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48th International
Chemistry Olympiad

Theoretical Problems

28 July 2016
Tbilisi, Georgia
48th IChO Theoretical Problems, Official English version

1

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Instructions

Begin only when the START command is given. You have 5 hours to work on the
problems.

Use only the pen and calculator provided.

The problem booklet has 23 pages, the answer sheet is 28 pages.

Make sure that your code is on every page of the answer sheet.

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Anything written elsewhere will not be graded. Use the reverse of the problem
pages if you need scratch paper.

Write relevant calculations in the appropriate boxes when necessary. If you
provide only correct end results for complicated questions, you will receive noscore.



Raise your hand if you need a restroom break.

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You can keep the problem booklet.

You must stop your work immediately when the STOP command is given. A delay
in doing this may lead to cancellation of your exam.



Do not leave your seat until permitted by the supervisors.

The official English version of this examination is available on request only forclarification.

48th IChO Theoretical Problems, Official English version

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Constants and formulae
Avogadro constant:

NA = 6.022·1023 mol–1

Zero of the Celsius
scale:

273.15 K

Gas constant:

R = 8.314 J K–1 mol–1

Faraday constant:

F = 96485 C mol–1

Ideal gas equation:

pV = nRT

Gibbs energy:

G = H – TS

Ionic product of water
𝐾𝑤 = 10−14
at 298.15K :𝑅𝑇 𝑐𝑜𝑥0.059 V𝑐𝑜𝑥
𝐸 = 𝐸𝑜 +ln= 𝐸𝑜 +
log
𝑛𝐹 𝑐𝑟𝑒𝑑𝑛𝑐𝑟𝑒𝑑𝑜∆𝑟 𝐺 𝑜 = −𝑅𝑇𝑙𝑛𝐾 = −𝑛𝐹𝐸𝑐𝑒𝑙𝑙

Nernst equation:

or 𝐸 = 𝐸 𝑜 −𝑅𝑇
0.059 Vln 𝑄 = 𝐸 𝑜 −
log 𝑄
𝑛𝐹
𝑛

𝐼0
= 𝜀𝑐𝑙
𝐼
Consider all gases ideal throughout the exam.
Beer-Lambert law:𝐴 = log

Periodic table with relative atomic masses
1

18

1

H

2

1.008

2

13

14

15

16

17

3

4

5

6

7

8

9

9.01

10.81

12.01

14.01

16.00

19.00

11

12

13

14

15

16

17

22.99

24.30

3

4

5

6

7

8

9

10

11

12

26.98

28.09

30.97

32.06

35.45

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

39.10

40.08

44.96

47.87

50.94

52.00

54.94

55.85

58.93

58.69

63.55

65.38

69.72

72.63

74.92

78.97

79.90

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

85.47

87.62

91.22

92.91

95.95

-

101.1

102.9

106.4

107.9

112.4

114.8

118.7

121.8

127.6

126.9

131.3

55

56

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

132.9

137.3

178.5

180.9

183.8

186.2

190.2

192.2

195.1

197.0

200.6

204.4

87

88

104

105

106

107

108

109

110

111

112

113

-

-

-

-

-

-

-

-

-

-

-

57

58

59

60

61

62

63

64

65

138.9

140.1

140.9

144.2

-

150.4

152.0

157.3

89

90

91

92

93

94

95

96

-

232.0

231.0

-

-

-

-

-

K

Ca Sc Ti

Rb Sr

Y

88.91

Zr

Cs Ba 57-71 Hf
Fr-

Ra 89103-

V

Cr Mn Fe Co

Ni

W

Re Os

Ir

Pt

Si

P

O
S

Cu Zn Ga Ge As Se

Nb Mo Tc Ru Rh Pd Ag Cd
Ta

Al

N

Au Hg

In

Tl

Rf Db Sg Bh Hs Mt Ds Rg Cn Nh

Sn Sb Te
Pb

F

10

Be

Na Mg

C

4.003

Li

6.94

B

He

Cl

Br
I

Ne

20.18
18

Ar

39.95
36

Kr

83.80
54

Xe

207.2

Bi

209.0

Po

At Rn
-

-

114

115

116

117

118

-

Fl

Mc Lv
-

-

Ts Og
-

-

66

67

68

69

70

71

158.9

162.5

164.9

167.3

168.9

173.0

175.0

97

98

99

100

101

102

103

-

-

-

-

-

La Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu

Ac Th PaU

238.0

Np Pu Am Cm Bk

Cf-

48th IChO Theoretical Problems, Official English version

Es Fm Md No

Lr3

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Problem 1

5% of the total

Nitrogen trifluoride is a surprisingly stable compound that was first prepared in themelt electrolysis of a mixture of ammonium fluoride and hydrogen fluoride.
1.1.

On which electrode does nitrogen trifluoride form? Write a balanced chemical
equation for the electrode half reaction for the formation of NF3.

Interestingly the related fluoroamine (NH2F) and difluoroamine (NHF2) are very
unstable materials; decomposition of either pure substance can even be explosive. Thisis dangerous as they are formed in the electrolysis as side products.
1.2.

Which of NF3, NHF2 or NH2F compound is expected to condense at the lowest
temperature?

The N-F bond lengths in these molecules were determined to be 136, 140 and 142 pm.
The change in the bond lengths can be explained with a simple electrostatic model
taking into account the partial charges on the atoms.
1.3.

Assign the N-F bond lengths (136, 140, 142 pm) to the molecules.

When NHF2 is bubbled through a solution of KF in HF, a binary nitrogen – fluorinecompound can be obtained as a mixture of two geometric isomers.
1.4.

Write a balanced chemical equation for the formation of the binary nitrogenfluorine compound.

Tetrafluoroammonium ion (NF4+) and its corresponding salt can form from NF3 withelementary fluorine in the presence of an appropriate reagent.
1.5.

Propose a suitable reagent and write a balanced chemical equation for thereaction.

NF4+ ions form stable salts with a number of anions. These are very sensitive tohumidity, because NF4+ ion hydrolyzes forming NF3 and O2. Interestingly nitrogen
trifluoride always forms quantitatively, while the quantity of oxygen is often less than
expected due to side reactions.
1.6.

Write a balanced chemical equation for the hydrolysis of NF4+. Write a balanced
chemical equation for a possible side reaction that can decrease thetheoretically expected O2:NF3 mole ratio.

Tetrafluoroammonium salts were investigated for use as solid rocket fuels, because NF3
and F2 are released from them on heating. One of them has a fluorine content of
65.6 m/m%, all of which is converted into NF3 and F2 upon decomposition. During thedecomposition 2.5 times as many moles of F2 are formed as of NF3.
1.7.

Determine the formula of the salt in question.

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Problem 2

8% of the total

One of the first materials used in solid state electronics was red copper(I) oxide. Interestis renewed nowadays because it could be a non-toxic and cheap component of solar
cells.
AB

The two figures above depict the cubic unit cell of the Cu2O crystal. The lattice constantof the structure is 427.0 pm.
2.1.1.

Which of the atoms (A or B) is copper?
Which basic structure (primitive cubic, face centered cubic, body centered cubic,
diamond) is formed by the A atoms and which structure is formed by the B
atoms?
What are the coordination numbers of the atoms?

2.1.2.

Calculate the smallest O-O, Cu-O and Cu-Cu distances in the structure.

2.1.3.

What is the density of pure copper(I) oxide?

A common defect in this crystal is some copper atoms missing with the oxygen latticeunchanged. The composition of one such crystal sample was studied, and 0.2% of all
copper atoms were found to be in oxidation state +2.
2.2.

What percentage of normal copper sites are empty in the crystal sample? Whatis x in the empirical formula Cu2-xO of the crystal?

Copper(I) oxide is insoluble in water. It is stable in dry air, but humidity in the air
catalyzes a transformation (Reaction 1).
When copper(I) oxide is dissolved in dilute sulfuric acid, a blue solution containing aprecipitate is formed without evolution of a gas (Reaction 2). When hot, concentrated
sulfuric acid is used, no precipitate remains, but an odorous gas forms (Reaction 3). The
same gas forms when the precipitate from reaction 2 is dissolved in hot concentrated
sulfuric acid.
2.3.

Write balanced chemical equations for reactions (1-3).

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Copper(I) oxide can be produced in a number of ways. Heating copper in air is acommon method in the synthesis of semiconductor Cu2O. In a pure oxygen atmosphere,
the three species containing copper (Cu(s), Cu2O(s) or CuO(s)) can potentiallyinterconvert.
Suppose that the ΔfHo and So data given for 105 Pa are independent of temperature:
ΔfHo / kJ mol−1 So/ J mol−1 K−1
Cu(s)
0
65
O2(g)
0
244
CuO(s)−156
103
Cu2O(s)
−170
180
2.4.

Determine the temperature ranges, if any, of thermodynamic stability of copperand its oxides between 500 and 1500 K in a 105 Pa oxygen atmosphere.

Important data are given for 298 K. Use this temperature in the following calculations:
Ksp(Cu(OH)2)=2∙10−19
Cu2O(s) + H2O(l) + 2 e− ⟶ 2 Cu(s) + 2 OH−(aq)Cu2+(aq) + e− ⟶ Cu+(aq)Cu2+(aq) + 2 e− ⟶ Cu(s)Eo= −0.360 V
Eo= +0.159 V
Eo= +0.337 V

One possibility for producing Cu2O is the anodic oxidation of copper. Electrolysis of anaqueous basic solution (e.g. NaOH) with a copper anode and platinum cathode can leadto formation of copper(I) oxide on the anode.
2.5.

Write the half reaction equations for the electrode processes during the anodic
production of Cu2O in NaOH solution with a platinum cathode and copper
anode.

Electrolytic reduction of copper(II) ions in solution is another possibility.
2.6.1.

Write the half reaction equation of the cathode process giving Cu2O in acidic
medium.

Let us use 0.100 mol dm−3 Cu2+ solution and carry out electrolysis with platinum
electrodes.
2.6.2.

What is the maximum pH at which the concentration of copper(II) can be
maintained at 0.100 mol dm−3?

If the pH is too low, reduction to metallic copper is preferred to the formation ofcopper(I) oxide.
2.6.3.

What is the minimum pH at which the cathodic production of Cu2O in a 0.100
mol dm−3 Cu2+ solution is still possible?

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Problem 3

9% of the total

Iodine deficiency is of special concern in Georgia because it occupies a region whereiodine is scarce in soil and water. Iodine deficiency can be effectively and inexpensively
prevented if salt for human consumption is fortified with small amounts of iodine.
Methods for analyzing salt for iodine content are thus important. Current regulations in
Georgia are that iodized salt must contain between 25-55 ppm iodine (1 ppm =1 mg iodine/kg salt).
Most salt is iodized by fortification with potassium iodate (KIO3). Iodate content can be
determined in salt samples using iodometric titration. In a typical procedure, 10.000 g ofan iodized salt sample is dissolved in 100 cm3 of 1.0 mol/dm3 aqueous HCl to which
1.0 g KI has been added. The solution is then titrated with 0.00235 mol/dm3 aqueous
sodium thiosulfate solution to a starch endpoint; this requires 7.50 cm3 of titrant.
3.1.1.

Write a balanced net ionic equation for the reaction when iodate reacts withexcess iodide in acidic solution.

3.1.2.

Write a balanced net ionic equation for the reaction taking place during thetitration with thiosulfate.

3.1.3.

Calculate the iodization level, in ppm, of this salt sample.

A less common agent for iodizing salt is potassium iodide, which cannot be easily
measured by iodometric titration.
One possible method for analyzing iodide in the presence of chloride is potentiometrictitration. However, this method is not very precise in the presence of large amounts ofchloride.
In this method, a silver wire is immersed in the solution (containing iodide and chloride)
to be analyzed and silver ion is gradually added to the solution. The potential of thesilver wire is measured relative to a reference electrode consisting of a silver wire in a
1.000 mol/dm3 solution of AgNO3. The measured potentials are negative and theabsolute values of these potentials are reported. The solution to be analyzed has avolume of 1.000 dm3 (which you may assume does not change as silver ion is added),
and T = 25.0°C.
The results of this experiment are governed by three equilibria: the solubility of AgI(s)[KspI] and AgCl(s) [KspCl] and the formation of AgCl2−(aq) [Kf]. (Iodide also forms complex
ions with silver but this may be neglected at the very low concentrations of iodide
present in this experiment).
AgI(s) ⇌ Ag+(aq) + I−(aq)KspI

AgCl(s) ⇌ Ag+(aq) + Cl−(aq)KspCl

Ag+(aq) + 2 Cl−(aq) ⇌ AgCl2−(aq)Kf48th IChO Theoretical Problems, Official English version

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Below are shown the results of two experiments measuring the observed potential as afunction of added number of moles of silver ion. Experiment A (solid circles) was carriedout with 1.000 dm3 of solution containing 1.00∙10−5 mol/dm3 iodide and no chloride ion.
Experiment B (open circles) was done using 1.000 dm3 of solution containing
1.00∙10−5 mol/dm3 iodide and 1.00∙10−1 mol/dm3 chloride.

μmol Ag+
added
1.00
3.00
5.00
7.00
9.00
10.0
11.0
12.0
13.0
15.0

|E|, Vexp. A
0.637
0.631
0.622
0.609
0.581
0.468
0.355
0.337
0.327
0.313

|E|, Vexp. B
0.637
0.631
0.622
0.610
0.584
0.558
0.531
0.517
0.517
0.517

μmol Ag+ added

3.2.1.

Select an appropriate data point from the experiments and use it to calculate thesolubility product of AgI (KspI).

3.2.2.

Select an appropriate data point from the experiments and use it to calculate thesolubility product of AgCl (KspCl).

3.2.3.

Select an appropriate data point from the experiments and use it to calculate Kf.
You may need to use values of KspI or KspCl to do this calculation. If you wereunable to carry out the calculations in 3.2.1. or 3.2.2., you may use the arbitraryvalues of KspI = 1.00∙10−15 and KspCl = 1.00∙10−9 without penalty.

48th IChO Theoretical Problems, Official English version

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An analytical method that is more practical, because it is not sensitive to the presence ofchloride, uses the Sandell-Kolthoff reaction. This is the reaction of H3AsO3 with Ce(IV) to
give Ce(III) in acidic solution, which is strongly catalyzed by iodide ion.
3.3.1.

Write balanced net ionic equations for the reaction of cerium(IV) with H3AsO3 inacidic solution, as well as reactions of cerium(IV) with a species containing theelement iodine and H3AsO3 with a species containing the element iodine, that
could reasonably account for the catalysis of the net reaction by iodide.

The reaction of Ce(IV) with H3AsO3 can be monitored by measuring the absorbance at
405 nm, as Ce(IV) is orange and absorbs significantly at 405 nm, while the otherreactants and products are colorless and do not absorb appreciably. Three runs werecarried out, all in 0.50 mol/dm3 H2SO4 at 25.0°C using the following initialconcentrations:
Run
1
2
3

[H3AsO3]0, [Ce(IV)]0,
mol dm−3 mol dm−3
0.012500.001200.006250.00060
0.012500.00120[I–]0,
mol dm−31.43∙10−6
1.43∙10−6
7.16∙10−7

An analyst initiated the reactions by mixing the reagents in a cuvette. After a shortvariable delay absorbance measurements were started, with the first measurement
recorded at t=0 s. The data obtained are shown below:

A405
Run 1
0
0.621
20 0.348
40 0.198
60 0.113
80 0.064
100 0.037
t, s

A405
Run 2
0.287
0.149
0.083
0.046
0.025
0.014

A405
Run 3
0.818
0.608
0.455
0.340
0.254
0.191

Under these conditions (0.5 mol/dm3 H2SO4, 25.0°C), the rate law for the reaction can be
written as
Rate = k[H3AsO3]m[Ce(IV)]n[I−]p3.3.2.

where m, n, and p are integers.

Determine the values of m, n, and p and calculate the value of k (be sure to
specify its units).

48th IChO Theoretical Problems, Official English version

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A 1.000 g sample of iodized salt is dissolved in water to give 10.00 cm3 of solution. A
0.0500 cm3 aliquot of this solution is added to a mixture of 1.000 cm3 0.025 mol/dm3
H3AsO3 in 0.5 mol/dm3 H2SO4 and 0.800 cm3 0.5 mol/dm3 H2SO4. To this mixture isadded 0.200 cm3 0.0120 mol/dm3 Ce(NH4)2(NO3)6 in 0.5 mol/dm3 H2SO4 and theabsorbance at 405 nm is measured as a function of time at 25.0°C:

t, s
0
20
40
60
80
100

3.3.3.

A405
0.756
0.531
0.373
0.262
0.185
0.129

Calculate the iodization level, in ppm, of this salt sample.

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Problem 4

8% of the total

Application of kinetic studies in water treatment
Industrial waste is a major cause of water pollution and kinetic studies are carried out ina laboratory to design effluent treatment. 1,4-dioxane, more commonly known as
dioxane (C4H8O2), an industrial solvent and by-product, is a significant watercontaminant. It can be oxidised to hazard free chemicals using oxidants such asperoxodisulfate, ozone or hydrogen peroxide.
The data obtained in the kinetic study of oxidation of dioxane with potassium
peroxodisulfate (K2S2O8) as oxidant and AgNO3 as catalyst at T = 303.15 K are given
below. The reaction was monitored by the estimation of unreacted peroxodisulfate. The
concentration of AgNO3 used in this study was 1.00∙10−3 mmol∙dm−3.
Trial
1
2
3
4

Dioxanemmol∙dm−3
0.01000.01000.00500
0.0110K2S2O8mmol∙dm−3
2.50
5.10
13.8
13.8

Initial rate
mmol∙dm−3∙min−1
1.661∙10−2
3.380∙10−2
9.200∙10−2
9.201∙10−2

In many countries the accepted maximum level of dioxane in drinking water is specified
as 0.35 μg dm−3.
A water sample contains an initial dioxane concentration of 40.00 μg dm−3. Assume that
1 mol dioxane requires 1 mol of peroxodisulfate for oxidation. The concentration of
AgNO3 used in this study was 1.00∙10−3 mmol∙dm−3.
4.1.1.

Calculate the time in minutes the oxidation process has to continue in order toreach the accepted level of dioxane at 303.15 K if the initial concentration of
K2S2O8 is 5.0∙10−6 mol dm−3. Assume that the rate law obtained from the data
above is valid under these conditions.

Various mechanisms have been proposed for the peroxodisulfate oxidation of dioxane.
Misra and Ghosh (1963) proposed the following mechanism:
S2O82− + Ag+

Ag3+ + D (dioxane)
4.1.2.

k1⇄
k2

Ag3+ + 2SO42−

k3⟶ D’ (dioxane oxidised) + 2H+ + Ag+

Assuming Ag(III) to be in steady state, deduce the rate equation for theoxidation of dioxane.

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4.1.3.

Which of the following is/are correct?
A) The rate equation based on the mechanism given in 4.1.2, at very highconcentrations of dioxane, is consistent with the experimental data in 4.1.1.
B) The rate equation based on the mechanism given in 4.1.2, at very low
concentrations of dioxane, is consistent with the experimental data in 4.1.1.
C) The units of the observed rate constant are dm3∙mol−1∙s−1 at very highconcentrations of dioxane.
D) The units of the observed rate constant are dm3∙mol−1∙s−1 at very low
concentrations of dioxane.

Degradation of pharmaceutical products – a kinetic overview
Kinetic studies are important in deciding the shelf life of a pharmaceutical product.
Several chemical reactions can affect the shelf life of pharmaceutical products and therates of these reactions depend on conditions such as pH, temperature, humidity.
Lysine acetylsalicylate (LAS) is prescribed as a pain killer and anti-inflammatory drug
under the brand name Aspegic. LAS on hydrolysis forms lysine salicylate and acetic acid.

Hydrolysis of LAS can proceed via three different pathways (a) acid catalysed,(b) uncatalysed and (c) base catalysed.
If [LAS] denotes the concentration of LAS at time ‘t’, the overall rate of the hydrolysis
reaction can be written as−𝑑[LAS]= k H [LAS][H + ] + k 0 [LAS] + k OH [LAS][OH − ]𝑑𝑡where kH, k0 and kOH are the rate constants of the acid catalysed, uncatalysed and basecatalysed pathways of hydrolysis, respectively. The observed rate constant is defined by:−
4.2.1.

𝑑[LAS]= k obs [LAS]𝑑𝑡

Write an expression for kobs in terms of kH, k0, kOH and [H+].

Hydrolysis of LAS was carried out at 298.15 K at various pH values (0.50 to 13.0). A very
low initial concentration of LAS ensured that the pH did not change during the course of
the reaction.

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The following graph shows the pH dependence of the hydrolysis of LAS.

4.2.2.

Which of the following is/are correct?
A) kobs ≌ k0 at pH = 12
B) kobs ≌ k0 at pH = 5.0
C) The rate of the reaction increases when the pH is changed from 0.50 to 1.0.
D) The rate of the reaction increases when the pH is changed from 10 to 12.

4.2.3.

Using the diagram and the data given below, calculate kH, k0 and kOH. Make sureto specify the units.
pH
1.300 5.300 12.180
log (kobs/min‒1) −3.886 −4.000 −1.726

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Acetylsalicylic acid, more commonly known as aspirin, is a medicine often used for
reducing fever, pain and inflammation. Like LAS, the hydrolysis of aspirin can also takedifferent pathways depending on the pH. The pH rate profile of aspirin hydrolysis at
333.15 K is given below:

The following are possible reactions for the hydrolysis of aspirin. Depending on the pH,
one or more of these reactions will predominate.
I.
CH3COOC6H4COOH + H3O+ ⟶ HOC6H4COOH + CH3COOH + H+
II.
CH3COOC6H4COOH + H2O ⟶ HOC6H4COOH + CH3COOH
III.
CH3COOC6H4COOH + OH− ⟶ HOC6H4COO− + CH3COOH
IV.
CH3COOC6H4COO− + H3O+ ⟶ HOC6H4COOH + CH3COOH
V.
CH3COOC6H4COO− + H2O ⟶ HOC6H4COO− + CH3COOH
VI.
CH3COOC6H4COO− + OH− ⟶ HOC6H4COO− + CH3COO−
4.3.1. Using the pH-rate profile diagram and the reactions given above, state which of
the following statements is/are correct. (pKa of aspirin = 3.57 at 333.15 K)a) In the region C-D, reaction IV is predominantb) In the region C-D, reaction V is predominantc) In the region D-E reaction VI is predominantd) In the region A-B, reaction II is predominant
A separate plot of kobs vs pH for the hydrolysis of aspirin has been confirmed to show aminimum at a particular pH. At 290.15 K the following rate constants for reactions I, II
and III were determined:kH (reaction I)
k0 (reaction II)
kOH (reaction III)
3
−1
−1
−4
−1
1.99 dm ∙mol ∙day
2.29∙10 day
3.18∙109 dm3∙mol−1∙day−1
The ionic product of water at 290.15 K can be taken as 1.0∙10−14.
4.3.2.

Assuming that only reactions I, II and III occur, calculate the value of the pH at
the minimum of kobs.

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Problem 5

8% of the total

5500 years ago in ancient Egypt people learned for the first time how to synthesize ablue pigment. Now we know this pigment as Egyptian blue. About 2000 years later inancient China another pigment was widely used, which is now referred to as Chinese
blue. The two pigments are similar in structure, but have different elemental
compositions.

Ushabti figurines from Egyptian pharaoh tomb covered with Egyptian blue anda Chinese blue soap dispenser sold at Alibaba
The ancient method of preparation for these pigments can be easily reproduced in amodern laboratory.
When considering the amounts, assume that all of the compounds in this task are pure,and the yields are quantitative.
To make Egyptian blue, one should heat 10.0 g of mineral A with 21.7 g of SiO2 and9.05 g of mineral B at 800–900°C for a prolonged time. 16.7 dm3 of a mixture of two
gaseous products are released (the volume is measured at 850°C and 1.013∙105 Pa(1.013 bar) pressure. In result, 34.0 g of the pigment was obtained. No other products
are formed. As the gas mixture is cooled, one component of the mixture condenses. As
the remaining gas is further cooled to 0°C, the gaseous volume reduces to 3.04 dm3.
5.1.1.

Find the mass of the gaseous mixture formed upon heating of A with B and SiO2.

5.1.2.

Determine the quantitative composition of this gas mixture.

When 10.0 g of mineral A is heated with 21.7 g of SiO2 in the absence of B, it forms
8.34 dm3 of gaseous products (measured at 850°C and 1.013∙105 Pa = 1.013 barpressure). Mineral A contains only one metal.
5.1.3.

Calculate the molar mass and determine the formula of mineral B. Hint: it is anionic solid insoluble in water and containing no water of crystallization.

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In order to obtain Chinese blue, one should take 17.8 g of mineral C instead of mineral
B (keeping the amounts of mineral A and SiO2 same as for Egyptian blue), and run thereaction at higher temperatures. Besides the pigment, the same gaseous products in thesame quantities are formed as in the preparation of Egyptian blue.
5.1.4.

Determine the formula of mineral C.

5.1.5.

Determine the formulae of Egyptian blue and Chinese blue.

5.1.6.

Determine the formula of mineral A.

Elemental analysis of some samples of Chinese blue shows traces of sulfur. This led to aconclusion that those were synthesized using another common mineral instead of C.
5.2.1.

Suggest a formula for the mineral used in place of C.

5.2.2.

Could the temperature of synthesis of Chinese blue be decreased if this mineralis used instead of C?

If during the synthesis of Chinese blue we take a smaller amount of silica than in theprocess above, we will obtain a purple pigment: Chinese violet. It was used, in particular,
for coloring the famous Terracotta army soldiers.

Terracotta army from Xian, China and reconstruction of its original coloring
5.3.

Write down the formula of a binary compound that forms under the conditions
required for Chinese violet and is responsible for the change of the color.

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Problem 6

7% of the total

Although there is currently no known cure for Alzheimer’s disease, there are
medications available to manage the neurodegenerative disorder. Among these areacetylcholinesterase inhibitors, of which galantamine 1 is an example. This molecule canbe isolated from the Caucasian snowdrop, a plant native to Georgia; however, the large
amounts needed for therapy require a synthetic route. Shown below is the route used toprepare galantamine industrially.

excessresolution

Notes about the synthesis:

1H NMR of A indicates 2 aromatic protons in a para arrangement.



C is labile in aqueous conditions, so it is not isolated, but rather reacted
immediately with NaBH4 to convert it to D.

6.1.1.

Suggest structures for A, B, C, D, F, and G. None of the reactions except for the
final transformation with L-selectride are stereoselective. Therefore,
stereochemistry does not need to be indicated in your answers.

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6.1.2.

Give the formula for a possible reagent, X, to convert compound D to E.

The optical rotation of the material obtained by resolution was –400° cm2 g-1, while thatof the enantiomerically pure compound is –415° cm2 g-1 when measured under the same
conditions. You may assume that the only optical impurity is the other enantiomer.
One way of describing optical purity is enantiomeric excess (ee). It is defined as thedifference in the percentages of the enantiomers in a mixture. For example in a mixtureof 70% R and 30% S, the ee is 40%.
6.2.1.

What is the enantiomeric excess of the resolved compound as prepared by theindustrial route?

L-selectride is a commercial reagent that performs the final reaction stereoselectively.
6.2.2.

Assign the labelled stereocentres () in (–)-1
as R or S.

6.2.3.

Give the formula for a reagent that carries out thesame reaction as L-selectride, converting H to 1.
You need not worry about stereoselectivity.

An alternative route to galantamine occurs with theseven-membered ring being the last ring to form.

(4 equivalents)

pH = 56.3.1.

Give the formula for compound Y to carry out the first step of the route.

6.3.2.

Suggest structures for J and K.

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Problem 7

8% of the total

This question looks at the synthesis of dolasetron
mesylate, Z (shown right), a drug sold under the
tradename Anzemet and used to treat postoperative nausea and vomiting.

The synthesis begins as shown below.

First cyclic compound A is made, which contains C, H, and O only. Compound G is achiraland can be prepared directly from D using ozone under reductive conditions, or via
stereoisomers E1 and E2 using OsO4, or via stereoisomers F1 and F2 using the peracid
shown.
7.1.

Determine the empirical formula of G from the percentage masses given.

7.2.

Give the structures of A, B, C, D, E1, E2, F1, F2 and G.

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Compound G is used in the next stage of the synthesis, under buffered conditions, toform H (as a mixture of two achiral diastereoisomers). Reduction of H with NaBH4 gives
alcohol I (as a mixture of four achiral diastereoisomers). I reacts with acidified
dihydropyran to form J (as a mixture of even more diastereoisomers). J is then treatedfirst with t-butoxide base, then refluxed with acid before finally extracting under weakly
basic conditions to form K as a mix of two diastereomers, K1 (major product) and K2(minor product). These could be separated, and K1 was used in the final stages of thesynthesis.

7.3.1.

Give the structures of H, I, and J. There is no need to show the different
diastereoisomers formed.

7.3.2.

Give the structures of diastereoisomers K1, and K2.

In the final stage of the synthesis, L and M react to form intermediate N. N then reacts
with K1 to form, after extraction, the neutral amine which gives the target compound
upon protonation with CH3SO3H.

7.4.

Give the structure of N.

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Problem 8

7% of the total

An exotic, but biologically relevant sugar analogue can be prepared from D-glucose in thefollowing manner. Heating a mixture of D-glucose and acetone with a few drops ofconcentrated acid results in the formation of a diacetonide A. Then A can be hydrolyzed
selectively to B.

Acetone/H+

D-glucose

8.1.1.

A

B

Which of the following sentences is true?
A is an α isomer.
A is a β isomer.

8.1.2.

X

A is neither α nor β.
A is a mixture of α and β isomers.

Which of the following sentences is true?
We can get product A only if we use α-D glucose as starting material.
We can get product A only if we use β-D glucose as starting material.
We can get product A either from α- or from β-D glucose as starting material.

8.1.3.

Which one of these reagents can be utilized as X for the selective hydrolysis ofA?
50% acetic acid6 M HCl in water
6 M HCl in acetic acid8.1.4.

concentrated H2SO4
1 M NaOH in water

Which is the stereochemically correct structure for compound B?

Neither of these

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B is treated with sodium metaperiodate to get C. C is then reacted with an aqueous
solution of NaCN, then heated with 10% NaOH solution to get a mixture of two
diastereomeric compounds D1 and D2. These compounds can be separated by column
chromatography.
1. NaCN
IO4−

2. NaOH/H2O

BC

D1 + D2

188.2 g/mol
Reaction of D1 with LiAlH4 followed by heating with 1M HCl solution gives sugar F that is
the hydrolysis product of the most abundant natural polysaccharide.
LiAlH4D1
8.2.1.

1M HCl/H2O
E

F

Draw the structures of C, D1, D2, E and F including stereochemical information.
Show F as the more stable 6-membered ring containing isomer using the ringskeleton. Indicate with a wavy line if absolute chirality around a carbon is notknown.

8.2.2. The reaction sequence from glucose to F does not seem to be useful. In somecases, however, this is the most economical way to produce F. In which case?
13C labelling at carbon 6 of F
13C labelling at carbon 5 of F
13C labelling at carbon 1 of F
15O labelling at glycosidic OH of F

synthesis of an uncommon isomer of F
Neutralization of D2 with HCl followed by heating in toluene results in dehydration andformation of G, which has a tricyclic structure in water-free solvents. Boiling G in 1M HCl
solution gives H (C6H10O7), which is a natural sugar derivative containing a 6 memberedring. H is a building block of heparin, an anticoagulant polysaccharide produced by ourbodies.
1. Equimolar HCl
2. Heat, toluene
D2

1M HCl/H2O
G

H

−H2O
8.3.1.

Draw the structure of G including the stereochemistry.
Draw H as the more stable 6-membered ring containing isomer using the ringskeleton. Indicate with a wavy line if absolute chirality around a carbon is notknown.

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8.3.2.

How are the rings of G fused together?
both junctions cis
one cis and one trans junction
both junctions trans8.3.3.

Which of the following is true for H? (You can choose more than one option.)
H is a reducing sugar (reacts with Fehling’s reagent)
H is an aldaric acid (dicarboxylic derivative of an aldose)
H is an aldonic acid (C1 is a carboxyl group)
H is a uronic acid (C5 or C6 is a carboxyl group)
H is a ketose
H is a lactone
H is a D sugar
H is achiral
H is a meso compound

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48th International
Chemistry Olympiad

Theoretical Problems
Answer sheets

28 July 2016
Tbilisi, Georgia
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Problem 1

5% of the total

1.1. 1.2. 1.3. 1.4. 1.5. 1.6. 1.7. Sum
3
1
2
2
2
4
3
17

1.1.

On which electrode does nitrogen trifluoride form?
CathodeAnodeWrite a balanced chemical equation for the electrode half reaction for theformation of NF3.
NH4+ + 3 F– ⟶ NF3 + 4 H+ + 6 e– or NH4F + 2 HF ⟶ NF3 + 6 H+ + 6 e– or equ.
3p (1p choosing anode, 1p species, 1p coefficients)
1.2.

Which of NF3, NHF2 or NH2F compound is expected to condense at the lowest
temperature?
NF3

1.3.

NH2F

NF3

Assign the N-F bond lengths (136, 140, 142 pm) to the molecules.
Molecule
N-F bond
length, pm

1.4.

NHF2

NH2F

NHF2

NF3

The partial positive charge on N
increases in this order, so the attraction
also increases between the N and F.

Write a balanced chemical equation for the formation of the binary nitrogen –fluorine compound.
2 NHF2 + 2 KF ⟶ N2F2 + 2 KHF2 or 2 NHF2 ⟶ N2F2 + 2 HF
2p (1p N2F2, 1p equation)

1.5.

Propose a suitable reagent for the formation of NF4+ and write a balanced
chemical equation for the reaction.
NF3 + F2 + SbF5 ⟶ NF4+ + SbF6−

any strong fluoride acceptor (AsF5, BF3)

2p (1p species, 1p coefficients)

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1.6.

Write a balanced chemical equation for the hydrolysis of NF4+.
2 NF4+ + 2 H2O ⟶ 2 NF3 + O2 + 2 HF + 2 H+ ⇒ n(O2):n(NF3) = 1:2
2p (1p species, 1p coefficients)

Write a balanced chemical equation for a possible side reaction that candecrease the theoretically expected O2:NF3 mole ratio.
e.g.: NF4+ + 2 H2O ⟶ NF3 + H2O2 + HF + H+

HOF, O3, OF2 also accepted.

2p (1p species, 1p coefficients)
1.7.

Determine the formula of the salt in question.

Your work:
From the NF3:F2 ratio it is clear that the anion also contains fluorine.
Moreover, the starting NF4+: liberated F2 ratio is 1:2, and all fluorine content
is released. With a (NF4)xAFy composition, x:y = 1:4.
8𝑥𝑀(F)
𝑀(A)= 0.656 ⇒= 65.7 g/mol
8𝑥𝑀(F)+𝑥𝑀(N)+𝑀(A)𝑥With x=1, the atomic mass of A is close to zinc, but not all fluorine would be
released with Zn. If x=2, then the element is xenon, and the formula is(NF4)2XeF8
3p (1.5p for Zn)
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Problem 2
2.1.1
5

2.1.2
3

8% of the total

2.1.3 2.2 2.3 2.4 2.5 2.6.1
2
2
3
6
2
1

2.1.1.

Which of the atoms (A or B) is copper?

Cu:

There are 2 A atoms and 4 B atoms in the cell. Cu: B

2.6.2
3

2.6.3 Sum
6
33

Which basic structure (primitive cubic, face centered cubic, body centered cubic,
diamond) is formed by the A atoms and which structure is formed by the B
atoms?
pr. cubic

fcc

bcc

diamondA
B

B: fcc A: bcc

What are the coordination numbers of the atoms?
A:
2.1.2.

B:

B: 2

A: 4

Calculate the smallest O-O, Cu-O and Cu-Cu distances in the structure.

Your work:
O-O:

O-O: half of the cell body diagonal 1/2∙427.0 pm∙√3 = 369.8 pm.
Cu-O: 1/4th of the cell body diagonal 1/4∙427.0 pm∙√3 = 184.9 pm.

Cu-O:

Cu-Cu: half of the face diagonal: 1/2∙427.0 pm∙√2 = 301.9 pm.
1p each, no penalty for Cu and O switched.

Cu-Cu:

2.1.3.

What is the density of pure copper(I) oxide?

Your work:
The volume of the unit cell is (427.0 pm)3.
The mass of a unit cell is (4MCu + 2 MO)/ NA. The density is 6.106 g/cm3.
2p

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density:
2.2.

What percentage of normal copper sites are empty in the crystal sample?

Your work:
From 1000 coppers 998 atoms are Cu(I) and 2 are Cu(II). To balance thecharge of the anions, there has to be 2 vacant Cu sites.
The percentage of empty sites is: 2/1002 ≈ 0.2%
1ppercentage:
What is x in the empirical formula Cu2-xO of the crystal?
Your work:
0.2% of 2 coppers is missing. That is 0.004.
1p
x:
2.3.

Write balanced chemical equations for reactions (1-3).

Reaction 1:(1) 2 Cu2O + O2 ⟶ 4 CuO (copper(II)-hydroxides and carbonates accepted)(2) Cu2O + 2 H+ ⟶ Cu + Cu2+ + H2OReaction 2:(3) Cu2O + H2SO4 + 4 H+ ⟶ 2 Cu2+ + SO2 + 3 H2O
3p (non-ionic equations also accepted)
Reaction 3:

2.4.

Determine the temperature ranges, if any, of thermodynamic stability of copperand its oxides between 500 and 1500 K in a 105 Pa oxygen atmosphere.

Your work:
ΔfG(CuO) = −156000 J mol−1 + 84 J mol−1 K−1 T
ΔfG(Cu2O) = −170000 J mol−1 + 72 J mol−1 K−1 T
Both Gibbs energies of formation are negative in the range 500-1500 K, so
the oxides are more stable than the elements. Cu is not stable.
Looking at the conversion process: 2 CuO = Cu2O + 0.5 O2
ΔrG = 142000 J mol−1 −96 J mol−1 K−1 T
The Gibbs energy of the reaction is negative above 1480 K.
CuO is stable below 1480 K, Cu2O above 1480 K.
6 p (1 for each ΔG or equivalent calculation, 1 for each conclusion)

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Temperature range between 5001500K

2.5.

The most stable form (Cu, Cu2O or
CuO)

Write the half reaction equations for the electrode processes during the anodic
production of Cu2O in NaOH solution with a platinum cathode and copper
anode.

Cathode:C: 2 H2O +2 e– ⟶ H2 + 2 OH–
A: 2 Cu + 2 OH– ⟶ Cu2O + H2O + 2e–

Anode:

2.6.1.

2p
Write the half reaction equation of the cathode process giving Cu2O in acidic
medium.
2 Cu2+ + H2O + 2 e– ⟶ Cu2O + 2 H+
1p

2.6.2.

What is the maximum pH at which the concentration of copper(II) can be
maintained at 0.100 mol dm−3?

Your work:
If the pH is too high, Cu(OH)2 precipitates.
Ksp ≥ 0.1[OH–]2⇒

pH ≤ 5.15

3 p (1p realizing that precipitation occurs, 1p equation, 1p result)maximum pH:

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2.6.3.

What is the minimum pH at which the cathodic production of Cu2O in a
0.100 mol dm−3 Cu2+ solution is still possible?

Your work:
The potential of the cathodic process (2 Cu2+ + H2O + 2 e– = Cu2O + 2 H+)
depends on the pH.
The standard potential of the cathodic process can be calculated from:
Cu2O(s) + H2O(l) + 2 e– ⟶ 2 Cu(s) + 2 OH– (aq) ΔGo1 = –2F(–0.36 V)
Cu2+(aq) + 2 e– ⟶ Cu(s)ΔGo2 = –2F(+0.337 V)
H2O(l) ⟶ H+(aq) + OH– (aq)
ΔGo3 = –RT lnKw
For 2 Cu2+(aq) + H2O(l) + 2 e–⟶ Cu2O(s) + 2 H+(aq):
ΔGo = –ΔGo1 + 2ΔGo2 + 2ΔGo3
Eo = –ΔGo /2F = 0.36 V + 2∙0.337 V – (RT/F) ∙lnKw = 0.208 V
The concentration dependence of the cathodic Cu2O production potential:
E = 0.208 V + 0.059/2 log ([Cu2+]2/[H+]2)
This potential has to be higher than the potential of the reduction of Cu2+.
0.337 + 0.059/2 log [Cu2+] = 0.208 + 0.059/2 log ([Cu2+]2/[H+]2)[H+]2 = [Cu2+]/23600
pH = 2.69

6 p (3p for standard potential, 3p for calculation)
Many other routes possible – all correct ones accepted.

minimum pH:

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Problem 3

9% of the total

3.1.1. 3.1.2. 3.1.3. 3.2.1. 3.2.2. 3.2.3. 3.3.1. 3.3.2. 3.3.3. Sum
2
2
6
5
4
7
6
11
5
48

3.1.1.

Write a balanced net ionic equation for the reaction when iodate reacts withexcess iodide in acidic solution.
IO3− + 8 I−+ 6 H+  3 I3−+ 3 H2O or IO3− + 5 I−+ 6 H+  3 I2+ 3 H2O
2p (1p species, 1p coefficients, 0.5p penalty if spectator ions are included.)

3.1.2.

Write a balanced net ionic equation for the reaction taking place during thetitration with thiosulfate.
I3− + 2 S2O32−  3 I−+ S4O62− or I2 + 2 S2O32−  2 I−+ S4O62−
2p (1p species, 1p coefficients, 0.5p penalty if spectator ions are included.)

3.1.3.

Calculate the iodization level, in ppm, of this salt sample.

Your work:(0.00750 dm3 titrant)∙(0.00235 mol dm−3 S2O32−) = 1.76∙10−5 mol S2O32−(1.76∙10−5 mol S2O32−)∙(1 mol IO3–/6 mol S2O32−) = 2.94∙10−6 mol IO3−
(2.94∙10−6 mol IO3–)∙(126.90 g/mol) = 3.73∙10−4 g iodine{(3.73∙10−4 g iodine)/(10.00 g salt)} ∙106 ppm = 37.3 ppm iodine
2p for calculating mol thiosulfate
2p for calculating mol iodate
2p for converting to ppm

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3.2.1.

Select an appropriate data point from the experiments and use it to calculate thesolubility product of AgI (KspI).

Your work:
In the experiments, |E| = –(RT/nF)ln([Ag+]cell/[Ag+]ref) = −0.0591∙log[Ag+]
There is a sharp endpoint at n(added Ag+) = n (I– initially present) inexperiment A ([Cl–] = 0), so precipitation of AgI(s) must be essentiallycomplete at any point in the titration curve. If one considers, for example,
5.0 mol added Ag+, then[I–] = [I–]0 − 5.0∙10–6 mol/dm3 = 5.0∙10–6 mol/dm3|E| = 0.622 V = −0.0591∙log[Ag+] ⇒ [Ag+] = 3.2∙10–11 mol/dm3
KspI = [Ag+][I–] = 1.6∙10–16
2p for relationship between E and [Ag+] (full credit if used even if not statedexplicitly)
1p for selecting a data point with n(Ag+) ≤ 10.0 μmol
2p for calculation of KspI

KspI:
3.2.2.

Full marks are awarded here and later for correct answers with numericaldifferences stemming from using different data points or minor rounding
errors.
Select an appropriate data point from the experiments and use it to calculate thesolubility product of AgCl (KspCl).

Your work:
In the titration with [Cl−] = 0.100 mol/dm3 (experiment B), the fact that thepotential stops changing at n(added Ag+) ≥ 11.8 μmol must be due to theprecipitation of AgCl(s) (the high concentration of chloride therefore
effectively fixes the [Ag+] in the solution). So in this regime:|E| = 0.517 V = −0.0591∙log[Ag+] ⇒ [Ag+] = 1.6∙10–9 mol/dm3
KspCl = [Ag+][Cl–] = 1.6∙10–10
1p for selecting a data point with n(Ag+) > 11.8 μmol
3p for calculation of KspCl

KspCl:

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3.2.3.

Select an appropriate data point from the experiments and use it to calculate Kf.
You may need to use values of KspI or KspCl to do this calculation. If you wereunable to carry out the calculations in 3.2.1. or 3.2.2., you may use the arbitraryvalues of KspI = 1.00∙10−15 and KspCl = 1.00∙10−9 without penalty.

Your work:
There are a number of good approaches to this problem.
One can use the point at which AgCl(s) first precipitates (estimated at
11.8 μmol Ag+ added) to calculate Kf. At this point, [Ag+] = 1.6∙10–9 mol/dm3,[Cl–] = 0.100 mol/dm3 (see above).
Almost all of the originally present 1.0∙10–5 mol/dm3 iodide has been
precipitated out as 9.9 μmol AgI, since [I−] = KspI/[Ag+] = 1.0∙10–7 mol/dm3
Total Ag in solution = 11.8 μmol – 9.9 μmol = 1.9 μmol
[AgCl2–] = 1.9∙10–6 mol/dm3 (since free [Ag+] is only 1.6∙10–9 mol/dm3)[AgCl2− ]1.9 ∙ 10−6
Kf === 1.2 ∙ 105 [1.9 ∙ 104 given const. ]
+
–
2
−9
2
[Ag ][Cl ]1.6 ∙ 10 ∙ 0.100
Same approach works for 11 μmol Ag+ added.
2p for selecting a data point between 10-12.0 μmol
3p for reasonable method of calculating Kf
2p for value of Kf
Full marks if errors in Ksp are propagated into an error in Kf
An alternative approach is to look at the equivalence point, where|E| = 0.558 V = −0.0591∙log[Ag+] ⇒ [Ag+] = 3.62∙10–10 mol/dm3
Since AgI(s) is present, [I–] = KspI/[Ag+] = 4.42∙10–7 mol/dm3
The amount of dissolved iodine and silver is equivalent:[I–] = [Ag+] + [AgCl2–] ⇒ [AgCl2–] = 4.42∙10–7 mol/dm3
Kf =[AgCl2− ]4.42 ∙ 10−7=
= 1.2 ∙ 105 [7.6 ∙ 105 given const. ][Ag + ][Cl– ]2 3.62 ∙ 10−10 ∙ 0.1002

Kf:

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3.3.1.

Write balanced net ionic equations for the reaction of cerium(IV) with H3AsO3 inacidic solution, as well as reactions of cerium(IV) with a species containing theelement iodine and H3AsO3 with a species containing the element iodine, that
could reasonably account for the catalysis of the net reaction by iodide.

Net reaction of cerium(IV) with H3AsO3 in acidic solution:
2 Ce4+ + H3AsO3 + H2O  2 Ce3+ + H3AsO4 + 2 H+
2p (1p for H3AsO3/H3AsO4 couple, 1p balanced)
Reaction of cerium(IV) with an iodine-containing species:
2 Ce4+ + 2 I–  2 Ce3+ + I2
2p (1p species, 1p balanced). Full marks for I3– or I∙ as products
Reaction of H3AsO3 with an iodine-containing species:
H3AsO3 + I2 + H2O  H3AsO4 + 2 I– + 2 H+
2p (1p species, 1p balanced). Full marks for I3– or I∙as products
Two iodine-containing reactions must add up to the net reaction, otherwise–2p for the iodine-containing reactions
3.3.2.

Determine the integer values of m, n, and p and calculate the value of k (be sureto specify its units).

Your work:
The limiting reactant is Ce(IV) which is < 10% of the concentration of
H3AsO3, so only the concentration of Ce(IV) changes appreciably over thecourse of the reaction. (I− is a catalyst and is not consumed.) So the order inCe(IV) can be judged by the time course of the reaction. By eye, it appears to
be first order.

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This can be verified by calculating –ln(A/A0)∙(1/t), which should be aconstant (kobs) if the reaction is first-order:t, s kobs, s−1, Run 1 kobs, s−1, Run 2 kobs, s−1, Run 3
20
0.02900.03280.014840
0.0286
0.03100.014760
0.02840.03050.014680
0.02840.03050.0146100
0.02820.03020.0145avg.
0.02850.03100.0146
So n = 1.
Since kobs is unchanged (within 10%) from run 1 to run 2 despite
decreasing [H3AsO3] by a factor of two, m = 0.
In contrast, decreasing [I–] by a factor of two from run 1 to run 3 results in adecrease in observed rate constant of a factor of two, so p = 1.
k = kobs/[I–], giving values of 1.99∙104, 2.17∙104, and 2.04∙104 dm3 mol−1 s−1
for runs 1-3; average k = 2.07∙104 dm3 mol−1 s−1.
4p for documenting 1st-order in Ce(IV), 2p each for m and p,
2p for value of k, 1p for unit of k consistent with given rate law
m=
3.3.3.

n=p=k=Calculate the iodization level, in ppm, of the salt sample.

Your work:t, s –ln(A/A0)∙(1/t), s−1
200.0177400.017760
0.017780
0.0176100
0.0177−1
So kobs = 0.0177 s = k[I–] = (2.07∙104 dm3 mol−1 s−1)[I–][I–] = 8.55∙10−7 mol/dm3
Since the salt solution was diluted by a factor of (2.05 cm3)/(0.050 cm3) =41, the concentration in the original salt solution was
41∙(8.55∙10−7 mol/dm3) = 3.51∙10−5 mol/dm3.
(3.51∙10−5 mol dm−3)(0.01000 dm3) = 3.51∙10−7 mol I in the salt sample(3.51∙10−7 mol iodine)(126.90 g/moI) = 4.45∙10−5 g iodine{(4.45∙10−5 g iodine)/(1.000 g salt)}∙106 ppm = 44.5 ppm I
2p for calculating kobs, 1p for [I–] from kobs, 2p for converting to ppm
ppm I =

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Problem 4

8% of the total

4.1.1 4.1.2 4.1.3 4.2.1 4.2.2 4.2.3 4.3.1 4.3.2 Sum
5
3
2
1
2
6
2
4
25

4.1.1.

Calculate the time in minutes the oxidation process has to continue in order toreach the accepted level of dioxane at 303.15 K if the initial concentration of
K2S2O8 is 5.0∙10−6 mol/dm3. Assume that the rate law obtained from the data
above is valid under these conditions.

Your work:
Trial1 and 2 ⇒the rate doubles when concentration of K2S2O8 is doubled.
Order w.r.t. K2S2O8 =1.
Trial3 and 4 ⇒the rate does not change with concentration of dioxane.
Order w.r.t. dioxane =0.
Initial rate = k[K2S2O8]

Average k = 6.65∙10−3 min−1

Change in dioxane concentration: (40.00 −0.35) μg∙dm−3 / 88.1 g∙mol−1 =0.450 μmol∙dm−3[K2S2O8]0 = 5.00 μmol dm−3; [K2S2O8]t = 4.55 μmol dm−3
4.55 = 5.00 exp(−6.65∙10−3 t)
t = 14.2 minutes
5p (1p for each order, 1 p for k, 2p for final answer)

Oxidation time:
4.1.2.

Assuming Ag(III) to be in steady state, deduce the rate equation for theoxidation of dioxane.

Your work:

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−2 2
+
+3
+3
𝑘1 [S2 O−2
8 ][Ag ] − 𝑘2 [SO4 ] [Ag ] − 𝑘3 [Ag ][D] = 0[Ag +3 ] =+𝑘1 [S2 O−2
8 ][Ag ]𝑘2 [SO−2
4 ] + 𝑘3 [D]+𝑘1 𝑘3 [S2 O−2
8 ][D][Ag ]𝑟𝑎𝑡𝑒 𝑜𝑓 𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛 =2
𝑘2 [SO−2
4 ] + 𝑘3 [D]

3p (1p for each step)

4.1.3.

A,C

2p, (1p penalty for every mistake)4.2.1.

Write an expression for kobs in terms of kH, k0, kOH and [H+].

Your work:𝑘obs = 𝑘H [H + ] + 𝑘0 + 𝑘OH

𝐾w[H + ]

1p, (0.5p if [OH−] is included in the expression)

B, D

4.2.2.
4.2.3.

2p, (1p penalty for every mistake)Using the diagram and the data in the table, calculate kH, k0 and kOH. Make sureto specify the units.

Your work:
At pH = 5.30, k0 is dominantlog(kobs/min‒1)= −4.000 ⇒ kobs = k0 = 1.00∙10−4 min−1
At pH = 12.18, [OH−] = 0.01514 mol dm−3log(kobs/min‒1)= −1.726 ⇒ kobs = 1.88∙10−2 min−1
kobs = k0 + kOH[OH−]k0 can be neglected
kOH = kobs/[OH−] = 1.24 dm3∙mol−1∙min−1
At pH = 1.30, [H+] = 0.0501 mol dm−3log(kobs/min‒1)= −3.886 ⇒ kobs = 1.30∙10−4 min−1
kobs = kH[H+] + k0
k0 cannot be neglected𝑘H =𝑘obs − 𝑘0
3.0 ∙ 10−5 min−1=[H + ]
0.0501mol∙dm−3

kH = 6.0∙10−4 dm3∙mol−1∙min−1
6p (1p for k0, 2p for kOH, 3p for kH, 0.5p penalty for wrong or no units)

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k0:

kOH:

kH:

4.3.1.

b, c

2p, (1p penalty for every mistake)4.3.2.

Assuming that only reactions I, II and III occur, calculate the value of the pH at
the minimum of kobs.

Your work:𝑘obs = 𝑘H [H + ] + 𝑘0 + 𝑘OH

𝐾w[H + ]

This is a minimum if
𝑑𝑘obs
𝐾w
=
𝑘
−
𝑘
=0
H
OH
[H + ]2𝑑[H + ]
[H + ]min = √𝑘OH 𝐾w𝑘H

𝑝Hmin =

1
1
𝑘H
𝑝K W + log= 2.402
2𝑘OH

or see alternative solution without calculus
4p (3p for expression of [H+] at minimum, 1p for numerical result)𝑘kobs is a minimum if 𝑘H [H + ] + 𝑘OH [Hw+] is minimal.
The minimum happens when the two terms are equal. So at minimum:𝑘H [H + ] = 𝑘OH[H + ]min = √𝐾w[H + ]𝑘OH 𝐾w𝑘H

𝑝Hmin =1𝑘H
𝑝K W + log= 2.402𝑘OH

pH at minimum:

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Problem 5

8% of the total

5.1.1. 5.1.2. 5.1.3. 5.1.4. 5.1.5. 5.1.6. 5.2.1. 5.2.2. 5.3. Sum
1
3
2
2
5
2
1
1
2
19

5.1.1.

Find the mass of the gaseous mixture formed upon heating of A with B and SiO2.

Your work:m = 10.0 + 21.7 + 9.05 – 34.0 = 6.75 g
1p

5.1.2.

Determine the quantitative composition of this gas mixture.

Your work:
At 850°C, the number of moles of gaseous products is:𝑛1 =𝑝𝑉101325 Pa ∙ 0.0167 m3== 0.181 mol
𝑅𝑇 8.314 Jmol−1 K −1 ∙ (850K + 273K)

while at 0°C
𝑛2 =𝑝𝑉 101325 Pa ∙ 0.00304 m3== 0.136 mol
𝑅𝑇 8.314 Jmol−1 K −1 ∙ 273 K

The difference of 0.045 moles is probably water that has condensed. Hence,
the mass of the gas at 0°C is m = 6.75–0.045·18 = 5.94 g, and the molar massis m/n2 ≈ 44 g/mol. Taken into account that we dealt with minerals, and thetemperature was high, we can conclude that the rest of the gas is CO2.
The gas formed at 850°C contains 0.045 mol of H2O and 0.136 mol of CO2.
3p (0.5p for finding each compound, 1p for the quantity of each compound)

The gaseous mixture formed at 850°C contains ……… mol of ………. and ……… mol of ………

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5.1.3.

Calculate the molar mass and determine the formula of mineral B. Hint: it is anionic solid insoluble in water and containing no water of crystallization.

Your work:
Heating of mineral A with SiO2 alone produces half of the gases evolved in
the presence of B. Thus, B should form the rest 0.181/2 = 0.0905 moles ofgases and is a carbonate of some metal. If it forms pure CO2, the molar massof B per carbonate group is 9.05 g / 0.0905 mol = 100 g mol−1. The molar
mass of CO3 group is
60 g mol−1, so the mass of metal per carbonate group is 40 g mol−1.
This corresponds to Ca, B – CaCO3
2p

B is:
5.1.4.

Determine the formula of mineral C.

Your work:
Similar to the previous question, the molar mass of C per carbonate group is
17.8 g / 0.0905 mol = 197 g mol−1. The mass of metal per carbonate group is
137 g mol−1.
This corresponds to Ba, C – BaCO3.
2p

C is:

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5.1.5.

Determine the formulae of Egyptian blue and Chinese blue.

Your work:
Taking into account the molar ratios of known compounds, we can write ageneral equation of formation of Egyptian blue:
А + 2 CaCO3 + 8 SiO2 = pigment + 3 CO2 + H2O
The composition of Egyptian blue can be written as 2CaO·8SiO2·nMexOy.
Oxide MexOy forms from mineral A. Let us find the molar mass of nMexOy:𝑀(pigment) = 𝑀(2CaO ∙ 8SiO2 ) + 𝑀(𝑛Me𝑥 O𝑦 )𝑀(pigment) = 2𝑀(CaCO3 )𝑚(pigment)𝑚(CaCO3 )𝑀(𝑛Me𝑥 O𝑦 ) = 2𝑀(CaCO3 )34.0 g− 𝑀(2CaO ∙ 8SiO2 ) ≈ 159 g/mol
9.05 gTrying different values (at least from 1 to 3) of n, x, and y, we ensure thatthe only possibility is n = 2, x = y = 1, Me is Cu. This is supported by a blue
color of a pigment.
The formula of Egyptian blue is then CaCuSi4O10.
The formula of Chinese blue is BaCuSi4O10.
3p for proving presence of Cu (1.5 p for correct molar mass of oxide without
further advances)
2p for formulas of the pigments
Any correct way of calculation is fully marked

Egyptian blue is:5.1.6.

Chinese blue is:

Determine the formula of mineral A.

Your work:
Upon heating A turns into 2 CuO, 1 CO2 and 1 H2O.
It means that A is malachite Cu2CO3(OH)2.
2p

Upon heating A turns into 2 CuO, 1 CO2 and 1 H2O.
It means that A is malachite Cu2CO3(OH)2.

A is:

2p

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5.2.1.

Suggest the formula of the mineral used in place of C.

BaSO4 (the most stable compound containing Ba and S, mineral barite)
1p (BaS is soluble and reactive, can not be a mineral.)

5.2.2.

Could the temperature of synthesis of Chinese blue be decreased if this mineralis used instead of C? No. BaSO4 is more stable than BaCO3.
Yes

5.3.

No

1p

Write down a formula of a binary compound that forms under the conditions
required for Chinese violet and is responsible for the change of the color.

Cu2O (mixing red with blue gives purple)
2p

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Problem 6

7% of the total

6.1.1. 6.1.2 6.2.1. 6.2.2. 6.2.3. 6.3.1 6.3.2.
24
4
3
6
2
2
8

Sum
49

6.1.1.

Suggest structures for A, B, C, D, F, and G. None of the reactions except for the
final transformation with L-selectride are stereoselective. Therefore,
stereochemistry does not need to be indicated in your answers.

AB

0p if lacks 2 para protons
C

D

F

G
2p penalty if Br
removed
2p penalty if Br
present

6.1.2.
X

Give the formula for a possible reagent, X, to convert compound D to E.
CH3CH2OCOH or any other reasonable reagent. 4p (3p for formyl chloride)
Formic acid = 0p

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6.2.1.

What is the enantiomeric excess of the resolved compound prepared by theindustrial route?

Your work:
If x is the percentage of the (−) enantiomer:x(−415) + (1 – x)(415) = −400
x = 815/830
ee = (815/830) – (15/830) = 800/830 = 0.964 or 96.4%
3p (2p enantiomer composition, 1p ee calculation)ee:
6.2.2.

Assign the labelled stereocentres (α, β, γ) in (–)-1 as R or S.

α

β

γ

S
R
S
6.2.3. Give the formula for a reagent that carries out the same reaction as L-selectride,the conversion of H to 1. You need not worry about stereoselectivity.
NaBH4, LiAlH4, etc. 2p

6.3.1.
Y

Give the formula for compound Y to carry out the first step of the route.
CH3NH2 2p

6.3.2.

Suggest structures for J and K.

J

K
4p

3p

2p
0p for J and K if not in agreementwith the formula, full marks foropen form of K.

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Problem 7

8% of the total

7.1. 7.2. 7.3.1. 7.3.2. 7.4. Sum
2
36
16
8
4
66

7.1.

Determine the empirical formula of G from the percentage masses given.

Your work:
C2H3O
2p (1p for molecular formula)

7.2.

Give the structures of A, B, C, D, E1, E2, F1, F2 and G.

AB

4p (2pt if no de-carboxylation)(2p in mono-ester)

4p (0pt if still contains Cl)(2p penalty if H− is used as reducing
agent)CD

4p (allow carry forward from B)

4p (0p if remade A)(1p if pyridinium intermediate)

E1 and E2

and4p each (1p if OHs on different sides)(2p if no de-carboxylation; 1p if osmate ester)

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F1 and F2

and4p each(1p if no de-carboxylation; 1p for trans-diol; 0p if no de-carboxylation and diol)G

4p(0p if does not fit formula)7.3.1.

Give the structures of H, I, and J. There is no need to show the different
diastereoisomers formed.

H

8p(points for fitting formula, sensible chemistry, correct number of possible
diastereomers, and being achiral)I

4p(points for reduction of ketone, sensible chemistry, correct number of possible
diastereomers, and being achiral)

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J

4p(points for fitting formula, sensible chemistry, forming acetal, sensible number ofpossible diastereomers)
7.3.2

Give the structures of diastereoisomers K1, and K2.

K1

4p (2p if answer is K2)(answer must fit formula and be sensible)

K2

4 pt (2pt if answer is K1)(answer must fit formula and be sensible)

7.4.

Give the structure of N.

N

4p(full marks for anhydride of L)

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Problem 8

7% of the total

8.1. 8.2.1. 8.2.2. 8.3.1. 8.3.2. 8.3.3. Sum
4
10
1
4
1
2
22

8.1.1.

Which of the following sentences is true?
(x) A is an α isomer.
 A is a β isomer.

8.1.2.

 A is neither α nor β.
 A is a mixture of α and β isomers.

Which of the following sentences is true?
 We can get product A only if we use α-D glucose as starting material.
 We can get product A only if we use β-D glucose as starting material.
(x) We can get product A either from α- or from β-D glucose as starting
material.

8.1.3.

Which one of these reagents can be utilized as X for the selective hydrolysis ofA?
(x) 50% acetic acid 6M HCl in water 6M HCl in acetic acid8.1.4.

 concentrated H2SO4
 1M NaOH in water

Which is the stereochemically correct structure for compound B?

Neither of these
8.2.1.

(x) Draw the structures of C, D1, D2, E and F including stereochemical information.
Show F as the more stable 6-membered ring containing isomer using the ringskeleton. Indicate with a wavy line if absolute chirality around a carbon is notknown.

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C

D1

D2

E

2p for each correct
structure
1p for structures with one
minor error (e.g. wrong
configuration)
0p for more minor
mistakes or a more serious
one (COOH->COH,
unidentifiable chiral
center)1-1p if D1 and D2 swapped
2p if –COOH in DF

8.2.2. The reaction sequence from glucose to F does not seem to be useful. In somecases, however, this is the most economical way to produce F. In which case?
(x) 13C labelling at carbon 6 of F
 13C labelling at carbon 5 of F
 13C labelling at carbon 1 of F
 15O labelling at glycosidic OH of F
 synthesis of an uncommon isomer of F

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8.3.1.

Draw the structure of G including stereochemistry.
Draw H as the more stable 6-membered ring containing isomer using the ringskeleton. Indicate with a wavy line if absolute chirality around a carbon is notknown.

G

H

8.3.2.

How are the rings of G fused together?
(x) both junctions cis one cis and one trans junction
 both junctions trans8.3.3.

Which of the following is true for H? (You can choose more than one option.)(x) H is a reducing sugar (reacts with Fehling’s reagent) H is an aldaric acid (dicarboxylic derivative of an aldose) H is an aldonic acid (C1 is a carboxyl group)(x) H is a uronic acid (C5 or C6 is a carboxyl group) H is a ketose H is a lactone
 H is a D sugar
 H is achiral H is a meso compound
2p, 1p with only one mistake(extra or missing letter), 0p
with 2 or more mistakes

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References
This appears only in the published version, it is not printed for the students.
Problem 1
Christe, Karl O., and William W. Wilson. 1982. “Perfluoroammonium and Alkali-Metal
Salts of the heptafluoroxenon(VI) and octafluoroxenon(VI) Anions.” Inorganic Chemistry
21 (12): 4113–17. doi:10.1021/ic00142a001.
Problem 2
Figure is from:
Gan, J., V. Venkatachalapathy, B.G. Svensson, and E.V. Monakhov. 2015. “Influence ofTarget Power on Properties of CuxO Thin Films Prepared by Reactive Radio Frequency
Magnetron Sputtering.” Thin Solid Films 594 (November): 250–55.
doi:10.1016/j.tsf.2015.05.029.
Problem 4
Figures are adapted from:
Kamal, Oussama, Abderazzak Benlyamani, Farid Serdaoui, Mohammed Riri, Abdelmjid
Cherif, and Miloudi Hlaïbi. 2012. “Stability Studies of Lysine Acetylsalicylate (Aspirin
Derivative): Mechanisms of Hydrolysis.” Open Journal of Physical Chemistry 2 (2): 81–87.
doi:10.4236/ojpc.2012.22011.
Marrs, Peter S. 2004. “Class Projects in Physical Organic Chemistry: The Hydrolysis of
Aspirin.” Journal of Chemical Education 81 (6): 870. doi:10.1021/ed081p870.
Problem 6
Küenburg, Bernhard, Laszlo Czollner, Johannes Fröhlich, and Ulrich Jordis. 1999.
“Development of a Pilot Scale Process for the Anti-Alzheimer Drug (−)-Galanthamine
Using Large-Scale Phenolic Oxidative Coupling and Crystallisation-Induced ChiralConversion.” Organic Process Research & Development 3 (6): 425–31.
doi:10.1021/op990019q.
Trost, Barry M., Weiping Tang, and F. Dean Toste. 2005. “Divergent Enantioselective
Synthesis of (-)-Galanthamine and (-)-Morphine.” Journal of the American Chemical
Society 127 (42): 14785–803. doi:10.1021/ja054449+.
Problem 8
Shafizadeh, F., and M. L. Wolfrom. 1955. “Synthesis1 of L-Iduronic Acid and an Improved
Production of L -Glucose-6-C14.” Journal of the American Chemical Society 77 (9): 2568–
69. doi:10.1021/ja01614a072.

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