condensed phases in a closed container. All liquids and solids have a
tendency to evaporate into a gaseous form, and all gases have a tendency to condense
back to their liquid or solid form
Compounds with high boiling points have low vapour pressures. Vapour pressure
decreases with increase in strength of intermolecular forces.
 Increase number of the same functional group, decreasing vapour pressure.
- The teacher must orientate the homework.
Homework:
Activity 1
The chart below shows the boiling points of various alcohols.
Alcohol
Methanol
Ethanol
Propan-1-ol
Butan-1-ol
Pentan-1-ol
Boiling point 0C
64.5
78.3
97
117
138
1.1
Define functional group.
1.2
What is the functional group of alcohols?
1.3
Plot a graph of n (the number of carbons in each alcohol) against the boiling point of each
alcohol.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 64
1.3.1 What trend does this graph show? Explain your observation. (This can be an
introduction for the next lesson).
1.4
The boiling point of propan-1,2-diol is 189 0C and that of propan-1,2,3-triol is 290 0C.
1.4.1 Draw the structural formulae for these two compounds.
ol
1.5
1.4.2 Compare the boiling point of these compounds with the boiling point of propan-1and explain the differences between these values.
Ethanol is the alcohol that is present in alcoholic drinks. These drinks have been used by
human beings for centuries, but excess alcohol consumption can be dangerous for you
and for other people. State at least 3 reasons why people should avoid excessive alcohol
consumption.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 65
Solution of the activities:
Functional group is a bond or an atom or a group of atoms that determine(s) the physical
and chemical properties of a group of organic compounds.
1.2
Hydroxyl (-OH) group
1.3
Boiling point Vs number of carbon atoms
Boiling point 0C
1.1
150
100
50
0
0
1
2
3
4
n (number of carbons)
5
6
1.3.1 The boiling point increases with increasing chain length. As the chain length increases the
intermolecular forces increases (the bigger the chain the bigger the intermolecular (Van
der Waals) forces. This means that more heat energy is required to separate the particles
and turn them into a gas.
1.4.1
propan-1,2-diol
Propan-1,2,3-triol
1.4.2 The difference in boiling points between these three compounds is due to the increase of
hydrogen bonding that occurs at the hydroxyl group. Hydrogen bonding increases the
boiling point as more energy is required to separate these bonds, so the compound with
the greatest number of hydroxyl groups in a particular chain will have the highest boiling
point.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 66
1.5
Excessive alcohol use can lead to:







Brain and liver damage (Physical damage)
Depression and emotional damage (mental damage)
Foetal alcohol syndrome if pregnant woman drinks heavily.
Increase of violence
Car accidents
Poverty
Unemployment
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 67
ORGANIC MOLECULES
Lesson 6
Topic: Physical properties and chain length. Physical properties and branched chains.
Objective:
Learners must be able to:
Recognize and apply to given examples the relationship between:
 Physical properties and chain length.
 Physical properties and branched chains
Initial activities
- Control of the attendance of the learners.
- The teacher must control and discuss (mark) the homework.
Base line assessment.
1. Define the following physical properties:
o boiling point.
- Boiling point is the temperature at which the vapour pressure of a substance
equals the pressure above it surface (atmospheric pressure)
o melting point.
- Melting point is the temperature at which a solid is converted to its liquid
phase.
o vapour pressure
- Vapour pressure (the pressure at which the vapour of a substance is in dynamic
equilibrium with its liquid or solid form.
OR
- Vapour pressure is the pressure exerted (at a specific temperature) on a solid
or liquid compound by molecules of that compound that are in the gas phase
Introduction:
- The teacher must summarise lesson 25 about intermolecular forces.
Summary:
Intramolecular forces occur within a molecule and include covalent bonds between
atoms.
Intermolecular forces are forces that act between molecules.
 Dipole-dipole forces
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 68
o Hydrogen bond
 Van der Waals forces
o Induced-dipole forces
o Dipole-induced-dipole forces
Physical properties of organic compounds are mainly determined by the strength of
intermolecular forces.
The more polar the functional group, the stronger the intermolecular forces.
Polarity ranking of functional groups (studied):
𝐂𝐚𝐫𝐛𝐨𝐧𝐱𝐢𝐥𝐢𝐜 𝐚𝐜𝐢𝐝 > 𝐴𝑙𝑐𝑜ℎ𝑜𝑙 > 𝐾𝑒𝑡𝑜𝑛𝐞 > 𝐴𝑙𝑑𝑒ℎ𝑦𝑑𝑒 > 𝐸𝑠𝑡𝑒𝑟 > 𝐻𝑎𝑙𝑜𝑎𝑙𝑘𝑎𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑦𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑎𝑛𝑒𝑠 > 𝐴𝑙𝑘𝑒𝑛𝑒
Boiling and melting points of organic compounds increase with increase in strength
of intermolecular forces - more energy is needed to break the intermolecular forces.
 If the number of functional groups increases, then it will be result in higher boiling and
melting points.
Vapour pressure is the pressure exerted (at a specific temperature) on a solid or
liquid compound by molecules of that compound that are in the gas phase.
Compounds with high boiling points have low vapour pressures. Vapour pressure
decreases with increase in strength of intermolecular forces.
 Increase number of the same functional group, decreasing vapour pressure
Remember that for compounds of the same functional group the intermolecular forces depend on:
Numbers of functional group.
Surface area (chain length).
Symmetry (Branched chain).
Polarise-ability (size of the atoms)
In lesson 28 we learned the relationship between the type and number of functional group and
some physical properties (melting point, boiling point, vapour pressure).
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must explain the relationship between some physical properties and chain length.
Remember that Boiling and melting points of organic compounds increase with increase in
strength of intermolecular forces - more energy is needed to break the intermolecular forces.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 69
BOILING POINTS:
SURFACE AREA:
For compounds of the same functional group, the larger the surface area (or the longer the
chain) the higher the boiling point.
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟑
propane
44 g·mol-1
bp = -44 °C
Wan der Waals forces
(only dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
butane
58 g·mol-1
bp = -0,5 °C
Wan der Waals forces
(only dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
pentane
72 g·mol-1
bp = 36 °C
Wan der Waals forces
(only dispersion)
Increase the surface area increase the boiling point
SYMMETRY
Similarly, the smaller the, surface area (or the more branching), the lower the boiling point.
Boiling points of isomers decrease with branching (as shown below).
Increase in branching, decrease surface area, decrease boiling point
𝐂𝐇𝟑
𝐂𝐇𝟑 − 𝐂 − 𝐂𝐇𝟑
𝐂𝐇𝟑
2,2-dimethylpropane
72 g·mol-1
bp = 10 °C
Wan der Waals forces
(only dispersion)
𝐂𝐇𝟑 𝐂𝐇𝐂𝐇𝟐 𝐂𝐇𝟑
𝐂𝐇𝟑
2-methylbutane
72 g·mol-1
bp = 30 °C
Wan der Waals forces
(only dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
pentane
72 g·mol-1
bp = 36 °C
Wan der Waals forces
(only dispersion)
Decrease in brunching increase the surface area increase the boiling point
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 70
Although the type of intermolecular force remains the same, there are more sites for these forces
with increasing surface area. The forces might be the same in strength, but because there are
more sites where they can act, the total effect results in stronger intermolecular forces.
POLARISABILITY
For compounds of the same functional group, the more polarisable the atoms, the higher the
boiling point. Larger atoms, e.g. iodine atoms, are easier to polarise and can form stronger
intermolecular forces than smaller atoms, e.g. fluorine atoms.
𝐂𝐇𝟑 𝐅
Fluoromethane
bp = -78 °C
F less polarisable; van der Waals
forces (dispersion & dipole-dipole)
𝐂𝐇𝟑 𝐈
iodomethane
bp = 42 °C
I more polarisable; van der Waals
forces (dispersion & dipole-dipole)
Increase polarise-ability of atoms, increase boiling point
MELTING POINT
SURFACE AREA
For compounds of the same functional group, the longer the chain the higher the melting point.
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟑
propane
44 g·mol-1
bp = -44 °C
mp = -190 °C
Wan der Waals forces (only
dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
butane
58 g·mol-1
bp = -0,5 °C
mp = -138 °C
Wan der Waals forces (only
dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
pentane
72 g·mol-1
bp = 36 °C
mp = 129,8 °C
Wan der Waals forces (only
dispersion)
Increase the surface area increase the boiling and melting point
SYMMETRY
For compounds of the same functional group and similar molecular masses, melting points
increase with increase in symmetry. A compact symmetrical molecule packs well into a crystalline
lattice.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 71
𝐂𝐇𝟑
𝐂𝐇𝟑 𝐂𝐇𝐂𝐇𝟐 𝐂𝐇𝟑
𝐂𝐇𝟑 − 𝐂 − 𝐂𝐇𝟑
𝐂𝐇𝟑
2-methylbutane
72 g·mol-1
mp = -160 °C
Wan der Waals forces
(only dispersion)
𝐂𝐇𝟑
2,2-dimethylpropane
72 g·mol-1
mp = -17 °C
Wan der Waals forces
(only dispersion)
More symmetrical molecule, increase in melting point
POLARISABILITY
For compounds of the same functional group, the more polarisable the atoms, the higher the
melting point. Larger atoms, e.g. iodine atoms, are easier to polarise and can form stronger
intermolecular forces than smaller atoms, e.g. fluorine atoms.
𝐂𝐇𝟑 𝐅
Fluoromethane
mp = -142 °C
F less polarisable; van der Waals
forces (dispersion & dipole-dipole)
𝐂𝐇𝟑 𝐈
iodomethane
mp = -66,5 °C
I more polarisable; van der Waals
forces (dispersion & dipole-dipole)
Increase polarise-ability of atoms, increase melting point
VAPOUR PRESSURE:
Remember that:
Vapour pressure is the pressure exerted (at a specific temperature) on a solid or liquid
compound by molecules of that compound that are in the gas phase.
Compounds with high boiling points have low vapour pressures. Vapour pressure decreases
with increase in strength of intermolecular forces.
SURFACE AREA
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 72
For compounds of the same functional group, the larger the surface area (or the longer the chain)
the lower the vapour pressure.
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟑
propane
44 g·mol-1
vp = 6 616 mmHg (21 °C)
Wan der Waals forces (only
dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
butane
58 g·mol-1
vp = 1 607 mmHg (21°C)
Wan der Waals forces (only
dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
pentane
72 g·mol-1
vp = 441 mmHg (21 °C)
Wan der Waals forces (only
dispersion)
Increase the surface area, decrease vapour pressure
SYMMETRY
The smaller the, surface area (or the more branching), the lower the boiling point and so the
higher the vapour pressure
𝐂𝐇𝟑
𝐂𝐇𝟑 − 𝐂 − 𝐂𝐇𝟑
𝐂𝐇𝟑
2,2-dimethylpropane
72 g·mol-1
vp = 1,29×10+3 mm Hg
(25°C) Wan der Waals
forces (only dispersion)
𝐂𝐇𝟑 𝐂𝐇𝐂𝐇𝟐 𝐂𝐇𝟑
𝐂𝐇𝟑
2-methylbutane
72 g·mol-1
vp = 689 mmHg (25 °C)
Wan der Waals forces
(only dispersion)
𝐂𝐇𝟑 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟐 𝐂𝐇𝟑
pentane
72 g·mol-1
vp = 514 mmHg (25 °C)
Wan der Waals forces
(only dispersion)
Decrease in brunching increase the surface area decrease vapour pressure
POLARISABILITY OF ATOMS
For compounds of the same functional group, the more polarisable the atoms, the lower the vapour
pressure. Larger atoms, e.g. iodine atoms, are easier to polarise and can form stronger
intermolecular forces than smaller atoms, e.g. fluorine atoms.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 73
𝐂𝐇𝟑 𝐅
Fluoromethane
vp = 24 750 mmHg (25 °C)
F less polarisable; van der Waals forces
(dispersion & dipole-dipole)
𝐂𝐇𝟑 𝐈
iodomethane
vp = 400 mmHg (25 °C)
I more polarisable; van der Waals forces
(dispersion & dipole-dipole)
Increase polarise-ability of atoms, decrease vapour pressure
Final activities:
- The teacher must summarise the work done about physical properties
Boiling points
Relationship
Explanation
The stronger the intermolecular The stronger the intermolecular forces the more energy that is
forces the higher the boiling point. required to separate the substance into its individual molecules
(particles), the higher the boiling point.
The longer the length the higher The longer the length of the chain the larger the surface area the
the boiling point and the melting stronger the intermolecular (Van der Waals) forces, the greater the
point
energy required to separate the substance into its individuals
molecules, the higher the boiling point.
The more the branches the lower The more the branches the more spherical the molecules, the
the boiling point.
smaller the surface area, the weaker the intermolecular (Van der
Waals) forces, the lower boiling point
The more number of halogen (Cl,
Br, I) or OH groups or COOH
groups or CO groups or CHO
groups the higher the boiling point.
The more the number of halogens (Cl, Br, I) or OH groups or COOH
groups or CO groups or CHO groups the stronger the
intermolecular forces (hydrogen bonds or Van der Waals forces)the
higher the boiling point
The stronger the intermolecular
forces the higher the melting point
The stronger the intermolecular forces the more the energy that is
required to separate the substance in to its individual molecules
(particles) the higher the melting point.
The stronger the intermolecular
forces the lower the vapour
pressure.
The stronger the intermolecular forces the fewer the individual
molecules the surface of the substance the lower the vapour
pressure
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 74
The longer the chain length the
lower the vapour pressure
The longer the length of the chain, the larger the surface area the
stronger the
Intermolecular (Van der Waals) forces ,the fewer the individual
molecules above
the surface of the substance the lower the vapour pressure.
The more number of halogen (Cl,
Br, I) or OH groups or COOH
groups or CO groups or CHO
groups the higher the melting
point
The more the number of halogen (Cl, Br, l) or OH groups or
COOH
Groups or CO groups or CHO groups the stronger the
intermolecular forces
(hydrogen bonds or Van der Waals forces) the higher the melting
point
The more number of halogen (Cl,
Br, I) or OH groups or COOH
groups or CO groups or CHO
groups the lower the vapour
pressure
The more number of halogen (Cl, Br, I) or OH groups or COOH
groups or CO groups or CHO groups the stronger the
intermolecular forces (hydrogen bonds or Van der Waals forces)
the less the number the individual molecules above the surface of
the substance the lower its vapour pressure
- The teacher must orientate the homework:
Homework:
Activity 1:
The table below shows the boiling points and melting points of five organic compounds.
A
B
C
D
E
1.1
1.2
1.3
1.4
1.5
1.6
Compound
Ethane
Pentane
2-methylbutane
propan-1-ol
Propane-1,2-diol
Boiling point 0C
-89
36
30
97
188
Melting point 0C
-183
-130
-160
-126
-59
Define the term boiling point.
Define the term melting point.
Define the term vapour pressure.
Which one of A or B has a lower vapour pressure. Explain.
B and C are structural isomers.
1.5.1 Define the term structural isomers.
Explain why B has a higher melting point than C.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 75
1.7
Expalin why compound E has higher boiling point and melting point than D.
SELF-ASSESSMENT
QUESTION 1
During a practical investigation the boiling points of the first six straight-chain ALKANES
were determined and the results were recorded in the table below.
ALKANE
MOLECULAR
BOILING
FORMULA
POINT (°C)
Methane
−164
CH4
Ethane
−89
C2H6
1.1
Propane
C3H8
−42
Butane
C4H10
−0,5
Pentane
C5H12
36
Hexane
C6H14
69
Write down the:
7.1.1 Most important use of the alkanes in the above table
(1)
7.1.2 General formula of the alkanes
(1)
Refer to the table to answer QUESTION 1.2 and QUESTION 1.3 below.
1.2
For this investigation, write down the following:
1.2.1 Dependent variable
(1)
2.2. 2 Independent variable
(1)
2.2.3 Conclusion that can be drawn from the above results
(2)
2.3
Alkanes burn readily in oxygen. Write down a balanced equation, using
molecular formulae, for the combustion of propane in excess oxygen.
(3)
2.4
Will the boiling points of the structural isomers of hexane be HIGHER THAN,
LOWER THAN or EQUAL TO that of hexane? Refer to MOLECULAR
STRUCTURE, INTERMOLECULAR FORCES and ENERGY NEEDED to
explain the answer.
(4)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
[13]
Page 76
Solution of the activities:
1.1
The boiling point is the temperature at which the vapour pressure of a substance equals
the pressure above it surface (atmospheric pressure).
1.2
Melting point is the temperature at which a solid changes to a liquid (or the temperature at
which solid and liquid phases are in equilibrium).
1.3
Vapour pressure (the pressure at which the vapour of a substance is in dynamic
equilibrium with its liquid or solid form.
1.4
Compound B. The longer the length of the chain, the larger the surface area the stronger
the Intermolecular (Van der Waals) forces, the fewer the individual molecules above the
surface of the substance the lower the vapour pressure.
1.5
1.5.1 Compounds with the same molecular formula but different structural formula.
1.6
Compound B is less branched and has stronger intermolecular forces. As a result more
energy is needed to overcome the intermolecular forces. Thus, it will have a higher
melting point.
1.7
The more the number of halogen OH groups the stronger the intermolecular forces
(hydrogen bonds) the higher the melting point.
MEMORANDUM SELF-ASSESSMENT
QUESTION 1
1.1
1.1.1 Fuels
(1)
1.1.2 CnH2n + 2
(1)
1.2.1 Boiling point 
(1)
1.2. 2 Chain length/Molecular size/Molecular mass 
Kettinglengte/Molekulêre grootte/Molekulêre massa 
1.2.3
(1)
(2)
Criteria for conclusion:
Dependent and independent variables correctly identified.
Mark

Relationship between the independent and dependent variables
correctly stated.
.

Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 77
Examples:
• Boiling point increases with increase in chain length/molecular size/molecular
mass.
• Boiling point decreases with decrease in chain length/ molecular size/molecular
mass.
• Boiling point is proportional to chain length/molecular size/molecular mass.
1.3
C3H8 + 5O2 → 3CO2 + 4H2O  bal 
(3)
1.4
Lower than 
• Structure:
Isomers have more branching/ more compact or spherical molecules / smaller
surface areas over which the intermolecular forces act. 
Intermolecular forces:
Weaker intermolecular forces/less intermolecular forces 
Energy:
Less energy needed to overcome intermolecular forces. 
(4)
[13]
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 78
ORGANIC MOLECULES
Lesson 7
Topic: Oxidation of alkanes. Exercises on balancing equations. Esterification.
Objective:
Learners must be able to:
 Explain the oxidation reaction of hydrocarbons.
 Identify the alcohol and carboxylic acid used to prepare a given ester and vice versa, and
write as equation to present this preparation.
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
Introduction:
Chemical reactions play an important role in Chemistry that is why they are considered the heart
of Chemistry. Some reactions like a forest fire and the explosion of dynamite are quite dramatic
and some others are not so obvious. However, all chemical reactions must involve detectable
change.
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must explain the oxidation reaction of hydrocarbons.
Fossil fuels are fuels formed by the natural process of the decomposition of organisms under
heat and pressure. They contain a high percentage of carbon and include fuels such as coal,
petrol, and natural gases. They are also a non-renewable.
Because alkanes contain only single bonds C-C and C-H, most alkanes are relatively unreactive,
at room temperature, for example they do not react with acids, bases or strong oxidizing agents.
Alkanes are not completely inert, however. One of the most commercially important reactions is
combustion (oxidation) in air, which is the basis of their uses as fuels.
Alkanes are our most important fossil fuels. The combustion (burning) of alkanes (also known as
oxidation) is highly exothermic.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 79
Remember that an exothermic reaction releases energy (∆H < 0), while an endothermic reaction
absorbs energy (∆H > 0). The fact that energy is released in a combustion reaction implies that
∆H < 0 and the reaction is exothermic.
In a combustion reaction a substance reacts with an oxidising agent (e.g. oxygen), and heat
and light are released.
In the complete combustion reaction of alkanes, carbon dioxide (CO 2) and water (H2O) are
released along with energy. Fossil fuels are burnt for the energy they release.
Carbon dioxide (CO2) is the primary greenhouse gas emitted through human activities. The main
human activity that emits CO2 is the combustion of fossil fuels (coal, natural gas, and oil) for energy
and transportation, although certain industrial processes and land-use changes (e.g. conversion
of forest into agricultural land) also emit CO2.
The general reaction for the combustion of an alkane as a fossil fuel is given as:
alkane + O2(g) → CO2(g) + H2O(g) + energy
For example:
a) The complete combustion reaction of methane:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + energy
b) The complete combustion reaction of propane:
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g) + energy
c) The complete combustion reaction for hexane:
2C6H14(ℓ) + 19O2(g) → 12CO2(g) + 14H2O(g) + energy
d) The complete combustion reaction for octane:
2C8H18(ℓ) + 25O2(g) → 16CO2(g) + 18H2O(g) + energy
- The teacher must explain how to balance equation for the reaction of alkanes.
Example 1 (Worked example 28 page 181, Siyavula book Grade 12)
Balance the following equation: C4H10(g) + O2(g) → CO2(g) + H2O(g)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 80
SOLUTION:
Step 1: Balance the carbon atoms
There are 4 carbon atoms on the left. There is 1 carbon atom on the right. Add a 4 in front
of the CO2 molecule on the right:
C4H10(g) + O2(g) → 4CO2(g) + H2O(g)
Step 2: Balance the hydrogen atoms
There are 10 hydrogen atoms on the left. There are 2 hydrogen atoms on the right. Add a
5 in front of the H2O molecule on the right:
C4H10(g) + O2(g) → 4CO2(g) + 5H2O(g)
Step 3: Balance the oxygen atoms
There are 2 oxygen atoms on the left. There are 13 oxygen atoms on the right (4 x 2 in
13
the CO2 and 5 in the H2O). Divide the number of O atoms on the right by 2 to get 2 , this
is the number of O2 molecules required on the left:
𝟏𝟑
C4H10(g) + 𝟐 O2(g) → 4CO2(g) + 5H2O(g)
This is acceptable but it is better for all numbers to be whole numbers.
Step 4: Make sure all numbers are whole numbers
𝟏𝟑
There is 𝟐 in front of the O2 while all other numbers are whole numbers. So multiply the
entire equation by 2:
𝟏𝟑
2C4H10(g) +𝟐 × 𝟐 O2(g) →2 x 4CO2(g) + 2 x 5H2O(g)
Hence:
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
Example 2 (Worked example 29 page 181, Siyavula book Grade 12)
Balance the equation for the complete combustion of heptane.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 81
SOLUTION:
Step 1: Write the unbalanced equation
The molecular formula for heptane is C7H16. Combustion always involves oxygen (O2).
The complete combustion of an alkane always produces carbon dioxide (CO2) and water
(H2O):
C7H16(ℓ) + O2(g) → CO2(g) + H2O(g)
Step 2: Balance the carbon atoms
There are 7 carbon atoms on the left. There is 1 carbon atom on the right. Add a 7 in front
of the CO2 molecule on the right:
C7H16(ℓ) + O2(g) → 7CO2(g) + H2O(g)
Step 3: Balance the hydrogen atoms
There are 16 hydrogen atoms on the left. There are 2 hydrogen atoms on the right. Add
an 8 in front of the H2O molecule on the right:
C7H16(ℓ) + O2(g) → 7CO2(g) + 8H2O(g)
Step 4: Balance the oxygen atoms
There are 2 oxygen atoms on the left. There are 22 oxygen atoms on the right (7 x 2 in
the CO2 and 8 in the H2O). Divide the number of O atoms on the right by 2 to get 11, this
is the number of O2 molecules required on the left:
C7H16(ℓ) + 11O2(g) → 7CO2(g) + 8H2O(g)
- The teacher must explain the esterification reaction.
As was discussed earlier in lesson 26 that one way to form an ester is through the reaction of an
alcohol and a carboxylic acid. This process is called an acid-catalysed condensation or
esterification of a carboxylic acid.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 82
In the general form above an alcohol and a carboxylic acid combine to form an ester and water.
Example:
The esterification of butanol and propanoic acid to form butyl propanoate, water is also formed in
this reaction.
A more general example is:
alcohol + carboxylic acid → ester + water
It is important to be able to identify what ester a specific alcohol and carboxylic acid will form.
Remember that the first part of the ester name takes its prefix from the alcohol with the suffix -yl.
The second part of the ester takes its prefix from the carboxylic acid with the ester suffix -oate.
- The teacher must explain example 3.
Example 3 Worked example 30 page 183, Siyavula book Grade 12) Adapted
What is the name of the ester that will form from hexanol and propanoic acid.
SOLUTION
Step 1: Which compound forms the first part of the ester name and which forms the second part
of the ester name?
The alcohol forms the first part of the ester name and takes the suffix -yl. The carboxylic
acid forms the second part of the ester name and takes the suffix -oate.
Step 2: Determine the first part of the ester name
The alcohol is hexanol, therefore there are 6 carbons and this will be hexyl.
Step 3: Determine the second part of the ester name
The carboxylic acid is propanoic acid, therefore there are 3 carbons and this will be
propanoate.
Step 4: Combine the first and second parts of the ester name
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 83
The ester will be hexyl propanoate.
It is also important to be able to determine which compounds were used to form an ester.
Example 4 (Worked example 31 page 183, Siyavula book Grade 12)
What compounds did the ester octyl heptanoate come from?
SOLUTION:
Step 1: What types of compounds are used to form esters?
Esters are formed from alcohols (which become the first part of the ester name) and
carboxylic acids (which become the second part of the ester name).
Step 2: Determine the prefix for the alcohol
The first part of the ester name comes from the alcohol (-ol). Therefore the prefix is oct-.
Step 3: Determine the prefix for the carboxylic acid
The second part of the ester name comes from the carboxylic acid (-oic acid). Therefore
the prefix is hept-.
Step 4: Determine the compounds use to form the ester
Octyl heptanoate was formed from octanol and heptanoic acid.
Some examples of esters are given in the following table.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 84
Final activities
- The teacher must summaryse the lesson.
Alkanes are our most important fossil fuels. The combustion (burning) of alkanes (also
known as oxidation) is highly exothermic.
In a combustion reaction a substance reacts with an oxidising agent (e.g. oxygen),
and heat and light are released.
The general reaction for the combustion of an alkane as a fossil fuel is given as:
alkane + O2(g) → CO2(g) + H2O(g) + energy
One way to form an ester is through the reaction of an alcohol and a carboxylic acid. This
process is called an acid-catalysed condensation or esterification of a carboxylic acid.
A more general example is:
alcohol + carboxylic acid → ester + water
- The teacher must orientate the homework.
Homework:
Activity 1 (Exercise 4 - 24.1 page 182, Siyavula book Grade 12):
Balance the following complete combustion equations:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 85
a)
b)
c)
C3H8(g) + O2(g) → CO2(g) + H2O(g)
C7H16(ℓ) + O2(g) → CO2(g) + H2O(g)
C2H6(g) + O2(g) → CO2(g) + H2O(g)
Activity 2 (Exercise 4 - 24.2 page 182, Siyavula book Grade 12):
Write the balanced equation for the complete combustion of:
a) pentane
b) butane
Activity 3 (Exercise 4 - 25.1 page 186, Siyavula book Grade 12):
Give the IUPAC name for the product in the esterification of ethanoic acid with:
a) methanol
b) octanol
c) hexanol
d) propanol
Activity 4 (Exercise 4 - 25.3 page 186, Siyavula book Grade 12):
Give the IUPAC name for the product in the reaction of butanol with:
a) ethanoic acid
b) pentanoic acid
c) heptanoic acid
d) methanoic acid
Activity 5 (Exercise 4 - 25.4 page 186, Siyavula book Grade 12):
Fill in the missing reactant or part of the product name in the reactions below:
a) octanol +_______________ → _______________ hexanoate
b) _______________+ propanoic acid → hexyl __________________
c) _______________ + butanol → _________________ pentanoate
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 86
Solution of the activities:
Activity 1:
a)
C3H8 (ℓ) + 5O2 (g) → 3CO2 (g) + 4H2O (g)
b)
C7H16 (ℓ) + 11O2 (g) → 7CO2 (g) + 8H2O (g)
c)
C2H6 (ℓ) + 2 O2 (g) → 2CO2 (g) + 3H2O (g)
7
we multiply the entire equation by 2
2C2H6 (ℓ) + 7O2 (g) → 4CO2 (g) + 6H2O (g)
Activity 2
a)
C5H12 (ℓ) + 8O2 (g) → 5CO2 (g) + 6H2O (g)
b)
C4H10 (ℓ) + 2 O2 (g) → 4CO2 (g) + 5H2O (g)
13
we multiply the entire equation by 2
2C2H6 (ℓ) + 13O2 (g) → 8CO2 (g) + 10H2O (g)
Activity 3:
a) methyl ethanoate
b) octyl ethanoate
c) hexyl ethanoate
d) propyl ethanoate
Activity 4:
a) butyl ethanoate
b) butyl pentanoate
c) butyl heptanoate
d) butyl methanoate
Activity 5 (Exercise 4 - 25.4 page 186, Siyavula book Grade 12):
a) octanol + hexanoic acid → octyl hexanoate
b) hexanol + propanoic acid → hexyl propanoate
c) pentanoic acid + butanol → butyl pentanoate
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 87
LESSON 8: Experiment
PREPARATION OF DIFFERENT ESTERS AND SMELL IDENTIFICATION (EXPERIMENT)
Example of experiment
GRADE 12
Experiment 1
TERM 1
Date:-------------Total Marks- 50
PREPARATION OF DIFFERENT ESTERS AND SMELL IDENTIFICATION
NAME OF SCHOOL
: …………………………………………………………….
NAME OF LEARNER
: …………………………………………………………….
CLASS
: …………………
Aim
1. To synthesize several esters using a range of carboxilic acids and alcohols.
2. To identify the esters formed by their smell.
Introduction
Esters are a group of organic compounds best known for their interesting odours and flavours.
Many natural odours and flavours were discovered to be esters and therefore, many synthesized
esters are used in perfumes and foods.
Esters can be synthesised by the reaction of a carboxilic acid and an alcohol. This reaction is
called esterification. This reaction can be catalysed by concentrated sulphuric acid.
In the laboratory, an ester is usually formed from the reaction of a carboxylic acid (RCOOH) and
an alcohol (R'OH) in the presence of an acid catalyst, giving an ester and water as the products.
Here were, R and R’ represent any alkyl group
The general equation
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 88
for the formation of esters as follows:
The first part of the name of an ester is derived from the alkyl group of the alcohol used and the
second part is from the carboxylic acid, using the ending -oate. If you are using ethanoic acid and
ethanol the equation would be;
The name of this ester is ethyl ethanoate
Other examples are: If ethyl alcohol (ethanol) combines with propanoic acid, the resulting ester is
named ethyl propanoate. The aroma of oranges is attributed to butyl ethanoate (formed from
butanol and ethanoic acid) and apricots have an aroma because of the presence of pentyl
butanoate (formed from pentanol and butanoic acid).
The reaction between the alcohol and acid is rather slow at room temperature. In order to speed
it up and get an appreciable yield in the time available, you will use a temperature of about 60°C
- 75°C and add sulphuric acid to act as a catalyst in the reaction. In this experiment, you will
prepare five esters and carefully smell (see safety precautions, below) them to see if there
are any odours you recognize.
Safety Precautions
Concentrated sulphuric acid is a strong oxidizing agent and highly corrosive. Sulfuric acid is
used as the catalyst for the esterification reactions. It is dangerous and can burn skin, eyes, and
clothing very badly. Avoid contact with skin or clothes.
It will start a fire if mixed incorrectly with any of the alcohol or other acids used in this experiment.
Use only in a fume hood. Use exactly as directed.
If it is spilled, wash immediately before the acid has a chance to cause a burn, and inform
the instructor.
Salicylic acid is dangerous if swallowed. It can irritate the respiratory system and the skin.
There is a risk that it can cause severe damage to the eyes.
Other than concentrated sulphuric acid, all (other) organic liquid acids used in this experiment are
toxic and corrosive to skin, eyes, and clothing. For examples, butanoic acid and propanoic acid.
Ethanoic acid is flammable and poisonous if swallowed. Both the liquid and vapour are irritating
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 89
to the skin and eyes and can cause burns and ulcers. While working on this experiment, ware
safety goggles, full face shield, gloves, and lab apron.
Wash spills and splashes off your skin and clothing immediately, using plenty of water.
Methanol is highly flammable. It is poisonous if swallowed. Its vapour is harmful to the eyes, lungs
and skin and other organs.
Ethanol is poisonous and its toxicity is increased by the presence of the denaturing substances
that are added to laboratory ethanol in order to reduce its illegal consumption.High concentrations
of ethanol vapour can be dangerous. It is highly flammable.
1-Propanol is harmful to the lungs, skin, eyes and other organs. Poisonous if swallowed.Highly
flammable. Use in a fume hood.
Smelling: The vapors of the esters produced in this experiment may be harmful. You should
always detect odours with caution.
When determining the odours of the esters produced in the experiment, do not deeply
inhale the vapors.
Breathing the vapours of some of these esters can cause sore throat, dizziness, headache, and
drowsiness. Hold the test tube 30 cm away and 15 cm below your nose. Waft a small amount of
vapor or odour from the ester toward your nose, sniffing cautiously, once or twice. Do not breathe
deeply while sniffing.
Waste disposal
Dispose of all materials down the sink with abundant amounts of water. If you have any extra
sulphuric acid at your bench remember to dilute it by adding the ACID to the beaker of WATER
before pouring it down the sink. (In order to prevent students from having concentrated sulphuric
acid on their benches, a beaker with the acid and a dropper can be placed at the front bench which
should not be removed).
Always remember to clean the surface of your bench and put away everything as you found it. All
organic wastes should be placed into the appropriate non-halogentated waste container.
Before leaving the laboratory, wash your hands thoroughly with soap and water.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 90
Structural formula of Salicylic acid
Apparatus and Materials:
Equipment/ Glassware
Chemicals / Reagents
Test tubes & test tube rack
Methanol
Water soluble marker
Ethanol
Test tube rack
Concentrated sulphuric acid
Eye dropper
Glacial Acetic acid (Ethanoic acid)
250 ml beakers
Salicylic acid(2-Hydroxybenzoic acid)
Scale
1-pentanol
Hot plate/ Bunsen burner
0,5 mol.dm-3 sodium carbonate solution
Tripod
Tongs
Lab apron & Safety goggles
Method (Procedure)
1.
Put on your lab apron and safety goggles.
2.
Put 1 drop of concentrated sulphuric acid in a test tube and add 10 drops of
ethanoic acid and 10 drops of ethanol in the same test tube.
3.
Pour about 100 cm3 of water into the 250 cm3 beaker.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 91
4.
Carefully lower the test tube into the beaker so that it stands upright.
5.
Heat the beaker using the hotplate or using Bunsen burner using a tripod.
6.
If the mixture in the test tube boils, use the tongs to lift it out of the water until the
boiling stops, and then return it to the hot water.
7.
After 1 minute, carefully remove the test tube and allow it to cool down.
8.
When cool, pour the mixture into a test tube half-full of 0,5 mol.dm-3 sodium carbonate
solution. There will be some effervescence. Mix well. A layer of ester
will separate
and float on top of the aqueous layer.
9.
Smell the product by gently wafting the odour towards your nose with your hand and
note smell in the relevant table.
10.
Repeat the steps 2 to 9 but use METHANOL and PROPANOL as the alcohol.
11.
Repeat steps 1 to 10 but use SALICYLIC ACID and METHANOL.
12
Repeat steps 1 to 10 but use SALICYLIC ACID and ETHANOL.
13. Results and interpretation of results
Complete the tables below for the report.
Choose ONE of the following to identify the ester formed by smell.

Paint

Pear

Pineapple

Strawberry

Ice cream

Nail polish remover

Wintergreen

Peppermint
IDENTIFICATION OF THE AROMA (SMELL)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 92
Experiment
Reactants
Smell
Marks
Number
1
Ethanol +Ethanoic acid
2 marks
2
Methanol +Ethanoic acid
2 marks
3
Propanol +Ethanoic acid
2 marks
4
Methanol +Salicylic acid
2 marks
5
Ethanol + Salicylic acid
2 marks
Total
10 marks
EXPERIMENT 1: ETHANOL+ETHANOIC ACID
Word equation
for the reaction
2 marks
Balanced
equation with
structural
formula
5 marks
Balanced
chemical
equation with
condensed
structural
formula
4 marks
Total
11 marks
EXPERIMENT 2: METHANOL+ETHANOIC ACID
Word equation
for the reaction
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
1 marks
Page 93
Balanced
equation with
structural
formula
3 marks
Balanced
chemical
equation with
condensed
structural
formula
2 marks
Total
6 marks
EXPERIMENT 3: PROPANOL+ETHANOIC ACID
Word equation
for the reaction
1 marks
Balanced
equation with
structural
formula
3 marks
Balanced
chemical
equation with
condensed
structural
formula
2 marks
Total
6 marks
14. Discussion of results
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 94
1
Which property of the sulphuric acid makes it suitable to use it as a catalyst for
the preparation of esters?
2
Why do we heat the test tube in water bath and not directly over a flame?
3
With reference to the characteristic smells of esters, name TWO examples
where esters are used in different industries.
4
Why do esters with high molecular weight not have strong fragrances?
(2)
(2)
(2)
(3)
15. Conclusion
Write a possible conclusion for the experiment
(2)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 95
The following rubric will be used to measure the skills necessary for this experiment.
Criteria
High (3)
Medium (2)
Low (1)
None (0)
Accurately
following a
Following a
sequence
sequence of
of instructions
Can complete
the experiment
by following a
sequence
Able to follow
some single
written,
Unable to
follow a single
written,
written/verbal
including
branched
diagrammatic
or verbal
instruction.
diagrammatic
or
instructions.
of instructions.
Able to use
only the
verbal
instruction.
Unable to use
even the
most basic
most basic
equipments
and
chemicals.
equipments
and chemicals.
instructions.
Manipulative
skills
include
correct
handling of
apparatus
and
Able to use all
apparatus
and chemicals
correctly.
Use most of
the apparatus
and chemicals
correctly.
Your
MARK
material.
6 marks
GRAND TOTAL - 50
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 96
MEMORANDUM
Criteria
High (3)
Medium (2)
Low (1)
None (0)
Accurately
following a
Following a
sequence
sequence of
of instructions
Can complete
the experiment
by following a
sequence
Able to follow
some single
written,
Unable to
follow a single
written,
written/verbal
including
branched
diagrammatic
or verbal
instruction.
diagrammatic
or
Able to use
only the
Unable to use
even the
most basic
most basic
equipments
and
chemicals.
equipments
and chemicals.
instructions.
Manipulative
skills
include
correct
handling of
instructions.

Able to use all
apparatus
and chemicals
correctly. 
of instructions.
Use most of
the apparatus
and chemicals
correctly.
apparatus
and
Your
MARK
3 marks
verbal
instruction.
3 marks
material.
6 marks
13. Results and interpretation of results
IDENTIFICATION OF THE AROMA (SMELL)
Experiment
Reactants
Smell
Marks
Number
1
Ethanol +Ethanoic acid
Nail polish or
2 marks
pear
2
Methahanol +Ethanoic acid
Paint
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
2 marks
Page 97
3
Propanol +Ethanoic acid
Pear
2 marks
4
Methanol +Salicylic acid
Wintergreen
2 marks
5
Ethanol + Salicylic acid
Peppermint
2 marks
Total
10 marks
EXPERIMENT 1: ETHANOL+ETHANOIC ACID
Word equation ethanol + ethanoicacid → ethylethanoate + water
for the reaction
Balanced
equation with
structural
formula
Balanced
chemical
equation with
condensed
structural
formula
Total-11 Marks
EXPERIMENT 2: METHANOL+ETHANOIC ACID
Word equation methanol + ethanoicacid → methylethanoate + water
for the reaction
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 98
Balanced
equation with
structural
formula
Balanced
chemical
equation with
condensed
structural
formula
Total-6 Marks
EXPERIMENT 3: PROPANOL+ETHANOIC ACID
Word equation propanol + ethanoicacid → propylethanoate + water
for the reaction
Balanced
equation with
structural
formula
Balanced
chemical
equation with
condensed
structural
formula
Total-6 Marks
14. Discussion of results
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 99
1
Dehydrating property
2
Most of the organic compounds are flammable and should not be left near an
open flame. 
3
4
(2)
(2)
ANY TWO 

Food industry- Flavouring of food/sweets.

Cosmetic industry- Perfumes/lotions

Alcohol industry- Smell

Medicine industry- Taste of children medicine
(2)
They have higher IMF
Higher boiling point
(3)
Lower vapour pressure 
15. Conclusion
CONSIDER LEARNERS' ANSWERS.
Possible answers could be:
ANY ONE
When an alcohol reacts with a carboxilic acid an ester and water are formed.
(2)
When an alcohol reacts with a carboxilic acid a compound of pleasant smell formed.
Esters have pleasant smell.
GRAND TOTAL - 50
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 100
INFORMATION REGARDING SALICYLIC ACID
Name:
 General name: Betula oil/ wintergreen oil
 Chemical name (non-IUPAC name): Salicylic acid
 Chemical name (IUPAC): 2-hydroxybenzoic acid
Formula:
 Molecular formula: C7H6O3
 Condensed formula: C6H4OHCOOH
 Structural formula:
Solubility:
 Poorly soluble in water – 2 g per litre at 20°C
Preparation:
 Dissolve 25 -35 g salicylic acid in 125 cm3 of water.
 Heat to improve the solubility.
 Soluble in benzene, propanol, ethanol, ether & acetone (Take care!!! All these solvents
are inflammable).
Ester name:
 Methanol + salicylic acid → methyl salicylate or methyl ester (non –IUPAC name)
 Methanol + 2-hydroxybenzoic acid → methyl-2-hydroxybenzoate (IUPAC name)
2-hydoxy
oate
methyl
benz(ene)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 101
 If ethanol is used with 2-hydroxybenzoic acid then methyl-2-hydroxybenzoate will be
formed.
- ORGANIC MOLECULES
Lesson 9
Topic: Addition reactions.
Objective:
Learners must be able to:
 Identify addition reactions
 Write down, using structural formulae, equations and reaction conditions for the following
addition reactions of alkenes:
 Hydrohalogenation:
The addition of a hydrogen halide to an alkene
 Halogenation:
The reaction of a halogen (Br2, Cℓ2) with a compound
 Hydration:
The addition of water to a compound
 Hydrogenation:
The addition of hydrogen to an alkene.
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
Introduction:
The most interesting and useful aspect of Organic Chemistry is the study of reactions. It is
practically impossible to remember all the organic reactions but we can organize the organic
reactions into logical groups based on how they are placed and what intermediates are involved.
We will study three main types of reactions - addition, elimination and substitution.
An addition reaction occurs when two or more reactants combine to form a single product.
This product will contain all the atoms that were present in the reactants. Addition ractions
occur with unsaturated compounds.
The general equation for an addition reaction: A + B → C
Notice that C is the final product with no A or B remaining as a residue.
An elimination reaction occurs when a reactant is broken up into two products. Elimination
reactions occur with saturated compounds.
The general equation for an elimination reaction: A → B + C
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 102
A substitution reaction occurs when an exchange of elements in the reactants takes place.
The initial reactants are transformed or swopped around to give a final product.
The general equation for a substitution reaction: AB + CD → AD + BC
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must explain the oxidation reaction of hydrocarbons.
Unsaturated compounds (alkenes, cycloalkenes, alkynes) undergo addition reactions in which the
double or triple bound breaks open, allowing for the addition of extra atoms.
An addition reaction is a reaction in which parts of the reactants are added to each carbon atom
of a carbon-carbon double or triple bond which converts to a carbon-carbon single bond or double
bond respectably.
In addition reactions two compounds join together to form a single compound.
During the reaction, the weaker bonds of the double or triple bond is broken and new atoms are
added onto the two carbons where the double or triple bond occurred. These types of reactions
are used as a test for distinguishing between saturated and unsaturated hydrocarbons.
Depending on the compounds added, addition reactions are given different names:
o
o
o
o
Hydrohalogenation – The addition of a hydrogen halide to an alkene.
Halogenation - The reaction of a halogen (Br2, Cℓ2) with a compound
Hydration- The addition of water (H2O). This reaction forms an alcohol for an alkene.
Hydrogenation-The addition of H2. This reaction is used to manufacture margarine from
oils.
HYDROHALOGENATION REACTIONS
Hydrohalogenation involves the addition of a hydrogen atom and a halogen atom to an
unsaturated compound (containing a carbon-carbon double bond).
The addition of HX to an alkene or alkyne where X can be fluorine (F), chlorine (Cl), bromine (Br)
or iodine (I).
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 103
Example 1(video animation https://www.youtube.com/watch?v=vH--iR5jwSk,)
Reaction conditions: HX (X = Cℓ, Br, I) added to alkene; no water must be present
(During addition of HX to unsaturated hydrocarbons, the H atom attaches to the C atom already
having the greater number of H atoms. The X atom attaches to the more substituted C atom.)
Unsymmetrical reagents, such as HCl and HBr, add to unsymmetrical alkenes to give two products
that are constitutional isomers. For example,
And
In one case, the hydrogen atom of HBr adds to carbon atom 1, giving 2-bromopropane as shown
below.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 104
In the other case, the hydrogen atom of HBr adds to carbon 2, giving 1-bromopropane as shown
below.
The two products are not form in the same amount. A Russian scientist, Vladimir Markovnikov,
studies these reactions and formulated the following rule.
Markovnikov’s rule: During the addition of HX (X=Cl,Br,I) to unsymmetrical alkene , the H atom
bonds to the less substituent C atom- that is the C atom that has more H atoms to begin with.
In other words.
Markovnikov’s rule is a generalization that states that the major product formed by the addition
of an unsymmetrical reagent such as H-Cl, H-Br, or H-OH is the one obtained when the H atom
of the reagent adds to the carbon atom of the multiple bond that already has the greater number
of hydrogen atoms attached to it.
If more than one product is possible the major product will be the compound where:
 the hydrogen atom is added to least substituted carbon atom
i.e. the carbon atom with the least number of carbon atoms bonded to it
 the halogen atom is added to the more substituted carbon atom
i.e. the carbon atom with the most number of carbon atoms bonded to it
The major product of a reaction is the product that is most likely to form. Minor products
are those that are less likely to form.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 105
HALOGENATION REACTIONS:
The addition of a halogen to an alkene or alkyne is called halogenation (addition of X2 (X = Cℓ, Br,
I, F) to alkenes or alkynes.
Example 1 (animation https://www.youtube.com/watch?v=8G1NXYZKGL0,)
CH2 = CH2 + Cℓ2 → CH2Cℓ - CH2Cℓ
Reaction conditions: X2 (X = Cℓ, Br) added to alkene
Example 2
HYDRATION REACTIONS:
A hydration reaction involves the addition of water (H2O) to an unsaturated compound. This is one
way of preparing an alcohol from the corresponding alkene
Example 1
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 106
Reaction conditions:
o water must be present in excess
o an acid catalyst is needed for this reaction to take place
o the catalyst that is most commonly used is phosphoric acid (H3PO4)
Markovnikov’s rule is also applicable here.
(During addition of H2O to unsaturated hydrocarbons, the H atom attaches to the C atom already
having the greater number of H atoms. The OH group attaches to the more substituted C-atom.)
If more than one product is possible the major product will be the compound where:
 the hydrogen atom is added to the least substituted carbon atom
 the hydroxyl anion (OH-) is added to the more substituted carbon atom
HYDROGENATION REACTIONS:
Hydrogenation involves adding hydrogen (H2) to alkenes or alkynes. During hydrogenation the
double bond or triple bonds is broken (as with hydrohalogenation and halogenation) and more
hydrogen atoms are added to the molecule.
Example 1
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 107
Reaction conditions:
 a catalyst such as platinum (Pt), palladium (Pd) or nickel (Ni) is needed for these reactions
 heating is required
 the reaction must be done under an inert atmosphere, not air (e.g. N2(g) atmosphere)
The hydrogenation of vegetable oils to form margarine is another example of this addition reaction.
Final activities:
- The teacher must summarise the lesson.
Summary:
 In addition reactions two compounds join together to form a single compound.
 Depending on the compounds added addition reactions are given different names.
 Hydrohalogenation – The addition of a compound consisting of hydrogen and a
halide ion such as I-, Cl-, Br-.
 Halogenation - The addition of a halogen molecule such as I2, Cl2, Br2.
 Hydration - The addition of water (H2O). This reaction forms and alcohol for and
alkene.
 Hydrogenation -The addition of H2. This reaction is used to manufacture
margarine from oils.
- The teacher must orientate the homework
Homework:
Activity 1
Multiple-choice questions
Four options are provided as possible answers to the following questions. Each question has
only one correct answer. Choose the answer and make a cross.
1.1
An example of an addition reaction in which 2 atoms of hydrogen are added to an alkene
(hydrogenation) is:
A)
C2H4 (g) + 2O2 (g) → 2 CO2 (g) + 2HO2(g)
B)
CH2 = CH2 + H – H
C)
CH2= CH2 + H – Br
CH3 – CH3
Ni
→ CH3 – CH2Br
Ni
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 108
D)
H – C ≡ C – H + H2
CH2 = CH2
1.2
The process of adding hydrogen to an unsaturated organic compound is called:
A
Halogenation
B
Oxidation
C
Hydration
D
Hydrogenation
Activity 2
What is the major product of the following reaction?
Activity 3
Consider the following incomplete chemical equations representing addition reactions.
A
CH2 = CH2 + H2 →
B
CH2 = CH2 + H2O→
C
CH2 = CH2 + I2 →
D
CH2 = CH2 + HCl→
3.1
Complete the chemical equation for each one.
3.2
Give the IUPAC name for the products.
3.3
Classify the addition reaction into:




Hydrohalogenation
Halogenation
Hydration
Hydrogenation
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 109
Solutions of the activities
Activity 1
1.1
1.2
B
D
Activity 2
2-chloro-2-methylbutane
Activity 3
Note: It is not necessary to use structural formulae.
1.2
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 110
A
B
C
D
Ethane
Ethanol
1,2-diiodoethane
1-chloroethane
1.3
A
B
C
D
Hydrogenation
Hydration
Halogenation
Hydrohalogenation
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 111
ORGANIC MOLECULES
Lesson 10
Topic: Elimination reactions.
Objective:
Learners must be able to:
 Identify elimination reactions
 Write down, using structural formulae, equations and reaction conditions for the following
elimination reactions:
 Dehydrohalogenation of haloalkanes:
The elimination of hydrogen and a halogen from a haloalkane
 Dehydration of alcohols:
Elimination of water from an alcohol.
 Cracking of alkanes:
The chemical process in which longer chain hydrocarbon molecules are broken down
to shorter more useful molecules.
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
Introduction:
- The learners must do the informal test 1 (only 10 minutes)
Informal test 1
Consider the following incomplete chemical equation representing an addition reaction.
1.1
Complete the chemical equation using structural formula.
(2)
1.2
Give the IUPAC name for the product.
(2)
1.3
Classify the addition reaction into:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 112




Hydrohalogenation
Halogenation
Hydration
Hydrogenation
(1)
[5]
Chemical reactivity of organic compounds is determined by their functional groups. Just like other
compounds, organic compounds can undergo acid-base and redox reactions. In addition to these,
organic compounds also undergo addition, elimination and substitution reactions. In previous
lesson we studied the addition reactions. Today we are going to study the elimination reactions.
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must explain elimination reactions of hydrocarbons.
Elimination involves the removal of a molecule from a larger one.
Saturated compounds (haloalkanes, alcohols, alkanes) undergo elimination reactions to form
unsaturated compounds.
Elimination reactions are the opposite (reverse) of addition reactions.
Elimination reaction is a reaction in which two adjacent groups on adjacent carbon atoms
are removed from a molecule and a double bound forms between the adjacent carbon
atoms.
General equation for elimination reactions
Depending on what is eliminated, the reaction is given a special name:




Dehydrohalogenation- the removal of hydrogen halide from a haloalkane, forming and
alkene.
Dehydration - the removal of water (H2O) from an alcohol, forming and alkene.
Dehydrogenation - The removal of hydrogen from an alkane, forming an alkene. This is a
very important reaction in the fuel industry and in the production of plastics.
Cracking - Breaking up large hydrocarbon molecules into smaller and more useful bits.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 113
DEHYDROHALOGENATION:
Elimination of HX from a haloalkane produces an alkene.
In dehydrohalogenation a haloalkane is exposed to a base, the base then helps the elimination of
the halogen and a hydrogen atom. A double bond is formed (alkane → alkene).
The elimination of iodine from iodoethane is an example of dehydrohalogenation:
Many compounds can eliminate in more than one way, to give mixture of alkenes. To predict
which elimination product will predominate we can use Saitsev’s rule: In elimination reactions,
the must substituted alkene usually predominates.
Or
When two alkenes can be formed in an elimination reaction, the alkene that dominates is the one
with more alkyls substituents on the 𝐶 = 𝐶 bond. This implies that the H atom will be removed
from the C atom already having the lesser number of H atoms.
Example 1
In order for elimination to occur the following reaction conditions must be used:



heat under reflux (approximately 70°C)
a concentrated, strong base (e.g. NaOH, KOH)
the base must be dissolved in pure ethanol (hot ethanolic base)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 114
DEHYDRATION OF ALCOHOLS:
During the dehydration of an alcohol the hydroxyl (-OH) group and a hydrogen atom are eliminated
from the reactant. A molecule of water is formed as a product in the reaction, along with an alkene.
This can be thought of as the reverse of a hydration (addition) reaction.
Reaction conditions:


an excess of a strong acid catalyst (generally H2SO4 or H3PO4)
high temperature (approximately 180°C)
If more than one elimination product is possible, the major product is the one where the H atom is
removed from the C atom with the least number of H atoms.

the hydrogen atom is removed from the carbon atom bonded to the most number of carbon
atoms (i.e., the more substituted carbon atom)
CRACKING OF HYDROCARBONS:
The cracking of alkanes is an example of an elimination reaction.
Breaking up large hydrocarbon molecules into smaller and more useful bits.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 115
Cracking is the chemical process in which longer chain hydrocarbons molecules are broken down
to shorter more useful molecules.
Cracking reactions form one of the following series of products:
 An alkane with a shorter chain than the original chain plus a short chain alkene.
 Two or more alkenes and hydrogen.
Cracking at high temperature and pressures is known as thermal cracking. Crakcing at lower
temperatures and pressures in the presence of a catalyst is known as a catalytic cracking.
In thermal cracking, high temperatures and pressures are used to break the large hydrocarbons
into smaller ones. Thermal cracking gives mixtures of products containing high proportion of
hydrocarbons with double bonds-alkenes. Thermal cracking takes place in absence of a catalyst.
The thermal cracking of ethane can be represented as follows
heat
C2H6 (g)
Ethane
CH2 = CH2(g) + H2 (g)
Ethene
hydrogen gas
This process of removing H2 from and alkane producing an alkene is called
DEHYDROGENATION:
In thermal cracking alkanes are simply passed through a chamber heated to a high temperature.
Large alkanes are converted into smaller alkanes. In a modification called steam cracking the
hydrocarbon is diluted with steam heated for a fraction of a second to 700-900 °C and rapidly
cooled. It has a great importance in the production of hydrocarbons as chemicals, including
ethylene, propylene, butadiene, isoprene, and cyclopentadiene.
Catalytic cracking
The alkane is brought into contact with the catalyst at lower temperature (than for thermal
cracking) and moderately low pressures. Catalytic cracking is used to produce fuels and it is also
use to improve the quality of the gasoline (petrol).
Final activities:
- The teacher must summarise the lesson.
Summary:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 116
 Elimination reaction is a reaction in which two adjacent groups on adjacent
carbon atoms are removed from a molecule and a double bound forms between
the adjacent carbon atoms.
 Depending on what is eliminated, the reaction is given a special name.
 Dehydrohalogenation – The elimination of hydrogen and a halogen from a
haloalkane.
 Dehydration - The elimination of water from an alcohol.
 Cracking - Cracking is the chemical process in which longer chain hydrocarbons
molecules are broken down to shorter more useful molecules.
- The teacher must orientate the homework
Homework:
Activity 1
The equation given below shows an elimination reaction.
1.1
What type of elimination reaction does the equation given above represent?
1.2
Name the products of this reaction.
1.3
The organic product formed then reacted with HCl.
1.3.1 Use structural formulae to show this reaction.
1.3.2 What type of reaction does it represent?
Activity 2
Complete the following chemical equations of elimination reactions and give the name of each.
2.1
𝐂𝐇𝟑 − 𝐂𝐇𝐁𝐫 − 𝐂𝐇𝟑
Heat
2.2
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 117
Solution of the activities:
Informal test
1.1

1.2
1,2- dibromopropane
1.3
Halogenation
Activity 1:
1.1
Dehydration
1.2
Ethene and water
1.3.1
1.3.2 Addition reaction/hydrohalogenation.
Activity 2:
2.1
Dehydrohalogenation
𝐂𝐇𝟑 − 𝐂𝐇𝐁𝐫 − 𝐂𝐇𝟑
2.2
Heat
𝐂𝐇𝟑 = 𝐂𝐇 − 𝐂𝐇𝟑 + 𝐍𝐚𝐁𝐫 + 𝐇𝟐 𝐎
Dehydration
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 118
ORGANIC MOLECULES
Lesson 11
Topic: Substitution reactions.
Objective:
Learners must be able to:
 Identify substitution reactions
 Write down, using structural formulae, equations and reaction conditions for the following
substitution reactions:
 Hydrolysis of haloalkanes:
The reaction of a compound with water.
 The reaction of HX (X = Cℓ, Br) with alcohols to produce haloalkanes:
 Halogenations of alkanes:
The reaction of a halogen (Br2, Cℓ2) with a compound.
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
Introduction:
In previous lessons we studied the addition and elimination reactions. Today we are going to study
a very important type of organic reaction called substitution reactions.
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must explain substitution reactions.
A substitution reaction is a reaction in which an atom or a group of atoms in a molecule is
replaced by another atom or group of atoms.
Substitution reactions require a good leaving group and an atom or group of atoms that can
replace it. This replacing group takes the place of the leaving group in the molecule. The rest of
the molecule does not change; there is just a substitution of the leaving group by the replacing
group.
HYDROLYSIS
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 119
In a previous section we saw the hydration of an ethene to form an alcohol. That is an addition
reaction. Alcohols can also be formed through a substitution reaction with a haloalkane.
Definition of hydrolysis: The reaction of compounds with water to form other compounds.
Haloalkanes can undergo substitution with water to form alcohols. Haloalkanes do not mix with
water.
In this substitution reaction (hydrolysis) the halogen atom is replaced by an OH group to form and
alcohol.
Moderate heating of a haloalkane with an aqueous solution of a base gives the corresponding
alcohol. The reaction is also known as hydrolysis-splitting of a chemical bond with water.
The haloalkane is heated under reflux with an aqueous solution of sodium or potassium hydroxide.
Heating under reflux means heating with condenser placed vertically in the flask to prevent loss
of volatile substances from the mixture.
Using a base (e.g. KOH, NaOH) dissolved in water and warming the solution would increase the
rate of the reaction. However, in order for substitution to occur the following reaction conditions
must be used:
low temperatures (around room temperature)
a dilute solution of a strong base (e.g. NaOH)
the solution must be aqueous (in water)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 120
FORMATION OF HALOALKANES:
Haloalkanes can be formed when the hydroxyl (-OH) group of an alcohol is replaced by a halogen
atom (X = Cl, Br). This reaction works best with tertiary alcohols where it can occur at room
temperature.
A tertiary carbon atom is attached to three other carbon atoms. A secondary carbon atom is
attached to two other carbon atoms while a primary carbon atom is attached to one other carbon
atom.
It is possible for a secondary or primary alcohol to form a secondary or primary haloalkane as well
but this requires high temperatures and the reaction is very slow.
Reaction conditions for the reaction of alcohol with hydrogen halides to for alkyl halides: For
tertiary alcohols: add the hydrogen halide (HBr) directly to the tertiary alcohol. For primary and
secondary alcohols: add hydrogen halide (HBr) and then heat the reaction mixture.
HALOGENATION OF ALKANES:
Under the right conditions, alkanes can react with other molecules. An important example is the
reaction of alkanes with the halogens F2, Cl2 and Br2. The reaction of alkanes with halogens
requires light or heat to take place.
Example 1(substitution reaction animation www.youtube.com/watch?v=MvWQOYLm1f4)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 121
Methane reacts with chlorine (Cl2) in presence of sunlight or high temperature. This is an example
of substitution reaction.
In this substitution reaction one or more chlorine atoms substitute hydrogen atoms of the methane
molecule. If there is excess of chlorine (Cl2) more atoms of hydrogen are replaced by chlorine
atoms.
When an alkane is mixed with halogen molecules and the mixture is heated or exposed to sunlight,
substitution can take place. The halogen atom replaces one of the hydrogen atoms attached to a
carbon atom. The name given to this type of substitution reaction is halogenation.
Conditions for the halogenations of alkanes: The reaction takes place in light or ultra-violet light or
by heating it.
- The learners must do activity 1 in class. Informal assessment
Activity 1:
Consider the following incomplete chemical equation representing substitution reactions
A
B
1.1
1.2
1.3
Complete the chemical equation using structural formulae.
Give the IUPAC name of the products.
What is the name given to the substitution reactions represented?
(4)
(4)
(2)
[10]
- The teacher must mark activity 1.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 122
Final activities:
- The teacher must orientate the homework.
Homework:
Activity 2: One word answer
Write only the word/term for each of the following descriptions next to the question number.
2.1
Reaction type in which two molecules join to form a single, new molecule.
2.2
A reaction in which an atom or a group of atoms in a molecule is replaced by another
atom or group of atoms.
2.3
A reaction in which two adjacent groups on adjacent carbon atoms are removed from a
molecule and a double bound forms between the adjacent carbon atoms.
Process of removing H2 from and alkane producing an alkene.
The addition of a halogen to an alkane.
2.4
2.5
Activity 3: Multiple-choice questions.
Four options are provided as possible answers to the following questions. Each question has only
one correct answer. Choose the answer and make a cross.
3.1
An example of an addition reaction in which 2 atoms of hydrogen are added to an alkene
(hydrogenation) is:
A)
C2H4 (g) + 2 O2 (g) → 2 CO2(g) + 2 HO2(g)
B)
CH2 = CH2 + H – H
C)
CH2 = CH2 + H – Br
D)
H–C≡C–H+H
Ni
CH3 – CH3
→ CH3 – CH2Br
Ni
CH2= CH2
Activity 4:
Write the chemical equations of the following reactions using structural formulae and name the
products.
4.1
Addition of chlorine to propene.
4.2
Hydrogenation of pent-2-ene (2- pentene).
4.3
Addition of hydrogen chloride to propene.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 123
Solution of the activities:
Activity 1:
1.1
A


B


1.2
A
B
ethanol and potassium chloride
1-bromoethane and hydrogen bromide 
1.3
A
B
Hydrolysis
Halogenation
[10]
Activity 2
2.1
2.2
2.3
2.4
2.5
Addition reaction
Substitution reaction
Elimination reactions
Dehydrogenation
Halogenation
Activity 3
2.1- B
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 124
Activity 4
1,2-dichlorepropane
4.1
4.2
pentane
4.3
2-chlorepentane
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
1-chlorepentane
Page 125
- ORGANIC MOLECULES
Lesson 12
Topic: Revision exercises.
Objective:
Learners must be able to:
 Identify different types of organic reactions.
 Write down, using structural formulae, equations and reaction conditions for different types
of organic reactions.
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
Introduction:
- The teacher must give learners a summary of organic reactions.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 126
HYDROHALOGENATION (addition of a hydrogen halide to an alkene
(X = Cℓ, Br, I)
HALOGENATION (addition of X2 (X = Cℓ, Br, I, F) to alkenes or
alkynes)
Example:
CH2 = CH2 + Cℓ2 → CH2Cℓ - CH2Cℓ
HYDRATION (addition of water to an unsaturated compound. This is
one way of preparing an alcohol from the corresponding alkene)
ADDITION
REACTION
Two compounds
join together to
form a single
compound
Reaction conditions:
o water must be present in excess
o an acid catalyst is needed for this reaction to take place (e.g.
H3PO4)
HYDROGENATION (adding hydrogen (H2) to alkenes or alkynes)
Reaction conditions:
 a catalyst such as platinum (Pt), palladium (Pd) or nickel (Ni) is needed.
 heating is required
 the reaction must be done under an inert atmosphere, not air (e.g. N 2(g)
atmosphere)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 127
DEHYDROHALOGENATION
alkene
(Elimination of HX from a haloalkane produces an
Reaction conditions must be used:
 heat under reflux (approximately 70°C)
 a concentrated, strong base (e.g. NaOH, KOH)
 the base must be dissolved in pure ethanol (hot ethanolic base)
ELIMINATION
REACTION
Two adjacent groups
on adjacent carbon
atoms are removed
from a molecule and
a
double
bound
forms between the
adjacent
carbon
atoms
CRACKING OF HYDROCARBONS
Cracking is the chemical process in which longer chain hydrocarbons
molecules are broken down to shorter more useful molecules.
 An alkane with a shorter chain than the original chain plus a short
chain alkene.
 Two or more alkenes and hydrogen.
Cracking at high temperature and pressures is known as thermal
cracking. Cracking at lower temperatures and pressures in the
presence of a catalyst is known as a catalytic cracking.
DEHYDROGENATION. This process of removing H2 from and
alkane producing an alkene is called
DEHYDRATION OF ALCOHOLS During the dehydration of an alcohol the hydroxyl (OH) group and a hydrogen atom are eliminated from the reactant. A molecule of water
is formed as a product in the reaction, along with an alkene.
Reaction conditions:
 an excess of a strong acid catalyst (generally H2SO4 or H3PO4)
Developed by: G. Izquierdo
& G.(approximately
Izquierdo Gómez
Page 128
 high Rodríguez
temperature
180°C)
 the hydrogen atom is removed from the carbon atom bonded to the most number
Copyright reserved©
of carbon atoms (i.e., the more substituted carbon atom)
HYDROLYSIS (The reaction of compounds with water to form other compounds)
Haloalkanes can undergo substitution with water to form alcohols
Reaction conditions must be used:
low temperatures (around room temperature)
a dilute solution of a strong base (e.g. NaOH)
the solution must be aqueous (in water)
SUBSTITUTION
REACTION
An atom or a group of
atoms in a molecule is
replaced by another atom
or group of atoms
HALOGENATION OF ALKANES
Reaction of alkanes with the halogens F2, Cl2 and Br2
Conditions for the halogenations of alkanes: The reaction takes place
in light or ultra-violet light or by heating it.
FORMATION OF HALOALKANES
Haloalkanes can be formed when the hydroxyl (-OH) group of an alcohol is replaced by a
halogen atom (X = Cl, Br).
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 129
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must orientate the class work.
Activity 1
Classify each of the following reactions as substitution, addition or elimination. Explain the criteria
used for your answer.
1.1
C3 H6OH + HBr → C3H6Br + H2O
1.2
CH3CH2OH → CH2 = CH2 + H2O
1.3
CH2 = CH2 (g) + H2(g) → CH3CH3(g)
1.4
CH3CH2Cl + NaOH → CH2 = CH2 + NaCl + H2O
1.5
C3H6Br + KOH → C3H6OH + KBr
𝑵𝒊
1.6
Br
Br Br
Br
H - C = C – H + Br - Br
H–C–C–H
Br Br
Activity 2
Consider the following incomplete chemical equations representing organic reactions.
A.
CH4 + Br2 →
B.
CH3 – CH3 + Cl2 →
C.
CH2 = CH2 + HBr →
D.
CH3 – CH2 – CH = CH2 + H2 →
E.
CH3 – CH3
F.
CH2 = CHCH3 + H2 →
G.
CH3CH2CH2OH →
2.1
Complete the chemical equation for each one of the above organic reactions.
Heat
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 130
2.2
2.3
2.4
2.5
Name the products.
Classify each chemical reaction as substitution, addition or elimination.
Classify the substitution reactions as Hydrogenation, halogenations, hydration or
hydrohalogenation.
Classify the elimination reactions as dehydration, dehydrohalogenation, dehydrogenation
or cracking.
Activity 3
Consider the following organic reaction:
3.1
3.2
3.3
3.4
Classify this reaction as oxidation, addition, elimination, substitution or esterification
reaction.
Name the organic product according to IUPAC rules.
The organic product then reacts with NaOH. Use structural formulae to show this
reaction.
Name the organic product formed.
Final activities:
- The teacher must mark the class work.
- The teacher must orientate the homework.
Homework:
Activity 4
Water is added to the following compound
4.1
4.2
4.3
Use structural formula to show this reaction
Give the IUPAC name of the major product formed in this reaction
What is the name given to this type of reaction?
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 131
Activity 5 (Provincial common paper 2012)
The letters A-E in the flow diagram below represents types of ORGANIC REACTIONS:
5.1
5.2
5.3
5.4
5.5
5.6
5.7
Write down the IUPAC name of 𝐶𝐻3 𝐶𝐻2 𝑂𝐻.
(1)
To which homologous series does pentanoic acid belongs?
(2)
Name the type of reactions represented by the following letters.
5.3.1 B
(1)
5.3.2 E
(1)
Name the type of addition reaction represented by A.
(1)
Reaction D represents the conversion of an alcohol to compound F.
Write down the . . .
5.5.1 name given to this type of reaction;
(1)
5.5.2 formula of the catalyst added;
(1)
5.5.3 balanced chemical equation for reaction D using STRUCTURAL FORMULAE;
(4)
5.5.4 IUPAC name of compound F.
(2)
Use MOLECULAR FORMULAE to write down the balanced equation for reaction E.
(3)
Write down a balanced equation for the reaction of ethane with oxygen.
(3)
[20]
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 132
Solutions of the activities:
Activity 1
1.1
Substitution. (Formation of haloalkanes). Reaction between an alcohol and HX. An
atom or atoms are replaced.
1.2
Elimination. (Dehydratation). Removal of water from an alcohol resulting in the
production of an alkene. Reaction in which two adjacent groups on adjacent carbons
atoms are removed from a molecule and a double bond forms between the adjacent
atoms.
1.3
Addition. (Hydrogenation). Hydrogen is added to an alkene.
1.4
Elimination. (Dehydrohalogenation). Removal of HX from a haloalkane producing
an alken.
1.5
Substitution reaction (hydrolysis of alkyl halides). Reaction in which two adjacent
groups on adjacent carbons atoms are removed from a molecule and a double bond
forms between the adjacent atoms.
1.6
Addition reaction (halogenation). Bromine is added to an alkene resulting in the
production of an alkane.
Activity 2
2.1
A.
CH4 + Br2 → CH3 Br + HBr.
B.
CH3 – CH3 + Cl2 → H3-CH2-Cl + HCl
C.
CH2 = CH2 + HBr → C2H5Br
D.
CH3 – CH2 – CH = CH2 + H2
E.
CH3 – CH3
F.
CH2 = CHCH3 + H2
G.
CH3CH2CH2OH
2.2
A.
Bromomethane and hydrogen bromide
Heat
CH3CH2 CH2 CH3
CH2=CH2 + H2
CH3CH2CH3
H2SO4
CH3CHCH2 + H2O
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 133
B.
C.
D.
E.
F.
G.
Ethane and hydrogen chloride
Bromoethane.
Butane.
Ethene and Hydrogen
Propane.
Propene and water
2.3.
A
B
C
D
E
F
G
Substitution
Substitution
Addition
Addition
Elimination
Addition
Elimination
2.4. and 2.5.
A.
B.
C.
D.
E.
F.
G.
Halogenation
Halogenation
Hydrohalogenation
Hydrogenation
Thermal cracking
Hydrogenation
Dehydration.
Activity 3
3.1
Substitution
3.2
2-chloropropane
3.3
3.4
propan-2-ol or 2- propanol
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 134
Activity 4:
4.1



(3 marks)
4.2
butan-2-ol / 2- butanol 
(2 marks)
4.3
hydration 
(1 mark)
[6 marks]
Activity 5
5.1
Ethanol 
(1)
5.2
Carboxylic acid
(2)
5.3.1
Elimination / dehydration
(1)
5.3.2
Substitution / hydrolysis
(1)
5.4
Hydration
(1)
5.5.1
Esterfication / condensation
(1)
5.5.2
𝑯𝟐 𝑺𝑶𝟒 
(1)
5.5.3
(REACTANTS ; PRODUCTS ; CORRECT FORMULAE  ; BALANCING /
(4)
5.5.4
Ethylpentanoate
(2)
5.6
𝑪𝟐 𝑯𝟓 𝑩𝒓 + 𝑯𝟐 𝑶 ⟶ 𝑪𝟐 𝑯𝟔 𝑶 + 𝑯𝑩𝒓 (reactants; products ; balancing)
(3)
5.7
𝟐𝑪𝟐 𝑯𝟔 + 𝟕𝑶𝟐 ⟶ 𝟒𝑪𝑶𝟐 + 𝟔𝑯𝟐 𝑶 (reagents; products;  balancing)
(3)
[20]
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 135
ORGANIC MOLECULES
Lesson 13
Topic: Polymer, macromolecule, monomer. Polymerisation (addition and condensation).
Objective:
Learners must be able to:
 Describe the following terms:
 Macromolecule.
 Polymer.
 Monomer.
 Polymerisation.
 Distinguish between addition polymerisation and condensation polymerisation.
 Identify monomers from given addition polymers.
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
Introduction
Celluloid, discovered around 1850, was the first commercial plastic. Initially it was used as a cheap
substitute for ivory in making billiard balls, but it soon became the material of choice for many
consumer items such as combs, toothbrushes and also for producing photographic films for motion
pictures.
Plastics such celluloid and bakelite are polymers, which are giant molecules constructed by
covalently bonding together many small molecules.
Many of the compounds we encounter in the world around us are composed of large molecules,
also known as polymers, which are made up of hundreds or even thousands of atoms.
For a long time humans processed naturally occurring polymers such as wool, leather, silk and
natural rubber into usable materials.
Over the last century, chemists have learned to produce synthetic polymers such as plastics and
synthetic fibres.
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 136
Development:
- The teacher must describe the terms (macromolecule, polymer, monomer, polymerisation).
Polymers are large molecules (macromolecules) that are made up of many repeating structural
units called monomers which have various functional groups. To put it more simply, a monomer
is like a building block.
Monomer are small organic molecules that can be covalently bonded to each other in a
repeating pattern.
When lots of monomers are joined together
by covalent bonds, they form a polymer.
Polymer is a large molecule composed of
smaller monomer units covalently
bonded to each other in a repeating
pattern.
Note that mono means one, while poly
means many. So a monomer is the single
unit, and a polymer is made from many
monomers.
In an organic polymer, the monomers are joined by the carbon atoms of the polymer backbone or
chain. A polymer can also be inorganic, in which case there may be atoms such as silicon in the
place of carbon atoms. We will look solely at organic polymers. Polymers are a specific group of
macromolecules. A macromolecule is any compound with a large number of atoms. A biological
macromolecule is one that is found in living organisms. Biological macromolecules include
molecules such as carbohydrates, proteins, nucleic acids and lipids. They are essential for all
known forms of life to survive.
Macromolecule is a molecule that consists of a large number of atoms
The key feature that makes a polymer different from other macromolecules is the repetition of
identical or similar monomers in the polymer chain. Polymers will contain chains of the same type
of functional group and that functional group is dependent on the monomer used. Plastics are a
group of polymers that can be molded during manufacture. They can be one polymer or a blend
(mixture) of polymers and may contain other substances as well. These other substances can be
inorganic (e.g. used for electronic packaging) or stabilising (e.g. used for increasing fire
resistance).
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 137
Plastic is a subgroup of organic polymers that can be moulded. Plastics may contain more than
one organic polymer as well as other additives.
Polymers are formed through a process called polymerisation, where monomers react together
to form a polymer chain.
Polymerisation is a chemical reaction in which monomer molecules join to form a polymer
- The teacher must distinguish between addition polymerisation and condensation
polymerisation.
Two of the types of polymerisation reactions are addition polymerisation and condensation
polymerisation.
ADDITION POLYMERISATION:
In this type of reaction, monomer molecules are added to a growing polymer chain one at a time.
(No small molecules are eliminated in the process).
Addition polymerisation is a reaction in which small molecules join to form very large
molecules by adding on double bonds
Addition polymer: A polymer formed when monomers (usually containing a double bond)
combine through an addition reaction
Addition polymers form:
when unsaturated carbon molecules (monomers) react to form a long chain polymer
molecule.
and no small molecules or atoms are eliminated during the reaction
-
The teacher must explain the three steps in the production of the chain- initiation,
propagation and termination.
Step 1: INITIATION:
The reaction is initiated by a free radical which is a molecular fragment with unpaired
electron R·.
Step 2: PROPAGATION:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 138
The new free radical adds to the double bond of another monomer molecule and creates another
free radical with a longer chain. Successful additions of this kind give a macromolecule with a
head to tail arrangement of monomer units. This process is very fast and a chain may grow to
1000 units or more within a second.
Step 3: TERMINATION:
This is the step which terminates the chain. The formation of the chain will end if:
 Two radicals join together
 One radical removes a hydrogen atom from another radical forming and alkane and an
alkene
-
The teacher must show some addition polymers.
Four major examples of addition polymers are polyethylene, polypropylene,
polyvinylchloride (PVC) and polystyrene. All four of these organic polymers are also
plastics.
Polyethene (polyethylene)
In lesson 24 we learned about the structure group of hydrocarbons called the alkenes, for example
the molecule ethene.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 139
When lots of ethene molecules bond together, a polymer called polyethene (commonly called
polyethylene) is formed. Ethene is the unsaturated monomer which, when joined to other ethene
molecules through an addition reaction, forms the saturated polymer polyethene.
A polymer may be a chain of thousands of monomers, and so it is impossible to draw the entire
polymer. Rather, the structure of a polymer can be condensed and represented:
The monomer is enclosed in brackets and the n represents the number of repeating units (the
saturated form of the monomer) in the polymer, where n is any whole number. What this shows is
that the monomer is repeated an indefinite number of times in a molecule of polyethene.
Polyethene is the most common plastic with over 80 million metric tons produced each year. It is
commonly known as polyethylene. It is cheap and is used to make squeeze bottles, plastic bags,
films, toys and moulded objects as well as electric insulation. It has a recycling number 4 which
means that it is easy to process, has strength, toughness, flexibility, is easy to seal and has a
barrier to moisture.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 140
Polypropene (polypropylene)
Another example of a polymer is polypropene. Polypropene (commonly known as polypropylene)
is also a plastic, but is stronger than polyethene and is used to make crates, fibres and ropes as
well as being used in textiles, stationery and car parts.
In this polymer, the monomer is the alkene called propene.
The polymerisation of a propene monomer to form polypropylene polymer.
A simple representation of polypropylene will be:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 141
Polyvinyl chloride (PVC)
PVC is used in construction, especially plastic piping. With the addition of a plasticiser it is also
used in clothing and upholstery and to replace rubber. The role of the plasticiser is to increase the
ability of a material to change shape without breaking.
Polyvinyl chloride or PVC is formed from the monomer chloroethene, which is commonly known
as vinyl chloride.
The polymerisation of a chloroethane monomer to form a polyvinyl chloride polymer.
A simplified representation of polyvinyl chloride.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 142
Polyvinyl acetate
Polyvinyl acetate or PVA is formed from the monomer ethenyl ethanoate, which is commonly
known as vinyl acetate.
The polymerisation of an ethenyl ethanoate monomer to form a polyvinyl acetate polymer
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 143
A simplified representation of polyvinyl acetate
PVA is used in various glues and adhesives (such as wood glue).
Polystyrene
Polystyrene is made from the monomer styrene which is a liquid petrochemical. Styrene consists
of a benzene ring (a six membered ring with three double bonds) bonded to an ethene chain.
The polymerisation of a styrene monomer to form a polystyrene polymer.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 144
A simplified representation of polystyrene will be:
Polystyrene is an aromatic polymer and has many uses including protective packaging, in trays,
as plastic lids and bottles.
CONDENSATION POLYMERISATION
In this type of reaction, two monomer molecules combine by means of a covalent bond, and a
small molecule such as water is lost in the bonding process. Nearly all biological macromolecules
are formed using this process. Polyesters are polymers that form through condensation
polymerisation.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 145
Condensation polymerisation: Molecules of two monomers with different functional
groups undergo condensation reactions with the loss of small molecules, usually water
Condensation polymer: A polymer formed by two monomers with different functional
groups that are linked together in a condensation reaction in which a small molecule,
usually water, is lost.
𝐇𝟐 𝐎
𝐀 − 𝐀 − 𝐀 − 𝐀 − 𝐀 − 𝐀 − 𝐎𝐇 𝐇 − 𝐁 − 𝐁 − 𝐁 − 𝐁 − 𝐁 − 𝐁
𝐀−𝐀−𝐀−𝐀−𝐀−𝐀−𝐁−𝐁−𝐁−𝐁− 𝐁−𝐁
Polyesters have a number of characteristics which make them very useful. They are resistant to
stretching and shrinking, they are easily washed and dry quickly, and they are resistant to mildew.
It is for these reasons that polyesters are being used more and more in textiles. Polyesters are
stretched out into fibres and can then be made into fabric and articles of clothing. In the home,
polyesters are used to make clothing, carpets, curtains, sheets, pillows and upholstery.
Polyesters are a group of polymers that contain the ester functional group in their main chain. This
bond is called an ester linkage. For a polyester however there needs to be continuation of the
chain. This requires a -diol (two alcohol functional groups) and a diacid (two carboxylic acid
functional groups), or a monomer that contains both a hydroxyl group and a carboxylic acid.
Final activities:
- The teacher must summarise the lesson.
Summary:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 146
Monomer are small organic molecules that can be covalently bonded to each other in a
repeating pattern.
Polymer is a large molecule composed of smaller monomer units covalently bonded to
each other in a repeating pattern.
Macromolecule is a molecule that consists of a large number of atoms
Polymerisation is a chemical reaction in which monomer molecules join to form a polymer
Addition polymerisation is a reaction in which small molecules join to form very large
molecules by adding on double bonds
Addition polymer: A polymer formed when monomers (usually containing a double bond)
combine through an addition reaction
Condensation polymerisation: Molecules of two monomers with different functional
groups undergo condensation reactions with the loss of small molecules, usually water.
Condensation polymer: A polymer formed by two monomers with different functional
groups that are linked together in a condensation reaction in which a small molecule,
usually water, is lost.
- The teacher must orientate the homework.
Homework:
Activity 1 (Exercise 4 - 30.2 page 202, Siyavula book, Grade 12):
Name the monomer (where polymer is given) or polymer (where monomer is given) for the
following:
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 147
Activity 2:
You are supplied with the formula of a complete product of a polymerisation reaction.
C6H5
C6H5
C6H5
C6H5
C6H5
C6H5
│
│
│
│
│
│
-CH2- CH- CH2- CH--CH2- CH- CH2- CH--CH2- CH- CH2- CH-
2.1
Identify the substance as either an addition or a condensation polymer.
2.2
Write down the name of the polymer.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 148
Solution of the activities:
Activity 1:
a)
This is a monomer called propene.
The polymerisation of a propene monomer will be polypropene or polypropylene polymer
b)
This is a polymer polyethene
The monomer for this polymer is ethene or ethylene.
c)
This is the polymer polyvinyl chloride.
The monomer will be chloroethane.
Activity 2:
2.1
Addiction polymer
2.2
Polystyrene
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 149
ORGANIC MOLECULES
Lesson 14
Topic: Revision exercises on polymers.
Objective:
Learners must be able to:




Distinguish between addition polymerisation and condensation polymerisation.
Identify monomers from given addition polymers.
Write down an equation for the polymerisation of ethene to produce polythene.
State the industrial uses of polythene
Initial activities:
- Control of the attendance of the learners.
- The teacher must control and discuss the homework.
Introduction:
- The teacher must summarise the most important terms of polymers
Summary:
Monomers are small organic molecules that can be covalently bonded to each other in a
repeating pattern.
Polymer is a large molecule composed of smaller monomer units covalently bonded to
each other in a repeating pattern.
Macromolecule is a molecule that consists of a large number of atoms
Polymerisation is a chemical reaction in which monomer molecules join to form a polymer
Addition polymerisationis a reaction in which small molecules join to form very large
molecules by adding on double bonds
Addition polymer:A polymer formed when monomers (usually containing a double bond)
combine through an addition reaction
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 150
Condensation polymerisation: Molecules of two monomers with different functional
groups undergo condensation reactions with the loss of small molecules, usually water.
Condensation polymer: A polymer formed by two monomers with different functional
groups that are linked together in a condensation reaction in which a small molecule,
usually water, is lost.
- The teacher must orientate the objectives of the lesson and write the topic on the board.
Development:
- The teacher must explain some examples.
It is important to be able to identify the monomer used to produce a polymer from the repeat unit
of the chain.
Example 1(Example 32 page 205, Grade 12, Siyabula book).
Which monomer was used to make this polymer (give the name and draw the structure):
SOLUTION:
Step 1: Identify the polymer through functional groups
This polymer contains a chlorine atom. Which polymers contain chlorine? Polyvinylchloride
contains chlorine.
Step 2: Consider what you know of this polymer
PVC forms through an addition reaction. There are two carbon atoms in the repeat
unit(ethane) and the polymer contains chlorine atoms.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 151
Step 3: Apply this knowledge to the monomer
An addition reaction indicates that the monomer must contain a double bond. As the
polymer contains units of ethane the monomer must be an ethene. The monomer must
contain a chlorine atom.
Step 4: Name the monomer
The monomer must be chloro-ethene also known as vinyl chloride
Step 5: Draw the structure of this monomer:
Example 2 (Example 33, page 206, Grade 12, Siyabula book).
Which monomer was used to make this polymer (give the name and draw the structure)?
SOLUTION:
Step 1: Identify the polymer through functional groups
This polymer contains a benzene ring connected to two carbon atoms. Which polymers
contain a benzene ring? Polystyrene contains a benzene ring connected to two carbon
atoms.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 152
Step 2: Consider what you know of this polymer
Polystyrene forms through an addition reaction. There are two carbon atoms attached to
the benzene ring.
Step 3: Apply this knowledge to the monomer
An addition reaction means that the monomer must contain a double bond. The polymer
contains two carbon atoms attached to the benzene ring which must be connected by a
double bond in the monomer.
Step 4: Name the monomer
The monomer must be styrene.
Step 5: Draw the structure of this monomer
Example 3 (Example 35 page 207, Grade 12 Siyabula book).
Which polymer is formed from propene? Give the name and repeat structural unit.
SOLUTION
Step 1: Draw the structural representation of the monomer
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 153
Step 2: Consider what you about this monomer
Propene contains a double bond between the first and second carbon atoms.
Step 3: Apply this knowledge to the polymer
The polymer must form through an addition reaction by breaking the double bond.There
must be a methyl group on every second carbon atom.
Step 4: Name the polymer
The polymer must be polypropene (polypropylene).
Step 5: Draw the repeat structural unit of this polymer.
Example 4 (Example 37 page 208, Grade 12, Siyabula book):
Was an addition or a condensation reaction used to make this polymer?
SOLUTION:
Step 1: Look at the types of bonds in the polymer repeat unit
There are carbon-carbon single bonds. If you expand the repeating unit you can see that
there is an ester linkage in the structural chain.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 154
Step 2: What type of monomer could this polymer have formed from?
The ester linkage requires an alcohol and a carboxylic acid monomer or a monomer that
contains both a hydroxyl and carboxyl functional group.
Step 3: Apply this knowledge to determine the type of reaction
The reaction of an alcohol and carboxylic acid to form an ester linkage (and resulting in the
loss of water) is a condensation reaction.(This is polylactic acid and was formed by the
condensation reaction of lactic acidmonomers)
-
The teacher must orientate the class work.
Activity 1 (Exercise 4 – 32. 1 page 212, Grade 12 Siyabula book)
The following monomer is a reactant in a polymerisation reaction:
a)
b)
c)
What is the IUPAC name of this monomer?
Give the structural representation of the polymer that is formed in this polymerisation
reaction.
Is the reaction an addition or condensation reaction?
Activity 2 (Exercise 4 – 32.2 page 213, Grade 12 Siyabula book)
The polymer below is the product of a polymerisation reaction.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 155
a)
b)
c)
d)
Give the structural formula of the monomer used to form this polymer.
What is the name of the monomer?
Draw the abbreviated structural formula for the polymer (the repeat unit).
Has this polymer been formed through an addition or condensation polymerisation
reaction?
Activity 3 (Exercise 4 – 32. 1 page 213, Grade 12 Siyabula book)
A polymerisation reaction takes place between two lactic acid monomers.
a)
b)
c)
Give the structural representation for:
i.
lactic acid
ii.
the repeat unit of the polymer product
What is the name of the product?
Was this polymer formed through an addition or a condensation reaction?
- The teacher must mark the class work.
Final activities:
-
The teacher must orientate the homework.
Homework:
Activity 4 (Question 2, Exemplar 2014 P2)
The letters A to G in the table below represent seven organic compounds.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 156
4.1
4.2
4.3
Write down the:
4.1.1 Name of the homologous series to which compound F belongs
(1)
4.1.2 Name of the functional group of compound D
(1)
4.1.3 Letter that represents a primary alcohol
(1)
4.1.4 IUPAC name of compound A
(2)
4.1.5 Structural formula of the monomer of compound B
(2)
4.1.6 Balanced equation, using molecular formulae, for the combustion of compound E in
excess oxygen
(3)
Briefly explain why compounds C and D are classified as POSITIONAL ISOMERS.
(2)
Compound G is prepared using an alcohol as one of the reactants. Write down the balanced
equation for the reaction using structural formulae for all the organic reagents.
(7)
[19]
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 157
SELF-ASSESSMENT
QUESTION 1
Consider the organic compounds represented by the letters A to H.
A
C
B
𝐶𝐻3 − 𝐶𝐻 − 𝐶𝐻3
Butane
𝐶𝐻3 − 𝑂 − 𝐻
D
𝐶𝐻3
𝐶𝐻3 − 𝐶 ≡ 𝐶 − 𝐶𝐻3
E
G
1.1
1.2
1.3
F
𝐶𝐻3 − 𝐶𝐻2 − 𝐶𝐻2 − 𝐶𝑙
H
Propanone
Write down the LETTER that represents a compound that:
1.1.1
is a ketone
(1)
1.1.2
is an alcohol
(1)
1.1.3
is an alkyne
(1)
Write down the IUPAC name of:
1.2.1
Compound C
(2)
1.2.2
Compound E
(2)
1.2.3
Compound F
(2)
1.2.4
The monomer of compound A
(1)
Write down the structural formula of:
1.3.1
Compound B
(2)
1.3.2
Compound H
(2)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 158
1.4
Name the homologous series to which compound G belongs.
1.5
The table contain compounds that are structural isomers.
(1)
1.5.1 Define the term structural isomer.
(2)
1.5.2 Write down the LETTERS that represent two compounds that are structural
isomers.
(1)
1.5.3 Are these structural isomers chain isomers, positional isomers or functional
isomers?
(1)
[19]
QUESTION 2
The following diagram below shows the convertion of propene to a secondary acohol.
Propene
HBr
Compound X
Substitution
Secondary alcohol
H2O
2.1
Use structural formulae to write a balanced equation for the formation of compound
X.
(4)
2.2
Name the type of reaction that takes place when propene is converted to compound X.
(1)
2.3
Write down the condensed structural formula and IUPAC name of the secondary
alcohol that is formed.
(2)
Name the type of substitution reaction that takes place when compound X is converted
to secondary alcohol.
(1)
2.4
2.5
2.6
With the aid of a catalyst such as H2SO4, propene can be converted directly to the
secondary alcohol, with out the formation of the intermediate compound X.
2.5.1 Beside propene, write down the NAME of the reactant needed for this direct
conversion.
(1)
2.5.2 Write down the common name of the catalyst.
(1)
2.5.3 Name the type of reaction that will take place during this direct conversion.
(1)
Instead of adding water to compound X, concentrated sodium hydroxide is added and
the mixture is heated. Name the type of reaction that takes place.
(1)
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
[12]
Page 159
Solution of the activities:
Activity 1
a) Propene
b)
c) Addition.
Activity 2
a)
b) The monomer is chloroethene or vinyl chloride.
c)
d) Addition.
Activity 3
a) i.
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 160
ii.
b) polylactic acid
c) condentation.
Activity 4:
4.1
4.1.1
4.1.2
4.1.3
4.1.4
Alkynes 
Hydroxyl group 
C
2-methylpentan-3-one 
(1)
(1)
(1)
(2)
4.1.5

4.1.6 2C4H10 + 13O2 → 8CO2 + 10H2O 
(2)
4.2
Bal. 
(3)
Same molecular formula, but different positions of the functional group. 
(2)
4.3
(7)
[19]
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 161
MEMOORANDUM OF SELF-ASSESSMENT
QUESTION 1
11
1.1.1
H
(1)
1.1.2
D
(1)
1.1.3
E
(1)
1.2
1.2.1
2-methylpropane/2-metielpropaan 
(2)
1.2.2
But-2-yne/ But-2-yn
(2)
1.2.3
1-chloropropane /1-chloropropaan 
(2)
1.2.4
Ethene/eteen
(1)
1.3
1.3.1
(2)

1.3.2
(2)


1.4
Carboxylic acids/Karboksielsure
(1)
1.5.1
(2)
1.5
Organic molecules with the same molecular formula, but different structural
formulae.
Organiese molekule met dieselfde molekulêre formule, maar verskillende
struktuurformules. 
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 162
1.5.2
B - C
(1)
1.5.3
Chain isomers/kettingisomere
(1)
[19]
QUESTION 2

2.1


(4)

2.2
2.3
Addition reaction (or hydrohalogenation or hydrobromination)
(1)
𝑂𝐻 
𝐶𝐻3 − 𝐶𝐻 − 𝐶𝐻3
Propan-2-ol
2.4
Hydrolisis
2.5.1
Water
2.5.2
Sulfuric acid
2.5.3
Addition or hydration
2.6
Elimination reaction or dihydrohalogenation
(2)
(1)
(1)
(1)
(1)
(1)
[12]
Developed by: G. Izquierdo Rodríguez & G. Izquierdo Gómez
Copyright reserved©
Page 163