TOPIC 10:
Organic Chemistry
10.1 Homologous Series (SL/HL)
This is the name given to a group of compounds that can be described by a
general formula.
Name
Alkanes
Alkenes
Alcohols
General
Formula
CnH2n+2
CnH2n
CnH2n+2O
Examples
CH3CH3
CH3CH2CH3
CH3CH2CH2CH3
CH2CH2
CH2CHCH3
CH2CHCH2CH3
CH3CH2OH
CH3CH2CH2OH
CH3CH2CH2CH2OH
 All members of a homologous series have similar chemical properties.
 The next member in a homologous series differs by a –CH2- group
(called a repeating unit).
 This causes a gradual change in physical properties as the chain length
increases. (see below)
10.1 Physical Properties in Homologous Series
 Boiling Point increases as chain length increases.
 This is due to an increase in Van De Waal forces between the molecules
as the molar mass (and surface area) increases.
10.1 Representing Organic Structures
Organic structures can be represented in a variety of ways:
Empirical Formula: Simplest ratio of atoms of each elements in a
compound.
E.g. Pentene: Empirical Formula = CH2
Molecular Formula: Actual numbers of atoms of each element present in
the compound.
E.g. Pentene: Molecular Formula = C5H10
Structural Formula: How the atoms are arranged in the molecule.
Condensed structural formula shows the position of the atoms but omits the
bonds.
E.g. Pentene:
CH2=CHCH2CH2CH3 or CH2CH(CH2)2CH3
Full structural formula/graphic formula/displayed formula = Shows every
atom and bond.
E.g. Pentene:
H
H H
H H
C=C - C - C - C - H
H
H
H
H
Show the condensed structural formula and the full structural of the
following molecule:
Hexane: Molecular formula = C6H14
TOK: Why is it necessary to use different types of formulas to represent
organic molecules?
10.1 Functional Groups (SL/HL)
 The functional group is the reactive part of an organic molecule that
largely dictates its chemical properties.
Homologous Series
Formula of Functional Group
Alkane
R-H
Alkene
C=C
Benzene ring
Halogenoalkane
(Halide)
R-X
Alcohol
R-OH
Aldehydes
Ketones
Carboxylic acids
Amines
Amides
Amino Acids
Esters
HL ONLY :Nitriles
R-NH2
Example
10.1 Naming Organic Compounds (SL/HL)
1. Find the longest chain
1 = meth
2 = eth
7 = hept
8 = oct
3 = prop
4 = but
5 = pent
6 = hex
2. Name any branches
CH3 = methyl
C2H5 = ethyl
C3H7 = propyl
3. Number the positions of the branches (keeping the numbers as
low as possible)
E.g. CH3CH2CH(CH3)CH3
2-methyl butane
When there are more than one side chains, these are indicated as di = 2
tri = 3 and tetra = 4
E.g. CH3C(CH3)2CH2CH3
CCl3H
2,2-dimethylbutane
trichloromethane
4. Name the Functional Group.
These are usually indicated by a suffix
Functional group
Alkane
-ane
Alkene
-ene
Alcohol
-anol
Aldehyde
-anal
Ketone
Carboxylic acid
Benzene
HL ONLY:
Suffix
-anone
-anoic acid
Benzene
Esters
- anoate
Amines
- amine
Amide
- amide
Nitriles
- nitrile
A few are indicated by a prefix
Functional group
Halogenoalkane
Benzene
Prefix
Chloro-, bromo-, iodo-.
Phenyl/benzene
TOK: Why are organic compounds named in this way?
What advantages are there internationally?
This system of naming has meant that many older names of substances
(chloroform, toluene, formaldehyde) have become obsolete. What are the
advantages and disadvantages of this process?
E.g.
CH3OH
methanol
CH3CH2Cl
chloroethane
CH3CH2CH3CHO
butanal
CH3COCH3
propanone
CH3CH2CO2H
Propanoic acid
CH2CHCH3
CH3CH2CH2OH
HCOOH
CH3CH2CH3I
CH3COCH2CH2CH3
CH3CHO
HL ONLY: CH3CH2CH2NH2
CH3(CH2)3CN
CH3COOCH3
HCOOCH2CH2CH3
CH2COONH2
Extension: C6H5CH3
CH3CH2COONHCH3
C6H5CH2COOH
C6H5OH
10.1 Structural Isomers (SL/HL)
 Two organic molecules can have the same molecular formula, but
have different structural formulas depending on how the atoms are
arranged.
10.1 Structural Isomers of Alkanes (SL/HL)
 This is called chain isomerism and is a type of structural
isomerism.
Draw and name the chain isomers of C4H10
Draw and name the chain isomers of C5H12 and C6H14.
Note:
 Straight chain isomers tend to have higher melting and boiling points
than branched chains.
 This is due to the reduction in surface area of the branched chains and
the reduced Van De Waal forces that this causes.
10.1 Structural Isomers in other Functional Groups (SL/HL)
 If the functional group can be in more than one position then it must
be numbered and named accordingly.
CH3CH2CH2OH propan-1-ol
CH3CH(OH)CH3 propan-2-ol
 This is a different type of isomerism called position isomerism.
CH2=CHCH2CH3 but-1-ene
CH3CH=CHCH3 but-2-ene
 The final type of structural isomerism is called functional group
isomerism.
 This occurs when a compound has the same molecular formula, but
has a different functional group.
E.g.
aldehyde/ketone:
C3H7O
aldehyde:
Ketone:
CH3CH2CHO (propanal)
CH3COCH3 (propanone)
carboxylic acids/esters:
C3H6O2
alcohols/ethers*:
C3H8O
10.1 Primary, Secondary and Tertiary Structures.
This indicates the number of carbon atoms attached to the functional group
in the organic molecule.
Primary, secondary and tertiary structures often react very differently (see
oxidation of alcohols and nucleophilic substitutions of halogenoalkanes)
Primary
Secondary
Tertiary
Draw the condensed structural formula of the isomers, name the molecules
and then decide if they are primary secondary or tertiary structures.
C4H10O
C4H10Br
10.1 Physical Properties of Functional Groups (SL/HL)
1. Boiling Points and Volatility
 The longer the carbon chain in an organic molecule, the higher the
boiling point. This is due to an increase in molar mass and consequent
increase in Van De Waals forces and occurs in each homologous series.
 The nature of the functional group plays a large effect in the type of
intermolecular force and therefore the boiling point of each series of
compounds.
 Hydrocarbons have only weaker Van De Waals forces. They tend to
have low boiling points.
 Aldehydes, ketones, halogenoalkanes and esters have stronger dipoledipole forces and therefore have higher boiling points.
 Amines, amides and more particularly alcohols and carboxylic acids
have very high boiling points due to the formation of hydrogen bonds
between the molecules.
 Volatility (ease of vaporisation) follows the same trends. Compounds
that are hydrogen bonded tend to be less volatile, and substances tend to
be less volatile as carbon chain increases.
2. Solubility in water
 Organic compounds that can form hydrogen bonds with water tend to
be soluble in water (carboxylic acids, alcohols, amines, amides.)
 The solubility decreases as the length of the non-polar carbon chain
increases.
 The most soluble compounds therefore tend to be short chain alcohols
and carboxylic acids.
 Carboxylic acids are more soluble than alcohols since they can form
stronger H-bonds due to the additional electronegativity of C=O group.
10.2 Reactions of Alkanes (SL/HL)
 Alkanes are relatively unreactive and inert.
 This is due to the strength and lack of polarity of the C-C and C-H bonds.
 The main type of reaction that they will undergo is combustion.
 The products of complete combustion when there is a plentiful supply of
oxygen are carbon dioxide and water.
E.g.
C3H8 +
5 O2
3 CO2
+
4 H2O
 Even though the energy required to break the bonds within the alkane are
very large, the C=O and O-H bonds formed are even stronger resulting in
a lot of energy release and a very exothermic reaction.
 The CO2 released can cause environmental problems. It will absorb IR
radiation reflected from the Earth and so contribute to global warming.
 Incomplete combustion will occur in a limited supply of oxygen. The
products may be carbon monoxide or carbon powder.
E.g.
C3H8 +
3.5 O2
3 CO +
4H2O
C3H8 +
2 O2
3C
4H2O
+
 All these combustion reactions are redox reactions. In all cases the
oxygen is reduced to O2- in H2O by gaining electrons.
 In complete combustion the carbon is oxidised to C+4 in CO2
 In incomplete combustion the carbon is only oxidised to C+2 in CO, or C0
in elemental carbon.
 The products of incomplete combustion can be serious pollutants. Carbon
monoxide is a poisonous gas that is absorbed by the blood instead of
oxygen leading to internal suffocation. Carbon powder in the air, if
breathed in, will settle on the lining of the lungs causing serious
breathing difficulties.
10.2 Free Radical Substitution Reactions of Alkanes (SL/HL)
 Alkanes will react with halogens (Cl2 or Br2) in the presence of ultra
violet light.
 The mechanism for reaction involves free radical substitution,
 Initiation: The bond between the halogen atoms is quite weak. The
UV light has sufficient energy to cause it to break so that each of the
atoms has an unpaired non-bonded electron.
Cl – Cl
Cl*
+
Cl*
 This is called homolytic fission.
 The highly reactive species created are called free radicals.
 Since this step in the reaction mechanism involves the generation of
reactive free radicals it is called the initiation step.
 Propagation: These free radicals will react with inert alkane
molecules to produce a halogenoalkane and hydrogen halide.
CH4 +
Cl*
CH3* +
CH3* + Cl2
CH3Cl +
HCl
Cl*
 Since these steps in the reaction mechanism involve both using and
generating free radicals, they are called propagation steps.
 Termination: The possibility of two free radicals being used up can
stop the reaction.
CH3* +
Cl*
CH3Cl
 This step in the mechanism is called the termination step.
 Show and label the steps in the mechanism for the free radical
substitution of ethane with bromine.
10.3 Reactions of Alkenes (SL/HL)
 The C=C bond in alkenes is said to be unsaturated.
 The C=C bond is rich in electrons and tends to be attacked by positive
species which are electron deficient (called electrophiles)
 The C=C bond is not twice as strong as the C-C and so the conversion
of a double into a single bond is relatively energetically favourable
with a reasonably low activation energy.
 The most common type of reactions of the alkenes involves the
(electrophillic) addition of a species to the double bond. (The bond is
then said to be saturated).
Common Addition Reactions
1. With Hydrogen (Hydrogenation)
E.g.
This reaction is industrially important: Hydrogen is added to unsaturated
vegetable fats during the manufacture of margarine. This removes some of
the C=C bonds and results in an increase in melting point, allowing the
margarine to become a solid at room temperature.
2. With Bromine (Bromination)
E.g.
When a solution of bromine is added to an alkene, the orange colour
disappears. This is used as a test for unsaturation (molecules containing
C=C bonds).
3. With Water (Hydration)
E.g.
This reaction is industrially important in the production of ethanol from
ethane (obtained by cracking crude oil) as an important alternative to the
formation of ethanol through fermentation.
4. With Hydrogen Halides
E.g.
Note: In examples 3 and 4 it is important to note the symmetrical nature of
the alkene. In an unsymmetrical alkene, there are two possible products. IB
(SL/HL) does not require this level of understanding.
10.3 Reactions with other Alkenes (Polymerisation) (SL/HL)
 An alkene (monomer) will undergo addition reactions with
itself to form a long chain (polymer) under heat and pressure.
 This process is called polymerisation.
 Man-made polymers are called plastics.
E.g. ethene forms polyethene.
E.g.
chloroethene forms polychloroethene. (PVC)
The reactions have huge industrial importance due to the desirable and
diverse properties of plastics.
Show an equation that illustrates the conversion of propene to polypropene
by polymerisation:
10.4 Alcohols (SL/HL)
 Alcohols can be classed according to their structure as primary,
secondary or tertiary.
 This class of structure plays an important role in the reactions of the
alcohols.
 What is the structural difference between a primary, secondary and
tertiary alcohol?
 Alcohols can be used as fuels and will combust readily in a supply of
oxygen.
E.g.
CH3OH +
1 ½ O2
CO2 +
2H2O
Give an equation that shows the complete combustion of ethanol.
10.4 Oxidation of Primary Alcohols (SL/HL)
 Primary alcohols can be easily oxidised using acidified potassium
dichromate solution [O] (a strong oxidising agent see Topic 9).
 This causes the chromium to change from +6 state (orange) to +3
(green)
 Give the ionic half equation showing the action of this oxidising
agent.
 The initial product of this oxidation reaction is an aldehyde.
E.g. Ethanol.
Where [O] is used as a symbol for the oxidising agent.
 To obtain the aldehyde in good yield it must be distilled from the
reaction mixture and away from the oxidising agent as soon as it is
produced. (aldehydes have a much lower boiling point than the
unreacted alcohol)
 Aldehydes can be further oxidised to carboxylic acid if they are kept
in the reaction mixture with the oxidising agent. This process is called
heating under reflux.
E.g. Ethanal.
Where [O] is used as a symbol for the oxidising agent.
Show all the steps, reactants and reaction conditions involved in the
complete oxidation of methanol.
10.4 Oxidation of Secondary and Tertiary Alcohols (SL/HL)
Secondary alcohols:
These are oxidised to ketones.
They cannot be further oxidised as there are no
more H atoms that can be replaced.
E.g. propan-2-ol
Where [O] is used as a symbol for the oxidising agent.
Tertiary alcohols :
Cannot be oxidised by a dichromate solution.
Show all the steps, reactants and reaction conditions involved in the
oxidation of butan-2-ol.
10.5 Reactions of the Halogenoalkanes (SL/HL)
 Halogenoalkanes can be classed according to their structure as
primary, secondary or tertiary.
 This class of structure plays an important role in the reactions of the
halogenoalkanes.
 Halogenoalkanes contain polar C-X bonds that make them easy to
convert to other products.
 Halogenoalkanes will undergo nucleophilic substitution (SN)
reactions.
 This involves the attack of a negatively charged species (called a
nucleophile) on the slightly positive carbon atom bonded to the
halogen.
 This causes the polar C-X bond to break so that the halogen is
substituted by the nucleophile.
E.g. 1.
Reaction with NaOH
 There are 2 mechanisms for nucleophilic substitution reactions,
called SN1 and SN2. This stands for substitution nucleophilic first
or second order.
SN1 Mechanism
 Tertiary halogenoalkanes react in a SN1 mechanism.
 The rate of this reaction is dictated only by one factor, the
concentration of the halogenoalkane.
Rate = k [halogenoalkane]
 Only the halogenoalkane is involved in the rate expression and
therefore involved in the rate determining step (see topic 7
kinetics mechanisms)

This is therefore a first order reaction that is unimolecular
process with a molecularity of one.
First stage (rate determining = slow)



The first stage of the mechanism involves the breaking of the
C-X bond to give an intermediate carbocation.
This is bond breaking is called heterolytic fission.
This is a slow process and so is the rate-determining step.
Second stage (Fast)
 The second stage involves the nucleophile bonding with the
carbocation intermediate. This happens very quickly. It therefore
does not appear in the rate expression.
SN2 Mechanism
 Primary halogenoalkanes react in this way.
 The rate of reaction is dictated by two factors. By both the
concentration of the halogenoalkane and the concentration of the
nucleophile.
Rate = [Halogenoalkane] [Nucleophile]
 Both the halogenoalkane and the nucleophile are involved in the
rate expression and therefore both are involved in the rate
determining step (see topic 7 kinetics mechanisms)

This is a second order reaction, a bimolecular process with a
molecularity of two.
 This mechanism involves the simultaneous attack of the nucleophile
and loss of the halogen.
 An activated complex is formed during the process.
E.g.
 Secondary halogenoalkanes react by a combination of both SN1
and SN2 mechanisms.
10.6 Reaction Pathways
Deduce the two stage reaction pathways for the following organic
conversions, given the starting materials and products:
Include the reagents and conditions required.
Starting Material
Product
Propene
Propanone
Methane
Methanol
Bromoethane
Ethanal
Extension:
Ethene
© Chemistry@ Shatin College, Hong Kong.
Tetrachloroethane