Chemistry for the gifted and talented
Olympiad past paper questions
Student worksheet: CDROM index 28SW
Discussion of answers: CDROM index 28DA
Topics
Topics met in post-16 chemistry.
Level
Post-16 students.
Prior knowledge
The content of post-16 courses.
Rationale
The questions on the CDROM (index 28SW) are a selection of the questions used in Round
1 of the selection process for the International Chemistry Olympiad. They are accompanied
by a Discussion of answers (index 28DA).
Further examples are to be found at www.rsc.org/olympiad (accessed July 2007).
Use
For revision or for preparation for the Olympiad selection procedure.
55
Student worksheet
Chemistry for the gifted and talented
Olympiad past paper questions
1. Heating a cup of coffee
Nescafé has recently launched a self-heating can of coffee. To heat up the coffee, a button is
pressed which mixes the heating ingredients – a very dilute solution of sodium / potassium
hydroxide and calcium oxide. The can then warms 210 cm3 of coffee by approximately
40 °C.
a) Write an equation for the reaction between calcium oxide and water.
b) The rate of the heating may be controlled by altering the pH of the solution.
How would you expect the rate of reaction to vary in acidic, basic and neutral conditions?
[Your answer should simply be the words ‘acidic’, ‘basic’ and ‘neutral’ in the order in
which they affect the rate of reaction, fastest first.]
c) Given the standard enthalpies of formation of calcium hydroxide, calcium oxide and water
are –1003, –635 and –286 kJ mol–1 respectively, calculate the standard molar enthalpy
change for the reaction in a).
d) Assuming that the heat capacity for the coffee is the same as that of water,
4.18 J K–1 g–1, calculate the energy needed to warm 210 cm3 of coffee by 40 °C.
e) Hence calculate the minimum mass of calcium oxide needed in the can to function as
specified.
continued on page 2
Student worksheet 28SW
Olympiad past paper questions
Page 1 of 11
Chemistry for the gifted and talented
2. Superconductors
Yttrium oxide, barium carbonate and copper oxide react in a solid state reaction at high
temperature (900 °C) to form the superconductor A which contains 13.4%Y, 41.2%Ba and
28.6%Cu.
a) Assuming that the only element unaccounted for in A is oxygen, determine the empirical
formula of A.
b) Given that the oxidation state of yttrium is +3 and barium is +2, calculate the average
oxidation state of copper.
c) Reducing A on a thermogravimetric analyser at 200 °C in hydrogen reduces all the copper
3+ in the material to copper 2+ and produces compound B. All the other elements
remain in the same oxidation state. Given that the starting mass is 84.2 mg, what is the
mass of the remaining compound B after reduction? (Hint: consider the change in oxygen
content for the two different compounds.)
You will need to refer to a Periodic Table for this question. (Ar yttrium = 88.9.)
continued on page 3
Student worksheet 28SW
Olympiad past paper questions
Page 2 of 11
Chemistry for the gifted and talented
3. Combining proportions of elements
Illustration from A compleat body of Chymistry by Le Febure (or Le Fevre) 1670 English edition, courtesy Peter Wothers.
Understanding the proportions in which the elements combine was a crucial step in developing
the atomic theory of matter. The picture above shows an experiment performed in the 1660s
in which antimony was heated, using the sun’s rays, to form an oxide.
In the experiment, ‘ye Artist’ reported that ‘12 grains of antimony increased to 15 grains of
calx’, (a grain is an old measure of mass). Given the crudeness of the experiment, this value is
remarkably close to the theoretical yield of 14.4 ‘grains’.
a) Calculate the formula of the oxide formed.
In another experiment published in 1673, Robert Boyle measured the increase in mass when
zinc metal is heated in air. He describes the experiment thus:
We took a Drachm of filings of Zink and kept it in a Cupelling-fire about three Hours. Then it
look’d as if the filings had been calcin’d. This being weigh’d in the same scales gain’d full six
grains.
b) Given that there are 60 grains in a Drachm, calculate the mass of product (in grains) that
would have been produced, assuming a yield of 100%.
Assuming that Boyle’s measurements are accurate, only a fraction (α) of the zinc must have
been converted to the oxide. (α is a fraction between 0 and 1; 0 meaning none of the zinc
reacted and 1 meaning all the zinc reacted).
c) Calculate the value of α and hence the masses of zinc oxide and unreacted zinc metal at
the end of the experiment.
continued on page 4
Student worksheet 28SW
Olympiad past paper questions
Page 3 of 11
Chemistry for the gifted and talented
4. Green chemistry
Increasing concerns over the use and generation of hazardous substances in chemical processes
has encouraged some chemists to look for more environmentally friendly ways to make
chemical products. To help evaluate a process environmentally, chemists often use the term
‘percentage atom economy’, where
% Atom economy
=
Mr of desired product x 100
Mr of all products
An environmentally friendly chemical process would normally be expected to have a high %
atom economy, indicating that a high proportion of the starting materials end up as part of the
final product, hence reducing the amount of waste. Efforts are constantly being made to
increase the % atom economy of chemical processes. As an example, the manufacture of
ethene oxide (C2H4O) for many years was via the classical chlorohydrin route:
C2H4
+
Cl2
ClCH2CH2OH +
+
H2O
Ca(OH)2
+
ClCH2CH2OH
HCl
C2H4O
+ HCl
+
CaCl2
+ 2H2O
a) i) Write a balanced equation for the overall reaction.
ii) Calculate the % atom economy for this process.
The modern petrochemical route involves the following reaction:
C2H4
+
1/2
O2
Ag
C2H4O
b) Calculate the % atom economy of this process.
Ibuprofen, a non-steroidal anti-inflammatory drug, was first synthesised by Boots using a sixstep process, with a % atom economy of 40%. When the patent expired in the 1980’s, several
companies began developing new methods for the preparation of ibuprofen. The BHC
Company synthesis, which proved highly successful, is shown below:
continued on page 5
Student worksheet 28SW
Olympiad past paper questions
Page 4 of 11
Chemistry for the gifted and talented
Step 1 involves using ethanoic anhydride, (CH3CO)2O.
c) i) Calculate the % atom economy of the BHC Company process.
ii) What does HF do in step 1?
iii) What happens to the % atom economy if the ethanoic acid is re-used?
continued on page 6
Student worksheet 28SW
Olympiad past paper questions
Page 5 of 11
Chemistry for the gifted and talented
5. Reinecke’s salt
When ammonium dichromate(VI) is added gradually to molten ammonium thiocyanate,
Reinecke’s salt is formed. It has the formula NH4[Cr(SCN)x(NH3)y] and the following composition
by mass.
Cr
15.5%
S
38.15%
N
29.2%
a) Calculate the values of x and y in the above formula.
b) Calculate the oxidation number of chromium in the complex.
c) Suggest a shape for the complex anion.
d) Draw two possible structures for the anion and state the type of isomerism it exhibits.
continued on page 7
Student worksheet 28SW
Olympiad past paper questions
Page 6 of 11
Chemistry for the gifted and talented
6. The NMR spectra of NanoPutians
In June 2003, a research paper was
published announcing the synthesis of
the smallest representations of the
human form: 2 nm tall anthropomorphic
molecules, nicknamed
‘NanoPutians’ by their creators.
The molecules synthesised included
‘NanoKid’, ‘NanoBaker’ and
‘NanoAthlete’. The compound shown to
the right was called ‘NanoBalletDancer’
and has the formula C41H50O2.
When assigning an NMR spectrum, the
first step is to identify how many atoms
there are in unique environments.
Both carbon atoms (13C) and hydrogen
atoms (1H) give NMR signals. Each atom
in a different environment will give rise to
one signal.
For example, in the structure of
NanoBalletDancer, carbon atoms 37 and
39 are equivalent; we may write
37 ≡ 39. Hence, although there are two
carbon atoms (37 and 39) which have
one oxygen atom attached, only one
signal would be observed in a
13C
NMR
spectrum due to these carbon atoms
since they are equivalent.
a) Which carbon atoms making up the benzene rings are equivalent?
Write down w ≡ x, y ≡ z etc for any equivalent atoms. How many signals in total would
be observed due to benzene-ring carbons in a carbon NMR spectrum of
NanoBalletDancer?
b) List the groups of equivalent triple bond carbons in NanoBalletDancer. How many signals
would be seen in total in the
13C
spectrum due to triple bond carbon atoms?
continued on page 8
Student worksheet 28SW
Olympiad past paper questions
Page 7 of 11
Chemistry for the gifted and talented
c) How many different methyl groups (–CH3 groups) are there in NanoBalletDancer? Again,
list them in groups.
d) How many different carbon environments are there in NanoBalletDancer – ie how many
signals would be seen in total in the
13C
NMR spectrum?
Similarly, in 1H NMR, the total number of signals depends on the number of different
environments of hydrogen atoms in a structure. There are 13 different environments of
hydrogens in NanoBalletDancer; their signals are labelled A–M in the spectrum below. The
numbers of hydrogen atoms in each unique environment is given under the label. Hydrogen
atoms in similar environments all have similar chemical shifts. For example, all the hydrogens
on the benzene rings occur in the same region of the spectrum, i.e. they have a similar chemical
shift.
However, 1H NMR is complicated by coupling. If a hydrogen is within three bonds of another
hydrogen which is in a different environment, instead of appearing as a single peak, its signal
is split into a number of peaks. In general, if the hydrogen under consideration is within three
bonds of n hydrogens in a different environment from the one under consideration, it will be
split into (n + 1) peaks. The ratio of the area under the peaks is given by Pascal’s triangle as
outlined below.
observed ratio
for a hydrogen coupling to
0 hydrogens
1 hydrogen
2 hydrogens
3 hydrogens
4 hydrogens
The signal for a given hydrogen is not split by any hydrogens which are in the same
environment.
e) Into how many peaks will the signal from a hydrogen that couples with 5 other hydrogens
be split? What will the ratio of the peaks be?
It is possible to assign the 1H NMR spectrum of NanoBalletDancer by considering the numbers
of hydrogens in different environments, their chemical shifts and their coupling patterns.
For example, the signal at 7.15 ppm (B) is due to the hydrogen atoms on carbons 19 and 23.
On the table below, assign (as far as possible) which signals are due to which hydrogen atoms.
The assignment for signal B has already been filled in. (For some signals, it might not be possible
to decide between two alternative assignments – in which case just write ‘... or ...’ on your
answer sheet.)
continued on page 9
Student worksheet 28SW
Olympiad past paper questions
Page 8 of 11
Chemistry for the gifted and talented
1H
NMR Signal
Hydrogen(s) on carbon(s)
A
B
19, 23
C
D
E
F
G
H
I
J
K
L
M
continued on page 10
Student worksheet 28SW
Olympiad past paper questions
Page 9 of 11
7.5
A
1H
B
2H
7.0
C
2H
6.5
6.0
D
1H
5.5
solvent
The 1H NMR spectrum of NanoBalletDancer
5.0
E
2H
4.5
F
2H
4.0
3.5
G
2H
3.0
2.5
H
4H
2.0
I
4H
1.5
J
3H
K
18H
1.0 ppm
M
3H
L
6H
Chemistry for the gifted and talented
continued on page 11
Student worksheet 28SW
Olympiad past paper questions
Page 10 of 11
Chemistry for the gifted and talented
7. Reactions of amines
Draw the structures for compounds A–G in the following sysnthesis:
A
B
C
D
E
F
G
Student worksheet 28SW
Olympiad past paper questions
Page 11 of 11
Discussion of answers
Chemistry for the gifted and talented
Olympiad past paper questions
1. Heating a cup of coffee
a) CaO + H2O
→ Ca(OH)2
b) Acidic, neutral, basic
In acidic conditions there are two reactions running in parallel.
1. CaO + 2H+ → Ca2+ +H2O
2. CaO + H2O
→ Ca(OH)2
Because the rate of reaction 1 depends on the concentration of H+, the acidic conditions will
react faster than the neutral conditions. Another factor may be the limited solubility of
Ca(OH)2. Solid Ca(OH)2 forming around the CaO may hinder the collision between water and
the surface of the CaO. Precipitation of Ca(OH)2 occurs fastest in basic conditions.
c) ∆rH = – 1003 + 635 + 286 = –82 kJ mol-1
d) To warm 1 g by 1 °C requires 4.18 J.
To warm 210 g by 40 °C requires 4.18 x 210 x 40 J = 35.1 kJ.
e) 1 mol CaO provides 82 kJ, but we need 35.1 kJ.
35.1 / 82 mol = 0.428 mol
Taking the Mr for CaO as 56, minimum mass required = 56 x 0.428 = 24.0 g
(The actual mass used in cans is 70 g)
continued on page 2
Discussion of answers 28DA
Olympiad past paper questions
Page 1 of 13
Chemistry for the gifted and talented
2. Superconductors
This question is interesting because it introduces a type of compound with a mixture of cations
in a fixed ratio. A simple example of this type of compound is Fe3O4. The average oxidation
number for iron, in the compound, is +2.66. Fe3O4 can be thought of as a 1:1 composite of the
iron(II) and iron(III) oxides, and might be written FeO.Fe2O3. The ‘.’ in the formula performs the
same role as in CuSO4.5H2O; it separates parts of the compound to indicate that the different
components that are associated in a fixed ratio.
The mixed oxide Fe3O4 has an electrical conductivity a million times greater than Fe2O3. This
seems to imply that the mixture of oxidation states is important in producing superconductivity.
a) The question starts off asking you to calculate the empirical formula of A. The first step is to
calculate the percentage of oxygen in A.
100% - 13.4% - 41.2% - 28.6% = 16.8% oxygen
Element
Y
Ba
Cu
O
%
13.4
41.2
28.6
16.8
Ar
88.9
137.3
63.5
16.0
100 g of superconductor A)
0.15
0.30
0.45
1.05
Moles in 100g/0.150
1.00
1.99
2.99
6.97
1
2
3
7
Percentage /Ar (moles in
Ratio
Therefore the empirical formula is YBa2Cu3O7.
b) The oxidation numbers have to balance for the compound overall. Oxygen’s contribution
will be -2 per atom, that is 7x-2=-14. The metals must contribute +14 between them
when their oxidation numbers are added together.
The average contribution per copper atom must be
14 – 3 – 2(2)
3
= 2.33
c) This part of the question leads us to consider the copper as being in a mixture of +2 and
+3 oxidation states. Given the average oxidation state was 2.33 the ratio of copper(II) :
copper(III) must be 2 : 1.
We can now treat superconductor A as if it were composed of the separate oxides in fixed
ratios Y2O3 : BaO : CuO : Cu2O3 of 1 : 4 : 4 : 1. Bear in mind that because the oxide of
yttrium(III) has two yttriums in it we have had to double up the empirical formula. This gives
us a suggested formula of superconductor A Y2O3.4BaO.4CuO.Cu2O3 (Mr = 1332 g mol-1).
The formula for compound B might be written Y2O3.4BaO.6CuO (Mr = 1316 g mol-1).
continued on page 3
Discussion of answers 28DA
Olympiad past paper questions
Page 2 of 13
Chemistry for the gifted and talented
So the mass left (in mg) is 84.2 x
658
666
= 83.19
Research/discussion suggestion:
How does the mixture of oxidation states increase conductivity?
continued on page 4
Discussion of answers 28DA
Olympiad past paper questions
Page 3 of 13
Chemistry for the gifted and talented
3. Combining proportions of elements
Some students had difficulty with this question because the Periodic Table supplied with the
Round 1 paper only had the symbols of the elements and the question gave only the name of
antimony so they needed to know what the symbol of antimony is to answer the question.
It is worth being familiar with the symbols of the elements.
This question shows how far you can go with the ratios of masses in determining formulae and
equations. You cannot convert grains to grams, but you do not need to.
a) Twelve grains of Sb give 14.4 grains of oxide.
Twelve grains of Sb combine with 2.4 grains of oxygen.
Method 1
These ratios would hold if we measured the masses in any other unit including grams.
The ratio of moles of Sb : O can be found by dividing the masses by the Ar of Sb and O
respectively.
The molar ratio Sb : O = (12/121.8) : (2.4/16.0) = 0.09852 : 0.15 = 1:1.5 = 2:3
Formula = Sb2O3 (this is an empirical formula as are all formulae of compounds with giant
structures such as NaCl or SiO2.)
Method 2
Suppose the conversion factor for grains to grams = k
Moles of Sb = 12k / 121.8
Moles of O = 2.4k / 16.0
The molar ratio Sb : O = (12/121.8) : (2.4/16.0) = 0.09852 : 0.15 = 1:1.5 = 2:3
Formula = Sb2O3
b) One mol of Zn forms one mol of ZnO
65.4 g of Zn forms (65.4 + 16.0) g ZnO = 81.4 g
The increase in mass is 81.4 / 65.4 so 60 grains should produce (60 x 81.4) / 65.4 grains =
74.6(8) grains.
c) The total mass at the end = unreacted Zn + ZnO
Sixty grains of Zn should give 74.68 grains ZnO
Method 1
If the fraction of Zn reacting is α, the amount of Zn used is 60 α grains which forms
74.68 α grains of ZnO.
The amount of Zn left = 60 – 60 α
The total mass at the end = 60 – 60 α + 74.68 α + 66 grains + 60 + 14.68 α
continued on page 5
Discussion of answers 28DA
Olympiad past paper questions
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α = (66 – 60) / 14.68 = 0.40(87)
Mass of unreacted Zn = 60 – 60 α = 35(•48) grains
Mass of ZnO = 74.68 α = 30(•52) grains.
Method 2
The expected increase for 100% conversion is 14.68 g. The actual increase is six grains, so
the fraction converted = 6 / 14.68 g = 0.41
The mass of Zn left = 60(1-0.41) = 35 grains and mass of ZnO = 66 – 35 = 31 grains.
continued on page 6
Discussion of answers 28DA
Olympiad past paper questions
Page 5 of 13
Chemistry for the gifted and talented
4. Green chemistry
a) i) C2H4 + Cl2 + Ca(OH)2 → C2H4O + CaCl2 + H2O
The first step in this process is electrophilic addition to ethene. The electrophile is Cl2 but
the major product is not CH2ClCH2Cl because H2O competes with Cl– as the nucleophile
which attacks the carbocation reactive intermediate CH2ClCH2+.
The second step possibly proceeds via an internal nucleophilic elimination:
ii) % atom economy =
44
x 100
(not 23.6)
44 + 111 + 18
= 25.4
b) 100%
c) i) Mr ibuprofen = 206
% atom economy =
206
x 100 = 77.4
206 + 60
ii) Catalyst. The HF can catalyse the reaction by protonating the ethanoic acid anhydride
molecule, making it even more susceptible to nucleophilic attack.
iii) It goes up to 100%
continued on page 7
Discussion of answers 28DA
Olympiad past paper questions
Page 6 of 13
Chemistry for the gifted and talented
5. Reinecke’s salt
a) There are at least two approaches to solving this problem.
1. Use the composition by mass of Cr to calculate the Mr of Reinecke’s salt because the
formula contains one Cr atom which is 15.5%.
15.5
100
=
52
Mr
or rearranged Mr = 52 x
100
15.5
= 335.5
Next it makes sense to calculate x because the % of S in Reinecke’s salt depends only on
x, but the % of N in Reinecke’s salt depends on x and y.
32x =
38.15
100
x 335.5 = 128 so x =
128
32
=4
Now considering N. There are 1 + x + y atoms of N in the formula.
14(1 + x + y) =
29.2
100
x 335.5 = 98 so y =
98
14
–5=2
2. Use the ratio of % composition divided by Ar.
a) The ratio of numbers of atoms in the formula Cr : S = 1 : x
This ratio can also be expressed
15.5 38.15
38.15 52
:
so x =
x
=4
52
32
15.5 32
The ratio Cr : N = 1 : (1 + x + y)
This ratio can also be expressed
15.5 29.2
29.2 52
:
so y =
x
–5=2
52
14
15.5 14
b) The sum of all the oxidation numbers in a neutral compound = 0 and it is convenient
in this case to use the charges of the different components ( NH4+, NH3 and SCN–)
rather than the individual oxidation numbers.
So, NH4 [Cr(SCN)4 (NH3)2] = 0
+ 1 + Cr + (4 x -1) + (2 x 0) = 0
Therefore the oxidation number for Cr = + 3
Many students will be unfamiliar with the thiocyanate ion SCN- so need to try out some
structures to find the charge (-1 or -2) on the ion. It must be negatively charged
becasue it forms a salt with ammonium ions. Sulfur commonly forms two bonds,
carbon four and nitrogen three. Carbon can make four covalent bonds out of the
continued on page 8
Discussion of answers 28DA
Olympiad past paper questions
Page 7 of 13
Chemistry for the gifted and talented
required five to satisfy the bonds needed to sulfur and nitrogen. This leaves one bond
which will need to be ionic so this implies a charge of -1. Possible structures might be
S–
C
N
or
S
C
N
c) There are six ligands so that implies an octahedral geometry.
d) Two octahedral structures, one with two NH3 groups adjacent, one with them opposite.
Geometrical isomerism for two shapes.
continued on page 9
Discussion of answers 28DA
Olympiad past paper questions
Page 8 of 13
Chemistry for the gifted and talented
6. The NMR spectra of NanoPutians
A molecular model of NanoBalletDancer
a) 6 ≡ 8 ( since both ‘legs’ are the same), 9 ≡ 11, 19 ≡ 23 (since the ‘waist’ of
NanoBalletDancer freely rotates and the ‘arms’ are the same), 20 ≡ 22.
Eight signals in total due to benzene ring carbons (the four above and 7,10,18 and 21).
b) 4 ≡ 13, 5 ≡ 12, 24 ≡ 30, 25 ≡ 31.
Four signals in total due to triple bond carbons.
c) 1 ≡ 16, 27 ≡ 28 ≡ 29 ≡ 33 ≡ 34 ≡ 35, 40 (unique), 41 (unique).
Four signals in total due to methyl group carbons.
On the CDROM Chemistry for the gifted and talented there is a three-dimensional model
called nanoput.mol (CDROM index 28MO) which you will be able to view in Internet
Explorer® if you have the Chime® plug-in.
A side on view of the head of NanoBalletDancer (with arm removed)
continued on page 10
Discussion of answers 28DA
Olympiad past paper questions
Page 9 of 13
Chemistry for the gifted and talented
The model shows that the two sides of the head are not the same, one side is closer to the body
than the other. This means that carbons 40 and 41 are unique (not equivalent) and that the two
hydrogen atoms on carbon 39 are not equivalent to each other, but that the hydrogens on
atoms 39 and 37 that are on the same side are equivalent.
d) Twenty three different environments and therefore twenty three different signals
(eight aromatic, four methyl, four triple bonds as well as 2 ≡ 15, 3 ≡ 14, 17, 36, 26 ≡ 32,
39 ≡ 37, 38).
e) The signal is split into 6, ratio 1 : 5 : 10 : 10 : 5 : 1
f)
1H
NMR Signal
Hydrogen(s) on Carbon(s)
A
7 (only 1H aromatic signal)
B
19, 23 (given in question)
C
9, 11 (other aromatic 2H signal)
D
36 (only non aromatic 1H signal)
E
17 (2H signal with no other Hs close enough to couple)
F
37 and 39 (1H from each of 37 and 39 both pointing up (or down))
G
37 and 39 (1H from each of 37 and 39 both pointing down (or up))
H
3, 14 (a 4H signal coupled to 2Hs)
I
2, 15 (a 4H signal coupled to 5Hs)
J
40 or 41 (a 3H singlet)
K
27, 28, 29, 33, 34, 35 (an 18H signal)
L
1, 16 (6H triplet)
M
41 or 40 (a 3H singlet)
continued on page 11
Discussion of answers 28DA
Olympiad past paper questions
Page 10 of 13
Chemistry for the gifted and talented
7. Reactions of amines
In tackling this type of question you need to be able to work forwards and backwards from a
structure that you are confident about.
a) and b)
This question centres on the given structure of 4-methylphenylamine. You need to work
backwards to A and B. B is hydrogenated to give 4-methylphenylamine. The hydrogenation
must involve a functional group being reduced to an amine or methyl because the
delocalised ring is still intact. We need a further clue to the structure of B before deciding
which functional group is reduced. B is formed by treating A with conc HNO3 and conc
H2SO4. This is, hopefully, recognisable as a nitrating mixture. The H2SO4 protonates the HNO3
which then forms the electrophile NO2+ (and water). The electrophile NO2+ is strong enough
to cause electrophilic substitution on the (substituted) benzene ring. Because the functional
group ends up at the 4 position in 4-methylphenylamine, the nitro group must have
substituted the hydrogen there.
This gives us
a)
and
b)
c) CH3I is susceptible to nucleophilic attack, the nitrogen in 4-methylphenylamine has a lone
pair of electrons and is nucleophilic. The nucleophilic substitution can occur a couple of times
because the nitrogen can lose H+ twice. This gives the tertiary amine:
One possibility you may have considered is that it could go one further and form the
quaternary ammonium ion:
continued on page 12
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Chemistry for the gifted and talented
This is less likely for an aromatic amine than an aliphatic (joined to an alkyl group) amine. In
the aromatic amine the lone pair is partially involved in the delocalised system.
d) The amine is basic and the reaction with HCl gives:
e) You may or may not recognise the reactants necessary to produce a diazonium ion,
R
+
N
N, from an amine. You should recognise however that E undergoes a
nucleophilic substitution on the benzene ring... quite an achievement. You need a very
good leaving group to achieve the nucleophilic substitution and N2 is a very stable leaving
group. The reaction to form the diazonium salt involves nucleophilic attack on the
protonated nitrate(III) ion.
continued on page 13
Discussion of answers 28DA
Olympiad past paper questions
Page 12 of 13
Chemistry for the gifted and talented
f) The amine is again acting as a nucleophile to give:
g) This is formed by reducing the nitrile to the amine.
Discussion of answers 28DA
Olympiad past paper questions
Page 13 of 13