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AS Chemistry Unit 2
AQA AS-LEVEL
Student Guide to Unit 2
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LUND Apr-17
1
AS Chemistry
Unit 2
AS Chemistry Unit 2
Chemical Energetics
Chemical bonding

a chemical bond is an electrostatic force of attraction between oppositely charged particles
Ionic
Metallic
Covalent

‘an electrostatic force of attraction between a lattice of oppositely charged ions
– cations (which are ‘pussytive’) and anions (which are negative)
‘an electrostatic force of attraction between delocalised valence electrons and a
lattice of positive ions’ (the positive centres are what remain behind as a
result of the valence electrons being delocalised)
‘a shared pair of electrons’
obviously the true nature of a covalent bond in terms of electrostatic interactions is not
fully explained in the description above but it is acceptable at this level (a simple
interpretation of the potential well concept can better explain the concepts of bond length,
bond energy and why helium does not form a diatomic molecule).
http://mysite.verizon.net/kdrews47/bonding/bonding3.html#Well

there is a general association between bond dissociation enthalpy and bond length in that
longer bonds (involving larger atoms) are typically weaker – see Coulomb's Law
the more able could possibly explain E(F-F) in the context of the trend in group 7

bond dissociation enthalpy of triple bonds > double bonds > single bonds although the
individual component bonds of a multiple bond are not of equal strength i.e. a double bond is
not twice the value of a single bond between the same atoms (the nature/efficiency of orbital
overlap plays a part – look up ‘hybridisation’ on chemguide if you want to understand more)

in the case of C=C the double bond is less than twice the value of the single (σ) bond as the
orbital overlap of the second bond (known as the π ‘pi’ bond) is less efficient

chemical reactions involve breaking bonds and then forming new bonds as all that has taken
place is a rearrangement of the participating particles (hence the underlying basis of a
balanced chemical equation)
LUND Apr-17
2
AS Chemistry Unit 2


breaking bonds (bond dissociation) requires energy
forming chemical bonds releases energy

the energy required to break a given bond is the opposite of that released when it is
formed under identical conditions (the dehydration and re-hydration of hydrated copper(II)
sulphate is a good example of this – and fun to do)

the overall outcome (exothermic or endothermic) of a chemical reaction is a consequence of
the energy input to break bonds and the energy output when new bonds are formed

the amount of energy exchanged with the environment varies pro-rata with the quantities of
chemicals reacted (somewhat obvious if you think about it … hotter fire = larger gas bill)

if a reaction is exothermic overall then energy will have been released to the environment as
the energy supplied to break the old bonds is less than the energy released when the new
bonds are formed
in effect the new bonds are stronger than the old bonds
exothermic examples:
combustion, respiration



ENDOTHERMIC
EXOTHERMIC
+H
-H
if a reaction is endothermic overall then energy will have been taken from the environment
(the degree of this will determine whether an additional energy source such as a Bunsen
burner is necessary) as the energy supplied to break the old bonds is more than the energy
released when the new bonds are formed
in effect the new bonds are weaker than the old bonds (at first sight this may seem
implausible as it suggest a less stable situation arises, however, there is another energetic
concept called Entropy which we’ll explain that to you next year)
endothermic examples:
thermal decomposition, electrolysis, photosynthesis
Summary Questions
How Science Works
Page 114
page 113
AS Chemistry (Nelson Thornes) AQA
Chemguide
1, 2, 3 and 4
‘The energy value of fuels’
112 -- 114
Energetics
s-cool: Chemical Energetics
LUND Apr-17
3
AS Chemistry Unit 2
Mean Bond Enthalpy

values for bond dissociation enthalpies represent average values in the gaseous state (this
will result in slightly different values to those obtained for specific reagents)

using average bond energies to do calculations is pretty much the same level as what you did
in GCSE Unit 3 so first of all lets re-visit those calculations
(i)
(ii)

How Science
Works: E
combustion of methane and other hydrocarbons
combustion of alcohols in the homologous series (where each alcohol differs by CH2
and so there will be extra bonds to break and an extra CO2 and H2O produced per
increment – perhaps you can work out the expected values starting from methanol)
representing the above as an energy cycle – see page 133 although thermochemical cycles will
be explained in the next section in more detail
Summary Questions
Exam Style Questions
Page 133
Page 134
AS Chemistry (Nelson Thornes) AQA
Chemguide
1, 2, 3, 4 and 5
1
131 - 133
Bond enthalpy (the first bit)
s-cool: Chemical Energetics
LUND Apr-17
4
AS Chemistry Unit 2
Enthalpy Change

enthalpy (H) is the total heat content of a system and cannot be determined experimentally

ALL chemical reactions are accompanied by energy changes (principally in the form of heat
energy but also as light, sound etc) which are transferred to/from the system/surroundings
(exothermic or endothermic)

note: heat and temperature are related they are not the same quantity e.g. consider the skin
damage caused by a kettle of boiling water at 100oC and sparks from a sparkler at ~2000 oC

reaction pathway diagrams can be used to show exothermic (-H) and endothermic (+H)
reactions.

even in an exothermic reaction some (but not all) bond breaking must occur first so some
energy must be initially supplied (e.g. a match, or being rather clumsy with nitro-glycerine!)
- this is the activation energy

once the reaction has started, energy released as new
bonds form is used to break more bonds hence an
exothermic reaction continues without a continuous
supply of energy.

this is not the case for an endothermic reaction where
energy must be continuously supplied for the reaction
to take place e.g. in photosynthesis


enthalpy changeH = heat exchange at constant pressure (open container)
enthalpy change cannot be measured directly but can be determined experimentally
(usually done by measuring temperature change at constant pressure in a calorimeter)

H varies with temperature and pressure so Standard Conditions (Ho298) are required:
temperature
pressure
physical state
298 K (25ºC i.e. nominal room temperature)
100kPa don’t put ‘1 atmosphere’ in the exam!!!
at room temperature and most stable allotrope (e.g. graphite)
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AS Chemistry Unit 2

you MUST learn the definition of the enthalpy of formation and the enthalpy of
combustion (one of them is a certainty – well almost – in the exam!)

the use of these values is considered in the next section

the enthalpy of formation of an element is by definition zero

there is a lot of commonality in the wording – hence the method used below to help you learn:
the enthalpy of
is the enthalpy change that occurs when one mole of
in their standard states at 298K and 100 kPa (i.e. standard conditions)
formation (Hof,298)
a compound is formed from its elements
combustion (oc,298)
an element or compound is completely combusted in xs oxygen
Summary Question
Page 116
AS Chemistry (Nelson Thornes) AQA
Chemguide
1
115 -- 117
Enthalpy change
s-cool: Chemical Energetics
LUND Apr-17
6
AS Chemistry Unit 2
Hess’s LAW
Hess’s Law: the enthalpy change for a reaction is dependent only on the initial and final
states of the system and is independent of the route taken.

this is a consequence of the first law of thermodynamics - energy can neither be created
or destroyed (Hess’s law is analogous to saying that the height of the mountain you climb
(hence the gain in potential energy) is independent of your route up it – even if you have a
rather cunning short cut – consider the consequences for an external observer watching
several mountaineers using different routes if this was not true!)

enthalpy of reaction (Hor) = enthalpy change for the specific reaction equation given under
the stated conditions
i.e. this is specific to the equation as it is written (again another obvious idea … you get twice
as much energy released from two moles of fuel than is released from one) – this will become
better understood after a few calculations in the next section


we use Hess’s Law to construct simple energy cycles (NOTE: inverse sign for reverse path –
again lets use our mountain analogy – if you climb a 1000m cliff half way up you gain
+500m, but if you fall of when you get there it’s -500m height change that will occur as you
are now travelling in the opposite direction to the data specified)

enthalpy changes that cannot be determined by direct measurement can be determined
indirectly using Hess’s Law e.g. Hof,298 (CO) or thermal decomposition reactions (the
reasons for that one should be obvious)
this is analogous to it being possible to determine the height of an impossible cliff by using a
GPS on a viable route – its still the same height irrespective of the route taken

Enthalpy Level Diagrams and Thermochemical Cycles – what’s the difference?






USING ENTHALPY LEVEL DIAGRAMS TO DO CALCULATIONS IS
MENTIONED HERE SO THAT YOU ARE AWARE OF AN ALTERNATIVE
APPROACH – EXAMPLES ARE SHOWN IN YOUR TEXTBOOK
you will most likely find that thermochemical cycles are easier to use to solve simple
problems
There will be situations in which energy level diagrams are required at A2 e.g. when you meet
the Born-Haber Cycle
thermochemical cycle diagrams have no energy scale y-axis so can be written in any
orientation (however writing the reaction you are investigating at the top of an inverted
triangle is best practise)
however, enthalpy level diagrams set the energy value of elements (in their most stable
allotropic form at room temperature) as a zero reference (just like sea level).
exothermic changes go down relative to the vertical axis and endothermic changes go up
LUND Apr-17
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AS Chemistry Unit 2
Using Enthalpy of Formation
Reactants
Products
1
Hess’s Law states that 1 = 2
A MARK IN THE EXAM !!

Hor,298 = Hof,298 (Products) - Hof,298 (Reactants)
2
Elements

Tips: Remember that the enthalpy of formation of an element is 0
Hof,298 (H2O) = Hoc,298 (H2) and Hof,298 (CO2) = Hoc,298 (C)
Summary Questions
Page 124
Page 130
Page 134
Page 232
Exam Style Questions
1
1
2
1
Using Enthalpy of Combustion
Reactants
s
Products
1
Hess’s Law states that 1 = 2
Hor,298 = Hoc,298 (Reactants) - Hoc,298 (Products)
2
Combustion
Products
Summary Question
Exam Style Questions
Page 126
Page 134
AS Chemistry (Nelson Thornes) AQA
Chemguide
1
4
127 - 130
Hess, Bond enthalpy (the rest of this section)
s-cool: Chemical Energetics
LUND Apr-17
8
AS Chemistry Unit 2
Calorimetry

it is not possible to measure heat exchange directly but it can be determined from temperature
changes

the usual way to determine an enthalpy change is to measure the temperature change of a
given mass of water (or solution)

q
=
m c 
m=
mass of water/solution
being heated/cooled

c is the specific heat capacity (the heat required to raise 1g of a substance by 1K – UNITS ?)

the next step is scaling to the enthalpy change that would have occurred had one mole of
reactant been used (avoid the common error of mixing up mass of water/solution heated with
mass of chemical used to do the heating)
SCALE up to one mole using:

Enthalpy change


n=
q
n
use
How Science
Works: C, D




moles of limiting
reagent i.e. the one not
in xs
solids
n = m/Mr
solutions
n = cV
Use common sense to make sure
that the sign of the FINAL
answer is correct by considering
the change in temperature.
e.g. –ve if it RISES
AND DON’T FORGET THE
UNITS!!!
A polystyrene cup is used for the determination of reactions in solution (e.g. neutralisation
and displacement)
ensure that you consider the total mass being heated if two solutions are mixed together and
don’t include the mass of any solid added in the total mass being heated
it is assumed that the density of dilute solutions approximates to that of water for converting
volume into mass
in a displacement reaction you must determine which reagent was in excess before
carrying out the scaling calculation
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AS Chemistry Unit 2


the energy released by fuels involves heating a water tank and measuring its temperature gain
investigating a homogenous series of fuels (incrementing by -CH2-) can provide an
experimental basis for mean bond enthalpy values
How Science
Works: E

experimentally determined values obtained in the school laboratory will inevitably differ from
those quoted in data books due to the limitations in the accuracy of the equipment and the
nature of method used

the thermometer will probably contribute the most significant apparatus error (particularly
if the temperature change is small)
however, even if the result lies within the range (from the data book value) associated with
instrumental (measurement) error there will always be experimental error due to heat loss
and this must be commented upon (errors can cancel one another out to give the impression
that the experiment is better than it was)
this is minimised if improved insulation is used


How science works Page 118 - 119
Questions 1 - 3
The Flame Calorimeter – missing diagram in textbook
How Science
Works: D


extrapolating a cooling curve back to the time where the reaction was started improves
accuracy also allowing time for the two solutions to stand while recording their temperature
prior to mixing (such measurements will be used to derive a line of best fit prior to the
reaction being initiated)
it must be remembered that the experiment you carried out was not done under standard
conditions (this is a secondary consideration to the far more significant heat loss)
Summary Questions
Exam Style Questions
Page 121
Page 134
AS Chemistry (Nelson Thornes) AQA
Chemguide
1, 2, 3 and 4
3
117 - 121
s-cool: Chemical Energetics
LUND Apr-17
10
AS Chemistry Unit 2
Reaction Kinetics
Reaction Rate = change of the concentration of products (or reactants) wrt time.
Measuring Reaction Rates

Continuous measurements by remote sensing
(i)
GAS PRODUCED
e.g. Mg + HCl(aq) or marble chips and acid
the volume of gas produced or mass loss as gas leaves the reaction vessel
(ii)
PRECIPITATION – a solid is formed in a reaction between two solutions
e.g. the reaction between hydrochloric acid and sodium thiosulphate (the ‘thiosulphate
cross experiment’) or analysing the relative rates of hydrolysis of alkyl halides using
silver halide precipitation
How Science
Works: F

(iii)
A COLOUR CHANGE IN SOLUTION
e.g. zinc and copper(II) sulphate where the blue colour fades as blue Cu2+ ions are
displaced from solution by the relatively more reactive zinc metal and precipitated as
metallic copper
e.g. the ‘iodine clock’ experiment
a colorimeter can be used for accurate measurements (i.e. less human error)
(iv)
CHANGES IN ELECTRICAL CONDUCTION
i.e. the number of ions in a reaction mixture changes as the reaction proceeds e.g.
hydrolysis of bromobutane by water
Direct Chemical Analysis
samples are taken at given time intervals, and the reaction rate in the sample slowed by rapid
cooling, dilution, removing the catalyst or one of the reactants – quenching before
subsequent analysis by, for example, titration
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11
AS Chemistry Unit 2
Collision Theory

this is basically the same as for GCSE (which means any GCSE book will be good for
learning the basics) i.e.
REACTION RATE VARIES WITH COLLISION FREQUENCY
NOT ALL COLLISIONS LEAD TO A REACTION (MOST DON’T)
THEY MUST HAVE ADEQUATE ENERGY UPON COLLISION TO OVERCOME
ACTIVATION ENERGY (KINETIC BARRIER)
COLLISIONS THAT DO RESULT IN A REACTION ARE DEEMED ‘SUCCESSFUL
COLLISIONS’

the effect of concentration, pressure and surface area can be adequately explained with this
theory on a pro-rata basis e.g.
DOUBLING A REAGENTS CONCENTRATION
→ DOUBLES THE NUMBER OF PARTICLES
→ DOUBLES THE NUMBER OF (SUCCESSFUL) COLLISIONS
→ DOUBLES THE RATE OF REACTION
Activation Energy


however temperature changes and the effect of catalysts require greater consideration
collision theory does not explain the significant effect of raising the temperature

activation energy (kinetic stability) must be considered:
the minimum collision energy between reactant particles per mole of collisions for a
reaction to occur to give products

the peak in an energy level diagram represents a ‘transition state’ or ‘activated complex’ –
some bonds have been broken and the product is now going to be formed

temperature increases mean more effective (with energy greater than the activation
energy) collisions (hence increased rate (e.g. food sell by dates and refrigeration)

same concept can explain the effect of light (e.g. in photosynthesis)
Summary Questions
Page 137
1, 2, 3
LUND Apr-17
12
AS Chemistry Unit 2
Maxwell-Boltzmann Distribution Curve
There is a distribution of
energies amongst particles
(i.e. they don’t all have the
same energy)
no molecules have
zero energy
Few molecules have
very high energy
The Effect of Temperature
Notice that the peaks shift to the right
AND get lower as temperature rises
BUT
EA
Those molecules that can engage in
successful collisions are represented by
the area under the curve where E > EA



You MUST know the effect of temperature change on the shape of the curve and take great
care to use precision when drawing it
implications for reaction rate
NOTE EA DOES NOT CHANGE
Summary Questions
Exam Style Questions
Page 137
Page 144
AS Chemistry (Nelson Thornes) AQA
Chemguide
1
2
136 - 139
Collision, Maxwell
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13
AS Chemistry Unit 2
Catalysis



you should know the correct definition of a catalyst
catalysts enable a different mechanism with a lower activation energy
NOTE CATALYSTS CHANGE THE VALUE EA NOT THE SHAPE OF THE CURVE.

you must ensure that you are able to clearly differentiate between the use of the MBDC
to explain the effect of increased temperature and how a catalyst works in the context of
why more particles find themselves with enough energy to react.
Homogeneous Catalysis

same phase as reactants (e.g. all in solution)
acid catalysed esterification
enzymes in biological systems
chlorine free radicals (formed by the action of UV
light on CFC’s) and ozone (O3) depletion
Heterogeneous Catalysis


different phase to the reactants
typically transition metals or their compounds are used e.g.
manufacture of ammonia
Haber Process
Fe
catalytic converters
Pt and Rh
hardening fats (making margarine) Hydrogenation
Ni
(adsorption onto the surface of the solid nickel catalyst weakens π bonds)
manufacture of nitric acid
Ostwald Process
Pt and Rh
manufacture of sulphuric acid
Contact Process
V2O5

other (non transition metal) examples include:
hydration of ethene to produce ethanol
cracking of hydrocarbons
silica/H3PO4
aluminium oxide/silicon dioxide
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AS Chemistry Unit 2
How Science
Works: L
Catalytic Converters




platinum and rhodium are coated
onto a honeycomb ceramic
material (large surface area =
increased rate) since adsorption
only occurs at the surface
(expensive metal
underneath
would be wasted)
the reactant gases form weak
bonds with the surface of the
catalyst (adsorption) this weakens their bonds thus lowering the activation energy
(additionally the catalyst also helps promote more favourable molecular orientation)
this is followed by desorption in which the products depart
the catalyst selected provides bonding strong enough to hold the reactant gases on the surface
whilst not preventing the products from leaving thus blocking an active site
Write equations
for these
CO and NO react to form CO2 and N2
NO also reacts with uncombusted hydrocarbons to produce CO2, H2O and N2
Hardening Fats – catalytic hydrogenation

the raw materials for the manufacture of margarine are vegetable oils e.g. sunflower, olive)
which contain triglyceride esters derived from propane-1,2,3-triol (glycerol), and polyunsaturated fatty acids such as linoleic acid.

adsorption onto the surface of a nickel catalyst at about 200oC weakens π
bonds allowing hydrogenation
the relative strength of the VdW increases as the removal of semi-rigid double bonds allows
more efficient overlapping thus increasing the relative (i.e. the oil becomes a fat)

Here’s a few links for those of you interested in the history and health aspects of margarine and
spreads (you ALL will eat this stuff at some point even though you might think that you don’t)
http://www.margarine.org.uk/whatisspread-history.html
How Science
http://www.margarine.org/historyofmargarine.html
Works: B
How science works
Summary Questions
Exam Style Questions
Make notes on Margarine and Zeolites (142)
Page 143
1
Page 144
1, 3, 4, 5
Page 234
8
AS Chemistry (Nelson Thornes) AQA
Chemguide
140 - 143
Catalyst, margarine
LUND Apr-17
15
I can’t believe
it’s not
Chemistry
AS Chemistry Unit 2
Chemical Equilibrium


reversible reactions can occur in a closed system e.g. in solution (where no gas is given off)
eventually a position of DYNAMIC equilibrium will be achieved (as opposed to static)
where the relative concentrations of the reactants and products will remain constant providing
that the reaction conditions remain unchanged
at chemical equilibrium the rate of the forward and reverse reaction are identical
this does not mean that the relative amount of reactants and products are identical


examples of a homogeneous equilibrium in which all the reactants and products are in the
same phase include the Haber Process and the Contact Process
note that the catalyst used can still be in a different phase (heterogeneous catalysis)
CATALYSTS DO NOT AFFECT THE POSITION OF EQUILIBRIUM (SINCE THE
FORWARD AND BACKWARD REACTIONS ARE BOTH SPEEDED UP) BUT DO ALLOW
IT TO BE ACHIEVED FASTER

equilibria can be monitored using similar techniques to those used to investigate rates of
reaction
Summary Questions
Page 147
1, 2
Le Chatelier’s Principle.

LCP is a predictive tool NOT an explanation of the reason why the position of equilibria
shifts when reaction conditions (concentration, temperature, pressure) are changed
Le Chatelier’s Principle states that if you change the reaction conditions then the position of
equilibrium changes in the direction that seems to oppose that change

LCP is NOT an explanation of WHY it happens so avoid statements such as ‘because of
LCP’, ‘LCP causes …’ and learn to state ‘LCP predicts that …..’

LCP is not suggesting that the system completely reverses the change imposed, e.g. reversing
a temperature increase when establishing a new equilibrium
it implies that the shift in the position (in terms of reactants and products) of equilibria is in
the direction that seems to minimize the effect of that change
a new position of equilibria in which the relative rates of the forward and backward
reaction are once again in balance under the new set of conditions is eventually arrived at




by custom the chemicals on the RHS of a chemical equation are deemed the products
a shift in equilibria to the right (i.e. the imposed constraint favoured the forward reaction)
increases the yield of the reaction
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AS Chemistry Unit 2
LCP and Concentration

the effect of concentration changes can be predicted by LCP and explained by relative
changes in the rate of the forward and backward reaction when a change in concentration is
made and therefore a change in collision frequency
LCP and Temperature

an increase in temperature favours the endothermic process (which uses energy and so
‘opposes’ that increase)
e.g. the equilibrium between dinitrogen tetroxide and nitrogen dioxide
e.g. the complex looking reactions below
[Cu(H2O)6]2+(aq) + 4Cl-(aq)
Blue

CuCl42-(aq) + 6H2O(l)
Yellow
[Co(H2O)6]2+(aq) + 4Cl-(aq)
CoCl42-(aq) + 6H2O(l)
Pink
Mauve
the reason this is so reflects the fact that EA will be larger for the endothermic process so it
will be relatively more favoured by a rise in temperature
LCP and Pressure

this is only applicable where there is an imbalance between the number of moles of
gaseous particles on either side of the equation

an increase in pressure favours the direction that reduces the total number of gaseous particles
in effect ‘opposing’ the pressure increase by reducing the number of gaseous particles in the
reaction vessel
e.g. the equilibrium between dinitrogen tetroxide and nitrogen dioxide and also the Haber
process
Summary Questions
Exam Style Question
Page 150
Page 154
1, 2
3
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17
AS Chemistry Unit 2
How Science
Works: A
Haber Process


know how we obtain nitrogen and hydrogen
nitrogen is un-reactive (consider the car engine
conditions)

have a look at this animation:
http://www.absorblearning.com/media/attachment.action?quic
k=128&att=2741

consider the opposing affect of temperature on rate and
equilibrium and the compromise made

recycling the un-reacted hydrogen and nitrogen (ammonia is
liquefied and removed) helps to compensate for the poorer
yield thus saving costs

whilst increased pressure provides both an increased yield
AND an increased rate this is limited by cost (energy for
pumps, high pressure vessels) and safety considerations

a catalyst (Fe in small
lumps – large surface
area) is used to increase
rate without requiring an even higher temperature
(energy cost) and of course it can be re-used (cost
saving)

compounds derived from ammonia are important – e.g. nitric acid, fertilizers, nylon, dyes,
explosives etc
How science works
Summary Questions
Exam Style Questions
Page 151 - 152
Page 153
Page 154
1-3
1
4, 5 and 6
Hydration of Ethene to form Ethanol


ethanol can be manufactured as a batch process using fermentation
it can also be made from the direct hydration of ethene obtained from the fractional
distillation of crude oil in a continuous process


the reaction involves the use of a catalyst – phosphoric acid on silica
the position of equilibrium for this process will be affected by concentration, temperature and
pressure
as with the Haber process there are trade offs of which you should be aware and practicalities
to be overcome.

Summary Question
Page 153
2
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18
AS Chemistry Unit 2
Carbon Monoxide and Hydrogen to form Methanol

methanol is used as a chemical feedstock and as an additive to petrol

it can be manufactured by the reversible reaction between carbon monoxide and hydrogen in
the presence of a copper catalyst
the reactants (‘synthesis gas’) are manufactured from the reaction of methane or propane with
steam
as with ethanol the temperature and pressure that are used represent a compromise


Summary Question
Exam Style Questions
Page 153
Page 154
AS Chemistry (Nelson Thornes) AQA
Chemguide
3
1, 2
146 - 153
Haber, Reversible, Le Chatelier’s Principle
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19
AS Chemistry Unit 2
Acid-Base Equilibria
This section is not officially part of Module 2, it is a revision of GCSE Chemistry but its contents
are vital to your understanding of the rest of the course so work through this yourselves.


Arrhenius definition of an acid – releases hydrogen ions into solution (alkali releases
hydroxide ions)
Bronsted-Lowry Theory of Acids and Bases - acids are proton donors, bases are proton
acceptors (e.g. hydroxide ions accept protons to form water).
Chemistry of Acids and Bases


when a substance dissolves in water it forms an aqueous solution which may be
(i)
acidic
(ii)
alkaline
(iii)
neutral.
e.g. soluble oxides of non-metals (for example, carbon dioxide,
sulphur dioxide and nitrogen dioxide)
e.g. soluble metal oxides and hydroxides (for example, the oxides
and hydroxides of sodium, potassium, and to some extent calcium)
e.g. salt water
pure water is neutral
(i)
(ii)
distilled water can be slightly acidic due to dissolved CO2
tap water is often slightly alkaline due to dissolved minerals


pH scale is used to show how acidic or alkaline a solution is
indicators (which are pH sensitive dyes) are different colours at different pH’s

acids (or alkalis) of the same concentration are not necessarily of the same strength

a strong acid or strong alkali is one that is 100% ionised in water.

HCl(aq)

H+(aq)
+
Cl-(aq)
NaOH (aq)

Na+(aq)
+
OH-(aq)
A weak acid or weak alkali is only partially ionised in water.
e.g.
CH3COOH(aq)
ethanoic acid
e.g.
NH3 (aq)
ammonia
+
H+(aq)
+
CH3COO-(aq)
ethanoate ions
NH4+(aq)
+
ammonium ion
H2O(aq)
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OH-(aq)
AS Chemistry Unit 2
Neutralisation Reactions

examples include: antacids; spreading lime on fields to reduce soil acidity, or in lakes to
reduce acidity caused by acid rain.
acid



+

base
salt
+
water
salt produced when an acid is neutralised depends on the metal in the substance neutralising
the acid, and the acid used
ammonia can also neutralise an acid (ammonium sulphate (a fertilizer) is produced with
sulphuric acid)
when neutralisation occurs between acids and alkalis, hydrogen ions react with hydroxide
ions forming water molecules
H+(aq) +

OH-(aq)
H2O(1)

other ions are ‘spectator ions’ ‘they do nowt’

sulphuric acid will require twice as much alkali for neutralisation as two hydrogen ions are
released per molecule. This means it will be twice as strong as hydrochloric acid of the same
concentration.

H2SO4(aq)

+
SO42-(aq)
metal carbonates and hydrogencarbonates react with acids to produce a salt, carbon
dioxide and water.
acid + metal (hydrogen)carbonate 

2H+(aq)
salt + water + carbon dioxide
when a reactive metal reacts with an acid:
metal + acid 
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salt + hydrogen
AS Chemistry Unit 2
REDOX REACTIONS
How Science
Works: A





simplest model involves the gain (oxidation) or loss (reduction) of oxygen
similarly the gain or loss of hydrogen can be used
these are still used to explain oxidation and reduction in organic chemistry
they do not however include all instances of redox reactions
this idea can be improved by considering redox reactions as an electron transfer process

oxidation is the LOSS of electrons, reduction is the gain of electrons OILRIG
oxidising agents (OXIDANTS) accept electrons and are themselves reduced
reducing agents (REDUCTANTS) donate electrons and are themselves oxidised
Half (ionic) equations






since oxidation and reduction MUST both occur in the same reaction in order for electrons to
be transferred then it is possible to write separate equations to show each in turn
you will already have encountered this in electrolysis
separate equations were written describing what takes place at the anode and cathode
cations arrived at the cathode an accepted electrons hence reduction
anions arrived at the anode and surrendered electrons hence oxidation
remember that charges on ions and electrons must be shown in addition to state symbols(note
electrons don’t have a state symbol)
e.g.
electrolysis of brine
ANODE
OXIDATION
2Cl
CATHODE REDUCTION
2H
Summary Question
-
(aq)
+
(aq)
Page 157
AS Chemistry (Nelson Thornes) AQA
Chemguide

Cl2(g) +
+
2e
1
156 - 157
Oxidation and reduction
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22
-

-
2e
H2(g)
AS Chemistry Unit 2
Oxidation Numbers

oxidation is the LOSS of electrons, reduction is the gain of electrons OILRIG
oxidising agents (oxidants) accept electrons and are themselves reduced
reducing agents (reductants) donate electrons and are themselves oxidised
Oxidation State

this is a ‘book keeping’ method of the effective control of electrons used in bonding
elements
=
0
oxidation state of elements in simple ions
=
charge on ion
 oxidation state of elements in polyatomic ions
=
charge on ion
 oxidation state of elements of a compound
=
0
the relatively more electronegative element is assigned the negative oxidation state
hydrogen
=
+1 (except in metal hydrides where it is -1)
oxygen
=
-2 (except in peroxide O22-) where it = -1)
group 1 metals
=
+1
group 2 metals
=
+2
fluorine
=
-1 (even with oxygen, which is +2 in OF2)
Aluminium
=
+3
metals are always positive in a compound or polyatomic ion

maximum possible oxidation state

oxidation numbers and nomenclature e.g. cobalt(II) nitrate(V), phosphorus(V) oxide
=
group number (note not always possible for
various reasons – see later)
take care not to mix up charge and oxidation numbers in polyatomic species e.g. a
sulphate(IV) ion does NOT have a 4- charge (the IV refers to the oxidation state of the
sulphur
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AS Chemistry Unit 2



changes in oxidation numbers can be used to identify redox reactions in inorganic and organic
reactions e.g. metal or halogen displacement reactions
we specifically refer to an element in a species (e.g. ‘the iron in Fe2O3 is reduduced’)
assume that multiple instances of an element in a species have the same value (e.g. both
carbons in ethene are -2)
OXIDATION
REDUCTION

oxidation number becomes relatively more positive
oxidation number becomes relatively more negative
in a disproportionation reaction atoms of the same element are oxidised and others are
reduced in the same reaction
e.g.
chlorine with water to form chlorate(I) and chloride
(note: chlorine with alkali goes to completion)
decomposition of hydrogen peroxide
Summary Questions
Page 159
AS Chemistry (Nelson Thornes) AQA
Chemguide
1-5
158 - 159
Oxidation number
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24
AS Chemistry Unit 2
Balancing Redox Equations

protocol for constructing half equations
1
2
3
4
5


get the formula correct (and stoichiometry – see example (iii) below)
balance the oxygen adding H2O(l)’s to the side with least O’s
balance the hydrogens using H+(aq)’s to the side with least H’s
balance the charge on each side by adding e- to relatively more positive side
add state symbols
(i)
Fe3+(aq)/ Fe2+(aq)
(ii)
MnO4-(aq)/Mn2+(aq)
(iii)
Cr2O72-(aq)/Cr3+(aq)
(iv)
S4O62- (aq)/ S2O32- (aq) (tetrathionate/thiosulphate)
combining two half equations:
as with any equation cancelling out/down should be undertaken
(i)
iodine and thiosulphate ions
(ii)
iron(II) ions and manganate(VII) ions
(iii)
dichromate(VI) ions and ethanol (to ethanal and to ethanoic acid)
(note that the use of [O] is acceptable as a simplification where the specific details of
the oxidising agent used are superfluous)
Summary Questions
Exam Style Questions
Page 163
Page 164 – 5
AS Chemistry (Nelson Thornes) AQA
Chemguide
1, 2 (2a is a little tricky at this stage)
1-8
160 - 163
redox equations
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25
AS Chemistry Unit 2
Group 7 Elements And Compounds
Physical Properties



colour of chlorine, bromine and iodine deepens down the
group.
trends in the volatility of the elements reflect increase in
Mr and increased Van der Waals’ forces
relative electronegativity decreases down the group as
atomic radii increases (proton increase is cancelled by a
similar increase in the number of screening electrons)
Summary Questions
Page 167
Group 7
1-3
Halogens as Oxidizing Agents






halogens are oxidising agents with decreasing oxidising ability (and hence relative reactivity)
down the group.
the increased number of protons are cancelled by more screening electrons, however as the
atoms are larger the halogen less readily gains an electron
the above idea is actually a simplification of the overall process but is adequate for chlorine,
bromine and iodine as this trend reflects the outcome although in reality there are other
contributory factors to consider
a more reactive halogen can displace a less reactive halide from a solution of its salt, in effect
oxidising it (the halogen is used in solution in practicals – less dangerous)
be able to write half equations for displacement reactions and associated observations
examples are the extraction of bromine from sea water using chlorine, and iodine from kelp
How science works
Summary Questions

Page 169
Page 169
1-2
1
test for iodine – turns starch black
(NOTE: IODIDE IONS DO NOT TURN STARCH BLACK !!!!!)
AS Chemistry (Nelson Thornes) AQA
Chemguide
166 - 169
Halogens
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26
AS Chemistry Unit 2
Halide Ions as Reducing Agents


halide ions become relatively stronger reducing agents down the group, themselves being
oxidised to the respective halogen
as halide ion size increases, the effective nuclear charge (i.e. actual nuclear charge diminished
by screening electrons and distance) decreases and the ease with which the outer electron is
lost increases.

remember starch is NOT a test for iodide ions !!!!!!
Reactions with conc. H2SO4

in all cases the hydrogen halide is produced when concentrated sulphuric acid is added to the
solid halide – this is a displacement reaction NOT REDOX (check the oxidation numbers)
e.g.
S= +6
+6
NaCl(s)
+
H2SO4(aq)

NaHSO4(aq) +
HCl(g)

bromides and iodides (but NOT chlorides) then undergo a redox reaction
+4


2HBr(aq)
+
H2SO4(aq)

Br2(l)
+
SO2(g) +
8HI(aq)
+
H2SO4(aq)

4I2(s)
+
H2S(g) +
H2O(l)
-2
4H2O(l)
iodides are a better reducing agent than bromide ions hence they reduce the sulphur initially
to SO2 (+4) as with bromide but subsequently further reduce it to S (O) and finally H2S (-2)
you should be able to write equations separately for each of these redox reactions
you must be aware of the associated observations for each of the above – covering colour
changes, gases evolved and precipitates (sulphur in this case)
Note: whilst in reality it may not be possible to see each one as it may be masked by other
observations you should still be able to describe what might possibly be observed as a
consequence of each step, particularly for iodide ions
Summary Questions
Page 172
AS Chemistry (Nelson Thornes) AQA
Chemguide
1
170 - 171
Halogens
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AS Chemistry Unit 2
Test for Halide Ions

halide ion solutions are all colourless
remember starch is NOT a test for iodide ions !!!!!!

add dilute nitric acid* followed by silver nitrate solution – which will produce a silver halide
precipitate
Silver Fluoride
Silver Chloride
Silver Bromide
Silver Iodide
No precipitate – it’s soluble
White
soluble in dilute ammonia solution
Cream
soluble in conc. ammonia
Yellow
insoluble in conc. Ammonia
* Nitric acid is used as the nitrate anion will not produce a precipitate with any metal ions
present. It also ensures that no silver oxide precipitate is formed.



silver salts decompose in strong light (hence they are stored in dark bottles)
two of the precipitates are quickly affected by sunlight - silver chloride turns purple/grey and
silver bromide turns green/yellow
silver is deposited as an opaque layer on black and white film
2AgCl(s)
2AgBr(s)



2Ag(s) +
2Ag(s) +
Cl2(g)
Br2(l)
photochromic sunglasses contain a mixture of silver chloride and copper(I) chloride – find
out how they work (and don’t work too well for driving)
http://www.explainthatstuff.com/photochromiclenses.html
http://www.district87.org/staff/sutterm/Chem%20Matters/Organized%20according%20to%
20topics/Acid%20Bases/Automatic%2520Sunglasses.pdf
Summary Questions
Exam Style Questions
Page 172
Page 174 - 175
AS Chemistry (Nelson Thornes) AQA
Chemguide
2
3, 5, 8, 9
171 - 172
Halogens
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AS Chemistry Unit 2
How Science
Works: I
Reaction of Chlorine with Water


chlorine gas is poisonous but its solution is used in water purification (drinking water
and swimming pools) to kill bacteria
chlorine reacts with water (‘chlorine water’) to produce a mixture of two acids hydrochloric and chloric(I) - by disproportionation (chlorine atoms are simultaneously
oxidised and reduced)
0
Cl2(g) +






H2O(l)
+
+1
HOCl(aq)
a similar reaction occurs but to a relatively lesser extent with other halogens down the group
moist blue litmus paper is first turned red, then bleached by the chloric(I) acid
in drinking water it prevents typhoid and cholera
in swimming pools the chloric(I) acid, HOCl, (an oxidising agent) kills bacteria by oxidation
and it is also a bleach (swimming costumes might fade over time as result)
in sunlight chlorine oxidises water to oxygen hence the need to add more chlorine,
particularly in shallow pools, but the concentration of the chlorine solution must be monitored
to ensure that toxic levels are avoided
alternatively sodium or calcium chlorate(I) can be added to water which reacts with it to yield
choric(I) acid
ClO-(g)

-1
HCl(aq)
+
H2O(l)
HOCl(aq)
+
OH-(aq)
the pH kept slightly acidic – use LCP to suggest WHY
Reaction of Chlorine with Sodium Hydroxide Solution

disproportionation reaction occurs to yield sodium chlorate(I) which is used to make bleach
0
Cl2(g) + 2OH-(aq)

-1
+1
Cl (aq) + ClO-(aq) + H2O(l)

adding acid to bleach is dangerous
Cl-(aq) + ClO-(aq) + 2H+ (aq)


Cl2(g) + H2O(l)
solutions of sodium chlorate(I) (i.e. bleaches) react with KI to liberate iodine How could
I2 can be estimated by titration against Na2S2O3(aq) using starch as an indicator this be used?
Summary Questions
Exam Style Questions
Page 173
Page 174 - 175
AS Chemistry (Nelson Thornes) AQA
Chemguide
1-3
1, 2, 7
173
Disproportionation, halogens
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AS Chemistry Unit 2
Group 2 Alkaline Earth Elements
Physical Properties




atomic radius increases down the group as there are more electron shells
both first AND second ionisation energy decrease down a group as the size of the atom and
number of shielding electrons increases (which offsets increasing nuclear charge).
increasing ease of ionisation contributes to the general increase in reactivity down the group
melting point decreases down the group as increasing size means that the distances between
the positive nuclei and delocalised electrons are greater hence the metallic bonding is weaker
(this also make them softer down the group)
Reactions of Group 2 Elements with Water



when metals react they form positive ion and therefore are oxidised hence the chemical with
which they react is reduced
metals are therefore potential reducing agents with the more reactive metals being better
reducing agents
magnesium reacts with steam to form magnesium oxide and hydrogen
Mg(s) + H2O(aq)



MgO(s) + H2(g)
others react with water to form metal hydroxides
Ca(s) + 2H2O(aq)

Ca(OH)2(aq) + H2(g)
Ba(s) + 2H2O(aq)

Ba(OH)2(aq) + H2(g)
increase in reactivity down the group reflects ease of removal of the two outer s sub-shell
electrons
Solubility of Group 2 Hydroxides and Sulphates

solubility of hydroxides increases down group while sulphates decrease

the alkalinity of solutions of the hydroxides thus increases down the group
LUND Apr-17
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AS Chemistry Unit 2
Uses of Group 2 Metals compounds
Magnesium hydroxide
very insoluble and used medicinally as a slurry (a solid suspension
known as milk of magnesia) and in toothpaste
in both cases it acts as an antacid and is better than a carbonates since
no gassiness results from the production of CO2 as when the latter
reacts with acid in the stomach
Magnesium Sulphate
very soluble, Epsom salts – a mild laxative
Calcium sulphate
Plaster of Paris – poorly soluble
Calcium hydroxide
(slaked lime) sparingly soluble, used in lime water



Ca(OH)2(aq + CO2(g) 
CaCO3(s) + H2O(l)
note that the cloudy precipitate re-dissolves in xs CO2
CaCO3(s) + CO2(g) + H2O(l) 
Ca(HCO3)2(aq)
this is in effect the same equation as that associated with the formation of hard water by the
action of rain water on limestone
the reverse process takes place when the product is evaporated, which is the cause of lime
scale deposits, stalagmites and stalactites
Ca(HCO3)2(aq)

CaCO3(s) + CO2(g) + H2O(l)
Barium sulphate
barium meal used in medicine
to outline the gut in X-rays
fairly safe - relatively low solubility – particularly with addition of
sodium sulphate (see ‘solubility product’ to know more) which ensures
that toxic barium ions do not dissolve into the body
Acidified Barium chloride = test for sulphate ions SO42- - white ppt
(acidified with HCl(aq) (or HNO3(aq) since all metal nitrates are soluble)
since it’s not going to introduce any additional anions whilst ensuring
no carbonate ions are present since barium carbonate will also give a
white precipitate)
Summary Questions
Exam Style Questions
Page 178
Page 179
Page 233
AS Chemistry (Nelson Thornes) AQA
Chemguide
1-6
1-8
4
176 - 178
Group 2
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AS Chemistry Unit 2
METAL EXTRACTION
Some environmental considerations
There will probably be a lot of ‘How science works’ type questions on this topic. Try to be
specific in your answer e.g. avoid statements such as ‘it causes pollution’ try to clearly say what
form of pollution and why e.g. ‘smelting the ore causes the release of carbon dioxide which is a
greenhouse gas and contributes to global warming.
Mining, transporting and extracting the ore






loss of landscape due to mining
air and noise pollution due to dust from mining
air pollution from transporting the ore – CO2 emissions – greenhouse effect
disposal of slag, some of which is sometimes just dumped
solid waste is called gangue, typically silica rocks and clays for which other uses would be
found to reduce ££££
air pollution from the extraction process, in particular carbon dioxide (greenhouse gas) and
sulphur dioxide (acid rain) in the roasting process but note the latter can be used in the
manufacture of sulphuric acid
Recycling



saving of raw materials and energy by not having to first extract the ore
avoiding the pollution problems in the extraction of the metal from its ore (see above)
not having to find space to dump the unwanted metal if it wasn't recycled
(offsetting these to a minor extent - energy and pollution ££££ in collecting and transporting
the recycled materials)
Rocks, Ores and Minerals





an ore is a rock containing an economically viable quantity of the metal to be extracted
a mineral is the particular compound in an ore that contains the metal
these minerals are typically metal oxides or metal sulphides
the latter are converted to an oxide by heating them in air prior to extraction – this is called
roasting
this produces an acidic gas SO2 which can cause acid rain and so it is used to make sulphuric
acid thus minimising waste and reducing pollution
→
SO2(g) + H2O(l) + ½ O2(l)
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H2SO4(aq)
AS Chemistry Unit 2
Extraction Method

metals are removed from their ore by reduction (either chemical or electrolytic depending
upon the relative position of the metal in the reactivity series)
Continuous
saves energy in high temperature processes (heating and cooling)
raw materials are constantly added and the products produced
are constantly removed which is a very efficient and ££££ saving way
to produce materials in large quantities
Batch
useful when small amounts are required
better when purity is important
involves extra ££££ due to equipment cleaning and non use between
batches
Method used depends on:
purity required
energy requirements
position of metal in reactivity series
££££ of reducing agent
Iron
metal oxide with carbon
Titanium
Metal chloride with a more reactive
metal egg sodium or magnesium
Aluminium
electrolysis of metal oxide
Tungsten
metal oxide with hydrogen
Carbon





coke (mainly carbon) is a low ££££ choice
it may require excessively high (££££) temperatures with some ores
the purity of the metal obtained must be considered as metal carbides can be formed that
weaken the metal structurally – it is thus unsuitable for extracting titanium , tungsten and
aluminium.
it is used for the extraction of iron, copper and manganese in preference to Pepsi
waste, particularly polluting gases such as CO2 when using coke, is an issue
Hydrogen



made from methane and water – see the Haber process
water is the co-product so pollution is relatively low
it has potential dangers (explosions) and ££££ more than coke
Electrolysis



used to extract metals high in the reactivity series
lots of energy is required to melt/dissolve the ore and also for the electrolysis process itself
plants are often associated with hydroelectric sites to minimise energy ££££
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AS Chemistry Unit 2
Continuous process
-
cheap

the most commonly used iron ores are haematite, Fe2O3, and magnetite, Fe3O4.

these can be reduced to iron by heating them with carbon in the form of coke which is cheap
and produced by heating coal in the absence of air (effectively distilling off the organics)

coke provides both the reducing agent for
the reaction and also the heat source - as
you will see below
iron ore is added to the blast furnace
with coke and limestone
coke is a solid fuel composed largely of
carbon (formed after the distillation of
coal, which removes volatile elements)
limestone is a sedimentary rock consisting
largely of calcium carbonate (CaCO3).
the mixture is heated by blasts of air,
preheated to about 750°C, blown in at the
bottom of the furnace (note this is
partially heated by hot waste gases from
the top to reduce energy ££££)
the coke is oxidised under the strong heat, the carbon reacts with oxygen to produce carbon
dioxide and this process is highly exothermic, raising the temperature to about 2000°C





C(s) + O2(g)  CO2(g)

the carbon dioxide formed then reacts (endothermically) with the coke to form carbon
monoxide
CO2(g) + C(s)  2CO(g)


the carbon monoxide produced then reacts with the iron ore acting as a reducing agent
CO is better for reduction than C as it is a gas hence more collisions hence faster rate
Fe2O3(s) + 3CO(g)  2Fe(l) + 3CO2(g)



the dense molten iron sinks to the bottom of the furnace and is tapped off regularly as cast
iron
cast iron is very runny when it is molten and doesn't shrink much when it solidifies so is ideal
for making castings - hence its name.
cast iron is very impure, containing about 4% of carbon which makes it very hard, but also
very brittle (if you hit it hard, it tends to shatter rather than bend or dent).

cast iron is used for things like manhole covers, guttering and drainpipes, cylinder blocks in
car engines, Aga-type cookers, and very expensive and very heavy cookware.

silica (silicon(IV) oxide SiO2) impurities that exist in the iron ore react with the limestone
added to form slag.
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AS Chemistry Unit 2

calcium carbonate undergoes thermal decomposition to form calcium oxide and carbon
dioxide.
CaCO3(s)  CaO(s) + CO2(g)

the calcium oxide (a basic oxide) then reacts with the silica (an acidic oxide) to form calcium
silicate (this is slag and is removed from the surface – relatively low density)
CaO(s) + SiO2(s)  CaSiO3(s)
CaO(s) + Al2O3(s)  CaAl2O4(s)

slag is used in road making and as "slag cement" - a finely ground slag which can be used in
cement, often mixed with Portland cement.
Batch process




resists corrosion hence useful for - cutlery, cooking utensils, kitchen sinks, industrial
equipment for food and drink processing
made from purified pig iron and scrap iron
chromium and nickel also added to make an alloy (this makes it more expensive than mild
steel) as well as small amounts of carbon
scrap iron used in the process because:
less mining – so less eyesore, traffic, noise, dust
less energy required than extracting the iron from its ore
recycles scrap (obviously) hence less landfill requirements
less pollutant gases (e.g. CO2, SO2)
economically sound (££££ of extraction and transportation of the ore as well as energy saved)

scrap iron is easy to recycle as it can be extracted with a magnet

scrap iron can also be used to extract copper from its solution since it is higher in the
reactivity series, much cheaper than copper
in the past copper was extracted from malachite which contains copper(II) carbonate
the carbonate was thermally decomposed to the oxide then the copper extracted with
carbon, both processes produced CO2 and require a lot of energy
nowadays it is possible to extract copper from low grade ores and mining waste by
firstly extracting Cu2+(aq) using dilute acid and bacteria then reducing it using the scrap
iron



How Science Works
Summary Questions
Exam Style Questions
Page 182
Page 182
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AS Chemistry (Nelson Thornes) AQA
Chemguide
1-5
4
180 - 182
Extraction iron
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AS Chemistry Unit 2
Continuous process






the main ore is bauxite that contains economically viable quantities of the mineral alumina
(Al2O3)
it cannot be extracted using carbon due to the relatively high position of aluminium in the
reactivity series.
alumina (an amphoteric oxide – reacts with acid and alkali) is extracted from its ore using
NaOH (details not required)
alumina has a high melting point >2000oC (strong electrostatic forces of attraction between
3+ cations and 2- anions) hence the direct electrolysis of molten alumina is not economically
viable.
alumina is dissolved in molten cryolite at 1000oC (compare this with the notion of melting
salt or dissolving it in water at room temperature to separate the ions)
electrolysis requires energy locating the smelting plant near a hydroelectric dam is
advantageous given the relatively low ££££ energy
recycling aluminium also saves energy



Al3+ + 3e- → Al

Al is formed at the cathode (pussytive
cations → -ve electrode i.e. the
cathode)
O2 is formed at the anode and reacts
with the graphite electrodes at high
temperatures producing CO2 hence they
must be replaced periodically - ££££
you MUST be able to write the
associated redox equations
2O2- → O2 + 4e-
note that 3 moles of electrons are required per mole of aluminium produced – energy ££££
Recycling:
melting of the metal is used
requires a fraction of the energy to extract the same quantity of aluminium
from its ore – huge ££££ saving
significantly reduces CO2 emissions
there are ££££ in sorting and transportation although specialised recycling
points (aluminium can banks) help to reduce this
How science works
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5
182 - 184
Extraction
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36
AS Chemistry Unit 2
Batch process





expensive
carbon could be used to extract Ti at very high temperatures (££££) but is NOT used because
titanium carbide is produced making the metal brittle.
titanium is abundant in the crust so this is NOT the reason for its ££££.
titanium ore (Rutile) contains an economically viable quantity of titanium(IV) oxide (TiO2)
this is converted to the chloride:
TiO2 + 2Cl2 + 2C

-
→
TiCl4 + 2CO (or TiO2 + 2Cl2 + C →
TiCl4 + CO2)
Titanium chloride is covalent so can be distilled off (strong bonds within TiCl4 but weak
IMF in between molecules) but it’s covalent nature rules out electrolysis to extract the metal.
titanium is extracted using a more reactive metal e.g. sodium at high temperature in an argon
atmosphere (to prevent oxidation of the hot metal)
once the reaction is complete, and everything has cooled (several days in total - an obvious
inefficiency of the batch process), the mixture is crushed and washed with dilute hydrochloric
acid to remove the sodium chloride.
TiCl4 + 4Na → 4NaCl + Ti

Titanium is expensive because of:
raw materials – chlorine, argon, and in particular sodium used in extraction
since it’s a batch process rather than continuous (like iron)
Batch process




expensive
a relatively rare metal – ££££ implications
used in incandescent light bulb filaments due to its high melting point although this will be
less common as energy saving light bulbs are becoming more widespread.
carbon could be used to extract W at very high temperatures (cost) but is NOT used because
tungsten carbide is produced making the metal brittle.
it is extracted from its oxide WO3 (oxidation state of tungsten) by reduction with hydrogen
(the relatively high ££££ of tungsten makes this method viable).
WO3 + 3H2

-
→
W + 3H2O
this must be carefully controlled because of the dangerous nature of hydrogen at high
temperatures.
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1, 2, 3, 6, 7
5
183 - 184
Extraction
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37
AS Chemistry Unit 2
Haloalkanes




haloalkanes are rarely found naturally but are important in many synthetic products
they are used extensively as solvents
2-bromo-2-chloro-1,1,1-trifluoroethane (halothane) is a modern anaesthetic
chloroethene and tetrafluoroethene are used to make PVC and PTFE
Nomenclature






general formula CnH2n+1X
prefixes chloro, bromo and iodo used e.g. 1-bromobutane
structure of primary, secondary and tertiary haloalkanes
as with alkyl branches substituents are named in alphabetical order NOT numerical order
(ignore di, tri etc. for order determination)
numbers are used to individually locate each and every substituent
e.g. 2-bromo-1,1-dichloroethane
Physical Properties




boiling points generally determined by VdW intermolecular forces and largely depend
on Ar of the halogen rather than dipole
1-iodobutane > 1-bromobutane > 1-chlorobutane >> butane (in the latter case there is
the double advantage of higher VdW and the presence of a dipole)
immiscible with water as they cannot form hydrogen bonds with it (unlike don’t mix)
hence a universal solvent such as ethanol is used when carrying out practical’s
they do mix with hydrocarbons (grease and oil) so can be used as dry cleaning fluids (being
fairly volatile also assists drying)
Summary Questions
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1
186 - 187
Haloalkanes
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38
AS Chemistry Unit 2
Hydrolysis of Bromoethane

hydrolysis is the breaking of bonds in a substance by reaction with water or its ions
(e.g. OH-).
bromoethane
+

OH-
ethanol
+
Br-
NS
dissolve in ethanol (universal solvent) then add dilute aqueous sodium hydroxide solution (alkaline
hydrolysis)
a nucleophile is an electron pair donor e.g. :OH-, :CN-, H2O:, :NH3


mechanism of nucleophilic substitution and the use of curly arrows
the halide ion is called the leaving group

haloalkanes are also hydrolysed by water, but more slowly since water is a weaker
nucleophile than hydroxide ions
Factors Affecting the Rate of Hydrolysis






relative polarity of C-X bond suggest reactivity of chloro > bromo > iodo haloalkanes as
greater polarity should more strongly attract the nucleophile
BUT actual relative rate is iodo > bromo > chloro
relative strengths of the C-Hal bond (idea of ‘good leaving group’) determine the rate
practically demonstrated by carrying out hydrolysis in the presence of aqueous silver nitrate
solution (and ethanol as a universal solvent)
rate of formation of silver halide precipitate is proportional to rate of displacement of the
halide ion which is proportional to the rate of substitution.
colour of silver halide precipitate can also be used to determine the halogen present in organic
analysis
chloroalkanes
bromoalkanes


white ppt of AgCl
pale yellow ppt of AgBr
iodoalkanes

yellow ppt of AgI
LUND Apr-17
39
soluble in dilute ammonia
soluble in conc. (or xs)
ammonia
insoluble in ammonia
AS Chemistry Unit 2
Nucleophilic Substitution with CN-
bromoethane
+
CN-

propanenitrile
boil under reflux with NaCN or KCN (not HCN)

+
Br-
NS
dissolved in ethanol
the product is called a nitrile
AN IMPORTANT REACTION SINCE IT INCREASES THE CARBON CHAIN LENGTH
BY ONE CARBON ATOM

the nitrile group is converted into a carboxylic acid by acid hydrolysis
propanenitrile
+
2H2O +

H+
propanoic acid
+
NH4+
boil under reflux with dilute sulphuric acid
Nucleophilic Substitution with NH3
bromoethane
+
2NH3 
ethylamine
+
NH4Br
NS
heat with xs ethanolic ammonia under pressure
NOTE: the HBr produced will then form the salt NH4Br in the presence of excess NH3

a primary amine is formed


xs ammonia ensures the primary amine is the main product
the ethylamine produced is also a nucleophile (stronger than ammonia due to the +ve
inductive effect of the alkyl group) so if excess bromoethane is used instead of xs ammonia
further substitution is possible to give diethylamine, triethylamine and finally
tetraethylammonium bromide (a quaternary ammonium salt)
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187 - 190
Nucleophilic, hydrolysis, silver nitrate
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40
AS Chemistry Unit 2
Elimination Reactions
bromoethane
+
OH-

ethene
+
H2 O
+
Br-
E
heat with a solution of sodium hydroxide dissolved in ethanol NOT water


notice that the conditions are different even though the same reagent that was used for
nucleophilic substitution is used
in this case the hydroxide ion is behaving as a base and abstracting a proton

in effect HBr is eliminated resulting in the formation of a double bond (evidence for this is
the fact that the gaseous product decolourises bromine water)

in reality elimination and substitution reactions compete with the major product determined
by the conditions
dissolving the hydroxide ions in water and at room temperature favours substitution
dissolving the hydroxide ions in hot ethanol favours elimination

the outcome is affected by the type of haloalkane
favours elimination
1o
2o
3o
favours substitution

2-bromobutane yields two products – but-2-ene (which can be present as two isomers – see
Alkenes) and but-1-ene
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2, 4
191 - 193
Elimination
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41
AS Chemistry Unit 2
The formation of Haloalkanes from Alkanes


alkanes are non-polar molecules with strong bonds between C-H and C-C hence are
relatively unreactive (so far we have only seen combustion)
they do however react with halogens in UV (or strong sunlight) (remember that ALKENES
react with bromine water even in the dark) to form a haloalkane by free radical substitution

Cl2


2Cl∙
INITIATION (UV breaks C-Cl bonds homolytically)
homolytic fission (rather than heterolytic fission) takes place as the UV light provides
adequate energy (which visible light would not) for the dissociation of chlorine molecules
the halogen bonds are broken first as they are the weakest
Cl∙ + CH4

∙CH3 + Cl2

∙CH3 + HCl
CH3Cl + Cl∙
PROPAGATION
Radical is regenerated hence a chain reaction takes place
∙CH3
∙CH3

+ ∙CH3 
CH3CH3
+ Cl∙ 
CH3Cl
overall equation – determined by the propagation cycle rather than termination since this
happens on numerous occasions compared to the termination
Cl2 + CH4


TERMINATION

HCl
+ CH3Cl
OVERALL
further substitutions are possible so these reactions typically produce a mixture of haloalkanes
(which will have yield and cost implications)
other halogenation reactions work the same way e.g. ethane + bromine, bromoethane +
bromine etc
Summary Questions
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3, 5, 6
194 - 195
Free radical
Look at s-cool also
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42
AS Chemistry Unit 2
How Science
Works: B, L
Haloalkanes – Environmental Issues



after the 30’s CFC’s replaced toxic and smelly ammonia in fridges
CFCs are unreactive so were also used extensively for – aerosol propellants
this relative lack of reactivity initially seemed like a good thing

however, their lack of reactivity meant that they did not decompose over a short timescale
hence were able to make their way into the upper atmosphere
concern was raise about CFC when it became apparent that they were acting as catalysts in
the destruction of the ozone layer
CFC’s interact with UV solar radiation in the upper atmosphere to release separate chlorine
atoms (chlorine free radicals)


CFC’s

Cl∙
INITIATION (UV breaks C-Cl bonds homolytically)
Cl∙ + O3

ClO∙ + O2
PROPAGATION
ClO∙ + O3

2O2 + Cl∙
Radical is regenerated hence a chain reaction takes place
2O3






3O2
OVERALL
this means that an awful lot of ozone can be destroyed by the presence of a relatively small
proportion of CFC’s – an ‘ozone hole’ came to be
the ozone layer is an important shield in our planets defence against harmful levels of UV
solar radiation overexposure to which can cause cancers and more worryingly a reduction in
the amount of Plankton in the oceans – removing a major component of the food chain
following the Montreal Protocol of 1987 it was agreed that they would be phased out.
Ozone does form naturally in the upper atmosphere, however, it will take time for the ozone
hole to be ‘refilled’ as there was a reservoir of CFC’s still present after 1987
Chemists have since developed safer alternatives to replace CFC’s (e.g.
hydrochlorofluorocarbons, HCFC’s and hydrofluorocarbons, HFC’s)
http://www.atm.ch.cam.ac.uk/tour/
http://www.theozonehole.com/
Summary Questions
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2
1
143, 193, 195
Ozone hole
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43
AS Chemistry Unit 2
Alkenes




nomenclature of alkenes (unsaturated hydrocarbons), general formula CnH2n
locant for the double bond
formula confusion with cycloalkanes
homologous series similar to alkanes
Isomerism in Alkenes

positional isomerism (a form of structural isomerism) also occurs in higher alkenes where
the name will be identical apart from different numbers being used e.g. but-1-ene, but-2-ene


alkenes can also exhibit a type of stereoisomerism called geometrical isomerism
this is consequential of the non-rotation of a double bond (unlike in alkanes) – which is
what you will state as the fundamental requirement in the exam
How Science
this lack of free rotation is consequential of the π bond present in the alkene Works: A


additionally for geometrical isomerism to be possible both carbon atoms on the double bond
must have different atoms/groups attached to themselves, however, the carbon atoms can still
both be identical in that respect
Geometrical isomers have the same molecular formula, same structural formula but a different
spatial arrangement of the atoms due to the non rotation of the carbon-carbon double bond





E and Z are used to distinguish between
the two isomers (its quite EZy to do will
a little practice)
E isomers have the main grouping
diagonally across the double bond
Z isomers have the main grouping on the
same side of the double bond
cis and trans were previously used and in most cases E corresponds
to trans and Z to cis BUT NOT ALWAYS
How Science
Works: H
the trick is to identify what the main groupings are:
where the two atoms directly bonded to the carbons of the double bond with the largest
atomic numbers (highest priority) are diagonally opposite then it is deemed an E isomer
where the two atoms directly bonded to the carbons of the double bond with the largest
atomic numbers (highest priority) on the same side then it is deemed an Z isomer
if on one of the carbons the atoms directly bonded are identical then to establish the
highest priority grouping a tie break situation arises in which you look at the next
highest priority atom attached to each of them e.g -CH2Br beats CH2Cl and so on
LUND Apr-17
44
AS Chemistry Unit 2
Example: but-2-ene
Step 1: split the alkene
Step 2: assign the relative
priorities.
The two attached atoms are C
and H, so since the atomic
numbers C > H then the -CH3
group is higher priority.
Step 3: look at the relative
positions of the higher priority
groups : same side = Z, hence
(Z)-but-2-ene.
The two attached atoms are C and H, so since the atomic numbers C > H then the -CH3 group is
higher priority.
Therefore the two high priority groups are on the opposite side, then this is (E)-but-2-ene.
Combustion of Alkenes


combustion reactions – same general idea as alkanes – practice a few just in case
since alkenes are an important chemical feedstock (used to make other stuff) they are not
economically used as fuels
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198 - 201
Alkenes, geometrical isomerism
Electrophilic Addition




alkenes are more reactive than alkanes, despite the fact that a double bond is stronger overall,
due to relatively weak  bond that constitutes the second bond
being weaker than a single bond it requires less energy to break
double bonds present a high electron density so are attractive to electron deficient species
called electrophiles
such species can add onto unsaturated compounds such a alkenes in an electrophilic addition
electrophiles are electron pair acceptors

reaction with HBr and Br2 (in the dark – unlike alkanes)
ethene
+
HBr

bromoethane
EA
ethene
+
Br2

1,2-dibromoethane
EA
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AS Chemistry Unit 2

bromine water is the standard test for alkenes (carbon-carbon double bond)


heterolytic fission takes place in these reactions
again, when outlining the mechanism of electrophilic addition take the trouble to learn the
correct use curly arrows
in these reactions a positive carbocation intermediate is generated


evidence for the intermediate carbocation can be obtained by carrying out the addition of
Br2 in the presence of chloride ions which then also produces 1-bromo-2-chloroethane
Other Addition Reactions
ethene
+
H2 O

ethanol
EA
1
2
cold cH2SO4 to produce ethyl hydrogensulphate THEN
warm with water to produce ethanol (overall sulphuric acid is a catalyst)

industrially alcohols are produced by reacting alkenes with steam and phosphoric acid – see
ALCOHOLS
Electrophilic Addition to Unsymmetrical Alkenes





reactions of HBr and H2SO4 with unsymmetrical alkenes (i.e. where the double bond is not in
the middle) e.g. propene yield two isomeric major and minor products
Markovnikov’s rule determines the outcome
relative stability 3o>2o>1o carbocation intermediate due to positive inductive effect (+I) of the
alkyl group
this arises because the alkyl group is a relatively better e- source than H which thus stabilises
the positive charge
this in turns means EA is relatively lower thus favouring that reaction path
Summary Questions
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1, 2, 3, 5, 7
202 - 205
Electrophilic addition, Markovnikov
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AS Chemistry Unit 2
Addition Polymerisation






monomers can join together to form polymers with a large Mr
the conditions are typically heat, pressure, catalyst
variations of the reaction conditions can yield different types of polymer – with variations in
length and the degree of branching
low density poly(ethene) utilises high temperature and pressure via a free radical mechanism
which increases the extent to which branching occurs (due to the random nature of the radical
mechanism) and reduces density
high density poly(ethene) uses lower temperature and pressure and a Ziegler catalyst resulting
in significantly reduced branching hence closer packing
you should be able to show (or identify) the repeating monomer units in a polymer
poly(ethene)
low density
high density
plastic bags, plastic bottles
stronger – milk crates
poly(phenylethene) normal
pipes, chairs
(Polystyrene)
poly(propene)
toys, mobile phones
poly(chloroethene)
‘imitation leather’, drain pipes
poly(methyl 2-methylpropenoate)
‘imitation glass’, lenses (cheap)
poly(propenenitrile)
acrylic fibre - clothing
poly(tetrafluoroethene)
lubrication, non-stick pans
(PVC)
(Perspex)
(PTFE aka Teflon)
problems
non biodegradability (since most these polymers are typically an
alkane hence unreactive) so take centuries to decompose in landfill
combustion yields toxic gases
solutions
mechanical recycling – after sorting out the different types of plastic the
thermosoftening types can be remoulded (but only a few times due to
degradation of the plastic as chains break down )
feedstock recycling – cracking is used to break down the polymers prior to
making new plastics
replace poly(alkenes) with biodegradable and photodegradable polymers
Summary Questions
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4, 6
7
206 - 209
Polymerisation, Ziegler
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47
AS Chemistry Unit 2
Alcohols
Nomenclature





-OH hydroxyl group present with general formula CnH2n+1OH (or ROH is its short form)
form a homologous series of chemicals with similar properties
names end in ‘ol’ but will be used as a ‘hydroxy’ prefix if a higher rated group e.g. one with a
C=O is involved e.g. 2-hydroxypropanoic acid CH3CH2(OH)COOH
primary, secondary and tertiary alcohols – same idea as haloalkanes
di-ols e.g. ethane-1,2-diol (used in antifreeze) (the ‘e’ is kept as the next letter is a consonant)
Physical Properties






alcohols are structurally similar to water i.e. R-O-H
105o bond angles exist on the oxygen as was the case for water
capable of hydrogen bonding in addition to VdW – strong intermolecular force hence boiling
points higher than alkanes with comparable Mr
tri-ols e.g. propane-1,2,3-triol (glycerol) have a particularly high viscosity due to extensive
hydrogen bonding
can form hydrogen bonds with water, hence lower members miscible with it
miscibility decreases with increasing chain length as contribution of –OH becomes
progressively less significant and stronger VdW between larger alcohol molecules make them
harder to separate
Summary Questions
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212 - 213
alcohol
Uses of Ethanol




ethanol burns with light blue flame with little smoke (this reflects the relatively low
percentage by mass of carbon) and is a constituent (along with methanol) of methylated
spirits (picnic stoves) – to which an unpalatable purple dye is added thus avoiding taxation
useful as a petrol substitute in countries like Brazil (Gasohol) with limited oil reserves where
up to 20% can be added to petrol without engine modification
DON’T FORGET THAT IT ALREADY HAS AN OXYGEN ATOM WHEN WRITING
A COMBUSTION EQUATION
presence of -OH group and hydrocarbon part C2H5 makes ethanol a useful universal solvent
– where ease of evaporation is also a benefit e.g. in perfumes
it is also used as a feedstock to manufacture other chemicals
LUND Apr-17
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AS Chemistry Unit 2
Production of Ethanol
Fermentation






carbohydrates (from sugar can or sugar beet – a renewable resource as they can be grown
annually) are broken down into simple sugars and then to alcohols by enzymes (present in
yeast)
the reaction is exothermic which provides energy for yeast metabolism and these warm
conditions provide a reasonable rate of reaction (NOTE: higher temperatures will not give a
higher rate as the enzymes become denatured)
an anaerobic respiration process takes place (air excluded to prevent oxidation of ethanol to
ethanoic acid)
the yield of ethanol is poor as the increase in alcohol inhibits the fermentation process
Ethanol produced this way can be used as a BIOFUEL
nominally carbon input (photosynthesis) = carbon output (fermentation + combustion)
but other factors such as distribution mean that it is not 100% CARBON NEUTRAL
C6H12O6

2CH3CH2OH
catalysed by yeast, warm conditions
BAD
BAD
GOOD
GOOD
+
2CO2
fermentation is an example of a batch process – high production costs
slow reaction, yielding impure product (distillation of the aqueous solution necessary)
uses renewable resources - sugar cane mostly
requires little energy
Industrial production
ethene
+
H2 O
o
H3PO4 catalyst, ~300 C, 70atm


ethanol
covered in detail in the section on equilibria
GOOD
GOOD
BAD
BAD
continuous process – lower production costs
fast reaction, good yield of relatively pure product
uses non-renewable resources (oil)
energy required (high temperature and pressure)
Summary Questions
Exam Style Questions
1–3
2 (if not already done)
1, 11, 12
Page 215
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152 – 153, 214 - 215
Ethanol, fermentation
LUND Apr-17
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EA
AS Chemistry Unit 2
Oxidation of Alcohols



primary, secondary can be easily oxidised using the reagents K2Cr2O7(aq)/dilute H2SO4(aq)
remember ‘H+(aq)’ is REACTANT, sulphuric acid is the REAGENT that provides it
tertiary alcohols cannot be easily oxidised (no C-H bond to break)
primary alcohol
+
[O]

aldehyde
+
H2 O
mild i.e. NO REFLUX conditions - K2Cr2O7(aq)/H2SO4(aq) distil off aldehyde as formed to prevent
further oxidation to a carboxylic acid

the aldehyde initially produced is relatively less volatile than the alcohol due to reduced
opportunity for hydrogen bonding hence the distillate will be rich in the aldehyde with most
of the alcohol remaining in the reaction vessel
primary alcohol
+
2[O[ 
REFLUX with K2Cr2O7(aq)/H2SO4(aq)
secondary alcohol +
[O]
reflux with K2Cr2O7(aq)/H2SO4(aq)





carboxylic acid
+
ketone
H2 O
+
H2 O
orange dichromate(VI) ions (Cr2O72-(aq)) are reduced to green chromium(III) ions
(Cr3+(aq)) – PLEASE avoid putting statements such as ‘it changes from orange to green’
important test to differentiate between primary/secondary and tertiary alcohols
cannot differentiate between primary and secondary as both give a positive result
however the products can be isolated and then tested with Tollens or Fehlings (see Module 4)

carbonyl compounds such as aldehydes and ketones have CnH2nO as their common
general formula (unsaturated due to carbon-oxygen double bond)

aldehydes and ketones can be functional group isomers (a form of structural isomerism)


aldehydes are RCHO
ketones are RCOR’

higher Ketones can also exhibit positional isomerism (another form of structural
isomerism)
LUND Apr-17
50
AS Chemistry Unit 2
Distinguishing Between Aldehydes and Ketones

tests are based on the fact that aldehydes can be easily oxidised to a carboxylic acid while
ketones cannot be – i.e. a redox reaction can occur
Aldehyde
(i)
+
[O]

Carboxylic Acid
orange dichromate(VI) ions (Cr2O72-(aq)) are
reduced to green chromium(III) ions (Cr3+(aq))
reflux with K2Cr2O7(aq)/H2SO4(aq)
NOTE THAT THE ABOVE REAGENT WILL OXIDISE PRIMARY AND SECONDARY
ALCOHOLS. HOWEVER, THE TWO REAGENTS BELOW ONLY OXIDISE
ALDEHYDES
(ii)
warm with Fehling’s solution
(an alkaline solution containing
complexed copper(II) ions)
blue Cu2+(aq) is reduced to brick red Cu2O(s)
(this is the basis of the test for reducing sugars)
(iii)
warm with Tollen’s reagent
(aqueous silver nitrate in xs ammonia)
Ag+(aq) is reduced to Ag(s),
hence the silver mirror effect
Dehydration of Alcohols
alcohol

alkene
heat with excess cH2SO4 at 170oC
+
H2 O
OR passing the vapour over HOT Al2O3

elimination can yield more than one product
positional isomers and/or geometrical (E-Z) isomers can be produced

alcohols produced from renewable sources can be used to produce alkenes which can then be
used to produce polymers
Summary Questions
Exam Style Questions
1–6
2 – 11
9, 10
Page 219
Page 220 - 3
Page 235
AS Chemistry (Nelson Thornes) AQA
Chemguide
216 - 219
Oxidation alcohol, Fehling, dehydration
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AS Chemistry Unit 2
Mass Spectrometry

this provides us with:
Mr, Relative abundance and (fragmentation
Patterns – see A2 Chemistry)

do you know how a mass spectrometer works:
vaporisation
ionisation
acceleration
deflection
detection


you must be able to interpret simple mass spectra to determine the elements present from their
relative isotopic mass and also their relative abundance
the value of the relative isotopic mass can be read directly from the m/z value of a given
peak on the spectra from a mass spectrometer
relative abundance is reflected in the heights of the peaks

when a molecule is subject to ionisation a molecular ion is formed

M(g)

M(g)+ +
e-

This molecular ion can also undergo fragmentation which results in lots of peaks
corresponding to many smaller positive ions which will help to identify the molecule

chloro-alkanes will produce two molecular ion peaks when mono substituted in a ratio
commensurate with halogen abundance.
e.g. chloromethane gives M(g)+ peaks at 50 and 52 in a 3:1 ratio
Di-substituted chloro-alkanes will yield three molecular ion peaks of relative intensity for
dichloro 9:6:1 – those who do Maths Statistics will be able to explain that:
e.g. dichloromethane gives M(g)+ peaks at 84, 86 and 88 in a 9:6:1 ratio


high resolution mass spectra can distinguish between compounds with similar Mr e.g.
~123 for C6H5NO2 and C7H7O2 however, they will not be 123.000 exactly and will differ
from one another as a consequence allowing for individual identification

Mr data is particularly useful with IR data to establish the maximum carbons in a molecule
known to contain one or two oxygen atoms
Summary Questions
Exam Style Questions
Page 225
Page 230
AS Chemistry (Nelson Thornes) AQA
Chemguide
1-3
1, 3, 4
224 - 225
Mass spectrometry
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AS Chemistry Unit 2
IR Spectroscopy


molecules absorb IR energy at values corresponding to a natural vibration frequency
associated with asymmetric stretching and bending of the covalent bonds present which is
dependent on the bond energy and mass involved
(the ability of CO2 to do this is why there is a relationship between atmospheric CO2 and
global warming)

a beam of variable IR frequencies is shone through a sample (after calibration) and the
transmittance is plotted against frequency (wavenumber is used - more user friendly)

above 1500 cm-1 two important peak values are:
1620 – 1680 cm-1 C=C
can distinguish between alkenes and isomeric cycloalkanes
1680 – 1750 cm-1 C=O
strong and sharp with slight variations in position depending on
the type of compound (found in acids, aldehydes and ketones)
3230 – 3550 cm-1 O-H
this is broad (due to hydrogen bonding) cf the narrow C– H of
2850 – 3300 cm-1
note that the O-H for alcohols lies further to
the left than for acids revealing the narrower
C-H absorption which might be partially or
totally obscured with an acid
A data sheet will be provided in the exam –
similar to table 1 on page 227




used with mass spec they will help establish a maximum number of carbon atoms in the
compound by indicating the presence of at least one or two oxygen atoms.
the fingerprint region 400 – 1500 cm-1 is unique to each organic molecule and allows
identification by matching against a computer database
the C-O peak lies in this region so may be hard to see clearly
impurities will yield additional peaks on the IR spectra – in particular water in a wet sample
Summary Questions
Exam Style Questions
Page 229
Page 230 - 1
Page 233
AS Chemistry (Nelson Thornes) AQA
Chemguide
1-6
2, 5 – 8
6
226 - 229
Infra red spectroscopy
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53
AS Chemistry Unit 2
Amine
AS ORGANIC PATHWAYS
Haloalkane
Alkane
Alkene
Nitrile
Carbonyl
Carboxylic
Acid
Alcohol
Reagents/conditions
Mechanism required
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AS Unit 2 Chemistry in Action
AS ORGANIC CONVERSIONS
You should be able to write full chemical equations, and identify the type for all the reactions listed:
Oxidation, Reduction, Addition, Elimination,
Hydrogenation, Dehydrogenation, Hydrolysis
Substitution,
Hydration,
Dehydration,
Mechanisms should be known for all those shaded: Nucleophilic Substitution NS, Electrophilic
Addition EA or Elimination E.
Alkenes
to:
haloalkane, alkane, alcohol, alkoxyalkane
Alkenes are obtained from alkanes via cracking
ethene
+
HBr

bromoethane
EA
ethene
+
Br2

1,2-dibromoethane
EA
ethene
+
H2

ethane
Ni catalyst, ~200oC (catalytic hydrogenation)
+
Method 1
Method 2
cH3PO4 on a silica support with high temperature and pressure,
first react with cold cH2SO4, then warm with water
Alkyl Halides
H2 O

ethene
to:
ethanol
EA
amine, alcohol, nitrile, alkene
Alkyl halides can be made from alkanes by free radical substitution
bromoethane
+
2NH3 
ethylamine +
NH4Br
heat with xs ammonia under pressure to minimise further substitution
NS
bromoethane
+
OH-(aq) 
ethanol
+
BrNS
dissolve in ethanol (universal solvent) then boil under reflux with dilute aqueous NaOH
bromoethane
+
CN- 
propanenitrile
boil under reflux with alcoholic NaCN or KCN (NOT HCN)
+
Br-
bromoethane
+
OH- 
ethene
reflux with an alcoholic solution of NaOH
H2 O
+
Mr Lund January Apr-17
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+
NS
Br-
E
AS Unit 2 Chemistry in Action
Alcohols
to:
carbonyl (aldehyde and ketone), carboxylic acid, alkene
ethanol
+
[O]

ethanal
+
H2 O
mild conditions - K2Cr2O7(aq)/H2SO4(aq) distil off aldehyde as formed to prevent further oxidation to a
carboxylic acid
ethanol
+
2[O]
reflux with K2Cr2O7(aq)/H2SO4(aq)

ethanoic acid
+
H2 O
propan-2-ol
+
[O]
reflux with K2Cr2O7(aq)/H2SO4(aq)

propanone
+
H2 O
ethanol

ethene
heat with excess cH2SO4 at 170oC
+
H2 O
Mr Lund January Apr-17
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AS Unit 2 Chemistry in Action
Testing for functional groups
Bromine Water (Br2(aq))
Test for alkenes (unsaturated hydrocarbons)
Add bromine water dropwise to ~1 cm3 of the unknown substance.
OBSERVATION
INFERENCE
EXPLANATION
orange  colourless
alkene
electrophilic addition.
product is colourless
Acidified Potassium Dichromate Solution
Test for 10 and 20 alcohols (CARE it is also
positive with an aliphatic aldehyde)
To ~ 1 cm3 of the substance under test add ~ 1 cm3 of acidified (1 mol dm-3 sulphuric acid) potassium
dichromate solution then heat gently in a water bath.
Acidified purple potassium
23+
manganate(VII) will be
Redox reaction in which orange Cr2O7 is reduced to green Cr
decolourised in a similar test
2+
3+
Cr2O7 (aq) + 14H (aq) + 6e  2Cr (aq) + 7H2O(l)
OBSERVATION
INFERENCE
orange  green
1o or 2o alcohol
EXPLANATION
1o alcohol  aldehyde  acid
2o alcohol  ketone
aldehyde
aldehyde  carboxylic acid
Methanoic acid can also be oxidized to carbon dioxide
NOT 3o alcohol or ketone
Tollens’ Reagent (Silver Mirror Test)
Test for aldehydes NOT ketone
Don’t confuse with the test for alkyl halides
To ~1 cm3 of your sample add ~ 1 cm3 of Tollen’s reagent* and warm in a hot water bath.
(*to ~2 cm3 of ~ 0.1 mol dm-3 silver nitrate solution add ~2.0 mol dm-3 sodium hydroxide solution 1
drop at a time until a brown precipitate just forms. Add ~2.0 mol dm-3 ammonia solution to it
dropwise until the precipitate just dissolves - (ammoniacal silver nitrate)).
OBSERVATION
INFERENCE
silver mirror formed
on test tube
aldehyde
EXPLANATION
+
Ag(NH3)2 (aq) + e-  Ag(s) + 2NH3(aq)
aldehyde  acid
NOTE THAT ALCOHOLS DO NOT GIVE A POSITIVE RESULT WITH THIS TEST
Mr Lund January Apr-17
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AS Unit 2 Chemistry in Action
Fehlings’ Test
Test for aldehydes NOT ketone NOT benzaldehyde
To ~1 cm3 of your sample add ~ 1 cm3 of Fehling’s solution* (BLUE) and warm in a water bath.
(*made by mixing equal volumes of:
Fehling’s A (dissolve 17g of CuSO4.5H2O in 250 cm3 of water) and;
Fehling’s B (dissolve 86 g of Rochelle salt (potassium sodium 2,3-dihydroxybutanedioate (tartrate)
and 30 g of NaOH in 250 cm3 of water with gentle warming).
OBSERVATION
INFERENCE
EXPLANATION
orange-red precipitate
aldehyde
reduction of copper(II)  copper (I)
Cu2O is precipitated
aldehyde  acid
NOTE THAT ALCOHOLS DO NOT GIVE A POSITIVE RESULT WITH THIS TEST
Test for carboxylic acids
Sodium Carbonate Solution
To a small quantity of the unknown substance in a boiling tube add ~1 cm3 of sodium carbonate
solution. Test any gas evolved using a drop of lime water at the end of a glass rod.
OBSERVATION
INFERENCE
EXPLANATION
effervescence
gas evolved turns lime water cloudy
carboxylic acid
acid + metal carbonate
 salt + water + CO2
Acidified Silver Nitrate Solution
Test for alkyl halides
Don’t confuse with Tollen’s!
To test for an alkyl halide, the halogen atom must first be released as a halide ion by hydrolysis.
Dissolve ~ 1 cm3 of the alkyl halide in ~1 cm3 of ethanol, then add ~ 1 cm3 of sodium hydroxide
solution and warm in a water bath.
Add ~ 1 cm3 of silver nitrate solution that has been acidified with dilute nitric acid (this removes
excess OH-(aq) which would otherwise precipitate out Ag2O(s) thus masking the test results).
OBSERVATION
white precipitate
soluble in ammonia
pale yellow precipitate
soluble in xs ammonia
yellow precipitate
insoluble in ammonia
INFERENCE
EXPLANATION
alkyl chloride
AgCl(s)
alkyl bromide
AgCl(s)
alkyl iodide
AgI(s)
Relative rates of hydrolysis of alkyl halides are: iodo > bromo > chloro
Mr Lund January Apr-17
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AS Unit 2 Chemistry in Action
Mr Lund January Apr-17
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AS Unit 2 Chemistry in Action
Mr Lund January Apr-17
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AS Unit 2 Chemistry in Action
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AS Unit 2 Chemistry in Action
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AS Chemistry Unit 2
63
AS Chemistry Unit 2
64