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A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
AQA A2-LEVEL
Student Guide to Unit 4
Kinetics, Equilibria and Organic Chemistry
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Mr Lund 08 May 2017
1
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
How Science
Works: A, B
Equilibria



dynamic (not static - do you know the difference!) equilibrium is achievable in a closed
system (e.g. solutions in a test tube if there are no gaseous reactants/products).
rates of the forward and reverse reactions at equilibrium are identical
concentrations are unlikely to be 50 – 50
Le Chatelier’s Principle and Qualitative Aspects







can you unambiguously write down what LCP states (see 148 of the AS textbook)?
remember this is a predictive tool used to determine the effect on the position of equilibria
when a change in concentration, temperature or pressure is made
it 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 a system completely reverses a temperature change when
establishing a new equilibrium as the new equilibrium will be that for the changed
temperature
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
the position of equilibria is changed by:
concentration (which can be easily understood using rates/collision theory)
temperature (EA will be larger for the endothermic process so it will be relatively more
favoured by a rise in temperature)
pressure (only applicable where there is an imbalance between the number of moles of
gaseous particles on either side of the equation)

catalysts - do not affect position of equilibrium, just the time to achieve it
Monitoring Equilibria

remote sensing – this is non intrusive (e.g. level of absorbance of a given wavelength of light
by a coloured solution) so will not effect the position of equilibrium

titrimetric analysis – this will effect the position of equilibria (since the concentration of one
of the reactants or products will be changed) so is only applicable to a system with a slow
response to a change in conditions or where quenching (dilution or cooling) is used to slow
down the rate of reaction and thus the effect of the investigative technique
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
20 and Chapter 9 in the AS Book
16 – 17
Q 1 - 3 on page 16 – 17
Q 1 on page 32
Chemical equilibrium
Mr Lund 08 May 2017
2
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
This equation can vary and
simply illustrates the principle
behind its construction
Equilibrium Law (i.e. equation)
[C]c
aA








+



cC
+
dD
Kc =
[A]a
eqm
eqm
[B]b
eqm
eqm
only applies to systems at equilibrium so don’t use initial concentrations
Kc is calculated from concentrations at equilibrium (note that the concentration of a solid is
constant and so solids do not appear in the equilibrium expression)
equilibrium constant (Kc) is related to reaction stoichiometry (a, b, c etc) and is a constant at
constant temperature
the value of Kc is an indicator of the position of equilibrium (reverse reaction = inverse value)
the value of Kc is not indicative of how fast the reaction proceeds
you must be able to calculate the numerical value of Kc (possibly using data from an
experiment you will carry out yourself)
note that concentrations are used i.e. moles/volume not moles although quite often V will cancel
down or cancel out completely (when the ∑powers is the same on the top as the bottom) –
but show ALL working in exams
determination of the units of Kc – you must show workings in the exam but …..
check using (moldm-3)∑top powers - ∑bottom powers
Summary Questions
Exam Style Questions

bB
[D]d
Page 21
Page 28
1-3
1
determination of the concentrations of reactants/products present at equilibrium, given
appropriate data including the numerical value of Kc
this may involve the determination of an unknown value of x to solve an equation (but not a
quadratic one on this syllabus)
you may be required to realise that the equation can be simplified by taking the square root of
both the top and the bottom terms and the value of Kc(see the example on page 22 – 23)
you might also be asked to determine the amount of given reactant required in order to
produce a given amount of product (see the example on page 23 – 24)
Summary Questions
Exam Style Questions
Page 24
Page 28
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1
2
20 – 24
18 – 20, 22 - 27
Q 4 – 6, 8 - 13 on pages 18 – 27
Q 3 on page 33
Equilibrium equation
Mr Lund 08 May 2017
3
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Changing Conditions and Equilibria
Pressure or Concentration Change



the value of Kc does not change with variations in concentration or pressure
to improve yield wrt a more expensive reactant a cheaper reactant could be added.
removing the product as it is formed (assuming that the response of the system is fast enough)
would also improve yield whist also allowing reactants to be recycled (see Haber)

pressure only has an effect IF gaseous particles are involved and in addition the
stoichiometric ratio of gaseous particles is unequal on either side of the equation
in all cases a pressure increase will increase the rate of reaction involving gaseous reactants

Kc and Temperature Change


value of Kc increases as temperature increases for endothermic reactions i.e. the equilibrium
shifts to the RHS i.e. more products
value of Kc decreases as temperature increases for exothermic reactions i.e. the equilibrium
shifts to the LHS i.e. less products
Temperature
increases
decreases
Exothermic reaction
Kc decreases
Kc increases
Endothermic reaction
Kc increases
Kc decreases
Kc and Catalysts




value of Kc DOES NOT CHANGE when a catalyst is used
therefore catalysts do not change the position of equilibria i.e. change the yield
HOWEVER, the rate of the reaction will be faster (for both the forward and backward
reaction) hence equilibrium will be achieved sooner
adding a catalyst to a system already at equilibrium will not change its position
How science works
Summary Questions
Exam Style Questions
Page 26
Page 27
Page 28-29
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
Development of the Haber Process
1-3
3–5
25 - 27
18 – 27
Q 4 – 13 on pages 18 – 27
Q 3, 5 on page 33
Equilibrium equation
Mr Lund 08 May 2017
4
How Science
Works: E, F
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Kinetics

review basic ideas of collision theory and activation energy from AS
Rate Equations



the rate of reaction is the rate of change of either the reactants or products in a chemical
reaction
the rate can be determined from concentration versus time graphs
the mathematical expression for the graph on page 4 of reaction A + 2B →C:
Rate =

Don’t panic !! – this can be explained
easily in words – ask your teacher
there are different ways to express rate
(i)
(ii)
(iii)

d [C ]  d [ A]
1 d [ B]


dt
dt
2 dt
average rate
instantaneous rate (this is what the equations above depict) which is the gradient of
the tangent of the curve at selected values of concentration or time
initial rate is the instantaneous rate right at the start of a reaction
initial rate (t0) – the gradient is easiest to determine with confidence since it goes
through zero
Measuring Reaction Rates

see AS Module 2 guide for general practical techniques for monitoring progress in a given
reaction

generally the main methods comprise two different strategies
1
following a single reaction – measuring colour change, change in conduction/pH of one
product or reactant
then plotting a concentration time graph and determining the rates at given concentrations
2
clock techniques – measuring the time to an observable event from known different
initial conditions e.g. the ‘thiosulphate cross’ (the one you did at GCSE) or the ‘iodine
clock’
this must be relatively early in the reaction so that the concentration of other reagents
(other than the one being varied) can be deemed unchanged.
initial rate 
How science works
Summary Questions
1
t
Page 5
Page 6
‘Damn fast reactions indeed’
1-5
Mr Lund 08 May 2017
5
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Rate Equations and Order of Reaction



each reactant may or may not affect reaction rate (so it’s not quite as clear cut as it seemed at
GCSE/AS)
ALSO it is not necessarily directly proportional when it does – we shall see why later
for a given reactant we can state
Rate  [reactant]n

for two reactants A and B the general rate expression is:
Rate 

[A]m[B]n
moldm-3s-1
Note that rate always has these same
units.
m and n are the order of reaction for each reagent (values are limited to 0, 1 or 2 at A-Level)
overall order = individual orders)

unlike with equilibria the orders of reaction and thus the overall rate equation cannot be
determined from reaction stoichiometry they can only by determined by experiments
e.g.
2H2(g) + 2NO(g)

2H2O(g) + N2(g)
doubling [NO] quadruples the rate i.e. double2 = 4
doubling [H2] doubles the rate
Rate  [H2][NO]2
i.e.
[NO]2
The effect on
rate of the
change
here…
NOT as suggested by the ratio of reactants rate  [H2]2[NO]2

k is the rate constant (units depend upon overall order)
rate = k[H2][NO]2


..is raised
to this
power i.e.
its second
order wrt
to [NO]
moldm-3s-1
units of k will vary depending upon the overall order of the reaction
determination of the units of k for 0, 1, 2 and 3rd order reactions overall
should be attempted as they are likely in an exam and here is
another ‘cheat’ for checking your answer:
units of k = (moldm-3)1 - overall order (s-1)

note that if the order is 0 this term will not appear in the rate equation
e.g. A2BC0 = A2B x 1 = A2B
Mr Lund 08 May 2017
6
Maths tip:
Any number raised to
the power zero = 1
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Catalysts and Rate Equations

catalysts will be involved in the rate expression (this might only be a modification of the
value of k itself for example a solid in heterogeneous catalysis)
H+
e.g.
CH3COCH3(aq) + I2(aq)

CH2ICOCH3(aq) + H+(aq) + I-(aq)
by experimentation it was found that the reaction is:
first order wrt [H+] and [CH3CO CH3]
zero order wrt [I2]
i.e. second order overall
rate  [CH3COCH3]1[ H+]1[I2]0
rate = k[CH3COCH3][ H+]
moldm-3s-1
what are the units of k?
Rate Determining Step








most reactions occur in several steps (this is exemplified by organic reaction mechanisms)
each step will take place at a different rate
the slowest step will determine the overall rate of the reaction and is known as the rate
determining step
the order of the reaction regarding each reagent can provide information regarding its
involvement in the rate determining step
obviously a reactant with 0 order will not be involved in the rate determining step
study of reaction kinetics can yield important information regarding the mechanism of a
multi-step reaction
in the reaction above iodine would not be involved in the rate determining (slow) step
at this point you might ask your teacher to explain why there are variations in the mechanism
for the hydrolysis of a haloalkanes as discussed on pages 15 – 16 (or perhaps research SN1
and SN2 yourself)
Summary Questions
Page 9
Page 16
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1–4
1
4 – 9, 14 - 16
4–6
Q1, 2, 3 on page 5-6
Order of reaction, rate expression, SN2
Mr Lund 08 May 2017
7
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Concentration/Time and Rate/Concentration Graphs

rate/concentration graphs can show the order with respect to a given reagent

it is glossed over at A-Level that the other reagents will have to be present in xs so that their
concentration can be deemed to be unchanged during the course of a reaction where a series
of measurements of the concentration of a given chemical are measured.
or – in a clock technique the event measured in relatively early in the reaction so better
reflects the known initial concentration of each reagent


zero order e.g. decomposition of ammonia (tungsten catalyst), rate is independent of
concentration i.e. graphically it is a flat horizontal line
2NH3(g)
W

3H2(g) +
Do you have an idea as to why it might be zero
order?
N2(g)

first order e.g. thermal decomposition of dinitrogen monoxide to nitrogen and oxygen (gold
catalyst), rate is directly proportional to concentration and will be a straight line gradient = k
(any points not on the straight line will be anomalies and require identification and
explanation (e.g. temperature variations))

second order e.g. thermal decomposition of ethanal to methane and carbon monoxide will
produce a graph that curves upwards (rate against concentration2 is a straight line)
Initial Rates Method





initial reaction rates are determined by plotting the tangent to the time/concentration graph for
different initial reagent concentrations at the start of the reaction (t = 0) when reaction
concentrations are accurately known (and at a fixed temperature/catalyst)
the gradient of this line is the initial rate
orders of reaction can be determined from initial rates data by inspection
the value of k can also be determined from this data
the best way to grasp this idea is to try examples
Summary Questions
Exam Style Questions
Page 13
1
Page 17 - 19 1, 3, 5
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
10 - 12
7 – 11
Q4 on page 8
Q1 on page 14
rate
Mr Lund 08 May 2017
8
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Variation of k with Temperature
Maxwell-Boltzmann Distribution Curve

distribution of energies amongst particles at different temperatures give rise to the MaxwellBoltzmann distribution curves based on the Arrhenius expression
k  Ae
A
R
T

=
=
=

EA
RT
Arrhenius constant (determined by collision frequency and orientation factor)
gas constant (8.31 JK-1mol-1)
absolute temperature (K)
the main significance of this equation (which you don’t need to know – unless you are trying
to understand the subject) is that a small rise in temperature has an exponential (i.e. big)
effect on rate
Extra info for
those who do
Maths – how
can we get
values for EA
and A
ln(A) –
ln(k) =
1
E
-  A 
 R  T
+
ln(A)
m
+
c
y
=
Plot ln(k) against 1/T gives gradient = -
Exam Style Questions
Page 17 - 18
Page 158
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
EA
RT
ln(k) =
x
EA
R
and intercept = ln(A)
2,4
1
10 - 12
12 – 13
Q2-4 on page 14 - 15
Q1 on page 150
Arrhenius expression
Mr Lund 08 May 2017
9
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Acid-Base Equilibria



Arrhenius definition of an acid - hydrogen ions and oxonium ions (H3O+(aq))
soluble base = alkali
alkaline solutions have relatively high [OH-(aq)]
Brønsted-Lowry Theory of Acids and Bases

Arrhenius limited to aqueous solutions but acid-base concept is broader e.g.:
NH3(g) +

HCl(g)
NH4Cl(s)


acid-base equilibria involve proton transfer
acids are proton donors, bases are proton acceptors



water is amphoteric
it behaves as an acid with NH3
and as a base HCl





conjugate acid-base pairs undergo proton exchange
competition occurs for protons between bases on either side of the equilibria
relative strength of the base determines the equilibria bias
a relatively strong base has a relatively weak conjugate acid and visa-versa
note that in the protonation of nitric acid in the nitration of benzene – nitric acid acts as a base
You must make the
effort to learn the
definition of a BrønstedLowry acid and base
Ionic Product of Water (Kw)


water undergoes slight auto-ionisation (dissociation of the water molecule)
ionic product of water – again the equation defines the concept – and don’t forget units
Kw

=
variation of Kw
Summary Questions
[H+(aq)][OH-(aq)]
1 x 10-14 mol2dm-6 at 298K
its value increases as temperature increases - LCP - (hence [H+(aq)]
increases and therefore the pH of a neutral solution decreases!)
Page 31
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
=
1-3
30 - 31
34 – 35, 36, 42 - 43
Q1 on page 35
Q6 on page 37
Brønsted, ionic product
Mr Lund 08 May 2017
10
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
pH Calculations

define pH using an equation NOT in words as it is more certain to get full marks.
pH = -log10[H+(aq)]
learn to use your calculator !

pH of monoprotic (release a single proton into aqueous solution) acids e.g. HCl (note: pH’s
lower than 1, including –ve values, are possible)

you should be able to calculate pH after a strong acid of known volume and concentration (or
pH) is diluted by a known volume of water

pH of diprotic acids

calculating [H+(aq)] from pH values
[H+(aq)] = 2 x [H2SO4(aq)]
NOTE:
[H+(aq)] = 10- pH



learn to use your calculator !!!!!!
calculating the pH of alkalis using Kw to calculate [H+(aq)]
note for Ca(OH)2 that [OH-(aq)] = 2 x [Ca(OH)2(aq)]
but why might it give a weaker acid than sodium hydroxide?
[OH-(aq)]
=
Kw
[H  (aq) ]
How science works
Summary Questions
Page 33
Page 31
Measuring pH
Questions 1 - 5
Exam Style Questions
Page 50
Questions 3
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
32 – 35
38 – 40, 42 - 45
Q7 - 10 on pages 39 - 40
Q13, 14 on pages 37 – 38
Q 3, 4 on page 55
Q 7 on page 56
Strong acids pH
Mr Lund 08 May 2017
11
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Weak Acids and Ka

weak acids are only partially ionised
HA(aq) +
Kc =

H3O+(aq)
H2O(l)
+
A-(aq)
[H 3 O  (aq)][A - (aq)]
[HA(aq)][H 2 O (l) ]
[H2O]
~
constant
[A-(aq)]
=
[H3O+(aq)]
=
[H+(aq)]
given that it is a weak acid we can assume the degree of dissociation is minimal hence:
Ka
=
[acid]equilibrium ~
[acid]initial
Kc[H2O]
=
[H  (aq)][A - (aq)]
[HA(aq)]
Ka
=
[H  (aq)] 2
[HA(aq)]
pKa
=
-log10Ka
Ka
=
10- pKa


relatively higher Ka / relatively lower pKa = stronger acid (given the same concentration)
value is independent of concentration and therefore more useful

you should be able to calculate the pH of a weak acid of known concentration using Ka and
calculate Ka from the pH of a weak acid of known concentration
Summary Questions
Exam Style Questions
Page 38
Page 51
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1-3
6
32 – 35
35 – 36, 40 – 42
Q11 - 12 on pages 41 - 42
Q 1 on page 55
Weak acid
Mr Lund 08 May 2017
12
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Titration Curves

as an alkali is added to an acid the pH increases and visa-versa

you should also be able to calculate the pH at any point in the addition of sodium
hydroxide to a monoprotic acid (and visa-versa) including weak acids

typical errors to avoid are not converting to moles and getting the stoichiometric ratio
wrong
most likely you will forget to use the total volume of the solution created and thus get
the concentration wrong and therefore the pH
another likely error with a weak acid where xs alkali has not been added is assuming
that the remaining acid is fully dissociated i.e. forgetting to use Ka to determine [H+(aq)]




equivalence point is where two solutions have reacted in stoichiometrically the correct molar
ratio – this will be the vertical point on the titration curve where the pH changes markedly
pH’s at equivalence point and appropriate curves for:
WASB
SASB
SAWB
WAWB
http://www.avogadro.co.uk/chemeqm/acidbase/titration/phcurves.htm




pay particular attention to the position of the initial pH for strong and weak acids and look
carefully at how it changes at the start
also carefully note the position of the equivalence point – the mid point of the vertical section
finally ensure that a sensible final pH is shown to reflect the use of a strong or a weak base.
you should be able to calculate concentrations of an unknown acid or alkali from the results
of a titration – pretty much as was the case for AS level – but there will be more likely hood
of diprotic acids cropping up (e.g. sulphuric acid)
Summary
Questions
Page 41
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1-2
39 – 41
44 – 48
Q15 on page 45
Q 16 on page 48
Q 2, 8 on pages 55 - 56
Titration curves
Mr Lund 08 May 2017
13
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Ka of Weak Acids from Titration Curves

at the half equivalence point (half neutralisation point as stated in your text book) given that:





[HA] =
Ka
=
pKa =
[A]
[H+(aq)]
pH
the half equivalence point can be determined practically by determining the pH at half the
equivalence point (half the volume) from a plotted titration curve
note that it is not half the pH value itself at the equivalence point that is used!
alternatively a titration can be repeated with half the volume of the already determined
equivalence point and the pH then measured using a pH meter
End Point of an Indicator




indicators can only be used for a titration curves with a vertical section of >2 pH units
suitable indicators for acid-base titration’s will have an end point and range that lie within
that vertical section i.e. will thus exhibit a sharply defined colour change
pH meters can be used on coloured solutions
indicators are weak acids
H+(aq) + In-(aq)
HIn(aq)
[H(aq) ][In-(aq) ]
[HIn(aq) ]
Ka (or Kin)
=
at end point
[In-(aq)] = [HIn(aq)]
Summary Questions
Exam Style Questions
Kin
=
[H+(aq)]

pKa
=
pKin
Page 41
Page 50
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide

=
1-2
4
39 - 41
49 – 51
Q17 - 18 on pages 50 – 51
Q 5 on page 56
Indicators
Mr Lund 08 May 2017
14
pH
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Buffer Solutions



the effect of pH changes e.g. lemon juice and the proteins in milk
buffers are designed to maintain pH stability
you must understand buffer solution in terms of the response of an equilibrium system to the
addition of hydrogen or hydroxide ions
Acidic Buffer
Added H+(aq) is removed by A-(aq) provided by
the salt as the equilibrium shifts to the left
HA
⇋
H+ + A-
reservoir of HA provided by acid
Added OH-(aq) is removed by
reacting with H+(aq) provided
by the acid as the equilibrium
shifts to the right
NaA

An acidic buffer (for pH <7) consists of a
weak acid and its soluble salt
Na+ + A-
reservoir of A- provided by the sodium
salt of the acid
Basic Buffer
Added H+(aq) is removed by NH3 as the
equilibrium shifts to the right
NH3
+ H+
⇋
NH4+
reservoir of NH3 provided by weak base
Added OH-(aq) is removed by
reacting with NH4+(aq)
provided by the salt and the
equilibrium shifts to the left
NH4Cl

An alkaline buffer (for pH >7) consists of a
weak alkali and its soluble salt
NH4+ + Cl-
reservoir of NH4+ provided by the
ammonium salt
Mr Lund 08 May 2017
15
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Important Buffer Solutions

Carbonic acid-Bicarbonate Buffer in the Blood is by far the most important process for
maintaining the acid-base balance (in our bodies there are also phosphate and protein
buffers).
Added H+(aq) is removed as the
equilibrium shifts to the right
Added OH-(aq) is removed by reacting with
H+(aq) shitting the equilibrium to the left

a pH change of over ± 0.5 can be fatal

buffer solutions are also used in:
 the food industry (‘acidity regulators’)
 fabric dyeing – where ‘AZO’ dyes are used
 hair care products – which are kept slightly acidic ‘pH 5.5’ as alkaline conditions
make hair look rough as microscopic scales on the surface of the hair are made to
stand up.
 calibrating pH meters
 biochemical research (enzymes are denatured by pH extremes)
 and there are numerous other areas
Preparation of Buffer Solutions

a buffer solution can be made in two different ways:
1
2
adding a suitable soluble salt to an acid
partially neutralising a weak acid with a strong base up to the required pH

when [HA] = [A-] the buffer solution is equally able to deal with the addition of acid and base
by equal sized reservoirs

for this reason a weak acid with a pKa relatively close to desired pH is selected for more
effective buffering (see the calculations in the next section)
Mr Lund 08 May 2017
16
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Calculating the pH of a Buffer Solution
HA ⇋ H+ + A-
but
[H  (aq)][A - (aq)]
[HA(aq)]
[HA] ~ [ACID]i
and
[A-] = [salt]
Ka
=
Ka
=
[H  (aq)][SALT ]
[ACID)]
[H+(aq)]=


[ACID]
[SALT]
it is assumed that the volume of the weak acid solution is unchanged by the addition of a
small quantity of its solid salt
since all particles are present in the same total volume of solvent we can make life easier by
appreciating that the acid to salt:
moles ratio

Ka
Will the dilution
of
a
buffer
solution change
the pH?
=
concentration ratio
the best buffer will be that obtained at half the equivalence point:
and so
[ACID]
=
[SALT]
[H+(aq)]
=
Ka
=
pKa
pH

equimolar amounts of acid and salt produces a buffer solution with a pH of the same
numerical value as pKa

you should be able to determine the required combination of acid and salt to produce a buffer
solution of a given pH
you should also be able to demonstrate by calculation that adding acid or alkali to a buffered
solution changes the pH by less than for an un-buffered solution

Summary Questions
Exam Style Questions
Page 48
Page 49
Page 51
Page 160
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1
1, 2
5
5
45 - 48
52 – 54
Q17 - 20 on pages 52 – 54
Q6, 9 on page 56
Buffer
Mr Lund 08 May 2017
17
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
How Science
Works: H
Organic Nomenclature




displayed formula – every bond and every atom should be shown in examinations!!!
structural formula e.g. CH3CH2CH2OH
skeletal formula may be useful for prospective medical etc students
it is important that you understand the difference between 2-D displayed formula and the
actual 3-D molecular shape as this will prove particularly relevant later


functional group – one or more reactive sites on a hydrocarbon skeleton
homologous series – same general formula + same functional group (hence same chemistry
as the increase in chain length has little effect on their chemical reactivity)

nomenclature is based on four criteria:
root
suffix
prefix
locant





longest unbranched hydrocarbon chain (including main functional group)
determines principal functional group (a pecking order exists)
other changes to root molecule (e.g. side chains, functional group)
position of branch or substituent (e.g. double bond) on the main chain
look for the longest chain NOT the longest ‘straight’ chain of carbons
start numbering at the end of the chain that results in the lowest numbers in the name or
for the primary functional group position
alphabetical order is used where more than one type of functional group or branch is
required in the prefix
mono, di, tri and tetra indicate multiple functional groups or branches of a given type (don’t
change alphabetical order)
commas and dashes are important - so learn how to do this correctly !!!!
An “e” is dropped if the next letter is a vowel: “propan-2-ol, propane-1,2-diol”
An “a” is added if inclusion of di, tri, etc., would put two consonants consecutively: “buta1,3-diene”, not “but-1,3-diene” “propanenitrile, not propannitrile or propanitrile.)

you will need to aware of nomenclature examples of the following (including cyclic
variations):
alkanes
alkenes
haloalkanes
alcohols
aldehydes and ketones
carboxylic acids
esters
acyl chlorides
amines
amides
amino acids benzene and its derivatives
Summary Questions
Exam Style Questions
Page 56
Page 64
A2 Chemistry (Nelson Thornes) AQA
Chemguide
1-3
1
52 - 56
nomenclature
Mr Lund 08 May 2017
18
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
How Science
Works: H
Isomerism


isomers have the same molecular formula but different chemical and/or physical
properties
remember that there are a number of ways in which isomerism exists that you have already
met:
UNIT 1
isomers
structural isomers
position
chain
stereoisomers
UNIT 2
functional
group
UNIT 4
geometrical
optical
Structural Isomerism
structural isomers have the same molecular formula but different structural formula

one form of structural isomerism is called chain isomerism – unbranched chain and
branched chain – i.e. different hydrocarbon skeleton
e.g. How many versions of C4H8 can you find that represent structural isomers?

functional groups that are present at different positions are called positional isomers
(there will be different numbers in the name)
e.g. propan-2-ol and propan-1-ol

functional group isomerism exists where the molecular formula is the same but different
functional groups (and therefore chemical properties exist)
e.g. propane-l-ol and methoxy ethane, but-2-ene and cyclobutane
Mr Lund 08 May 2017
19
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Stereoisomerism
Stereoisomers have the same molecular and structural formula but differ in the spatial
arrangement of their atoms.
How Science
Works: A
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
this lack of free rotation is consequential of the π bond present in the alkene
a single carbon-carbon bond cannot give rise to this type of isomerism as (unless a chain is
present) there is unrestricted rotation
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

whilst the minimum requirement is the presence of restricted rotation consideration should
also be given to the substituent’s on each carbon

the two molecules on the left above are identical even though there is a carbon-carbon double
bond as simply flipping vertically makes them super imposable
across a carbon-carbon double bond each carbon in turn must have different substituent’s
it doesn't matter whether the two groups are the same e.g. in the example on the right no
amount of flipping or rotating makes them super imposable.


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.

How Science
Works: H



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
How Science
Works: L
Find out about the food industry and trans fats and the
hydrogenation of vegetable oils.
Mr Lund 08 May 2017
20
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry

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

take care here when using older text books as some molecules deemed ‘trans’ in the old
system would actually be ‘Z’ in the new system i.e. across the double bond in one system
does not directly yield across the double bond in the other e.g. 3-bromobut-2-ene
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.

E-Z transformations are possible given an energy source e.g. photochemistry and eyesight:
nerve impulse
to the brain
CH3
CH3
CH3
O
light
to eye
CH3
CH3
CH3
CH3
CH3
CH3
CH3
How many functional
groups can you see
here?
How can you test to
provide evidence for
each one?
O
Mr Lund 08 May 2017
21
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Optical Isomerism
Optical isomerism exists where there is an asymmetric (i.e. chiral) carbon with four different
groups attached.
Can you give the
systematic name for this
molecule?
Optical isomers are non-super imposable molecules (enantiomers) which are mirror images of
one another.







don’t just put ‘mirror images’ as they can sometimes be superimposable where a plane
of symmetry exists
if two groups are the same i.e. there is a plane of symmetry then a simple rotation yields the
same spatial arrangement so they are not enantiomers
many chemicals synthesised in the lab produce equal amounts of both enantiomers (see
lactic acid later) which is called a racemic mix
this is because the reagents are not stereo specific (rather like a left handed screwdriver)
however, living organisms tend to manufacture one enantiomer in preference to the other as
determined by the reactive sites of the optically active enzyme used to construct it
enzymes are stereo specific reagents
the way that these enantiomeric molecules interact with biological systems can be different,
for example:
carvone
limonene
one enantiomer tastes of spearmint the other caraway
one smells of lemons the other oranges (see if you can spot the chiral centre in
these enantiomers of limonene)
Mr Lund 08 May 2017
22
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Enantiomers rotate plane polarised light in opposite directions (clockwise and
anticlockwise).



this physical property is the means by which they can be distinguished
one enantiomer rotates the polarised light clockwise (to the right) and is the (+) enantiomer;
the other rotates the polarised light anticlockwise (to the left) and is called the (–) enantiomer.

a racemic mix consists of a 50:50 mix of both isomers and this will therefore NOT rotate
plane polarised light as the two enantiomers cancel one another out
How Science
Works: L
How science works
Summary Questions
Exam Style Questions
Page 62
Page 59
Page 64
A2 Chemistry (Nelson Thornes) AQA
Chemguide
The thalidomide tragedy
1-4
2, 4, 5
57 - 59
isomerism
Mr Lund 08 May 2017
23
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Aldehydes and Ketones
Physical properties


lower Mr aldehydes and ketones are miscible with water due to polar C=O bonds ability to
hydrogen bond with water molecules
higher Mr molecules have greater VdW with one another due to increasing size of
hydrocarbon tail hence miscibility is reduced for energetic reasons
Preparation
primary alcohol

aldehyde
heat with K2Cr2O7(aq)/H2SO4(aq), distil off
aldehyde (boiling point lower than alcohol)
secondary alcohol

ketone
reflux under heat with K2Cr2O7(aq)/H2SO4(aq)

orange dichromate(VI) ions (Cr2O72-aq)) are reduced to green chromium(III) ions
(Cr3+(aq))
Distinguishing Between Aldehydes and Ketones
ACIDIFIED POTASSIUM DICHROMATE SOLUTION

tests are based on the fact that aldehydes can be easily oxidised to a carboxylic acid while
ketones cannot be.
Aldehyde
+
[O]

reflux with K2Cr2O7(aq)/H2SO4(aq)
Carboxylic Acid
orange dichromate(VI) ions (Cr2O72-(aq)) are reduced to
green chromium(III) ions (Cr3+(aq))
Care: The above test is only applicable if it is clear that the unknown sample is not a primary
or secondary alcohol.
Mr Lund 08 May 2017
24
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
NOTE: PRIMARY AND SECONDARY ALCOHOLS CANNOT BE OXIDISED BY
FEHLINGS’ OR TOLLENS’ HENCE DO NOT GIVE POSITIVE RESULTS WITH
THE TESTS BELOW.

writing balanced redox equations under alkaline conditions is a little more involved than
acidic conditions but you will find an excellent strategy on CHEMGUIDE – see the link
below
FEHLINGS’ TEST
warm with Fehlings’ solution
(an alkaline solution of a complexed
copper ion)
blue complexed Cu2+(aq) is reduced to brick
red Cu2O(s)
(this is the basis of the test for reducing sugars)
RCHO(aq) + 2Cu2+(aq) + 4OH-(aq)

RCOOH(aq) + Cu2O(s) + 2H2O(l)
TOLLENS’ REAGENT
complex ion [Ag(NH3)2]+(aq) is reduced to Ag(s),
hence the silver mirror effect
warm with Tollens’ reagent
(aqueous silver nitrate in xs ammonia)
RCHO(aq) + 2[Ag(NH3)2]+(aq) + 2OH-(l)  RCOOH(aq) + 2Ag(s) + 4NH3(aq) + H2O

strictly speaking you will get the carboxylate anion RCOO-(aq) under alkaline conditions rather
than the carboxylic acid itself

additionally it’s worth knowing that methanoic acid (which has a hydrogen present HCOOH)
can be oxidised to carbon dioxide via carbonic acid H2CO3 which then easily breaks down
into CO2 and H2O (see if you can work out the equations)

you should be aware of changes in the IR spectra during the oxidation reactions of
compounds containing one or more oxygen atoms (and also the reduction reactions of said
molecules)
Summary Questions
Page 67
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1-4
66 – 67, 69
65 – 67
Q 1 on page 66
Ionic alkaline, Tollens’, Fehlings’
Mr Lund 08 May 2017
25
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Nucleophilic Addition
Reduction of Aldehydes and Ketones

NA
polar nature of the carbonyl group - you would be wise to revise electro negativity and the
nature of nucleophiles
aldehyde + 2[H]

primary alcohol
ketone + 2[H]

secondary alcohol



sodium tetrahydridoborate(III) (NaBH4)
in water (+ ethanol as a universal solvent
for higher (longer chain) members)
BH4- provides the hydride ion, H- which acts as a nucleophile
mechanism is required (nucleophilic attack by hydride ion)
intermediate ion then gains H+ from the water present in the aqueous solvent – just show H+
in the mechanism itself
NA
Reaction with Hydrogen Cyanide
ethanal + HCN

2-hydroxypropanenitrile



HCN is made ‘in situ’ using acidified sodium or potassium cyanide
HCN is a toxic gas so there are health and safety issues in its use
cyanide ions are toxic so this will not be done in the lab


note: an extra carbon is introduced into the chain so this is an important synthesis step
mechanism for reaction = nucleophilic addition


trigonal planar carbonyl group can be attacked from either side
hence products from aldehydes other than methanal exhibit optical isomerism i.e. there is
an asymmetric (i.e. chiral) carbon with four different groups attached hence two nonsuper imposable molecules (enantiomers) exist which are mirror images of one another.

a racemic mix is produced – i.e. one that is 50:50 of both isomers and this will therefore
NOT rotate plane polarised light in this case as the two enantiomers cancel one another out
symmetrical ketones do not yield enantiomers as no chiral centre is present




hydroxynitriles can be converted to a carboxylic acid by undergoing acid hydrolysis
this involves refluxing with a dilute acid (below I have shown the organic as named for
clarity - although you MUST give its formula - as the other particles will always be the same
irrespective of the number of carbons in the chain)
the optical properties are preserved in the carboxylic acid
2-hydroxypropanenitrile + 2H2O + H+

2-hydroxypropanoic acid
(lactic acid)
Mr Lund 08 May 2017
26
+
NH4+(aq)
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry

lactic acid synthesised this way will exist as racemates BUT that produced biologically will
not be since enzymes are stereo specific yielding only one optical isomer hence this DOES
rotate plane polarised light.
Summary Questions
Exam Style Questions
Page 63
Page 70
Page 83
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1–3
1-4
2
60 – 63, 68, 70
66 – 70, 57 – 62
Q 2 – 3 on pages 68
Q 4 -6 on page 70
Q 1, 3, 5 on page 82
Q 4 on page 149
Aldehydes, Isomerism
Mr Lund 08 May 2017
27
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Carboxylic Acids and Esters
Nomenclature





be aware of different ways to write the acid functional groups RCOOH, RCO2H etc
benzenecarboxylic acid C6H5CO2H
general structural formula of esters: RCO2R` e.g. CH3CO2CH2 CH2CH3
you should be able to work out the acid/alcohol used to make an ester and visa versa
esters and carboxylic acids are functional group isomers (easily distinguished by IR or nmr
– see later - or a simple chemical test using sodium carbonate and testing for evolved CO2)
Summary Questions
Page 73
1–4
Physical properties




lower members of the carboxylic acids and esters are miscible with water due to hydrogen
bonding with water
higher members are less miscible with water as the extent of VdW with themselves becomes
prevalent (they are more soluble in sodium hydroxide solution – do you know why?)
most carboxylic acids are crystalline solids (hydrogen bonding) whilst esters are typically oils
and fats (no hydrogen bonding) compared to similar sized hydrocarbons
melting points can be used to identify and determine the purity (to some extent) of organic
solids
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
71 - 73
71 – 75
Q 7 – 10 on pages 71 - 74
Carboxylic acids, Esters
Synthesis of carboxylic acids
primary alcohol or aldehyde


carboxylic acid
reflux under heat with xs
K2Cr2O7(aq) and H2SO4(aq)
orange dichromate(VI) ions (Cr2O72-aq)) are reduced to green chromium(III) ions (Cr3+(aq))
A word about REAGENTS: When the examiner asks for a reagent then it is the name on the bottle
NOT the active particle introduced e.g. H+(aq) is not a reagent but H2SO4(aq) is.
Mr Lund 08 May 2017
28
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Reactions of Carboxylic Acids and Esters


carboxylic acids are weak acids – understanding required in terms of equilibria
chlorine-substituted ethanoic acids are more acidic due to negative inductive (–I) effect of
chlorine atoms due to their relatively high electronegativity
ethanoic acid
ACID
+
+
sodium hydroxide
BASE

sodium ethanoate
SALT
ethanoic acid
+
sodium carbonate

sodium ethanoate + water + carbon
dioxide

+
+
water
WATER
the latter is a useful test for the presence of –COOH for which you will OBSERVE
effervescence and that the gas evolved turns limewater (Ca(OH)2(aq)) cloudy and thus
INFER that CO2(g) was produced suggesting a carboxylic acid.
Ca(OH)2(aq)


+

CO2(g)
CaCO3(s)
+
H2O(l)
acid catalysed esterification with an alcohol is relatively slow and gives a poor yield
ethanol + ethanoic acid
ethyl ethanoate + water
acid
ester
+
alcohol
Summary Questions
Page 77
reflux under heat with cH2SO4
catalyst
+ water
1–4
Hydrolysis of Esters
ethyl ethanoate + water
ethanoic acid + ethanol
sulphuric acid catalyst

initiated by nucleophilic attack by the water molecule on the Cδ+ of the carbonyl group

acid catalysed hydrolysis not complete due to an equilibrium being established


alkali (hot NaOH) catalysed (saponification) hydrolysis is quicker and goes to completion
sodium salt of the carboxylic acid is produced since the acid produced reacts with the sodium
hydroxide which will drive the equilibrium  RHS as acid is removed from the system
adding xs sulphuric acid protonates the alkanoate anion  carboxylic acid

A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
74 - 76
71 – 74
Q 11 on page 74
Carboxylic acids, esters hydrolysis
Mr Lund 08 May 2017
29
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Uses of Esters


solvents – e.g. ethyl ethanoate for nail varnish
they evaporate relatively easily since there is no hydrogen bonding

esters have pleasant smells (unlike carboxylic acids which typically have unpleasant smells –
rancid fats – ask to smell some butanoic acid – you’ll get the idea) so are used in perfumes
and in the food industry as flavourings:
butyl butanoate


ethyl pentanoate
3-methylbutyl ethanoate
The cost of synthesising esters is often far less expensive than extracting them from
natural sources
2-methoxyphenol is a waste product from the paper industry and can be used to make methyl
vanillin (4-hydroxy-3-methoxybenzaldehyde) an artificial vanilla – by the end of module 4
you might be able to draw its structure and suggest a possible synthesis strategy (although the
industrial process is more complex)

plasticizers – added to plastics (e.g. PVC) to make them softer and more flexible as they
weaken the IMF between polymer strands allowing them to slide over each other more readily
(loss over time makes the plastic brittle)

some phthalate based plasticizers have a possible association with birth
defects although this is still a subject of some disagreement
How Science
Works: I
Fats and Oils








animal fats + vegetable oils are triesters of propane-1,2,3-triol (glycerol) and fatty acids
(long chained carboxylic acids)
they are triglycerides - three alcohol groups esterified by up to three different carboxylic
acids (take care with the hydrolysis equation regarding the actual products e.g. under alkaline
conditions the anion of each acid group will be formed and each of these could be different)
oils have a higher degree of unsaturation cf fats (research E-Z isomerism and trans fats)
a greater number of double bonds reduces the relative flexibility (due to restricted rotation) of
the molecule which in turn reduces the overlap efficiency of intermolecular forces
if a triglyceride undergoes alkaline hydrolysis (NaOH(aq))(saponification) then the salt of a
fatty acid is produced e.g. sodium octadecanoate which is also known as sodium stearate and
better known as a soap
being ionic, soaps are soluble
the carboxylate anion RCOO- released is miscible
with water (due to the hydrophilic carboxylate
group) and miscible with grease (due to its
hydrophobic tail) hence can solvate grease into
water allowing its removal
soap is precipitated out of solution by adding xs
common salt (salting out) – which can be
understood from an equilibria point of view
Mr Lund 08 May 2017
30
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Glycerol (glycerine)

the co-product of soap making, propane-1,2,3-triol (glycerol) has numerous uses:
it is a useful solvent e.g. in medicines, food colouring
its extensive hydrogen bonding makes it good at retaining water thus preventing drying out
(e.g. facial creams)
and can be used to make nitro-glycerine (which mixed with finely divided silicon(iv) oxide =
dynamite)
How Science
Works: K
Biodiesel

a renewable fuel made from oils obtained from vegetable matter (e.g. rape seed) which
generally consist of a combination of any three of five common carbon chains linked by a
glycerol structure


it can also be made from animal fat and waste oil
small industrial plants tend to be batch processes but larger plants can use continuous flow
methods which are more economical

methyl esters are produced by reacting these oils with methanol and a strong alkali at around
60oC
this is called base-catalysed transesterification
the methyl groups replace the glycerol structure on each of the fatty acids
Note: there may be three different methyl esters produced but the general formula of each is:



CH3OOCCxHy where x and y depend on chain length and degree of unsaturation
(this can be written the other way around CxHyCOOCH3)





the ester produced does not readily mix with the propane-1,2,3-triol co product so can be
separated using a separating tank or a centrifuge
any remaining glycerol can be extracted using water (hydrogen bonding)
there may be some soap bi-product so further processing will be necessary to achieve a level
of purity acceptable for a biofuel
rape seed (the yellow stuff you see in fields) produces rape methyl ester (RME) which is very
similar to the diesel obtained from crude oil
this can be used directly or as a small % of filling station diesel
Exam Style Questions
Page 83
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
4
76 - 77
75
Fats and oils, soap, glycerol
Mr Lund 08 May 2017
31
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Acylation using Acyl Chlorides or Acid Anhydrides

nomenclature – named after the parent carboxylic acid hence deemed acid derivatives (as are
esters and amides)
O
R
C
X

Acyl chlorides:
X = Cl
-oic acid replaced with:
-oyl chloride
e.g. ethanoyl chloride = an acyl chloride
Acid anhydrides
X = OR
e.g. ethanoic anhydride

both are readily attacked by nucleophiles (book error on 79) due to very polar carbonyl group
the polarity of which is increased by the electron withdrawing effect of X
thus this group of compounds are more useful than carboxylic acids in synthesis due to their
high reactivity due to the enhanced δ+ of the carbonyl carbon and since -X is a good leaving
group cf –OH

used to join an acyl group
O
R
C
:Nu
to the oxygen of water, alcohol or phenol
or the nitrogen of ammonia or an amine
(ethanoylation is specifically when R = CH3)
H+ is eliminated in each case in the final step
Hydrolysis
NAE
ethanoyl chloride
+
water

ethanoic acid +
hydrogen chloride

very exothermic reaction, steamy fumes of hydrogen chloride are produced even when
exposed to air (due to the reaction between HCl and water vapour), hence anhydrous
conditions essential with acyl chlorides and must be stated in exams


reaction is faster than with haloalkanes due to additional polarising effect of C=O
hydrogen chloride fumes can be tested for using:
conc. ammonia
a drop at the end of a glass rod will create a white
smoke of ammonium chloride
silver nitrate solution
a drop at the end of a glass rod will go cloudy as white
silver chloride is precipitated

mechanism is nucleophilic addition-elimination (condensation)

NOTE: H is not abstracted by Cl- (think why HCl is a strong acid!)
Ethanoic anhydride +
water

ethanoic acid
Mr Lund 08 May 2017
32
+
ethanoic acid
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Reaction with Alcohols and Phenols
ethanoyl chloride

ethanol

ethyl ethanoate
+
HCl
yield of ester better than with carboxylic acid since reaction goes to completion
NOTE:



+
NAE
acyl chlorides form esters with the phenol group unlike carboxylic acids. The
lone pair of the O in phenol is less readily available since these electrons are
delocalised into the ring system hence reducing the electron density and the thus
the effectiveness of phenol as a nucleophile (higher activation energy)
mechanism for reaction with an alcohol - ester formation is very similar to that with water
treat the alcohol as RO-H cf water as HO-H so RO- is added to the carbon rather than HONOTE: H is not abstracted by Cl-
ethanoic anhydride
+

ethanol
ethyl ethanoate +
ethanoic acid
Reactions with Ammonia and Amines
NAE
ethanoyl chloride + ammonia  ethanamide (a primary amide) + hydrogen chloride


violent reaction with aqueous ammonia at room temperature
ethanamide is the only product as further substitution does not occur due to strong electron
withdrawing effect of C=O which makes the lone pair of the nitrogen less readily available
than with amines (cf haloalkanes + ammonia)
NAE
ethanoyl chloride + ethylamine
(primary amine)

N-ethylethanamide + hydrogen chloride
(secondary amide)
ethanoyl chloride + phenylamine

N-phenylethanamide + hydrogen chloride
(i)
(ii)
the product (an acyl derivative) is a white crystalline solid with a sharp melting point
- can be recrystallised and used in the identification of the original amine
suggest reagents and mechanism for the synthesis of paracetamol N-(4hydroxyphenyl)ethanamide
ethanoic anhydride
+
Summary Questions
Exam Style Questions
Page 82
Page 83
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide

ammonia
ethanamide +
1–4
1, 3
78 - 81
76 – 83
Q 12 - 15 on pages 77 – 81
Q 1 – 8 pages 82 - 83
Acyl
Mr Lund 08 May 2017
33
ethanoic acid
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
How Science
Works: I, J
The Synthesis of Aspirin
(aspirin is 2-ethanoyloxybenzoic acid – don’t panic you wont be asked for this on the exam)


the benefits of willow bark, which contains salicylic acid (2-hydroxybenzoic acid) a similar
compound to aspirin, have been known for millennia e.g. Hippocrates (~460 B.C - 377 B.C.),
African Hottentots and North American Indians
it acts as an analgesic (pain killer) and has an anti-pyretic effect (body temperature)
salicylic acid was first isolated around 1829
the next step was to find a way to synthesise it rather than rely on extraction from a natural
source as this can be problematic:
the source might be rare, or seasonal, or have a low concentration, or have harmful
contaminants
in 1860 it was synthesised from phenol (a by product of the production of town gas from
coal) using the Kolbe process (NaOH and high pressure CO2)
but the problem was that it was tough on stomachs so alternatives with a similar structure
(hence retaining the benefits) were searched for
aspirin itself had been synthetically produced in 1853 by a French chemist named Charles
Frederic Gerhardt but he didn’t realise its potential and took it no further
in 1898, a German chemist named Felix Hoffmann rediscovered Gerhardt's formula
he gave it to his father who was suffering from the pain of arthritis and with good results (he
had tried other formulations before that!!) so convinced the German pharmaceutical company
Bayer to patent it in 1900 (the patent was ignored by the allies during WW1 and thereafter
– along with the patent they held for heroin!)
its sales increased dramatically during the Spanish Flu epidemic of 1918

aspirin can be synthesised from salicylic acid using ethanoyl chloride or ethanoic anhydride








+
2-hydroxybenzoic acid
(salicylic acid)

aspirin
+
Ethanoic anhydride +
2-hydroxybenzoic acid
(salicylic acid)

aspirin +
ethanoic acid
ethanoyl chloride


HCl
NAE
both are more readily attacked by nucleophiles than the corresponding acid
acid anhydrides offer certain advantages over acyl chlorides, despite being less reactive, in
that they are:
cheaper, less corrosive (no HCl liberated), less readily hydrolysed
How science works
Exam Style Questions
Page 82
Page 83
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
Aspirin
1
82
81
Q 14 on page 81
Aspirin
Mr Lund 08 May 2017
34
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Aromatic Chemistry


involves compounds containing a benzene ring (aka arenes look out for C6H5-)
empirical formula CH, Mr =78, molecular formula C6H6
Structure and Stability of Benzene

hydrogenation with 3 moles of H2 suggests the equivalence of 3 carbon-carbon double bonds
How Science
How science works
Page 85 Kekule’s dream
Works: A

Kekule did propose a cyclic structure – but it could not account for some major aspects of the
chemistry of Benzene:
1. no electrophilic addition reactions (e.g. with Br2(aq) in the dark) unlike alkenes
2. you don’t get two isomeric (1,2) disubstituted compounds
3. X-ray diffraction studies found intermediate – between double and single – and equal C-C
bond length i.e. a symmetrical structure
4. enthalpy of hydrogenation (208 kjmol-1) is less than 3 x cyclohexene (360 kjmol-1)
Why is it drawn like this?


How Science
Works: F

delocalised electrons increase relative stability (less
electron - electron repulsion)
Summary Questions
Exam Style Questions
Page 86
Page 93
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1–3
6
84 - 86
84 - 87
Q 1 – 2 on page 87
Bonding benzene
Mr Lund 08 May 2017
35
initially a resonance hybrid
structure was suggested
the true structure and shape of
benzene can be explained in
terms of the delocalisation of
the  electrons of 6 x 2p
orbital’s (whilst its not on the
syllabus, knowledge of orbital
hybridisation would be helpful
– have a look on Chemguide or
in an older A-level book)
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Physical properties


non-polar colourless liquid and does not mix with water (no hydrogen bonding)
boiling point similar to 6 carbon aliphatic hydrocarbons but melting point is higher a planar
structure allows better packing therefore more effective VdW.
Nomenclature



C6H5- phenyl group and simple monosubstituted aromatic compounds (arenes)
normally named as derivatives of benzene so ‘benzene’ often forms the root of the name
mono-substituted arenes are generally of formula C6H5X e.g. benzaldehyde C6H5CHO and
yield a peak of 77 on mass spectra due to the fragment C6H5+ (see later)
methylbenzene (toluene)
chlorobenzene
benzenecarboxylic acid (benzoic acid)


nitrobenzene
ethylbenzene
benzaldehyde
(chloromethyl)benzene
some names also use ‘phenyl’ or variations of it when the benzene is regarded as a side chain
phenol (instead of hydroxybenzene)
phenylamine (aniline) (instead of aminobenzene)
4-hydroxyphenyl ethanoate
phenylethene (instead of ethenylbenzene)
aromatic compounds with more than one substituent
1
2
lowest numbering possible
alphabetical order for substituent’s (ignore di, tri etc)
2-hydroxybenzoic acid
2,4,6-trinitrotoluene
2,4,6-trinitrophenol (picric acid)
benzene-1,4-dicarboxylic acid (Terephthalic acid)

take heart – naming aromatic compounds is complex but you will only have to deal with
simple examples as it is far more important that you understand the chemistry!
Summary Questions
Page 88
A2 Chemistry (Nelson Thornes) AQA
Chemguide
1–4
87 - 88
Naming aromatic
Mr Lund 08 May 2017
36
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Reactions of aromatic compounds
Electrophilic Substitution




high electron density in the ring attracts electrophiles
Br2(aq) is not decolourised (in the dark) – electrophile must be powerful (+ve not just +)
why benzene resists attack by poor electrophiles – stability of benzene compared to alkenes
electrophilic substitution rather than addition (c.f. alkenes) since this retains the relatively
stable benzene ring structure hence is energetically more favourable
Nitration of Benzene and Methylbenzene
ES
benzene


nitrobenzene (a yellow oil)
cH2SO4/cHNO3 refluxed at 50oC
(nitrating mixture)
mechanism of formation of NO2+ the nitronium (old name = nitryl) cation (sulphuric acid
acts as a homogeneous catalyst)
ES
methylbenzene

2(and 4)-nitromethylbenzene
cH2SO4/cHNO3 refluxed at 50oC

the methyl group is 2, 4, and 6 directing and activates the ring towards electrophilic
substitution (hence faster rate) since it donates electron density into the ring thus making it
relatively less stable and more susceptible to attack by electrophiles.
ES

further substitution requires more vigorous conditions (higher acid conc. and temperature) as
the nitro group deactivates the ring towards electrophilic substitution by withdrawing electron
density from the ring (i.e. increasing the extent of delocalisation thus further stabilising the
ring) – TNT is an explosive

nitrobenzene (a yellow oil)
phenylamine
cHCl/Sn or H2/Ni
Don’t use H2 for cHCL/Sn
C6H5NO2


+

6[H]
C6H5NH2
+
2H2O
the above reaction is much simplified – (whilst this equation is acceptable do you see why the
initial product would not be phenylamine? – a possible A* question perhaps?)
phenylamine is important for the production of AZO dyes since the end of 19th century (these
replaced the old technique of mordant dying – why not read about this – it is interesting)
How Science Works
Exam Style Questions
Page 90
Page 92
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
TNT
1, 3
89 - 90
88 - 89
Q 3, 4 on page 89
Electrophilic substitution, nitration
Mr Lund 08 May 2017
37
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Friedel-Crafts Acylation


role of the AlCl3 - halogen carrier – why is it a lewis acid
anhydrous conditions - AlCl3 readily hydrolysed by water before it does its job

electrophilic substitution mechanism proceeds via an acylium ion intermediate

you should be able to write a full balanced equation for the above
benzene
+
ethanoyl chloride
Summary Questions
Exam Style Questions
phenylethanone
Page 91
Pages 92 – 93
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide

warm with AlCl3 catalyst
anhydrous conditions
1-4
2-5
91
91, 184
Q 7, 8 on page 91
Q 1 – 8 on pages 94 - 95
Acylation
Mr Lund 08 May 2017
38
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Amines


nomenclature of primary, secondary and tertiary amines (by the way amine has ONE m in
it!!!)
note how this differs from alcohols and haloalkanes

amino can sometimes be used (see amino acids later on)
Physical properties







boiling points are elevated by the ability to hydrogen bond but are lower than similar sized
alcohols due to the relative electronegativity of O and N compared (ASK if you don’t
understand the significance of this)
lower members are gases
liquid amines smell like rotting (fishy) flesh – adding acid removes this smell – WHY?
smaller primary amines are water soluble (hydrogen bonding) producing alkaline solutions
solubility decreases with chain length (as with alcohols) due to increased mutual VdW
phenylamine is not very soluble in water as the VdW between the rings is significant
compared to hydrogen bonding between the amine group and water
produce alkaline solutions in water when they dissolve
Summary Questions
Page 95
1-4
Basic properties


lone pair on N

can be a nucleophile, base or ligand depending on the context
how good it is depends on the availability of that lone pair i.e. what the N is bonded to

Brønsted-Lowry bases – proton acceptors, Lewis base = lone pair donor

ethylamine is more basic than ammonia due to +I inductive effect of the alkyl group (its
also a better nucleophile than ammonia - see alkyl halides)
this explains the relatively more basic nature of small secondary amines but this does not hold
true for tertiary amines where reduced solubility is a factor




phenylamine is less basic than ammonia since the lone pair on the nitrogen is less available
due to delocalisation in the ring structure – diagram
(phenylmethyl)amine is as basic as primary amines as the N is NOT bonded directly to the
ring (note brackets in name – why ?)
acid amides are relatively poor bases/nucleophiles/ligands as the strongly electronegative
oxygen causes the lone pair on the nitrogen to be more withdrawn
Mr Lund 08 May 2017
39
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry

pKa value of conjugate acid increases (i.e. is poorer) with increased basicity of the conjugate
amine

amines react with acids to form salts in a similar manner to ammonia
ethylamine

+
hydrochloric acid

ethylammonium chloride + water
solvation of insoluble phenylamine achieved by the addition of HCl to form a soluble salt
phenylammonium chloride (reversed by adding NaOH)
phenylamine +
hydrochloric acid
Summary Questions
Exam Style Questions
phenylammonium chloride + water
Page 97
Pages 108 – 9
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide

1-3
1
94 - 97
96 – 97, 99 - 100
Q 1, 2 on pages 96 – 97
Q 5, 6 on pages 99 - 100
Q 4 on page 104
Amine name, amine base
Mr Lund 08 May 2017
40
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Preparation
From haloalkanes
NS
bromoethane

ethylamine
alcoholic solution of NH3 under pressure
(lots of by-products so not a good method)


mechanism – NH3 acts as nucleophile
excess ammonia is used to improve the yield of the primary amine

if xs bromoethane is used, since the ethylamine produced is also a nucleophile (stronger than
ammonia due to the +I inductive effect of the alkyl group) it can react with the xs
bromoethane to give diethylamine
further substitution can then occur to produce: triethylamine and tetraethylammonium
bromide (a quaternary ammonium salt cf ammonium ions)


note that acyl chlorides only yield primary amide (mechanism reminder) – lone pair
withdrawn by strong d+ on C due to the polarity of C=O caused by the electronegativity
of the O
From nitriles
Ethanenitrile
RCN +

ethylamine
4[H]

reduction by H2/Ni
RCH2NH2
(note: at AS you were also told that acid hydrolysis of nitrile yields a carboxylic acid)
LiAlH4 but not NaBH4 (not a powerful enough reducing agent) can also be used (don’t put H2 in
balanced equation in this case!!)
Aromatic Amines
benzene

nitrobenzene (a yellow oil)

*
nitrobenzene (a yellow oil)

phenylamine
cH2SO4/cHNO3 refluxed at 50oC
(nitrating mixture)
reduction cHCl/Sn (or H2/Ni)
phenylammonium chloride* is initially produced, and addition of NaOH yields
phenylamine (via deprotonation) which is then obtained by steam distillation
strictly speaking its actually phenylammonium hexachlorostannate(IV)
Mr Lund 08 May 2017
41
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Uses of Amines






quaternary ammonium salts are used as cationic
surfactants (long carbon chains help) in fabric
conditioners and hair conditioning products
the positive charges present will repel and add body
to the hair/fabric whilst the ‘tail’ section associates
with the fabric
if you are particularly interested in laundry see:
http://www.scienceinthebox.com/en_UK/glossary/s
urfactants_en.html
http://www.chemistryquestion.com/English/Questions/ChemistryInDailyLife/27c_nonionic_s
urfactant.html
aryl amines are used in synthetic dyes
aryl amines are used to make certain drugs e.g.
paracetamol
amines are used to make polymers e.g. polyurethane’s
(cavity wall insulation) and polyamides (nylons)
How Science Works
Summary Questions
Exam Style Questions
Pages 100 - 1
Page 101
Pages 108
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
Sulfa drugs
1, 2
2-4
99 - 101
97 – 98, 101 - 105
Q 7 on page 103
Q 13, 5-8 on pages 104 - 105
Amine preparation, amine uses
Mr Lund 08 May 2017
42
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Amino acids

there are 20 important naturally occurring  amino acids (amine is on the C next to the acid
group –CO2H)
R can vary
R=H
glycine
R = CH3
alanine
R = CO(OH)CH2
aspartic acid
etc

you should recognise that there is a CHIRAL centre hence amino acids exhibit optical
isomerism (name the exception)

some amino acids have been identified in space (those of you who are interested in science
might read the next link)
http://www.newscientist.com/article/dn7895-space-radiation-may-select-amino-acids-for-life.html

the amino acid proline is a 2o amine all the others
are primary amines
CH
2
NH
CH
2
C
COOH
CH2
H
Extra Info for potential medical/biochemistry/pharmacy students:

amino acids in proteins are all L-isomers

this does not necessarily mean plane polarised light is rotated the same way

if L(+) then D(-) for a given amino acid and L(-)/D(+) don’t exist in our natural system

CORN law – I have provided some info about this on my site – it is not on the exam – but it
may be helpful at your interview for medical school given the biochemical significance
Summary Questions
Page 103
Page 107
1, 2
1
Mr Lund 08 May 2017
43
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Zwitterions







the amino group and the carboxyl group of each
amino acid are both ionisable
the acidic carboxyl group (with a pKa of about 3 –
it’s a weak acid) is deprotonated
the basic amino group (with a pKa of around 9 –
it’s a weak base) is protonated
amino acids thus exist as zwitterions with both a
positive and negative charge present
Zwitterions are amphoteric i.e. they exhibit both acidic and basic properties in solution
because of the two functional groups
they thus form salts with both acids and bases (note all similar groups ionised as
appropriate)
amino acids can thus act as buffers (hence regulate pH)
Adding H+
pH decreasing



Adding OHpH increasing
charge on the zwitterion ion depends on pH,
-ve at high pH and +ve at low pH
at a given pH, the isoelectric point of the
amino acid there will be no overall charge
as the + and – cancel
this pH varies from amino acid to amino
acid and provides a means of separating
them
using
a
technique
called
electrophoresis.
Extra Info for potential medical/biochemistry/pharmacy students:

you might be interested to find out how electrophoresis works by looking at this link
http://www.saburchill.com/IBbiology/chapters01/003.html




Amino acids exist as zwitterions in the solid state and thus have strongly ionic character
this explains their high solubility in polar solvents e.g. water
it also explains the high MPt (white crystalline solid when pure)
MPt too high to be accounted for by hydrogen bonding alone - supporting existence of
Zwitterions and ionic nature of amino acids in the solid state
Exam Style Questions
Pages 108
5, 8, 11
Mr Lund 08 May 2017
44
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Condensation Polymerisation and hydrolysis

peptide (amide) links are formed between 2 amino acids by condensation reactions (where
a small molecule such as water is eliminated) to form a dipeptide

Polypeptide (~50 amino acids) are formed
by condensation polymerisation and
catalysed by enzymes

sequence of amino acids in a protein is
called its primary protein structure

hydrogen bonding between C=O and N-H
controls shape - secondary protein
structure - -helix (coiled) and -pleated
sheet (folded)

tertiary structure involves further
bending and twisting (a good analogy is a
knotted, multicoloured telephone coil)

the stretching of wool is dependent on
hydrogen bonding the length of which can
be reversibly increased up to a limit (very hot water can break these ruining the fluffyness)





hydrolysis of proteins is achieved by refluxing with acid, base or enzyme catalyst (cf
hydrolysis of an amide) in effect reversing the process shown in the diagram above
the liberated amino acids can then be separated by paper chromatography (developed by
treating with ninhydrin which colours amino acids violet
some enzymes are selective and only partially hydrolyse certain proteins enabling amino acid
sequences to be identified
enzymes are themselves proteins and are very specific in what they catalyse (substrate) due
to their shape
their shape is dependent on hydrogen bonding hence their activity is sensitive to elevated
temperatures where they are denatured
How Science Works
Summary Questions
Exam Style Questions
Pages 107
Page 107
Pages 109
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
Robots in the lab
1, 2
6, 8 - 10
102 - 107
111 – 114, 57 – 62
Q 8 - 12 on pages 111 – 113
Q 2 – 4, 5 - 6 on pages 115 - 116
Zwitterions, peptide
Mr Lund 08 May 2017
45
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Polymerisation
Addition Polymerisation




addition across the carbon-carbon double bond of a single unsaturated monomer
occurs by a free radical mechanism (details not required) started by an initiator (e.g. a
peroxide) which is incorporated at the start/end of the polymer chain
you should be able to work out monomer from polymer and visa-versa (tip use the >C=C<
form of the monomer i.e. rewrite it)
if asked for a single repeating unit don’t show brackets otherwise use –[C-C]-n format
ethene

poly(ethene)
propene

poly(propene)
heat, pressure, catalyst
in all cases
chloroethane

poly(chloroethene) (PVC)
phenylethene

poly(phenylethene) (polystyrene)
tetrafluoroethene 
poly(tetrafluoroethene) (PTFE or Teflon)
methyl 2-methylpropenoate

Summary Questions
Exam Style Questions
Page 114
Pages 122
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
‘perspex’
1-4
4, 5
112 - 114
106 -108
Q 1-3 on page 106
Q 7, 8 on page 116
Addition polymerisation
Mr Lund 08 May 2017
46
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Condensation Polymerisation
Polyesters






two different types of monomers, one a diol,
the other a dicarboxylic acid
this polymer is connected by ester linkages
you should be able to draw repeating units and
determine the monomers used to make a given
polymer – colour coding diagrams helps here
condition usually involve heat + catalyst
look for –CO2- in structural formula
balanced equation (don’t forget 2n H2O !!!!)
ethane-1,2-diol
(ethylene glycol)
+
benzene-1,4-dicarboxylic acid
(terephthalic acid)

PET

the chain repeat unit of PET is:

commonly just called ‘polyester’ PET, poly(ethylene terephthalate) was initially used as a
fibre (e.g. Terylene and Dacron)
it is now used extensively in plastic containers e.g. for fizzy drink bottles – it does not smash
on impact





you can read about the invention of polyester at:
http://inventors.about.com/library/inventors/blpolyester.htm
those of you doing textiles might also like to visit the ‘polyester story’ at
http://schwartz.eng.auburn.edu/polyester/polyester.home.html

poly(2-hydroxypropanoic acid) – used in surgery as its broken down by
enzymes/body fluids over a number of days (why is this good?)
how would you synthesise it from ethene?

How Science
Works: I
You will find it useful to look at the reaction
pathways template that I have put on line


you can find out more about plastics from renewable raw materials and biologically
degradable plastics at this site (its also a good example of (or idea for) an EPQ project!):
http://www.rsc.org/education/teachers/learnnet/green/docs/plastics.doc
Mr Lund 08 May 2017
47
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Polyamides





two different types of monomers, one a
diamine, the other a dicarboxylic acid
this polymer is connected by amide linkages
an diacyl chloride can be used instead of the
acid but what would be the pros and cons of this
nylon-6,6 is so called as both monomers have 6
carbons
nylon is a polyamide (just like proteins)
1,6-diaminohexane +

hexanedioic acid
nylon-6,6
How Science
Works: I

you might be interested in the history of nylon – it’s an ideal stocking filler for Xmas see:
http://www.cha4mot.com/p_jc_dph.html and http://inventors.about.com/od/nstartinventions/a/nylon.htm



polyamides have extensive
hydrogen bonding between
parallel strands (cf protein
structure)
when nylon is spun into
fibres, amide groups on
adjacent
chains
form
hydrogen bonds making
nylon yarn strong.
kevlar is an example of an aromatic polyamide and is made from the monomers benzene-1,4dicarboxylic acid and benzene-1,4-diamine
Medieval chainmail similarity?




the molecule is flat because of the aromatic groups
the uses of Kevlar are related to its strength (its several times stronger than
steel)
this is related to the packing together of sheets of molecules held together
by hydrogen bonds formed between N – H groups and C = O groups on
adjacent molecules
here is a scuba diving site that has an excellent overview of natural and synthetic polymers
– have a look it is very good: http://njscuba.net/artifacts/matl_polymers.html
Mr Lund 08 May 2017
48
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Environmental Issues

How Science
Works: J

polyalkenes are saturated, have no polar bonds and have strong C-C and C-H bonds hence
they are relatively unreactive e.g. with acids, alkali or oxidants
this in turn makes them difficult to dispose of

non biodegradability means they last a long time but you can use landfill – not ideal though

combustion yields toxic and greenhouse gases e.g CO and carbon particulates, NO2 and
HCN from polyurethane in older upholstery or HCl and dioxins released through combustion
of halogenated plastics such as PVC and of course there will always be CO2 produced
recycling is expensive as the plastics must be identified and separated from other waste
energy production is an option but as this involves combustion there will be CO2 produced
using as chemical feedstock after cracking






use biodegradable/photodegradable polymers instead is an option
polyesters and polyamides both have polar bonds hence they are potentially biodegradable
they are broken down by (catalytic) hydrolysis in acid/alkaline solution or with enzymes
although this can take a long time

recycling these materials e.g. terylene will save on natural resources and energy in their initial
production (as well as reducing the need for landfill), however, as they are biodegradable
their structural integrity will diminish after repetitive usage
as with polyalkanes the cost of collection, separation and transportation (require energy and
are labour intensive) must be taken into account

How Science Works
Summary Questions
Exam Style Questions
Pages 120 - 1
Page 119
Page 121
Pages 121 - 123
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
Hermann Staudinger + Q 1,2
1, 2
1-4
1, 2, 3, 6, 7
115 - 120
108 -110
Q 4 - 8 on page 109 - 110
Q 1, 5 on pages 115 - 116
Condensation polymerisation
Mr Lund 08 May 2017
49
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Aliphatic synthesis
You should be able to write full chemical equations, and identify the type for all the reactions listed:
Oxidation, Reduction, Addition, Elimination, Addition-Elimination (Condensation),
Substitution, Hydration, Dehydration, Hydrogenation, Dehydrogenation, Hydrolysis
Mechanisms: Nucleophilic Addition NA, Nucleophilic Substitution NS, Electrophilic Addition EA,
Nucleophilic Addition-Elimination NAE, Free Radical Substitution FRS or Elimination E.
Alkanes
haloalkane, alkene
methane
+

Cl2
chloromethane
FRS
alkanes can be used to produce alkenes by thermal cracking
Alkenes
haloalkane, alkane, alcohol, alkoxyalkane
ethene
+
HBr

bromoethane
EA
ethene
+
Br2

1,2-dibromoethane
EA
ethene
+
H2

ethane
o
Ni catalyst, ~200 C (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
ethanol
EA
amine, alcohol, nitrile, alkene
bromoethane
+
NH3 
ethylamine +
HBr
heat with xs ammonia under pressure to minimise further substitution
NS
bromoethane
+
OH-(aq) 
ethanol
reflux with dilute NaOH dissolved in water
NS
+
Br-
bromoethane
+
CN- 
propanenitrile
+
boil under reflux with alcoholic NaCN or KCN (NOT HCN)
propanenitrile
+
4[H] 
propylamine
reduction by H2/Ni
Br-
bromoethane
+
OH- 
reflux with NaOH dissolved in ethanol
+
ethene
Mr Lund 08 May 2017
50
+
H2 O
NS
Br-
E
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Alcohols
carbonyl (aldehyde and ketone), carboxylic acid, ester, haloalkane, 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) under heat
ethanoic acid
+
H2 O
propan-2-ol
+
[O]

reflux with K2Cr2O7(aq)/H2SO4(aq) under heat
propanone
+
H2 O
ethanol
+
ethanoic acid
ethyl ethanoate
+
reflux with glacial ethanoic acid and concentrated sulphuric acid catalyst
H2 O
ethanol

ethene
heat with excess cH2SO4 at 170oC
+
Aldehydes and Ketones
alcohol, carboxylic acid, hydroxynitrile
H2 O
ethanal
+
2[H]

ethanol
sodium tetrahydridoborate(III) (NaBH4) in aqueous ethanol
propanone
+
2[H]

NA
propan-2-ol
NA
ethanal
+
HCN

2-hydroxypropanenitrile
in alkaline solution to increase [CN ] as HCN is a weak acid
2-hydroxypropanenitrile + 2H2O + H+
Carboxylic Acids

NA
2-hydroxypropanoic acid
(lactic acid)
+
NH4+(aq)
water
WATER
a salt, ester
ethanoic acid
ACID
+
+
sodium hydroxide
BASE

sodium ethanoate
SALT
+
+
ethanoic acid
+
sodium carbonate

sodium ethanoate
ACID
+
METAL CARBONATE
SALT
+ water + carbon
dioxide
+ WATER + CO2
ethanol
ACID
+
+
ethanoic acid
ALCOHOL
ethyl ethanoate
ESTER
+
+
Mr Lund 08 May 2017
51
water
WATER
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Esters
carboxylic acids (or its salt – look for OH-(aq)) and alcohol
ethyl ethanoate
+

OH-(aq)
ethanoate ion
+
ethanol
ethanoate ion
+
H+(aq)

ethanoic acid
(protonation of the anion of a weak acid by a stronger acid – think about the equilibria here)
Acyl Chlorides
carboxylic acid, ester, amide
ethonoyl chloride
+ H2 O

ethanoic acid
+
hydrogen chloride NAE
ethanoyl chloride
+ ethanol

ethyl ethanoate
+
HCl
ethanoyl chloride
+ 2-hydroxybenzoic acid
(salicylic acid)
ethanoyl chloride
+ NH3


aspirin
ethanamide
+
NAE
HCl
NAE
+
HCl
NAE
ethanoyl chloride + ethylamine 
(primary amine)
N-ethylethanamide +
(secondary amide)
HCl
NAE
ethanoyl chloride + phenylamine 
N-phenylethanamide +
HCl
NAE
Acid Anhydrides
carboxylic acid, ester, amide
ethanoic anhydride +
H2 O

2 ethanoic acid
ethonoic anhydride +
ethanol

ethyl ethanoate + ethanoic acid
ethonoic anhydride + 2-hydroxybenzoic acid
(salicylic acid)
warm with conc. sulphuric acid catalyst

aspirin
ethonoic anhydride +
ammonia

ethanamide
ethonoic anhydride +
4-aminophenol
(primary amine)

paracetamol +
(secondary amide)
Mr Lund 08 May 2017
52
+
ethanoic acid
+
ethanoic acid
ethanoic acid
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Aromatic Synthesis
Electrophilic Addition EA, Electrophilic Sustitution ES, Nucleophilic Addition-Elimination NAE
Substitution on the ring
benzene
+
HNO3

nitrobenzene
cH2SO4/cHNO3 refluxed at 50oC (nitrating mixture)
+
H2 O
ES
methylbenzene
+
HNO3 
cH2SO4/cHNO3 refluxed at 50oC
2(and 4)-nitromethylbenzene + H2O
ES
benzene
+
ethanoyl chloride
AlCl3 catalyst anhydrous conditions

phenylethanone
+
HCl
ES

phenylamine
+
2H2O
phenylammonium chloride
+
H2 O
N-phenylethanamide +
HCl
NAE
HCl
NAE
Modification of substituents
nitrobenzene
cHCl/Sn or H2/Ni
+
6[H]
phenylamine +
HCl

phenylamine +
ethanoyl chloride

Polymerisation
phenylethene

free radical peroxide initiator, high pressure
phenylamine +
ethanoyl chloride
ethane-1,2-diol
(ethylene glycol)
+
1,6-diaminohexane +
poly(phenylethene)

N-phenylethanamide +
benzene-1,4-dicarboxylic acid
(terephthalic acid)

hexanedioic acid
Mr Lund 08 May 2017
53

nylon-6,6
Terylene  (a polyester)
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Amine
Nitrile
(hydroxynitrile)
Haloalkane
Alkane
Alkene
Carbonyl
Alcohol
Amide
(polyamide)
Carboxylic
Acid
Acyl chloride
(Acid anhydride)
Ester
(polyester)
Reagents/conditions
Mechanism
required
also2017
Mr Lund
08 May
54
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
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 08 May 2017
55
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
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)
AgCl(s) + 2NH3(aq)  Ag(NH3)2+(aq) + 2Cl-(aq)
AgCl(s)
AgBr(s) + 2NH3(aq)  Ag(NH3)2+(aq) + 2Br-(aq)
AgI(s)
alkyl bromide
alkyl iodide
Relative rates of hydrolysis of alkyl halides are: iodo > bromo > chloro
Mr Lund 08 May 2017
56
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Mass Spectrometry

this provides us with:
Mr, Relative abundance and fragmentation
patterns

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 (parent ion), which is a radical
cation, is formed (note BOTH + must be shown on the molecular ion)

M(g)

e-
This molecular ion can then undergo fragmentation which produces a radical (not detected)
and a positive ion (detected) which are shown with a + only

M(g)+


M(g)+ +
X(g)+
+
Y(g) 
can be either way around – so both give m/z peaks – but not necessarily in equal proportions
higher peaks correspond to more stable ions and the highest is called the base peak and this
determines the 100% benchmark on the spectra
m/z
Interpretation
value
15
CH3+
Notes
Not much use without other data
+
Look for a 2H quartet and 3H triplet on 1H nmr
29
CH3CH2
31
CH3O+
Look for a 3H singlet on 1H nmr
43
HIGH peak = CH3CO+
CH3CH2CH2+
57
CH3CH2CO+
43 and 29 could be butanone
43 and 15 could be propanone
43 with no 15 ethanal or acyl chloride possible
Check IR for a carbonyl group
Check IR for a carbonyl group
77
C6H5+
105
C6H5CO+
77 + 29 consider ethylbenzene
77 + 43 propylbenzene or phenylethanone
Benzaldehyde or Benzoic acid
Mr Lund 08 May 2017
57
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry



fragmentation occurs most easily at the weakest bond
Carbonyl compounds have a tendency to break adjacent to the C=O yielding the relatively
stable, hence high abundance, acylium ion RCO+ - look for m/z 43
the stability of a fragment determines its abundance on the spectra e.g. primary, secondary and
tertiary carbocations – the positive inductive effect (+I) explains their relative stability
http://www.chem.arizona.edu/massspec/example_html/examples.html

Have a look at
these examples
isomers can be determined from fragmentation patterns:
o this is particularly useful in resolving aldehydes and ketones which will both give
similar >C=O wave numbers on IR
o isomeric acids and esters should be obvious from IR (look for OH peak) data but
again will give different fragmentation patterns
o isomeric esters will give similar IR and nmr and may need fragmentation patterns to
resolve – these are harder to do


M+ + 1 peak (satellite peak) due to carbon-13
height = 1% of M+ per carbon in the compound e.g. 5% suggests 5 carbons
7% (or higher) is likely to involve a substituted benzene derivative
note: the number carbons can be used, in conjunction with Mr to make an estimate of the
possible number of oxygen’s for example

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 85, 87 and 89 in a 9:6:1 ratio


high resolution mass spectra can distinguish between compounds with similar M r e.g.
~123 for C6H5NO2 and C7H7O2

when dealing with mass spec in conjunction with either/both IR and nmr it is sensible to use
mass spec initially only to establish the Mr from which a reasonable estimate of the number of
carbons can be made
this is particularly useful with IR data to indicate whether there are any oxygen atoms in the
structure thus giving a maximum for the number of carbon atoms

Summary Questions
Exam Style Questions
Page 136
Page 159
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1, 2
4
132 - 136
117 – 123
Q2 on page 138
Q1 – 6 on pages 118 - 123
Mass spectrometry
Mr Lund 08 May 2017
58
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
IR Spectroscopy

used to identify certain functional groups in particular –OH and C=O in carbonyl
compounds, acids, alcohols and esters

beam splitting occurs in the IR spectrometer through a reference and the organic sample and
comparison determines the relative % transmission creating troughs in the spectra
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)



the spectra produced uses a scale of % transmission vs. wave number (cm-1)

above 1500 cm-1 two important absorption values are:
1680 – 1750 cm-1 C=O
strong and sharp with slight variations in position depending on
the type of compound (see table 1 page 139)
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 138





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 oxygen.
significant spectral changes involving the above upon redox or esterification can be revealing
regarding the functional groups present
the fingerprint region 400 – 1500 cm-1 is unique to each organic molecule
determination of the sample uses computerised comparisons with a database
impurities will yield additional peaks in this region
Summary Questions
Page 138
Page 141
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1, 2
1-3
137 - 141
124 – 127
Q7 – 10 on pages 124 – 127
Q 1 page 137
Q 5 on page 149
Infra red spectroscopy
Mr Lund 08 May 2017
59
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Nuclear Magnetic Resonance Spectroscopy



nmr can be used on small samples to provide details of structure
cf wet chemical tests - limited primarily to identifying functional groups
nmr can be applied where there is an odd nucleon number e.g. 1H and 13C

these nuclei has a small magnetic field
associated with spin
an external magnetic field results in either
alignment with ( - lower energy) or against ( higher energy)
resonance between these two states occurs when
radio waves of appropriate energy are applied



this results in energy being absorbed which is
detected by the instrumentation
Carbon-13 (13C) nmr



organic chemicals will contain 1% carbon-13
their resonance field strength will vary
depending on the number of surrounding
electrons
the field strengths of other carbon
environments only differ slightly so the  scale
is measured in ppm

electrons shield the nucleus thereby reducing the effective magnetic field and requiring an
applied radio frequency of a lower energy to cause resonance

when electrons are withdrawn from a nucleus, the nucleus is deshielded and feels a stronger
magnetic field requiring more energy (higher frequency) to cause resonance as the energy gap
is greater
NOTE: If the radio frequency is kept constant and the magnetic field varied instead to achieve
resonance then a relatively weaker magnetic field will be required for more deshielded carbon
atoms

the nmr spectra is based on the  scale of chemical shifts relative to a standard reference
peak – tetramethylsilane

TMS has 4 carbon atoms in the same environment which produces a single strong peak

Si is relatively less electronegative than C so the electrons in the C-Si bonds are closer to the
carbons than in other organic compounds resulting in a peak on the spectrum at the extreme
right-hand side
Mr Lund 08 May 2017
60
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry

 for TMS = 0 by definition

thus, nmr provides information about a
carbon’s electronic environment.
carbons attached to electron withdrawing
atoms/groups such as oxygen tend to
resonate at higher frequencies yielding
peaks downfield from TMS i.e. a higher 
(chemical shift)

carbonyl carbons
aromatic carbons
alcohols, esters, ethers
unsaturated aliphatic
saturated aliphatic


Note: 3 carbons
but 2 peaks
41
170
 160-220
 110-160
 50 - 90
 90-150
 5-40
C=O
C-O
C=C
C-C
170
ppm
ppm
ppm
ppm
ppm
the major differences that you will notice in 13C nmr in comparison to 1H-nmr spectra
include:
1.
no integration of carbon spectra
2.
wide range (0-220 ppm) of resonances for common carbon atoms (typical range for
protons 0-12 ppm)
CDCl3. (i.e. hydrogen has been replaced by its isotope, deuterium) is commonly used as a
solvent (also see proton-NMR) and the line for this carbon is removed from the final
spectrum
Summary Questions
Page 145
A2 Chemistry (Nelson Thornes) AQA
Chemguide
1, 2
143 - 145
Carbon 13 nmr
Mr Lund 08 May 2017
61
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Low Resolution 1H Spectra




1
H is far more abundant in organic compounds than 13C hence the spectra are much easier to
obtain
their resonance field strength will vary depending on the number of surrounding electrons
electrons shield the nucleus thereby reducing the effective magnetic field and requiring an
applied radio frequency of a lower energy to cause resonance
when electrons are withdrawn from a nucleus, the nucleus is deshielded and feels a stronger
magnetic field requiring more energy (higher frequency) to cause resonance
NOTE: If the radio frequency is kept constant and the magnetic field varied instead to achieve
resonance then a relatively weaker magnetic field will be required for more deshielded carbon
atoms

the nmr spectra is based on the  scale of chemical shifts relative to a standard reference
peak – tetramethylsilane

TMS is used as
o it provides a single strong signal from 12 equivalent H atoms
o it is chemically inert
o the peak will be up field of most other peaks given the relatively low electronegativity of
silicon compared to carbon

 for TMS = 0 by definition

hydrogen’s on the same atom in a molecule will experience an identical electromagnetic
environment and will thus yield a single peak at a given value on a low resolution spectra

greater electron density results in relatively more shielding hence the requirement for a higher
external field to cause resonance
electronegative atoms e.g. oxygen (or unsaturated regions e.g. a benzene ring) cause
deshielding hence a lower external energy requirement




number of peaks = number of non equivalent groups of H atoms
chemical shift identifies adjacent environment e.g. nearby oxygen
integrated spectra indicate the relative proportion of equivalent H atoms i.e. those in an
identical environment
Mr Lund 08 May 2017
62
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Below are common situations:





1 hydrogen strongly suggests an alcohol, acid or aldehyde
9 equivalent H’s suggests –C(CH3)3
6 equivalent H’s suggests –C(OH)(CH3)2 – NOT –CH(CH3)2 as this would be outside the
limitations for the AQA syllabus (see later)
3 equivalent H’s suggests –CH3 e.g. –COC H3 or -CH(OH)CH3 etc
2 equivalent H’s suggests –CH2-

a proton free solvent is used e.g. CCl4, D2O CDCl3 (2H gives no peak)
Summary Questions
How Science Works
Page 147
Pages 151-2
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
1
The birth of nmr
146 - 148
128 – 132
Q11 – 13 on pages 130 – 132
nmr spectroscopy
Mr Lund 08 May 2017
63
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Interpretation of High Resolution Spectra




non equivalent hydrogen’s on adjacent carbons will themselves behave as little magnets and
their effect can be seen on high resolution nmr
each creates a small magnetic field aligned with or against the external field – spin coupling
this creates small differences in the local field which modifies the radio frequency energy
required for resonance
hydrogen’s on the same atoms are identical and will not modify one another
NOTE:



hydrogen’s bonded to an oxygen are not split and do not cause others to split
the more adjacent hydrogen’s there are, the more possible permutations exist
peaks are thus split into sub sets by neighbouring non equivalent H atoms
n+1 rule – if there are n H atoms in total on adjacent C atoms then the peak is split into n+1
signals of relative intensities derived from Pascal’s triangle:
these two together
are strongly
indicative of an
ethyl group
NOTE:
A2 Chemistry (Nelson Thornes) AQA
A2 Chemistry (Heinemann) AQA
Chemguide
Your syllabus is limited to samples
yielding singlet’s, doublets, triplets and
quartets
Summary Questions
Page 152
1, 2
Exam Style Questions
Pages 155-7
Pages 158 – 161
1–6
2, 3, 5, 6
148 - 152
128 – 136
Q11 – 15 on pages 130 – 136
Q 1 – 4 on pages 137 -138
Q 6,11 on pages 151 - 153
Proton nmr
Mr Lund 08 May 2017
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A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
INTEGRATED SPECTROSCOPY



you may be required to use data from more than one spectra for identification
on my CHEMISTRY - READ site I have placed a suggested flow chart
it would be useful for you to think about and develop your own strategy


IR will enable you to determine if there is/are one or two oxygen’s present
coupled with Mass Spectrometry this will allow you to establish a maximum number of
carbon atoms in the structure (which may also be derived from the M+ + 1 peak (satellite
peak) due to carbon-13) and may also indicate the number of hydrogen’s hence give clues to
the degree of saturation in the compound

fragmentation patterns in Mass Spectrometry are best left until needed or as a confirmatory
check

at this point you might want to write down the pieces of the jigsaw that are present and then
use nmr to work out what is connected to what

be alert to the possibility of carbons/hydrogen’s in identical environments e.g. in propanone
or in branched alkyl components
Sample spectra: http://home.clara.net/rod.beavon/spectra.htm
Here are some useful tables:
Proton nmr
MS data
m/z
value
Interpretation
Notes
15
CH3+
Not much use without other data
29
CH3CH2+
31
CH3O+
Look for a 2H quartet and 3H triplet on 1H
nmr
Look for a 3H singlet on 1H nmr
43
HIGH peak = CH3CO+
CH3CH2CH2+
57
CH3CH2CO+
77
C6H5+
105
C6H5CO+
Infra-red absorption data
Bond Wavenumber
43 and 29 could be butanone
43 and 15 could be propanone
43 with no 15 ethanal or acyl chloride possible
Check IR for a carbonyl group
Check IR for a carbonyl group
77 + 29 consider ethylbenzene
77 + 43 propylbenzene or phenylethanone
Benzaldehyde or Benzoic acid
Mr Lund 08 May 2017
65
C—H
C—C
C=C
C=O
C—O
O—H
O—H
2850–3300
750–1100
1620–1680
1680–1750
1000–1300
(alcohols) 3230–
3550
(acids) 2500–3000
A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
CHROMATOGRAHY






essentially the physical separation of the constituents of a mixture
a mobile phase (solution or gases called the eluent) flows through a stationary phase (a solid
or liquid supported on a solid) and in so doing the components of a mixture are separated as
they travel at different rates.
the more soluble a solid, the further (faster) it moves in a given time
the stationary phase will also determine the rate of progress of the components as a function
of its affinity for each component
hence the trick is to use a suitable combination of solvent and stationary phase to achieve the
separation of similar components
these can then be analysed separately using other techniques where required
Column Chromatography



the solid phase is a powder e.g. silica or alumina packed into a tube
a solvent (the eluent) is added at the top
the components of a mixture are separated as they travel at different rates down the column
so can be collected separately for further analysis

the separation depends on the balance between
the solubility in the mobile phase and the
retention in the stationary phase

as in TLC (felt tip pens and paper is the
simplest form), there is an equilibrium
established between the solute adsorbed on the
silica gel or alumina and the eluting solvent
flowing down through the column.
this method is effective at separating mixtures
of amino acids
it can effectively separate fairly large amounts
of substance



different eluents can be used to extract different components

here is a practical guide to how it is done:
http://www.wfu.edu/academics/chemistry/courses/CC/index.htm

Here is a video clip of some people in white coats doing Chemistry:
Video clip
Mr Lund 08 May 2017
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A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Gas-liquid Chromatography GC



this is particularly
useful in the
separation
of
volatile liquids
a
sample
is
injected into the
device in which
there is a long
(typically 100 m)
thin (0.5 mm)
tube (coiled for
compactness)
containing the stationary phase (an inert powder coated with oil)
a carrier gas (unreactive e.g. N2 or He) is used as the eluent to provide the mobile phase






the carrier gas transports the sample
along the tube separating the
mixture in the process
separation depends on rate of
movement with the gas and
retention by the coil
the time taken for each component
to complete its journey through he
coil is called it’s retention time
data will also be collected regarding the relative amount of each component
the output samples can be analysed by MS, IR or nmr under computer control
this method is effective even with small sample sizes/trace amounts due to its high sensitivity
(e.g. food additives, drug usage detection)
Here is a ‘How Science Works’ link that
considers the controversy surrounding the
interpretation of GC/MS data regarding
whether life has/has not been detected on
Mars – there are many like this
http://www.msss.com/http/ps/life/life.html
GC/MS equipment on Mars lander
Summary Questions
Page 154
A2 Chemistry (Nelson Thornes) AQA
Chemguide
1-3
153 - 154
Chromatography
Mr Lund 08 May 2017
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A2 Unit 4 Kinetics, Equilibria and Organic Chemistry
Mr Lund 08 May 2017
68
AS Chemistry Unit 2
69
AS Chemistry Unit 2
70