AS Revision PowerPoint
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Atomic Structure
Amount of Substance
Bonding
Periodicity
Introduction to Organic Chemistry
Alkanes
Fundamental Particles
Protons, Neutrons and Electrons
Mass Number and Isotopes
Electron Arrangement
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Atoms are the fundamental building blocks of
matter. However, even atoms are constructed
of smaller, sub-atomic particles.
The fundamental sub-atomic particles are
protons, neutrons and electrons. The protons
and neutrons are held together by strong
nuclear binding forces in the nucleus of the
atom. The electrons may be considered to be
tiny particles that exist in regions of space
known as orbitals around the atom.
Mass
Charge
Proton
1
+1
Neutron
1
0
Electron
1/1836
-1
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Isotopes are atoms of a specific element that have a
definite number of neutrons and consequently a different
mass. In effect all atoms are isotopes of one element or
another.
Most elements have several isotopes, some of which are
stable, and others that spontaneously break apart
releasing radioactivity.
The percentage of an isotope occurring in a natural sample
of an element is called its natural abundance. The natural
abundance of each isotope on the earth are usually very
similar wherever the sample is obtained.
The 'relative abundance' of an isotope means the
percentage of that particular isotope that occurs in nature.
Most elements are made up of a mixture of isotopes.
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The mass spectrometer is an instrument used for analysing samples of elements
and compounds. It consists of six basic stages.
The sample is injected into the instrument. It may need vaporising by heating in
which case the initial stage also has an oven. The sample in gaseous state then
passes into an ionising beam of electrons which knock electrons off the sample
creating positive ions.
These positive ions are then accelerated by some electrostatically charged plates
into magnetic field, which then deflects the particles according to their
mass/charge ratio.
The deflected ions then arrive at the detector.
The five stages are then:
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Vaporisation
Ionisation
Acceleration
Deflection
Detection
(VIADD)
The mass spectrometer is a very sensitive instrument and can detect particles with
very small differences in relative mass. Isotopes have mass numbers that differ by
at least one atomic mass unit and so are easily differentiated. The intensity of the
signal in a spectrum is directly proportional to the amount of that species in the
sample. Hence, the spectrum tells us both which isotopes are present and their
relative proportions.
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This is the arrangement of electrons around the atom. They
are arranged in electron shells and within each shell there are
sets of sub-shells. The sub-shells themselves are made up of
orbitals each of which can hold a maximum of two electrons,
each with opposite spin.
The order in which the electrons fill up the shells is also
called the Aufbau principle.
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The first ionisation energy is defined as the energy required to
remove one mole of electrons from one mole of gaseous atoms
to provide one mole of gaseous single charged ions.
◦ Na(g)  Na+(g) + e¯
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Subsequent ionisation energies are defined in a similar way only
by removing electrons from already charged ions.
The second ionisation energy
◦ Na+(g)  Na2+(g) + e¯
If the energy required to achieve these ionisations is plotted on a
graph against the ionisation number, the 'jumps' in the required
energy clearly show the main and sub energy levels.
Relative Atomic and Molecular Mass
The Mole and Avogadro’s Constant
The Ideal Gas Equation
Empirical and Molecular Formula
Balanced Equations and Calculations
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The atomic mass number is represented by the
symbol (letter) 'A'.
The mass number gives the integral number of
nucleons, protons and neutrons found in the
nucleus of an atom.
The relative mass is a value that is not
necessarily integral that compares a mass to the
mass of a carbon isotope, assigned a value of
exactly 12 units.
The atomic number is represented by the symbol
(letter) 'Z'. It shows us the number of protons in
an atom (and the number of electrons) in a
neutral atom.
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Avogadro's number, or constant, is the number to which
the mass of an atom must be multiplied to give a mass in
grams numerically equal to its relative atomic mass.
Avogadro's constant (L) = 6.02 x 1023
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Carbon has a relative mass of 12 therefore 6.02 x
1023 carbon atoms have a mass of 12g
Magnesium has a relative atomic mass of 24 therefore
6.02 x 1023 magnesium atoms have a mass of 24g
This gives rise to two important definitions
◦ The amount of any substance containing an Avogadro number of
particles of that substance is called a mole.
◦ 1 mole of any substance has a mass equal to its relative mass
expressed in grams
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Example
1 mole of Mg contains 6.02 x 1023 Mg atoms
1 mole of Mg = 24g
12g of Mg = 0.5 moles of Mg
0.5 x 6.02 x 1023 = 3.01x 1023 magnesium atoms
The relationship between moles, mass and number of
particles can be expressed by simple formulae:
no. of particles = moles x Avogadro's constant
mass in grams = moles x relative mass
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These formulae can be used to find any quantity when the
other two quantities are known.
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The equation of state refers to a fixed mass of gas. From Avogadro's law we know
that the same volume of all gases contain the same number of moles and from
this, it follows that the volume is proportional to the number of moles.
Volume ∝ number of moles (n)
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These two equations can be combined to obtain an expression involving all the
quantities:
PV = nRT
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where:
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P = pressure (Pa)
V = volume (m3)
n = number of moles of gas
R = Universal Gas constant = 8.314 J K-1 mol-1
T = temperature (K)
It is often more convenient to express the pressure in kPa and the volume in litres
(dm3).
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Example: Calculate the number of moles of gas
present in 2.6 dm3 at a pressure of 1.01 x 105 Pa
and 300 K.
PV = nRT
2.6 dm3 = 0.0026 m3
0.0026 x 1.01 x 105 = n x 8.314 x 300
n = 0.0026 x 1.01 x 105 / 8.314 x 300
n = 0.105 moles
1 litre = 1 dm3 = 1000cm3
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The empirical formula is the simplest ratio of atoms within a
chemical compound.
Example: Find the empirical formula of the compound which has the
following percentage by mass composition: Carbon 12.12%; Oxygen
16.16%; Chlorine 71.17%.
Steps
Carbon
Oxygen
Chlorine
Percentage
12.12
16.16
71.17
Ar
12
16
35.5
%/Ar = X
1.01
1.01
2
X/(smallest
X value)
1
1
2
Formula
C
O
Cl₂
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The molecular formula is the number and type of atoms that
are present in a single molecule of a substance.
Determine the molecular mass in grams.
Divide the molecular mass of the compound by the molecular
mass of the empirical formula.
Round the quotient to the closest integer.
Multiply the rounded number by all the subscripts, using the
product as the new subscripts.
Example:
A compound has an empirical formula of CH₂ and a molecular
mass of 42. Determine its molecular formula.
◦ 12 + (2x1) = 14
◦ 42 ÷ 14 = 3
◦ CH₂ x 3 = C₃H₆
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To balance a chemical equation it is important to remember that the
formula of the reactants and products cannot be changed and that
coefficients may only be placed before the formulae, multiplying them
by whole numbers.
These equations are constructed by writing the formula of each of the
compounds in the reaction, and then by counting up the number of
atoms on each side to make sure they are equal. If they are not equal,
balancing numbers (coefficients) are added in front of each chemical
formula (where needed), so that the numbers of each type of particle on
each side of the equation are the same.
Step 1 - write the chemical equation
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Step 2 - write the formula of each of the reaction components
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Step 3 - add coefficients in front of the formulae to balance the equation
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◦ ammonia + oxygen nitrogen monoxide + water
◦ NH3 + O2 NO + H2O
◦ 4NH3 + 5O2 4NO + 6H2O
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Avogadro suggested that "equal volumes of all gases contain
equal number of molecules (the gases being measured at the
same temperature and pressure).
This means that in the reaction:
N2 + 3H2  2NH3
For a given volume of nitrogen, three times the volume of
hydrogen is needed for complete reaction.
Example: Find the volume of hydrogen required to react
completely with 200 cm3 of nitrogen according to the equation:
N2 + 3H2  2NH3
The equation tells us that 1 volume of nitrogen reacts completely
with 3 volumes of hydrogen:
Therefore volume of hydrogen = 3 x volume of nitrogen
Therefore volume of hydrogen = 3 x 200 cm3 = 600 cm3
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The concentration of a solution is the quantity of solute that it
contains per unit volume.
This may be given in grams per 100cm3 or grams per litre, but it
is usually given in terms of molarity as this gives a direct
measure of the number of solute particles contained by the
solution.
Molarity = number of moles of solute per litre of solution
Molarity = moles/litres
The molarity is denoted by the capital letter M, or the units mol
dm¯³
Example: Calculate the molarity of a solution containing 0.15
moles of potassium nitrate in 100cm3 of solution.
Molarity = moles/litres
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100cm3 = 0.1dm³
Molarity = 0.15/0.1 = 1.5 M
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A titration of a 25ml sample of a
hydrochloric acid solution of
unknown molarity reaches the
equivalence point when 38.28ml of
0.4370M NaOH solution has been
added.
What is the molarity of the HCl
solution?
HCl(aq) + NaOH(aq)  NaCl(aq) +
H₂O(l)
Use the volume and molarity of the
calculate the molarity of the NaOH
to calculate the number of HCl
solution? moles of NaOH that
reacted. Use the mole ratio
between base and acid to
determine the moles of HCl that
reacted. Use the volume of the acid
to calculate molarity.
=0.6691M
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This term refers to the efficiency of a
chemical process in terms of the atoms that
are lost as by-products to the intended
product.
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The yield of a reaction is the actual mass of
product obtained. The percentage yield can
be calculated:
Ionic, Covalent and Metallic Bonding
Bond Polarity
Forces Acting Between Molecules
States of Matter
Shapes of Simple Molecules and Ions
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Ions are charged atoms or groups of atoms. Atoms form ions to gain a noble gas
configuration and become more stable. Atoms form ions by losing or gaining
electrons.
Positive ions are formed when atoms lose electrons. Negative ions are formed
when atoms gain electrons
Ionic bonds never exist in isolation, instead they form part of a giant ionic lattice
in which each positive ion attracted to negative ions which surround the positive
ion in a regular arrangement. The positive ions are electrostatically attracted to
the negative ions.
They have high melting and boiling points; the attraction between the positive and
negative ions is very strong and thus a great deal of energy is needed to break the
bonds.
They are solid at room temperature (due to high melting and boiling points)
They are brittle; when a force is applied it knocks the regular pattern and positive
ions end up next to positive which cause repulsion.
Soluble in water
Conduct electricity when molten or in aqueous solution, but not when solid. The
ions are fixed in position in the solid, but in the liquid state, or in solution, they
are free to move and carry the current.
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In a covalent bond, two atoms share a pair of
electrons. The electron pair creates a ‘bond’
between the two atoms because it attracts the
nucleus of each atom and therefore resists the
separation of the two atoms.
Simple molecular: low melting and boiling points,
do not conduct electricity
Giant molecular: high melting and boiling points,
do not conduct electricity (except carbon)
A co-ordinate (dative) bond is formed when both
electrons in the shared pair are from one atom
(from the lone pair of electrons on the donor
atom)
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Metallic bonds in solids do not exist in isolation. They
make up a giant metallic lattice which comprises of
closely packed metal ions surrounded by delocalised
electrons which are free to move through the lattice.
They have high melting and boiling points (as a large
amount of energy is needed to remove a metal atom
from the attraction of the delocalised electrons)
They can conduct electricity (the sea of delocalised
electrons can carry an electrical charge)
They are malleable and ductile (because the forces in
a metallic lattice are strong, but not directed at one
particular atom, the layers of atoms can slide over
each other – known as slip)
Metal lustre (the delocalised electrons also cause the
meta to reflect light)
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Electronegativity is the power of an atom to
withdraw electron density from a covalent bond.
It can be calculated using at various methods and
is usually given a value between 0.7 and 0.4
Small atoms with a large numbers of protons in
the nucleus attract electron density more
strongly. Thus electronegativity increases from
left to right across a period of the Periodic Table
and from the bottom to the top of a group.
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When a covalent bond exists between atoms of
different electronegativity, the shared pair of
electrons is displaced towards the more
electronegative atom.
The displacement of electron density makes the
less electronegative atom slightly electron
deficient (δ⁺) and the more electronegative atom
has a slight excess of electron density (δ⁻)
This charge separation creates an electric ‘dipole’
and the molecule is said to be polar.
The polarity of a bond can be measured in a unit
called the debye. Its size depends on the
difference in electronegativity between the
elements involved.
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Covalent bonds are attracted to each other by
intermolecular forces. They are much weaker
than intramolecular forces.
The three types of intermolecular force are
van der Waal’s (temporary induced dipoledipole), permanent dipole-dipole and
hydrogen bonding.
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Permanent dipole-dipole interactions occur
between molecules that have permanent
dipoles.
The permanent dipole consists of regions of
partial positive charge and regions of partial
negative charge within the same molecule.
By convention we use the small Greek letter
delta, ∂, to represent a partial charge. Hence,
∂⁺ means a partial positive charge and ∂⁻
means a partial negative charge.
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Van der Waals' forces of attraction are temporary,
induced dipole-dipole attractions
At any moment in time, the electron distribution
in a non-polar covalent molecule may be unequal
due to the fluctuating movement of electrons.
This leads to a temporary dipole which induces
an opposite dipole on an adjacent. The second
molecule is thus attracted to the first one
The size or magnitude of van der Waal’s forces
increases with the size of molecules and also
depends upon their shape. Branched chain
hydrocarbons have weaker intermolecular forces
than straight chain molecules because they are
less polarisable.
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Hydrogen bonding is the strongest type of
intermolecular forces in neutral molecules. It is a
special case of dipole-dipole force which exists
between a lone pair of electrons on a N, O or F
atom and a hydrogen atom that has a strong
partial charge (∂⁺) because it is attached to a
highly electronegative atom e.g. N, O or F
The electronegative atom pulls electrons away
from hydrogen atom so that, on the opposite
side to the bond, the hydrogen atom appears
almost like an unshielded proton
State
Particles/Motion
Forces
Solid
Fixed in position
Vibrate
Ionic attractions; covalent
bonds; metallic bonds;
permanent dipole-dipole
forces; all van der Waal’s
forces
Liquid
Resembles a disorded solid
Some movement
Separation is 10% more than solids
Similar to a solid
Rapid, random motion
Intermolecular forces can
be ignored unless the
particles are close together
at high pressure or
temperature
Gas
SOLID
MELTING
LIQUID
SOLIDIFYING
EVAPORATING
GAS
CONDENSING
SUBLIMATION
SOLID  LIQUID
Heat energy is required to reduce forces
which hold the particles together. This heat
energy is called the enthalpy of fusion.
LIQUID  GAS
More energy is required than to change
from solid to liquid. The energy is used to
overcome the forces which hold the
particles together so that the particles can
be completely separated. This heat energy
is called the enthalpy of vaporisation.
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A crystal is a solid with a regular shape, containing particles
arranged in a regular structure. Crystals can be classified
according to the type of bonding between the particles.
Crystal Type
Melting and
Boiling Points
Ionic
Electrical Conductivity
Solubility in
Water
Solid
Molten
High
None
Good
Variable but
often good
Macromolecular
Very High
None – except
graphite
None
Insoluble
Molecular
Low
None
None
Variable
Metallic
Usually High
Good
Good
Insoluble
1.
2.
3.
The number of outer shell electrons
originally in the centre atom
The number of additional shared electrons
in covalent or dative bonds
The loss or gain of additional electrons if
the species is a positive or negative ion
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The final shape of a molecule is also altered if some
of the electron pairs are lone (non-bonding) pairs
Lone pairs are more compact than bonding pairs so
they repel more strongly, leading to bond angles
between bonding pairs which are smaller than those
found in totally symmetrical shapes
Lone pair-lone pair repulsions are stronger than lone
pair-bonding pair repulsions. These principles can be
used to predict different shapes.
Increasing
Repulsion
Lone Pair – Lone Pair
Lone Pair – Bonding Pair
Bonding Pair – Bonding Pair
Classification of Elements in the s, p
and d blocks
Properties of the Elements of
Period 3 and their Periodic
Trends
In these blocks, the elements have their highest
energy electrons in s, p, d or f electronic sub-levels.
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The atomic radii decrease because, across the period, the nuclear charge increases and the
outer electrons are attracted more strongly. Thus, they are drawn closer to the nucleus,
without any additional shielding of the nuclear charge by the added electrons.
The atomic radius of these elements is measured in suitable compounds of the elements is
measured in suitable compounds of the element. Argon forms no compounds and thus is not
possible to measure its atomic radius in a comparable way to the other elements.
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A general increase because nuclear charge increases
Mg-Al: removing electrons from 3s-orbital to 3p-orbital
P-S: pairs in orbitals in sub-levels
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The electronegativity of an element is the power of an atom to
attract electrons in a covalent bond. Electronegativity increases
across a period because the number of protons increases and
thus the attraction between the nucleus and a pair of electrons in
a covalent bond increases.
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The elements Na-Al are good electrical conductors because they
have metallic bonding, with delocalised electrons which are free
to move through the whole structure. The elements Si-Cl have
covalent bonding with electrons which are not free to move from
one atom to another. Their conductivities are almost zero.
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The variation in the melting and boiling points is related to the bond strengths and structures of the elements.
Argon has very low melting and boiling points because it exists as individual atoms. The atoms are not very
polarisable and thus van der Waal’s forces are very weak.
Na-Al are metallic elements. The melting and boiling points increase because from Na-Al the atoms are
smaller and have an increasing nuclear charge. Thus, the strength of the metal-metal bonds increase.
Si has high melting and boiling points because it is macromolecular with a diamond structure and strong
covalent bonds link all the atoms in three dimensions. A large amount of energy is needed to break these
bonds.
P, S and Cl are all molecular structures. The melting points are determined by the strength of the van der
Waal’s forces between molecules which, in turn, are determined by the size of the molecules (P₄ S₈ Cl₂). Each of
these elements has a low melting and boiling point because the van der Waal’s forces are weak and easily
broken.
Nomenclature
Isomerism
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When naming an alkene derivative, look for the
longest carbon chain in the skeleton. This
determines the stem of the name.
Many compounds have a branched carbon
skeleton, in which case the name of the side
chain also depends on the number of carbon
atoms in them e.g. a single carbon branch is CH₃
The position of the branch also has to be made
clear which is done by numbering the carbon
atoms in the skeleton. The numbers used are
kept as low as possible.
1.
2.
3.
4.
Use the name to determine the number of
carbon atoms in the longest chain
Draw this carbon skeleton and number the
carbon atoms
Add any functional groups in the correct
positions
Add hydrogen atoms until each carbon
atom has 4 bonds
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Isomerism occurs where molecules with the
same molecular formula have their atoms
arranged in different ways. Isomerism is
divided into main types:
◦ Structural Isomerism
◦ Stereoisomerism
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This occurs when there are two or more ways
of arranging the carbon skeleton of a
molecule.
e.g. C₄H₁₀ can be butane or 2-methylpropane
e.g. C₅H₁₂ can be pentane, 2-methylbutane or
2,2-dimethylpropane
These isomers have similar chemical
properties but slightly different physical
properties.
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These isomers have the same carbon skeleton
and the same functional groups but the
functional group is joined in a different place.
e.g. 1-bromopropane and 2-bromopropane
e.g. but-1-ene and but-2-ene
These isomers have similar chemistry
because they have the same functional group
but the different positions can cause different
properties
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These isomers have different functional
groups so different chemical and physical
properties.
e.g. aldehydes and ketones: C₃H₆O can be
propanal or propanone
e.g. carboxylic acids and esters: C₃H₆O₂ can be
propanoic acid, methyl ethanoate or ethyl
methanoate
e.g. alcohols and ethers: C₃H₆O can be ethanol
or methoxymethane
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Stereoisomerism have the same structural
formula but the bonds are arranged differently in
space.
Due to the π electron clouds present above and
below the plane of the bond, carbon-carbon
double bonds cannot rotate. If an alkene has two
different groups on each end of the double bond
(E/Z) isomerism occurs.
If the groups (often identical) are on the same
side, then they show Z-isomerism. If they are on
different sides, then they show E-isomerism.
One way to remember this is that Enemies are on
different sides, and people must be on ZeeZame-Zide (with a really fake German accent)
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Z-isomers usually have higher boiling points, as
they will have some polarity, whereas E-isomers
are less polar. For example, Z-but-2-ene has a
boiling point of 4˚C and E-but-2-ene is 1˚C.
E-isomers however have higher melting points
because they pack together more closely. For
example, E-but-2-ene’s melting point is -106˚C
while Z-but-2-ene’s is -139˚C
NB: molecules with two identical groups on the
same carbon atom of the double bond do not
show geometric isomerism.
Fractional Distillation of Crude
Oil
Cracking of Alkanes
Combustion of Alkanes
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Crude oil is a mixture of hydrocarbons,
mainly alkanes, formed by the decay of
remains of sea creatures and plants over
millions of years.
Crude oil is a complex mixture of
hydrocarbons which has no use in its raw
form. To provide useful products, the
components must be partly separated and, if
necessary, modified.
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The complex mixture of hydrocarbons in crude oil is separated
into less complicated mixtures or fractions by fractional
distillation (primary distillation). The crude oil is heated and the
vapour/liquid mixture is passed into a tower.
The top of the tower is cooler than the bottom i.e. there is a
mixture into fractions depending on those boiling points of the
hydrocarbons present. Only those with low boiling points (more
volatile) reach the top; others condense in trays at different
levels up the tower, and are drawn off.
The residue from the primary distillation still contains useful
materials such as lubricating oil and waxes which boil above
350˚C at atmospheric pressure. At such high temperatures some
of the components of the residue decompose. To avoid this, the
residue is further distilled under reduced pressure (vacuum
distillation). The reduced pressure lowers the boiling points, so
they can be distilled without decomposing.
Thus, fractional distillation is a physical process
Name of Fraction
Boiling Range
(˚C)
Uses
Length of Carbon
Chain (approx)
LPG
<25
Camping gas
1-4
Petrol
40-100
Fuel for cars
4-12
Naphtha
100-150
Petro chemicals
7-14
Paraffin
150-250
Aviation fuels
11-15
Diesel
220-350
Central heating
fuel
15-19
Lubricating Oil
350<
Lubricating oil
20-30
Fuel Oil
400<
Fuel for ships
30-40
Wax, Grease
400<
Candles
40-50
Bitumen
400<
Roofing
50<
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At high temperatures, the atoms in an alkane molecule vibrate
rapidly. If this temperature is high enough, the vibration
becomes sufficiently vigorous for chemical bonds to break. This
breakage of the C-C bonds in alkanes leads to the formation of
smaller hydrocarbon fragments and is called cracking. The
temperature required for cracking can be reduced by using a
solid catalyst.
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Thermal cracking results in the production of a high proportion
of alkenes. The energy required for bond breaking is provided by
heat, with temperatures used in the range 400-900˚C and
pressure of up to 700kPa
At the lower end of the temperature range, carbon chains break
preferentially part way along the carbon chain of the molecule,
leading to a greater proportion of low Mr alkenes
The process is initiated by homolytic fission of C-C bonds
forming two alkyl radicals. Each radical can then abstract a
hydrogen atom from an alkane molecule to produce a different
alkyl radical and a shorter alkane.
Alternatively, the alkyl radical can undergo further bond breaking
to give an alkene and a shorter alkyl radical.
Dehydrogenation, isomerism and cyclisation can also occur. The
mixtures of products are separated by fractional distillation.
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Catalytic cracking uses zeolite catalysts, a slight excess of
pressure and a temperature of about 450˚C. Large alkanes
are converted mainly into branched alkanes, cycloalkanes
and aromatic hydrocarbons.
The proportion of alkenes is small so catalytic cracking is
primarily used to produce motor fuels. Branched chain
alkanes burn more smoothly than unbranched chains. In
an engine, due to the pressures involved, the fuel-air
mixture may ignite before the ‘spark’ is produced, causing
‘knocking’. Using branched-chain alkanes prevents this
problem.
The mechanism for catalytic cracking involved the
formation of carbocations wit the catalyst acting as a Lewis
acid.
Benzene (and its methyl and dimethyl derivatives) arise
through cyclisation and dehydrogenation.
Thermal
Catalytic
Conditions
High temperature: 400-900˚C
High pressure: 700kPa
Short residence time
Lower temperature: 450˚C
Zeolite catalyst
Mechanism
Homolytic fission forming free
radicals
Heterolytic fission forming
carbocations
Products
Low Mr alkenes
Shorter alkanes
Some hydrogen
Branched chain alkanes
Cycloalkanes
Aromatics
Product Uses
Chemical industry (including
polymers)
Motor fuel
Other Products
Branched chain alkanes
Cycloalkanes
Aromatics
Low Mr alkenes
Shorter alkanes
Some hydrogen
Complete Combustion
 In the presence of a plentiful supply of oxygen,
complete combustion of alkanes occurs
exothermically to produce carbon dioxide and
water.
 As the number of carbon atoms increases, more
oxygen is required per mole of hydrocarbon for
complete combustion, and more energy is
released. Sulphur-containing impurities in the
hydrocarbons produce oxides of sulphur e.g.
sulphur di- and trioxide which are toxic and
dissolve in water to form acid rain.
Incomplete Combustion
 If insufficient oxygen is available, incomplete
combustion occurs and water forms with
either carbon monoxide or carbon.






Carbon monoxide is also formed by the incomplete combustion
of petrol vapour (C₈H₁₈) in a car engine.
Such engines produce other pollutants, notably oxides of
nitrogen, unburned hydrocarbons and, when leaded petrol is
used, compounds of lead. Cars with diesel engine produce
carbon, not carbon monoxide.
Oxides of nitrogen are also formed when the air/petrol mixture
is sparked and explodes. The temperature can reach values of
2500˚C and this provides sufficient activation energy for
nitrogen (in air) to react with oxygen (in air), forming nitrogen
monoxide.
On cooling, nitrogen monoxide easily reacts with more oxygen to
form nitrogen dioxide.
Nitrogen dioxide reacts with water and more oxygen to produce
nitric acid, which can lead to acid rain
Nitrogen dioxide also reacts with oxygen or hydrocarbons, in the
presence of sunlight, to form an irritating photochemical smog.




Catalytic converters help to remove harmful
gases from car exhausts. The converter contains
a honeycomb of ceramic material on which
metals e.g. platinum, palladium and rhodium are
spread in a thin layer. The metals catalyse
reactions between the pollutants, and help to
remove up to 90% of the harmful gases.
2CO + 2NO  2CO₂ + N₂
C₈H₁₈ + 25NO  8CO₂ + 12.5N₂ + 9H₂O
Thus the pollutant gases (CO,NO) and
hydrocarbons are replaced by the harmless
products (O₂ , N₂ , H₂O)











Energetics
Kinetics
Equilibria
Redox Reactions
Group 7: the Halogens
Group 2: the Alkaline Earth Metals
Extraction of Metals
Haloalkanes
Alkenes
Alcohols
Analytical Techniques
Enthalpy Change
Calorimetry
Hess’s Law
Bond Enthalpies




If a reaction produces heat (i.e. increases the
temperature of the surroundings) it is
exothermic. If the temperature of the reaction
mixture decreases (i.e. heat is absorbed) then
the reaction is endothermic.
Exothermic: a reaction which produces heat
(ΔH has a negative value by convention, -ve)
Endothermic: a reaction which absorbs heat
(ΔH has a positive value by convention, +ve)
Enthalpy of reaction: The change in internal
(chemical) energy (H) in a reaction = ΔH






Standard state: Pressure = 100 kPa,
Temperature = 298 K
The standard state of an element or compound is the form in which it
exists under standard conditions (not to be confused with STP)
Standard enthalpies of combustion (ΔHc )
The standard enthalpy of combustion is the energy released when 1
mole of a substance is burnt in excess air or oxygen, all quantities being
measured under standard conditions.
Standard enthalpies of formation (ΔHf )
The enthalpy change when 1 mole of a substance is made from its
elements in their standard states. There MUST be 1 mole of the
substance formed AND all of the elements forming it MUST be in their
standard states.
Change in energy = mass x specific heat
capacity x change in temperature
q = m x c x ΔT




Experimental
A known mass of solution should be placed in a container, as insulated
as possible, to prevent as much heat as possible from escaping. The
temperature is measured continuously, the value used in the equation is
the maximum change in temperature from the initial reading.
The result will be a change in temperature. This can be converted into a
change in heat (or energy) by using the above equation q = m x c x ΔT.
Δ-H may then be calculated for the amount of reactants present, and
then this can be used to calculate for a given number of mols.



Hess' Law states that the total enthalpy change
between given reactants and products is that same
regardless of any intermediate steps (or the reaction
pathway).
This basically means that energy changes can be
calculated from other energy changes which may be
found experimentally. It is the fundamental basis of
nearly all the thermodynamic data that we have
available today.
It is not possible in the majority of cases to calculate
the formation enthalpy of a substance, whereas it is
possible to find the combustion enthalpy by
experimentation. In many cases historically it was
possible to calculate the enthalpy of formation from
the combustion enthalpies using Hess' law.



Bond enthalpy (mean bond energy) is the average value of a particular
type of bond which has been measured over a range of molecules.
Example:
CH4 has four C-H bonds, and so will have four different bond
dissociation enthalpies corresponding to the following bonds breaking:
◦
◦
◦
◦


CH4  CH3 + H
CH3  CH2 + H
CH2  CH + H
CH  C + H
The C-H bond enthalpy is the average value of the four bond
dissociation enthalpies.
Bond energies (enthalpies) can be used to calculate unknown enthalpy
changes in reactions where only a few bonds are being formed/broken.
Bonds broken (left hand side) - bonds formed (right hand side)
= enthalpy change for the reaction.
Collision Theory
Maxwell–Boltzmann Distribution
Effect of Temperature on Reaction Rate
Effect of Concentration
Catalysts



Collision theory - reactions take place as a result
of particles (atoms or molecules) colliding and
then undergoing a reaction. Not all collisions
cause reaction, however, even in a system where
the reaction is spontaneous.
The particles must have sufficient kinetic energy,
and the correct orientation with respect to each
other for them to react. This energy is needed to
cause bonds to break, which is essential if
reaction is to take place.
Activation Energy: This is the minimum energy
that particles colliding must have in order to
produce successful reaction. It is given the
symbol Ea (Energy of Activation).

The Maxwell Boltzmann distribution is a
statistically derived function showing how the
energy of particles is distributed over all of the
particles with some particles having no energy to
particles with a great deal of energy. The majority
of particles have energies intermediate between
these two situations.


Increasing the temperature of a substance increases the average speed
(energy) of the particles and consequently the number of particles
colliding with sufficient energy (Ea) to react. At higher temperatures
there are more successful collisions and therefore a faster reaction.
Increasing temperature has two effects on a reaction:
◦ There are more collisions because the particles are moving faster on average
◦ The collisions have more energy

Increasing the concentration of a reactant
increases the number of particles in a given
volume and increases the chance of
successful collisions and in turn increases the
rate of reaction.



A catalyst is a substance which alters the rate of
reaction without being chemically changed itself
The catalyst provides an alternative reaction route of
lower activation energy than the uncatalysed one.
The catalyst has no effect on the overall enthalpy
change for the reaction.
The Dynamic Nature of Equilibria
Qualitative Effects of Changes of Pressure,
Temperature and Concentration on a
System in Equilibrium
Importance of Equilibria in Industrial
Processes


If a body of liquid is placed in a closed container, the higher
energy particles will continue to leave but now they will be held
in the same place (the air space above the liquid surface).
Occasionally, due to collisions, some of these vapour particles
will lose energy and rejoin the body of liquid. As more particles
gather in the air space there will be more possibility of particles
losing energy and returning to the liquid.
We now have two processes happening. Liquid particles turning
to vapour and vapour particles turning to liquid. Eventually the
rate at which the particles move from the liquid to the vapour
phase will equal the rate at which the particles move from the
vapour phase to the liquid. When this situation is arrived at the
concentration of vapour particles in the air space will be constant
even though there is movement in both directions. We call this
Dynamic Equilibrium




In reactions where gases are produced (or there are more mols of
gas on the left), and increase in pressure will force the reaction
to move to the left (in reverse). If pressure is decreased, the
reaction will proceed forward to increase pressure. If there are
more mols of gas on the left of the equation, this is all reversed.
Pressure change only affects gaseous equilibria and only then
when there are an unequal number of moles of gas on either side
of the equilibrium.
When the pressure is increased on a gaseous mixture at
equilibrium the equilibrium will respond so as to release the
applied pressure. In other words the equilibrium will move
towards the side of the fewer number of moles.
It should be remembered that the pressure of a gas is directly
proportional to the number of moles of gas and that this does
not depend on the nature of the gas in question - all gases
behave in the same way in terms of their physical properties.






The effect of a change of temperature on a reaction will depend on whether the
reaction is exothermic or endothermic.
When the temperature increases, Le Chatelier's principle says the reaction will
proceed in such a way as to counteract this change, i.e. lower the temperature.
Therefore, endothermic reactions will move forward, and exothermic reactions will
move backwards (thus becoming endothermic). The reverse is true for a lowering
of temperature.
If a reaction is exothermic in the forward direction in an equilibrium it will be
endothermic in the reverse direction and vice-versa (law of conservation of
energy)
Increasing the temperature of an equilibrium mixture will always favour the
endothermic direction over the exothermic process. The rate of the endothermic
process will increase more than the exothermic direction of change and the
equilibrium will be re-established with new concentrations based on more of the
endothermic product (i.e. the product of the endothermic direction of change)
To express this in non-scientific terms, it can be considered that the endothermic
direction is absorbing heat therefore giving it more heat by increasing the
temperature will favour it.
The reverse argument is true for the direction of exothermic change. The
exothermic reaction is giving out heat and therefore applying more will hinder this
process. Raising the temperature retards the reaction in the exothermic direction.


When the concentration of a product is increased, the reaction proceeds in reverse
to decrease the concentration of the products. When the concentration of a
reactant is increased, the reaction proceeds forward to decrease the concentration
of reactants.
Some general rules:
◦
◦
◦
◦



If we add to the left hand side of an equilibrium we make more of the right hand side
If we add to the right hand side of an equilibrium we make more of the left hand side
If we remove from the left hand side the equilibrium makes more of the left hand side
component
If we remove from the right hand side the equilibrium makes more of the left hand side
component
When we study rates of reaction one of the first conclusions drawn is that the rate
of a chemical reaction depends on the concentrations of the reactants. The greater
the concentration the more collisions occur and the faster the rate of the reaction.
When we add more reactant, the forward rate will now be greater than the back
rate. The reaction is now not at equilibrium and the forward reaction proceeds
faster than the back reaction until the equilibrium conditions are re-established.
Similarly, removal of a component from one side will reduce the rate of its
reaction and case the equilibrium to make more of it.

The Haber Process
N2(g) + 3H2(g)  2NH3(g)





ΔH = -92.4 kJ mol-1
There are more moles of gas on the left than the right, so a greater yield will be
produced at high pressure.
The reaction is exothermic, therefore it will give a greater yield at low
temperatures.
In practice, if low temperatures are used the time taken for the reaction to attain
equilibrium becomes unfeasibly long. An intermediate temperature is chosen
(450ºC) which allows the reaction to get to an equilibrium in a reasonable time
and still has enough of the products in the equilibrium mixture.
A catalyst of finely divided iron is also used to help speed the reaction (finely
divided to maximise the surface area).
To make the process more efficient the ammonia produced at equilibrium is
removed by first cooling the mixture when the ammonia turns into a liquid which
can be tapped off. The unreacted gases in the process are then mixed with fresh
reactants and returned to the reaction chamber to re-establish the equilibrium
again and the cycle is repeated continuously.


Manufacturing of Ethanol
The reaction between ethene and steam in the presence of
a phosphoric acid catalyst at 300ºC and 60 atmospheres
pressure is used in industry to manufacture ethanol.
C2H4(g) + H2O(g)  C2H5OH(g)


This is a reversible reaction - alcohols can be dehydrated
to alkenes - so to encourage the forward reaction high
pressure is used.
You can see that in the equation there are two moles of
gas on the left hand side and only one mole of gas on the
right hand side. According to Le Chatelier the increased
pressure pushes the reaction towards the side of fewer
moles of gas, in this case the products.


Manufacturing of Methanol
Synthesis gas, a mixture of carbon monoxide, carbon dioxide and
hydrogen, is first produced in a reformer. This is carried out by passing
a mixture of a hydrocarbon feedstock and steam through a heated
tubular reformer. The ratio of hydrogen and carbon oxides in the syngas
may need to be adjusted by purging excess hydrogen or adding carbon
dioxide.
H2O(g) + C(g)  CO(g) + H2(g)

The syngas is cooled and then compressed before being fed to the
methanol converter. The methanol synthesis takes place in the presence
of copper-based catalysts at 250-260ºC. The crude methanol is
recovered and purified by distillation.
2H2(g) + CO(g)  CH3OH(g)

The reaction is reversible. The reaction is exothermic, so the
temperature is kept to a minimum, high enough to ensure rapid reaction
but low enough to ensure a high percentage of product at equilibrium.
High pressure is also used to move the equilibrium to the side of fewer
gaseous moles.
Oxidation and Reduction
Oxidation States
Redox Equations


Oxidation occurs when the oxidation state of
an element increases (becomes more positive)
Reduction occurs when the oxidation state of
an element decreases (becomes more
negative)

Oxidation states help to improve the definitions of oxidation and
reduction.
Rules for Assigning Oxidation States
1.
2.
3.
4.
5.
6.
7.
8.
The oxidation state of an atom in its element is always zero
For a simple ion, the oxidation state is the same as the charge on the
ion; Na⁺, K⁺ are +1; Mg²⁺, Ca²⁺ are +2; F⁻ and Cl⁻ are -1; O²⁻ and S²⁻
are -2
The oxidation state of fluorine is always -1 because it is the most
electronegative element
The oxidation state of oxygen is -2, except for peroxides and F₂O
The oxidation state of chlorine is usually -1, except when combined
with F or O
The oxidation state of hydrogen is +1, except when bonded to a
metal ion, when it is -1
The sum of all the oxidation states of the atoms or ions in a
compound is always zero
The sum of the oxidation states in a polyatomic ion is always the
charge on the ion






Many compounds have traditional names e.g. water and
ammonia which tell you nothing about the elements in it
As some elements can have different oxidation states in
their compounds (e.g. iron oxide can be FeO or Fe₂O₃)
using the oxidation state can help with naming them.
If a metal can only have one oxidation state in its
compounds, the oxidation state is not put it i.e. NaCl is
sodium chloride, not sodium (I) chloride
Example: HNO₃
Using the rules, H=+1 and O=-2, making the current
oxidation state -5.
As there is no overall charge on the compound, N must be
+5 and therefore this is nitric (V) acid



Oxoanions are negative ions that contain oxygen.
Oxoanions containing a metal need a systematic
name to show the oxidation state of the metal.
Acids of oxoanions are named after their anions
e.g. phosphoric acid, H₃PO₄
To name, first calculate the oxidation state of
phosphorus in PO₄³⁻
◦
◦
◦
◦
◦
Find out number of oxygen atoms (4)
Find out total oxidation number due to oxygen (-8)
Find out overall charge on the ion (-3)
Find the oxidation state of the central atom (+5)
Therefore the name of the species is Phosphoric (V) acid

The equation: Mg + 2HCl  MgCl₂ + H₂ is a
redox reaction
Oxidation state of H
Oxidation state of Mg
Combined in HCl = +1
Uncombined metal = 0
Uncombined element = 0
Combined in MgCl₂=+2
The hydrogen ion is
reduced by the
magnesium metal
The magnesium metal is
oxidised by the
hydrochloric acid

The equation: MnO₂ + 4HCl  MnCl₂ + Cl₂ +
2H ₂ O is a redox reaction
Oxidation state of Mn
Oxidation state of Cl
Combined in MnO₂ = +4
Combined in HCl = -1
Combined in MnCl₂ = +2
Uncombined element=0
The manganese is
reduced by the chloride
ions in HCl
The chloride is oxidised
by manganese (VI) oxide

The equation: MgO + 2HCl  MgCl₂ + H ₂ O is
NOT a redox reaction
Oxidation state of Mg
Oxidation state of Cl
Combined in MgO = +2
Combined in HCl = -1
Combined in MgCl₂ = +2
Combined in MgCl₂ =-1
The oxidation state does
not change
The oxidation state does
not change
1.
2.
3.
4.
5.
1.
2.
3.
4.
Identify the oxidation states of the elements (only one
will be changing in a half equation)
Put in the correct number of electrons on the
appropriate side of the equation
If necessary, add water to balance oxygen atoms
If necessary, add hydrogen ions to balance hydrogen
atoms
Check that the equation balances for atoms and charge
Cr₂O₇²⁻ 2Cl³⁺; oxidation state of Cr (+6), O (-2), Cl (+3)
Cr₂O₇²⁻ +6e⁻  2Cl³⁺
Cr₂O₇²⁻ +6e⁻  2Cl³⁺ +7H₂O
Cr₂O₇²⁻ +6e⁻ +14H⁺  2Cl³⁺ +7H₂O
Trends in physical properties
Trends in the oxidising abilities of the halogens
Trends in the reducing abilities of the halide ions
Identification of halide ions using silver nitrate
Uses of chlorine and chlorate(I)
Element
Symbol
Colour
State
Atomic
No.
Outer
electrons
Atomic
Radius
(nm)
Radius
f X⁻ ion
(nm)
Boiling
Point (K)
Electro
negativity
Electron
Affinity
(kJmol⁻¹)
Fluorine
F
Pale
yellow
Gas
9
5
0.071
0.133
85
4.0
-328
Chlorine
Cl
Light
green
Gas
17
5
0.099
0.181
239
3.5
-349
Bromine
Br
Reddish
-brown
Liquid
35
5
0.114
0.196
332
2.8
-324.6
Iodine
I
Violet
Solid
53
5
0.133
0.220
457
2.5
-295.2
Trend
-
-
-
-
Astatine: Extremely rare (estimated that
there is only 0.029g present in Earth’s
crust), and radioactive, thus chemistry
has not been studied.





The gaseous halogens vary in appearance (see table) and they all have a
characteristic ‘swimming-bath’ smell
A number of the properties of fluorine are untypical. Many of these
untypical properties stem from the fact that the F-F bond is
unexpectedly weak, compared with the trend for the rest of the
halogens. The small size of the F atom leads to repulsion between nonbonding electrons because they are so close together
Size of atoms: the atoms get bigger as we go down the group because
each element has one extra filled main level of electrons compared with
the one above it
Electronegativity: The shared electrons in the H-X bond get further away
from the nucleus as the atoms get larger going down the group. This
makes the shared electrons further from the halogens nucleus and
increases are more important than the increasing nuclear charge, so the
electronegativity decreases as we go down the group
Melting and boiling points: These increase as we go down the group.
This is because larger atoms have more electrons and this makes the
van der Waal’s forces between the molecules stronger. The lower the
boiling point, the more volatile the element. So, chlorine, which is a gas
at room temperature, is more volatile than iodine, which is a solid.
The Reactions of Cl₂ (aq) with Br¯ (aq) and I¯ (aq)
Halide Ion
Observations
Conclusion
Equation
Br¯ (aq)
Yellow/Brown
Cl₂ displaced
2Br¯ + Cl₂  2Cl¯ + Br₂
I¯ (aq)
Brown Colour and/or
Black Precipitate
I₂ displaced
2I¯ + Cl₂  2Cl¯ + I₂
The Reactions of Br₂ (aq) with Cl¯ (aq) and I¯ (aq)
Halide Ion
Observations
Conclusion
Equation
Cl¯ (aq)
No change
Cl₂ not displaced
No reaction
I¯ (aq)
Brown Colour and/or
Black Precipitate
I₂ displaced
2I¯ + Br₂  2Br¯ + I₂
The Reactions of I₂ (aq) with Br¯ (aq) and Cl¯ (aq)
F₂ > Cl ₂ > Br ₂ > I ₂
Halide Ion
Observations
Conclusion
Equation
Br¯ (aq)
No change
Cl₂ not displaced
No reaction
I¯ (aq)
No change
I₂ not displaced
No reaction

The trend in the reducing power of the halide ions is shown in the
reaction of solid halide salts with concentrated sulphuric acid. The
oxidation state of sulphur in sulphuric acid is +6. This can be reduced to
+4, 0 or -2 depending on the reducing power of the halide ion.
NaX
Observations
Products
Type of Reaction
NaF
Steamy fumes
HF
Acid-base (F⁻ acting as a base)
NaCl
Steamy fumes
HCl
Acid-base (Cl¯ acting as a base)
NaBr
Steamy fumes
Colourless gas
Brown fumes
HBr
SO₂
Br₂
Acid-base (Br¯ acting as a base)
Redox (reduction product is H₂SO₄)
Redox (oxidation product of Br¯)
NaI
Steamy fumes
Colourless gas
Yellow solid
Smell of bad eggs
Black solid, purple fumes
HI
SO₂
S
H₂S
I₂
Acid-base (I¯ acting as a base)
Redox (reduction product is H₂SO₄)
Redox (reduction product is H₂SO₄)
Redox (reduction product is H₂SO₄)
Redox (oxidation product of I¯)
These results indicate that:
•Iodide ions can reduce the sulphur in H₂SO₄ from oxidation state +6 to +4 in
SO₂, then to 0 as the element sulphur, and finally to -2 as H₂S
•Bromide ions can reduce the in H₂SO₄ from oxidation state +6 to +4 in SO ₂
•Fluoride and chloride ions cannot reduce the sulphur in H₂SO₄ under these
conditions
Results found by using silver nitrate and ammonia solution
Halide Ion
Precipitate
Colour
Solubility of precipitate in ammonia solution
F⁻
None
-
Precipitate doesn’t form
Cl¯
AgCl
White
Soluble in dilute NH₃ (aq)
Br¯
AgBr
Cream
Sparingly soluble in dilute NH₃ (aq)
Soluble in concentrated NH₃ (aq)
I¯
AgI
Yellow
Insoluble in concentrated NH₃ (aq)
These results show that the solubility of the silver halides in
ammonia solution decreases in the following order;
AgF > AgCl > AgBr > AgI

Chlorine is a pale green toxic gas that dissolves in water giving a pale green solution. However,
chloride is very reactive and undergoes reaction with the water molecules forming a mixture of
hydrochloric acid and hypochlorous acid (chlorate (I) acid). This solution is usually called
chlorine water:
Cl2 + H2O HCl + HOCl


This is an example of a disproportionation reaction in which the chlorine gets simultaneously
oxidised and reduced.
Chlorine as an element is in the zero oxidation state. In hydrogen chloride it is in the -1
oxidation state and in chlorate (I) acid it is in the +1 oxidation state.
1½Cl2 + e-  Cl2½Cl2 + H2O  HOCl + H+ + e-





The first equation shows chlorine being reduced and the second shows chlorine being oxidised
If chlorine water is tested with indicator paper, it first registers the red colour of the
hydrochloric acid, but then is rapidly decolourised by the action of the hypochlorous acid,
which is a good bleaching agent.
The poisonous nature of chlorine is taken advantage of in water treatment. Humans are
tolerant to small doses of chlorine, while small microorganisms are killed at even low chlorine
concentrations.
This allows water treatment plants to add chlorine to the process, killing microorganisms and
making the water fit for human consumption.
Ozone gas is also used for water treatment, but it has the disadvantage of being difficult to
store and more expensive. It does however have the advantage of not adding taste to the
water.

As a solution of chlorine water is a mixture of acids it is logical that it should react with bases.
Sodium hydroxide neutralises chlorine water forming salts from the two acids present:
2NaOH + HOCl + HCl  NaCl + NaOCl + 2H2O





If chlorine is bubbled through cold sodium hydroxide solution a different reaction occurs:
3Cl2 + 6NaOH --> NaClO3 + 5NaCl + 3H2O
This is an example of a disproportionation reaction in which the chlorine is simultaneously
oxidised and reduced.
Chlorine as an element is in the zero oxidation state. In sodium chlorate(V) it is in the +5
oxidation state and in sodium chloride it is in the -1 oxidation state.
The two half equations representing each process are:
½ Cl2 + e-  Cl½ Cl2 + 6OH-  ClO3- + 3H2O + 5e-
To add the two half-equations together the electrons must first be equalised by multiplying
the first equation by five:
2½Cl2 + 5e- 5Cl½ Cl2 + 6OH-  ClO3- + 3H2O + 5e3Cl2 + 6OH-  ClO3- + 3H2O + 5Cl-

Sodium chlorate(V) is a useful weedkiller and oxidising agent.
Trends in Physical Properties
Trends in Chemical Properties
Outer
electrons
Metallic
Radius
(nm)
Melting
Point
(K)
Boiling
Point
(K)
1st + 2nd IEs
(kJmol⁻¹)
Density
(gcm⁻³)
Element
Symbol
Atomic
No.
Magnesium
Mg
12
2
0.160
922
1380
738+1451=2189
1.74
Calcium
Ca
20
2
0.197
1112
1757
590+1145=1735
1.54
Strontium
Sr
38
2
0.215
1042
1657
550+1064=1614
2.60
Barium
Ba
56
2
0.224
998
1913
503+965=1468
3.51
Trend
-
-




Electron arrangement: the elements all have two electrons in an outer s-orbital.
This s-orbital becomes further away from the nucleus as we go down the group
The sizes of the atoms: the atoms get bigger as we go down the group. The
atomic (metallic) radii increase because each element has an extra filled main level
of electrons compared with the one above it
Melting points: Group 2 elements are metals with high melting points, typical of a
giant metallic structure. As we go down the group, the electrons in the ‘sea’ of
delocalised electrons are further away from the positive nuclei. As a result, the
strength of the metallic bonds decreases as we go down the group. For this
reason, the melting points of Group 2 elements decrease slightly as we go down
the group, starting with Ca. Magnesium, with the lowest melting point, does not
fit this trend. This is because the lattice arrangement of atoms is different from
that of the elements below, which makes them slightly easier to separate
Ionisation energies: in all their reactions, atoms of elements in Group 2 lose their
outer electrons to form ions with two positive charges.
M  M¹⁺ + 2e⁻
So, an amount of energy equal to the sum of the first and the second ionisation
energies is needed for complete ionisation
M  M¹⁺ + e⁻ + M¹⁺  M²⁺ + e⁻
Both the first ionisation energy and the second ionisation energy decrease as we
go down the group; it takes less energy to remove the electrons as they become
further and further away from the positive nucleus. The nucleus is more effectively
shielded by more inner shells of electrons. In all their reactions, the metals get
more reactive as we go down the group.



Oxidation is the loss of electrons so in all their reactions, the Group 2 metals are
oxidised. The metals go from oxidation state 0 to oxidation state +2. These are
redox reactions
Reaction with water
With water, we see the same trend in reactivity; the metals get more reactive as we
go down a group. The basic reaction is as follows, where M is any Group 2 metal:
M(s) + 2H₂O(l)  M(OH)₂(aq) + H₂(g)


Magnesium hydroxide is ‘milk of magnesia’ and is used in indigestion remedies to
neutralise excess stomach acid.
Magnesium reacts very slowly with cold water but rapidly with steam to form an
alkaline oxide and hydrogen.
Mg(s) + H₂O(g)  MgO(s) + H₂(g)

Calcium reacts in the same way but more vigorously, even with cold water.
Strontium and barium react more vigorously still. Calcium hydroxide is
sometimes called ‘slaked lime’ and is used to treat acidic soil. Most plants
have an optimum level of acidity or alkalinity in which they thrive. For
example, grass prefers a pH of around 6 so if the soil has a pH much below
this, then it will not grow as well as it could. Crops such as wheat, corn, oats
and barley prefer soil that is nearly neutral.





There is a clear trend in the solubilities of the
hydroxides, as we go down the group: they
become more soluble. The hydroxides are all
white solids.
Magnesium hydroxide, Mg(OH)₂ (milk of
magnesia), is almost insoluble. It is solid as a
suspension in water, rather than a solution.
Calcium hydroxide, Ca(OH)₂, is sparingly soluble
and a solution is used as lime water.
Strontium hydroxide, Sr(OH)₂, is more soluble
Barium hydroxide, Ba(OH)₂, dissolves to produce
a strongly alkaline solution:
Ba(OH)₂(s) + aq  Ba²⁺(aq) + 2(OH)⁻(aq)



The solubility trend in the sulphates is exactly the opposite, they
become less soluble as we go down the group. So, barium
sulphate is virtually insoluble. This means that it can be taken by
mouth as a ‘barium meal’ to outline the gut in medical X-rays.
(The heavy barium atoms is very good at absorbing X-rays). This
test is safe, despite the fact that barium compounds are very
toxic, because barium is so insoluble.
The insolubility of barium sulphate is also used in a simple test
for sulphate ions in solution. The solution is first acidified with
nitric or hydrochloric acid. Then, barium chloride solution is
added to the solution under test and if a sulphate is present a
white precipitate of barium sulphate formed.
Ba²⁺(aq) + SO₄²⁻(aq)  BaSO₄(s)
The addition of acid removes carbonate ions as carbon
dioxide. (Barium carbonate is also a white insoluble
solid, with would be indistinguishable from barium
sulphate)
Compound
Uses
Magnesium hydroxide
Used in medicine.
Milk of magnesia which is used to neutralise
excess stomach acid and treat indigestion
Calcium hydroxide
Barium sulphate
Used in agriculture to neutralise acidity of soil
“Barium meal”
Used in medicine.
Very insoluble & appears opaque on X-Rays so
is used to diagnose problems with the digestive
system
Principles of Metal Extraction
Environmental Aspects of
Metal Extraction




The production of metals is an important part of the chemical
industry. The way in which metals are produced from natural
resources involves an understanding of both social and economic
aspects of the processes, as well as an appreciation of the
underlying chemistry.
Of the possible engineering materials, only Al and Fe are very
abundant, with Ti being the next most common. These three
metals are also widely distributed, and so are metals we might
expect to be used whenever possible.
Some metals in everyday use, e.g. Cu and Ni, are relatively rare in
the Earth’s crust. Fortunately, they sometimes occur in highgrade ores in a few specific locations. To be economically viable,
ores should have a high concentration of the desired metal and
be free of impurities.
It should be noted that ores of expensive commercial metals,
such as copper, may have a low concentration of the metal
whereas ores of lower value metals, such as iron, must contain
very high concentrations of the metal to be economically viable.

Metals usually occur in combination with oxygen or sulphur. The process of
extraction is therefore one of reduction – often of oxides.
As these oxides are very stable compounds, energy usually has to be put in to
reduce them to the metal i.e. extraction reactions are endothermic (unless a very
powerful reducing agent such as Al metal is used)
Sulphide ores are not usually reduced directly to the metals. They are first roasted
in air to produces oxides; this liberates sulphur dioxide and so gives rise to the
potential pollution hazard of acid rain.
Sulphur dioxide dissolves in water in the clouds to form sulphurous acid

Some of the sulphur dioxide is oxidised to sulphur trioxide

This gas dissolves in water to form sulphuric acid





SO₂ (g) + H₂O (l)  H₂SO₄ (aq)
2SO₂ (g) + O₂ (g)  2SO₃ (g)
SO₃ (g) + H₂O (l)  H₂SO₄ (aq)
These acids can then fall as acid rain, damaging plants and polluting lakes
Oxide ores are often reduced directly to the metal. However, the use of fossil fuels
e.g. coke from coal, may give a rise to environmental problems through emission
of carbon dioxide, a greenhouse gas

The method chosen for industrial scale for each metal depends
on:
◦ The cost of the reducing agent: a reducing agent that is naturally available
such as carbon (from coal) or hydrogen (from natural gas) is cheaper than
one, such as aluminium, which has to be prepared by a separate and often
more costly process
Carbon is usually cheaper than hydrogen and easier to store. Carbon is
readily available as coke while hydrogen has to be prepared from methane
or water.
◦ The energy costs for the process: a process, which requires less energy by
operating at a lower working temperature usually, has an economical
advantage over a process that requires greater energy input
◦ The required purity of the metal.

The costs of the reductant and the energy required have to be
weighed against each other. If a metal is required with very high
purity (and in small quantity) for a particular use, then a
relatively expensive process may be used. The extra cost being
necessary to satisfy demand for high purity and justified by the
product’s market value. If high purity is not essential, then the
cheapest method of producing saleable metal is used.



Carbon is a cheap and plentiful reducing agent –
it occurs in coal (which when heated in the
absence of air gives coke, a solid with a very high
carbon content) and in charcoal (which is
obtained from wood, a renewable resource)
All metal oxides in theory can be reduced by
carbon if the temperature is high enough. In
practice, temperatures above 2000˚C are
uneconomic and impractical
The most important example of carbon reduction
is the manufacture of iron from high quality
haematite, Fe₂O₃



Iron oxides are reduced in the blast
furnace using coke.
This is a continuous process
(minimises costs) in which iron (III)
oxide, Fe₂O₃ ,coke and limestone are
fed in at the top of the furnace and
hot air is blown in near the bottom
Molten iron collect at the bottom of
the furnace and is run off. The
product is very impure iron which
typically contains ~4% carbon, as well
as manganese, silicon, phosphorus
and sulphur in amounts which
depend on the operating conditions
and ore used.
1.
2.
Coke reacts with the hot air blast in a strongly exothermic reaction
C (s) + O₂ (g)  CO₂ (g)
This reaction produces the heat energy needed for the reduction of the iron (III)
oxide
The carbon dioxide formed reacts at lower temperatures with unreacted coke to
form carbon monoxide
CO₂ (g) + C (s)  2CO (g)
3.
The carbon monoxide reduces most of the iron (III) oxide at around 1200˚C
4.
In the hotter part of the furnace, coke also reacts directly with the iron oxide

Fe₂O₃ (s) + 3CO (g)  2Fe (l) + 3CO₂ (g)
Fe₂O₃ (s) + 3C (s)  2Fe (l) + 3CO (g)
The other product in the process is slag; this contains impurities from the ore
combined with lime, and is largely calcium silicate, CaSiO₃; it decomposes in the
heat of the furnace to form calcium oxide (lime) and carbon dioxide
CaCO₃ (s)  CaO (s) + CO₂ (g)

Calcium oxide is a basic oxide and combines with the acidic oxides in the
furnace to form the slag. Sand and soil contain the acidic silica, which reacts to
give calcium silicate
CaO (s) + SiO₂ (s)  CaSiO₃ (l)

Slag is used to make ‘breeze blocks’ for use in the construction industry



Many metals are extracted by reduction of their
oxides with carbon. However, some metals form
metal carbides rather than the metal itself, so it
is not a practical method for extracting these
metals
This route can be used to make carbides, which
are important engineering materials and
potential catalysts. When titanium (IV) oxide is
heated with carbon, the following reaction
occurs:
TiO₂ + 3C  TiC + 2CO
Oxides of tungsten give carbides when they are
heated with carbon. Tungsten is prepared by
reduction of its oxides with hydrogen




The manufacture of aluminium is carried out by the electrolysis of purified
bauxite, Al₂O₃, which is dissolved in molten cryolite, Na₃AlF₆. The melting point of
aluminium oxide is >2000˚C, but by dissolving it in molten cryolite, the
temperature of the melt is reduced to about 970˚C. The electrodes are made of
carbon and the reactions at the electrodes are:
◦ At the cathode: Al³⁺ + 3e⁻  Al
◦ At the anode: 2O²⁻  O₂ + 4e⁻
Some of the oxygen evolved reacts with the anodes at high temperature, forming
carbon monoxide and carbon dioxide:
2C + O₂  2CO and C + O₂  CO₂
The process uses large amounts of electricity (because electricity is needed to
melt the cryolite and decompose the Al₂O₃) and is only economic when electricity
is relatively cheap. The process is continuous but regular additions of aluminium
oxide are needed, and the carbon electrodes need replacing as they are
consumed. There is potential environmental problems through waste cryolite
causing fluoride pollution
Aluminium is strong, light and can be made into thin sheets. It has a low density,
and is an excellent conductor of heat and electricity. Although reactive, it does not
corrode easily. If exposed to air, a clean surface reacts quickly with oxygen
forming a protective coating of aluminium oxide.



When the purity of the metal is of prime importance, and contamination by carbon or
oxygen cannot be tolerated, then reduction of a metal halide with a reactive metal
becomes a desirable route despite the costs associated
Titanium is very abundant in the Earth’s crust. It is also a very desirable engineering
metal, having a low density, high strength and high resistance to corrosion. However,
unlike iron, traces of carbon, oxygen or nitrogen have unwelcome effects (e.g. the metal
is brittle). Titanium is thus extracted from its chloride by reduction with an active metal
The ore rutile, impure titanium (IV) oxide, is converted into titanium (IV) chloride using
chlorine and coke at about 900˚C
TiO₂ + 2C + 2Cl₂  TiCl₄ + 2CO

Titanium (IV) chloride is a colourless liquid which fumes in moist air because of
hydrolysis. It is purified from other chlorides (e.g. those of iron, silicon and chromium)
by fractional distillation under argon or nitrogen. In the UK, the chloride is then reduced
by sodium in the exothermic reaction
TiCl₄ + 4Na  Ti + 4NaCl

The sodium is initially held at about 550˚C but the temperature rises to ~1000˚C during
the reaction. An inert atmosphere of argon is used to prevent contamination of the
metal with oxygen or nitrogen. The sodium chloride by-product is washed out, leaving
titanium as a granular powder.

Elsewhere in the world, magnesium is used as the reducing agent in a similar reduction
process (called the Kroll process). The magnesium chloride by-product is removed from
the titanium by vacuum distillation at high temperature.
TiCl₄ + 2Mg  Ti + 2MgCl₂

This is a batch process – one batch at a time, which is more costly than a continuous
process. Costs are high because:
◦
◦
◦
◦




Chlorine and sodium have to be produced first (reducing agents not readily available)
High temperatures are involved in both stages of production
Precautions have to be taken in handling TiCl₄, which reacts violently even with cold water
An argon atmosphere has to be maintained to prevent oxidation
Despite extensive searches, other methods for producing titanium have not yet been
able to compete because of cost and purity considerations. Thus use is limited, despite
its desirable properties and a high natural abundance, due to the high cost of
production.
Tungsten cannot be extracted using carbon because a carbide is formed
It is useful because of its high boiling point
It is reduced using hydrogen to extract it
WO₃ + 3H₂  W + 3H₂O






Recycling of metals is carried out extensively, and would be environmentally friendly
if all scrap was returned to the metal works and recycled. Unfortunately, because
scrap metal is often widely spread and many miles from metal-producing plants, the
scrap must be collected and transported. This procedure creates an energy cost
which must be carefully calculated and offset against the savings in extraction
Iron (as steel) is the most extensively recycled metal with as much as 40% of the
world’s iron and steel production coming from scrap metal. The world reserved of
iron ore are vast and commercial ores have high iron content – so why recycle?
Percentage iron: the scrap iron contains a higher percentage of iron than a
commercial ore. Many alloy steels are produced using only scrap metal in the Electric
Arc Process. Also, in the BOS steel making process, about 30% of scrap metal is
added before adding the impure liquid iron
Environmental issues: without recycling, a serious environmental would result from
all the scrap metal that is discarded. About 1.5 million cars a year are discarded in
the UK; with about 75% of a car’s mass is recycled and only about 0.3% going to
landfill. Steel cans are also recycled as they are magnetic and easily separated from
other waste.
The extraction methods for both iron and aluminium produce carbon dioxide a
‘greenhouse gas’; re-melting does not produce this. Unlike extraction of iron, the
extraction of aluminium is very expensive because electrolysis consumes vast
amount of energy. The re-melting of aluminium cans, which are high-quality scrap,
saves 95% of this extraction energy.
It should be noted that there are considerable energy costs in collecting and
separating the scrap aluminium and these costs must be carefully balanced against
the savings made in extraction.
Synthesis of Chloroalkanes
Nucleophilic Substitution
Elimination

1.
Methane does not react with chlorine at room temperature, or in the dark. In the presence of
ultraviolet light, however, a mixture of chlorine and methane will explode, forming hydrogen
chloride and a mixture of chlorinated methanes. This is a free radical substitution reaction
which occurs in steps:
Initiation
Cl₂  2Cl∙
The ultraviolet light provides the energy needed to start the reaction i.e. to split some chlorine
molecules into atoms (radicals). This occurs first because the Cl-Cl bond in chlorine is weaker than the
C-H bond in methane
◦
◦
Propagation
2.
Cl∙ + CH₄  CH₃∙ + HCl
CH₃∙ + Cl₂  CH₃Cl + Cl∙
Overall reaction: CH₄ + Cl₂  CH₃Cl + HCl
In each step, a radical is used and a radical is formed so the process continues and leads to a chain
reaction. Each step is also exothermic, so the chain reaction can produce an explosion
◦
◦
◦
◦
Termination
3.
If two radicals combine (i.e. unpaired electrons pair up to form a covalent bond) they form a stable
molecule, and the sequence of reaction stops. Possible termination steps include:
◦


◦
CH₃∙ + Cl∙  CH₃Cl
CH₃∙ + CH₃∙  CH₃CH₃
Such termination steps lead to trace amounts of impurities, such as ethane, in the chloromethane








The reaction of a chlorine radical with methane
extracts a hydrogen radical to form HCl
Chloromethane still contains three hydrogen
atoms, so further pairs of propagation steps are
possible, leading to CH₂Cl₂, dichloromethane,
CHCl₃, trichloromethane and CCl₄,
tetrachloromethane
CH₃Cl + Cl∙  ∙CH₂Cl + HCl
∙CH₂Cl + Cl₂  CH₂Cl₂ + Cl∙
CH₂Cl₂ + Cl∙  ∙CHCl₂ + HCl
∙CHCl₂ + Cl∙  CHCl₃ + Cl∙
CHCl₃ + Cl∙  ∙CCl₃ + HCl
CCl₃ + Cl₂  CCl₄ + Cl∙
When haloalkanes are
warmed with aqueous
solution or potassium
hydroxide, alcohols
are formed
When haloalkanes are
warmed with
aqueous/alcoholic
solutions of potassium
cyanide, nitriles (RCN) are
formed.
Note: an extra carbon has
been added to the chain
Nitriles are hydrolysed to
carboxylic acids by
heating under reflux
either with aqueous
alkali, or mineral acid; an
amide intermediate forms
When haloalkanes are warmed
with excess ammonia in a
sealed container, primary
amines are formed
Since the acid HBr will
immediately react with the base
ammonia
The excess ammonia minimises
the chance of further reaction
of the primary amines to form
secondary, tertiary or
quaternary ammonium salts
Reduction of nitriles, using
hydrogen, in the presence of a
nickel catalyst, also produces
primary amines
Structure, Bonding and Reactivity
Addition Reactions
Polymerisation


Alkenes contain 2 hydrogen atoms fewer than
their parent alkanes, and are said to be
unsaturated. This is because they C=C double
bond
The double bond results from overlap of the
spare unbonded, singly-filled p-orbital present
on each carbon atom in the bond. This overlap
produces a cloud of electrons above and below
the molecule (a Π bond). The two carbon atoms
of the double bond, plus the fur atoms attached
to the double bond must lie in the same plane,
thus ethene is a planar molecule.



If the alkene is unsymmetrical and the
molecule being added is also unsymmetrical
then two possible products can form. The
major product is the one formed via the more
stable carbocation.
Tertiary > Secondary > Primary
This is due to the inductive (electron
releasing) effect of the attached alkyl groups.
The more alkyl groups around the
carbocation, the more stable it is and more
likely it is to be formed.
A curly arrow
shows the
movement of
a pair of
electrons
Nomenclature
Ethanol Production
Classification and Reactions
Elimination


When naming an alkene derivative, look for
the longest carbon chain in the skeleton
Look at slide 50 for a recap.

Alcohol is produced by the fermentation process that uses
living yeast cells to convert sugars (e.g. glucose) into
carbon dioxide and ethanol
C₆H₁₂O₆  2C₂H₅OH + 2CO₂


The reaction is slow at low temperature as the enzymes in
yeast are inactivated. At higher temperatures, the yeast is
killed as the enzymes are denatured. The process is
therefore usually carried out at ~35˚C
Fermentation produces an aqueous solution of ethanol of
concentration between 3-15%. Beers are usually about 37% and wines about 9-13%. Fermentation rarely produces
higher ethanol concentrations as these kill the yeast. More
concentrated alcohol solutions i.e. spirits such as whisky,
brandy or gin, are about 40% alcohol, and are produced by
fractional distillation of the fermentation products.


Ethanol is also produced industrially by the direct hydration of ethene using steam
and a phosphoric acid catalyst at 300˚C and 6.5x10³ kPa (65 atmospheres)
pressure
The direct hydration method is preferred at present for the production of ethanol
for industrial use in the UK. However, as the method uses ethene as a raw
material, it may become less popular compared with fermentation in the future as
oil supplies begin to run out
Method
Hydration
Fermentation
Rate of Reaction
Fast
Slow
Quality of
Product
Raw Material
Type of Process
Pure
Ethene from Oil
(finite resource)
Continuous
(manpower cheap)
(equipment expensive)
Impure
Sugars
(renewable resource)
Batch
(manpower expensive)
(equipment cheap)

Alcohols are oxidised by strong oxidising
agents. Potassium dichromate(VI) in the
presence of dilute acid is the reagent of
choice. The reaction is useful because the
different types of alcohol behave differently.
◦ 1º alcohols get oxidised to aldehydes
◦ 2º alcohols are oxidised to ketones
◦ 3º alcohols are not oxidised






Alcohol: ethanol
1st Step: H⁺/K₂Cr₂O₇; acidified potassium dichromate (VI); orange
Product: ethanal (aldehyde); goes green
2nd Step: Tollen’s reagent/Fehlings solution
Product: ethanoic acid (carboxylic acid); Tollen’s reagent  silver mirror;
Fehlings solution  brick red precipitate
Methods:
Distillation
(aldehyde)
Refluxing
(carboxylic acid)
CH₃CH₂OH + [O]  CH₃COH + H₂O
CH₃CH₂O + [O]  CH₃COOH
CH₃CH₂OH + 2[O]  CH₃COOH + H₂O






Alcohol: propan-2-ol
1st Step: H⁺/K₂Cr₂O₇; acidified potassium dichromate (VI); orange
Product: propanone (ketone); goes green
2nd Step: Tollen’s reagent/Fehlings solution
Product: no change; no silver mirror; no brick-red precipitate; no further
oxidation
Methods: use refluxing
CH₃CHCH ₃OH + [O]  CH₃COH₃ + H₂O



Alcohol: 2-methylpropan-2-ol
1st Step: H⁺/K₂Cr₂O₇; acidified potassium
dichromate (VI); orange
Product: no change; no oxidation state; stays
orange



The products formed depend on the conditions used:
Alkenes are formed in the presence of H2SO4 (or
H3PO4 better, as it doesn't produce as many byproducts) and the correct temperature (hot for
primary, warm for secondary and cool for tertiary)
alcohols lose a water molecule.
One of the advantages of this method of preparing
alkenes is that the starting materials, the alcohols,
can be produced by biological means such as
fermentation. This means that plastic manufacture
does not have to depend on the petrochemicals
industry.
Fermentation  alcohols  dehydration
 alkenes  polymerisation  polymers
Mass Spectrometry
Infrared Spectroscopy



Mass spectrometry is the main method for
finding the relative molecular mass of organic
compounds.
See slide 7 for a recap
The strength of the magnetic field required to
deflect the ion into the detector depends on the
mass and charge of the ion. The output is then
presented as a graph of relative abundance
against mass/charge ratio. However, since the
charge on the ions is usually +1, the
mass/charge ratio is effectively relative mass.
This graph is called a mass spectrum.




When ethanol is ionised it forms the ion CH₃CH₂OH⁺. This is called the molecular ion. Many of
these ions will then break up; some of their bonds break as they are ionised, so we have other
ions of smaller molecular mass. This process is called fragmentation. Each of these produces a
line in the mass spectrum. These can provide information that will help to deduce the structure
of the compound. They also act as a ‘fingerprint’ to help identify.
However there are normally a few ionised molecules remaining intact to give a peak
corresponding to the relative molecular mass of the compound.
The main peak furthest to the right of the mass spectrum, corresponds to the molecular ion (it
has the highest mass). The molecular ion peak for ethanol is at mass 46; this tells us the
relative molecular mass of ethanol.
Mass spectrometry is the most important technique for measuring the relative molecular mass
of organic compounds.
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Many mass spectrometers can measure masses to
three or sometimes four decimal places. This method
allows us to work out the molecular formula of the
parent ion. It makes use of the fact that isotopes of
atoms do not have exactly whole number atomic
masses (except for carbon-12).
For example, a parent ion of mass 200, to the nearest
whole number could have the following molecular
formulae: C₁₀H₁₆O₄;C₁₁H₄O₄;C₁₁H₂₀O₃
Adding up the accurate atomic masses gives the
following Mr’s:
◦ C₁₀H₁₆O₄ = 200.1049
◦ C₁₁H₄O₄ =200.0110
◦ C₁₁H₂₀O₃ =200.1413
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These can easily be distinguished by high resolution
mass spectrometry
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A pair of atoms joined by a chemical bond is always
vibrating. Stronger bonds vibrate faster (at higher
frequency) and heavier atoms make the bond vibrate
more slowly (at lower frequency). Every bond has its
own unique natural frequency that is in the infra-red
region of the electromagnetic spectrum
When we shine a beam of infra-red radiation (heat
energy) through a sample, the bonds in the sample
can absorb energy from the radiation and vibrate
more. However, any particular bond can only absorb
radiation that has the same frequency as the natural
frequency of the missing frequencies that correspond
to the bonds in the sample.
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A beam of infra-red radiation containing a
frequencies is passed through a sample
The radiation that emerges is missing the frequencies
that correspond to the types of bonds found in the
sample
The instrument plots a graph of the intensity of the
radiation emerging from the sample, called the
transmittance, against the frequency of radiation
The frequency is expressed as a wavenumber,
measured in cm⁻¹
Source of
infra-red
radiation
Sample
Detector
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The dips in the infra-red spectrum (peaks) represent particular bonds.
These can help us identify the functional groups present in a compound
Bond
Wavenumber/cm⁻¹
C-H
2850-3300
C-C
750-1100
C=C
1620-1680
C=O
1680-1750
C-O
1000-1300
O-H (alcohols)
3230-3550
O-H (acids)
2500-3000
N-H
3300-3500
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The area of an infra-red spectrum below about 1500cm usually has many peaks caused by a
complex vibrations of the whole molecule. This shape is unique for any particular substance.
It can be used to identify the chemical, just as people can be identified by their fingerprints. It
is therefore called the fingerprint region.
We can use a computer to match the fingerprint region of a sample with those on a database of
compounds. An exact match confirms the identification of the sample.
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Infra-red spectra can also be used to show up the
presence of impurities. These may not be revealed by
peaks that should not be there in the pure compound.
In the spectra below, we can compare pure and impure
caffeine. The broad peak at around 3000cm in the impure
sample is an O-H stretch caused by water in the sample
that has not completely dried. Notice that there are no OH bonds in caffeine.