ADVANCED
LEVEL
CHEMISTRY
Cartwright n.Nduwa
The more energy you put into a bond the
harder it is to break
LATTICE ENERGY
5.1.1 Lattice Energy & Enthalpy Change of
Atomisation
DOWNLOAD PDF
Lattice Energy & Enthalpy Change of Atomisation

Enthalpy change (ΔH) refers to the amount of heat energy transferred during a
chemical reaction, at a constant pressure
Enthalpy change of atomisation





The standard enthalpy change of atomisation (ΔHatꝋ) is the enthalpy
change when 1 mole of gaseous atoms is formed from its element under standard
conditions
o Standard conditions in this syllabus are a temperature of 298 K and a
pressure of 101 kPa
The ΔHatꝋ is always endothermic as energy is always required to break any bonds
between the atoms in the element, to break the element into its gaseous atoms
o Since this is always an endothermic process, the enthalpy change will always
have a positive value
Equations can be written to show the standard enthalpy change of atomisation
(ΔHatꝋ) for elements
For example, sodium in its elemental form is a solid
The standard enthalpy change of atomisation for sodium is the energy required to
form 1 mole of gaseous sodium atoms:
Na(s) → Na(g) ΔHatꝋ = +107 kJ mol -1
Worked example: Writing equations for the standard enthalpy change of
atomisation
1|Page
Answer
Answer 1: Potassium in its elemental form is a solid, therefore the standard enthalpy
change of atomisation is the energy required to form 1 mole of K(g) from K(s)
K(s) → K(g)
Answer 2: Mercury in its elemental form is a liquid, so the standard enthalpy change of
atomisation of mercury is the energy required to form 1 mole of Hg(g) from Hg(l)
Hg(l) → Hg(g)
Lattice energy






The lattice energy (ΔHlattꝋ) is the enthalpy change when 1 mole of an ionic
compound is formed from its gaseous ions (under standard conditions)
The ΔHlattꝋ is always exothermic, as when ions are combined to form an ionic solid
lattice there is an extremely large release of energy
o Since this is always an exothermic process, the enthalpy change will always
have a negative value
o Because of the huge release in energy when the gaseous ions combine, the
value will be a very large negative value
The large negative value of ΔHlattꝋ suggests that the ionic compound is much
more stable than its gaseous ions
o This is due to the strong electrostatic forces of attraction between the
oppositely charged ions in the solid lattice
o Since there are no electrostatic forces of attraction between the ions in the
gas phase, the gaseous ions are less stable than the ions in the ionic lattice
o The more exothermic the value is, the stronger the ionic bonds within the
lattice are
The ΔHlattꝋ of an ionic compound cannot be determined directly by one single
experiment
Multiple experimental values and an energy cycle are used to find the ΔHlattꝋ of ionic
compounds
The lattice energy (ΔHlattꝋ) of an ionic compound can be written as an equation
2|Page
o
o
For example, magnesium chloride is an ionic compound formed from
magnesium (Mg2+) and chloride (Cl-) ions
Since the lattice energy is the enthalpy change when 1 mole of magnesium
chloride is formed from gaseous magnesium and chloride ions, the equation
for this process is:
Mg2+(g) + 2Cl-(g) → MgCl2(s)
Worked Example: Writing equations for lattice energy
Answer
Answer 1: Mg2+(g) + O2-(g) → MgO(s)
Answer 2: Li+(g) + Cl-(g) → LiCl(s)
Exam Tip
Make sure the correct state symbols are stated when writing these equations – it is crucial
that you use these correctly throughout this entire topic.
1. PHYSICAL CHEMISTRY2. INORGANIC CHEMISTRY3. ORGANIC CHEMISTR6. INORGANIC CHEMISTRY (A
LEVEL ONLY)7. ORGANIC CHEMISTRY (A LEVEL ONLY)
5.1.2 Electron Affinity & Trends of
Group 16 & 17 Element
8. ANALYSIS (A
Electron Affinity
Electron affinity
3|Page

The first electron affinity (EA1) is the enthalpy change when 1 mole of electrons is
added to 1 mole of gaseous atoms, to form 1 mole of gaseous ions each with a
single negative charge under standard conditions
X(g) + e- → X-(g)



EA1 is usually exothermic, as energy is released
o Since this is generally an exothermic process, then the value for EA1 will usually
be a negative number
An element can also accept more than one electron, in which case successive electron
affinities are used
o For example, the second electron affinity (EA2) and third electron affinity (EA3)
of an element represent the formation of 1 mole of gaseous ions with 2- and 3charges respectively
The second and third electron affinities are endothermic, as energy is absorbed
o This is because the incoming electron is added to an already negative ion
o Energy is required to overcome the repulsive forces between the incoming
electron and negative ion
o Since these are endothermic processes, the values will be positive
Second & third electron affinity table
Factors affecting electron affinity

The value of the electron affinity depends on how strongly the incoming electron is
attracted to the nucleus
o The greater the attractive forces between the electron and nucleus, the more
energy is released and therefore the more exothermic (more negative)
the EA1 value will be
4|Page


The factors affecting the electron affinity of an element are the same as those that affect
the ionisation energy (the formation of positive ions via the loss of electrons)
These are:
o Nuclear charge: the greater the nuclear charge, the stronger the attractive forces
between an incoming electron and the nucleus
o Distance: the greater the distance between the nucleus and the outermost
shell/orbital where the electron is added, the weaker the force of attraction
o Shielding: the greater the number of shells, the greater the shielding effect and the
weaker the force of attraction
Trends in electron affinity of Group 16 & Group 17 elements


Electron affinities of non-metals become more exothermic across a period, with a
maximum at Group 17
There is generally a downwards trend in the size of the electron affinities of the elements
in Group 16 and 17
o The electron affinities generally become less exothermic for each successive
element going down both Groups, apart from the first member of each Group
(oxygen and fluorine respectively)
Electron affinity table


An atom of chlorine has a greater nuclear charge than an atom of sulfur
Chlorine will therefore have a greater attractive force between its nucleus and its outer
electrons
5|Page
o


More energy is released upon adding an electron to chlorine, so the EA1 of Cl
is more exothermic than for S
Going down Group 16 and 17:
o The outermost electrons are held less tightly to the nucleus as they are further
away
o The number of electron shells increases causing an increased shielding of the
outermost electrons
o It gets more difficult to add an electron to the outer shell
o Less energy is released upon adding an electron to the outer shell
o So generally, the EA1 becomes less exothermic
Fluorine is an exception and has a lower EA1 than chlorine
o Fluorine has a very small atomic radius
o This means that the electron density of fluorine is high
o There is more repulsion between the incoming electron and the electrons that are
already present in fluorine
o These repulsive forces reduce the attractive forces between the incoming
electron and nucleus
o As a result, the EA1 of fluorine is less exothermic than expected
LEVEL ONLY)
5.1.3 Constructing Born-Haber Cycles
DOWNLOAD PDF
Constructing Born-Haber Cycles




To calculate the lattice energy (ΔHlattꝋ) of an ionic compound, a Born-Haber cycle is
used, which is a special type of energy cycle
To find the ΔHlattꝋ of an ionic compound, the following values must be known:
o Enthalpy change of formation (ΔHfꝋ)
o The various enthalpy changes involved when going from elements in
their standard states to their gaseous ions (the sum of all of these can be
referred to as ΔH1ꝋ)
The various enthalpy changes involved when going from elements in their standard
states to their gaseous ions include:
o Enthalpy change of atomisation of each element
o First ionisation energy of the metal
o Successive ionisation energies of the metal if applicable
o First electron affinity of the non-metal
o Successive electron affinities of the non-metal if applicable
The order that these are written in the Born-Haber cycle is important - they must be
done in the order written above
o First ionisation energy cannot take place before atomisation, because first
ionisation energy is the process of turning gaseous atoms into gaseous
ions
6|Page
o




Electron affinity cannot take place before first ionisation energy, since the
electrons must be lost from the metal first during ionisation, to be present in
order for the non-metal to gain them
Hess’s law states that ‘the enthalpy change in a chemical reaction is the same
regardless of the route taken, as long as the final and initial conditions and states of
reactants and products are the same for each route’
Although a Born-Haber cycle is usually constructed and used to calculate the lattice
energy of an ionic compound, the cycle could be used to calculate any stage in the
cycle, since an energy change is the same regardless of the route taken
For example, the energy cycle shows that there are two routes to go from the
elements in their standard states to the ionic compound:
o Each route begins at the same stage of the cycle and ends at the same stage
of the cycle
According to Hess’s law, the enthalpy change for both routes is the same, so the
cycle can be used to calculate an unknown value within the cycle
The green arrow shows the indirect route, which is equal in energy to the direct route
of the element in the standard states changing into the ionic compound (blue arrow)
Worked example: Constructing a Born-Haber cycle for KCl
7|Page
8|Page
Worked example: Constructing a Born-Haber cycle for MgO5.1.4
9|Page
Calculations using Born-Haber Cycles
DOWNLOAD PDF
Calculations Using Born-Haber Cycles

Once a Born-Haber cycle has been constructed, it is possible to calculate the lattice
energy (ΔHlattꝋ) by applying Hess’s law and rearranging:
ΔHfꝋ = ΔHatꝋ + ΔHatꝋ + IE + EA + ΔHlattꝋ


If we simplify this into three terms, this makes the equation easier to see:
o ΔHlattꝋ
o ΔHfꝋ
o ΔH1ꝋ (the sum of all of the various enthalpy changes necessary to convert the
elements in their standard states to gaseous ions)
The simplified equation becomes
ΔHfꝋ = ΔH1ꝋ + ΔHlattꝋ
So, if we rearrange to calculate the lattice energy, the equation becomes
ΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ



When calculating the ΔHlattꝋ, all other necessary values will be given in the question
A Born-Haber cycle could be used to calculate any stage in the cycle
o For example, you could be given the lattice energy and asked to calculate the
enthalpy change of formation of the ionic compound
o The principle would be exactly the same
o Work out the direct and indirect route of the cycle (the stage that you are being
asked to calculate will always be the direct route)
o Write out the equation in terms of enthalpy changes and rearrange if necessary to
calculate the required value
Remember: sometimes a value may need to be doubled or halved, depending on the
ionic solid involved
o For example, with MgCl2 the value for the first electron affinity of chlorine would
need to be doubled in the calculation, because there are two moles of chlorine
atoms
o Therefore, you are adding 2 moles of electrons to 2 moles of chlorine atoms, to
form 2 moles of Cl- ions
10 | P a g e
Worked example: Calculating the lattice energy of KCl
Answer

Step 1: The corresponding Born-Haber cycle is:
11 | P a g e

Step 2: Applying Hess’ law, the lattice energy of KCl is:
ΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ
ΔHlattꝋ = ΔHfꝋ - [(ΔHatꝋ K) + (ΔHatꝋ Cl) + (IE1 K) + (EA1 Cl)]

Step 3: Substitute in the numbers:
ΔHlattꝋ = (-437) - [(+90) + (+122) + (+418) + (-349)] = -718 kJ mol-1
Worked example: Calculating the lattice energy of MgO
12 | P a g e
Answer

Step 1: The corresponding Born-Haber cycle is:
13 | P a g e

Step 2: Applying Hess’ law, the lattice energy of MgO is:
ΔHlattꝋ = ΔHfꝋ - ΔH1ꝋ
ΔHlattꝋ = ΔHfꝋ - [(ΔHatꝋ Mg) + (ΔHatꝋ O) + (IE1 Mg) + (IE2 Mg) + (EA1 O) + (EA2 O)]

Step 3: Substitute in the numbers:
ΔHlattꝋ = (-602) - [(+148) + (+248) + (+736) + (+1450) + (-142) + (+770)]
= -3812 kJ mol-1
Exam Tip
Make sure you use brackets when carrying out calculations using Born-Haber cycles as you may
forget a +/- sign which will affect your final answer!
14 | P a g e
5.1.5 Factors Affecting Lattice Energy
DOWNLOAD PDF
Lattice Energy: Ionic Charge & Radius

The two key factors which affect lattice energy, ΔHlattꝋ, are the charge and radius of the
ions that make up the crystalline lattice
Ionic radius





The lattice energy becomes less exothermic as the ionic radius of the ions increases
This is because the charge on the ions is more spread out over the ion when the ions
are larger
The ions are also further apart from each other in the lattice
o The attraction between ions is between the centres of the ions involved, so the
bigger the ions the bigger the distance between the centre of the ions
Therefore, the electrostatic forces of attraction between the oppositely charged ions in
the lattice are weaker
For example, the lattice energy of caesium fluoride (CsF) is less exothermic than the
lattice energy of potassium fluoride (KF)
o Since both compounds contain a fluoride (F-) ion, the difference in lattice energy
must be due to the caesium (Cs+) ion in CsF and potassium (K+) ion in KF
o Potassium is a Group 1 and Period 4 element
o Caesium is a Group 1 and Period 6 element
o This means that the Cs+ ion is larger than the K+ ion
o There are weaker electrostatic forces of attraction between the Cs+ and F- ions
compared to K+ and F- ions
o As a result, the lattice energy of CsF is less exothermic than that of KF
15 | P a g e
The lattice energies get less exothermic as the ionic radius of the ions increases
16 | P a g e
Ionic charge





The lattice energy gets more exothermic as the ionic charge of the ions increases
The greater the ionic charge, the higher the charge density
This results in stronger electrostatic attraction between the oppositely charged ions in
the lattice
As a result, the lattice energy is more exothermic
For example, the lattice energy of calcium oxide (CaO) is more exothermic than the
lattice energy of potassium chloride (KCl)
o Calcium oxide is an ionic compound which consists of calcium (Ca2+) and oxide
(O2-) ions
o Potassium chloride is formed from potassium (K+) and chloride (Cl-) ions
o The ions in calcium oxide have a greater ionic charge than the ions in potassium
chloride
o This means that the electrostatic forces of attraction are stronger between the
Ca2+ and O2- compared to the forces between K+ and Clo Therefore, the lattice energy of calcium oxide is more exothermic, as more
energy is released upon its formation from its gaseous ions
o Ca2+ and O2- are also smaller ions than K+ and Cl-, so this also adds to the value for
the lattice energy being more exothermic
17 | P a g e
18 | P a g e
Exam Tip
When constructing Born-Haber cycles, be sure to use ionisation
energies (IE) for metals (as metals lose electrons) and electron affinities (EA) for nonmetals (as non-metals gain electrons).Remember that the order that the steps go in is
important!
19 | P a g e
5.1.6 Enthalpies of Solution & Hydration
DOWNLOAD PDF
Enthalpy Change of Hydration & Solution
Enthalpy change of solution


The standard enthalpy change of solution (ΔHsolꝋ) is the enthalpy change when 1
mole of an ionic substance dissolves in sufficient water to form a very dilute solution
The symbol (aq) is used to show that the solid is dissolved in sufficient water
o For example, the enthalpy changes of solution for potassium chloride are
described by the following equations:
KCl(s) + aq → KCl(aq)
OR
KCl(s) + aq → K+(aq) + Cl-(aq)

ΔHsolꝋ can be exothermic (negative) or endothermic (positive)
Enthalpy change of hydration





The lattice energy (ΔHlattꝋ) of KCl is -711 kJ mol-1
o This means that 711 kJ mol-1 is released when the KCl ionic lattice is formed
o Therefore, to break the attractive forces between the K+ and Cl- ions, +711 kJ
mol-1 is needed
However, the ΔHsolꝋ of KCl is +26 kJ mol-1
This means that another +685 kJ mol-1 (711 - 26) is required to break the KCl lattice
This is compensated for by the standard enthalpy change of hydration (ΔHhydꝋ)
The standard enthalpy change of hydration (ΔHhydꝋ) is the enthalpy change when 1
mole of a specified gaseous ion dissolves in sufficient water to form a very
dilute solution
o For example, the enthalpy change of hydration for magnesium ions is described
by the following equation:
Mg2+(g) + aq → Mg2+(aq)

o

Hydration enthalpies are the measure of the energy that is released when there is
an attraction formed between the ions and water molecules
o Hydration enthalpies are exothermic
When an ionic solid dissolves in water, positive and negative ions are formed
20 | P a g e



Water is a polar molecule with a δ- oxygen (O) atom and δ+ hydrogen (H) atoms which
will form ion-dipole attractions with the ions present in the solution
The oxygen atom in water will be attracted to the positive ions and the hydrogen atoms
will be attracted to the negative ions
Since the ΔHhydꝋ of KCl is -685 kJ mol-1, 685 kJ mol-1 is released in forming these iondipole attractions when KCl dissolves in water
o This compensates for the remaining +685 kJ mol-1 which was needed to break
down the KCl lattice
The polar water molecules will form ion-dipole bonds with the ions in solution (a) causing the
ions to become hydrated (b)
21 | P a g e
5.1.7 Constructing Energy Cycles using
Enthalpy Changes & Lattice Energy
DOWNLOAD PDF
Energy Cycle Using Enthalpy Changes & Lattice Energy

The standard enthalpy change of hydration (ΔHhydꝋ) can be calculated by
constructing energy cycles and applying Hess’s law
Energy cycle involving enthalpy change of solution, lattice energy, and enthalpy change of
hydration
22 | P a g e

The energy cycle shows that there are two routes to go from the ionic lattice to the
hydrated ions in an aqueous solution:
o Route 1: going from ionic solid → ions in aqueous solution (this is the direct
route)
ΔHsolꝋ= Enthalpy of solution

Route 2: going from ionic lattice → gaseous ions → ions in aqueous solution (this is
the indirect route)
-ΔHlattꝋ + ΔHhydꝋ = reverse lattice enthalpy + hydration enthalpies
Lattice enthalpy usually means Lattice formation enthalpy, in other words bond forming. If we
are breaking the lattice then this is reversing the enthalpy change so a negative sign is added in
front of the term (alternatively it is called lattice dissociation enthalpy)

According to Hess’s law, the enthalpy change for both routes is the same, such that:
ΔHsolꝋ = -ΔHlattꝋ + ΔHhydꝋ
ΔHhydꝋ = ΔHsolꝋ + ΔHlattꝋ

Each ion will have its own enthalpy change of hydration, ΔHhydꝋ, which will need to be
taken into account during calculations
o The total ΔHhydꝋ is found by adding the ΔHhydꝋ values of both anions and cations
together
Worked example: Constructing an energy cycle and energy level
diagram of KCl
Answer
Energy cycle:
23 | P a g e
Energy level diagram:
24 | P a g e
25 | P a g e
Worked example: Constructing an energy cycle and energy level
diagram of MgCl2
26 | P a g e
Answer
Energy cycle:
Energy level diagram:
27 | P a g e
28 | P a g e
5.1.8 Energy Cycle Calculations
DOWNLOAD PDF
Energy Cycle Calculations


The energy cycle involving the enthalpy change of solution (ΔHsolꝋ ), lattice energy
(ΔHlattꝋ), and enthalpy change of hydration (ΔHhydꝋ) can be used to calculate the different
enthalpy values
According to Hess’s law, the enthalpy change of the direct and of the indirect route will
be the same, such that:
ΔHhydꝋ = ΔHlattꝋ + ΔHsolꝋ


This equation can be rearranged depending on which enthalpy value needs to be
calculated
For example, ΔHlattꝋ can be calculated using:
ΔHlattꝋ = ΔHhydꝋ - ΔHsolꝋ


Remember: the total ΔHhydꝋ is found by adding the ΔHhydꝋ values of both anions and
cations together
Remember: take into account the number of each ion when completing calculations
o For example, MgCl2 has two chloride ions, so when completing calculations this
will need to be accounted for
o In this case, you would need to double the value of the hydration enthalpy, since
you are hydrating 2 moles of chloride ions instead of 1
Worked example: Calculating the enthalpy change of hydration of
chloride
29 | P a g e
Answer
Step 1: Draw the energy cycle of KCl

Step 2: Apply Hess’s law to find ΔHhydꝋ [Cl-]
ΔHhydꝋ = (ΔHlattꝋ[KCl]) + (ΔHsolꝋ[KCl])
(ΔHhydꝋ[K+]) + (ΔHhydꝋ[Cl-]) = (ΔHlattꝋ[KCl]) + (ΔHsolꝋ[KCl])
(ΔHhydꝋ[Cl-]) = (ΔHlattꝋ[KCl]) + (ΔHsolꝋ[KCl]) - (ΔHhydꝋ[K+])

Step 3: Substitute the values to find ΔHhydꝋ [Cl-]
ΔHhydꝋ [Cl-] = (-711) + (+26) - (-322) = -363 kJ mol-1
30 | P a g e
Worked example: Calculating the enthalpy change of hydration of
magnesium
Answer

Step 1: Draw the energy cycle of MgCl2
31 | P a g e

Step 2: Apply Hess’s law to find ΔHhydꝋ [Mg2+]
ΔHhydꝋ = (ΔHlattꝋ[MgCl2]) + (ΔHsolꝋ [MgCl2])
(ΔHhydꝋ[Mg2+]) + (2ΔHhydꝋ [Cl-]) = (ΔHlattꝋ [MgCl2]) + (ΔHsolꝋ [MgCl2])
(ΔHhydꝋ[Mg2+]) = (ΔHlattꝋ[MgCl2]) + (ΔHsolꝋ[MgCl2]) - (2ΔHhydꝋ[Cl-])

Step 3: Substitute the values to find ΔHhydꝋ [Mg2+]
ΔHhydꝋ[Mg2+] = (-2592) + (-55) - (2 x -363) = -1921 kJ mol-1
32 | P a g e
5.1.9 Factors Affecting Enthalpy of Hydration
Enthalpy of Hydration: Ionic Charge & Radius


The standard enthalpy change of hydration (ΔHhydꝋ) is affected by the amount that the
ions are attracted to the water molecules
The factors which affect this attraction are the ionic charge and radius
Ionic radius


ΔHhydꝋ becomes more exothermic with decreasing ionic radii
o Smaller ions have a greater charge density resulting in stronger ion-dipole
attractions between the water molecules and the ions in the solution
o Therefore, more energy is released when they become hydrated and
ΔHhydꝋ becomes more exothermic
For example, the ΔHhydꝋ of magnesium sulfate (MgSO4) is more exothermic than the
ΔHhydꝋ of barium sulfate (BaSO4)
o Since both compounds contain a sulfate (SO42-) ion, the difference in ΔHhydꝋ must
be due to the magnesium (Mg2+) ion in MgSO4 and barium (Ba2+) ion in BaSO4
o Magnesium is a Group 2 and Period 3 element
o Barium is a Group 2 and Period 6 element
o This means that the Mg2+ ion is smaller than the Ba2+ ion
o The attraction is therefore much stronger for the Mg2+ ion
o As a result, the standard enthalpy of hydration of MgSO 4 is more
exothermic than that of BaSO4
Ionic charge


ΔHhydꝋ is more exothermic for ions with larger ionic charges
o Ions with large ionic charges have a greater charge density resulting
in stronger ion-dipole attractions between the water molecules and the ions in
the solution
o Therefore, more energy is released when they become hydrated and
ΔHhydꝋ becomes more exothermic
For example, the ΔHhydꝋ of calcium oxide (CaO) is more exothermic than the ΔHhydꝋ of
potassium chloride (KCl)
o Calcium oxide is an ionic compound that consists of calcium (Ca2+) and oxide
(O2-) ions
o Potassium chloride is formed from potassium (K+) and chloride (Cl-) ions
o Both of the ions in calcium oxide have a greater ionic charge than the ions in
potassium chloride
o This means that the attractions are stronger between the water molecules and
Ca2+ and O2- ions upon hydration of CaO
33 | P a g e
o
o
The attractions are weaker between the water molecules and K+ and Cl- ions
upon hydration of KCl
Therefore, the ΔHhydꝋ of calcium oxide is more exothermic as more energy is
released upon its hydration
The enthalpy of hydration is more exothermic for smaller ions and for ions with a
greater ionic charge
34 | P a g e
5.2.1 Entropy & Entropy Change
Defining Entropy




The entropy (S) of a given system is the number of possible arrangements of the
particles and their energy in a given system
o In other words, it is a measure of how disordered a system is
When a system becomes more disordered, its entropy will increase
An increase in entropy means that the system becomes energetically more stable
For example, during the thermal decomposition of calcium carbonate (CaCO3) the
entropy of the system increases:
CaCO3(s) → CaO(s) + CO2(g)





In this decomposition reaction, a gas molecule (CO2) is formed
The CO2 gas molecule is more disordered than the solid reactant (CaCO3), as it is
constantly moving around
As a result, the system has become more disordered and there is an increase in entropy
Another typical example of a system that becomes more disordered is when a solid
is melted
For example, melting ice to form liquid water:
H2O(s) → H2O(l)





The water molecules in ice are in fixed positions and can only vibrate about those
positions
In the liquid state, the particles are still quite close together but are arranged more
randomly, in that they can move around each other
Water molecules in the liquid state are therefore more disordered
Thus, for a given substance, the entropy increases when its solid form melts into a liquid
In both examples, the system with the higher entropy will
be energetically the most stable (as the energy of the system is more spread out when it
is in a disordered state)
35 | P a g e
Melting a solid will cause the particles to become more disordered resulting in a more
energetically stable system
Exam Tip
Make sure you don’t confuse the system with your surroundings!The system consists of the
molecules that are reacting in a chemical reaction.The surroundings are everything else such as
the solvent, the air around the reaction, test-tube, etc.
Entropy Changes


All elements have positive standard molar entropy values
The order of entropy for the different states of matter are as follows:
gas > liquid > solid

o


There are some exceptions such as calcium carbonate (solid) which has a higher
entropy than mercury (liquid)
Simpler substances with fewer atoms have lower entropy values than complex
substances with more atoms
o For example, calcium oxide (CaO) has a smaller entropy than calcium carbonate
(CaCO3)
Harder substances have lower entropy than softer substances of the same type
o For example, diamond has a smaller entropy than graphite
36 | P a g e
Change in state




The entropy of a substance changes during a change in state
The entropy increases when a substance melts (change from solid to liquid)
o Increasing the temperature of a solid causes the particles to vibrate more
o The regularly arranged lattice of particles changes into an irregular arrangement
of particles
o These particles are still close to each other but can now rotate and slide over each
other in the liquid
o As a result, there is an increase in disorder
The entropy increases when a substance boils (change from liquid to gas)
o The particles in a gas can now freely move around and are far apart from each
other
o The entropy increases significantly as the particles become very disordered
Similarly, the entropy decreases when a substance condenses (change
from gas to liquid) or freezes (change from liquid to solid)
o The particles are brought together and get arranged in a more regular arrangement
o The ability of the particles to move decreases as the particles become more
ordered
o There are fewer ways of arranging the energy so the entropy decreases
37 | P a g e
The entropy of a substance increases when the temperature is raised as particles become more
disordered




The entropy also increases when a solid is dissolved in a solvent
The solid particles are more ordered in the solid lattice as they can only slightly vibrate
When dissolved to form a dilute solution, the entropy increases as:
o The particles are more spread out
o There is an increase in the number of ways of arranging the energy
The crystallisation of a salt from a solution is associated with a decrease in entropy
o The particles are spread out in solution but become more ordered in the solid
38 | P a g e
When a solid is dissolved in a solvent to form a dilute solution, the entropy increases as the
particles become more disordered
Entropy changes in reactions




Gases have higher entropy values than solids
So, if the number of gaseous molecules in a reaction changes, there will also be a change
in entropy
The greater the number of gas molecules, the greater the number of ways of arranging
them, and thus the greater the entropy
For example the decomposition of calcium carbonate (CaCO3)
CaCO3(s) → CaO(s) + CO2(g)

o


The CO2 gas molecule is more disordered than the solid reactant (CaCO3) as it can
freely move around whereas the particles in CaCO3 are in fixed positions in which
they can only slightly vibrate
o The system has therefore become more disordered and there is
an increase in entropy
Similarly, a decrease in the number of gas molecules results in a decrease in entropy
causing the system to become less energetically stable
For example, the formation of ammonia in the Haber process
N2(g) + 3H2(g) ⇋ 2NH3(g)
39 | P a g e

o
o
o
o
o
o
In this case, all of the reactants and products are gases
Before the reaction occurs, there are four gas molecules (1 nitrogen and 3
hydrogen molecules) in the reactants
After the reaction has taken place, there are now only two gas molecules (2
ammonia molecules) in the products
Since there are fewer molecules of gas in the products, there are fewer ways of
arranging the energy of the system over the products
The system has become more ordered causing a decrease in entropy
The reactants (N2 and H2) are energetically more stable than the product (NH3)
5.2.2 Calculating Entropy Changes
DOWNLOAD PDF
Calculating Entropy Changes


The standard entropy change (ΔSsystemꝋ ) for a given reaction can be calculated using the
standard entropies (Sꝋ ) of the reactants and products
The equation to calculate the standard entropy change of a system is:
ΔSsystemꝋ = ΣΔSproductsꝋ - ΣΔSreactantsꝋ
(where Σ = sum of)

For example, the standard entropy change for the formation of ammonia (NH3) from
nitrogen (N2) and hydrogen (H2) can be calculated using this equation
N2(g) + 3H2(g) ⇋ 2NH3(g)
ΔSsystemꝋ = (2 x ΔSꝋ(NH3)) - (ΔSꝋ(N2) + 3 x ΔSꝋ(H2))
Worked example: Calculating entropy changes
40 | P a g e
Answer
ΔSsystemꝋ = ΣΔSproductsꝋ - ΣΔSreactantsꝋ
ΔSsystemꝋ = (2 x 38.20) - (2 x 32.60 + 205.0)
= -193.8 J K-1 mol-1
Exam Tip
Use the stoichiometry of the equation and the correct state of the compounds when calculating
the entropy change of a reaction.
41 | P a g e
5.2.3 Gibbs Free Energy Change & Gibbs
Equation
DOWNLOAD PDF
The Gibbs Equation
Gibbs free energy




The feasibility of a reaction does not only depend on the entropy change of the
reaction, but can also be affected by the enthalpy change
Therefore, using the entropy change of a reaction only to determine the feasibility of a
reaction is inaccurate
The Gibbs free energy (G) is the energy change that takes into
account both the entropy change of a reaction and the enthalpy change
The Gibbs equation is:
ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ




The units of ΔGꝋ are in kJ mol-1
The units of ΔHreactionꝋ are in kJ mol-1
The units of T are in K
The units of ΔSsystemꝋ are in J K-1 mol-1 (and must therefore be converted to kJ K-1 mol-1 by
dividing by 1000)
The Gibbs Equation: Calculations

The Gibbs equation can be used to calculate the Gibbs free energy change of a
reaction
ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ


The equation can also be rearranged to find values of ΔHreactionꝋ, ΔSsystemꝋ or the
temperature, T
For example, if for a given reaction, the values of ΔGꝋ, ΔHreactionꝋ and ΔSsystemꝋ are given,
the temperature can be found by rearranging the Gibbs equation as follows:
42 | P a g e
Worked example: Calculating Gibbs free energy
Answer

Step 1 - Calculate ΔSsystemꝋ
ΔSsystemꝋ = ΣΔSproductsꝋ - ΣΔSreactantsꝋ
ΔSsystemꝋ = (ΔSꝋ [CH3Br(g)] + ΔSꝋ [H2O(l)]) - (ΔSꝋ [CH3OH(l)] + ΔSꝋ [HBr(g)])
= (246 + 70.0) - (240 + 99.0)
= -23.0 J K-1 mol-1

Step 2 - Convert ΔSꝋ into kJ K-1 mol-1
= -0.023 kJ K-1 mol-1

Step 3 - Calculate ΔGꝋ
ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ
= -47 - (298 x -0.023)
= -40.146 kJ mol-1
= -40.1 kJ mol-1
43 | P a g e
5.2.4 Reaction Feasibility
The Gibbs Equation: Reaction Feasibility

The Gibbs equation can be used to calculate whether a reaction is feasible or not
ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ


When ΔGꝋ is negative, the reaction is feasible and likely to occur
When ΔGꝋis positive, the reaction is not feasible and unlikely to occur
Worked Example
Worked example: Determining the feasibility of a reaction
Answer

Step 1: Calculate ΔSsystemꝋ
ΔSsystemꝋ = ΣΔSproductsꝋ - ΣΔSreactantsꝋ
ΔSsystemꝋ = (2 x ΔSꝋ [CaO(s)]) - (2 x ΔSꝋ [Ca(s)] + ΔSꝋ [O2(g)])
= (2 x 40.00) - (2 x 41.00 + 205.0)
= -207.0 J K-1 mol-1

Step 2: Convert ΔSꝋ to kJ K-1 mol-1
44 | P a g e
= -0.207

Step 3: Calculate ΔGꝋ
ΔGꝋ = ΔHreactionꝋ - TΔSsystemꝋ
= -635.5 - (298 x -0.207)
= -573.8 kJ mol-1

Step 4: Determine whether the reaction is feasible
Since the ΔGꝋ is negative the reaction is feasible and likely to occur
5.2.5 Reaction Feasibility: Temperature
Changes
DOWNLOAD PDF
45 | P a g e
Feasibility & Temperature Changes


The feasibility of a reaction can be affected by the temperature
The Gibbs equation will be used to explain what will affect the feasibility of a reaction
for exothermic and endothermic reactions
Exothermic reactions




In exothermic reactions, ΔHreactionꝋ is negative
If the ΔSsystemꝋ is positive:
o Both the first and second term will be negative
o Resulting in a negative ΔGꝋ so the reaction is feasible
o Therefore, regardless of the temperature, an exothermic reaction with a positive
ΔSsystemꝋ will always be feasible
If the ΔSsystemꝋ is negative:
o The first term is negative and the second term is positive
o At high temperatures, the -TΔSsystemꝋ will be very large and positive and will
overcome ΔHreactionꝋ
o Therefore, at high temperatures ΔGꝋ is positive and the reaction is not feasible
o The reaction is more feasible at low temperatures, as the second term will not be
large enough to overcome ΔHreactionꝋ resulting in a negative ΔGꝋ
This corresponds to Le Chatelier’s principle which states that for exothermic
reactions an increase in temperature will cause the equilibrium to shift position in favour
of the reactants, i.e. in the endothermic direction
o In other words, for exothermic reactions, the products will not be formed at high
temperatures
o The reaction is not feasible at high temperatures
46 | P a g e
The diagram shows under which conditions exothermic reactions are feasible
Endothermic reactions




In endothermic reactions, ΔHreactionꝋ is positive
If the ΔSsystemꝋ is negative:
o Both the first and second term will be positive
o Resulting in a positive ΔGꝋ so the reaction is not feasible
o Therefore, regardless of the temperature, endothermic with a negative
ΔSsystemꝋ will never be feasible
If the ΔSsystemꝋ is positive:
o The first term is positive and the second term is negative
o At low temperatures, the -TΔSsystemꝋ will be small and negative and will not
overcome the larger ΔHreactionꝋ
o Therefore, at low temperatures ΔGꝋ is positive and the reaction is less feasible
o The reaction is more feasible at high temperatures as the second term will
become negative enough to overcome the ΔHreactionꝋ resulting in a negative ΔGꝋ
This again corresponds to Le Chatelier’s principle which states that for endothermic
reactions an increase in temperature will cause the equilibrium to shift position in favour
of the products
47 | P a g e
o
o
In other words, for endothermic reactions, the products will be formed at high
temperatures
The reaction is therefore feasible
The diagram shows under which conditions endothermic reactions are feasible
48 | P a g e
5.3.1 Electrolysis
DOWNLOAD PDF
Products of Electrolysis


Electrolysis is the breaking down of a compound into its elements using an electric
current
For example, the electrolysis of zinc chloride (ZnCl2) into its elements zinc and chlorine
ZnCl2(s) → Zn(s) + Cl2(g)


This method is often used to:
o Extract metals from their metal ores when the metals cannot be extracted by
heating their ores with carbon
o Purify metals
o Produce non-metals such as fluorine
Electrolysis is carried out in an electrolysis cell which consists of:
o An electrolyte - this is the compound that is broken down during electrolysis and
it is either a molten ionic compound or a concentrated aqueous solution of ions
o Two electrodes - these are metal or graphite rods conduct electricity to the
electrolyte and away from the electrolyte
 The positive electrode is called the anode
 The negative electrode is called the cathode
o The power supply, which is direct current
Electrolysis takes place in an electrochemical cell;
this type of electrochemical cell is called an electrolytic cell
49 | P a g e
Electrolysis of molten electrolytes

Cations (positively charged ions) move to the negatively charged cathode where
they gain electrons
o Reduction takes place at the cathode
o If a metal is formed, a layer of metal is deposited on a cathode or it forms
a molten layer in the cell
o If hydrogen gas is formed, bubbles are seen
o For example, silver and hydrogen both form positively charged ions which would
be reduced at the cathode as follows:
Ag+ + e- → Ag
2H+ + 2e- → H2

Anions (negatively charged ions) move to the positively charged anode where they lose
electrons
o Oxidation takes place at the anode
o For example, bromine forms negatively charged ions which would be oxidised at
the anode as follows:
2Br- → Br2 + 2e-
Products formed by electrolysis when a pure molten ionic compound containing two simple
ions is electrolysed table
50 | P a g e
Electrolysis of aqueous solutions


Aqueous solutions have more than one cation and anion in solution due to the presence
of water
Water contributes H+ and OH- ions to the solution, which makes things more complicated
o Water is a weak electrolyte and splits into H+ and OH- ions as follows:
H2O ⇌ H+ + OH-

The actual ions that are discharged during electrolysis will depend on:
o The relative electrode potential of the ions
o The concentration of the ions
Relative electrode potential of ions


The relative electrode potential (Eꝋ) of ions describes how easily an ion is discharged
during electrolysis
The positively charged cation with the most positive Eꝋ will be discharged at
the cathode as this is the cation that is most easily reduced
o For example, a concentrated aqueous solution of NaF will contain hydrogen (H+)
and sodium (Na+) ions
o The half-equations for the reduction of these ions and their Eꝋ values are as
follows:
2H+(aq) + 2e- ⇌ H2(g)
Na+(aq) + e- ⇌ Na(s)


Eꝋ = 0.00 V
Eꝋ = -2.71 V
Since H+ ions have a higher Eꝋ value, hydrogen gas (H2) is formed at the cathode instead
of sodium (Na)
The negatively charged anion with the most negative Eꝋ will be discharged at
the anode, as this is the anion that is most easily oxidised
o For example, a concentrated aqueous solution of NaF will contain hydroxide (OH) and fluoride (F-) ions
o The half-equations for the oxidation of these ions and their Eꝋ values are as
follows:
4OH-(aq) → O2(g) + 2H2O(l) + 4e- Eꝋ = -0.40 V
2F-(aq) → F2(g) + 2e- Eꝋ = -2.87 V

Since F- ions have a lower Eꝋ value than OH- ions, fluorine (F2) gas is formed at the anode
Concentration of ions

Ions that are present in higher concentrations are more likely to be discharged
51 | P a g e


For example, when a concentrated solution of NaF is electrolysed, there are far
more fluoride ions which are discharged at the anode, instead of the hydroxide ions as
the fluoride ions are in higher concentration
o So, mainly fluorine will form at the electrode
However, if a very dilute solution of NaF is electrolysed, there will be much
more oxygen and much less fluorine gas formed at the anode
o In reality, a mixture of both oxygen and fluorine gas is formed
Exam Tip
Electrolysis is a redox reaction as a reduction reaction takes place at one electrode and an
oxidation reaction at the other electrode.When writing the overall redox equation make sure that
the electrons lost at the anode balance the electrons gained at the cathode.



Cathode: Cu2+ + 2e- → Cu
Anode: 2Cl- → Cl2 + 2eOverall: CuCl2 → Cu + Cl2
52 | P a g e
5.3.2 Faraday's Law & Avogadro
DOWNLOAD PDF
Faraday's Law

The amount of substance that is formed at an electrode during electrolysis is
proportional to:
o The amount of time where a constant current to passes
o The amount of electricity, in coulombs, that passes through the electrolyte
(strength of electric current)
o The relationship between the current and time is:
Q=Ixt
Q = charge (coulombs, C)
I = current (amperes, A)
t = time, (seconds, s)


The amount or the quantity of electricity can also be expressed by the faraday (F) unit
o One faraday is the amount of electric charge carried by 1 mole of electrons or 1
mole of singly charged ions
o 1 faraday is 96 500 C mol-1
Thus, the relationship between the Faraday constant and the Avogadro constant (L) is:
F=Lxe
F = Faraday’s constant (96 500 C mol-1)
L = Avogadro’s constant (6.022 x 1023 mol-1)
e = charge on an electron
Worked example: Determining the amount of electricity required
53 | P a g e
Answer
One Faraday is the amount of charge (96 500 C) carried by 1 mole of electrons
Answer 1
As there is one mole of electrons, one faraday of electricity (96 500 C) is needed to deposit one
mole of sodium.
Answer 2
Now, there are two moles of electrons, therefore, two faradays of electricity (2 x 96 500 C) are
required to deposit one mole of magnesium.
Answer 3
Two moles of electrons are released, so it requires two faradays of electricity (2 x 96 500 C) to
form one mole of fluorine gas.
Answer 4
54 | P a g e
Four moles of electrons are released, therefore it requires four faradays of electricity (4 x 96
500 C) to form one mole of oxygen gas.
Determining Avogadro's Constant by Electrolysis


The Avogadro’s constant (L) is the number of entities in one mole
o L = 6.02 x 1023 mol-1
o For example, four moles of water contains 2.41 x 1024 (6.02 x 1023 x 4) molecules
of H2O
The value of L (6.02 x 1023 mol-1) can be experimentally determined by electrolysis using
the following equation:
Finding L experimentally

The charge on one mole of electrons is found by using a simple electrolysis experiment
using copper electrodes
55 | P a g e
Apparatus set-up for finding the value of L experimentally



Method
o The pure copper anode and pure copper cathode are weighed
o A variable resistor is kept at a constant current of about 0.17 A
o An electric current is then passed through for a certain time interval (e.g. 40
minutes)
o The anode and cathode are then removed, washed with distilled water, dried with
propanone, and then reweighed
Results
o The cathode has increased in mass as copper is deposited
o The anode has decreased in mass as the copper goes into solution as copper ions
o Often, it is the decreased mass of the anode which is used in the calculation, as the
solid copper formed at the cathode does not always stick to the cathode properly
o Let’s say the amount of copper deposited in this experiment was 0.13 g
Calculation:
o The amount of charge passed can be calculated as follows:
Q=Ixt
56 | P a g e
= 0.17 x (60 x 40)
= 408 C


To deposit 0.13 g of copper (2.0 x 10-3 mol), 408 C of electricity was needed
The amount of electricity needed to deposit 1 mole of copper can therefore be calculated
using simple proportion using the relative atomic mass of Cu
Calculating the amount of charge required to deposit one mole of copper table


Therefore, 199 292 C of electricity is needed to deposit 1 mole of Cu
The half-equation shows that 2 mol of electrons are needed to deposit one mol of copper:
Cu2+(aq) + 2e- → Cu(s)

So, the charge on 1 mol of electrons is:
= 99 646 C

Given that the charge on one electron is 1.60 x 10-19 C, then L equals:
57 | P a g e
= 6.23 x 1023 mol-1

The experimentally determined value for L of 6.23 x 1023 mol-1 is very close to
the theoretical value of 6.02 x 1023 mol-1
5.3.3 Standard Electrode & Cell Potentials
DOWNLOAD PDF
Standard Electrode & Standard Cell Potentials
Electrode potential






The electrode (reduction) potential (E) is a value which shows how easily a substance
is reduced
These are demonstrated using reversible half equations
o This is because there is a redox equilibrium between two related species that are
in different oxidation states
o For example, if you dipped a zinc metal rod into a solution which contained zinc
ions, there would be zinc atoms losing electrons to form zinc ions and at the same
time, zinc ions gaining electrons to become zinc atoms
o This would cause a redox equilibrium
When writing half equations for this topic, the electrons will always be written on the
left-hand side (demonstrating reduction)
The position of equilibrium is different for different species, which is why different
species will have electrode (reduction) potentials
The more positive (or less negative) an electrode potential, the more likely it is for that
species to undergo reduction
o The equilibrium position lies more to the right
For example, the positive electrode potential of bromine below, suggests that it is likely
to get reduced and form bromide (Br-) ions
Br2(l) + 2e- ⇌ 2Br-(aq)


voltage = +1.09 V
The more negative (or less positive) the electrode potential, the less likely it is that
reduction of that species will occur
o The equilibrium position lies more to the left
For example, the negative electrode potential of sodium suggests that it is unlikely that
the sodium (Na+) ions will be reduced to sodium (Na) atoms
Na+(aq) + e- ⇌ Na(s)
voltage = -2.71 V
Standard electrode potential
58 | P a g e









The position of equilibrium and therefore the electrode potential depends on factors such
as:
o Temperature
o Pressure of gases
o Concentration of reagents
So, to be able to compare the electrode potentials of different species, they all have to be
measured against a common reference or standard
Standard conditions also have to be used when comparing electrode potentials
These standard conditions are:
o Ion concentration of 1.00 mol dm-3
o A temperature of 298 K
o A pressure of 1 atm
The electrode potentials are measured relative to something called a standard hydrogen
electrode
The standard hydrogen electrode is given a value of 0.00 V, and all other electrode
potentials are compared to this standard
This means that the electrode potentials are always referred to as a standard electrode
potential (Eꝋ)
The standard electrode potential (Eꝋ) is the voltage produced when a standard halfcell is connected to a standard hydrogen cell under standard conditions
For example, the standard electrode potential of bromine suggests that relative to the
hydrogen half-cell it is more likely to get reduced, as it has a more positive Eꝋ value
Br2(l) + 2e- ⇌ 2Br-(aq)
2H+(aq) + 2e- ⇌ H2(g)

Eꝋ = +1.09 V
Eꝋ = 0.00 V
The standard electrode potential of sodium, on the other hand, suggests that relative to the
hydrogen half-cell it is less likely to get reduced as it has a more negative Eꝋ value
Na+ (aq) + e- ⇌ Na(s)
Eꝋ = -2.71 V
2H+ (aq) + 2e- ⇌ H2(g)
Eꝋ = 0.00 V
Standard cell potential


Once the Eꝋof a half-cell is known, the voltage of an electrochemical cell made up of
two half-cells can be calculated
o These could be any half-cells and neither have to be a standard hydrogen
electrode
This is also known as the standard cell potential (Ecellꝋ)
o The standard cell potential is the difference in Eꝋ between two half-cells
o For example, an electrochemical cell consisting of bromine and sodium half-cells
has an Ecellꝋ of:
59 | P a g e
Ecellꝋ = (+1.09) - (-2.71)
= +3.80 V
Standard Hydrogen Electrode

When a metal rod is placed in an aqueous solution, a redox equilibrium is established
between the metal ions and atoms
o For example, the copper atoms get oxidised and enter the solution as copper ions
Cu(s) → Cu2+(aq) + 2e-
Oxidation of copper ions

o
60 | P a g e
The copper ions gain electrons from the metal rod and deposit as metal atoms on
the rod
Cu2+(aq) + 2e- → Cu(s)
Reduction of copper ions

o

When equilibrium is established, the rate of oxidation and reduction of copper
is equal
The position of the redox equilibrium is different for different metals
o Copper is more easily reduced, thus the equilibrium lies further over to the right
Cu2+ (aq) + 2e- ⇌ Cu (s)

o
Vanadium is more easily oxidised, thus the equilibrium lies further over to
the left
V2+ (aq) + 2e- ⇌ V(s)



The metal atoms and ions in solution cause an electric potential (voltage)
This potential cannot be measured directly however the potential between the
metal/metal ion system and another system can be measured
This value is called the electrode potential (E) and is measured in volts
o The electrode potential is the voltage measured for a half-cell compared to
another half-cell
o Often, the half-cell used for comparison is the standard hydrogen electrode
Standard hydrogen electrode
61 | P a g e

The standard hydrogen electrode is a half-cell used as reference electrodes and
consists of:
o Hydrogen gas in equilibrium with H+ ions of concentration 1.00 mol dm-3 (at 1
atm)
2H+ (aq) + 2e- ⇌ H2 (g)

o

An inert platinum electrode that is in contact with the hydrogen gas and H+ ions
When the standard hydrogen electrode is connected to another half-cell, the standard
electrode potential of that half-cell can be read off a voltmeter
The standard electrode potential of a half-cell can be determined by connecting it to a
standard hydrogen electrode
5.3.4 Measuring the Standard Electrode
Potential
DOWNLOAD PDF
Measuring the Standard Electrode Potential

There are three different types of half-cells that can be connected to a standard hydrogen
electrode
o A metal / metal ion half-cell
o A non-metal / non-metal ion half-cell
o An ion / ion half-cell (the ions are in different oxidation states)
62 | P a g e
Metal/metal ion half-cell
Example of a metal / metal ion half-cell connected to a standard hydrogen electrode


An example of a metal/metal ion half-cell is the Ag+/ Ag half-cell
o Ag is the metal
o Ag+ is the metal ion
This half-cell is connected to a standard hydrogen electrode and the two half-equations
are:
Ag+ (aq) + e- ⇌ Ag (s)
Eꝋ = + 0.80 V
2H+ (aq) + 2e- ⇌ H2 (g)
Eꝋ = 0.00 V



Since the Ag+/ Ag half-cell has a more positive Eꝋ value, this is the positive pole and the
H+/H2 half-cell is the negative pole
The standard cell potential (Ecellꝋ) is Ecellꝋ = (+ 0.80) - (0.00) = + 0.80 V
The Ag+ ions are more likely to get reduced than the H+ ions as it has a greater Eꝋ value
o Reduction occurs at the positive pole
o Oxidation occurs at the negative pole
Non-metal/non-metal ion half-cell



In a non-metal/non-metal ion half-cell platinum wire or foil is used as an electrode to
make electrical contact with the solution
o Like graphite, platinum is inert and does not take part in the reaction
o The redox equilibrium is established on the platinum surface
An example of a non-metal/non-metal ion is the Br2/Br- half-cell
o Br is the non-metal
o Br- is the non-metal ion
The half-cell is connected to a standard hydrogen electrode and the two half-equations
are:
63 | P a g e
Br2 (l) + 2e- ⇌ 2Br- (aq)
2H+ (aq) + 2e- ⇌ H2 (g)



Eꝋ = +1.09 V
Eꝋ = 0.00 V
The Br2/Br- half-cell is the positive pole and the H+/H2 is the negative pole
The Ecellꝋ is: Ecellꝋ = (+ 1.09) - (0.00) = + 1.09 V
The Br2 molecules are more likely to get reduced than H+ as they have a greater Eꝋ value
Example of a non-metal / non-metal ion half-cell connected to a standard hydrogen electrode
Ion/Ion half-cell



A platinum electrode is again used to form a half-cell of ions that are in different
oxidation states
An example of such a half-cell is the MnO4-/Mn2+ half-cell
o MnO4- is an ion containing Mn with oxidation state +7
o The Mn2+ ion contains Mn with oxidation state +2
This half-cell is connected to a standard hydrogen electrode and the two half-equations
are:
MnO4- (aq) + 8H+ (aq) + 5e- ⇌ Mn2+ (aq) + 4H2O (l)
2H+ (aq) + 2e- ⇌ H2 (g)



Eꝋ = +1.52 V
Eꝋ = 0.00 V
The H+ ions are also present in the half-cell as they are required to convert MnO4- into
Mn2+ ions
The MnO4-/Mn2+ - half-cell is the positive pole and the H+/H2 is the negative pole
The Ecellꝋ is Ecellꝋ = (+ 1.09) - (0.00) = + 1.09 V
64 | P a g e
Ions in solution half cell
Standard Cell Potential: Direction of Electron Flow & Feasibility
Direction of electron flow

The direction of electron flow can be determined by comparing the Eꝋ values of two
half-cells in an electrochemical cell
2Cl2 (g) + 2e- ⇌ 2Cl- (aq)
Cu2+ (aq) + 2e- ⇌ Cu (s)



Eꝋ = +1.36 V
Eꝋ = +0.34 V
The Cl2 more readily accept electrons from the Cu2+/Cu half-cell
o This is the positive pole
o Cl2 gets more readily reduced
The Cu2+ more readily loses electrons to the Cl2/Cl- half-cell
o This is the negative pole
o Cu2+ gets more readily oxidised
The electrons flow from the Cu2+/Cu half-cell to the Cl2/Cl- half-cell
o The flow of electrons is from the negative pole to the positive pole
65 | P a g e
The electrons flow through the wires from the negative pole to the positive pole
Feasibility




The Eꝋ values of a species indicate how easily they can get oxidised or reduced
The more positive the value, the easier it is to reduce the species on the left of the halfequation
o The reaction will tend to proceed in the forward direction
The less positive the value, the easier it is to oxidise the species on the right of the halfequation
o The reaction will tend to proceed in the backward direction
o A reaction is feasible (likely to occur) when the Ecellꝋ is positive
For example, two half-cells in the following electrochemical cell are:
Cl2 (g) + 2e- ⇌ 2Cl- (aq)
Eꝋ = +1.36 V
Cu2+ (aq) + 2e- ⇌ Cu (s)
Eꝋ = +0.34 V


Cl2 molecules are reduced as they have a more positive Eꝋ value
The chemical reaction that occurs in this half cell is:
Cl2 (g) + 2e- → 2Cl- (aq)


Cu2+ ions are oxidised as they have a less positive Eꝋ value
The chemical reaction that occurs in this half cell is:
66 | P a g e
Cu (s) → Cu2+ (aq) + 2e
The overall equation of the electrochemical cell is (after cancelling out the electrons):
Cu (s) + Cl2 (g) → 2Cl- (aq) + Cu2+ (aq)
OR
Cu (s) + Cl2 (g) → CuCl2 (s)


The forward reaction is feasible (spontaneous) as it has a positive Eꝋ value of +1.02 V
((+1.36) - (+0.34))
The backward reaction is not feasible (not spontaneous) as it has a negative Eꝋ value of
-1.02 ((+0.34) - (+1.36))
A reaction is feasible when the standard cell potential Eꝋ is positive
67 | P a g e
Exam Tip
Remember that the electrons only move through the wires in the external circuit and not through
the electrolyte solution.
Redox Equations

The redox equations of an electrochemical cell can be constructed using the relevant
half-equations of the two half-cells
Constructing redox equations

Step 1: Determine in which half-cell the oxidation and in which half-cell the
reduction reaction takes place
Cl2 (g) + 2e- ⇌ 2Cl- (aq)
Eꝋ = +1.36 V
Zn2+ (aq) + 2e- ⇌ Zn (s)
Eꝋ = -0.76 V

o
o

Reduction occurs in the Cl2/Cl- half-cell as it has the more positive Eꝋ value
Oxidation occurs in the Zn+/Zn half-cell as it has the least positive Eꝋ value
Step 2: Write down the half equations for each half-cell
o Half-equation of the Cl2/Cl- half-cell
Cl2 (g) + 2e- → 2Cl- (aq)

o
Half-equation of the Zn+/Zn half-cell
Zn (s) → Zn2+ (aq) + 2e-

Step 3: Balance the number of electrons in both half-equations
o The number of electrons is already balanced in both half-equations as they both
contain two electrons

Step 4 - Add up the two half-equations
Cl2 (g) + 2e- → 2Cl- (aq)
Zn (s) → Zn2+ (aq) + 2e______________________________________ +
Cl2 (g) + Zn (s) + 2e → 2Cl- (aq) + Zn2+ (aq) + 2e68 | P a g e

Step 5 - Cancel out the electrons (and H+ ions and H2O molecules if any present) to find
the overall redox reaction
Cl2 (g) + Zn (s) → 2Cl- (aq) + Zn2+ (aq)
OR
Cl2 (g) + Zn (s) → ZnCl2 (s)
Worked example: Constructing redox reactions
Answer

Step 1:
Determine in which cell oxidation and in which cell reduction takes place
Reduction occurs in the Ag+/Ag half-cell as it has the most positive Eꝋ value
Oxidation occurs in the MnO4-/ Mn2+ half-cell as it has the least positive Eꝋ value

Step 2:
Write down the half-equations for each cell
The half-equation for the Ag+/Ag half-cell is:
Ag+ (aq) + e- → Ag (s)
The half-equation for the MnO4-/ Mn2+ half-cell is:
Mn2+ (aq) + 4H2O (l) → MnO4- (aq) + 8H+ (aq) + 5e
Step 3:
69 | P a g e
Balance the number of electrons in both half-equations
Multiply the half-equation for the Ag+/Ag half-cell by 5 so that both half-equations
contain 5 electrons
This gives: 5Ag+ (aq) + 5e- → 5Ag (s)

Step 4:
Add the two half-equations
5Ag+ (aq) + 5e- + Mn2+ (aq) + 4H2O (l) → 5Ag (s) + MnO4- (aq) + 8H+ (aq) + 5e
Step 5:
Cancel out the electrons to find the overall redox equation
5Ag+ (aq) + 5e- + Mn2+ (aq) + 4H2O (l) → 5Ag (s) + MnO4- (aq) + 8H+ (aq) + 5eThe fully balanced redox equation is:
5Ag+ (aq) + Mn2+ (aq) + 4H2O (l) → 5Ag (s) + MnO4- (aq) + 8H+ (aq)
NEXT TOPIC
5.4.1 Electrolysis: Calculations
DOWNLOAD PDF
Calculations in Electrolysis

Faraday’s constant can be used to calculate:
o The mass of a substance deposited at an electrode
o The volume of gas liberated at an electrode
Calculating the mass of a substance deposited at an electrode

To calculate the mass of a substance deposited at the electrode, you need to be able to:
o Write the half-equation at the electrode
o Determine the number of coulombs needed to form one mole of substance at the
specific electrode using Faraday’s constant
o Calculate the charge transferred during electrolysis
o Use simple proportion and the relative atomic mass of the substance to find its
mass
70 | P a g e
Worked example: Calculating the mass of a substance deposited at an
electrode
Answer
The magnesium (Mg2+) ion is a positively charged cation that will move towards the cathode.

Step 1: Write the half-equation at the cathode
Mg2+(aq)
+
1 mol

2e2 mol
→
Mg(s)
1 mol
Step 2: Determine the number of coulombs required to deposit one mole of magnesium
at the cathode
For every one mole of electrons, the number of coulombs needed is 96 500 C mol-1
In this case, there are two moles of electrons required
So, the number of coulombs needed is:
F = 2 x 96 500
F = 193 000 C mol-1

Step 3: Calculate the charge transferred during the electrolysis
Q=Ixt
Q = 2.20 x (60 x 15)
= 1980 C

Step 4: Calculate the mass of magnesium deposited by simple proportion using the
relative atomic mass of Mg
71 | P a g e
Calculating the mass of a substance deposited at an electrode table
Therefore, 0.25 g of magnesium is deposited at the cathode
Calculating the volume of gas liberated at an electrode

To calculate the volume of gas liberated at an electrode, you need to be able to:
o Write the half-equation at the electrode
o Determine the number of coulombs needed to form one mole of substance at the
specific electrode using Faraday’s constant
o Calculate the charge transferred during electrolysis
o Use simple proportion and the relationship 1 mol of gas occupies 24.0 dm3 at
room temperature
Worked example: Calculating the volume of a gas produced at an
electrode
Answer
The oxygen gas is formed from the oxidation of negatively charged hydroxide (OH-) ions at the
anode
Step 1: Write the half-equation at the anode
72 | P a g e
4OH-(aq)
→
4 mol

O2(g)
1 mol
+
2H2O(l)
+
2 mol
4e4 mol
Step 2: Determine the number of coulombs required to form one mole of oxygen gas at
the anode
For every one mole of electrons, the number of coulombs needed is 96 500 C mol-1
So, for four moles of electrons, the number of coulombs needed is:
F = 4 x 96 500
F = 386 000 C mol-1

Step 3: Calculate the charge transferred during the electrolysis
Q=Ixt
Q = 0.75 x (60 x 35)
= 1575 C

Step 4: Calculate the volume of oxygen liberated by simple proportion using the
relationship 1 mol of gas occupies 24.0 dm3 at room temperature
Calculating the volume of a gas liberated at an electrode table
Therefore, 0.0979 dm3 of oxygen is formed at the anode
73 | P a g e
Worked example: Calculating the volume of hydrogen gas produced at
an electrode
The hydrogen gas is formed from the reduction of positively charged hydrogen (H+) ions at the
cathode

Step 1: Write the half-equation at the cathode
2H+ (aq) +
2e-
2 mols
2 mols

→
H2 (g)
1 mol
Step 2: Determine the number of coulombs required to form one mole of hydrogen gas at
the cathode
For every one mole of electrons, the number of coulombs needed is:
F = 96 500 C mol-1
F = 1 x 96 500
F = 96 500 C
So, for two moles of electrons, the number of coulombs needed is:
F = 2 x 96 500
F = 193 000 C

Step 3: Calculate the charge transferred during the electrolysis
Q=Ixt
Q = 3.25 x (60 x 17.5)
= 3 413 C
74 | P a g e

Step 4: Calculate the volume of hydrogen liberated by simple proportion using the
relationship 1 mol of gas occupies 24.0 dm3 at room temperature
Calculating the volume of hydrogen gas produced at an electrode table
Therefore, 0.42 dm3 of hydrogen is formed at the cathode
5.4.2 Standard Cell Potential: Calculations,
Electron Flow & Feasibility
DOWNLOAD PDF
Calculating Standard Cell Potential

Once the standard electrode potentials (Eꝋ) of the half-cells are determined, the standard
cell potential (Ecellꝋ) can be calculated by subtracting the less positive Eꝋ from the more
positive Eꝋ value
o The half-cell with the more positive Eꝋ value will be the positive pole
o The half-cell with the less positive Eꝋ value will be the negative pole
75 | P a g e
Worked example: Calculating the standard cell potential
Answer

Step 1: Calculate the standard cell potential
Ecellꝋ = (+0.34) - (-0.76)
= +1.10 V
The voltmeter will therefore read off a value of 1.10 V

Step 2: Determine the positive and negative poles
The Cu2+/Cu half-cell is the positive pole as its Eꝋ is more positive than the Eꝋ value of the
Zn2+/Zn half-cell
76 | P a g e
5.4.3 Electrochemical Series & Redox
Equations
Electrochemical Series





The Eꝋ values of a species indicate how easily they can get oxidised or reduced
In other words, they indicate the relative reactivity of elements, compounds and ions
as oxidising agents or reducing agents
The electrochemical series is a list of various redox equilibria in order of
decreasing Eꝋ values
More positive (less negative) Eꝋ values indicate that:
o The species is easily reduced
o The species is a better oxidising agent
Less positive (more negative) Eꝋ values indicate that:
o The species is easily oxidised
o The species is a better reducing agent
Example of an electrochemical series in which the equilibria are arranged in order
of decreasing Eꝋ values
77 | P a g e
5.4.4 Non-Standard Conditions
Effect of Concentration on Electrode Potential


Changes in temperature and concentration of aqueous ions will affect the standard
electrode potential (Eꝋ) of a half-cell
Under these non-standard conditions, E is used as a symbol for the electrode potential
instead of Eꝋ
Increasing the concentration of the species on the left




If the concentration of the species on the left is increased, the position of equilibrium will
shift to the right
This means that the species on the left gets more easily reduced
The E value becomes more positive (or less negative)
Let’s look at the half-cell below as an example
Zn2+ (aq) + 2e- ⇌ Zn (s)




If the concentration of Zn2+ (species on the left) is increased, the equilibrium position
will shift to the right
The species on the left (Zn2+) will get more easily reduced
Therefore, the E value becomes less negative and will change too, for example, -0.50 V
instead
This principle can also be applied to a half-cell with a positive Eꝋ value such as:
Fe3+ (aq) + e- ⇌ Fe2+ (aq)



Eꝋ = -0.76 V
Eꝋ = +0.77 V
If the concentration of Fe3+ (species on the left) is increased, the equilibrium position
will shift to the right
The species on the left (Fe3+) will get more easily reduced
Therefore, the E value becomes more positive and will change too, for example, +0.89 V
instead
Increasing the concentration of species on the right




If the concentration of the species on the right is increased, the position of equilibrium
will shift to the left
This means that the species on the left gets less easily reduced
The E value becomes less positive (or more negative)
Let’s look again at the half-cell below
Zn2+ (aq) + 2e- ⇌ Zn (s)
78 | P a g e
Eꝋ = -0.76 V




If the concentration of Zn (species on the right) is increased, the equilibrium position
will shift to the left
The species on the left (Zn2+) will get less easily reduced
Therefore, the E value becomes more negative and will change too, for example, -0.82 V
instead
This principle can, again, also be applied to a half-cell with a positive Eꝋ value:
Fe3+ (aq) + e- ⇌ Fe2+ (aq)



Eꝋ = +0.77 V
If the concentration of Fe2+ (species on the right) is increased, the equilibrium position
will shift to the left
The species on the left (Fe3+) will get less easily reduced
Therefore, the E value becomes less positive and will change too, for example, +0.56 V
instead
Effect of concentration on the electrode potential
79 | P a g e
5.4.5 Nernst Equation
DOWNLOAD PDF
The Nernst Equation


Under non-standard conditions, the cell potential of the half-cells is shown by the
symbol Ecell
The effect of changes in temperature and ion concentration on the Ecell can be deduced
using the Nernst equation
E = electrode potential under nonstandard conditions
E⦵ = standard electrode potential
R = gas constant (8.31 J K-1 mol-1)
T = temperature (kelvin, K)
z = number of electrons transferred in the reaction
F = Faraday constant (96 500 C mol-1)
ln = natural logarithm

This equation can be simplified to

o
o


At standard temperature, R, T and F are constant
ln x = 2.303 log10 x
The Nernst equation only depends on aqueous ions and not solids or gases
The concentrations of solids and gases are therefore set to 1.0 mol dm-3
Applying Nernst equation

The concentrations of ions for the Fe3+/Fe2+ half-cell are as follows:
Fe3+ (aq) + e- ⇌ Fe2+ (aq)
80 | P a g e
[Fe3+] = 0.034 mol dm-3
[Fe2+] = 0.64 mol dm-3

The Nernst equation for this half-reaction is, therefore:

o
o
o

The oxidised species is Fe3+ as it has a higher oxidation number (+3)
The reduced species is Fe2+ as it has a lower oxidation number (+2)
z is 1 as only one electron is transferred in this reaction
An example of a half-cell in which two electrons are transferred is the Cu2+/Cu half-cell
Cu2+ (aq) + 2e- ⇌ Cu (s)
[Cu2+] = 0.0010 mol dm-3

The Nernst equation for this half-reaction is:

o
o
o
o
81 | P a g e
The oxidised species is Cu2+ as it has a higher oxidation number (+2)
The reduced species is Cu as it has a lower oxidation number (0)
Cu is a solid and is not included in the Nernst equation (its concentration doesn’t
change)
z is 2 as 2 electrons are transferred in this reaction
Worked example: Calculating the electrode potential of a Fe3+/Fe2+ halfcell
Answer
= (+0.77) + (-0.075)
= +0.69 V
82 | P a g e
Worked example: Calculating the electrode potential of a Cu2+/Cu halfcell
Answer
= (+0.34) + (-0.089)
= +0.25 V
Exam Tip
Make sure you always check what the temperature is. If the temperature is not 298 K (or 25 oC)
the full Nernst equation should be used.You don’t need to know how to simplify the Nernst
equation to
You are only expected to use the equation when the temperature is 298 K (or 25 oC).
83 | P a g e
5.4.6 Standard Electrode Potentials: Free
Energy Change
Calculating Free Energy Change Using Standard Electrode Potentials

The standard free energy change can be calculated using the standard cell potential of
an electrochemical cell
ΔGꝋ = - n x Ecellꝋ x F
ΔGꝋ = standard Gibbs free energy
n = number of electrons transferred in the reaction
Ecellꝋ = standard cell potential (V)
F = Faraday constant (96 500 C mol-1)
84 | P a g e
Worked Example: Calculating the standard Gibbs free energy change
Answer


Step 1: Determine the two half-equations and their Eꝋ using the Data booklet
Fe3+ (aq) + e- ⇌ Fe2+ (aq)
Eꝋ = +0.77 V
Cu2+ (aq) + 2e- ⇌ Cu (s)
Eꝋ = +0.34 V
Step 2 : Calculate the Ecellꝋ
Ecellꝋ = Eredꝋ - Eoxꝋ
= (+0.77) - (+0.34)
= +0.43 V

Step 3: Determine the number of electrons transferred in the reaction
The Cu2+/Cu has a smaller Eꝋ value which means that it gets oxidised
It transfers two electrons to two Fe3+ ions
Each Fe3+ ion accepts one electron so the total number of electrons transferred is two

Step 4: Substitute the values in for the standard Gibbs free energy equation
ΔGꝋ = - n x Ecellꝋ x F
= -2 x (+0.43) x 96 500
= -82 990 J mol-1
= -83 kJ mol-1
85 | P a g e
EQUILIBRIA
5.5.1 Acids & Bases
Conjugate Acids & Bases

A Brønsted acid is a species that can donate a proton
o For example, hydrogen chloride (HCl) is a Brønsted acid as it can lose a proton to
form a hydrogen (H+) and chloride (Cl-) ion
HCl (aq) → H+ (aq) + Cl- (aq)

A Brønsted base is a species that can accept a proton
o For example, a hydroxide (OH-) ion is a Brønsted base as it can accept a proton to
form water
OH- (aq) + H+ (aq) → H2O (l)









In an equilibrium reaction, the products are formed at the same rate as the reactants are
used
This means that at equilibrium, both reactants and products are present in the solution
For example, ethanoic acid (CH3COOH) is a weak acid that partially dissociates in
solution
When equilibrium is established there are CH3COOH, H2O, CH3COO- and H3O+ ions
present in the solution
The species that can donate a proton are acids and the species that can accept a proton
are bases
The reactant CH3COOH is linked to the product CH3COO- by the transfer of
a proton from the acid (CH3COOH) to the base (CH3COO-)
Similarly, the H2O molecule is linked to H3O+ ion by the transfer of a proton
These pairs are therefore called conjugate acid-base pairs
A conjugate acid-base pair is two species that are different from each other by an H+ ion
o Conjugate here means related
o In other words, the acid and base are related to each other by one proton
difference
86 | P a g e
Defining Conjugate Acids & Bases





Conjugate acid-base pairs are a pair of reactants and products that are linked to each
other by the transfer of a proton
For example, in the equilibrium reaction below, the propanoic acid (CH3CH2COOH)
partially dissociates in solution to form propanoate (CH3CH2COO-) and hydrogen (H+)
ions
When equilibrium is established there are CH3COOH, H2O, CH3COO- and H3O+ ions
present in the solution
The species that can donate a proton are acids and the species that can accept a proton
are bases
In the forward reaction:
o The acid CH3CH2COOH is linked to the base CH3CH2COOo CH3CH2COO- is, therefore, the conjugate base of CH3CH2COOH
CH3CH2COOH - CH3CH2COO- = conjugate acid-base pair

o
o
The base H2O is linked to the acid H3O+
H3O+ is, therefore, the conjugate acid of H2O
H2O - H3O+ = conjugate acid-base pair

In the reverse reaction
o The base CH3CH2COO- is linked to the acid CH3CH2COOH
o CH3CH2COOH is therefore conjugate acid of CH3CH2COO-
CH3CH2COO- - CH3CH2COOH = conjugate acid-base pair

o
o
The acid H3O+ is linked to the base H2O
H2O is, therefore, the conjugate base of H3O+
H3O+ - H2O = conjugate acid-base pair
Worked Example: Identifying conjugate acid-base pairs
87 | P a g e
Answer


In the forward reaction:
o NH4+ is the conjugate acid of the base NH3
o OH- is the conjugate base of the acid H2O
In the reverse reaction
o NH3 is the conjugate base of the acid NH4+
o H2O is the conjugate acid of the base OH-
88 | P a g e
5.5.2 pH, Ka, pKa & Kw Calculations
Calculating pH, Ka, pKA & Kw
pH



The pH indicates the acidity or basicity of an acid or alkali
The pH scale goes from 0 to 14
o Acids have pH between 0-7
o Pure water is neutral and has a pH of 7
o Bases and alkalis have pH between 7-14
The pH can be calculated using: pH = -log10 [H+]
where [H+] = concentration of H+ ions (mol dm-3)

The pH can also be used to calculate the concentration of H+ ions in solution by
rearranging the equation to:
[H+] = 10-pH
Worked Example: Calculating the pH of acids
Answer
pH = -log [H+]
= -log 1.32 x 10-3
= 2.9
89 | P a g e
Ka & pKa


The Ka is the acidic dissociation constant
o It is the equilibrium constant for the dissociation of a weak acid at 298 K
For the partial ionisation of a weak acid HA the equilibrium expression to find Ka is as
follows:
HA (aq) ⇌ H+ (aq) + A- (aq)



When writing the equilibrium expression for weak acids, the following assumptions are
made:
o The concentration of hydrogen ions due to the ionisation of water is negligible
o The dissociation of the weak acid is so small that the concentration of HA is
approximately the same as the concentration of AThe value of Ka indicates the extent of dissociation
o A high value of Ka means that:
 The equilibrium position lies to the right
 The acid is almost completely ionised
 The acid is strongly acidic
o A low value of Ka means that:
 The equilibrium position lies to the left
 The acid is only slightly ionised (there are mainly HA and only a few
H+ and A- ions)
 The acid is weakly acidic
Since Ka values of many weak acids are very low, pKa values are used instead to compare
the strengths of weak acids with each other
pKa = -log10 Ka

The less positive the pKa value the more acidic the acid is
90 | P a g e
Worked Example: Calculating the Ka & pKa of weak acids
Answer

Step 1: Write down the equation for the partial dissociation of ethanoic acid
CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

Step 2: Write down the equilibrium expression to find Ka

Step 3: Simplify the expression
The ratio of H+ to CH3COO- is 1:1
The concentration of H+ and CH3COO- is, therefore, the same
The equilibrium expression can be simplified to:

Step 4: Substitute the values into the expression to find Ka
= 1.74 x 10-5
91 | P a g e

Step 5: Determine the units of Ka
= mol dm-3
The value of Ka is therefore 1.74 x 10-5 mol dm-3

Step 6: Find pKa
pKa = - log10 Ka
= - log10 (1.74 x 10-5)
= 4.76
92 | P a g e
Kw


The Kw is the ionic product of water
o It is the equilibrium constant for the dissociation of water at 298 K
o Its value is 1.00 x 10-14 mol2 dm-6
For the ionisation of water the equilibrium expression to find Kw is as follows:
H2O (l) ⇌ H+ (aq) + OH- (aq)



As the extent of ionisation is very low, only small amounts of H+ and OH- ions are
formed
The concentration of H2O can therefore be regarded as constant and removed from
the Kw expression
The equilibrium expression therefore becomes:
Kw = [H+] [OH-]

As the [H+] = [OH+] in pure water, the equilibrium expression can be further simplified
to:
Kw = [H+]2
Worked Example: Calculating the concentration of H+ of pure water
Answer

Step 1: Write down the equation for the partial dissociation of water
In pure water, the following equilibrium exists:
H2O (l) ⇌ H+ (aq) + OH- (aq)

Step 2: Write down the equilibrium expression to find Kw
93 | P a g e

Step 3: Simplify the expression
Since the concentration of H2O is constant, this expression can be simplified to:
Kw = [H+] [OH-]

Step 4: Further simplify the expression
The ratio of H+ to OH- is 1:1
The concentration of H+ and OH- is, therefore, the same and the equilibrium expression can be
further simplified to:
Kw = [H+]2

Step 5: Rearrange the equation to find [H+]
Kw = [H+]2

Step 6: Substitute the values into the expression to find Kw
= 1.00 x 10-7 mol dm-3
Exam Tip
Remember:The greater the Ka value, the more strongly acidic the acid is.The greater the
pKa value, the less strongly acidic the acid is.Also, you should be able to rearrange the following
expressions:
pH = -log10 [H+] TO [H+] = 10-pH
pKa = - log10 Ka TO Ka = 10-pKa
94 | P a g e
5.5.3 pH & [H+] Calculations
Calculating [H+] & pH

If the concentration of H+ of an acid or alkali is known, the pH can be calculated using
the equation:
pH = -log [H+]

Similarly, the concentration of H+ of a solution can be calculated if the pH is known by
rearranging the above equation to:
[H+] = 10-pH
Strong acids

Strong acids are completely ionised in solution
HA (aq) → H+ (aq) + A- (aq)



Therefore, the concentration of hydrogen ions ([H+]) is equal to the concentration of acid
([HA])
The number of hydrogen ions ([H+]) formed from the ionisation of water is very
small relative to the [H+] due to ionisation of the strong acid and can therefore
be neglected
The total [H+] is therefore the same as the [HA]
95 | P a g e
Worked Example: pH calculations of a strong acid
Answer
Hydrochloric acid is a strong monobasic acid
HCl (aq) → H+ (aq) + Cl- (aq)
Answer 1
The pH of the solution is:
pH = -log [H+]
= -log 1.6 x 10-4
= 3.80
Answer 2
The hydrogen concentration can be calculated by rearranging the equation for pH
pH = -log [H+]
[H+] = 10-pH
= 10-3.1
= 7.9 x 10-4 mol dm-3
Strong alkalis

Strong alkalis are completely ionised in solution
96 | P a g e
BOH (aq) → B+ (aq) + OH- (aq)


Therefore, the concentration of hydroxide ions ([OH-]) is equal to the concentration of
base ([BOH])
o Even strong alkalis have small amounts of H+ in solution which is due to the
ionisation of water
The concentration of OH- in solution can be used to calculate the pH using the ionic
product of water
Kw = [H+] [OH-]

Since Kw is 1.00 x 10-14 mol2 dm-6

Once the [H+] has been determined, the pH of the strong alkali can be founding using pH
= -log[H+]
Similarly, the ionic product of water can be used to find the concentration of OH- ions in
solution if [H+] is known

97 | P a g e
Worked Example: pH calculations of a strong alkali
Answer
Sodium hydroxide is a strong base which ionises as follows:
NaOH (aq) → Na+ (aq) + OH- (aq)
Answer 1
The pH of the solution is:
pH = -log [H+]
= -log 3.5 x 10-11
= 10.5
Answer 2

Step 1: Calculate hydrogen concentration by rearranging the equation for pH
pH = -log [H+]
= 10-pH
= 10-12.3
= 5.01 x 10-13 mol dm-3

Step 2: Rearrange the ionic product of water to find the concentration of hydroxide
ions
Kw = [H+] [OH-]
98 | P a g e

Step 3: Substitute the values into the expression to find the concentration of hydroxide
ions
Since Kw is 1.00 x 10-14 mol2 dm-6
= 0.0199 mol dm-3
Weak acids

The pH of weak acids can be calculated when the following is known:
o The concentration of the acid
o The Ka value of the acid
Worked Example: pH calculations of weak acids
99 | P a g e
Answer
Ethanoic acid is a weak acid which ionises as follows:
CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

Step 1: Write down the equilibrium expression to find Ka

Step 2: Simplify the expression
The ratio of H+ to CH3COO- ions is 1:1
The concentration of H+ and CH3COO- ions are therefore the same
The expression can be simplified to:

Step 3: Rearrange the expression to find [H+]

Step 4: Substitute the values into the expression to find [H+]
= 1.32 x 10-3 mol dm-3

Step 5: Find the pH
pH = -log10 [H+]
= -log10 1.32 x 10-3
= 2.88
100 | P a g e
5.5.4 Buffers
Buffers

A buffer solution is a solution in which the pH does not change a lot when small
amounts of acids or alkalis are added
o A buffer solution is used to keep the pH almost constant
o A buffer can consists of weak acid - conjugate base or weak base - conjugate
acid
Ethanoic acid & sodium ethanoate as a buffer


A common buffer solution is an aqueous mixture of ethanoic acid and sodium
ethanoate
Ethanoic acid is a weak acid and partially ionises in solution to form a relatively low
concentration of ethanoate ions

Sodium ethanoate is a salt which fully ionises in solution

There are reserve supplies of the acid (CH3COOH) and its conjugate base (CH3COO-)
o The buffer solution contains relatively high concentrations of CH3COOH (due to
ionisation of ethanoic acid) and CH3COO- (due to ionisation
of sodium ethanoate)
In the buffer solution, the ethanoic acid is in equilibrium with hydrogen and ethanoate
ions

101 | P a g e





When H+ ions are added:
The equilibrium position shifts to the left as H+ ions react with CH3COO- ions to form
more CH3COOH until equilibrium is re-established
As there is a large reserve supply of CH3COO- the concentration of CH3COO- in solution
doesn’t change much as it reacts with the added H+ ions
As there is a large reserve supply of CH3COOH the concentration of CH3COOH in
solution doesn’t change much as CH3COOH is formed from the reaction of
CH3COO- with H+
As a result, the pH remains reasonable constant
When hydrogen ions are added to the solution the pH of the solution would decrease;
However, the ethanoate ions in the buffer solution react with the hydrogen ions to prevent this
and keep the pH constant


When OH- ions are added:
The OH- reacts with H+ to form water
OH- (aq) + H+ (aq) → H2O (l)


The H+ concentration decreases
The equilibrium position shifts to the right and more CH3COOH molecules ionise to form
more H+ and CH3COO- until equilibrium is re-established
CH3COOH (aq) → H+ (aq) + CH3COO- (aq)



As there is a large reserve supply of CH3COOH the concentration of CH3COOH in
solution doesn’t change much when CH3COOH dissociates to form more H+ ions
As there is a large reserve supply of CH3COO- the concentration of CH3COO- in solution
doesn’t change much
As a result, the pH remains reasonable constant
102 | P a g e
When hydroxide ions are added to the solution, the hydrogen ions react with them to form
water; The decrease in hydrogen ions would mean that the pH would increase however the
equilibrium moves to the right to replace the removed hydrogen ions and keep the pH constant
Uses of buffer solutions in controlling the pH of blood



In humans, HCO3- ions act as a buffer to keep the blood pH between 7.35 and 7.45
Body cells produce CO2 during aerobic respiration
This CO2 will combine with water in blood to form a solution containing H+ ions
CO2 (g) + H2O (l) ⇌ H+ (aq) + HCO3- (aq)




This equilibrium between CO2 and HCO3- is extremely important
If the concentration of H+ ions is not regulated, the blood pH would drop and cause
‘acidosis’
o Acidosis refers to a condition in which there is too much acid in the body fluids
such as blood
o This could cause body malfunctioning and eventually lead to coma
If there is an increase in H+ ions
The equilibrium position shifts to the left until equilibrium is restored
H+ (aq) + HCO3- (aq) → CO2 (g) + H2O (l)


This reduces the concentration of H+ and keeps the pH of the blood constant
If there is a decrease in H+ ions
o The equilibrium position shifts to the right until equilibrium is restored
CO2 (g) + H2O (l) → H+ (aq) + HCO3- (aq)

This increases the concentration of H+ and keeps the pH of the blood constant
Exam Tip
Remember that buffer solutions cannot cope with excessive addition of acids or alkalis as their
pH will change significantly.The pH will only remain relatively constant if small amounts of
acids or alkalis are added.
103 | P a g e
5.5.5 Buffer Calculations
Calculating pH of Buffer Solutions


The pH of a buffer solution can be calculated using:
o The Ka of the weak acid
o The equilibrium concentration of the weak acid and its conjugate base (salt)
To determine the pH, the concentration of hydrogen ions is needed which can be found
using the equilibrium expression

To simplify the calculations, logarithms are used such that the expression becomes:

Since -log10 [H+] = pH, the expression can also be rewritten as:
Worked Example: Calculating the pH of a buffer solution
Answer
Ethanoic acid is a weak acid that ionises as follows:
CH3COOH (aq) ⇌ H+ (aq) + CH3COO- (aq)

Step 1: Write down the equilibrium expression to find Ka
104 | P a g e

Step 2: Rearrange the equation to find [H+]

Step 3: Substitute the values into the expression
= 8.39 x 10-6 mol dm-3

Step 4: Calculate the pH
pH = - log [H+]
= -log 8.39 x 10-6
= 5.08
105 | P a g e
5.5.6 Solubility Product
Solubility Product

Solubility is defined as the number of grams or moles of compound needed
to saturate 100 g of water, or it can also be defined in terms of 1 kg of water, at a given
temperature
o For example, sodium chloride (NaCl) is considered to be a soluble salt as a
saturated solution contains 36 g of NaCl per 100 g of water
o Lead chloride (PbCl2) on the other hand is an insoluble salt as a saturated solution
only contains 0.99 g of PbCl2 per 100 g of water
Solubility product

The solubility product (Ksp) is:
o The product of the concentrations of each ion in a saturated solution of a
relatively soluble salt
o At 298 K
o Raised to the power of their relative concentrations
C (s) ⇌ aAx+ (aq) + bBy- (aq)
Ksp = [Ax+ (aq)]a [By- (aq)]b


When an undissolved ionic compound is in contact with a saturated solution of its
ions, an equilibrium is established
The ions move from the solid to the saturated solution at the same rate as they move from
the solution to the solid
o For example, the undissolved magnesium chloride (MgCl2) is in equilibrium with
a saturated solution of its ions
MgCl2 (s) ⇌ Mg2+ (aq) + 2Cl- (aq)
106 | P a g e
When the undissolved MgCl2 salt gets in contact with its ions in a saturated solution, an
equilibrium between the salt and ions is established

o
The solubility product for this equilibrium is:
Ksp = [Mg2+ (aq)] [Cl- (aq)]2


The Ksp is only useful for sparingly soluble salts
The smaller the value of Ksp, the lower the solubility of the salt
Expressing Ksp

The general equilibrium expression for the solubility product (Ksp) is:
C (s) ⇌ aAx+ (aq) + bBy- (aq)
Ksp = [Ax+ (aq)]a [By- (aq)]b
Worked Example: Expressing Ksp of ionic compounds
107 | P a g e
Answer
Expressing Ksp of ionic compounds answers table
5.5.7 Solubility Product Calculations
Calculating the Solubility Product

Calculations involving the solubility product (Ksp) may include::
o Calculating the solubility product of a compound from its solubility
o Calculating the solubility of a compound from the solubility product
Worked example: Calculating the solubility product of a compound from
its solubility
108 | P a g e
Answer

Step 1: Write down the equilibrium equation
PbBr2 (s) ⇌ Pb2+ (aq) + 2Br- (aq)

Step 2: Write down the equilibrium expression
Ksp = [Pb2+(aq)] [Br- (aq)]2

Step 3: Calculate the ion concentrations in the solution
[PbBr2(s)] = 1.39 x 10-3 mol dm-3
The ratio of PbBr2 to Pb2+ is 1:1
[Pb2+(aq)] = [PbBr2(s)] = 1.39 x 10-3 mol dm-3
The ratio of PbBr2 to Br- is 1:2
[Br-(aq)] = 2 x [PbBr2(s)] = 2 x 1.39 x 10-3 mol dm-3
= 2.78 x 10-3 mol dm-3

Step 4: Substitute the values into the expression to find the solubility product
Ksp = (1.39 x 10-3) x (2.78 x 10-3)2
= 1.07 x 10-8

Step 6: Determine the correct units of Ksp
Ksp = (mol dm-3) x (mol dm-3)2
= mol3 dm-9
The solubility product is therefore 1.07 x 10-8 mol3 dm-9
109 | P a g e
Worked example: Calculating the solubility of a compound from its
solubility product
Answer

Step 1: Write down the equilibrium equation
CuO (s) ⇌ Cu2+ (aq) + O2- (aq)

Step 2: Write down the equilibrium expression
Ksp = [Cu2+ (aq)] [O2- (aq)]

Step 3: Simplify the equilibrium expression
The ratio of Cu2+ to O2- is 1:1
[Cu2+(aq)] = [O2-(aq)] so the expression can be simplified to:
Ksp = [Cu2+ (aq)]2

Step 4: Substitute the value of Ksp into the expression to find the concentration
5.9 x 10-36 = [Cu2+ (aq)]2
= 2.4 x 10-18 mol dm-3
Since [CuO(s)] = [Cu2+ (aq)] the solubility of copper oxide is 2.4 x 10-18 mol dm-3
Exam Tip
Remember that the solubility product is only applicable to very slightly soluble salts and cannot
be used for soluble salts such as:





Group 1 element salts
All nitrates salts
All ammonium salts
Many sulfate salts
Many halide salts (except for lead(II) halides and silver halides)
110 | P a g e
5.5.8 The Common Ion Effect
Solubility Product & the Common Ion Effect





A saturated solution is a solution that contains the maximum amount of dissolved salt
If a second compound, which has an ion in common with the dissolved salt, is added to
the saturated solution, the solubility of the salt reduces and a solid precipitate will be
formed
This is also known as the common ion effect
For example, if a solution of potassium chloride (KCl) is added to a saturated solution
of silver chloride (AgCl) a precipitate of silver chloride will be formed
o The chloride ion is the common ion
The solubility product can be used to predict whether a precipitate will actually form or
not
o A precipitate will form if the product of the ion concentrations is greater than the
solubility product (Ksp)
Common ion effect in silver chloride


When a KCl solution is added to a saturated solution of AgCl, an AgCl precipitate forms
In a saturated AgCl solution, the silver chloride is in equilibrium with its ions
AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)

When a solution of potassium chloride is added:
o Both KCl and AgCl have the common Cl- ion
o There is an increased Cl- concentration so the equilibrium position shifts to the
left
o The increase in Cl- concentration also means that [Ag+ (aq)] [Cl-(aq)]
is greater than the Ksp for AgCl
o As a result, the AgCl is precipitated
The addition of potassium chloride to a saturated solution of silver chloride results in the
precipitate of silver chloride
111 | P a g e
Worked Example: Calculations using the Ksp values and the
concentration of the common ion
Answer

Step 1: Determine the equilibrium reaction of CaSO4
CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)

Step 2: Write down the equilibrium expression for Ksp
Ksp = [Ca2+ (aq)] [SO42- (aq)]

Step 3: Determine the concentrations of the ions
There are equal volumes of each solution
This means that the total solution was diluted by a factor of 2
The new concentrations of the ions are there halved
= 5.0 x 10-4 mol dm-3

Step 4: Substitute the values into the expression
Product of the ion concentrations = [Ca2+ (aq)] x [SO42- (aq)]
= (5.0 x 10-4) x (5.0 x 10-4)
= 2.5 x 10-7 mol2 dm-6

Step 5: Determine if a precipitate will form
As the product of the ion concentration (2.5 x 10-7 mol dm-3 ) is smaller than the Ksp value (2.0 x
10-5 mol2 dm-6), the CaSO4 precipitate will not be formed
112 | P a g e
5.5.9 Partition Coefficients
Partition Coefficient & Calculations




The partition coefficient (Kpc) is the ratio of the concentrations of a solute in two
different immiscible solvents in contact with each other when equilibrium has been
established (at a particular temperature)
For example, methylamine (CH3NH2) is dissolved in two immiscible solvents:
o Water
o An organic solvent
A separating funnel is shaken with the organic solvent and aqueous methylamine
The methylamine is soluble in both solvents, so when the mixture is left to settle
an equilibrium is established
o The rate of methylamine molecules moving from the organic layer into the
aqueous layer is equal to the rate of molecules moving from the aqueous layer to
the organic layer
CH3NH2(aq) ⇌ CH3NH2(organic solvent)

The value of its equilibrium constant is also called the partition coefficient
The partition coefficient is the ratio of methylamine molecules in the organic and aqueous
layer once equilibrium has been established
113 | P a g e
Calculating Partition Coefficients

The partition coefficient (Kpc) for a system in which the solute is in the same physical
state in the two solvents can be calculated using the equilibrium expression
Worked example: Calculating the partition coefficient
Answer

Step 1: Write down the equilibrium equation
CH3NH2(aq) ⇌ CH3NH2(organic solvent)

Step 2: Write down the equilibrium expression

Step 3: Determine how many moles of CH3NH2 has reacted with HCl at the end-point
At the end-point, all CH3NH2 (aq) has been neutralised by HCl (aq)
CH3NH2 (aq) + HCl (aq) → CH3NH3Cl (aq)
CH3NH2 and HCl react in a ratio of 1:1
Mol (HCl) = mol (CH3NH2) = 0.225 x 0.0141
= 3.18 x 10-3 mol
114 | P a g e

Step 4: Determine the number of moles of CH3NH2 present in the aqueous layer
Only 50.0 cm3 of the aqueous layer was used to titrate against HCl
Thus, 3.18 x 10-3 mol of CH3NH2 was present in only 50.0 cm3 of the aqueous layer
The number of moles of CH3NH2 in 100 cm3 aqueous layer is, therefore:
Mol (CH3NH2 aqueous layer) = 3.18 x 10-3 x 2 = 6.34 x 10-3 mol

Step 5: Determine the number of moles of CH3NH2 in the organic layer
Mol CH3NH2 (organic layer) = mol CH3NH2 (total) - mol CH3NH2(aqueous layer)
Mol CH3NH2 (total) = 0.100 x 0.150 = 0.015 mol
Mol CH3NH2 (organic layer) = 0.015 - 6.34 x 10-3 = 8.67 x 10-3 mol

Step 6: Change the number of moles into concentrations
= 0.063 mol dm-3
= 0.116 mol dm-3

Step 7: Substitute the values into the Kpc expression
= 1.83
Since the value of Kpc is larger than 1, methylamine is more soluble in the organic solvent than
in water
115 | P a g e
Factors Affecting the Partition Coefficient






The partition coefficient (Kpc) depends on the solubilities of the solute in the two solvents
The degree of solubility of a solute is determined by how strong the intermolecular
bonds between solute and solvent are
The strength of these intermolecular bonds, in turn, depends on the polarity of
the solute and solvent molecules
For example, ammonia is more soluble in water than in an organic solvent such as carbon
tetrachloride (CCl4)
o Ammonia and water are both polar molecules that form hydrogen bonds with
each other
o Ammonia forms permanent dipole-induced dipole forces with the non-polar
CCl4 molecules
o Since these forces are much weaker than hydrogen bonding, ammonia is less
soluble in CCl4
When Kpc is < 1 the solute is more soluble in water than the organic solvent
When Kpc is > 1 the solute is more soluble in the organic solvent than the water
116 | P a g e
REACTION KINETICS
5.6.1 Basics of Kinetics
Kinetics: Basics



The rate of reaction refers to the change in the amount or concentration of a reactant or
product per unit time and can be found by:
Measuring the decrease in the concentration of a reactant OR
Measuring the increase in the concentration of a product over time
o The units of rate of reaction are mol dm-3 s-1
Rate equation

The thermal decomposition of calcium carbonate (CaCO3) will be used as an example to
study the rate of reaction
CaCO3 (s) → CaO (s) + CO2 (g)

The rate of reaction at different concentrations of CaCO3 is measured and tabulated
Rate of reactions table
117 | P a g e

A directly proportional relationship between the rate of the reaction
and concentration of CaCO3 is observed when a graph is plotted
Rate of thermal decomposition of CaCO3 over the concentration of CaCO3

The rate of reaction for the thermal decomposition of CaCO3 can also be written as:
Rate of reaction = k x [CaCO3]


The proportionality constant k is the gradient of the graph and is also called the rate
constant
The rate equation is the overall expression for a particular reaction without the ‘x’ sign
Rate of reaction = k [CaCO3]

Rate equations can only be determined experimentally and cannot be found from
the stoichiometric equation
Rate of reaction = k [A]m [B]n
[A] and [B] = concentrations of reactants
m and n = orders of the reaction

For example, the rate equation for the formation of nitrogen gas (N2) from nitrogen
oxide (NO) and hydrogen (H2) is rate = k [NO]2 [H2]
118 | P a g e
2NO (g) + 2H2 (g) → N2 (g) + 2H2O (g)
rate = k [NO]2 [H2]

As mentioned before, the rate equation of the reaction above cannot be deduced from the
stoichiometric equation but can only experimentally be determined by:
o Changing the concentration of NO and determining how it affects the rate while
keeping [H2] constant
o This shows that the rate is proportional to the square of [NO]
Rate = k1 [NO]2


Then, changing the [H2] and determining how it affects the rate while keeping [NO]
constant
This shows that the rate is proportional to [H2]
Rate = k2 [H2]

Combining the two equations gives the overall rate equation (where k = k1 + k2)
Rate = k [NO]2 [H2]
Order of reaction



The order of reaction shows how the concentration of a reactant affects the rate of
reaction
o It is the power to which the concentration of that reactant is raised in the rate
equation
o The order of reaction can be 0, 1,2 or 3
o When the order of reaction of a reactant is 0, its concentration is ignored
The overall order of reaction is the sum of the powers of the reactants in a rate equation
For example, in the following rate equation, the reaction is:
Rate = k [NO2]2[H2]

o
o
o
119 | P a g e
Second-order with respect to NO
First-order with respect to H2
Third-order overall (2 + 1)
Half-life

The half-life (t1/2) is the time taken for the concentration of a limiting reactant to become
half of its initial value
Rate-determining step & intermediates



The rate-determining step is the slowest step in a reaction
If a reactant appears in the rate-determining step, then the concentration of that reactant
will also appear in the rate equation
For example, the rate equation for the reaction below is rate = k [CH3Br] [OH-]
CH3Br + OH- → CH3OH + Br
o


This suggests that both CH3Br and OH- take part in the slow rate-determining
step
This reaction is, therefore, a bimolecular reaction
o Unimolecular: one species involved in the rate-determining step
o Bimolecular: two species involved in the rate-determining step
The intermediate is derived from substances that react together to form it in the ratedetermining step
o For example, for the reaction above the intermediate would consist of CH3Br and
OH-
The intermediate is formed from the species that are involved in the rate-determining step (and
thus appear in the rate equation)
120 | P a g e
5.6.2 Kinetics Calculations
Kinetics: Calculations
Order of reaction

The order of reaction shows how the concentration of a reactant affects the ra of
reaction
Rate = k [A]m [B]n




When m or n is zero = the concentration of the reactants does not affect the rate
When the order of reaction (m or n) of a reactant is 0, its concentration is ignored
The overall order of reaction is the sum of the powers of the reactants in a rate equation
For example, in the reaction below, the overall order of reaction is 2 (1 + 1)
Rate = k [NO2] [Cl2]
Order of reaction from concentration-time graphs

In a zero-order the concentration of the reactant is inversely proportional to time
o This means that the concentration of the reactant decreases with increasing
time
o The graph is a straight line going down
Concentration-time graphs of a zero-order reaction
121 | P a g e

In a first-order reaction the concentration of the reactant decreases with time
o The graph is a curve going downwards and eventually plateaus
Concentration-time graphs of a first-order reaction

In a second-order reaction the concentration of the reactant decreases more steeply with
time
o The concentration of reactant decreases more with increasing time compared to in
a first-order reaction
o The graph is a steeper curve going downwards
Concentration-time graphs of a second-order reaction
122 | P a g e
Order of reaction from initial rate



The progress of the reaction can be followed by measuring the initial rates of the
reaction using various initial concentrations of each reactant
These rates can then be plotted against time in a rate-time graph
In a zero-order reaction the rate doesn’t depend on the concentration of the reactant
o The rate of the reaction therefore remains constant throughout the reaction
o The graph is a horizontal line
o The rate equation is rate = k
Rate-time graph of a zero-order reaction

In a first-order reaction the rate is directly proportional to the concentration of a reactant
o The rate of the reaction decreases as the concentration of the reactant decreases
when it gets used up during the reaction
o The graph is a straight line
o The rate equation is rate = k [A]
123 | P a g e
Rate-time graph of a first-order reaction

In a second-order reaction, the rate is directly proportional to the square of concentration
of a reactant
o The rate of the reaction decreases more as the concentration of the reactant
decreases when it gets used up during the reaction
o The graph is a curved line
o The rate equation is rate = k [A]2
Rate-time graphs of a second-order reaction
Order of reaction from half-life




The order of a reaction can also be deduced from its half-life (t1/2 )
For a zero-order reaction the successive half-lives decrease with time
o This means that it would take less time for the concentration of reactant to halve
as the reaction progresses
The half-life of a first-order reaction remains constant throughout the reaction
o The amount of time required for the concentration of reactants to halve will be the
same during the entire reaction
For a second-order reaction, the half-life increases with time
o This means that as the reaction is taking place, it takes more time for the
concentration of reactants to halve
124 | P a g e
Half-lives of zero, first and second-order reactions
Calculating the initial rate


The initial rate can be calculated by using the initial concentrations of the reactants in
the rate equation
For example, in the reaction of bromomethane (CH3Br) with hydroxide (OH-) ions to
form methanol (CH3OH) the reaction equation and rate are as follows:
CH3Br + OH- → CH3Br + Br- (aq)
Rate = k [CH3Br] [OH-]
Where k = 1.75 x 10-2 dm-2 mol-1 s-1

If the initial concentrations of CH3Br and OH- are 0.0200 and 0.0100 mol dm3
respectively, the initial rate of reaction is:
Rate = k [CH3Br] [OH-]
Initial rate = 1.75 x 10-2 x (0.0200) x (0.0100)
Initial rate = 3.50 x 10-6 mol dm-3 s-1
125 | P a g e
5.6.3 Half-Life
First Order Reaction Half-life


The half-life of a first-order reaction is independent of the concentration of reactants
o This means that despite the concentrations of the reactants decreasing during
the reaction
o The amount of time taken for the concentrations of the reactants to halve will
remain the same throughout the reaction
o The graph is a straight line going downwards
The rearrangement of the methyl group (CH3) in ethanenitrile (CH3CN) is an example
of a first-order reaction with rate equation rate = k [CH3CN]
CH3CN (g) → CH3NC (g)
Rearrangement of the CH3 group in CH3CN


Experimental data of the changes in concentration over time suggests that the half-life is
constant
o Even if the half-lives are slightly different from each other, they can still be
considered to remain constant
This means that no matter what the original concentration of the CH3CN is, the half-life
will always be around 10.0 minutes
Half-life table
126 | P a g e
In a first-order reaction, the time taken for the concentration to halve remains constant
Worked Example: Using the half-life of first-order reactions in
calculations
127 | P a g e
Answer

Step 1: Plot the concentration-time graph using appropriate scales and labels for the axis

Step 2: Find the first and second half-lives by determining when the concentrations halve
using the graph
Step 2 table

Step 3: Determine the reaction order
It is s a first-order reaction as the successive half-lives remain reasonably constant
(around 450 seconds) throughout the reaction
128 | P a g e
5.6.4 Rate Constant Calculations
Calculating the Rate Constant

The rate constant (k) of a reaction can be calculated using:
o The initial rates and the rate equation
o The half-life
Calculating the rate constant from the initial rate


The reaction of calcium carbonate (CaCO3) with chloride (Cl-) ions to form calcium
chloride (CaCl2) will be used as an example to calculate the rate constant from the initial
rate and initial concentrations
The reaction and rate equation are as follows:
CaCO3 (s) + 2Cl- (aq) + 2H+ (aq) → CaCl2 (aq) + CO2 (g) + H2O (l)
Rate = k [CaCO3] [Cl-]

The progress of the reaction can be followed by measuring the initial rates of the
reaction using various initial concentrations of each reactant
Experimental results of concentrations & initial rates table

To find the rate constant (k):
o Rearrange the rate equation to find k
129 | P a g e

Substitute the values of one of the experiments to find k (for example measurement 1)
= 1.40 x 10-2 dm3 mol-1 s-1

The values of measurement 2 or 3 could also have been used to find k (they all give the
same result of 1.40 x 10-2 dm3 mol-1 s-1)
Calculating the rate constant from the half-life



The rate constant (k) can also be calculated from the half-life of a reaction
You are only expected to deduce k from the half-life of a first-order reaction as the
calculations for second and zero-order reactions are more complicated
For a first-order reaction, the half-life is related to the rate constant by the following
expression:

Rearranging the equation to find k gives:

So, for a first-order reaction such as
the methyl (CH3) rearrangement in ethanenitrile (CH3CN) with a half-life of 10.0
minutes the rate constant is:
= 1.16 x 10-3 dm3 mol-1 s-1
130 | P a g e
5.6.5 Multi-Step Reactions
Kinetics of Multi-Step Reactions


The reaction mechanism of a reaction describes how many steps are involved in
the making and breaking of bonds during a chemical reaction
It is the slowest step in a reaction and includes the reactants that have an impact on the
reaction rate when their concentrations are changed
o Therefore, ll reactants that therefore appear in the rate equation will also appear
in the rate-determining step
o This means that reactants that have a zero-order and intermediates will not be
present in the rate-determining step
Predicting the reaction mechanism



The overall reaction equation and rate equation can be used to predict a possible
reaction mechanism of a reaction
For example, nitrogen dioxide (NO2) and carbon monoxide (CO) react to form nitrogen
monoxide (NO) and carbon dioxide (CO2)
The overall reaction equation is:
NO2 (g) + CO (g) → NO (g) + CO2 (g)

The rate equation is:
Rate = k [NO2]2


From the rate equation it can be concluded that the reaction is zero order with respect to
CO (g) and second order with respect to NO2 (g)
This means that there are two molecules of NO2 (g) involved in the rate-determining
step
A possible reaction mechanism could therefore be:

Step 1:

2NO2 (g) → NO (g) + NO3 (g)

slow (rate-determining step)
Step 2:
NO3 (g) + CO (g) → NO2 (g) + CO2 (g)
131 | P a g e
fast

Overall:
2NO2 (g) + NO3 (g) + CO (g) → NO (g) + NO3 (g) + NO2 (g) + CO2 (g)
=
NO2 (g) + CO (g) → NO (g) + CO2 (g)
Predicting the reaction order & deducing the rate equation


The order of a reactant and thus the rate equation can be deduced from a reaction
mechanism given that the rate-determining step is known
For example, the reaction of nitrogen oxide (NO) with hydrogen (H2) to form nitrogen
(N2) and water
2NO (g) + 2H2 (g) → N2 (g) + 2H2O (l)

The reaction mechanism for this reaction is:

Step 1:
NO (g) + NO (g) → N2O2 (g)

fast
Step 2:
N2O2 (g) + H2 (g) → H2O (l) + N2O (g)

Step 3:
N2O (g) + H2 (g) → N2 (g) + H2O (l)




slow (rate-determining step)
fast
The second step in this reaction mechanism is the rate-determining step
The rate-determining step consists of:
o N2O2 which is formed from the reaction of two NO molecules
o One H2 molecule
The reaction is, therefore, second order with respect to NO and first order with respect
to H2
So, the rate equation becomes:
Rate = k [NO]2 [H2]

The reaction is, therefore, third order overall
132 | P a g e
Identifying the rate-determining step



The rate-determining step can be identified from a rate equation given that the reaction
mechanism is known
For example, propane (CH3CH2CH3) undergoes bromination under alkaline solutions
The overall reaction is:
CH3CH2CH3 + Br2 + OH- → CH3CH2CH2Br + H2O + Br-

The reaction mechanism is:
Reaction mechanism of the bromination of propane under alkaline conditions

The rate equation is:
Rate = k [CH3CH2CH3] [OH-]


From the rate equation, it can be deduced that only CH3COCH3 and OH- are involved in
the rate-determining step and not bromine (Br2)
Since only in step 1 of the reaction mechanism are CH3COCH3 and OH- involved,
the rate-determining step is step 1 is the case for step 1 of the reaction mechanism
Identifying intermediates & catalyst



When a rate equation includes a species that is not part of the chemical reaction equation
then this species is a catalyst
For example, the halogenation of butanone under acidic conditions
The reaction mechanism is:
133 | P a g e

The reaction mechanism is:
Reaction mechanism of the halogenation of butanone under acidic conditions

The rate equation is:
Rate = k [CH3CH2COCH3] [H+]



The H+ is not present in the chemical reaction equation but does appear in the rate
equation
o H+ must therefore be a catalyst
Furthermore, the rate equation suggest that CH3CH2COCH3 and H+ must be involved in
the rate-determining (slowest) step
The CH3CH2COCH3 and H+ appear in the rate-determining step in the form of
an intermediate (which is a combination of the two species)
Intermediate is formed in the rate-determining step from the reaction of CH3CH2COCH3 and
H+
134 | P a g e
5.6.6 Factors affecting Rate Constant
Effect of Temperature on the Rate Constant



At higher temperatures, a greater proportion of molecules have energy greater than
than the activation energy
Since the rate constant and rate of reaction is directly proportional to the fraction
of molecules with energy equal or greater than the activation energy, then at higher
temperatures:
o The rate constant increases
o The rate of reaction increases
The relationship between the rate constant and the temperature is given by the
following equation:
ln k = natural logarithm of the rate constant
A = constant related to the collision frequency and orientation of the molecules
Ea = activation energy (joules, J)
R = gas constant (8.31 J K-1 mol-1)
T = temperature (kelvin, K)


A varies only a little bit with temperature, it can be considered a constant
Ea and R are also constants

The equation shows that an increase in temperature (higher value of T) gives
a greater value of ln k (and therefore a higher value of k)
135 | P a g e

Since the rate of the reaction depends on the rate constant (k) an increase in k also
means an increased rate of reaction
The graph of ln k over 1/T is a straight line with gradient -Ea/R
Exam Tip
You are not required to learn this equation however it is helpful in understanding the effects
of temperature on the rate constant.
136 | P a g e
5.6.7 Homogeneous & Heterogeneous
Catalysts
Homogeneous & Heterogeneous Catalysis




Catalysts increase the rate of reaction by providing an alternative pathway which has a
lower activation energy
Catalysts can be either homogeneous or heterogeneous
Homogeneous catalysts are those that are in the same phase as the reaction mixture
For example, in the esterification of ethanoic acid (CH3COOH) with ethanol
(CH3CH2OH) to form ethyl ethanoate (CH3COOCH2CH3) under acidic conditions

The H+ is a homogeneous catalyst as like the reactants and product it is in
the aqueous phase

Heterogeneous catalysts are those that are in a different phase to the rest of the reaction
mixture
For example, in the Born-Haber process to form ammonia (NH3) from nitrogen (N2) and
hydrogen (H2) an iron (Fe) catalyst is used


The Fe catalyst is a heterogeneous catalyst as it is in the solid phase whereas the reactants
and products are all in the gas phase
Heterogeneous Catalysis




In heterogeneous catalysis, the molecules react at the surface of a solid catalyst
The mode of action of a heterogeneous catalyst consists of the following steps:
Adsorption (or chemisorption) of the reactants on the catalyst surface
o The reactants diffuse to the surface of the catalyst
o The reactant is physically adsorbed onto the surface by weak forces
o The reactant is chemically adsorbed onto the surface by stronger bonds
o Chemisorption causes bond weakening between the atoms of the reactants
Desorption of the products
137 | P a g e
o

The bonds between the products and catalyst weaken so much that the products
break away from the surface
For example, the adsorption of hydrogen molecules onto a palladium (Pd) surface
The reactants are adsorbed on the catalyst surface causing bond weakening and eventually
desorption of the products
Iron in the Haber process




In the Haber process ammonia (NH3) is produced from nitrogen (N2) and hydrogen (H2)
An iron catalyst is used which speeds up the reaction by bringing the reactants close
together on the metal surface
This increases their likelihood to react with each other
The mode of action of the iron catalyst is as follows:
o Diffusion of the nitrogen and hydrogen gas to the iron surface
o Adsorption of the reactant molecules onto the iron surface by forming bonds
between the iron and reactant atoms
 These bonds are so strong that they weaken the covalent bonds between
the nitrogen atoms in N2 and hydrogen atoms in H2
 But they are weak enough to break when the catalysis has been completed
o The reaction takes place between the adsorbed nitrogen and hydrogen atoms
which react with each other on the iron surface to form NH3
o Desorption occurs when the bonds between the NH3 and iron surface are
weakened and eventually broken
o The formed NH3 diffuses away from the iron surface
138 | P a g e
Iron brings the nitrogen and hydrogen closer together so that they can react and hence
increases the rate of reaction
139 | P a g e
Heterogeneous catalyst in catalytic converters




Heterogeneous catalysts are also used in the catalytic removal of oxides of nitrogen
from the exhaust gases of car engines
The catalysts speed up the conversion of:
o Nitrogen oxides (NOy) into harmless nitrogen gas (N2)
o Carbon monoxide (CO) into carbon dioxide (CO2)
The catalytic converter has a honeycomb structure containing small beads coated
with platinum, palladium, or rhodium metals which act as heterogeneous catalysts
The mode of action of the catalysts is as following:
o Adsorption of the nitrogen oxides and CO onto the catalyst surface
o The weakening of the covalent bonds within nitrogen oxides and CO
o Formation of new bonds between:
 Adjacent nitrogen atoms to form N2 molecules
 CO and oxygen atoms to form CO2 molecules
o Desorption of N2 and CO2 molecules which eventually diffuse away from the
metal surface
The metals in catalytic converters speed up the conversion of nitrogen oxides and CO into
N2 and CO2 respectively
Homogeneous Catalysis


Homogeneous catalysis often involves redox reactions in which the ions involved in
catalysis undergo changes in their oxidation number
o As ions of transition metals can change oxidation number they are often good
catalysts
Homogeneous catalysts are used in one step and are reformed in a later step
The iodine-peroxydisulfate reaction

This is a very slow reaction in which the peroxydisulfate (S2,O82- )
ions oxidise the iodide to iodine
S2O82- (aq) + 2I- (aq) → 2SO42- (aq) + I2 (aq)
140 | P a g e



Since both the S2O82- and I- ions have a negative charge, it will require a lot of energy for
the ions to overcome the repulsive forces and collide with each other
Therefore, Fe3+ (aq) ions are used as a homogeneous catalyst
The catalysis involves two redox reactions:
o First, Fe3+ ions are reduced to Fe2+ by I2Fe3+ (aq) + 2I- (aq) → 2Fe2+ (aq) + I2 (aq)

o
Then, Fe2+ is oxidized back to Fe3+ by S2O822Fe2+ (aq) + S2O82- (aq) → 2Fe3+ (aq) + 2SO42- (aq)


By reacting the reactants with a positively charged Fe ion, there are no repulsive forces,
and the activation energy is significantly lowered
The order of the two reactions does not matter
o So, Fe2+ can be first oxidised to Fe3+ followed by the reduction of Fe3+ to Fe2+
The catalysed reaction has two energy ‘humps’ because it is a two-stage reaction
Nitrogen oxides & acid rain

As fossil fuels contain sulfur, burning the fuels will release sulfur dioxide which oxidises
in air to sulfur trioxide, and then dilute sulfuric acid (H2SO4) is formed by reaction with
water. The result is acidification of rain:
SO3(g) + H2O(l) → H2SO4(aq)
141 | P a g e

Nitrogen oxides can act as catalysts in the formation of acid rain by catalysing the
oxidation of SO2 to SO3
NO2(g) + SO2(g) → SO3(g) + NO(g)

The formed NO gets oxidised to regenerate NO2
NO(g) + ½ O2(g) → NO2(g)

The regenerated NO2 molecule can again oxidise another SO2 molecule to SO3 which will
react with rainwater to form H2SO4 and so on
142 | P a g e
INORGANIC CHEMISTRY
6.1 GROUP 2
6.1.1 Similarities, Trends & Compounds of
Magnesium to Barium
Effect of Ionic Radius on Thermal Stability of Group 2 Nitrates &
Carbonates


The Group 2 nitrates and carbonates become more thermally stable going down the
group
The charge density of the cation (Group 2 metal ion) and
the polarisation of the anion (the nitrate and carbonate ion) attribute towards this
increased stability
Trends in thermal stability going down the group







All Group 2 metals form 2+ ions as they lose two electrons from their valence shells
The metal cations at the top of the group are smaller in size than those at the bottom
o For example, the atomic radius of beryllium (the first element in Group 2) is 112
pm whereas the atomic radius of calcium (further down the group) is 197 pm
The metal cations at the top of Group 2, therefore, have the greatest charge density as
the same charge (2+) is packed into a smaller volume
As a result, smaller Group 2 ions have a greater polarising effect on neighbouring
negative ions
When a carbonate or nitrate ion approaches the cation, it becomes polarised
o This is because the metal cation draws the electrons in the carbonate or nitrate ion
towards itself
The more polarised the anion is, the less heat is required to thermally decompose them
Therefore, the thermal stability increases down the group
o As down the group, the cation becomes larger
o Thus has a smaller charge density
o And a smaller polarising effect on the carbonate or nitrate anion
o So the anion is less polarised
o Therefore, more heat is required to thermally decompose them
Trends in Solubility & Enthalpy Change of Solution of Group 2
Hydroxides & Sulfates

The solubility of Group 2 hydroxides increases down the group
143 | P a g e



In contrast, the Group 2 sulfates show a decrease in solubility going down the group
Compounds that have very low solubility are said to be sparingly soluble
o For example, calcium sulfate (CaSO4) has low solubility as only 0.21 g of
CaSO4 dissolves in 100 g of water
Most of the sulfates are soluble in warm water with the exception of barium
sulfate which is insoluble
Solubility of Group 2 elements table
Enthalpy change of hydration and lattice energy


The standard enthalpy of solution (ΔHsolꝋ) is the energy absorbed or released when 1
mole of ionic solid dissolves in enough water to form a dilute solution (under standard
conditions)
o The ΔHsolꝋ can be either exothermic or endothermic
For example, the ΔHsolꝋ of sodium chloride (NaCl) is +3.9 kJ mol-1
NaCl (s) + aq → NaCl (aq) OR
NaCl (s) + aq → Na+ (aq) + Cl- (aq)


This means, that 3.9 kJ mol-1 of energy is absorbed when one mole of NaCl is dissolved
in enough water to form a dilute solution
The ΔHsolꝋ is the sum of the lattice energy (ΔHlattꝋ) and the standard enthalpy change of
hydration (ΔHhydꝋ)
ΔHsolꝋ = ΔHlattꝋ + ΔHhydꝋ

The lattice (formation) energy is the energy released when gaseous ions combine to
form one mole of an ionic compound under (standard conditions)
144 | P a g e
o
o
Since energy is released when an ionic compound is formed, the ΔHlattꝋ is
always exothermic
For example, the ΔHlattꝋ of NaCl is -787 kJ mol-1
Na+ (g) + Cl- (g) → NaCl (s)


This means, that 787 kJ mol-1 of energy is released when NaCl is formed from its gaseous
ions
The standard enthalpy of hydration is the energy released when gaseous ions dissolve
in enough water to form a dilute solution (under standard conditions)
o Since energy is released when gaseous ions become hydrated, the ΔHhydꝋ is
always exothermic
o For example, the ΔHhydꝋ of the sodium (Na+) ion is -406 kJ mol-1
Na+ (g) → Na+ (aq)

This means, that 406 kJ mol-1 of energy is released when Na+ ions become hydrated
Trends of enthalpy change of solution





Going down the group, the ΔHlattꝋ of the ionic compounds decreases
o Going down the group, the positively charged cations become larger
o There is more space between the negatively and positively charged ions in the
ionic compound so there are weaker attractive forces between the ions
o As there are weaker electrostatic forces between the ions, there is less energy
released upon formation of the ionic compound from its gaseous ions
o Therefore, the ΔHlattꝋ becomes less exothermic
Going down the group, the ΔHhydꝋ also decreases
o Again, the positively charged ions become larger going down the group
o As a result, the ion-dipole bonds between the cations and water molecules
get weaker
o This means that less energy is released when the gaseous Group 2 ions become
hydrated
o The ΔHhydꝋ , therefore, becomes less exothermic
For Group 2 hydroxides:
o Hydroxide ions are relatively small ions
o The ΔHlattꝋ falls faster than the ΔHhydꝋ
o The enthalpy change of solution is, therefore, more exothermic going down the
group
For Group 2 sulfates:
o Sulfate ions are relatively large ions
o The ΔHlattꝋ falls slower than the ΔHhydꝋ enthalpy
o The ΔHsolꝋ will become more endothermic going down the group
The more exothermic the ΔHsolꝋ the more soluble the compound
o This is why the sulfates become less soluble going down the group and
the hydroxides more soluble
145 | P a g e
TRANSITION ELEMENTS
6.2.1 General Properties of the Transition Elements: Titanium
to Copper
Define Transition Element





Transition elements are d-block elements which form one or more stable ions with
an incomplete d subshell
They are all metals found in the d-block of the Periodic Table, between Groups 2 and 13
o Sometimes they are referred to as transition metals
Not all d-block elements are classed as transition elements: scandium and zinc, for
example, are not classed as transition elements, despite being in the d-block
Scandium is not classed as a transition element because:
o It only forms one ion, Sc3+
o The Sc3+ion has no electrons in its 3d subshell; it has the electronic configuration
of [Ar]
Zinc is also not classed as a transition element because:
o It also forms only one ion, Zn2+
o The Zn2+ ion has a complete 3d subshell; it has the electronic configuration
[Ar]3d10
The transition elements on the periodic table
Shape of 3d_xy & 3d_z2 Orbitals
146 | P a g e




The transition elements all have incomplete d subshells
There are five orbitals in a d subshell. Some of these orbitals may have
similar shapes but different orientations, whereas others may have completely
different shapes
The five orbitals are
o 3dyz
o 3dxz
o 3dxy
o 3dx2 - y2
o 3dz2
Note that students are required to sketch the shapes of the 3dxy and 3dz2 orbitals only
Shapes of the 3d orbitals

The 3dyz, 3dxz, and 3dxy orbitals are orbitals which lie in the y-z, x-z and x-y plane
respectively
o They all have four lobes that point between the two axes
The 3dyz, 3dxz, and 3dxy orbitals all have four lobes which are similar in shape but point between
different axes


The 3dx2 - y2 orbital looks like the 3dyz, 3dxz, and 3dxy orbitals, as it also consists of four lobes
The difference is that these lobes point along the x and y axes and not between them
147 | P a g e
The four lobes in a 3dx2-y2 orbital point along the axes


The 3dz2 orbital is different from the other orbitals, as there are two main lobes which
form a dumbbell shape
The two main lobes point along the z-axis and there is a “doughnut” ring around the
centre
The 3dz2 orbital has a dumbbell shape with a ring around the centre
Properties of the Transition Elements

Although the transition elements are metals, they have some properties unlike those of
other metals on the periodic table, such as:
o Variable oxidation states
o Behave as catalysts
o Form complex ions
o Form coloured compounds
Ions of transition metals




Like other metals on the periodic table, the transition elements will lose electrons to form
positively charged ions
However, unlike other metals, transition elements can form more than one positive ion
o They are said to have variable oxidation states
Because of this, Roman numerals are used to indicate the oxidation state on the metal ion
o For example, the metal sodium (Na) will only form Na+ ions (no Roman numerals
are needed, as the ion formed by Na will always have an oxidation state of +1)
o The transition metal iron (Fe) can form Fe2+ (Fe(II)) and Fe3+ (Fe(III)) ions
The table below shows the most common oxidation states of a few transition metals
148 | P a g e
Oxidation states of transition elements table
Coloured complex

Another characteristic property of transition elements is that their compounds are
often coloured
o For example, the colour of the [Cr(OH)6]3- complex (where oxidation state of Cr is
+3) is dark green
o Whereas the colour of the [Cr(NH3)6]3+ complex (oxidation state of Cr is still +3)
is purple
Examples of some transition metal ions and their coloured complexes
149 | P a g e
Transition elements as catalysts



Since transition elements can have variable oxidation states, they make
excellent catalysts
During catalysis, the transition element can change to various oxidation states by gaining
electrons or donating electrons from reagents within the reaction
o For example, iron (Fe) is commonly used as a catalyst in the Haber Process,
switching between the +2 and +3 oxidation states
Substances can also be adsorbed onto their surface and activated in the process
Complex ions




Another property of transition elements caused by their ability to form variable oxidation
states, is their ability to form complex ions
A complex ion is a molecule or ion, consisting of a central metal atom or ion, with a
number of molecules or ions surrounding it
The molecules or ions surrounding the central metal atom or ion are called ligands
Due to the different oxidation states of the central metal ions, a different number and
wide variety of ligands can form bonds with the transition element
o For example, the chromium(III) ion can form [Cr(NH3)6]3+, [Cr(OH)6]3- and
[Cr(H2O)6]3+ complex ions
150 | P a g e
6.2.2 Oxidation States of Transition Metals
Effects of the 3d & 4s Subshells on Oxidation States of the Transition
Elements





Transition elements can have variable oxidation states
These variable oxidation states can be formed as the 3d and 4s atomic orbitals are similar
in energy
This means that a similar amount of energy is needed to remove a different number of
electrons
When the transition elements form ions, the electrons of the 4s subshell are lost first,
followed by the 3d electrons
The most common oxidation state is +2, which is usually formed when the two 4s
electrons are lost
Oxidation number at the start of the 3d transition elements





At the start of the period, it is easier for the transition elements to lose
the maximum number of electrons
The maximum oxidation number of these transition elements involves all the 4s and 3d
electrons in the atom
For example, the maximum oxidation state of a titanium (Ti) ion is +3 or +4, as two 4s
electrons and either 1 or 2 3d electrons are lost
Ti atom = 1s2 2s2 2p6 3s2 3p6 3d2 4s2
Ti3+ ion = 1s2 2s2 2p6 3s2 3p63d1
o Ti4+ ion = 1s2 2s2 2p6 3s2 3p6
Oxidation number at the end of the 3d transition elements




Towards the end, the 3d transition elements are more likely to adopt the +2 oxidation
state
This is because across the d block, the 3d electrons become slightly harder to remove as
the nuclear charge increases
o The 3d electrons are attracted more strongly to the nucleus
o The higher oxidation states become less stable
Therefore, the elements are more likely to lose their 4s electrons only
For example, nickel (Ni) is a transition element at the end of the period which only forms
ions with oxidation state +2, due to loss of the 4s electrons only
o Ni atom = 1s2 2s2 2p6 3s2 3p6 3d8 4s2
o Ni2+ ion = 1s2 2s2 2p6 3s2 3p63d8
151 | P a g e
Transition Elements: Catalysts

Transition elements are often used as catalysts due to their ability to form ions with more
than one stable oxidation state, and the fact that they contain vacant d orbitals
Oxidation states




Transition element ions can adopt more than one stable oxidation state
This means that they can accept and lose electrons easily to go from one oxidation state
to another
They can therefore catalyse redox reactions, by acting as both oxidising
agents and reducing agents
For example, iron (Fe) is often used as a catalyst due to its ability to form Fe(II) and
Fe(III) ions, acting as an oxidising agent and a reducing agent
o When Fe(II) acts as a reducing agent, it will reduce another species and become
oxidised itself
Fe2+ → Fe3+ + e-

The Fe3+ formed in the catalytic cycle, can then also act as an oxidising agent by oxidising
another species and getting reduced itself to reform the Fe2+ ion
Fe3+ + e- → Fe2+

Transition element ions with high oxidation states make powerful oxidising agents,
because they will readily accept electrons
o A common example of this is potassium permanganate (VII), where manganese
has an oxidation state of +7
Vacant d orbitals




When transition elements form ions, they have vacant d orbitals which
are energetically accessible
o The orbitals are not too high in energy
This means that dative bonds can be formed between the transition element ion
and ligands
o Each ligand provides the pair of electrons required for the formation of a bond
between the ion and the ligand
o This pair of electrons is donated into the ion’s vacant d orbital
The table below shows the electron configuration of the transition element atoms
When they form ions, empty d orbitals are obtained which can be filled by the pairs of
electrons donated by the ligands
Electronic configuration of transition elements table
152 | P a g e
Transition Metals: Complex Ions




A complex is a molecule or ion formed by a central metal atom or ion surrounded by one
or more ligands
o A complex can have an overall positive or negative charge, or it can be neutral
o If a complex is charged overall, it is often called a complex ion
Transition elements can easily form complex ions, because they have empty d
orbitals that are energetically accessible
o The empty d orbitals are therefore not too high in energy and can accommodate a
lone pair of electrons
The transition element in the centre will accept pairs of electrons from the ligands into
their empty d orbitals, forming dative bonds
o The transition element in the centre is often referred to as the central metal ion,
as all transition elements are metals, and it is often an ion in the centre
For example, the titanium(III) (Ti3+) ion, has an electronic configuration of
1s2 2s2 2p6 3s2 3p63d1
o This means that there are vacant d orbitals that can be occupied by electrons, from
ligands such as H2O for example, to form a [Ti(H2O)6]3+ complex ion
o 6 water ligands have each donated a pair of electrons, to form 6 dative bonds with
the central metal ion
153 | P a g e
6.2.3 Transition Metal Complexes
Transition Elements: Ligands & Complex Formation


Transition element ions can form complexes which consist of a central metal
ion and ligands
Copper(II) and cobalt(II) ions will be used as examples of the central metal ions, in the
complex formation with water (H2O), ammonia (NH3), hydroxide (OH-), and chloride (Cl) ligands
o A copper(II) ion has an electronic configuration of 1s22s22p63s23p63d9
o A cobalt(II) ion has an electronic configuration of 1s22s22p63s23p63d7
Complexes with water & ammonia molecules






Water and ammonia molecules are examples of neutral ligands
Both ligands contain a lone pair of electrons which can be used to form a dative covalent
bond with the central metal ion
o In water, this is the lone pair on the oxygen atom
o In ammonia, it is the lone pair on the nitrogen atom
Since water and ammonia are small ligands, 6 of them can fit around a central metal ion,
each donating a lone pair of electrons, forming 6 dative bonds
o The coordination number of a complex is the number of dative bonds formed
between the central metal ion and the ligands
o Since there are 6 dative bonds, the coordination number for the complex is 6
Complexes with a coordination number of 6 have an octahedral shape
The overall charge of a complex is the sum of the charge on the central metal ion, and the
charges on each of the ligands
A complex with cobalt(II) or copper(II) as a central metal ion, and water or ammonia
molecules as ligands, will have an overall charge of 2+
o The central metal ion has a 2+ charge and the ligands are neutral
154 | P a g e
Cobalt(II) and copper(II) form octahedral complexes with ammonia and water ligands
Complexes with hydroxide & chloride ions







Hydroxide and chloride ions are examples of negatively charged ligands
Both ligands contain a lone pair of electrons which can be used to form a dative covalent
bond with the central metal ion
Hydroxide ligands are small, so 6 of them can fit around a central metal ion and the
complex formed will have a coordination number of 6
Chloride ligands are large ligands, so only 4 of them will fit around a central metal ion
Complexes with 4 chloride ligands will have a coordination number of 4
o Complexes with 4 chloride ligands will form tetrahedral complexes
o Whereas hydroxide ligands will form octahedral complexes
A complex with cobalt(II) or copper(II) as a central metal ion and chloride ions as
ligands, will have an overall charge of 2o The central metal ion has a charge of 2+
o Each chloride ligand has a charge of 1o There are 4 chloride ligands in the complex, so the overall negative charge is 4o The overall positive charge is 2+
o Therefore, the overall charge of the complex is 2A complex with cobalt(II) or copper(II) as a central metal ion and hydroxide ions as
ligands, will have no overall charge
o The central metal ion has a charge of 2+
o Each hydroxide ligand has a charge of 1o There are 2 hydroxide ligands in the complex, so the overall negative charge is 2o The overall positive charge is 2+
o Therefore, the overall charge on the complex is 0
155 | P a g e
Cobalt(II) and copper(II) form tetrahedral complexes with chloride and octahedral complexes
with water and hydroxide ligands
6.2.4 Ligands
Define Ligand


A ligand is a molecule or ion that has one or more lone pairs of electrons
These lone pairs of electrons are donated by the ligand, to form dative covalent bonds to
a central metal atom or ion
Examples of ligands table
156 | P a g e
Example of a complex formed between a transition metal ion (Fe2+) and a ligand (H2O) by
dative covalent bonds
Types of Ligands

Different ligands can form different numbers of dative bonds to the central metal ion in a
complex.
o Some ligands can form one dative bond to the central metal ion
o Other ligands can form two dative bonds, and some can form multiple dative
bonds
Monodentate ligands


Monodentate ligands can form only one dative bond to the central metal ion
Examples of monodentate ligands are:
o Water (H2O) molecules
o Ammonia (NH3) molecules
o Chloride (Cl-) ions
o Cyanide (CN-) ions
Examples of complexes with monodentate ligands
157 | P a g e
Bidentate ligands



Bidentate ligands can each form two dative bonds to the central metal ion
This is because each ligand contains two atoms with lone pairs of electrons
Examples of bidentate ligands are:
o 1,2-iaminoethane (H2NCH2CH2NH2) which is also written as ‘en’
o Ethanedioate ion (C2O42- ) which is sometimes written as ‘ox’
Examples of complexes with bidentate ligands
Polydentate ligands



Some ligands contain more than two atoms with lone pairs of electrons
These ligands can form more than two dative bonds to the and are said to
be polydentate ligands
An example of a polydentate ligand is EDTA4- , which is a hexadentate ligand as it forms
6 dative covalent bonds to the central metal ion
Example of a polydentate ligand complex
158 | P a g e
6.2.5 Defining Complexes
Define Complex



A complex is a molecule or ion formed by a central metal atom or ion surrounded by
one or more ligands
A ligand is a species that contains one or more lone pairs of electrons
The ligand forms a dative covalent bond with the central metal atom or ion, by donating
its lone pair of electrons
Example of a complex ion formed by a central aluminium(III) ion and six water molecule
ligands
6.2.6 Geometry of Complexes
Geometry of the Transition Element Complexes

Depending on the size of the ligands and the number of dative bonds to the central
metal ion, transition element complexes have different geometries
o Dative bonds can also be referred to as coordinate bonds, especially when
discussing the geometry of a complex
Linear



Central metal atoms or ions with two coordinate bonds form linear complexes
The bond angles in these complexes are 180o
The most common examples are a copper (I) ion, (Cu+), or a silver (I) ion, (Ag+), as the
central metal ion with two coordinate bonds formed to two ammonia ligands
159 | P a g e
Example of a linear complex
Tetrahedral


When there are four coordinate bonds the complexes often have a tetrahedral shape
o Complexes with four chloride ions most commonly adopt this geometry
o Chloride ligands are large, so only four will fit around the central metal ion
The bond angles in tetrahedral complexes are 109.5o
Example of a tetrahedral complex
Square planar


Sometimes, complexes with four coordinate bonds may adopt a square
planar geometry instead of a tetrahedral one
o Cyanide ions (CN-) are the most common ligands to adopt this geometry
o An example of a square planar complex is cisplatin
The bond angles in a square planar complex are 90o
Cisplatin is an example of a square planar complex
160 | P a g e
Octahedral





Octahedral complexes are formed when a central metal atom or ion forms six
coordinate bonds
This could be six coordinate bonds with six small, monodentate ligands
o Examples of such ligands are water and ammonia molecules
and hydroxide and thiocyanate ions
It could be six coordinate bonds with three bidentate ligands
o Each bidentate ligand will form two coordinate bonds, meaning six coordinate
bonds in total
o Examples of these ligands are 1,2-diaminoethane and the ethanedioate ion
It could be six coordinate bonds with one polydentate ligand
o The polydentate ligand, for example EDTA4-, forms all six coordinate bonds
The bond angles in an octahedral complex are 90o
Examples of octahedral complexes
161 | P a g e
Types of ligands table
Coordination Number & Predicting Complex Ion Formula & Charge



The coordination number of a complex is the number of coordinate bonds that are
formed between the ligand(s) and the central metal atom or ion
Some ligands can form only one coordinate bond with the central metal ion
(monodentate ligands), whereas others can form two (bidentate ligands ) or more
(polydentate ligands)
It is not the number of ligands which determines the coordination number, it is the
number of coordinate (dative) bonds
Predicting complex ion formula & charge

The formula and charge of a complex ion can be predicted if the following are known:
o The central metal ion and its charge/oxidation state
o The ligands
o The coordination number/geometry
162 | P a g e
6.2.7 Degenerate & non-Degenerate d
Orbitals
Define Degenerate & non-Degenerate d Orbitals





There are five d orbitals in an isolated transition element atom or ion
o An isolated transition element is one that is not bonded to anything else
These d orbitals are all at the same energy level (they are equal in energy) and are
therefore said to be degenerate orbitals
When ligands are attached, the transition element ion is not isolated anymore
The dative bonding from the ligands causes the five d orbitals to split into two sets
These two sets are not equal in energy and are described as being non-degenerate
orbitals
Upon binding to ligands, the d orbitals of the transition element ion split into two nondegenerate sets of orbitals
Degenerate d Orbital Splitting


An isolated transition element has five degenerate 3d orbitals
Upon dative covalently bonding to a ligand, these d orbitals are split into two sets of nondegenerate orbitals
Splitting in octahedral complexes

In octahedral complexes, there are six ligands arranged around the central metal ion
163 | P a g e






The lone pairs of the ligands repel the electrons in the x2-y2 and z2 orbitals of the metal
ion more than they repel the electrons in the 3dyz, 3dxz, and 3dxy orbitals
This is because the 3dx2-y2 and 3dz2 orbitals line up with the dative bonds in the complex’s
octahedral shape
This is because the ligands are attached to or approaching the central metal ion along the
x, y and z axes, and the 3dx2-y2 and 3dz2 orbitals have lobes along these axes
The electrons in these two orbitals are closer to the bonding electrons, so there is
more repulsion
This means that when the d orbitals split, the 3dx2-y2 and 3dz2 orbitals are at a slightly
higher energy level than the other three
The difference in energy between the non-degenerate d orbitals is labelled as ΔE
Splitting of 3d orbitals in an octahedral complex
Splitting in tetrahedral complexes





In tetrahedral complexes, there are four ligands arranged around the central metal ion
The bonding pair of electrons from the four ligands now line up with the 3dyz, 3dxz, and
3dxy orbitals of the central metal ion
Now, the 3dx2-y2 and 3dz2 orbitals lie between the metal-ligand bonds
Therefore, there is less repulsion with the 3dx2-y2 and 3dz2 orbitals
When the d orbitals split this time, the 3dx2-y2 and 3dz2 orbitals are
at lower and more stable energy level than the other three
164 | P a g e
Splitting of 3d orbitals in a tetrahedral complex
6.2.8 Coloured Complexes
Coloured Compounds & Electron Promotion



Most transition element complexes are coloured
A transition element complex solution which is coloured, absorbs part of the
electromagnetic spectrum in the visible light region
The observed colour is the complementary colour which is made up of light with
frequencies that are not absorbed
o For example, copper(II) ions absorb light from the red end of the spectrum
o The complementary colour observed is therefore pale blue (cyan)
165 | P a g e
The visible light region of the electromagnetic spectrum
Electron promotion





In an isolated transition element ion (which is not bonded to any ligands), all of the 3d
orbitals are degenerate
However, when ligands are attached to the central metal ion through dative covalent
bonds, these orbitals are split into two sets of non-degenerate orbitals
The difference in energy between these two sets of orbitals is ΔE
When light shines on a solution containing a transition element complex, an electron will
absorb this exact amount of energy (ΔE)
The amount of energy absorbed can be worked out by the equation:
ΔE = h x v
h = Planck's constant (6.626 x 10-34 m2 kg s-1)
v = frequency (Hertz, Hz or s-1)



The electron uses the energy from the light to jump into a higher, non-degenerate energy
level
o This is also called electron promotion
The other frequencies of light which are not absorbed combine to make
the complementary colour
The diagram below shows an example of electron promotion in an octahedral complex of
a nickel(II) Ni2+ ion
166 | P a g e
Electron promotion in a Ni(II) complex when light shines on the solution
Effects of Ligands on Complementary Colour




Transition element complexes absorb the frequency of light which corresponds to the
exact energy difference (ΔE) between their non-degenerate d orbitals
The frequencies of light which are not absorbed combine to make the complementary
colour of the complex
It is the complementary colour which is seen
However, the exact energy difference (ΔE) is affected by the different ligands which
surround the transition element ion
167 | P a g e





Different ligands will split the d orbital by a different amount of energy
This depends on the repulsion that the d orbital experiences from these ligands
Therefore, the size of ΔE and thus the frequency of light absorbed by the electrons will
be slightly different
As a result, a different colour of light is absorbed by the complex solution and a
different complementary colour is observed
This means that complexes with similar transition
elements ions, but different ligands, can have different colours
o For example, the [Cu(H2O)6]2+ complex has a light blue colour
o Whereas the [Cu(NH3)4 (H2O)2]2+ has a dark blue colour
o Despite the copper ion having an oxidation state of +2 in both complexes
o This is evidence that the ligands surrounding the complex ion affect the colour of
the complex
Ligand Exchange in Copper(II) & Cobalt(II) Complexes


Different ligands may affect the complementary colour of a transition ion complex
solution
This is shown by ligand exchange reactions in copper(II) and cobalt(II) complexes, as
this causes a change in colour of the complexes
Copper(II) & cobalt(II) ions




The ligand exchange of [Cu(H2O)6]2+ and [Co(H2O)6]2+ by NH3 ligands causes a change in
the colour of the solutions
o [Cu(H2O)6]2+ is light blue in colour whereas [Cu(NH3)4(H2O)2)]2+ is deep blue in
colour
o [Co(H2O)6]2+ is a pink solution whereas [Co(NH3)6]2+ is a brown solution
The colour change results from the ammonia ligands, which cause the d orbitals to split
by a different amount of energy (ΔE)
Therefore, the size of ΔE and the frequency of light absorbed by the electrons will be
slightly different
As a result, a different colour of light is absorbed and thus a different complementary
colour is observed
Ligand exchange of the water ligands by ammonia ligands causes a change in colour of the
copper(II) complex solution
168 | P a g e
Ligand exchange of the water ligands by ammonia ligand causes a change in colour of the
cobalt(II) complex solution

Similarly, full ligand exchange by chloride ions in copper(II) and cobalt(II) complexes
results in a change in complementary colour
Ligand exchange by chloride ligands causes a change in colour of the copper(II) complex
solution
Ligand exchange by chloride ligands causes a change in colour of the cobalt(II) complex
solution

As before, this suggests that different ligands will split the d orbitals differently
169 | P a g e
6.3.1 Ligand Exchange
Ligand Exchange





Ligand exchange (or ligand substitution) is when one ligand in a complex is replaced
by another
Ligand exchange forms a new complex that is more stable than the original one
The ligands in the original complex can be partially or entirely substituted by others
There are no changes in coordination number, or the geometry of the complex, if the
ligands are of a similar size
But, if the ligands are of a different size, for example water ligands and chloride ligands,
then a change in coordination number and the geometry of the complex will occur
Substitution in copper(II) complexes








When a transition element ion is in solution, it can be assumed that it exists as
a hexaaqua complex ion (i.e. it has six water ligands attached to it)
o For example, Cu2+(aq) is [Cu(H2O)6]2+(aq)
The [Cu(H2O)6]2+ (aq) complex ion is blue in colour
Upon dropwise addition of sodium hydroxide (NaOH) solution, a light blue
precipitate is formed
Partial ligand substitution of two water ligands by two hydroxide ligands has occurred
Upon addition of excess concentrated ammonia (NH3) solution, the pale blue precipitate
dissolves to form a deep blue solution
Again, partial ligand substitution has occurred
If you were to add concentrated ammonia (NH3) solution dropwise to the
[Cu(H2O)6]2+ (aq), rather than sodium hydroxide (NaOH) solution, the same light blue
precipitate would form
Again, the pale blue precipitate will dissolve to form a deep blue solution,
if excess ammonia solution is then added
170 | P a g e
Water ligands are exchanged by hydroxide and ammonia ligands in the copper(II) complex






The water ligands in [Cu(H2O)6]2+ can also be substituted by chloride ligands, upon
addition of concentrated hydrochloric acid (HCl)
The complete substitution of the water ligands causes the blue solution to turn yellow
The coordination number has changed from 6 to 4, because the chloride ligands are
larger than the water ligands, so only 4 will fit around the central metal ion
The geometry of the complex has also changed from octahedral to tetrahedral
This is a reversible reaction, and some of the [Cu(H2O)6]2+ complex ion will still be
present in the solution
o The mixture of blue and yellow solutions in the reaction mixture will give it
a green colour
Adding water to the solution will cause the chloride ligands to be displaced by the
water molecules, and the [Cu(H2O)6]2+ (aq) ion and blue solution will return
171 | P a g e
Water ligands are exchanged by chloride ligands in the copper(II) complex
Substitution in cobalt(II) complexes






The [Co(H2O)6]2+(aq) complex ion is pink in colour
Upon dropwise addition of sodium hydroxide (NaOH) solution, a blue precipitate is
formed
Partial ligand substitution of two water ligands by two hydroxide (OH-) ligands has
occurred
o If the alkali is added in excess, the blue precipitate will turn red when warmed
If excess concentrated ammonia solution is added to [Co(H2O)6]2+, a brown solution will
also be formed
o There will be no precipitate formed in this instance, as the ammonia has been
added in excess and not dropwise
Complete ligand substitution of the water ligands by ammonia ligands has occurred
The ammonia ligands make the cobalt(II) ion so unstable that it readily gets oxidised in
air to cobalt(III), [Co(NH3)6]3+
172 | P a g e
Water ligands are exchanged by hydroxide and ammonia ligands in the cobalt(II) complex





The water ligands in [Co[H2O)6]2+ can also be substituted by chloride ligands, upon
addition of concentrated hydrochloric acid
The complete substitution of the water ligands causes the pink solution to turn blue
Like with [Cu(H2O)6]2+ above, the coordination number has changed from 6 to 4,
because the chloride ligands are larger than the water ligands, so only 4 will fit around
the central metal ion
The geometry of the complex has also changed from octahedral to tetrahedral
Adding water to the solution will cause the chloride ligands to be displaced by the
water molecules, and the [Co(H2O)6]2+ (aq) ion and pink solution will return
173 | P a g e
Water ligands are exchanged by chloride ligands in the cobalt(II) complex
6.3.2 Predicting Feasibility of Redox
Reactions
DOWNLOAD PDF
Feasibility of Redox Reactions Using Standard Electrode Values





Transition elements can form ions with various oxidation states
The change in their oxidation states involves the transfer of electrons
Transition elements are often involved in redox reactions
A redox reaction is a reaction in which one species is oxidised (loses electrons) and
another is reduced (gains electrons)
The standard electrode potentials (Eꝋ) of the two species can be used to predict
the feasibility of redox reactions involving transition elements and their ions
Predicting feasibility of redox reactions





The standard electrode potential (Eꝋ) of a species gives an indication of how well it can be
reduced
In the exam, you will be provided with a half equation and the standard electrode
potential (Eꝋ)
The half equations are always written as a reduction equation
o They are equilibrium reactions, as they demonstrate the equilibrium reached when
the species in the equation gains electrons at the same rate as it is losing electrons
The more positive the standard electrode potential (Eꝋ) of a species is, the more readily
that element will be reduced (gain electrons)
o This is always when compared to the standard hydrogen electrode
o The opposite is of course true; the more negative the standard electrode potential
(Eꝋ) of a species is, the more readily that element will be oxidised (lose electrons)
The feasibility of a reaction can be predicted using these values
174 | P a g e

For example, the feasibility of Fe3+ being reduced to Fe2+ when reacted with Cu2+, can be
predicted using their standard electrode potentials
Standard electrode potentials of Fe(III) & Cu(II) table






The table above shows that yes, the reaction is feasible and Fe3+ is more likely to
get reduced to Fe2+
o Fe3+ has a more positive standard electrode potential
o Fe3+ will gain electrons more readily than Cu2+
o Therefore, Fe3+ is the better oxidising agent
o The reaction for this half equation will therefore proceed in the forward direction
(reduction)
Since it is feasible that the Fe3+ will be reduced and this half equation will move in the
forward direction, this means that the half equation for copper will move in the backward
direction (oxidation)
o Cu2+ equation has a less positive (or more negative) standard electrode potential
o The Cu+ will therefore be oxidised to Cu2+
o The reaction for this half equation will therefore be in the reverse direction
Combining these two half-equations to get the overall equation gives (after cancelling the
electrons on both sides):
The positive value of Ecellꝋ (+0.62 V) suggests that the reaction is likely to proceed
The changes in the transition element ions’ oxidation states are therefore feasible
Standard electrode potentials (Eꝋ) are only predictions about the feasibility of a reaction;
they do not guarantee that a reaction will definitely occur
o For example, a reaction may be feasible according to these rules, but have a very
large activation energy barrier meaning that in reality it will not occur
175 | P a g e
6.3.3 Redox Systems
Redox Systems: Ferrous & Permanganate



The oxidation states of transition element ions can change during redox reactions
o A species will either be oxidised or reduced, depending on what reaction is
occurring
To find the concentration of specific ions in solution, a titration can be performed
There are three particular redox titrations that need to be learnt:
o Iron (II) (Fe2+) and permanganate (MnO4-) in acid solution given suitable data
o Permanganate (MnO4-) and ethanedioate (C2O42-) in acid solution given suitable
data
o Copper (II) (Cu2+) and iodide (I-) given suitable data
Reaction of MnO4- & Fe2+ in acid


The concentration of Fe2+ ions can be determined by titrating a known volume of Fe(II)
ions with a known concentration of MnO4- ions
During the reaction of MnO4- with Fe2+, the purple colour of the manganate(VII) ions
disappears
The end-point is when all of the Fe2+ ions have reacted with the MnO4- ions, and the first
trace of a permanent pink colour appears in the flask
o At this point, the MnO4- is very slightly in excess
The two half-reactions that are involved in this redox reaction are as following:

The half equations are combined to get the overall equation


176 | P a g e
Redox Systems: Permanganate & Oxalate

The second redox titration involving transition element ions that needs to be learned, is
the titration of permanganate (MnO4-) and ethanedioate, sometimes known as oxalate
(C2O42-) in acid solution given suitable data
Reaction of MnO4- & C2O42- in acid




The reaction of MnO4- with ethanedioate, C2O42- is an example of a redox reaction in
which the ethanedioate ions (C2O42-) get oxidised by manganate(VII) (MnO4-) ions
A titration reaction can be carried out to find the concentration of the toxic ethanedioate
ions
As before, the end point is when all of the ethanedioate ions have reacted with the
MnO4- ions, and the first permanent pink colour appears in the flask
o At this point, the MnO4- is very slightly in excess
The two half-reactions that are involved in this redox reaction are as following:
177 | P a g e

The half equations are combined to get the overall equation




This is an example of an autocatalysis reaction
This means that the reaction is catalysed by one of the products as it forms
In this reaction, the Mn2+ ions formed act as the autocatalyst
The more Mn2+ formed, the faster the reaction gets, which then forms even more
Mn2+ ions and speeds the reaction up even further
Transition element ions can act as autocalysts as they can change their oxidation states
during a reaction

178 | P a g e
Redox Systems: Cupric & Iodide

The third redox titration involving transition metal ions that needs to be learnt is the
titration between copper (II) ions (Cu2+) - sometimes known as cupric ions - and iodide
ions (I-)
Reaction of Cu2+ & I

The reaction of Cu2+ with I- is an example of a redox reaction in which the copper ions
(Cu2+) oxidise the iodide ions (I-) and as a result are themselves reduced
The two half-reactions that are involved in this redox reaction are as follows:

The half equations could be combined to get this overall equation
179 | P a g e

When excess iodide ions are reacted with Cu(II), a precipitate of copper(I) iodide
and iodine is formed
2Cu2+ (aq) + 4I-(aq) → I2(aq) + 2CuI (s)




A titration reaction can be carried out to find an unknown concentration of the copper(II)
solution
This is done by finding the amount of iodine which is liberated during the reaction,
through a titration
Step 1: A known concentration of sodium thiosulfate solution is added to the mixture
formed in Reaction 1 from a burette
Step 2: The I2 formed in Reaction 1 will react with the thiosulfate ions
I2 (aq) + 2S2O32- (aq) → 2I- (aq) + S4O62- (aq)







Reaction 1
Reaction 2
Step 3: As the iodine is used, up the brownish colour of the solution gets lighter
Step 4: When most of the iodine colour is gone, starch is added, which turns
deep blue/black with the remaining I2 (aq)
Step 5: Titrate further until the blue/black colour disappears, i.e. when all of the iodine
has reacted
By knowing the number of moles of thiosulfate ions added in the titration, you can use
the molar ratios from the reaction equations and work backwards to calculate the number
of moles of Cu(II)
Step 6: Look at Reaction 2: it can be concluded that half the number of moles of I2 reacts
when compared to the moles of thiosulfate that react
Step 7: Now look at Reaction 1: the number of moles of I2 which reacts in Reaction 2, is
the moles formed in Reaction 1. The number of moles of Cu(II) is twice that of I2 (aq)
(i.e. the same number of moles as thiosulfate ions added in the titration)
Step 8: Divide the number of moles of Cu(II) by the volume in dm3 to get the
concentration of Cu(II)
180 | P a g e
6.3.3 Redox Systems
Redox Systems: Ferrous & Permanganate



The oxidation states of transition element ions can change during redox reactions
o A species will either be oxidised or reduced, depending on what reaction is
occurring
To find the concentration of specific ions in solution, a titration can be performed
There are three particular redox titrations that need to be learnt:
o Iron (II) (Fe2+) and permanganate (MnO4-) in acid solution given suitable data
o Permanganate (MnO4-) and ethanedioate (C2O42-) in acid solution given suitable
data
o Copper (II) (Cu2+) and iodide (I-) given suitable data
Reaction of MnO4- & Fe2+ in acid




The concentration of Fe2+ ions can be determined by titrating a known volume of Fe(II)
ions with a known concentration of MnO4- ions
During the reaction of MnO4- with Fe2+, the purple colour of the manganate(VII) ions
disappears
The end-point is when all of the Fe2+ ions have reacted with the MnO4- ions, and the first
trace of a permanent pink colour appears in the flask
o At this point, the MnO4- is very slightly in excess
The two half-reactions that are involved in this redox reaction are as following:
181 | P a g e

The half equations are combined to get the overall equation
Redox Systems: Permanganate & Oxalate

The second redox titration involving transition element ions that needs to be learned, is
the titration of permanganate (MnO4-) and ethanedioate, sometimes known as oxalate
(C2O42-) in acid solution given suitable data
Reaction of MnO4- & C2O42- in acid




The reaction of MnO4- with ethanedioate, C2O42- is an example of a redox reaction in
which the ethanedioate ions (C2O42-) get oxidised by manganate(VII) (MnO4-) ions
A titration reaction can be carried out to find the concentration of the toxic ethanedioate
ions
As before, the end point is when all of the ethanedioate ions have reacted with the
MnO4- ions, and the first permanent pink colour appears in the flask
o At this point, the MnO4- is very slightly in excess
The two half-reactions that are involved in this redox reaction are as following:
182 | P a g e

The half equations are combined to get the overall equation




This is an example of an autocatalysis reaction
This means that the reaction is catalysed by one of the products as it forms
In this reaction, the Mn2+ ions formed act as the autocatalyst
The more Mn2+ formed, the faster the reaction gets, which then forms even more
Mn2+ ions and speeds the reaction up even further
Transition element ions can act as autocalysts as they can change their oxidation states
during a reaction

183 | P a g e
Redox Systems: Cupric & Iodide

The third redox titration involving transition metal ions that needs to be learnt is the
titration between copper (II) ions (Cu2+) - sometimes known as cupric ions - and iodide
ions (I-)
Reaction of Cu2+ & I

The reaction of Cu2+ with I- is an example of a redox reaction in which the copper ions
(Cu2+) oxidise the iodide ions (I-) and as a result are themselves reduced
The two half-reactions that are involved in this redox reaction are as follows:

The half equations could be combined to get this overall equation
184 | P a g e

When excess iodide ions are reacted with Cu(II), a precipitate of copper(I) iodide
and iodine is formed
2Cu2+ (aq) + 4I-(aq) → I2(aq) + 2CuI (s)




A titration reaction can be carried out to find an unknown concentration of the copper(II)
solution
This is done by finding the amount of iodine which is liberated during the reaction,
through a titration
Step 1: A known concentration of sodium thiosulfate solution is added to the mixture
formed in Reaction 1 from a burette
Step 2: The I2 formed in Reaction 1 will react with the thiosulfate ions
I2 (aq) + 2S2O32- (aq) → 2I- (aq) + S4O62- (aq)







Reaction 1
Reaction 2
Step 3: As the iodine is used, up the brownish colour of the solution gets lighter
Step 4: When most of the iodine colour is gone, starch is added, which turns
deep blue/black with the remaining I2 (aq)
Step 5: Titrate further until the blue/black colour disappears, i.e. when all of the iodine
has reacted
By knowing the number of moles of thiosulfate ions added in the titration, you can use
the molar ratios from the reaction equations and work backwards to calculate the number
of moles of Cu(II)
Step 6: Look at Reaction 2: it can be concluded that half the number of moles of I2 reacts
when compared to the moles of thiosulfate that react
Step 7: Now look at Reaction 1: the number of moles of I2 which reacts in Reaction 2, is
the moles formed in Reaction 1. The number of moles of Cu(II) is twice that of I2 (aq)
(i.e. the same number of moles as thiosulfate ions added in the titration)
Step 8: Divide the number of moles of Cu(II) by the volume in dm3 to get the
concentration of Cu(II)
6.3.4 Calculations of Other Redox Systems
Calculations of Other Redox Systems


You are required to perform calculations involving redox reactions of transition elements
These include:
o Constructing redox equations
o Calculating oxidation states
o Selecting suitable oxidising agents and reducing agents
o Calculating cell potentials
185 | P a g e
6.3.5 Stereoisomerism in Transition Element
Complexes
Geometrical & Optical Stereoisomerism in Complexes

Transition element complexes can exhibit stereoisomerism
Geometrical (cis-trans) isomerism



Even though transition element complexes do not have a double bond, they can still
have geometrical isomers
Square planar and octahedral complexes with two pairs of different ligands
exhibit cis-trans isomerism
An example of a square planar complex with two pairs of ligands is the anti-cancer
drug cis-platin
o Whereas cis-platin has beneficial medical effects by binding to DNA in cancer
cells, trans-platin cannot be used in cancer treatment
Cis-platin is an example of a square planar transition element complex that exhibits
geometrical isomerism



As long as a complex ion has two ligands attached to it that are different to the rest, then
the complex can display geometric isomerism
Examples of octahedral complexes that exhibit geometrical isomerism are the
[Co(NH3)4(H2O)2]2+ and [Ni(H2NCH2CH2NH2)2(H2O)2]2+ complexes
o [Ni(H2NCH2CH2NH2)2(H2O)2]2+ can also be written as [Ni(en)2(H2O)2]2+
Like in the square planar complexes, if the two ‘different’ ligands are next to each other
then that is the ‘cis’ isomer, and if the two ‘different’ ligands are opposite each other then
this is the ‘trans’ isomer
o In [Co(NH3)4(H2O)2]2+, the two water ligands are next door to each other in the cis
isomer and are opposite each other in the trans isomer
186 | P a g e
Octahedral transition metal complexes exhibiting geometrical isomerism
Optical isomerism




Octahedral complexes with bidentate ligands also have optical isomers
This means that the two forms are non-superimposable mirror images of each other
o They have no plane of symmetry, and one image cannot be placed directly on top
of the other
The optical isomers only differ in their ability to rotate the plane of polarised light in
opposite directions
Examples of octahedral complexes that have optical isomers are the
[Ni(H2NCH2CH2NH2)3]2+and [Ni(H2NCH2CH2NH2)2(H2O)2]2+ complexes
o The ligand H2NCH2CH2NH2 can also be written as ‘en’ instead
187 | P a g e
Octahedral transition metal complexes exhibiting optical isomerism
Polarity of Complexes


The isomers of transition elements complexes may be polar or non-polar
This is caused by differences in electronegativity of the atoms in the ligands that form
the dative bond to the complex ion
Polarity in square planar complexes







In cis-platin, the two chlorine atoms are on the same side
These atoms have a stronger pull on the electrons in the dative bond and will carry
a partial negative charge
As a result, there is an imbalance of charge causing the complex to become polar
In trans-platin, the same ligands are on opposite sides of each other
The pull on electrons in the dative bonds to the complex ion are therefore balanced
The overall charge is balanced and the complex is non-polar
Trans-platin does not have the same benefits medically as cis-platin does
188 | P a g e
Example of a square planar complex and its polar and non-polar geometric isomers
Polarity in octahedral complexes






Again, the trans-isomer in octahedral complexes is non-polar whereas the cis-isomer is
slightly polar
In cis-[Co(NH3)4(H2O)2]2+ for example, the oxygen atoms in the H2O ligands are more
electronegative than the nitrogen atoms in the NH3 ligands
This causes the side of the water ligands to be partially negative
Resulting in a charge imbalance causing the complex to become polar
The symmetrical arrangement in the trans isomers means that the charge is evenly
distributed in the complex
Trans-isomers are therefore non-polar
Example of an octahedral complex and its polar and non-polar geometric isomers
189 | P a g e
6.3.6 Stability Constants, Kstab
Define & Write a Stability Constant for a Complex




When transition element ions are in aqueous solutions, they will automatically become
hydrated
o Water molecules will surround the ion and act as ligands by forming dative
covalent bonds to the central metal ion
When there are other potential ligands present in solution, there is a competing
equilibrium in ligand exchange and the most stable complex will be formed
For example, a Co(II) ion in solution will form a [Co(H2O)6]2+ complex
Adding ammonia results in the stepwise substitution of the water ligands by ammonia
ligands until a stable complex of [Co(NH3)4(H2O)2]2+ is formed
[Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O

For the substitution reaction above, there are four stepwise constants:
[Co(H2O)6]2+ + NH3 ⇌ [Co(NH3)(H2O)5]2+ + H2O
K1
[Co(NH3)(H2O)5]2+ + NH3 ⇌ [Co(NH3)2(H2O)4]2+ + H2O
K2
[Co(NH3)2(H2O)4]2+ + NH3 ⇌ [Co(NH3)3(H2O)3]2+ + H2O
K3
[Co(NH3)3(H2O)3]2+ + NH3 ⇌ [Co(NH3)4(H2O)2]2+ + H2O
K4


These stepwise constants are summarised in the overall stability constant, Kstab
The stability constant is the equilibrium constant for the formation of the complex ion
in a solvent from its constituent ions or molecules
Expression of Kstab


The expression for Kstab can be deduced in a similar way as the expression for the
equilibrium constant (Kc)
For example, the equilibrium expression for the substitution of water ligands by ammonia
ligands in the Co(II) complex is:
[Co(H2O)6]2+ + 4NH3 ⇌ [Co(NH3)4(H2O)2]2+ + 4H2O
190 | P a g e





The concentration of water is not included in the expression as the water is in excess
Therefore, any water produced in the reaction is negligible compared to the water that is
already present
The units of the Kstab can be deduced from the expression in a similar way to the units
of Kc
The stability constants can be used to compare the stability of ligands relative to
the aqueous metal ion where the ligand is water
The larger the Kstab value, the more stable the complex formed is
Calculations Involving Stability Constants


If the concentrations of the transition element complex and the reacting ligands are
known, the expression for the stability constant (Kstab) can be used to determine which
complex is more stable
The greater the value of Kstab the more stable the complex is
6.3.7 Effect of Ligand Exchange on Stability
Constant
Effect of Ligand Exchange on Stability Constant



The stability constants (Kstab) of ligands are often given on a log10 scale so that it becomes
easier to compare them with each other
Ligand exchange in a complex occurs to form a more stable complex with a larger Kstab
The stability constants can be used to explain the substitution of ligands in a copper
complex
Ligand substitution in a Co(II) complex

When excess ammonia is added to the [CoCl4]2- complex a blue solution is obtained
191 | P a g e
The chloride ligands are substituted by the ammonia ligands to form the more stable ammonia
complex

The formation of the ammonia complex could be explained by comparing the stability of
the chloride and ammonia ligands
Stability of chloride and ammonia ligands table



The stability constant of the ammonia ligand is greater than that of the chloride ligands
The blue ammonia complex is therefore more stable
As a result, the position of the equilibrium is shifted to the right
192 | P a g e
ORGANIC CHEMISTRY
7.1.1 Functional Groups
Functional Groups of Organic Compounds (AL)


Many organic compounds contain one or more functional groups
A functional group is an atom or group of atoms in an organic molecule, that determines
its characteristic chemical and physical properties
Arenes



Arenes are aromatic compounds that contain a benzene ring
Chemical properties
o Due to the delocalised electron ring (π system of electrons), these compounds
are electron-rich and therefore can undergo electrophilic attack under the right
conditions
o However, because the delocalised electron ring system makes benzene so stable,
it is resistant to addition reactions
o This is very different to alkenes, which are very reactive and readily undergo
addition reactions
Physical properties
o Benzene has van der Waals dispersion forces of attraction between the molecules
and has a boiling point of 80 C
o The presence of the non-polar hydrocarbon part in the arene functional group
means that these compounds are often insoluble in water
o Benzene would have to break many hydrogen bonds between the water molecules
to be soluble in water, which does not happen as it is not energetically favourable
Arene functional group
193 | P a g e
Halogenoarenes




These are aromatic compounds that contain a halogen bonded to a benzene ring
They are also known as aryl halides
Chemical properties
o These compounds are also prone to electrophilic attack because of the π system
of delocalised electrons
o The halogens can also take part in substitution reactions
Physical properties
o Chlorobenzene, bromobenzene and iodobenzene are all liquid at room
temperature with an oily texture
o As you might expect, the boiling points increase as the size of the halogen
attached increases, because the number of electrons within the molecule increases
o Like other arenes, halogenoarenes are insoluble in water because of the nonpolar hydrocarbon part of the ring
o These molecules are large relative to the size of water molecules, and as with the
arenes it is not energetically favourable for the halogenoarene molecules to break
the hydrogen bonds between the water molecules, so it does not happen
Halogenoarene functional group
Phenols



Phenols are another type of aromatic compounds containing a hydroxide bonded to
a benzene ring
Chemical properties
o The -OH group in phenols is more acidic than in alcohols as the oxygen donates
one of its lone pairs of electrons into the ring system
o This causes an increased electron density of the ring, causing it to become much
more reactive than benzene itself
o It also makes it easier for the hydrogen of the -OH group to be donated
o Phenols can also react with reactive metals such as sodium to form alkoxide ions
Physical properties
o Phenol is a white, crystalline solid, and it has a disinfectant like smell
o Due to the -OH group in phenols, they can form hydrogen bonds with water
molecules, and therefore to a degree phenol is soluble in water
194 | P a g e
Phenol functional group
Acyl chlorides




Acyl chlorides are (carboxylic) acid derivatives containing:
o A chlorine atom attached to a C=O group (replacing what would have been the OH group of a carboxylic acid)
o An acyl (hydrocarbon) group attached to a C=O group
Acyl chlorides are also known as 'acid' chlorides
Chemical properties
o They are fuming liquids and are colourless, with a strong smell
o Acyl chlorides are extremely reactive and readily take part
in substitution reactions in which the chlorine atom is substituted by other
species
o This reactivity is why they are fuming liquids and why they have such a strong
smell - they react with any water vapour in the air
Physical properties
o Acyl chlorides react violently with water, so we cannot say whether or not they
would be soluble in water
Acyl chloride functional group
Amines



Amines are compounds with the -NH2 (primary amine), -NH (secondary amine) or -N
(tertiary amine) group
Classification of amines
o In primary amines, the N of the amine group is bonded to one R group (and two
hydrogen atoms)
o In secondary amines, the N of the amine group is bonded to two R groups (and
one hydrogen atom)
o In tertiary amines, the N of the amine group is bonded to three R groups
Chemical properties
o Due to the lone pair of electrons on the nitrogen, amines are basic compounds
195 | P a g e

Physical properties
o The lone pair on the N of the amine group means that they can form hydrogen
bonds
o They are often soluble in water because they form hydrogen bonds with water
molecules
o The smaller amines are very soluble in water, but their solubility decreases as the
non-polar hydrocarbon chain gets longer
o They often have a fishy smell, especially as the size of the amines increases
Amine functional group
Amides



Amides are compounds containing:
o An amine (-NH2) group
o A carbonyl group (C=O)
o The amide group is -CONH2
Chemical properties
o Amides are less basic than amines, as the lone pair of electrons on the nitrogen
is delocalised
Physical properties
o Amides are often soluble in water as they can form hydrogen bonds with water
molecules
o The smaller amides are very soluble in water, but their solubility decreases as the
non-polar hydrocarbon chain gets longer
Amide functional group
Amino Acids


Amino acids are the building blocks of proteins and consists of:
o An amine (-NH2) group
o A carboxyl (-COOH) group
Chemical properties
o Amino acids react with bases to form salts
196 | P a g e
o
o

They also react with alcohols to form esters
The reaction of amino acids with amines gives amides
Physical properties
o Most of the amino acids are soluble in water but insoluble in organic solvents
o Amino acids have chiral centres and exhibit optical isomers (except for glycine)
Amino acid functional group
7.1.2 Formulae of Functional Groups
Formulae of Organic Compounds (AL)

Students are expected to interpret and use the following formulae of organic compounds
o General formula
o Structural formula
o Displayed formula
o Skeletal formula
Formulae of organic compounds table
197 | P a g e
198 | P a g e
7.1.3 Nomenclature of Compounds
Nomenclature of Aliphatic Compounds (AL)


Students are required to use systematic nomenclature of simple aliphatic compounds
The following method can be applied when naming organic compounds:
o Identify the longest carbon chain containing the functional group
o Identify the functional group on the chain to determine the suffix or prefix on the
compound
o Count along the carbon chain such that the functional group has
the lowest number
o Add any side chains or lower priority functional groups as prefixes to the
beginning of the name in alphabetic order
o Use the prefixes di-, tri- and tetra- if there are two or more identical functional
groups or side chains
Nomenclature of simple aliphatic organic molecules with functional groups table
199 | P a g e
Nomenclature of Aromatic Compounds


The method used to name aromatic compounds is similar to that of aliphatic
compounds
Students are required to use systematic nomenclature of simple aromatic molecules with
one benzene ring and one or more simple substituents
Nomenclature of simple aromatic organic molecules with functional groups table
200 | P a g e
7.1.4 Organic Chemistry Terminology
Terminology Used in Reaction Mechanisms (AL)


Organic reactions are often associated with terminology students should be familiar with
Two of these important terms include:
o Electrophilic substitution
o Addition-elimination
Electrophilic substitution





Electrophiles are species that are electron deficient and can act as an electron pair
acceptor
o Electrophiles are ‘electron loving’ species
Substitution reactions are reactions that involve the replacement of one atom or group
of atoms by another
Electrophilic substitution reactions are therefore reactions in which an atom or group of
atoms are replaced by an electrophile after initial attack by the electron-deficient species
An example of an electrophilic substitution reaction is the reaction
of benzene with bromine in the presence of anhydrous aluminium bromide catalysts
o The bromine acts as an electrophile and attacks the electron-rich benzene ring
o A hydrogen atom is substituted by a bromine atom to form bromobenzene and
hydrogen bromide
Benzene undergoes substitution reactions rather than addition reactions because of the
stability of the benzene ring
The hydrogen atom in benzene is substituted by the bromine which acts as an electrophile
Addition- elimination



Other common organic reactions include addition and elimination reactions
In an addition reaction, two or more molecules combine to give a single product only
An example of an addition reaction is the hydrogenation of alkenes
o Hydrogen (H2) is reacted with an alkene to form an alkane only with no byproducts
201 | P a g e
Hydrogenation of an alkene to give an alkane is an example of an addition reaction


Elimination reactions are the reverse of addition reactions; a small molecule (such as
H2O or HCl) is removed or ‘eliminated’ from an organic molecule
An example of such a reaction is the elimination of HCl from an alkane to form an alkene
Elimination of HCl from an alkyl halide results in the formation of an alkene
7.1.5 Shape of Aromatic Molecules
DOWNLOAD PDF
Shape of Benzene & Aromatic Molecules



Aromatic molecules consist of one or more rings with conjugated π systems
Conjugated π systems arise from alternating double and single bonds in which the
electrons are delocalised
Aromatic compounds are called ‘aromatic’ as they often have pleasant odours
Examples of aromatic compounds including benzene table
202 | P a g e
Shape of benzene & aromatic compounds


Benzene and other aromatic compounds contain sp2 hybridised carbons as two of
their p orbitals have mixed with an s orbital
This means that each carbon atom in benzene and other aromatic compounds has one p
orbital
The carbon atoms in aromatic compounds are sp2 hybridized as two of their p orbitals mix
with an s orbital



Each carbon atom in the ring forms three σ bonds using the sp2 orbitals
The remaining p orbital overlaps laterally with p orbitals of neighbouring carbon atoms
to form a π bond
This extensive sideways overlap of p orbitals results in the electrons being delocalised
and able to freely spread over the entire ring
203 | P a g e



Benzene and other aromatic compounds are regular and planar compounds with bond
angles of 120o
The delocalisation of electrons means that all of the carbon-carbon bonds in these
compounds are identical and have both single and double bond character
The bonds all being the same length is evidence for the delocalised ring structure of
benzene
Like other aromatic compounds, benzene has a planar structure due to the sp2 hybridisation of
carbon atoms and the conjugated π system in the ring
7.1.6 Optical Isomers
Properties of Enantiomers


Stereoisomers are molecules that have the same structural formula but have the atoms
arranged differently in space
There are two types of stereoisomerism
o Geometrical (cis/tras)
o Optical
Optical isomerism


A carbon atom that has four different atoms or groups of atoms attached to it is called
a chiral carbon or chiral centre
Compounds with a chiral centre (chiral molecules) exist as two optical isomers which
are also known as enantiomers
204 | P a g e
A molecule has a chiral centre when the carbon atom is bonded to four different atoms or
group of atoms; this gives rises to enantiomers


The enantiomers are non-superimposable mirror images of each other
Their physical and chemical properties are identical but they differ in their ability to
rotate plane polarised light
o Hence, these isomers are called ‘optical’ isomers
o One of the optical isomers will rotate the plane of polarised in
the clockwise direction
o Whereas the other isomer will rotate it in the anti-clockwise direction
205 | P a g e
When unpolarised light is passed through a polariser, the light becomes polarised as the waves
will vibrate in one plane only
Biological activity of enantiomers





Enantiomers also differ from each other in terms of their biological activity
Enzymes are chiral proteins that speed up chemical reactions by binding substrates
They are very target-specific as they have a specific binding site (also called active site)
and will only bind molecules that have the exact same shape
Therefore, if one enantiomer binds to a chiral enzyme, the mirror image of this
enantiomer will not bind nearly as well if at all
It’s like putting a right-hand glove on the left hand!
Enantiomers differ from each other in their biological activity
206 | P a g e
Optically Active Compounds & Racemic Mixtures


Enantiomers are optical isomers that are mirror images of each other and are nonsuperimposable
They have similar chemical properties but differ from each other in their ability to
rotate plane polarised light and in their biological activity
Optical activity





Let’s suppose that in a solution, there is 20% of the enantiomer which rotates the plane
polarised light clockwise and 80% of the enantiomer which rotates the plane of polarised
light anticlockwise
There is an uneven mixture of each enantiomer, so the reaction mixture is said to
be optically active
o The net effect is that the plane of polarised light will be rotated anticlockwise
Similarly, if the percentages of the enantiomers are reversed, the reaction mixture is
still optically active but now the plane of polarised light will be rotated clockwise
o In this case, there is 20% of the enantiomer, which rotates the
plane anticlockwise
o And 80% of the enantiomer, which rotates the plane clockwise
A racemic mixture is a mixture in which there are equal amounts of enantiomers
present in the solution
A racemic mixture is optically inactive as the enantiomers will cancel out each others
effect and the plane of polarised light will not change
207 | P a g e
When one of the enantiomers is in excess, the mixture is optically active; when there are equal
amounts of each enantiomer the mixture is optically inactive
Effect of Optical Isomers on Plane Polarised Light





Molecules with a chiral centre exist as optical isomers
These isomers are also called enantiomers and are non-superimposable mirror images of
each other
The major difference between the two enantiomers is that one of the enantiomers rotates
plane polarised light in a clockwise manner and the other in an anticlockwise fashion
o The enantiomer that rotates the plane clockwise is called the R enantiomer
o The enantiomer that rotates the plane anticlockwise is called the S enantiomer
These enantiomers are therefore said to be optically active
Therefore, the rotation of plane polarised light can be used to determine the identity of an
optical isomer of a single substance
o For example, pass plane polarised light through a sample containing one of the
two optical isomers of a single substance
o Depending on which isomer the sample contains, the plane of polarised light will
be rotated either clockwise or anti-clockwise
o No effect will be observed when the sample is a racemic mixture
Each enantiomer rotates the plane of polarised light in a different direction
208 | P a g e
7.1.7 Chirality & Drug Production
Chirality & Drug Production




Most of the drugs that are used to treat diseases contain one or more chiral centres
These drugs can therefore exist as enantiomers which differ from each other in their
ability to rotate plane polarised light
Another crucial difference between the enantiomers is in their
potential biological activity and therefore their effectiveness as medicines
Drug compounds should be prepared in such a way that only one of the optical isomer is
produced, in order to increase the drugs’ effectiveness
o Some drug enantiomers can have very harmful side effects
Potential biological activity of enantiomers





If conventional organic reactions are used to make the desired drug, a racemic
mixture will be obtained
o In a racemic mixture, there are equal amounts of the two enantiomers
The physical and chemical properties of the enantiomers are the same, however, they
may have opposite biological activities
For example, the drug naproxen is used to treat pain in patients that suffer from arthritis
o One of the enantiomers of naproxen eases the pain, whereas another enantiomer
causes liver damage
One enantiomer of a drug used to treat tuberculosis is effective whereas another
enantiomer of this drug can cause blindness
Thalidomide is another example of a drug that used to be used to treat morning sickness,
where one of the enantiomers caused very harmful side effects for the unborn baby
Separating racemic mixtures



Due to the different biological activities of enantiomers, it is very important to separate a
racemic mixture into pure single enantiomers which are put in the drug product
This results in reduced side-effects in patients
o As a result, it protects pharmaceutical companies from legal actions if the side
effects are too serious
It also decreases the patient’s dosage by half as the pure enantiomer is more potent and
therefore reduces production costs
o A more potent drug has a better therapeutic activity
209 | P a g e
Chiral catalysts


In order to produce single, pure optical isomers, chiral catalysts can be used
The benefits of using chiral catalysts are that only small amounts of them are needed and
they can be reused
o For example, an organometallic ruthenium catalyst is used in the production
of naproxen which is used in the treatment of arthritis
The organometallic ruthenium catalyst is a chiral catalyst which ensures that only one of the
enantiomers is formed which can be used in treating arthritis





Enzymes are excellent biological chiral catalysts that promote stereoselectivity and
produce single-enantiomer products only
o Stereoselectivity refers to the preference of a reaction to form one enantiomer
over the other
Due to the specific binding site of enzymes, only one enantiomer is formed in the
reaction
The enzymes are fixed in place on inert supports so that the reactants can pass over
them without having to later separate the product from the enzymes
The disadvantage of using enzymes is that it can be expensive to isolate them from
living organism
o Therefore, more research has recently been carried out into designing synthetic
enzymes
Although using enzymes to produce pure enantiomers in drug synthesis takes longer than
conventional synthetic routes, there are many advantages to it in the long run
o For example, using enzymes to synthesise drugs is a greener process as fewer
steps are involved compared to conventional synthetic routes
210 | P a g e
7.2.1 Reactions of Arenes
Reactions of Arenes




Arenes are very stable compounds due to the delocalisation of π electrons in the ring
o This is because the negative charge is spread out over the molecule instead of
being confined to a small area
During chemical reactions such as substitution reactions, this delocalised ring is
maintained
Addition reactions however, disrupt the aromatic stabilisation
Arenes undergo a series of reactions including:
o Substitution
o Nitration
o Friedel-Crafts alkylation
o Friedel-Crafts acylation
o Complete Oxidation
o Hydrogenation
Substitution


Halogenation reactions are examples of electrophilic substitution reactions
Arenes undergo substitution reactions with chlorine (Cl2) and bromine (Br2) in the
presence of anhydrous AlCl3 or AlBr3 catalyst respectively to form halogenoarenes (aryl
halides)
o The chlorine or bromine act as an electrophile and replaces a hydrogen atom on
the benzene ring
o The catalyst is required for the reaction to take place, due to the stability of the
benzene structure
Arenes undergo substitution reactions with halogens to form aryl halides



Alkylarenes such as methylbenzene undergo halogenation on the 2 or 4 positions
This is due to the electron-donating alkyl groups which activate these positions
o Phenol (C6H5OH) and phenylamine (C6H5NH2) are also activated in the 2 and 4
positions
The halogenation of alkylarenes therefore result in the formation of two products
211 | P a g e
Alkylarenes are substituted on the 2 or 4 position

Multiple substitutions occur when excess halogen is used
In the presence of excess halogen, multiple substitutions occur
Nitration



Another example of a substitution reaction is the nitration of arenes
In these reactions, a nitro (-NO2) group replaces a hydrogen atom on the arene
The benzene is reacted with a mixture of concentrated nitric acid (HNO3) and
concentrated sulfuric acid (H2SO4) at a temperature between 25 and 60 oC
212 | P a g e
Nitration of benzene

Again, due to the electron-donating alkyl groups in alkylarenes, nitration of
methylbenzene will occur on the 2 and 4 position
Nitration of alkylarenes
Friedel-Crafts reactions





Friedel-Crafts reactions are also electrophilic substitution reactions
Due to the aromatic stabilisation in arenes, they are often unreactive
To use arenes as starting materials for the synthesis of other organic compounds, their
structure, therefore, needs to be changed to turn them into more reactive compounds
Friedel-Crafts reactions can be used to substitute a hydrogen atom in the benzene ring for
an alkyl group (Friedel-Crafts alkylation) or an acyl group (Friedel-Crafts acylation)
Like any other electrophilic substitution reaction, the Friedel-Crafts reactions consist of
three steps:
o Generating the electrophile
o Electrophilic attack on the benzene ring
o Regenerating aromaticity of the benzene ring
213 | P a g e
Examples of Friedel-Crafts alkylation and acylation reactions
Friedel-Crafts alkylation



In this type of Friedel-Crafts reaction, an alkyl chain is substituted into the benzene ring
The benzene ring is reacted with a chloroalkane in the presence of an AlCl3 catalyst
An example of an alkylation reaction is the reaction of benzene with chloropropane
(CH3CH2CH2Cl) to form propylbenzene
214 | P a g e
Example of a Friedel-Crafts alkylation reaction
Friedel-Crafts acylation



In the Friedel-Crafts acylation reaction, an acyl group is substituted into the benzene ring
o An acyl group is an alkyl group containing a carbonyl, C=O group
The benzene ring is reacted with an acyl chloride in the presence of an AlCl3 catalyst
An example of an acylation reaction is the reaction of methylbenzene with propanoyl
chloride to form an acyl benzene
o Note that the acyl group is on the 4 position due to the -CH3 group on the benzene
215 | P a g e
Example of a Friedel-Crafts acylation reaction
216 | P a g e
Complete oxidation



Normally, alkanes are not oxidised by oxidising agents such as potassium
manganate(VII) (KMnO4)
However, the presence of the benzene ring in alkyl arenes affect the properties of the
alkyl side-chain
The alkyl side-chains in alkyl arenes are oxidised to carboxylic
acids when refluxed with alkaline potassium manganate(VII) and
then acidified with dilute sulfuric acid (H2SO4)
o For example, the complete oxidation of ethylbenzene forms benzoic acid
The complete oxidation of alkyl side-chains in arenes gives a carboxylic acid
Hydrogenation


The hydrogenation of benzene is an addition reaction
Benzene is heated with hydrogen gas and a nickel or platinum catalyst to
form cyclohexane
Hydrogenation of benzene

The same reaction occurs when ethylbenzene undergoes hydrogenation to
form cycloethylbenzene
217 | P a g e
Hydrogenation of methylbenzene
Summary of reactions of arenes table
218 | P a g e
7.2.2 Electrophilic Substitution of Arenes
Arenes: Electrophilic Substitution

The electrophilic substitution reaction in arenes consists of three steps:
o Generation of an electrophile
o Electrophilic attack
o Regenerating aromaticity
Mechanism of electrophilic substitution

The halogenation and nitration of arenes are both examples of electrophilic
substitution reactions
o A hydrogen atom is replaced by a halogen atom or a nitro (-NO2) group
The overall reaction of halogenation of arenes
The overall reaction of nitration of arenes

In the first step, the electrophile is generated
o For the halogenation reaction, this is achieved by reacting the halogen with
a halogen carrier
o The halogen molecules form a dative bond with the halogen carrier by donating a
lone pair of electrons from one of its halogen atoms into an empty 3p orbital of
the halogen carrier
219 | P a g e
o
In the nitration reaction, the electrophile NO2+ ion is generated by reacting it with
concentrated nitric acid (HNO3) and concentrated sulfuric acid (H2SO4)
Step 1 of the halogenation reaction of arenes
Step 1 of the nitration reaction of arenes


Once the electrophile has been generated, it will carry out an electrophilic attack on the
benzene ring
o The nitrating mixture of HNO3 and H2SO4 is refluxed with the arene at 25 - 60 oC
A pair of electrons from the benzene ring is donated to the electrophile to form
a covalent bond
o This disrupts the aromaticity in the ring as there are now only four π electrons and
there is a positive charge spread over the five carbon atoms
Step 2 of the halogenation reaction of arenes
220 | P a g e
Step 2 of the nitration reaction of arenes

In the final step of the reaction, this aromaticity is restored by heterolytic cleavage of
the C-H bond so that bond electrons in this bond go into the benzene π bonding system
Step 3 of the halogenation reaction of arenes
Step 3 of the nitration reaction of arenes
221 | P a g e
Addition reactions of arenes



The delocalisation of electrons (also called aromatic stabilisation) in arenes is the main
reason why arenes predominantly undergo substitution reactions over addition reactions
In substitution reactions, the aromaticity is restored by heterolytic cleavage of the C-H
bond
In addition reactions, on the other hand, the aromaticity is not restored and is in some
cases completely lost
o The hydrogenation of arenes is an example of an addition reaction during which
the aromatic stabilisation of the arene is completely lost
o The cyclohexane formed is energetically less stable than the benzene
7.2.3 Location of Halogenation on Arenes
Arenes: Location of Halogenation


Arenes will undergo substitution reactions with halogens to form aryl halides
o This reaction is also called a halogenation reaction
Depending on the reaction conditions, halogenation can occur:
o In the aromatic ring
o In the side chain
Halogenation in the aromatic ring

Halogenation of alkylarenes in the aromatic ring will occur when a halogen and
anhydrous halogen carrier catalyst (such as AlBr3 or AlCl3) is used
Halogenation of alkylarenes in the aromatic ring

Aryl halides are less reactive than halogenoalkanes as the carbon-halogen bond in aryl
halides is stronger
222 | P a g e


This is due to the partial overlap of the lone pairs on the halogen atom with the π system
in the benzene ring
The carbon-halogen bond, therefore, has a partial double bond character
Aryl halides are unreactive due to the partial double bond character of the carbon-halogen
bond
Halogenation in the side chain

Halogenation of alkylarenes in the side chain will occur when the halogen is passed
into boiling alkylarene in the presence of ultraviolet (UV) light
o This is a free-radical substitution reaction
Halogenation of alkylarenes in the side chain is an example of a free-radical substitution
reaction

If excess halogen is used, all hydrogen atoms on the alkyl side-chain will be substituted
by the halogen atoms
223 | P a g e
In excess halogen, all hydrogen atoms on the alkyl side-chain will be replaced

Note that no substitution into the benzene ring occurs under these conditions
7.2.4 Directing Effects of Substituents on
Arenes
Arenes: Directing Effects of Substituents


Arenes readily undergo electrophilic substitution of one of their hydrogen atoms with
another species
Substituents that are already present on the arenes can affect where the substitution of
the hydrogen atom on the arene takes place
o These groups are said to direct substitution reactions to different ring positions
Electron-withdrawing & -donating groups


The substituents on the arenes can either be electron-withdrawing or electron-donating
groups
Electron-withdrawing substituents remove electron density from the π system in the
benzene ring making it less reactive
o These groups deactivate attack by electrophiles and direct the incoming
electrophile to attack the 3 and/or 5 positions
o For example, the nitro group in nitrobenzene is an electron-withdrawing group
 Upon bromination of nitrobenzene, the bromine electrophile will be
directed to the 3 and/or 5 position
 The products are 3-chloronitrobenzene and 5-chloronitrobenzene
o Electron-donating substituents donate electron density into the π system of the
benzene ring making it more reactive
224 | P a g e


These groups activate attack by electrophiles and direct the incoming
electrophile to attack the 2, 4 and/or 6 positions
For example, the methyl group in methylbenzene is an electron-donating
group
 Upon bromination of methylbenzene, the bromine electrophile will
be directed to the 2 and/or 4 position
 The products are 2-chloromethylbenzene and 4chloromethylbenzene
Electron-withdrawing & -donating substituents table
225 | P a g e
7.3.1 Production of Halogenoarenes
Production of Halogenoarenes


Halogenoarenes are arenes which are bonded to halogen atoms
They can be prepared from substitution reactions of arenes with chlorine or bromine in
the presence of an anhydrous catalyst
Substitution of benzene to form halogenoarenes






Chlorine gas is bubbled into benzene at room temperature and in the presence of an
anhydrous AlCl3 catalyst to form chlorobenzene
The AlCl3 catalyst is also called a halogen carrier and is required to generate
the electrophile (Cl+)
This electrophile attacks the electron-rich benzene ring in the first stage of the reaction
which disrupts the delocalised π system in the ring
To restore the aromatic stabilization, a hydrogen atom is removed in the second
stage of the electrophilic substitution reaction to form chlorobenzene
When this happens, the delocalised π system of the ring is restored
The same reaction occurs with benzene and bromine in the presence of an AlBr3 catalyst
to form bromobenzene
Halogenoarenes can be formed from the electrophilic substitution reaction of arenes with
halogens
226 | P a g e
Substitution of methylbenzene to form halogenoarenes






The electrophilic substitution of methylbenzene with halogens results in the formation
of multiple halogenoarenes as products
This is because the methyl group (which is an alkyl group) in methylbenzene is electrondonating and pushes electron density into the benzene ring
This makes the benzene ring more reactive towards electrophilic substitution reactions
The methly group is said to be 2,3-directing and as a result,
the 2 and 4 positions are activated
Electrophilic substitution of methylbenzene with chlorine and anhydrous AlCl3 catalyst,
therefore, gives 2-chloromethylbenzene and 4-chloromethylbenzene
The reaction mechanism is the same as the substitution mechanism of benzene
The methyl group on methylbenzene directs the incoming halogen on the 2 and 4 position

In the presence of excess chlorine, substitution on the 6 position will also occur
227 | P a g e
7.3.2 Reactivity of Halogenoarenes
Difference in Reactivity of Halogenoalkanes & Halogenoarenes


Halogenoarenes are very unreactive compared to halogenoalkanes
The difference in reactivity between the two compounds is because of the carbon-halogen
bond strengths
Halogenoalkanes




The halogenoalkane chloroethane can take part in nucleophilic substitution reactions
A nucleophile, such as a hydroxide (OH-) ion, will attack the slightly positive
carbon atom
A covalent bond is formed between that carbon atom and the nucleophile which causes
the carbon-halogen bond to break
Overall, the halogen is replaced by the nucleophile
Halogenoalkanes readily undergo nucleophilic substitution reactions
Halogenoarenes


Halogenoarenes, such as chlorobenzene, do not readily undergo nucleophilic
substitution reactions
o Only under extremely harsh conditions, such as temperatures of 200 oC and a
pressure of 200 atmospheres, will the chlorine in chlorobenzene get replaced by a
nucleophile such as a hydroxide (OH-) ion
This is because the carbon-chlorine bond is very strong and cannot be easily broken
o One of the lone pairs of electrons on the chlorine will interact with the π system
of the ring
o This causes the carbon-chlorine bond to have a partial double-bond
character, which strengthens the bond
228 | P a g e
The carbon-chlorine bond is very strong, as it has partial double-bond character



The unreactivity of halogenoarenes can therefore be explained by the delocalisation of a
lone pair on the halogen over the benzene
This causes additional stabilisation of the system and strengthens the carbon-halogen
bond, which affects the reactions that halogenoarenes will undergo
It gets harder to break the carbon-halogen bond in halogenoarenes, which decreases
reactivity
229 | P a g e
7.4.1 Alcohols
Reaction of Alcohols With Acyl Chlorides




Acyl chlorides are reactive organic compounds with a -COCl functional group
The carbonyl carbon is electron-deficient and has a partial positive charge
It is therefore susceptible to nucleophilic attack
The carbon-chlorine bond breaks and white fumes of hydrogen chloride, HCl, are
formed
Reaction with alcohols and phenols



Acyl chlorides react with alcohols and phenols to form esters in a nucleophilic
substitution reaction
The -OH group acts as a nucleophile and attacks the carbonyl carbon to substitute the
chlorine atom
Forming esters from acyl chlorides rather than carboxylic acids is more
effective because::
o Acyl chlorides are more reactive (so it produces the ester faster)
o Acyl chloride reactions go to completion (so more of the ester is produced)
Reaction with alcohols

The reaction of acyl chlorides with alcohols is vigorous and white fumes of HCl gas are
formed
Acyl chlorides react vigorously with alcohols to form esters
Reaction with phenols

For the reaction of acyl chlorides with phenols to occur, heat and a base are required
o The base is needed to deprotonate the phenol and form a phenoxide ion
o The phenoxide ion is a better nucleophile than the original phenol molecule and
will be able to attack the carbonyl carbon
230 | P a g e
Acyl chlorides react with phenols when heated and in the presence of a base to form esters
A base is needed to form a phenoxide ion which is a better nucleophile than phenol; now,
nucleophilic attack on the carbonyl carbon can more readily occur
7.4.2 Production of Phenol
Production of Phenol


Phenols are organic compounds characterised by the presence of an -OH group which is
attached to a benzene ring
Phenols can be produced by the reaction of phenylamine with nitrous acid (HNO2)
Production of phenol


Phenols can be prepared from phenylamines under the following reaction conditions:
o NaNO3 with dilute acid (to form HNO2)
o Ice to keep the temperature below 10 oC (step 1)
o Heat (step 3)
This reaction involves three steps:
o Step 1 - The HNO2 is so unstable that it needs to be prepared in a test-tube by
reacting sodium nitrate (NaNO3) and dilute hydrochloric acid (HCl) while keeping
the temperature below 10 oC using ice
231 | P a g e
Nitrous acid can be prepared in a test-tube by reacting sodium nitrate with dilute hydrochloric
acid

o
Step 2 - Phenyl amine is then reacted with the HNO2 to form an unstable
diazonium salt
Formation of an unstable diazonium salt

o
232 | P a g e
Step 3 - The diazonium salt is so unstable that it will thermally
decompose when heated to form a phenol
A phenol is formed upon the thermal decomposition of the diazonium salt
7.4.3 Reactions of Phenol
Reactions of Phenol


Phenols can undergo many types of reactions as both the electron-rich benzene ring and
the polar -OH group can participate in chemical reactions
Some of the reactions of phenols include:
o With bases
o With reactive metals
o With diazonium salts
o Nitration
o Bromination
Reactions of the -OH group in phenols


The -OH group in phenols has a slightly acidic character
It can therefore act as an acid and take part in acid-base reactions
Reaction with bases


Phenols are only slightly soluble in water due to the large non-polar benzene ring
However, they do dissolve in alkaline solutions and undergo acid-base reactions
with bases to form a soluble salt and water
233 | P a g e
Phenols are weak acids and undergo acid-base reactions in alkaline solutions
Reaction with reactive metals



Molten phenols react vigorously with reactive metals such as sodium (Na)
This is also an acid-base reaction
Now, a soluble salt is formed and hydrogen gas is given off
Molten phenols react vigorously with reactive metals to form a soluble salt and hydrogen gas
Reaction with diazonium ions





Diazonium ions are very reactive compounds containing an -N2+ group
When phenols are dissolved in sodium hydroxide (NaOH), a solution of sodium
phenoxide is obtained
This solution is cooled in ice and cold diazonium ion is added to the sodium phenoxide
After the reaction has occurred, a yellow-orange solution or precipitate of an azo
compound is formed
These are compounds in which two benzene rings are linked by a nitrogen bridge
234 | P a g e
Azo compounds are formed from the reaction of phenols with diazonium ions
Reactions of the aromatic ring in phenols




Phenols react more readily with electrophiles compared to benzene
This is because one of the lone pairs of electrons on the oxygen atom in -OH overlaps
with the π bonding system
This increases the electron density of the benzene ring making it
more susceptible to electrophilic attack
The -OH group in phenols is activating and directs incoming electrophiles to the 2, 4,
and 6 positions
Nitration



Phenols can undergo electrophilic substitution reactions when reacted with dilute nitric
acid (HNO3) at room temperature to give a mixture of 2-nitrophenol and 4nitrophenol
o When concentrated HNO3 is used, the product will be 2,4,6-trinitrophenol
instead
A hydrogen atom in the benzene ring is substituted by a nitro (-NO2) group
This is also known as the nitration of phenol
235 | P a g e
Phenols undergo nitration when reacted with dilute HNO3 at room temperature
Bromination



Phenols also undergo electrophilic substitution reactions when reacted with bromine
water at room temperature
Phenol decolourises the orange bromine solution to form a white precipitate of 2,4,6tribromophenol
This is also known as the bromination of phenol
Phenols undergo bromination when reacted with bromine water at room temperature
236 | P a g e
7.4.4 Acidity of Phenols
Acidity of Phenols



Although phenol compounds contain an alcohol (-OH) group, they are weakly acidic
This is due to the delocalisation of one of the lone pairs from the oxygen atom into the
aromatic ring
This increases the electron density of the ring and increases the acidic behaviour
Delocalisation of charge density






The conjugate base of phenol is the phenoxide ion
In the phenoxide ion, the negative charge on the oxygen is spread out over the entire ion
This is possible as one of the lone pairs on the oxygen atom overlaps with the
delocalised π system of the ring
Because of this delocalisation, there is less charge density on the oxygen atom
The H+ ions are therefore not strongly attracted to the phenoxide ion and are less likely
to reform the phenol molecule
This means that phenol is more likely to lose a proton (and act as an acid) rather than
to gain a proton (and act as a base )
The negative charge is spread over the ion, causing the electrons to become less available for
bonding with an incoming proton
Stability of the conjugate base




Phenol ionises to form a more stable negative phenoxide ion with its negative
charge spread out
This means that phenol is more likely to undergo ionisation
The equilibrium position, therefore, lies further to the right and a higher proportion of
phenol molecules donate a proton compared to for example water and ethanol
The phenol compound is, therefore, more likely to act as an acid rather than a base
237 | P a g e
Since the phenoxide ion formed from the ionisation of phenol is more stable than phenol
itself, the equilibrium position lies further to the right-hand-side and phenol is more likely to
act as an acid rather than a base
Relative Acidities of Water, Phenol & Ethanol


The pKa is a measure of the acidity of a substance
The values of water, phenol, and ethanol show that phenol is a stronger acid than ethanol
and water
Relative acidity of ethanol, water & phenol table

The order of acidity can be explained by looking at their conjugate bases which are
formed from the dissociation of the compounds
Delocalisation of charge density





In the phenoxide ion (which is the conjugate base of phenol) the charge density on the
oxygen atom is spread out over the entire ion
As a result, the electrons on the oxygen atom are less available for bond formation with a
proton (H+ ion)
The conjugate base of ethanol is the ethoxide ion
The ethyl group in the ion is an electron-donating group that donates electron density to
the oxygen atom
As a result, the electron density on the oxygen atom is more readily available for bond
formation with an H+ ion
238 | P a g e
The electron-donating alkyl group in the ethoxide ion concentrates charge density on the
oxygen atom which can more easily bond an H+ ion



The conjugate base of water is the hydroxide ion
Since the charge density of the oxygen atom cannot become delocalised over a ring, the
hydroxide ion more readily accepts an H+ ion compared to the phenoxide ion
o Water is, therefore, a stronger base compared to phenol
However, as there are no electron-donating alkyl groups, less negative charge is
concentrated on the oxygen atom which therefore less readily accepts an H+ ion compared
to the ethoxide ion
o Water is, therefore, a weaker base compared to ethanol
The hydroxide ion lacks an aromatic ring and electron-donating alkyl groups so water is a
stronger base than phenol but a weaker base than ethanol

Therefore, the position of equilibrium lies:
o Further to the right-hand side favouring the dissociated phenoxide ions
o Further to the left-hand side favouring the undissociated ethoxide and
hydroxide ions
Relative equilibrium positions for the dissociation of ethanol, water, and phenol
239 | P a g e
7.4.5 Nitration & Bromination of Phenol
Nitration & Bromination of Phenol





Compared to benzene, phenol reacts more readily with electrophiles
This is because one of the lone pairs of electrons on the oxygen atom in
phenol overlaps with the π bonding system of the benzene ring
As a result, there is now an increased electron density in the ring
The electron-donating -OH group in phenol, therefore, activates the benzene ring
and directs incoming electrophiles to the 2, 4, and 6 positions
The increased reactivity of phenol means that different reagents and conditions are used
for electrophilic substitution reactions of phenols compared to benzene
Nitration



Nitration is an example of an electrophilic substitution reaction
The nitration of benzene requires a mixture of concentrated nitric acid (HNO2) and
sulfuric acid (H2SO4) refluxed with benzene between 25 oC and 60 oC
Since phenol is more reactive, nitration can occur under milder conditions by reacting it
with dilute nitric acid at room temperature
o If concentrated nitric acid is used, 2,4,6-trinitrophenol is formed
Bromination



Bromination is another example of an electrophilic substitution reaction
Benzene will undergo bromination only when reacted with pure bromine (not a
solution) and in the presence of an anhydrous aluminium bromide
(AlBr3) catalyst at room temperature
Phenol on the other hand readily reacts with bromine water in the absence of a catalyst
Reagents & conditions for nitration and bromination of phenol & benzene table
240 | P a g e
Directing Effects of Hydroxyl Group on Phenol






Phenols consist of a hydroxyl (-OH) group attached to a benzene ring
The oxygen atom in this hydroxyl group donates electron density into the ring
One of the lone pairs of the oxygen atom overlaps with the π system of the benzene ring
and become delocalised causing an increased electron density in the aromatic ring
Due to the increased electron density, the benzene ring is now more likely to
undergo electrophilic attack and becomes activated
The incoming electrophiles are directed by the hydroxyl group of the phenol to the 2, 4,
and 6 positions
An example is the bromination of phenol
o The bromine acts as an electrophile and substitutes a hydrogen atom in the
benzene ring
o The substitution of the hydrogen atom can occur on the 2, 4, or 6 positions
The hydroxyl group in phenol directs bromination in the 2, 4, and 6 positions
241 | P a g e
7.4.6 Reactions of Other Phenolic
Compounds
Reactions of Other Phenolic Compounds


Phenolic compounds are those that contain a phenol functional group
An example of a phenolic compound is 1-naphthol
1-Naphthol is a phenolic compound





1-Naphthol contains a phenol group attached to another benzene ring
Just like with phenol, the -OH group in 1-naphthol is also electrondonating and activates the benzene ring to electrophilic substitution reactions
The electrophiles are directed to the 2 and/or 4 positions
Substitution at the 6 position is not possible as there is no hydrogen atom on this carbon
o This carbon is bonded to a carbon atom of the second benzene ring
1-Naphthol and other phenolic compounds react in a similar way as phenol
242 | P a g e
7.5.1 Production of Benzoic Acid
Production of Benzoic Acid

Benzoic acids are the simplest aromatic carboxylic acids with the molecular formula
of C6H5COOH
Benzoic acids and their derivatives are often used as reagents in the synthesis
of esters
The compounds can be produced from the oxidation of alkylbenzenes


Oxidation of alkylbenzenes

The alkyl side-chain in alkylbenzenes, such as methylbenzene, can be oxidised to
a carboxylic acid
The alkylbenzene is heated under reflux with a solution of hot alkaline KMnO4 (this
is the oxidising agent)
o The purple colour of the Mn7+ ions disappears as they are reduced to Mn4+ ions
o A brown precipitate of MnO2 is formed
The mixture is then acidified with dilute acid (such as hydrochloric acid)
to protonate the organic product form and produce a benzoic acid


Alkylbenzenes such as methylbenzene undergo oxidation to form benzoic acid
7.5.2 Reactions of Carboxylic Acids
Reactions of Carboxylic Acids to Produce Acyl Chlorides



Acyl chlorides are compounds with the functional group -COCl
They look similar in structure to carboxylic acids but have a Cl atom instead of an -OH
group attached to the carbonyl (C=O)
Acyl chlorides are more reactive than their corresponding carboxylic acids and are
therefore often used as starting materials in the production of organic compounds such
as esters
243 | P a g e


They can be prepared from the reaction of carboxylic acids with:
o Solid phosphorus(V) chloride (PCl5)
o Liquid phosphorus(III) chloride (PCl3) and heat
o Liquid sulfur dichloride oxide (SOCl2)
For example, the acyl chloride ethanoyl chloride can be formed from ethanoic acid in
the above reactions
Production of acyl chlorides from their corresponding carboxylic acids
Further Oxidation of Carboxylic Acids



Carboxylic acids can be formed from the oxidation of primary alcohols
The primary alcohols are firstly oxidised to aldehydes and then further oxidised
to carboxylic acids
Some carboxylic acids can get even further oxidised
Methanoic acid


Methanoic acid is a strong reducing agent and gets further oxidised to carbon
dioxide (CO2)
The oxidation of methanoic acid can occur by:
244 | P a g e
o
o
Warming methanoic acid with mild oxidising agents such
as Fehling’s or Tollens’ reagent
 In a Fehling’s solution, the Cu2+ ion is reduced to Cu+ ion
which precipitates as red Cu2O
 With Tollens’ reagent, the Ag+ is reduced to Ag
Using stronger oxidising agents such as acidified KMnO4 or acidified K2Cr2O7
 The purple KMnO4 solution turns colourless as Mn7+ ions are reduced to
Mn2+ ions
 The orange K2Cr2O7 solution turns green as the Cr6+ ions are reduced to
Cr3+ ions
Ethanedioic acid


Another carboxylic acid that can get further oxidised is ethanedioic acid
A strong oxidising agent such as warm acidified KMnO4 is required for the oxidation
of ethanedioic acid to carbon dioxide
Ethanedioic acid is a dicarboxylic acid that can get further oxidised to carbon dioxide
245 | P a g e
7.5.3 Relative Acidities of Carboxylic Acids,
Phenols & Alcohols
Relative Acidities of Carboxylic Acids, Phenols & Alcohols


Carboxylic acids are compounds with a -COOH functional group
They can act as acids and lose a proton (H+ ion) in an aqueous solution to
form carboxylate salts and water
Carboxylic acids dissociate in aqueous solutions to form carboxylate salts and water


However, carboxylic acids are only weak acids as the position of equilibrium lies well
over to the left-hand side
The pKa values of carboxylic acids, phenols, and alcohols suggest that carboxylic acids
are stronger acids than alcohols and phenols
o The pKa is a measure of the relative strength of a species as an acid
o The smaller the pKa value, the stronger the acid
Relative acidity of ethanol, phenol & carboxylic acids table
246 | P a g e

This order of relative acidities can be explained by looking at the strength of the O-H
bond and the stability of the conjugate bases of the acids
Strength of O-H bond




In carboxylic acids, the electrons in the O-H bond are drawn towards the C-O bond
The electrons in the C-O bond are drawn towards the C=O bond
Overall, the O-H bond is weakened due to the carbonyl (C=O) group removing electron
density from it and drawing it towards itself
Carboxylic acids can therefore more easily lose a proton compared to phenols and
alcohols which lack this electron-withdrawing carbonyl group
The carbonyl group in carboxylic acids draws the electrons away from the O-H bond causing
it to become weaker compared to the O-H bond in phenols and alcohols
Stability of carboxylate ions





The conjugate base of carboxylic acids is the carboxylate ion
The charge density on the oxygen atom is spread out over the carboxylate ion
This is because the charge is delocalised on an electronegative carbonyl oxygen atom
As a result, the electrons on the oxygen atom are less available for bond formation with
an H+ ion to reform the undissociated acid molecule with -COOH group
The position of the dissociation equilibrium lies more to the right compared to alcohols
and phenols
247 | P a g e
The carboxylate ion is stable due to delocalisation of the charge density on the electronegative
oxygen
Stability of alkoxide ions







The conjugate base of alcohols is the alkoxide ion
The alkyl group in the ion is an electron-donating group that donates electron density to
the oxygen atom
As a result, the electron density on the oxygen atom is more readily available for bond
formation with an H+ ion
Alkoxide ions also lack the ability to delocalise the charge density on the entire ion
The conjugate bases of alcohols are therefore less stable than the alcohols themselves and
are more likely to reform the alcohol
This means that alcohols are weaker acids compared to carboxylic acids and phenols
The position of the dissociation equilibrium lies more to the left
The electron-donating alkyl groups in alkoxide ions increase the electron density on the
oxygen atom which is, therefore, more likely to bond an H+ ion and reform the alcohol
Stability of phenoxide ions



In the phenoxide ion (which is the conjugate base of phenol) the charge density on the
oxygen atom is spread out over the entire ion
o This delocalisation of electrons stabilises the phenoxide ion
As a result, the electrons on the oxygen atom are less available for bond formation with a
proton (H+ ion)
The conjugate base of phenols is therefore more stable than phenol
248 | P a g e



However, since the delocalisation of charge density is on carbon atoms and not on
electronegative oxygen atoms like in the carboxylate ion, phenoxide ions are less
stable than carboxylate ions
Therefore, phenols are weaker acids relative to carboxylic acids
The position of the dissociation equilibrium lies more to the right compared to alcohols
and more to the left compared to carboxylic acids
The charge density is delocalised on the entire benzene ring in the phenoxide ions
7.5.4 Relative Acidities of Chlorinesubstituted Carboxylic Acids
Relative Acidities of Chlorine-Substituted Carboxylic Acids





Electron-withdrawing groups bonded to the carbon attached to the -COOH group make
the carboxylic acids stronger acids
This is because the O-H bond in the undissociated acid molecule is even further
weakened as the electron-withdrawing group draws even more electron density away
from this bond
Furthermore, the electron-withdrawing groups extend the delocalisation of the negative
charge on the -COO- group of the carboxylate ion
The -COO- group is now even more stabilised and is less likely to bond with an H+ ion
Chlorine-substituted carboxylic acids are examples of carboxylic acids with electronwithdrawing groups
pKa values of ethanoic acid and chlorine-substituted derivatives table
249 | P a g e

The pKa values of ethanoic acid and chloro-substituted derivatives show that
the more electron-withdrawing groups there are on the carbon attached to the -COOH
group, the stronger the acid
7.5.5 Production of Esters
Production of Esters by Reacting Alcohols With Acyl Chlorides






Esters are organic compounds with an -COR functional group
They have characteristic smells and are used in perfumes, cosmetics and as solvents
Esters can be prepared from the condensation reaction between alcohols and carboxylic
acids
o This is also called an esterification reaction
A more effective way of preparing esters is from the condensation reaction between
alcohols and acyl chlorides instead
Unlike the reactions with carboxylic acids, acyl chlorides are more reactive (so the
reactions happen faster) and their reactions go to completion (so no equilibrium mixture
is formed and the yield of the ester is maximum)
Examples of esterification reactions include:
o Formation of ethyl ethanoate from ethanol and ethanoyl chloride
o Formation of phenyl benzoate from phenol and benzoyl chloride
Formation of esters from the reaction of alcohols with acyl chlorides
250 | P a g e
7.5.6 Production & Reactions of Acyl
Chlorides
Production of Acyl Chlorides



Due to the increased reactivity of acyl chlorides compared to carboxylic acids, they are
often used as starting compounds in organic reactions
Acyl chlorides are compounds that contain an -COCl functional group and can be
prepared from the reaction of carboxylic acids with:
o Solid phosphorus(V) chloride (PCl5)
o Liquid phosphorus(III) chloride (PCl3) and heat
o Liquid sulfur dichloride oxide (SOCl2)
Propanoyl chloride can this way be prepared from propanoic acid using the reactions
above
Formation of acyl chlorides from their corresponding carboxylic acids
Reactions of Acyl Chlorides


Acyl chlorides are reactive organic compounds that undergo many reactions such
as addition-elimination reactions
In addition-elimination reactions, the addition of a small molecule across the C=O bond
takes place followed by elimination of a small molecule
251 | P a g e

Examples of these addition-elimination reactions include:
o Hydrolysis
o Reaction with alcohols and phenols to form esters
o Reaction with ammonia and amines to form amides
Hydrolysis



The hydrolysis of acyl chlorides results in the formation of a carboxylic
acid and HCl molecule
This is an addition-elimination reaction
o A water molecule adds across the C=O bond
o A hydrochloric acid (HCl) molecule is eliminated
An example is the hydrolysis of propanoyl chloride to form propanoic acid and HCl
Acyl chlorides are hydrolysed to carboxylic acids
Formation of esters




Acyl chlorides can react with alcohols and phenols to form esters
o The reaction with phenols requires heat and a base
Esters can also be formed from the reaction of carboxylic acids with phenol and alcohols
however, this is a slower reaction as carboxylic acids are less reactive and the reaction
does not go to completion (so less product is formed)
Acyl chlorides are therefore more useful in the synthesis of esters
The esterification of acyl chlorides is also an addition-elimination reaction
o The alcohol or phenol adds across the C=O bond
o A HCl molecule is eliminated
Acyl chlorides undergo esterification with alcohols and phenols to form esters
252 | P a g e
Formation of amides




Acyl chlorides can form amides from their condensation
reaction with amines and ammonia
The nitrogen atom in ammonia and amines has a lone pair of electrons which can be used
to attack the carbonyl carbon atom in the acyl chlorides
The product is a non-substituted amide (when reacted with ammonia)
or substituted amide (when reacted with primary and secondary amines)
This is also an example of an addition-elimination reaction as
o The amine or ammonia molecule adds across the C=O bond
o A HCl molecule is eliminated
Acyl chlorides undergo condensation reactions with ammonia and amines to form amides
253 | P a g e
The more chlorine atoms there are in the carboxylic acids, the stronger the acid is

Trichloroethanoic acid is the strongest acid as:
o The O-H bond in CCl3COOH is the weakest since there are three very strong
electronegative Cl atoms withdrawing electron density from the -COOH group
o When the O-H is broken to form the carboxylate (-COO-) ion, the charge density
is further spread out by the three electron-withdrawing Cl atoms
o The carboxylate ion is so stabilised that it is less attracted to H+ ions
Relative acidity of trichloroethanoic acid
254 | P a g e

Ethanoic acid is the weakest acid as:
o It contains an electron-donating methyl group which strengthens the O-H bond
o The methyl group donates negative charge towards the -COO- group which
becomes more likely to accept an H+ ion
Relative acidity of ethanoic acid
7.5.7 Addition-Elimination Reactions of Acyl
Chlorides
Mechanism of Addition - Elimination in Acyl Chloride Reactions


Acyl chlorides undergo addition-elimination reactions such as hydrolysis,
esterification reactions to form esters, and condensation reactions to form amides
The general mechanism of these addition-elimination reactions involve two steps:
o Step 1 - Addition of a nucleophile across the C=O bond
o Step 2 - Elimination of a small molecule such as HCl or H2O
Mechanism of hydrolysis of acyl chlorides

In the hydrolysis of acyl chlorides, the water molecule acts as a nucleophile
o The lone pair on the oxygen atoms carry out an initial attack on the carbonyl
carbon
o This is followed by the elimination of a hydrochloric acid (HCl) molecule
255 | P a g e
Reaction mechanism of the hydrolysis of acyl chlorides
Formation of esters: reaction mechanism



In the esterification reaction of acyl chlorides, the alcohols or phenols act as
a nucleophile
o The lone pair on the oxygen atoms carry out an initial attack on the carbonyl
carbon
o This is again followed by the elimination of an HCl molecule
With phenols, the reaction requires heat to proceed and needs to be carried out in the
presence of a base
The base deprotonates the phenol to form a phenoxide ion which is a better
nucleophile than the phenol molecule
o The phenoxide ion carries out an initial attack on the carbonyl carbon
o A small molecule of NaCl is eliminated
256 | P a g e
Reaction mechanism of the esterification of acyl chlorides with alcohols
Reaction mechanism of the esterification of acyl chlorides with phenols
Formation of amides: reaction mechanism


The nitrogen atom in ammonia and primary/secondary amines act as a nucleophile
o The lone pair on the nitrogen atoms carry out an initial attack on the carbonyl
carbon
o This is followed by the elimination of an HCl molecule
Both reactions of acyl chlorides with ammonia and amines are vigorous however there
are also differences
o With ammonia - The product is a non-substituted amide and white fumes of
HCl are formed
257 | P a g e
o
With amines - The product is a substituted amide and the HCl formed reacts
with the unreacted amine to form a white organic ammonium salt
Reaction mechanism of the formation of amides from acyl chlorides with ammonia
258 | P a g e
Reaction mechanism of the formation of amides from acyl chlorides with primary amines
Reaction mechanism of the formation of amides from acyl chlorides with secondary amines
259 | P a g e
7.5.8 Relative Ease of Hydrolysis
Hydrolysis of Acyl Chlorides, Alkyl Chlorides & Halogenoarenes




Hydrolysis is the breakdown of a compound using water
The ease of hydrolysis for different organic compounds may differ
For example, the ease of hydrolysis, starting with the compounds most readily broken
down, is: acyl chloride > alkyl chloride > aryl chloride
This trend can be explained by looking at the strength of the C-Cl
Strength of C-Cl bond in acyl chlorides




Acyl chlorides are hydrolysed most readily at room temperature
This is because the carbon bonded to the chlorine atom is also attached to an oxygen
atom
There are two strong electronegative atoms pulling electrons away from the carbonyl
carbon, leaving it very δ+
The C-Cl bond is therefore weakened and nucleophilic attack of the carbonyl carbon is
much more rapid
The hydrolysis of acyl chlorides occurs most readily
Strength of C-Cl bond in alkyl chlorides



The carbonyl carbon in alkyl chlorides is only attached to one electronegative atom
which pulls electrons away from it
This carbon atom is therefore not very δ+ and the C-Cl bond is stronger than the C-Cl
bond in acyl chlorides
The hydrolysis of alkyl chlorides, therefore, requires a strong alkali (such as OH-) to
be refluxed with it
260 | P a g e

An OH- ion will hydrolyse the alkyl chloride as it a stronger nucleophile than H2O
The hydrolysis of alkyl chlorides requires a strong nucleophile
Strength of C-Cl bond in aryl chlorides




In aryl chlorides, the carbon atom bonded to the chlorine atom is part of the delocalised π
bonding system of the benzene ring
One of the lone pairs of electrons of the Cl atom overlaps with this delocalised system
The C-Cl bond, therefore, has some double-bond character causing it to
become stronger
As a result, the C-Cl bond is difficult to break and hydrolysis will not occur
Due to the strong C-Cl bond in aryl chlorides, these compounds will not undergo hydrolysis
261 | P a g e
7.6.1 Production of Amines
Production of Primary & Secondary Amines


Primary amines are organic compounds that have an -NH2 functional group attached to
an alkyl or aryl group
Secondary amines have two alkyl or aryl groups attached to an -NH group
Primary and secondary amines

Primary and secondary amines can be prepared from different reactions including:
o The reaction of halogenoalkanes with ammonia
o The reaction of halogenoalkanes with primary amines
o The reduction of amides
o The reduction of nitriles
Reaction of halogenoalkanes with ammonia


This is a nucleophilic substitution reaction in which the nitrogen lone pair in ammonia
acts as a nucleophile and replaces the halogen in the halogenoalkane
When a halogenoalkane is reacted with excess, hot ethanolic ammonia under
pressure a primary amine is formed
Formation of primary amine
262 | P a g e
Reaction of halogenoalkanes with primary amine


This is also a nucleophilic substitution reaction in which the nitrogen in the primary
amine acts as a nucleophile and replaces the halogen in the halogenoalkane
When a halogenoalkane is reacted with a primary amine in ethanol and heated in a
sealed tube, under pressure a secondary amine is formed
Formation of secondary amine
Reduction of amides


Amines can also be formed from the reduction of amides by LiAlH4 in dry ether
Whether a primary or secondary amine is formed depends on the nature of the amide
Amides can be reduced by LiAlH4 to form amines
263 | P a g e
Reduction of nitriles


Nitriles contain a -CN functional group which can be reduced to an -NH2 group
The nitrile vapour and hydrogen gas are passed over a nickel catalyst or LiAlH4 in dry
ether can be used to form a primary amine
Nitriles can be reduced with LiAlH4 or H2 and Ni catalyst
264 | P a g e
7.6.2 Production of Amides
Production of Amides




Amides are organic compounds with an -CONR2 functional group
They can be prepared from the condensation reaction between an acyl
chloride and ammonia or amine
In a condensation reaction, two organic molecules join together and in the
process eliminate a small molecule
In this case, the acyl chlorides and ammonia or amine join together to form
an amide and eliminate an HCl molecule
Condensation reaction







The chlorine atom in acyl chlorides is electronegative and draws electron density from
the carbonyl carbon
The carbonyl carbon is therefore electron-deficient and can be attacked by nucleophiles
The nitrogen atom in ammonia and amines has a lone pair of electrons which can act as
a nucleophile and attack the carbonyl carbon
As a result, the C-Cl bond is broken and an amide is formed
Whether the product is a substituted amide or not, depends on the nature of
the nucleophile
o Primary and secondary amines will give a substituted amide
o The reaction of acyl chlorides with ammonia will produce a non-substituted
amide
Note that amides can also be formed from the condensation reaction between carboxylic
acids and ammonia or amines
However, this reaction is slower as carboxylic acids are less reactive than acyl
chlorides and the reaction doesn’t go to completion
Acyl chlorides undergo condensation reactions with ammonia and amines to form amides
265 | P a g e
7.6.3 Basicity of Amines
Basicity of Aqueous Solutions of Amines


The nitrogen atom in ammonia and amine molecules can accept a proton (H+ ion)
They can therefore act as bases in aqueous solutions by donating its lone pair of
electrons to a proton and form a dative bond
o For example, ammonia undergoes an acid-base reaction with dilute hydrochloric
acid (HCl) to form a salt
NH3 + HCl → NH4+Clbase
acid
salt
The nitrogen atom in ammonia and amines can donate its lone pair of electrons to form a
bond with a proton and therefore act as a base
266 | P a g e
Strength of ammonia and amines as bases




The strength of amines depends on the availability of the lone pair of electrons on the
nitrogen atom to form a dative covalent bond with a proton
The more readily this lone pair of electrons is available, the stronger the base is
Factors that may affect the basicity of amines include:
o Positive inductive effect - Some groups such as alkyl groups donate electron
density to the nitrogen atom causing the lone pair of electrons to become more
available and therefore increasing the amine’s basicity
o Delocalisation - The presence of aromatic rings such as the benzene ring causes
the lone pair of electrons on the nitrogen atom to be delocalised into the benzene
ring
o The lone pair becomes less available to form a dative covalent bond with
ammonia and hence decreases the amine’s basicity
For example, ethylamine (which has an electron-donating ethyl group) is more
basic than phenylamine (which has an electron-withdrawing benzene ring)
7.6.4 Production & Reactions of Phenylamine
Preparation of Phenylamine


Phenylamine is an organic compound consisting of a benzene ring and an amine (NH2)
functional group
It can be produced in a three-step synthesis reaction followed by the separation of
phenylamine from the reaction mixture
o Step 1- Benzene undergoes nitration with concentrated nitric acid (HNO3) and
concentrated sulfuric acid (H2SO4) at 25 to 60 oC to form nitrobenzene
o Step 2 - Nitrobenzene is reduced with hot tin (Sn) and concentrated
hydrochloric acid (HCl) under reflux to form an acidic mixture that contains the
organic product C6H5N+H3
o Step 3 - Sodium hydroxide (NaOH) is added to the acidic reaction mixture to
form phenylamine
o Step 4 - The phenylamine is separated from the reaction mixture by steam
distillation
The overall reaction of formation of phenylamine from benzene
267 | P a g e
Multi-step synthesis of phenylamine from benzene
268 | P a g e
Reactions of Phenylamine


Both the benzene ring as well as the -NH2 group in phenylamine can take part in
chemical reactions
These reactions include
o The bromination of phenylamine
o Formation of a diazonium salt
Bromination of phenylamine






Phenylamines react in electrophilic substitution reactions in a similar way as phenols
The lone pair of electrons on the nitrogen atom in phenylamines donate electron density
into the benzene ring
o In phenols, the oxygen atom donates its lone pair of electrons instead
The delocalisation of the electrons causes an increased electron density in the benzene
ring
The benzene ring, therefore, becomes activated and becomes more readily attacked
by electrophiles
The incoming electrophiles are directed to the 2,4 and 6 positions
Phenylamines, therefore, react under milder conditions with aqueous bromine at room
temperature to form 2,4,6-tribromophenylamine
Bromination of phenylamines gives 2,4,6-tribromophenylamine
Formation of diazonium salt



Diazonium compounds are very reactive compounds containing an -N2+ group
The amine (-NH2) group of phenylamines will react with nitric(III) acid (HNO3) at a
temperature below 10 °C to form diazonium salts
o Since nitric(III) acid is unstable, it has to be made in the test-tube by reacting
sodium nitrate (NaNO2) and dilute acid (such as HCl)
These diazonium salts are so unstable that they will upon further warming with water to
form a phenol
269 | P a g e
Phenylamine can form an unreactive diazonium salt which thermally decomposes to a phenol
270 | P a g e
7.6.5 Relative Basicity of Ammonia,
Ethylamine & Phenylamine
Relative Basicity of Aqueous Ammonia, Ethylamine & Phenylamine




Ammonia and amines act as bases as they can donate their lone pair of electrons to form
a dative covalent bond with a proton
The basicity of the amines depends on how readily available their lone pair of electrons
is
Electron-donating groups (such as alkyl groups) increase the electron density on the
nitrogen atom and cause the lone pair of electrons to become more available for dative
covalent bonding
o The amine becomes more basic
Delocalisation of the lone pair of electrons into an aromatic ring (such as a benzene ring)
causes the lone pair of electrons to become less available for dative covalent bonding
o The amine becomes less basic
Comparing basicity of ammonia, ethylamine & phenylamine

The order of basicity of ammonia, ethylamine and phenylamine is as follows:
Ethylamine > ammonia > phenylamine
STRONGEST BASE




WEAKEST BASE
This trend can be explained by looking at the groups attached to the amine (-NH2) group
In ethylamine, the electron-donating alkyl group donates electron density to the nitrogen
atom causing its lone pair to become more available to form a dative covalent bond with
a proton
Ammonia lacks an electron-donating group hence it is less basic than ethylamine
however it is more basic than phenylamine as the lone pair on the nitrogen is not
delocalised
In phenylamine the lone pair of electrons overlap with the conjugated system on the
benzene ring and become delocalised; As a result, the lone pair of electrons become less
readily available to form a bond with a proton
271 | P a g e
Trends in the basicity of ammonia, ethylamine, and phenylamine
272 | P a g e
7.6.6 Azo Compounds
Azo Compounds


Azo (or diazonium) compounds are organic compounds that have an R1-N=N-R2 group
They are often used as dyes and are formed in a coupling reaction between
the diazonium ion and an alkaline solution of phenol
Azo compounds are characterized by the presence of an R1-N=N-R2 group
Coupling of benzenediazonium chloride with phenol in NaOH


Azo compounds can be formed from the coupling reaction of a benzenediazonium
chloride salt with alkaline phenol
Making an azo dye is a multi-step process
Formation of azo compounds table
273 | P a g e
Reaction mechanism of the formation of azo compounds


The delocalised electrons in the π bonding systems of the two benzene rings
are extended through the -N=N- which acts as a bridge between the two rings
As a result of the delocalisation of electrons throughout the compound, azo compounds
are very stable
274 | P a g e
Making other azo dyes


Other dyes can be formed via a similar route as described above
For example, the yellow dye can be formed from the coupling
reaction between benzenediazonium chloride and C6H5N(CH3)2 instead of phenol
(C6H5OH)
The yellow azo dye is formed via a coupling reaction between benzenediazonium chloride and
C6H5N(CH3)2
7.6.7 Reactions of Amides
Reactions of Amides


Amides are formed from the condensation reaction of carboxylic acids or acyl
chlorides with ammonia or amines
The amide group (CONR2) in these compounds can undergo reactions including
o Hydrolysis with aqueous alkali or aqueous acid
o Reduction with LiAlH4
Hydrolysis of amides




The -CON- group in substituted amides links two hydrocarbon sections of their
molecules together
This amide link can be broken down by hydrolysis by refluxing it with an acid or alkali
The products of a non-substituted amide are:
o Carboxylic acid
o Ammonia
The products of a substituted amide are:
o Carboxylic acid
o Primary amine
275 | P a g e
Hydrolysis of substituted and non-substituted amides


Ammonia will react in excess acid to form an ammonium salt
Carboxylic acid will get deprotonated in excess base to form a carboxylate ion
Amides are hydrolysed to carboxylic acids and ammonia or primary amine when refluxed with
acid or alkali
276 | P a g e
Reduction of amides



The C=O group in amides can be reduced by the strong reducing agent LiAlH4 to form
an amine
The products of a non-substituted amide are:
o A primary amine and water
The products of a substituted amide are:
o A secondary amine and water
Amides can be reduced to amines using LiAlH4
277 | P a g e
7.6.8 Relative Basicity of Amides & Amines
Relative Basicity of Amides & Amines


A base is a species that can donate its lone pair of electrons to form a dative covalent
bond with another species
Amines are basic as the nitrogen atom has a lone pair of electrons which can form a
dative covalent bond with an
electron-deficient species (such as an H+ ion)




The basicity of the amine depends on the availability of this lone pair of electrons
o The more readily available the lone pair of electrons is for dative covalent
bonding, the stronger the base
o The less readily available the lone pair of electrons is, the weaker the base
Electron-donating groups such as alkyl groups increase the electron density on the
nitrogen atom causing the lone pair to become more available
Electron-withdrawing groups such as aromatic benzene rings, cause delocalisation of
the lone pair of electrons which become less readily available
This is why phenylamine (which contains an electron-withdrawing benzene ring) is
a weaker base than propylamine (which contains an electron-donating alkyl group)
Basicity of amides





Amides also contain a nitrogen atom with a lone pair of electrons
Again, the basicity of the amide depends on the availability of this lone pair for dative
covalent bonding
Due to the presence of the electron-withdrawing oxygen atom in the amide group,
electron density is removed from the nitrogen atom
The lone pair on the nitrogen atom, therefore, becomes less readily available and is not
available to donate to an electron-deficient species
Since this electron-withdrawing oxygen is characteristic of amides and is not present in
amines, amides are much weaker bases than amines
7.6.9 Amino Acids
Acid / Base Properties of Amino Acids, Zwitterions & the Isoelectric
Point


Amino acids are organic compounds that contain two functional groups:
o A basic amino (-NH2) group
o An acidic carboxylic acid (-COOH) group
Due to the presence of both a basic and acidic group in amino acids, they are said to
be amphoteric
278 | P a g e
o
They can act as both acids and bases
Naturally occurring amino acids



2-aminocarboxylic acids are a type of amino acids in which the amine (-NH2) group is
bonded to the carbon atom next to the -COOH group
These type of amino acids form the ‘building blocks’ that make up proteins
There are 20 naturally occurring amino acids with the general structural formula
of RCH(NH2)COOH
General structural formula of amino acids

The R group varies in different amino acids and can be:
o Acidic
o Basic
o Neutral
The R group varies in different amino acids
Acid / base properties of amino acids

Amino acids will undergo most reactions of amines and carboxylic acids including acidbase reactions of:
279 | P a g e
o
o



Amines with acids
Carboxylic acids with bases
However, they can also interact intramolecularly (within themselves) to form
a zwitterion
A zwitterion is an ion with both a positive (-NH3+) and a negative (-COO-) charge
Because of these charges in a zwitterion, there are strong intermolecular forces of
attraction between amino acids
o Amino acids are therefore soluble crystalline solids
An amino acid molecule can interact within itself to form a zwitterion
Isoelectric point




A solution of amino acids in water will exist as zwitterions with
both acidic and basic properties
They act as buffer solutions as they resist any changes in pH when small amounts of
acids or alkali are added
If an acid is added (and thus the pH is lowered):
o The -COO- part of the zwitterion will accept an H+ ion to reform the -COOH
group
o This causes the zwitterion to become a positively charged ion
If a base is added (and thus the pH is raised):
o The -NH3+ part of the zwitterion will donate an H+ ion to reform the -NH2 group
o This causes the zwitterion to become a negatively charged ion
280 | P a g e
A solution of amino acids can act as a buffer solution by resisting any small changes in pH

The pH can be slightly adjusted to reach a point at which neither the negatively
charged or positively charged ions dominate and the amino acid exists as a neutral
zwitterion
o This is called the isoelectric point of the amino acid
The isoelectric point of amino acids is the pH at which the amino acid exists as a neutral
zwitterion
281 | P a g e
7.6.10 Peptide Bonds
Formation of Peptide Bonds




Each amino acid contains an amine (-NH2) and carboxylic acid (-COOH) group
The -NH2 group of one amino acid can react with the -COOH group of another amino
acid in a condensation reaction to form a dipeptide
o The new amide bond between two amino acids is also called a peptide
link or peptide bond
Since this is a condensation reaction, a small molecule (in this case H2O) is eliminated
The dipeptide still contains an -NH2 and -COOH group at each end of the molecule
which can again participate in a condensation reaction to form a tripeptide
A peptide bond is an amide bond between two amino acids

A polypeptide is formed when many amino acids join together to form a long chain of
molecules
282 | P a g e
A polypeptide is a long chain of amino acid molecules joined together
7.6.11 Electrophoresis
Electrophoresis




Electrophoresis is an analytical technique which separates ions by placing them in an
electrical field
o This method is often used in biochemical analysis to identify and purify proteins
A sample of amino acids is placed between two oppositely charged electrodes
o The positively charged ions will move towards the negative electrode
o The negatively charged ions will move towards the positive electrode
The rate (how fast) at which the ions move towards the electrodes depends on:
o The size of the ions: larger ions move more slowly
o The charge of the ions: highly charged ions move more quickly
An electropherogram is the series of bands which are observed on the paper or gel
after electrophoresis has occurred
o Each band in the electropherogram corresponds to a particular species
Separating mixtures of amino acids by varying the pH



The charge on the amino acid ions depends on the pH of the solution
The movement of the ions to the electrodes during electrophoresis will therefore be
affected by the pH
Consider a sample which consists of a mixture of three amino acids at pH 7
o Amino acid A: lysine, side-chain is positively charged
o Amino acid B: glycine, side chain is neutral
o Amino acid C: glutamic acid, side chain is negatively charged
283 | P a g e
The sample consists of a mixture of three amino acids which are separated using
electrophoresis


The amino acids in this mixture can be separated by electrophoresis
o Amino acid C will move towards the positive electrode
o Amino acid B will remain in the well where the sample is applied to the gel
o Amino acid A will move towards the negative electrode
Since glutamic acid is larger than lysine, it will travel towards the positive electrode at
a slower rate compared to lysine
Separation of a mixture of amino acids by electrophoresis
284 | P a g e
POLYMERISATION
7.7.1 Formation of Polyesters
Formation of Polyesters


Addition polymerisation has been covered in reactions of alkenes
o They are made using monomers that have C-C double bonds joined together to
form polymers such as (poly)ethene
Condensation polymerisation is another type of reaction and is used in the making of
polyesters
o A small molecule (eg. a water molecule) is lost when the monomers join together
to form a polyester
o Polyesters contain ester linkages
This polymer structure shows an ester functional group linking monomers together
Formation of polyesters

A diol and a dicarboxylic acid are required to form a polyester
o A diol contains 2 -OH groups
o A dicarboxylic acid contains 2 COOH groups
285 | P a g e
The position of the functional groups on both of these molecules allows condensation
polymerisation to take place effectively


When the polyester is formed, one of the -OH groups on the diol and the hydrogen atom
of the -COOH are expelled as a water molecule (H2O)
The resulting polymer is a polyester
Expulsion of a water molecule in this condensation polymerisation forms the polyester called
Terylene (PET)
286 | P a g e
Hydroxycarboxylic acids




So far the examples of making polyesters have focused on using 2 separate monomers for
the polymerisation
There is another route to making polyesters
A single monomer containing both of the key functional groups can also be used
These monomers are called hydroxycarboxylic acids
o They contain an alcohol group (-OH) at one end of the molecule while the other
end is capped by a carboxylic acid group (-COOH)
Both functional groups are needed to make a polyester are from the same monomer
Exam Tip


Polyesters can be made using condensation polymerisation
The monomers needed are diols and dicarboxylic acids/dioyl chlorides or a single
hydroxycarboxylic acid monomer
287 | P a g e
7.7.2 Formation of Polyamides
Formation of Polyamides
Amide link

Polyamides are also formed using condensation polymerisation
An amide link - also known as a peptide link - is the key functional group in a polyamide
Monomers



A diamine and a dicarboxylic acid are required to form a polyamide
o A diamine contains 2 -NH2 groups
o A dicarboxylic acid contains 2 -COOH groups
Dioyl dichlorides can also used to react with the diamine instead of the acid
o A dioyl chloride contains 2 -COCl groups
This is a more reactive monomer than dicarboxylic acid. However, a more expensive
alternative
288 | P a g e
The monomers for making polyamides
Formation of polyamides
This shows the expulsion of a small molecule as the amide link forms
289 | P a g e


Nylon 6,6 is a synthetic polyamide
Its monomers are 1,6-diaminohexane and hexane-1,6-dioic acid
o The ‘6,6’ part of its name arises from the 6 carbon atoms in each of Nylon 6,6
monomers
Nylon 6,6 is a synthetic polyamide made using diamine and dicarboxylic acid monomers
Kevlar






Kevlar is another example of a polymer formed through condensation polymerisation
The polymer chains are neatly arranged with many hydrogen bonds between them
This results in a strong and flexible polymer material with fire resistance properties
These properties also lend Kevlar to a vital application in bullet-proof vests
The monomers used to make Kevlar
o 1,4-diaminobenzene
o Benzene-1,4-dicarboxylic acid
As seen with Nylon, a dioyl chloride can be used instead of the acid as well (benzene-1,4dioyl chloride)
290 | P a g e
Kevlar is made using a diamine and dicarboxylic acid monomers
291 | P a g e
Aminocarboxylic acids





So far, condensation polymerisation has covered the use of monomers that contain 2 of
the same functional group (eg. diamine, Diol etc.)
It is possible to carry out a condensation polymerisation where one monomer provides
both of the function groups necessary for an amide/peptide link
For example 6-aminohexanoic acid has an amino group and a carboxylic acid group on
the same molecule
Molecules like this are called amino carboxylic acids
They are able to polymerise to form a structure similar to Nylon 6,6
6-aminohexanoic acid can be polymerised to make the synthetic polymer Nylon 6,6
Making Proteins




Proteins are vital biological molecules with varying functions within the body
They are essentially polymers made up of amino acid monomers
Amino acids have an aminocarboxylic acid structure
Their properties are governed by a branching side group - the R group
292 | P a g e
Amino acids contain an amine group, an acid group and a unique R group





Different amino acids are identified by their unique R group
The names of each amino acid is given using 3 letters
For example Glutamine is known as ‘Gln’
Dipeptides can be produced by polymerising 2 amino acids together
o The amine group (-NH2) and acid group (-COOH) of each amino acid is used to
polymerise with another amino acid
Polypeptides are made through polymerising more than 2 amino acids together
Dipeptides and polypeptides are formed by polymerising amino acid molecules together
293 | P a g e
Protein hydrolysis


Proteins (polypeptides) can be broken down into its constituent amino acids
This process occurs through a hydrolysis reaction
Proteins are broken down by hydrolysis
Exam Tip
Become familiar with the structures of the different monomers that can be used to make
condensation polymers.Also, remember that exam questions will require you to identify the key
functional groups and also draw small sections of polymers.
294 | P a g e
7.7.3 Repeat Units & Monomers
DOWNLOAD PDF
Deducing the Repeat Unit of a Condensation Polymer


In condensation polymerisation the monomers either contain 2 of the same functional
group or one single monomer has both functional groups needed for polymerisation
o For example Diamine and dicarboxylic acid
o Or an aminocarboxylic acid
When presented with 2 monomers there are steps to take in order to deduce the repeat
unit of a condensation polymer
Exam Tip



Remember: in condensation polymerisation, a small molecule is expelled as a result of
the 2 monomers joining together.
When a dioic acid and diamine polymerise, a water molecule is expelled
o OH from acid and H from the amine
When a dioyl acid and diamine are polymerised, a hydrochloric acid molecule is expelled
o Cl from the dioyl acid and H from the amine
Identifying Monomers in Condensation Polymers




When a section of polymer is presented, the monomers can be identified by considering
the small molecules expelled from the monomers
If a water molecule is expelled, the -OH must have been from an acid group
The hydrogen atom may be from an amine group of a monomer.
If the molecule was hydrochloric acid (HCl), a dioyl chloride monomer may have been
used
295 | P a g e
7.7.4 Predicting & Deducing the Type of
Polymerisation
DOWNLOAD PDF
Predicting Type of Polymerisation


When a set of monomers are given in an exam question, the type of polymerisation can
be determined
Firstly, it’s important to identify the key functional groups in the monomers
Addition polymerisation



If the monomer/s contain a C=C double bond, they will polymerise through addition
polymerisation
The double bond can open up in order to add more monomers either side of the starting
monomer
This type of polymerisation makes (poly)alkenes
Addition polymerisation of one alkene monomer is polymerised, a (poly)alkene is formed

(Poly)alkenes can be produced if there are 2 or more alkene monomers as well
296 | P a g e

When more than one monomer is used for addition polymerisation, the resulting product
is known as a copolymer
Two or more different alkene monomers can also be polymerised in Addition polymerisation.
This gives a co-polymer
Condensation polymerisation


Condensation polymerisation makes polyamides and polyesters
When looking to identify this type of polymerisation, there are some key functional
groups to be aware of
297 | P a g e
Monomers for condensation polymers table
Exam Tip



As well as the functional groups to be aware of, know that a small molecule is expelled
when the polymer is formed
Identify 2 functional groups that can react together to produce either a polyamide or a
polyester
There are instances where both of the functional groups are on the same monomer
molecule
o For example amino acid molecules contain an amine group (-NH2) and a
carboxylic acid group (-COOH) therefore it can polymerise to produce a
polyamide
298 | P a g e
Deducing Type of Polymerisation

The type of polymerisation can be determined by considering the structure of the polymer
backbone
Identifying addition polymerisation




The polymer backbone of an addition polymer does not contain functional groups
The backbone of the polymer is generally a chain of carbon atoms
There may be sidechains branching off from the backbone
Some examples of side chains are benzene rings, nitrile groups (-CN) and halogen atoms
(-F/-Cl/-Br/-I)
Addition polymers are identified using the plain carbon chain as the polymer backbone
Identifying condensation polymerisation


A condensation polymer can be identified by functional groups on the polymer backbone
Polyesters contain ester links and polyamides contain amide/peptide link on the backbone
itself
Condensation polymers are identified using function groups that form parts of the polymer
backbone
299 | P a g e
Exam Tip



Different sections of polymer chains may be formed using various type of polymerisation
In an exam, you may be given a section of a polymer and asked to determine the type of
polymerisation used to form that section
Firstly, look at the polymer backbone
o If there are functional groups along the backbone, that section was made using
condensation polymerisation
o If there are no functional groups along the backbone, addition polymerisation was
used
7.7.5 Degradabiity of Polymers
DOWNLOAD PDF
Poly(alkenes) & Biodegradability





Many of the polymers in use have been produced through addition polymerisation of
alkenes
The (poly)alkene chains are non-polar and saturated
This makes them chemically inert and therefore non-biodegradable
(poly)alkenes can be melted and recycled into new uses
o However, even in the new applications, the (poly)alkenes are not biodegradable
Recycling plants can burn used plastic materials
o The energy released from burning can be used to generate electricity
o Burning plastics in oxygen releases carbon dioxide and water (complete
combustion) which can contribute to global warming
Photodegradation of Polymers




Polyesters and polyamides are biodegradable polymers for a number of reasons
One such reason is their ability to breakdown with the use of light
Carbonyl groups (C=O) along polymer chains are able to absorb energy from the
Electromagnetic Spectrum
o In particular Ultraviolet (UV) light
Absorbing UV light weakens the carbonyl areas of polymers and breaks them down into
smaller molecules
Disadvantages of photo degradability

Despite this ability being a great advantage of polyesters and polyamides, it may pose a
problems when the polymers are repurposed
300 | P a g e


When applied to a new use, the biodegradability could give a weaker polymer
Breaking down polymers also poses another challenge
o Once used, polymeric materials are taken to landfill sites where many other
materials are piled on top of each other
o This could mean that photodegradable polyesters or polyamides do not have
access to UV light in order to break down naturally
Biodegrading Polyesters & Polyamides
Biodegradable polymers



Both polyesters and polyamides can be broken down using hydrolysis reactions
This is a major advantage over the polymers produced using alkene monomers
(polyalkenes)
When polyesters and polyamides are taken to landfill sites, they can be broken down
easily and their products used for other applications
Hydrolysis of polyamides



Hydrolysis is a breaking up of a molecules using water
In acidic hydrolysis, acid (such as hydrochloric acid) acts as the catalyst
o Polyamides are heated with dilute acid
o This reaction breaks the polyamide into carboxylic acid molecules and
ammonium chloride ions
Alkaline hydrolysis
o The polyamide is heated with a species containing hydroxide ions (eg. sodium
hydroxide)
o This breaks the polymer into the sodium salts of its monomers (dicarboxylic acids
and diamines)
o If the poly amide link used an aminocarboxylic acid as the monomer, then a
sodium salt of the original amino acid is reformed
301 | P a g e
When polyamides are degraded by hydrolysis, carboxylic acids and amines are formed
302 | P a g e
ORGANIC SYNTHESIS

Students should be able to identify organic functional groups and recall their
properties and the reactions that they undergo
Properties of functional groups

In addition to the functional groups mentioned in the AS course, students should also
be familiar with additional functional groups and their properties including:
o Arenes
o Halogenoarenes
o Carboxylic acids (and derivatives)
o Phenols
o Amides
o Amino acids
o Acyl chlorides
Overview of additional functional groups and their properties
303 | P a g e
Reactions of functional groups

Students should also be able to recall:
o The reactions by which these functional groups can be produced
o The reactions that these functional groups undergo
Reactions by which functional groups can be produced
304 | P a g e
305 | P a g e
306 | P a g e
Hydrolysis of polyesters


Ester linkages can also be degraded through hydrolysis reactions
Acid hydrolysis forms the alcohols and carboxylic acids that were used to form the
polyesters
When polyesters are degraded by hydrolysis, carboxylic acids and alcohols are formed
307 | P a g e
ANALYTICAL CHEMISTRY
8.1.1 Thin-Layer Chromatography
Thin Layer Chromatography: Basics






Thin Layer Chromatography (TLC) is a technique used to analyse small samples via
separation
o For example, we could separate a dye out to determine the mixture of dyes in a
forensic sample
There are 2 phases involved in TLC - stationary phase and mobile phase
Stationary phase
o This phase is commonly thin metal sheet coated in alumina (Al2O3) or silica (SiO2)
o The solute molecules adsorb onto the surface
o Depending on the strength of interactions with the stationary phase, the separated
components will travel particular distances through the plate
o The more they interact with the stationary phase, the more they will 'stick' to it
Mobile phase
o Flows over the stationary phase
o It is a polar or nonpolar liquid (solvent) or gas that carries components of the
compound being investigated
o Polar solvents - water or alcohol
o Non-polar solvents - alkanes
If the sample components are coloured, they are easily identifiable
We can examine the plate under UV light using ninhydrin to identify uncoloured
components
Conducting a TLC analysis

Step 1:
Prepare a beaker with a small quantity of solvent

Step 2:
On a TLC plate, draw a horizontal line at the bottom edge (in pencil)
This is called the baseline
308 | P a g e

Step 3:
Place a spot of pure reference compound on the left of this line, then a spot of the sample
to be analysed to the right of the baseline and allow to air dry
The reference compounds will allow identification of the mixture of compounds in the
sample

Step 4:
Place the TLC plate inside the beaker with solvent - making sure that the pencil baseline
is lower than the level of the solvent - and place a lid to cover the beaker
The solvent will begin to travel up the plate, dissolving the compounds as it does

Step 5:
As solvent reaches the top, remove the plate and draw another pencil line where the
solvent has reached, indicating the solvent front
The sample’s components will have separated and travelled up towards this solvent front
A dot of the sample is placed on the baseline and allowed to separate as the mobile phase flows
through the stationary phase; The reference compound/s will also move with the solvent
309 | P a g e
Rf values

A TLC plate can be used to calculate Rf values for compounds

These values can be used alongside other analytical data to deduce composition of
mixtures
Rf values can be calculated by taking 2 measurements from the TLC plate
Exam Tip
The baseline on a TLC plate must be drawn in pencil. Any other medium would interact with the
sample component and solvents used in the analysis process.
310 | P a g e
Interpreting & Explaining Rf Values in TLC




The less polar components travel further up the TLC plate
o Their Rf values are higher than those closer to the baseline
o They are more soluble in the mobile phase and get carried forwards with the
solvent
More polar components do not travel far up the plate
o They are more attracted to the polar stationary phase
The extent of separating molecules in the investigated sample depends on the solubility in
the mobile and stationary phases
Knowing the Rf values, of compounds being analysed, helps to compare the polarity of
various molecules
8.1.2 Gas/Liquid Chromatography: Basics
Gas/Liquid Chromatography: Basics



Gas-Liquid Chromatography (GLC) is used for analysing:
o Gases
o Volatile liquids
o Solids in their vapour form
The stationary phase:
o This method uses a column for the stationary phase
o A non-polar, long-chain, non-volatile hydrocarbon with a high boiling point is
mounted onto a solid support
o Small silica particles can be packed into a glass column to offer a large surface
area
o Sample gas particles travel through this phase and are able to separate well due to
the large surface area
The Mobile phase
o An inert carrier gas (eg. Helium, Nitrogen) moves the sample molecules through
the stationary phase
Retention times



Once sample molecules reach the detector, their retention times are recorded
o This is the time taken for a component to travel through the column
The retention times are recorded on a chromatogram where each peak represents a
volatile compound in the analysed sample
Retention times are then compared with data book values to identify unknown molecules
311 | P a g e
A gas chromatogram of a volatile sample compound has six peaks. Depending on each
molecule’s interaction with the stationary phase, each peak has its own retention time
312 | P a g e
8.1.3 Interpreting Rf Values in GL
Chromatography
Interpreting Rf Values in GL Chromatography
Features of a gas-liquid chromatogram


Peaks represent different molecules from the sample - each roughly taking the shape of a
triangle
The area under each peak is the relative concentration of each component (the peak
integration value)
Area under the peak = ½ x base x height

If the area under each peak is very small or too difficult to decipher, the height of peaks
are used for further analysis
To find the area under each peak, treat each peak as a triangle - see the examples shown
using blue triangles in the diagram
313 | P a g e
Percentage composition of a mixture


We can calculate the amount of a particular molecule in a sample by using an expression
If a chromatogram shows peaks for alcohols A, B, C and D, to calculate the %
composition of alcohol C, use this expression:
Explain Retention Times




Retention time is the time taken for a sample molecule to travel through the column, from
the time it is inserted into the machine to the time it is detected
Molecules in the gaseous mixture travel at different rates, therefore giving rise to
different retention times
Longer retention times are associated with:
o Non-polar components in the mixture
o They are more attracted to the non-polar liquid in the stationary phase
o So non-polar molecules travel slower through the column
Shorter retention times are associated with:
o Polar components in the mixture that prefer to interact with the carrier gas
o They are less attracted to the non-polar liquid in the stationary phase
o So polar molecules travel faster through the column
o These molecules may have lower boiling points, therefore are vapourised more
readily
314 | P a g e
8.1.4 Interpreting & Explaining Carbon-13
NMR Spectroscopy
Interpreting & Explaining Carbon-13 NMR Spectra





Nuclear Magnetic Resonance (NMR) spectroscopy is used for analysing organic
compounds
Atoms with odd mass numbers usually show signals on NMR
o For example isotopes of atoms
o Many of the carbon atoms on organic molecules are carbon-12
o A small quantity of organic molecules will contain the isotope carbon-13 atoms
o These will show signals on a 13C NMR
In 13C NMR, the magnetic field strengths of carbon-13 atoms in organic compounds are
measured and recorded on a spectrum
Just as in 1H NMR, all samples are measured against a reference compound –
Tetramethylsilane (TMS)
On a 13C NMR spectrum, non-equivalent carbon atoms appear as peaks with different
chemical shifts
Chemical shift values (relative to the TMS) for 13C NMR analysis table
315 | P a g e
Features of a 13C NMR spectrum



C NMR spectrum displays sharp single signals – there aren’t any complicated spitting
pattern as seen with 1H NMR spectra
The height of each signal is not proportional to the number of carbon atoms present in a
single molecular environment
CDCl3 is used as a solvent to dissolve samples for 13C NMR
o On spectra, a single solvent peak appears at 80 ppm caused by 13C atoms in the
CDCl
o This can be ignored when interpreting 13C spectra
13
Identifying 13C molecular environments


On an organic molecule, the carbon-13 environments can be identified in a similar way to
the proton environments in 1H NMR
For example propanone
o There are 2 molecular environments
o 2 signals will be present on its 13C NMR spectrum
There are 2 molecular environments in propanone
316 | P a g e
The 13C NMR of propanone showing 2 signals for the 2 molecular environments
8.1.6 Use of Tetramethylsilane (TMS)
Use of Tetramethylsilane (TMS)





In NMR spectroscopy, Tetrametylsilane (TMS) is used as a reference compound
The organic compound is dissolved in TMS before being introduced to the magnetic field
of the spectrometer
It is an ideal chemical to use as a reference
o TMS is inert and volatile
o This reduces undesirable chemical reactions with the compound to be analysed
o It also mixes well with most organic compounds
TMS gives a single sharp peak on the NMR spectrum and is given a value of zero
The molecular formula of TMS is Si(CH3)4
o There are 12 hydrogens in this molecule
o All of the protons are in the same molecular environment. Therefore gives rise to
just one peak
o This peak has a very high intensity as it is accounting for the absorption of energy
from 12 1H nuclei
317 | P a g e
Tetramethylsilane (TMS) – Si(CH3)4


When peaks are recorded from the sample compound, they are measured and recorded by
their shift away from the sharp TMS peak
This gives rise to the chemical shift values for different 1H environments in a molecule
H NMR spectrum for TMS showing it’s signal at 0 ppm
1
8.1.7 Deuterated Solvents in Proton NMR
Deuterated Solvents in Proton NMR




When samples are analysed through NMR spectroscopy, they must be dissolved in a
solvent
Tetramethylsilane (TMS) is a commonly used solvent in NMR
Despite TMS showing one sharp reference peak on NMR spectra, the proton atoms can
still interfere with peaks of a sample compound
To avoid this interference, solvents containing Deuterium can be used instead
o For example CDCl3
o Deuterium (2H) is an isotope of hydrogen (1H)
318 | P a g e


Deuterium nuclei absorb radio waves in a different region to the protons analysed in
organic compounds
Therefore, the reference solvent peak will not interfere with those of the sample
Use of D20 in Identifying O-H & N-H Protons
Identifying -OH & -NH signals on a 1H NMR spectrum



The proton in -OH is not affected by its neighbouring molecular environments
As a result, the hydrogen atom of -OH group appears as a singlet
This is due to the hydrogen atom readily exchanging with hydrogens atoms of water
molecules or any acid that may be present

When interpreting 1H NMR spectra of amines and amides, the same exchanging
phenomenon can be seen
Protons of these functional groups exchanging leads to changes in their chemical shift
ranges
This table shows the range of chemical shifts for -OH and -NH- protons
Their surrounding molecular environment has a direct impact on this range



The range of chemical shifts for -OH & -NH- protons table
Using D2O


Deuterium oxide (D2O) can be added to correctly identify -OH and -NH- protons
Adding a small quantity of this solvent ‘removes’ the peaks from the spectrum
The -OH proton and the -NH proton both undergo the same exchanging process as seen
before. This time with a deuterium atom (D) of D2O
319 | P a g e
8.1.5 Proton (1H) NMR Spectroscopy
Interpreting & Explaining Proton (¹H) NMR Spectra





Nuclear Magnetic Resonance (NMR) spectroscopy is used for analysing organic
compounds
Atoms with odd mass numbers usually show signals on NMR
In 1H NMR, the magnetic field strengths of protons in organic compounds are measured
and recorded on a spectrum
Protons on different parts of a molecule (in different molecular environments) emit
different frequencies when an external magnetic field is applied
All samples are measured against a reference compound – Tetramethylsilane (TMS)
o TMS shows a single sharp peak on NMR spectra, at a value of zero
o Sample peaks are then plotted as a ‘shift’ away from this reference peak
o This gives rise to ‘chemical shift’ values for protons on the sample compound
o Chemical shifts are measured in parts per million (ppm)
Features of a NMR spectrum




NMR spectra shows the intensity of each peak against their chemical shift
The area under each peak gives information about the number of protons in a particular
environment
The height of each peak shows the intensity/absorption from protons
A single sharp peak is seen to the far right of the spectrum
o This is the reference peak from TMS
o Usually at chemical shift 0 ppm
A low resolution 1H NMR for ethanol showing the key features of a spectrum
320 | P a g e
Molecular environments



Hydrogen atoms of an organic compound are said to reside in different molecular
environments
o Eg. Methanol has the molecular formula CH3OH
o There are 2 molecular environments: -CH3 and -OH
The hydrogen atoms in these environments will appear at 2 different chemical shifts
Different types of protons are given their own range of chemical shifts
Chemical shift values for 1H molecular environments table
321 | P a g e


Protons in the same molecular environment are chemically equivalent
Each peak on a NMR spectrum relates to protons in the same environment
Low resolution 1H NMR

Peaks on a low resolution NMR spectrum refers to molecular environments of an organic
compound
o Eg. Ethanol has the molecular formula CH3CH2OH
o This molecule as 3 separate environments: -CH3, -CH2, -OH
o So 3 peaks would be seen on its spectrum at 1.2 ppm (-CH3), 3.7 ppm (-CH2) and
5.4 ppm (-OH)
A low resolution NMR spectrum of ethanol showing 3 peaks for the 3 molecular environments
High resolution 1H NMR



More structural details can be deduced using high resolution NMR
The peaks observed on a high resolution NMR may sometimes have smaller peaks
clustered together
The splitting pattern of each peak is determined by the number of protons on
neighbouring environments
The number of peaks a signal splits into = n + 1
322 | P a g e
(Where n = the number of protons on the adjacent carbon atom)
High resolution 1H NMR spectrum of Ethanol showing the splitting patterns of each of the 3
peaks. Using the n+1, it is possible to interpret the splitting pattern

Each splitting pattern also gives information on relative intensities
o E.g. a doublet has an intensity ratio of 1:1 – each peak is the same intensity as the
other
o In a triplet, the intensity ratio is 1:2:1 – the middle of the peak is twice the
intensity of the 2 on either side
H NMR peak splitting patterns table
1
323 | P a g e
324 | P a g e