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SCI 1054 CHEMISTRY SEMESTER 1
B) Bomb Calorimeter

Figure 6.6 Bomb calorimeter
 By using a bomb calorimeter, heat is absorbed by both the water and the metal parts

of the calorimeter.
 Heat released by reactions = heat absorbed by water (mw cw ΔT) + heat absorbed by

calorimeter (mc cc ΔT)

where q = mw cw ΔT + mc cc ΔT
mc
mw = mass of calorimeter
cw = mass of water
cc = specific heat of water
T = specific heat of calorimer
= temperature change

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Example 6.3.2

Calculate the amount of the heat released from the reaction which is carried out in an
aluminium calorimeter with a mass of 3087.0 g and contains 1700.0 mL of water. The initial
temperature of the calorimeter is 25.0°C and at the end of the reaction, the temperature
increased to 27.8°C.

Given,

Specific heat capacity of aluminium = 0.553 Jg-1 °C-1

Specific heat capacity of water = 4.18 Jg-1 °C-1

Water density = 1.0 g mL-1

Solution:

ΔT = (27.8 -25.0 )°C = 2.8°C

Heat released heat absorbed by + heat absorbed by aluminium
= water calorimeter

q = mwcwΔT + mcccΔT
= (1700.0 g)(4.18 Jg-1 °C-1)(2.8 °C) + (3087.0 g)(0.553 Jg-1 °C-1)(2.8°C )

= 19896.8 + 4779.91
= 24676.71 J
= 24.68 kJ

= 24.7 kJ

Heat released, q = - 24.7 kJ

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Example 6.3.3
In an experiment, 0.100 g of H2 and 0.800 g of O2 were compressed into a 1.00 L bomb, which
was then placed into a calorimeter that has a heat capacity of 9.08 x 104 JoC1. The initial
temperature of the calorimeter was measured to be 25.000 oC and after the reaction took place,
the final temperature of the calorimeter was 25.155 oC. Calculate the amount of heat given off
in the reaction of H2 and O2 to form H2O, expressed in kJ per mole.
Solution:
Heat given off by the reactants = Heat absorbed by the calorimeter

q = C∆T = mole of H2 used
= (9.08 X 104 J0C-1) X ( 0.1550C) = 0.100 / 2.016
= 1.41 X 104 J = 0.0496 mol
= 14.1 kJ
Heat released, q = - 14.1 kJ

H2(g) + ½O2(g) → H2O(c)
Amount of H2O produced in the reaction

Heat of reaction = - 14.1 kJ/ 0.0496 mol
= - 284 kJ mol1

6.4 STANDARD CONDITIONS IN THERMOCHEMICAL MEASUREMENTS
The enthalpy change of a chemical reaction depends on:

 Pressure
 Temperature
 Concentration of solutions
 Physical state of the substances
 Type of allotropes
The standard conditions used in thermochemistry measurement are:
 Pressure is fixed at 1 atm or 101 kPa

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 Temperature is fixed at 25C or 298 K
 The concentration of solutions is at 1.0 mol dm-3
 The most stable allotrope at 298 K and 1 atm is used.

The enthalpy change measured under these conditions is called the standard enthalpy
change of reaction, and is represented by ΔH or ΔH(298).

6.5 TYPES OF ENTHALPY

Standard enthalpy change (H) is measured under standard condition.
Standard condition: Pressure 1 atm, temperature 25C.

Enthalpy change (H) is measure under non-standard condition.

C) Standard Enthalpy of Formation (Hf)

 The heat change when 1 mole of a compound is formed from its elements in their stable
or standard states at a constant pressure of 1 atm and fixed temperature of 298 K.

H2 (g) + ½ O2(g) → H2O (l) H  = 296 kJ mol1
f

 Every different compound will has their respective Hf value. But the standard enthalpy
of formation of any element in its most stable state form is ZERO.

Hf (O2 ) = 0 Hf (Cl2) = 0

 The value of standard enthalpy of formation of a substance can be either positive or
negative.

2 C (s) + H2 (g) → C2H2 (g) H  = +237 kJ mol1
f

C(graphite) + O2(g) → CO2(g) H  = 394 kJ mol1
f

iii) Finding Hf of substance

 The standard enthalpy of formation of some substances cannot be directly measured
from experiments.

 For example, carbon does not combine directly with hydrogen, under any conditions, to
produce ethyne, C2H2. The standard enthalpy of formation of ethyne is calculated from
other known enthalpy values using Hess’ Law.

 In the case below, given the standard enthalpy of combustion for CO2, H2O and C2H2.

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 From these values, we can find standard enthalpy of formation Hf for ethyne.

C(graphite) + O2(g) → CO2(g) Hc = 394 kJ
H2 (g) + ½ O2(g) → H2O (l) Hc = 296 kJ
C2H2 (g) + 5/2 O2 (g) → 2 CO2 (g) + H2O (l) Hc = 1 311 kJ

A thermochemical cycle is then constructed.

2 C (graphite) + H2 (g) C2H2 (g)

(-394 x 2) -296 5/2 O2
-1311

2 CO2 (g) + H2O (l)

By Hess‟ Law: x + (-1311) = 2 (-394) + (-286)
x = + 237 kJmol-1

Note : Algebraic method is the other alternative method in finding Hf of ethyne.
iv) Finding standard enthalpy change of reaction by using Hf value

 The standard enthalpy of formation can be used to calculate the enthalpy change of a
reaction.

 The enthalpy change of a reaction  ΔHf (product) -  ΔHf (reactants)
For a general reactions:

A+B → C+D

ΔHreaction =  ΔHf ( C + D) -  ΔHf (A + B)

Example: 6.6.1

The ΔHf for the following reaction is – 566 kJ.
2 CO (g) + O2 (g) → 2 CO2 (g)

Given that ΔHf of CO2 = - 394 kJ mol-1, calculate the ΔHf of CO.

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Solutions:

Lets the standard enthalpy of formation of CO = x kJ mol-1

2 CO (g) + O2 (g) → 2 CO2 (g)
ΔHºf : 2x 0 (- 394 x 2)

 - 566 = (- 394 x 2 ) – (2x + 0)
x = - 111 kJ mol-1

B) Standard Enthalpy of Combustion (Hc)

 the heat released (evolved) when 1 mole of substance is burnt completely in
excess oxygen at a standard reference state of 1 atm and 25°C.

C(s) + O2(g) → CO2(g) Hc = 393 kJ mol1
C3H8(g) + 5 O2(g) → 3 CO2 (g) + 4 H2O (l) Hc =  2220 kJ mol1

 Hc value is always negative.
Example 6.6.2
The heat given off when 1.5 g propanol, C3H7OH, was burned in excess oxygen raises the
temperature of 1 500 g of water by 9.2 °C. Calculate the standard enthalpy of combustion of
propanol.
(Specific heat capacity of water = 4.2 J g -1K-1.)

Solution:
Heat given off , q = m c ΔT

= 1 500 x 4.2 x 9.2 x 10-3
= 57.96 kJ
Moles of propanol = 1.5 g/88 g mol-1
= 0.017 mol
The Standard enthalpy of combustion = - 57.96 kJ / 0.017 mol = - 3 409 kJmol-1.
The standard enthalpy of combustion of a substance can be accurately measured using „bomb
calorimeter‟.

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Example 6.6.3

Combustion of 3.05 g of benzoic acid (Mr = 122) in a „bomb calorimeter‟ raises the temperature
of the water by 7.68C. Combustion of a substance A (Mr = 79) in the same calorimeter
increases the temperature by 6.92°C. Calculate the standard enthalpy of combustion of A if the
standard enthalpy of combustion of benzoic acid is -3230 kJ/mol. (mass of A=2.15g)

Solution:

Number of moles of benzoic acid = 3.05 g / 122 g mol-1 = 0.025 mol

The heat liberated = 0.025 mol x 3 230 kJ
= 80.75 kJ.

80.75 kJ increases the temperature of the calorimeter by 7.68C.

The heat capacity of the calorimeter = 80.75 / 7.68
= 10.51 kJ K-1 ( or kJC-1)

The heat evolved by burning of 2.15 g of A = 10.51 x 6.92 kJ
= 72.73 kJ
Moles of A used = 2.15 / 79 = 0.0272 mol

Heat evolved per mole of A = 72.73 / 0.0272 = 2 673.9 kJ

The standard enthalpy of combustion of A = - 2 673.9 kJ mol-1.

C) Standard Enthalpy of Neutralization (ΔHneut)

 the heat evolved (released) when 1 mole of water, H2O is formed from the neutralization
of acid and base at 1 atm and 25C.

 the value of ΔHneutralisation is always negative. ΔHneutralisation = -ve
 Acid + Base → salt + 1 mole of water
ΔH = 66.5 kJmol1
 ½ H2SO4 (aq) + NaOH (aq) → ½ NaSO4(aq) + H2O(l)

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Example 6.6.4

250 cm3 of 0.5 M NaOH initially at 27.5°C is added to 250 cm3 of 0.5 M HCl, also at 27.5°C, in a
plastic cup. The maximum temperature recorded is 30.9°C. Calculate the standard enthalpy of
neutralization between NaOH and HCl.
[Density of solution = 1.00 gcm-3 ; specific heat capacity of solution = 4.2 J g-1K-1]

Solution:

HCl (aq) + NaOH (aq)  NaCl(aq) + H2O(l)

Heat evolved during the experiment = m c t
= (250 + 250) x 4.2 x (30.9 – 27.5)
= 7 140 J

So, heat evolved = 7140 J

No. of moles of NaOH = No. of moles of HCl = 250/1000 x 0.5
= 0.125 mol

From balanced chemical equation, 1 mol NaOH @ 1 mol HCl  1 mol water,
No. of moles of water produced = 0.125 mol

Heat evolved per mol of water = - 7 140 x 10-3 kJ
0.125

= - 57.12 kJ
 The standard enthalpy of neutralization = - 57.12 kJ mol-1.

D) Standard Enthalpy of Atomization (Ha)

 The energy required to form 1 mole of gaseous atoms from the element under standard
conditions (1 atm and 25C or 298K)

Na (s)  Na(g) ΔH°a = 109 kJ mol1

½ Cl2 (g)  Cl(g) ΔH°a = 123 kJ mol1

 The standard enthalpy of atomization of the noble gases (Group 18) is zero, because all
of them exist as monoatomic gas under standard conditions.

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E) Standard Enthalpy of Sublimation (Hsubl )
 The heat needed (absorbed) when one mole of a substance sublimes (change phase
from solid to gas).

I2 (s) → I2(g) Hsubl = + 62.42 kJ mol1

F) Standard Ionization Energy (IEo)

 First Ionisation energy
o Is the minimum heat/energy required to remove 1 mole of electron from 1 mole
of gaseous atom at 1 atm and 298 K.

Mg (g) → Mg+ (g) + e IE1o = +740 kJ mol1
(1st ionization energy of Mg)

 Second Ionisation energy
Is the minimum heat/energy required to remove 1 mole of electron from every unipositive
ion in 1 mole of unipositive gaseous ions under standard conditions ( 1 atm , 298 K)

Mg+ (g) → Mg2+ (g) + e IE2o = +1400 kJ mol1
(2nd ionization energy of Mg)

Generally the ionization energy increases across a Period and decreases down a Group. The
bigger the size of atom, the lower the ionization energy.

G) Standard Electron Affinity (EAo)

 The first electron affinity:
The heat liberated when 1 mole gaseous atom receives 1 mole electron to form
uninegative charged ion, per mole of gaseous ions, under standard conditions (1 atm,
298K)

O (g) + e → O (g) EA1 = 142 kJ mol1

(First electron affinity of O)

 Second and higher electron affinities are positive.

O (g) + e → O2 (g) EA2 = +844 kJ mol1

(Second electron affinity of O)

This is due to the repulsion between the negative charged ion and the added electron.
Thus, energy has to be supplied to overcome the repulsive force.
Generally, the smaller the atomic size, the more negative is the electron affinity.

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H) Lattice Energy (Hlattice )

the heat change when 1 mole of solid (ionic compound) is formed from its constituent
gaseous ions.

Na+ (g) + Cl(g) → NaCl (s) ΔHolattice = - 771 kJ mol1

As the –ve value of lattice energy increase, electrostatic forces between cations and anions
also increase.

Lattice dissociation energy. (- ΔHolattice )
 Energy needed to break down the lattice of 1 mole ionic crystalline solid to form its
gaseous ions.

NaCl (s) → Na+ (g) + Cl- (g) ΔH = + 771 kJ mol1
 The heat released when 1 mole of solid (ionic compound) is formed from its constituent

gaseous ions.
 The lattice energy is measurement of the strength of the ionic bond in an ionic

compound.

 The lattice energy becomes more negative as
 the ionic charges increase

 the ionic radii decrease

 There is a stronger attraction between small, highly charged ions so the ΔH is more
negative.

 MgO has a more negative ΔHlattice than Na2O because Mg2+ is smaller in size and has
bigger charge than Na+. For instance, Hlattice (MgO) > Hlattice (Na2O)

I) Standard Bond Energy (ΔHoB)
 The amount of energy necessary (required) to brake 1 mole of bonds in a gaseous
covalent substance to form products in the gaseous state at constant temperature and
pressure (1 atm , 298 K)

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 The greater the bond energy, the more stable (stronger) the bond is, and the harder it
is to break. Thus bond energy is a measure of bond strengths.

H2 (g) → 2 H (g) ΔHorxn = ΔHoH-H = + 436 kJ / mol H-H bonds.

 The bond energy of the hydrogen-hydrogen bond is 436 kJmol-1 of bonds. In other
words 436 kJ of energy must be absorbed for every mole of H-H bonds that are broken.
This endothermic reaction (ΔHorxn is positive) can be written as
H2 (g) + 436 kJ  2 H (g)

6.6 HESS’S LAW

a) Hess’s Law

Hess’s Law states that when reactants are converted to products, the change in
enthalpy is the same whether the reaction takes place in one step or in the series of
steps.

The enthalpy change depends only on the nature of the reactants and products and is
independent of the route taken. So, the enthalpy change for two routes from the same
reactants A to the same products B will be equal.

H1
AB

H  H3
2
C

By Hess’s Law : H1 = H2 + H3
Hess‟s Law is useful in calculating enthalpy changes that cannot be measured directly from
experiments.

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The enthalpy changes by applying Hess‟s Law can be calculated by using:
a) Algebraic method
b) Energy cycle method

Example 6.5.1:

Find the standard enthalpy of formation of ethane C2H6(g). Given the
following data:

Hc C(graphite) = 393 kJmol1

Hc H2(g) = 286 kJmol1

Hc C2H6(g) = 1560 kJmol1

Solution:

METHOD 1 Algebraic method

Write the balanced chemical equations for the given enthalpy change (combustion equation)

C (s) + O2 (g) → CO2 (g) ΔH° = 393 kJ ------------- a
H2 (g) + ½ O2 (g) → H2O(l) ΔH° = 286 kJ ------------- b

C2H6 (g) + 7/2 O2 → 2CO2 (g) + 3H2O (l) ΔH° = 1560 kJ --------------- c

Chemical equation for the formation of ethane from its element:

2C (s) + 3H2 (g) → C2H6 (g) Hf = ?

Since we want to obtain one equation containing C and H2 as reactants and C2H6 as product, O2
, CO2 and H2O are need to be eliminated from equations given. Reverse the equation c (as well
as the sign of ΔH°), then multiply a by 2 and b by 3 (so as the ΔH°). Next, total up all the
enthalpies

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2CO2(g) + 3H2O(l) → C2H6(g) + 7/2 O2(g) ΔH° = +1560 kJ

2C(s) + 2O2(g) → 2CO2(g) ΔH°= 2(393) kJ

3H2(g) + 3/2 O2(g) → 3H2O(l) ΔH°= 3(286) kJ

2C (s) + 3H2 (g) → C2H6 (g) H  = +1560-786-858
f

= 84 kJ

So, standard enthalpy of formation of ethane is - 84 kJ/ mol.

METHOD 2 Energy cycle method

Write the chemical equation for the required enthalpy change. Look at the data given, if

H  values are given, complete the cycle by writing the combustion equation. If ΔH  values
c f

are given, elements must make up the cycle.
Draw the cycle and add the values (multiplying by a factor where necessary).
Draw the energy cycle and apply Hess‟s Law to calculate the unknown value.

HOf

2C (s) + 3H2 (g) C2H6 (g)
2O2(g)
H O = 3(-286) 7/2 O2 (g)
2 HO3 = - (-1560)

HO1 3/2 O2(g)
= 2(-393)

2CO2 (g) + 3H2O (l )

ΔH°f = 2( ΔH1° ) + 3(ΔH°2 ) + ΔH°3
= 786  858 + 1560
= 84 kJ mol1

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6.7 BORN-HABER CYCLE
Lattice energy cannot be measured directly from experiment. It can only be obtained indirectly
by using Hess‟ Law and a thermochemical cycle known as the Born Haber cycle.
The process of ionic bond formation can be broken down into stages. At each stage the
enthalpy changes are considered.

The Born-Haber cycle is a cycle that involves the enthalpy of formation of a compound, the
ionization energy of a metal, the electron affinity of the non-metal, the enthalpy of sublimation
or atomization of the metal, the bond dissociation energy (the atomization energy of non-metal)
and the lattice energy of the compound.

Example 6.7.1
Consider the enthalpy changes in the formation of sodium chloride. Calculate lattice energy of
NaCl.

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METHOD 1 Energy Cycle H1 = 411.3 kJ
Na(s) +
½Cl2(g) NaCl(s)

ΔH°2= +107.8 kJ ΔH°4= +121.3 kJ H6
Na(g) Cl(g) ΔH°6= ?

ΔH=°3 + 495.4 kJ ΔH5°= 348.8 kJ

Na+(g) + Cl(g)

ΔH1° = Enthalpy of formation NaCl
ΔH°2 = Enthalpy of atomization of Na
ΔH3° = First ionization energy of Na
ΔH°4 = Enthalpy of atomization of Cl
ΔH°5 = Electron affinity of Cl
ΔH°6 = Lattice energy of NaCl

ΔH1° = ΔH°2 + ΔH°3 + ΔH ° + ΔH°5 + ΔH°6
4
411.3 = +107.8 + 495.4 + 121.3 + (348.3) + ΔH6o

ΔH°6 = 787 kJ mol1

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METHOD 2 Energy level diagram

In the presentation of Born-Haber cycle by energy diagram, by convention,

 Positive values are denoted as going upwards 
 Negative values as going downwards 

Na+(g) + NCla(g+)(g) + e- + Cl (g)

Energy Energy ΔHIEo HIE HEA ΔHEAo
NNaa+(+g)(g+) +ClC(gl-)(g)
NaN(ga)(g+) C+l(gC)l (g)
ΔHao Ha (Cl) HlattΔicHe Lattice

NaN(ag)(g+) +½C½l2(Cgl)2 (g)

Ha (Na) NNaaCCl(ls(s))
NaN(as)(s+) +½C½l2(Cgl)2 (g)

ΔHHfof

Figure 6.7 Born-Haber Cycle for the formation of NaCl
ΔH°f = ΔH°a (Na) + ΔH°a (Cl) + ΔH°IE + ΔH°EA + ΔH°lattice

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CHAPTER 7 INTRODUCTION TO
ORGANIC CHEMISTRY

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7.0 INTRODUCTION TO ORGANIC CHEMISTRY
7.1 What is Organic Chemistry?

Organic chemistry is the chemistry of carbon compounds.
Carbon compounds constitute the central chemicals of all things on this planet. Carbon
compounds include deoxyribonuicleic acids (DNAs) the giant molecules that contain the
genetic information for all living species

Carbon compounds make up the proteins of our blood, muscle and skin. They make up the
enzymes that catalyze the reactions that occur in our bodies. Together with oxygen in the air
we breathe, carbon compounds in our diets furnish the energy that sustains life.

Some examples of carbon compounds in our daily lives:
OCOCH3

CH4 COOH
methane methyl salicylic acid (aspirin-a
(a component of natural gas)
drug)

H3C CHCOOH N CH3
CO2CH3
NH2
alanine OCO
(amino acid - a protein component) cocaine (a pain killer)

NN OH CH2CH
Cl
4-hydroxyphenylazobenzene
(a kind of dye n

polivinyl chloride (PVC)
(a type of synthetic leather)

O Cl CH Cl
CCl3
CH2 C NH S
Dichlorodiphenyltrichloroetane
N (DDT- a pesticide component)
O

COOH

penicillin (an antibiotic)

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(CH3)2N NN SO3 -Na+

methyl orange (an indicator)

O CH3

H3C N N

ON N

CH3

caffeine (found in coffee and tea)

7.2 FUNCTIONAL GROUPS AND HOMOLOGOUS SERIES
A functional group is an atom or group of atoms in an organic molecule which
characterized the molecule and enables the molecule to react in specific ways
(determines its chemical properties) .
Functional groups are important for three reasons:
a) They are the units by which we divide organic compounds into classes.
b) They are sites of chemical reaction; a particular functional group, in whatever
compound it is found, undergoes the same types of chemical reactions.
c) Functional groups serve as a basic for naming organic compounds.
Homologous series is series of compounds where each member differs from the next
member by a constant –CH2 unit (14 mass unit).
Members of a homologous series are called homologs.
A homologous series has four features:

1. All the members of a particular homologous series have the same
general formula.
For example, the members of the saturated aliphatic alcohol series have the
general formula, CnH2n+1OH, where n = 1, 2, 3, etc. For n = 1, the formula would
be CH3OH (methanol); for n = 2, C2H5OH (ethanol); and so on.

2. All the members of a particlar homologous series have the same
functional group. Thus, they have the same chemical reactions and can
be made by the same general methods.

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3. For example, all alcohols contain the –OH group; they react with carboxylic acids
4. to give esters. They can be prepared, for instance, by heating dilute sodium
hydroxide solution with an appropriate alkyl halide.
The successive members of any homologous series differ by –CH2.
For example, the first few alcohols are CH3OH (methanol); CH3CH2OH (ethanol);
CH3CH2CH2OH (propanol).
There is a trend in the physical properties of the members of any
homologous series.
As the molecules increase in size, there is a gradual change in physical
properties, e.g. their boiling points increase.

Some important functional groups in organic compounds :-

Homologou Functional group General IUPAC
s Series Formula nomenculature

alkane C=C none CnH2n+2 -prefix
alkene C≡C (double bond) CnH2n -ane
alkyne (triple bond) CnH2n-2 -ene
-yne

arene (aromatic ring) CnH2n-6 -benzene

alcohol –OH (hydroxyl) CnH2n+1 OH alkanol
ether CnH2n+2O alkoxyalkane
haloalkane –OR (alkoxy) CnH2n+1X haloalkane
aldehyde CnH2nO
–X (halogen) alkanal
ketone O CnH2nO
CH alkanone

(carbonyl)
O
C

(carbonyl)

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carboxylic alkanoic acid
acid O CnH2nO2 alkanoyl chloride
CnH2n-1 OCl alkyl alkanoate
acyl chloride C OH
ester (carboxyl) CnH2nO2 -amide
CnH2n+1ON -amine
amide O CnH2n+1 NH2
amine C Cl
(acyl)
O
COC
(carboalkoxy)
O

C NH2
(carboxamide)

-NH2
(amino)

Practice Exercise 7.2.1:

1. Write the general formulae and state the homologous series to which these compounds
belong.

a. C4H6 (cyclic compound) b. C4H10O

c. C2H5ON d. C2H7N

e. C3H6O2 f. C3H6O

2. Penicilin is used as an antibiotic in a medicine. The structure formula of peniciline is
shown below.

Identify all the functional groups in the penicilin molecule.

O S
CH2 C NH N

O COOH

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7.3 Reagents and Sites of Organic Reactions

a) Electrophile

 Literally translated means „electron loving’.
 An electron-deficient species and electron-pair acceptor that attacks a part of a molecule

where the electron density is high.
 An electrophile can be either neutral or positively charged.
 Examples of electrophiles:

1. cations such as H+, H3O+, NO2+, Br+ etc.
2. carbocations.
3. Lewis acids such as AlCl3, FeCl3, BF3 etc.
4. oxidizing agents such as Cl2, Br2 and etc.

 Examples of electrophilic sites in organic molecules :-
molecules with low electron density around a polar bond such as,
d+ d- d+ d- d+ d-
C = O (carbonyl) ; C – X (haloalkanes) ; C – OH (hydroxyl
compounds)
b) Nucleophile

 Literally translated means „nucleus loving’
 An electron-rich species and electron-pair donor that attacks a part of a molecule

where the electron density is low.
 A nucleophile can be either neutral or negatively charged.
 Examples of nucleophiles :

1. anions such as OH-, RO-, Cl-, CN- etc.
2. carbanions ( species with a negative charge on carbon atoms ).
3. Lewis bases which can donate lone pair electrons such as NH3, H2O, H2S etc.

 Examples of nucleophilic sites in organic molecules :-
molecules with high electron density around the carbon-carbon multiple bond such
as -C=C- (alkenes) , -CºC-(alkynes) , benzene ring and etc.

c) Free Radical
 A very reactive species with an unpaired electron.
 Formed in homolytic cleavage.
 Example:

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free radicals uv
Cl Cl
Cl + Cl

CC uv C+C

H CH3 uv H + CH3

7.3.1 Relative Stabilities of Free Radicals
 Free radical can be primary, secondary or tertiary, depending on the number of

carbon atoms directly bonded to the carbon atom with unpaired electron ( for free
radical )
 The stability of free radical increases as more alkyl groups are attached to the carbon
atom with unpaired electron.
Free Radical Stability :

HH RR
HCH < R C H < RCH<R C R

methyl primary secondary tertiary
radical (1o) (2o) (3o)

Increasing stability

7.4 REACTIONS OF ORGANIC COMPOUNDS

7.4.1 Types of Covalent Bond Cleavage/Fission

All chemical reactions involved bond breaking and bond making.

Two types of covalent bond cleavage :-
 Homolytic cleavage
 Heterolytic cleavage

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a) Homolytic Cleavage
 Occurs in a non-polar bond involving two atoms of similar electronegativity.
 A single bond breaks symmetrically into two equal parts, leaving each atom with one

unpaired electron.
 Free radicals are formed in homolytic cleavage.

XX X +X @ 2X
free radicals

 Type of arrow: Half head arrow showing transferring of 1 electron.

b) Heterolytic cleavage
 Occurs in a polar bond involving unequal sharing of electron pair between two atoms of

different electronegativities.
 A single bond breaks unsymmetrically and both the bonding electrons are transferred to

the more electronegative atom.
 Cation and anion are formed in heterolytic cleavage.

Aa-nion+ caBti+on A is more
electronegati

A : B ve

A+ + B- B is more

cation anion electronegativ

e

 Carbocations and free radicals are intermediates in organic reactions. They are unstable
and highly reactive.
 Type of arrow: Full head arrow showing transferring of 2 electrons.

7.4.2 Types of Organic Reactions

The four main types of organic reactions are:
I. Addition
II. Substitution

III. Elimination
IV. Rearrangement

I) Addition Reaction
 A reaction in which atoms or groups add to adjacent atoms of a multiple bond.
 Two types of addition:


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a) Electrophilic Addition
 Initiated by an electrophile, which attacks a nucleophilic site of a molecule.
 Typical reaction of unsaturated compounds such as alkenes and alkynes.
 Example:

CH3CH CH2 + Br2 room CH3CH(Br)CH2Br
temperature

electrophile

b) Nucleophilic Addition
 Initiated by a nucleophile, which attacks an electrophilic site of a molecule.
 Typical reaction of carbonyl compounds.
 Example:

δ- O OH
CH3 C CH3
CH3 C + HCN
CH3 CN

+

II) Substitution Reaction
 A reaction in which an atom or group in a molecule is replaced by another atom or

group.
 Three types of substitution :-

a) Free-radical Substitution
 Substitution which involves free radicals as intermediate species.
 Example :
uv light
CH3CH3 + Cl2 CH3CH2Cl + HCl

b) Electrophilic Substitution
 Typical reaction of aromatic compounds.
 The aromatic nucleus has high electron density, thus it is nucleophilic and is prone to

electrophilic attack.
 Example :

Fe

+ Br2 catalyst + HBr
electrophile

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c) Nucleophilic Substitution
 Typical reaction of saturated organic compounds bearing polar bond as functional
group, such as haloalkane and alchohol.
 Example:

δ+ δ- CH3CH2OH + Br
CH3CH2
Br + OH-
nucleophile

III) Elimination Reaction
 A reaction in which atoms or groups are removed from adjacent carbon atoms of a
molecule to form a multiple bond (double or triple bond).
 Elimination reaction results in the formation of unsaturated molecules.

 Example : conc. H2SO4

CH3CH2OH CH2= CH2 + H2O

IV) Rearrangement Reaction
 A reaction in which atoms or groups in a molecule change position.
 Occurs when a single reactant reorganizes the bonds and atoms.
 Example :
H

HC C R tautomerisme HC C R

H OH HO

Practice Exercise 7.4.1 :
1. Write an equation for the bromine-bromine bond cleavage in the bromination of

methane. State the type of bond cleavage.

2. Which would you expect to be the most stable free radical ?Explain

CH2CH3 , (CH3)2C , CH3 , CH3

H

3. Which of the following species is likely to be an electrophile, and which a nucleophile ?

i) AlCl3 ii) CH3OCH2CH3 iii) CH3OH iv) C6H5N2+

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SCI 1054 CHEMISTRY SEMESTER 1
v) NH3 vi) (CH3)2CH+
vii) viii) OHCH3CH=CH2

4. a) Write an equation to show the reaction between the ammonium and hydroxide ions
to form ammonia and water.
b) Identify the nucleophile and electrophile in the reaction.
c) Use curved arrows to show the transfer of an electron pair when the ammonium and
hydroxide ions react.

5. Write the structural formulae of the species formed in each of the following reactions.

a) Homolysis of CH4
b) Homolysis of C2H6
c) Heterolysis of (CH3)3CCl
d) Heterolysis of CH3Li

6. Identify the types of reactions shown by the following equations:
Br2/CCl4
i. CH3CH2CH=CH2 CH3CH2CHBrCH2Br

ii. CH3CH(OH)CH3 H+ CH3CH=CH2 + H2O

heat

iii. CH3CH2OH + HCl ZnCl2 CH3CH2Cl + H2O

iv. CH3CH2C(OH)=CH2 CH3CH2C CH3
O
AlCl3
C6H5CH3 + HCl
v. C6H6 + CH3Cl

vi. CH3CH2CH2CH3
vii. CH3C≡ CH + Br2 → CH3CBr=CHBr

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SCI 1054 CHEMISTRY SEMESTER 1

viii. + CH3COCl → + HCl

ix. CH2BrCH2Br + Zn → CH2=CH2 + ZnBr2

7.5 Structural Formulae
Representation of structural formula

 Structural formula shows the order in which atoms are bonded together.
 Structural formula can be drawn in the form of condensed, expanded and

skeletal structure.
a) Condensed Structure

 In condensed formulae all the hydrogen atoms that are attached to a particular carbon
are usually written immediately after that carbon.

 In fully condensed formulae all atoms that are attached to a carbon are written
immediately after that carbon

Example: C4H9Cl

CH3CH(Cl)CH2CH3
Condensed Structure

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SCI 1054 CHEMISTRY SEMESTER 1

b) Expanded Structure

Expanded structures indicate the way in which the atoms are attached to each other and
are not representations of the actual shapes of the molecules.

Example: HHHH
C4H9Cl HCCCCH

H Cl H H

Expanded structure

c) Skeletal Structure

This structure shows only the carbon skeleton
The hydrogen atoms that are assumed to be present, are not written.
Other atoms such as O, Cl, N and etc. are shown.

Examples:

Cl

i) CH3CH(Cl)CH2CH3 =

H2C CH2 =
ii) CH2

H2C

iii) CH2 = CHCH2 OH = OH

Practice Exercise 7.5.1:

Rewrite each of the following structures using skeletal formula:
i. CH3CH2CH2CCH3 iii. (CH3 )2 CH CH2 CH2 CH (CH3) CH2 CH3

O

ii. CH3CH2CH(CH3)CH2COH iv. CH2=CHCH2CH2CH=CHCH3

O

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7.6 CLASSIFICATION OF CARBON AND HYDROGEN ATOMS IN ORGANIC
MOLECULES

A carbon atom is classified as primary (1o), secondary (2o), tertiary (3o) or quarternary
(4o) depending on the number of carbon atoms bonded to it.

 a primary (1°) carbon is directly bonded to one other carbon atom (has 1
adjacent carbon atom).

 a secondary (2°) carbon is directly bonded to two other carbon atoms (has 2
adjacent carbon atoms).

 a tertiary (3°) carbon is directly bonded to three other carbon atoms (has 3
adjacent carbon atoms) .

 a quarternary (4°) carbon is directly bonded to four other carbon atoms (has 4
adjacent carbon atoms).

H H CH3 CH3
HCH H C CH3 H C CH3 H3C C CH3

CH3 CH3 CH3 CH3

1° carbon 2° carbon 3° carbon 4° carbon

Similarly, a hydrogen atom is also classified as primary, secondary or tertiary depending
on the type of carbon to which it is bonded.
1o hydrogen atom is bonded to a 1o carbon atom.
2o hydrogen atom is bonded to a 2o carbon atom.
3o hydrogen atom is bonded to a 3o carbon atom.

a) Classification of haloalkanes (alkyl halides)
Alkyl halides are classified based on the carbon atom to which the halogen (X) is directly
attached.

1o alkyl halide – the X atom is bonded to a primary carbon atom.
2oalkyl halide – the X atom is bonded to a secondary carbon atom.
3o alkyl halide – the X atom is bonded to a tertiary carbon atom.

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2o C 3o C

1o C HH H H CH3 H

HH

H C C Cl H C C CH H C C CH

HH H Cl H H Cl H

1o alkyl chloride 2o alkyl chloride 3o alkyl chloride

b) Classification of alcohols
Alcohols are classified based on the carbon atom to which the hydroxyl group is directly
attached.

1o alcohol – the hydroxyl group is attached to a 1o carbon atom.
2o alcohol – the hydroxyl group is attached to a 2o carbon atom.
3o alcohol – the hydroxyl group is attached to a 3o carbon atom.

1o C 2o C 3o C
HH HH H H CH3 H

H C C OH H C H C C CH
CH
C

HH H OH H H OH H

1o alcohol 2o alcohol 3o alcohol

c) Classification of amines

Amines are classified based on the number of alkyl groups or carbon atoms that are
directly attached to the nitrogen atom.

1o amine – N is bonded to one alkyl group.

2o amine – N is bonded to two alkyl groups.

3o amine – N is bonded to three alkyl groups.

HNH H N CH3 H3C N CH3

CH3 CH3 CH3
Primary (1°) amine Secondary (2°) amine Tertiary (3°) amine

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Practice Exercise 7.6.1:

1.
Classify the cabons labelled w , x and y as primary, secondary or
tertiary carbon atoms.

wx

CH3CH2CHCH3
Cy H3

2. A tertiary alcohol has the molecular formula C4H10O. Write the condensed structural formula
of this alcohol.

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SCI 1054 CHEMISTRY SEMESTER 1
7.7 ISOMERISM

Isomerism is the existence of different compounds with the same molecular formula but
different structural formulae.
Different compounds that have the same molecular formula are called isomers.
Two types of isomerism:
a) structural isomerism – isomerism resulting from different order of

attachment of atoms.
b) stereoisomerism – isomerism resulting from different spatial arrangement of atoms in

molecules.
Structural isomers are different compounds with the same molecular formula but differ in
the order of attachment of atoms.
Stereoisomers are isomers that differ only in the arrangement of atoms in space.

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Three types of structural isomerism:

a) Chain/skeletal isomerism

The isomers differ in the carbon skeleton (different carbon chain).
They possess the same functional group and belong to the same homologous series.
Example:

CH3

C5H12 : CH3CH2CH2CH2CH3 CH3CHCH2CH3 CH3CCH3

CH3
CH3

b) Positional isomerism

These isomers have a substituent group in different positions in the same carbon skeleton.

Examples :
i)

C3H7 : CH3CH2CH2Cl CH3CHCH3
1-chloropropane Cl

2-chloropropane

ii) C4H8 :CH2=CHCH2CH3 CH3CH=CHCH3
1-butene 2-butene

c) Functional group isomerism
These isomers have different functional groups and belong to different homologous
series with the same general formula.
Different classes of compounds that exhibit functional group isomerism:

General formula Classes of compounds
CnH2n+2O alcohol and ether
CnH2nO
CnH2n aldehyde and ketone
CnH2nO2 alkene and cycloalkane
carboxylic acid and ester

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SCI 1054 CHEMISTRY SEMESTER 1
Examples:
: CH3CH2OH CH3OCH3
i) C2H6O : ethanol dimethyl ether
CH3CH2CH
ii) C3H6O CH3CCH3
O
iii) C3H6O2 : propanal O
CH3CH2C OH propanone
CH3C O CH3
O
Propanoic acid O
Methyl ethanoate

Practice exercise 7.7.1:

1. There are seven structural isomers of C4H10O. Draw a pair of skeletal(chain), positional
and functional group isomers.

2. For the formula C4H9Cl, draw a representative pair of isomers illustrating each of the
following types of isomerism:
a) chain b) positional

3. Write the sructural formulae of two isomers with the molecular formula C2H4O2 to
illustrate functional group isomerism.

Stereoisomerism : Geometric isomerism

Results from rigidity in molecules and occurs only in two classes of compounds, namely
alkenes and cyclic compounds.
Geometric isomers (also called cis-trans isomers) are stereoisomers that differ by groups
being on the same side (cis-isomer) or opposite sides (trans-isomer) of a site of rigidity
in a molecule.
The requirements for geometric isomerism :

i. restricted rotation about a C=C,double bond, in alkenes or a C-C single bond
in cyclic compounds.

ii. each carbon atom of a site of restricted rotation has two different groups
attached to it.

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Examples :

i) H H H CH3
C C
CC

H3C CH3 H3C H
cis-2-butene trans-2-butene

ii) H3C CH3 H3C CH2CH3
C C
CC

H CH2CH3 H CH3
trans-3-methyl-2-pentene cis-3-methyl-2-pentene

iii) H H H CH3

CH3 CH3 CH3 H
cis-1,2- trans-1,2-
dimethylcyclohexane dimethylcyclohexane
Cl H
iv) Cl Cl

HH H Cl
cis-1,3- trans-1,3-

dichlorocyclopentane dichlorocyclopentane

“Don’t Talk, Just Act. Don’t Say, Just Show. Don’t Promise, Just prove.”
Pixelsquote.net

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SCI 1054 CHEMISTRY SEMESTER 1

If one of the doubly bonded carbons has 2 identical groups, geometric isomerism is not
possible. Observe the following examples:

i) CH3 ii) CH3
H H

CC CC

H CH2CH3 Cl CH3
2-methyl-2-butene 1-chloro-2-methylpropene

cis-trans isomers have similar chemical properties but different physical properties. They
differ in melting and boiling points and solubility due to different polarity of the
molecules. cis-isomers are polar molecules while trans-isomers are non-polar.

 Melting point: trans- isomer > cis-isomer
 Boiling point: cis-isomer > trans- isomer
 Stability: trans-isomer > cis-isomer

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Practice exercises 7.7.2:

1) Describe the structural features which a molecule must possess for it to exist as cis-
trans isomers.

2) Which of the following compounds can exist as a pair of cis-trans isomers? Draw each
cis-trans pair and indicate the geometry of each isomer.

a) CH3CH=CH2 f) CH3CH2CH=CHCH3

b) H3CHC CH g) CH3

OH

c) ClCH=CHBr h) (CH3)2C=CHCH3
d) CH3CH2C(CH3)=CHCH3 i) CH3
j) COOH
e)

CH3 COOH
3) Label each of the following pairs of compounds as structural isomers, geometric

isomers or the same compound:

a) Cl H H CH2Cl
CC CC

H CH2Cl Cl H

b) H3CH2C CH2CH3 H3C H
C C
CC
H CH2CH3
HH

c) H H H2C=HC H
C C
CC
H CH3
H2C=HC CH3

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SCI 1054 CHEMISTRY SEMESTER 1

Answer: b. CnH2n+2O (alcohol)
Exercise 7.2.1: d. CnH2n+1NH2 (amine)
f. CnH2nO (ketone/aldehye)
1.
a. CnH2n-2 (alkene)
c. CnH2n+1ON (amide)
e. CnH2nO2

2. O S carboxyl
aromatic ring CH2 C NH N

O COOH

amide

Exercise 7.4.1
1.
type of cleavage: homolytic cleavage

H3 , (CH3)2C , CH3 , CH3

H 2.
Explanation: the single electron/free radical carbon is stabilized by 3 methyl group
(electron donor)

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3.

i) AlCl3 ii) CH3OCH2CH3 iii) CH3OH iv) C6H5N2+
electrophile electrophile
Nucleophile nucleophile

v) NH3 vi) (CH3)2CH+ vii) viii) OHCH3CH=CH2
nucleophile electrophile nucleophile nucleophile

4.
a. NH4+ + OH-  NH3 + H2O
b. (E) (N)

c. NH3-H+ + OH-  NH3 + H2O

5.
a. CH3. + H.
b. .C2H5 + H.
c. (CH3)3C+ + Cl-
d. CH3- + Li+

6.
a. Addition
b. Elimination
c. Substitution
d. Rearrangement
e. Substitution
f. Rearrangement
g. Addition
h. Substitution
i. Elimination

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SCI 1054 CHEMISTRY SEMESTER 1
Exercise 7.5.1 iii.
i.

ii. iv.

Exercise 7.6.1
1. W - 2O
X-3O
Y- 1O
2. C(CH3)2(OH)CH3

Exercise 7.7.1
1.

Skeletal isomer :
1-butanol & 2-methyl-2-
propanol
Positional isomer :
1-butanol & 2-butanol
FG isomer:
Diethyl ether & 1-butanol

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SCI 1054 CHEMISTRY SEMESTER 1
2.
(a) chain isomer

(b) positional isomer

3.

Exercise 7.7.2
1. Cis-trans requirement

a. Molecules that has restricted rotation carbon (C=C or cyclic)

b. Each carbon of a site of restricted rotation has two different atoms/groups of
atoms attached to it.
2.

a. No f. Yes

Identical atoms (H) attached to the
same carbon double bonded

Trans Cis

b. Yes g. Yes

Trans cis

Cis trans

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SCI 1054 CHEMISTRY SEMESTER 1

c. Yes trans h. No
Identical group of atoms (CH3)
Cis attached to the same carbon double
d. Yes bonded

Trans i. No
e. Yes Identical atoms (H) attached to the
same carbon cyclic (restricted carbon)

cis
j. Yes

Cis trans Cis trans

3.
a. Same compound
b. Cannot compare (diff molecular formula compound)
c. Cis-trans @ geometric isomer

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CHAPTER HYDROCARBON
8 AND
HALOALKANES

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SCI 1054 CHEMISTRY SEMESTER 1

8.0 HYDROCARBON

INTRODUCTION
HYDROCARBONS

(made up of carbon and hydrogen only)

ALIPHATIC AROMATIC
Contains one or
more benzene ring

ALKANES CYCLOALKANES ALKENES CYCLOALKENES ALKYNES
(saturated) (saturated) (unsaturated) (unsaturated) (unsaturated)

Compounds such as the alkanes and cycloalkanes, whose molecules contain only single
bonds are referred to as saturated hydrocarbons. They contain the maximum
number of hydrogen atoms that the carbon compound can possess.
Compounds with multiple bonds such as alkenes, cycloalkenes, alkynes and aromatic
hydrocarbons are called unsaturated hydrocarbons. They posses fewer than the
maximum number of hydrogen atoms.

8.1 ALKANES
8.1.1 INTRODUCTION

Alkanes are known as saturated hydrocarbon which contains only single covalent bonds.
General formula of alkanes is CnH2n+2 where n = 1, 2, …….
Each carbon atom in alkanes is sp3 hybridised and tetrahedral with four sigma bond
formed by the four sp3 hybrid orbitals. All bond angles are close to 109.5o
Alkanes IUPAC names have the –ane suffix.

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SCI 1054 CHEMISTRY SEMESTER 1

Table 1 The First Ten Unbranched Alkanes

Name Number of Structure Molecular
Carbon Formula
Atoms

Methane 1 H-CH2-H CH4
Ethane 2 H-(CH2)2-H C2H6
Propane 3 H-(CH2)3-H C3H8
Butane 4 H-(CH2)4-H C4H10
Pentane 5 H-(CH2)5-H C5H12
Hexane 6 H-(CH2)6-H C6H14
Heptane 7 H-(CH2)7-H C7H16
Octane 8 H-(CH2)8-H C8H18
Nonane 9 H-(CH2)9-H C9H20
Decane 10 H-(CH2)10-H C10H22

8.1.2 IUPAC NOMENCLATURE
Branched-chain alkanes are named according to the following rules:
① Locate the longest continuous chain of carbon atoms. This chain determines the parent

name for alkanes.

Examples: Parent name: hexane
CH3CH2CH2CH2CHCH3
CH3

CH3CH2CH2CH2CHCH3 Parent name: heptane
CH2
CH3

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SCI 1054 CHEMISTRY SEMESTER 1

② Number the longest chain beginning with the end of the chain nearer the substituent.
Examples:

6 5 4 3 21 substituent
substituent
CH3CH2CH2CH2CHCH3

CH3

76 543

CH3CH2CH2CH2CHCH3

2 CH2

1 CH3

③ Use the numbers obtained by application of rule 2 to designate the location of the
substituent group.
 The parent name is placed last, and the substituent group preceded by the number
designating its location on the chain, is placed first.
 Numbers are separated from words by a hyphen.

Examples:

6 5 4 3 21

CH3CH2CH2CH2CHCH3

CH3 2-methylhexane
3-methylheptane
76 543

CH3CH2CH2CH2CHCH3

2 CH2

1 CH3

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SCI 1054 CHEMISTRY SEMESTER 1

Table 2 Some Common Substituent Groups

Substituent Name Substituent Name
CH3 tert-pentyl
-CH3 methyl |
C-CH2CH3 pentyl
CH2CH3 ethyl |
CH3
CH2CH2CH2CH2CH3

CH2CH2CH3 propyl CH2CH2CHCH3 isopentyl
CH-CH3 isopropyl | neopentyl
CH3
| cyclopropyl
CH3 CH3
|
CH2CH2CH2CH3 butyl CH2-C-CH3
|
CH3

CH2-CH-CH3 cyclobutyl
|
CH3 isobutyl

CHCH2CH3 sec-butyl or C6H5 phenyl
| nitro NH2 amino
CH3

NO2

OH hydroxyl

F fluoro

Cl chloro CH2 benzyl
Br bromo

I iodo

④ When two or more substituents are present, give each substituent a number
corresponding to its location on the longest chain.
 the substituent groups should be listed alphabetically.
 In alphabetizing, the prefixes i.e di, tri, tetra, sec-, tert-, etc. are ignored except iso
and neo.

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SCI 1054 CHEMISTRY SEMESTER 1
Example:

1 23 4 5 6

CH3CHCH2CHCH2CH3

CH3 CH2

CH3 4-ethyl-2-methylhexane

⑤ When two substituents are present on the same carbon atom, use that number twice.
Example:

CH3

1 2 34 5 6

CH3CH2CCH2CH2CH3

CH2CH3 3-ethyl-3-methylhexane

⑥ When two or more substituents are identical, indicate this by the use of prefixes di-,tri-
, tetra-, and so on. Commas are used to separate numbers from each other.

Example: 2,3-dimethylbutane
CH3

CH3CHCHCH3
CH3

 When two chains of equal length compete for selection as the parent chain, choose the
chain with the greater number of substituents.

CH3 CH3

7 6 5 43 2 1

CH3CH2CHCHCHCHCH3

CH2 CH3

CH2CH3

2,3,5-trimethyl-4-propylheptane
(four substituents)

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SCI 1054 CHEMISTRY SEMESTER 1

CH3 CH3

7 6 5 43 2 1

CH3CH2CHCHCHCHCH3

5 CH2 CH3

NOT CH2CH3 4-sec-butyl-2,3-dimethylheptane
(three substituents)
67

⑧ When branching first occurs at an equal distance from either end of the longest chain,
choose the name that gives the lower number at the first point of difference.

CH3 CH3

6 5 4 3 21

CH3CHCH2CHCHCH3

CH3 2,3,5-trimethylhexane
(NOT 2,4,5-trimethylhexane)

8.1.3 CYCLOALKANES
 Cycloalkanes – alkanes which carbon atoms are joined in rings.
 Cycloalkanes are known as saturated hydrocarbon, because it has the maximum
number of bonded hydrogen (only has single bonds).
 General formula: CnH2n where n = 3, 4, 5, ……

8.1.3.1 NOMENCLATURE OF CYCLOALKANES

1. Cycloalkanes with only one ring are named by attaching the prefix cyclo- to the names
of the alkanes possessing the same number of carbon atoms.

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