Problems

CHEMISTRY FOR A BETTER TOMORROW

July 25, 2020

52nd International Chemistry Olympiad, Istanbul, TURKEY

Instructions

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
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








Only write with pen. Your calculator should be non-programmable.
This examination has 9 problems.
You will have 5 hours to solve the exam.
Begin only when the START command is given.
All results must be written in the appropriate boxes in pen in the designated areas on theanswer sheets. Use the back of the exam sheets if you need scratch paper. Rememberthat answers written outside the answer boxes will not be graded.
Write relevant calculations in the appropriate boxes when necessary. Full marks will begiven for correct answers only when your work is shown.
The invigilator will announce a 30-minute warning before the Stop command.
You must stop working when the STOP command is given. Failure to stop writing willlead to the nullification of your exam.
The official English version of this examination is available on request only forclarification.

You are not allowed to leave your working place without permission. If you need any
assistance (broken calculator, need to visit a restroom, etc), raise hand and wait until an
invigilator arrives.

GOOD LUCK!
1

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems & Grading Information
ProblemTitle

1

Two Beauties of Turkey: the Van Cat andthe Ankara Cat
A Tale of a Reactive Intermediate(±)-Coerulescine
Symmetry Does Matter!
Konya, Carrot, Beta-Carotene, Vitamin-A,
Immune System, Vision
Thermodynamics through an Interstellar
Journey
Phthalocyanines
Boron Compounds and Hydrogen Storage
Quantification of Heavy Metal Ions
TOTAL

2
3
4
5
6
7
8
9

Total
Score
24

% of
Total
Score
8

77
51
66
100

10
8
10
14

80

12

85
58
100
641

12
14
12
100

2

52nd International Chemistry Olympiad, Istanbul, TURKEY

Table of Contents
Authors ....................................................................................................................................... 4
Physical Constants and Equations .............................................................................................. 5
Periodic Table of Elements ........................................................................................................ 6
1

H-NMR Chemical Shifts ........................................................................................................... 7

Typical Coupling Constants ....................................................................................................... 7
13

C-NMR Chemical Shifts .......................................................................................................... 8

IR Absorption Frequency Table ................................................................................................. 8
Problem 1. Two Beauties of Turkey: the Van Cat and the Ankara Cat ................................... 10
Problem 2. A Tale of a Reactive Intermediate ......................................................................... 13
Problem 3. (±)-Coerulescine..................................................................................................... 17
Problem 4. Symmetry Does Matter! ......................................................................................... 20
Problem 5. Konya, Carrot, Beta-Carotene, Vitamin-A, Immune System, Vision.................... 24
Problem 6. Thermodynamics through an Interstellar Journey ................................................. 28
Problem 7. Phthalocyanines ..................................................................................................... 32
Problem 8. Boron Compounds and Hydrogen Storage ........................................................... 36
Problem 9. Quantification of Heavy Metal Ions ...................................................................... 40

3

52nd International Chemistry Olympiad, Istanbul, TURKEY

Authors
ALANYALIOĞLU, Murat, Atatürk UniversityAYDOĞAN, Abdullah, İstanbul Technical University
BURAT, Ayfer Kalkan, İstanbul Technical University
DAĞ, Ömer, Bilkent UniversityDAŞTAN, Arif, Atatürk UniversityKILIÇ, Hamdullah, Atatürk UniversityMETİN, Önder, Koç UniversityÖZTÜRK, Turan, İstanbul Technical UniversitySARAÇOĞLU, Nurullah, Atatürk UniversityTÜRKMEN, Yunus Emre, Bilkent University
ÜNLÜ, Caner, İstanbul Technical University
YILMAZ, İsmail, İstanbul Technical University
YURTSEVER, Mine, İstanbul Technical University

Editor: SARAÇOĞLU, Nurullah, Atatürk University

4

52nd International Chemistry Olympiad, Istanbul, TURKEY

Physical Constants and Equations
Avogadro's number, 𝑁𝐴 = 6.0221 × 1023 𝑚𝑜𝑙 −1
Boltzmann constant, 𝑘𝐵 = 1.3807 × 10−23 𝐽𝐾 −1
Universal gas constant, 𝑅 = 8.3145 𝐽𝐾 −1 𝑚𝑜𝑙 −1 = 0.08205 𝑎𝑡𝑚 𝐿 𝐾 −1 𝑚𝑜𝑙 −1
Speed of light, 𝑐 = 2.9979 × 108 𝑚𝑠 −1
Planck's constant, ℎ = 6.6261 × 10−34 𝐽 𝑠
Faraday’s constant, 𝐹 = 9.6485 × 104 𝐶 𝑚𝑜𝑙 −1
Mass of electron, 𝑚𝑒 = 9.1093 × 10−31 𝑘𝑔
Standard pressure, 𝑃 = 1 𝑏𝑎𝑟 = 105 𝑃𝑎
Atmospheric pressure, 𝑃𝑎𝑡𝑚 = 1.01325 × 105 𝑃𝑎 = 760 𝑚𝑚𝐻𝑔 = 760 𝑡𝑜𝑟𝑟
Zero of the Celsius scale, 273.15 𝐾
1 picometer (pm) = 10−12 𝑚; 1Å = 10−10 𝑚; 1 nanometer (nm) = 10−9 𝑚
1 𝑒𝑉 = 1.6 × 10−19 𝐽
1 𝑐𝑎𝑙 = 4.184 𝐽
1 𝑎𝑚𝑢 = 1.6605 × 10−27 𝑘𝑔
Charge of an electron: 1.6 × 10−19 𝐶
Ideal gas equation: 𝑃𝑉 = 𝑛𝑅𝑇
Enthalpy:
𝐻 = 𝑈 + 𝑃𝑉
Gibbs free energy: 𝐺 = 𝐻 − 𝑇𝑆
𝛥𝐺 = ∆𝐺 0 + 𝑅𝑇𝑙𝑛𝑄0∆𝒓 𝑮𝟎 = −𝑅𝑇𝑙𝑛𝐾 = −𝑛𝐹𝐸𝑐𝑒𝑙𝑙
Entropy change:𝑞∆𝑆 = 𝑟𝑒𝑣, where qrev is heat for the reversible process
𝑇𝑉∆𝑆 = 𝑛𝑅𝑙𝑛 𝑉2 (for isothermal expansion of an ideal gas)
1

Nernst equation:

𝑅𝑇

0

𝐶

𝐸 = 𝐸 + 𝑛𝐹 𝑙𝑛 𝐶𝑜𝑥𝑖𝑑𝑎𝑡𝑖𝑜𝑛

𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛ℎ𝑐Energy of a photon: 𝐸 = 𝜆Integrated rate law
Zeroth-order:First-order:
Second order:[𝐴] = [𝐴]0 − 𝑘𝑡𝑙𝑛[𝐴] = 𝑙𝑛[𝐴]0 − 𝑘𝑡
1
[𝐴]

1

= [𝐴] + 𝑘𝑡
0−𝐸𝑎 /𝑅𝑇

Arrhenius equation: 𝑘 = 𝐴𝑒
Equation of linear calibration curve: 𝑦 = 𝑚𝑥 + 𝑛
Lambert–Beer equation: 𝐴 = 𝜀𝑙𝑐

552nd International Chemistry Olympiad, Istanbul, TURKEY

Periodic Table of Elements

1

18

1

H

atomic number

2

Symbol

2

1.008

13

14

15

16

17

atomic weight

He
4.003

3

4

5

6

7

8

9

10

Li

Be

B

C

N

O

F

Ne

6.94

9.01

10.81

12.01

14.01

16.00

19.00

20.18

11

12

13

14

15

16

17

18

Al

Si

P

S

Cl

Ar

26.98

28.09

30.97

32.06

35.45

39.95

31

32

33

Na Mg

3

4

5

6

7

8

9

10

11

12

25

26

30

22.99

24.31

19

20

21

22

23

24

27

28

29

34

35

36

K

Ca Sc

Ti

V

Cr Mn Fe Co

Ni

Cu Zn Ga Ge As Se

Br

Kr

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

83.80

37

38

39

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

Rb

Sr

Y

Zr

Nb Mo Tc Ru Rh Pd Ag CdIn

Sn Sb Te

I

Xe

85.47

87.62

88.91

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

Cs Ba 57-71

Hf

Ta

W

Re Os

Ir

Pt

Au Hg

Tl

Pb

Bi

Po

At

Rn

132.9

137.3

178.5

180.9

183.8

186.2

190.2

192.2

195.1

197.0

200.6

204.4

207.2

209.0

-

-

-

87

88

104

105

106

107

108

109

110

111

112

113

114

115

116

117

118

Fr

Ra 89-103 Rf

Db Sg Bh Hs Mt Ds Rg Cn Nh

Fl

Mc Lv

-

-

-

-

57

-

-

-

-

-

-

-

-

-

61

62

63

64

65

66

67

Ts Og

-

-

-

-

68

69

70

71

58

59

60

La Ce

Pr

Nd Pm Sm Eu Gd Tb Dy Ho

Er Tm Yb Lu

138.9

140.1

140.9

144.2

-

150.4

152.0

157.3

158.9

162.5

164.9

167.3

168.9

173.0

175.0

89

90

91

92

93

94

95

96

97

98

99

100

101

102

103

Ac Th PaU

Np Pu Am Cm Bk

Cf

Es Fm Md No

Lr

-

232.0

231.0

238.0

-

-

-

-

-

-

-

-

-

-

-

Copyright © 2018 International Union of Pure and Applied Chemistry
Reproduced by permission of the International Union of Pure and Applied Chemistry652nd International Chemistry Olympiad, Istanbul, TURKEY

1

H-NMR Chemical Shifts

Typical Coupling Constants

7

52nd International Chemistry Olympiad, Istanbul, TURKEY

13

C-NMR Chemical Shifts

IR Absorption Frequency Table
Functional
Group

Type of Vibration

Absorption Frequency
Region (cm–1)

Intensity(stretch, H-bonded)(stretch, free)(stretch)

3600–3200
3700–3500
1150–1050

strong, broadstrong, sharp
strongstretchbending

3000–2850
1480–1350

strongvariable

stretchbending
stretch3100–3010
1000–675
1680–1620

mediumstrongvariable

stretchstretchstretchstretch1400–1000
800–600
600–500
500

strongstrongstrongstrongstretch3300

stretch2260–2100

strong, sharp
variable, not present insymmetricalalkynes

Alcohol
O–H
C–O
AlkaneC–H
Alkene=C–H
C=C
Alkyl Halide
C–F
C–Cl
C–Br
C–I
Alkyne
C–H
CC
Amine
N–H

stretch3500–3300

medium (primary amines have
two bands; secondary amines
have one band, often very
weak)

8

52nd International Chemistry Olympiad, Istanbul, TURKEY

C–N
N–H
Aromatic
C–H
C=C
Carbonyl
C=O
Acid
C=O
O–H
C–O
Aldehyde
C=O
C–H
Amide
C=O
N–H
Anhydride
C=O
Ester
C=O
C–O
Ketoneacyclic

cyclic,unsaturated
aryl ketone
Ether
C–O
Nitrile
CN
Nitro
N–O

stretchbending

1360–1080
1600

medium-weakmediumstretchstretch3100–3000
1600–1400

mediummedium-weak, multiple bands

stretch1820–1670

strongstretchstretchstretch1725–1700
3300–2500
1320–1210

strongstrong, very broadstrongstretchstretch1740–1720
2850–2820 & 2750–2720

strongmedium, two peaks

stretchstretchbending

1690–1640
3500–3100
1640–1550

strongunsubstituted have two bands

stretch1830–1800 &1775–1740

two bands

stretchstretch1750–1735
1300–1000

strongtwo bands or morestretchstretchstretchstretchstretchstretch1725–1705
3-membered - 1850
4-membered - 1780
5-membered - 1745
6-membered - 1715
7-membered - 1705

strongstrongstrongstrongstrongstrongstretch1685–1665

strongconjugation moves absorptions to lower wavenumbers
stretch1700–1680
strongstretch1300–1000 (1150–1070)

strongstretch2260–2210

mediumstretch1560–1515 &
1385–1345

strong, two bands

9

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 1
8% of the total

Problems

Question

1.1

1.2

1.3

Total

Points

14

4

6

24

Problem 1. Two Beauties of Turkey: the Van Cat and theAnkara Cat

The most beautiful of cats, the Van cat is a pure breed living only in Lake Van basin. Anotherendemic cat breed is the Ankara cat. They are called Angora cats. Their most important featureis their two different eye colors.
Van catAnkara cat

Nepeta cataria (catnip)

Just like people, cats can sometimes be stressed and angry. Just as people are made happy bymelatonin, the stress of cats can be reduced and they can be made happy thanks to a natural
product. Nepetalactone is an organic compound isolated from the plant catnip (Nepeta cataria),
which acts as a cat attractant. Nepetalactone is a ten-carbon bicyclic monoterpenoid compound
derived from isoprene with two fused rings: a cyclopentane and a lactone.

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52nd International Chemistry Olympiad, Istanbul, TURKEY

Cat eating catnip in the garden

Problems

Cat's dream

Nepetalactone
Total synthesis of nepetalactone:

1.1. The above scheme describes the total synthesis of nepetalactone. Draw structures of A–G,
without stereochemical details.


Hints:
Compound A has strong and sharp band at 3300 cm−1 in the IR spectrum.
A, B, and F are monocyclic, while C, D, E, and G are bicyclic compounds.
F has one doublet at ~ 9.8 ppm in the 1H-NMR spectrum.

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52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

Reactions of nepetalactone:

The above scheme includes a few reactions of one of the enantiopure nepetalactone 1 isomers.
Three of the reaction products (5, 6, and J) are used as insect repellents in industry.
1.2. For the relationship between 5 and 6, which of the following is/are true? Tick the box next
to the correct answer(s) on your answer sheets.
☐☐☐☐

Enantiomers
Diastereomers
Identical
Stereoisomers

Reaction of 1 with DDQ gives highly conjugated compound H. Also, thermal reaction ofcompound H with p-quinone gives I with molar mass of 226.28 g/mol.
1.3. Draw the structures of H, I, and J indicating stereochemistry.

Hints:
During the formation of I, sequential pericyclic reactions and an oxidation reaction(due to the presence of O2) take place, and a well-known gas forms during the reaction.
J has a strong and very broad band between 3300 and 2500 cm–1 in the IR spectrum.

12

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 2
10% of the total

Problems

Question

2.1

2.2

2.3

2.4

2.5

2.6

2.7

Total

Points

7

9

8

16

5

4

28

77

Problem 2. A Tale of a Reactive Intermediate
Arynes constitute a special class of reactive intermediates. The first experimental evidence forthe structure of an aryne (benzyne) was demonstrated in 1953 via the elegant labelingexperiments by John D. Roberts and coworkers.
In one such experiment, chlorobenzene, whose carbon at position 1 was labeled with radioactive
14
C, was reacted with KNH2 in liquid NH3 to give nearly equal amounts of isotopic isomers A
and B along with the inorganic salt C. This reaction proceeds via the formation of aryne
intermediate D.

2.1. Draw the structure of A, B and D, and provide the formula of C. Indicate the position(s) of
14
C-labeled carbon(s) with an asterisk (*) whenever applicable.”
Analysis of the 14C-labeled product(s) was achieved via degradation experiments (the 14Clabeled carbons are not shown on the structures). Radioactivities of the intermediates and final
products were examined.

13

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

2.2. Tick the appropriate boxes on the answer sheet for the intermediates and products thatyou expect to exhibit radioactivity.
Considering only A:

Considering only B:☐ Compound 1☐ BaCO3 (Batch 1)☐ Compound 2☐ BaCO3 (Batch 2)☐ Compound 1☐ BaCO3 (Batch 1)☐ Compound 2☐ BaCO3 (Batch 2)With the aim of facilitating aryne formation, Kobayashi and co-workers developed a fluorideinduced aryne generation protocol. Using this method, benzene derivative 3 is reacted withfuran (4) in the presence of CsF, resulting in the formation of E, F, and G.

14

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

Combustion analysis of E revealed the following atom content: 75.8% carbon, 5.8%hydrogen, and 18.4% oxygen.
E does not have a proton that is exchangeable with D2O in 1H-NMR spectroscopy.
F is an ionic compound.

2.3. Determine the structures of E, F, and G (without stereochemical details).
In the absence of a nucleophile or a trapping agent, arynes can undergo [2+2]-type
cyclodimerization or [2+2+2]-type cyclotrimerization reactions under suitable conditions. The
aryne derivative that is obtained when 3 is treated with one equivalent of CsF in MeCN cangive, in principle, four different dimerization and trimerization products (H–K).


H has two planes of symmetry.
I is expected to exhibit 21 signals in its 13C-NMR spectrum.
I and J both exhibit an m/z value of 318.1 in their mass spectra.

2.4. Determine the structures of H–K.
When 5 is reacted with -ketoester 6 in the presence of 2 equivalents of CsF at 80 °C, L is
obtained as the major product. The 1H-NMR and 13C-NMR data for L, in CDCl3, are as follows:


1



13

H-NMR: δ 7.79 (dd, J = 7.6, 1.5 Hz, 1H), 7.47–7.33 (m, 2H), 7.25–7.20 (m, 1H), 3.91
(s, 2H), 3.66 (s, 3H), 2.56 (s, 3H) ppm.
C-NMR: δ 201.3, 172.0, 137.1, 134.4, 132.8, 132.1, 130.1, 127.5, 51.9, 40.2, 28.8
ppm.

2.5. Determine the structure of L.
2.6. In the reaction shown in task 2.5, which of the statement(s) in the answer sheet describe(s)the function of CsF?
The pKa values of HF and -ketoester 6 in dimethyl sulfoxide (DMSO) are about 15 and14, respectively.

15

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems☐ F– hydrolyzes the trifluoromethanesulfonate (O3SCF3) group of 5.
☐ F– attacks the –SiMe3 group of 5.
☐ F– acts as a base to deprotonate 6.
☐ F– acts as a nucleophile and attacks the ester group of 6.
Diazapyrone derivative 8 was shown to be a useful reactant for the construction of a variety ofcyclic frameworks. Its preparation from phenylglyoxylic acid (7) and its use in two different
reactions are described below.






Q and T are gases under ambient conditions.
O and P are constitutional isomers.
Q does not have any signals in its IR spectrum.
Heating 1 mol of R at 85 °C generates 1 mol of reactive intermediate S.
Reaction of 8 with two equivalents of S gives U, Q, and T.

Note:equiv= equivalentcat= catalyst
2.7. Determine the structures of M–U.

16

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 3
8% of the total

Question
Points

3.1
8

3.2
16

Problems

3.3
4

3.4
8

3.5
5

3.6
5

3.7
5

Total
51

Problem 3. (±)-Coerulescine
A spiro compound is typically an organic compound containing rings linked together by one
common atom (spiroatom) as carbon atom with bold in figure below. The spiro[pyrrolidin-3,3′oxindole] ring system is a structural framework incorporated in several cytostatic alkaloids andunnatural compounds. Coerulescine (1) and horsfiline are the simplest prototype members of
this subfamily that can be synthesized by the route shown below.

Claisen rearrangement, which is a [3,3]-sigmatropic rearrangement, is a powerful carbon–
carbon bond-forming reaction in which an allyl vinyl ether is converted thermally to an
unsaturated carbonyl compound as shown in the Scheme below. When compound A is heated,
it undergoes Claisen rearrangement to give carbonyl compound B.
For this entire task, your answers can be given without any stereochemical details.

17

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

3.1. Draw the structures of A and B.


A is an inseparable mixture of cis/trans isomers.
B has IR absorption at 1726 cm–1.

3.2. Draw structures for C, D, E, and F.


D–F have a bicyclic structure.

18

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

3.3. Choose the correct order of steps for the transformation of F to G.

☐ Imine formation, then reduction, then amidation☐ Amidation, then imine formation, then reduction☐ Reduction, then amidation, then imine formation

3.4. Draw structures for G and H (both spiro compounds).
3.5. Draw the structure of the intermediate produced by treatment with n-BuLi in the step H→ coerulescine (1).
Coerulescine (1), on treatment with N-bromosuccinimide (NBS), gives the bromo derivative,
which upon heating with sodium methoxide in the presence of cuprous iodide gives horsfiline(I) in 60% yield.
3.6. Choose the correct structure for compound I consistent with the following selected 1HNMR data: δ 7.05 (d, J = 1.4 Hz, 1H), 6.78 (d, J = 8.0 Hz, 1H), 6.72 (dd, J = 8.0, 1.4 Hz, 1H)
ppm.

☐

☐

☐

☐

3.7. When the allyl ether of 2 naphthol is heated a sigmatropic rearrangement is initiated. Write
the structure of major product isolated from this reaction.

19

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 4
10% of thetotal

Problems

Question

4.1

4.2

4.3

4.4

4.5

4.6

4.7

Total

Points

12

6

6

16

9

9

8

64

Problem 4. Symmetry Does Matter!
There are numerous reactions in organic chemistry that proceed through cyclic transition states
and these are classified as pericyclic reactions. Woodward–Hoffmann rules, developed byRobert B. Woodward and Roald Hoffmann, are used to rationalize stereochemical aspects andthe activation energy of pericyclic reactions.

Number ofelectrons
4n(n = 1, 2, ..)4n+2(n = 1, 2, ..)Woodward–Hoffmann rules
Electrocyclic reactions
Cycloadditions
Photochemical
Photochemical
Thermal ()Thermal ()
(h
(h

Conrotatory (con)Disrotatory

DisfavoredFavored

Disrotatory (dis)Conrotatory

FavoredDisfavored

4.1. Fill in the table for reactions (i)–(iii) or products 2–5:
Reaction
i
ii
iii

Product234
5[? + ?] cycloaddition

 or h20

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

There are three possible benzotropone isomers. Although two of the benzotropone isomers wereisolated, 3,4-benzotropone (1) has not been isolated. Its instability is attributed to the oquinoidal structure of 1 because it has no sextet electron system in the benzene ring.
4.2. Draw the structures of stable benzotropone isomers A (with 6 signals in its 13C-NMR) andB (with 11 signals in its 13C-NMR).
4.3. When the following tetraene is reacted under photochemical conditions, symmetry-allowed
product(s) of three different ring sizes can form according to the Woodward–Hoffmann rules.
Tick the correct answer in each row.

☐

☐

☐

☐

☐

☐

Prof. Dr. Aziz Sancar
The Nobel Prize in Chemistry 2015 was awarded jointly to the Turkish scientist Aziz Sancar,
Swedish scientist Tomas Lindahl, and American scientist Paul Modrich for their “mechanistic
studies of DNA repair”. Pyrimidine bases found in DNA may undergo a photochemical [2+2]cycloaddition reaction (see above figure) with UV light that reaches a person’s skin, causing

21

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

damage to DNA, which may ultimately lead to skin cancer. The research by Professor Aziz
Sancar focused on the DNA repair mechanism for this type of damage.
Thymine (T) is one of the nucleobases that can undergo such a photochemical reaction with UVlight. Let us assume that we have a solution of free thymine that was subjected to UV irradiation.

4.4. Considering stereochemistry, draw the structures of all possible products of this reaction
between two free thymine (T) molecules. Circle the compound(s) which is/are chiral. Drawing
only one enantiomer of an enantiomeric pair is sufficient. Please note that only C=C bonds
participate in this reaction.
A broad range of halogenated derivatives of norbornadiene (N) are known in the literature.
Tribromo-norbornadiene (C7H5Br3) has six achiral isomers. Three of these isomers (6, 7, and 8)
are given below.

4.5. How many signals do you expect from the 13C-NMR spectra of 6, 7, and 8? Fill in thefollowing boxes.

6

7

8

22

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

4.6. Draw structures of the remaining achiral tribromo-norbornadiene (C7H5Br3) isomers (C,D, and E) in addition to 6–8 over the given figures in the boxes.

C

D

E

The NMR spectrum of ether 9 is complex. Two MeO– groups are different as are all thehydrogen atoms on the rings. However, diphenol 10 has a very simple NMR spectrum and there
are only three types of protons (marked as a, b, and c). A reasonable average structure
responsible for all resonance structures and its symmetry is shown as 11.

4.7. How many signals do you expect from the 13C- and 1H-NMR spectra of 12 and 13?

23

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 5
14% of the total

Question
Points

5.1
13

5.2
13

5.3
15

Problems

5.4
8

5.5
12

5.6
10

5.7
17

5.8
12

Total
100

Problem 5. Konya, Carrot, Beta-Carotene, Vitamin-A, Immune
System, Vision

Mevlana (Rumi) was a great mystic and Sufi poet who lived out his days in Konya in the 13 th
century. The indirect relevance of Konya to chemistry is that the city provides 65% of thecountry’s carrot production, from which one of the essential vitamins (vitamin A) is obtained.
Carrot is an important source of β-carotene, which gives the vegetable its orange color. This
molecule is a red-orange pigment naturally found in plants and fruits and is a provitamin A
carotenoid. It is converted to vitamin A, which is essential for normal growth and development,
the immune system, and vision function.

Figure 1. Ball and stick representation of the structure of β-carotene. The gray and whitespheres represent the carbon and hydrogen atoms, respectively. The numbered carbon atoms
belong to the linear conjugated π-segment of the molecule.
β-Carotene has an extended polyene chain of 22 carbon atoms. It is a conjugated π-system,
having alternating single and double bonds. Its experimental maximum absorption wavelength(λmax) is 455 nm. We assume that all the bonds between C1 and C22 are conjugated bonds. Thereare 22 π-electrons in the molecule (Figure 1).

24

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

To a crude approximation, the electrons in the C-2pz orbitals, which are perpendicular to themolecular plane, are assumed to move along the entire molecule, without interacting with eachother. They are like independent particles confined in a molecule moving along the x-axis in
one dimension. These characteristics of π-electrons make them eligible for being treated by thesimplest model called the particle in one-dimensional box model.
The wave function and the energies of the quantized levels for an electron moving in a onedimensional box with infinite potential walls are given as follows:
2

𝑛𝜋𝑥

𝐿

𝐿

𝛹𝑛 (𝑥) = √ sin

Eq.1

where n is the quantum number, n=1,2,3,4,…. ∞, and L is the box length.

𝐸𝑛 =𝑛2 ℎ 2Eq.2

8𝑚𝑒 𝐿2In two dimension, within the framework of independent particle approximation, thewavefunction is expressed as a product of one-dimensional wavefunctions, and the energy isexpressed as a sum of one-dimensional energies. The energy levels of the two dimensional
rectangular box is given as follows:
𝑛 2

𝑛𝑦 2

𝐿𝑥

𝑦

𝐸𝑛𝑥,𝑛𝑦 = [ 𝑥2 +

ℎ2

]{ }
𝐿 2 8𝑚Eq. 3

𝑒where nx, ny are the quantum numbers and they are positive integers. Lx, Ly are the dimensionsof the box in the 2D model. They are positive numbers.
5.1. Which two of the sentences given below are correct? Tick only one answer which includes
correct sentences in your answer sheet.
The β-carotene molecule is orange in color becausei)it absorbs in the visible region of the electromagnetic spectrum.

ii)

HOMO →LUMO transition occurs by absorption of IR photon.

iii)

the spacing between the 22nd and the 23rd energy levels is equal to the energy of the IR
photon at the orange wavelength.

iv)

it absorbs green/blue light and it transmits red/yellow color.

v)

it absorbs in the UV-Vis region since the molecule has no net dipole moment.

25

52nd International Chemistry Olympiad, Istanbul, TURKEY☐ a) i and ii☐ e) ii and iii☐ j) iii and v☐ b) i and iii☐ f) ii and iv☐ k) iv and v☐ c) i and iv☐ g) ii and v

Problems☐ d) i and v☐ h) iii and iv

Although it is highly unrealistic, let us assume that the conjugated segment of the molecule islinear and treated with the particle in a one-dimensional box model as shown in figure 2. In
this case, the length of the box can be approximated as L=1.40×nC (in Å), where nC is thenumber of carbon atoms in the conjugated segment.
Use this information to answer the questions 5.2–5.6.

Figure 2. Schematic representation of the conjugated line segment made up carbon atoms of βcarotene in a one-dimensional box of length L.
5.2. Calculate the energies (in J) of the lowest two energy levels
5.3. Draw the wavefunctions of the lowest two energy levels with proper labelling the x-axis.
5.4. Sketch the energy level diagram up to n = 4 showing the relative spacing.
5.5. What is the total π-energy (in J) of the molecule?
5.6. Calculate the wavelength (in nm) at which the transition between the highest occupied
and lowest unoccupied energy levels occurs.
Use the particle in a two-dimensional box model to answer questions 5.7–5.8.

26

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

Figure 3. Schematic representation of the conjugated carbon atoms of β-carotene in a twodimensional box.
Assume that the conjugated segment is made up of conjugated bonds that are all-trans to eachother. The motion of the π-electrons is studied in the two-dimensional rectangular box with thedimensions Lx = 26.0 Å, Ly = 3.0 Å (Figure 3).
5.7. Calculate the energies (in J) of the highest occupied and the lowest unoccupied energy
levels and the wavelength (in nm) at which the transition between these energy levels occurs.
5.8. What should be the Lx value (in Å) in order for the molecule to absorb light at theexperimental λmax=455 nm if Ly is kept constant at 3.0 Å. (Assume that the quantum numbers
for homo and lumo are the same as in the question 5.7.)

27

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 6
12% of the total

Question
PointsQuestion
Points

6.1
5
6.9
8

6.2
12
6.10
6

6.3
6
6.11
6

Problems

6.4
3
6.12
8

6.5
6
6.13
6

6.6
6

6.7
5

6.8
3

Total
80

Problem 6. Thermodynamics through an Interstellar JourneyPart 1
In a hypothetical universe, an unknown amount of diborane participates in the following
reaction:
B2H6(g) + 6 H2O(l) → 2 H3BO3(s) + 6 H2(g)Assume that in this hypothetical universe, H3BO3(s) obtained from this reaction was completely
sublimed at 300 K. The necessary energy for sublimation was obtained through work released
by one cycle of an ideal heat engine in which one mole of monoatomic perfect gas flows throughthe cycle described in the pressure (p) – volume (V) diagram below:A→B; isothermal reversible expansion receiving 250 J by heat transfer (qH) at atemperature of 1000 K (TH) from a hot source.
B→D; reversible adiabatic expansion.
D→C; isothermal reversible compression at a temperature of 300 K (TC) releasing someamount of heat (qC) to a cold sink.
C→A; reversible adiabatic compression.

28

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

After heat transfers, the remaining energy is released as work (w). Also, qH and qC are related
to TC and TH as follows:
|𝑞𝐻 |
𝑇𝐻
=
|𝑞𝐶 |
𝑇𝐶
The efficiency of the cycle can be found by work released by cycle (w) divided by heat absorbed
by cycle (qH).
You are provided with the change in enthalpies of the following reactions at 300 K.

(1) B2H6(g) + 6 Cl2(g) → 2 BCl3(g) + 6 HCl(g)𝚫𝐫 𝐇(𝟏) = −𝟏𝟑𝟐𝟔 𝐤𝐉 𝐦𝐨𝐥−𝟏(2) BCl3(g) + 3 H2O(l) → H3BO3(g) + 3 HCl(g)𝚫𝐫 𝐇(𝟐) = −𝟏𝟏𝟐. 𝟓 𝐤𝐉 𝐦𝐨𝐥−𝟏(3) B2H6(g) + 6 H2O(l) → 2 H3BO3(s) + 6 H2(g)𝚫𝐫 𝐇(𝟑) = −𝟒𝟗𝟑. 𝟒 𝐤𝐉 𝐦𝐨𝐥−𝟏11(4) 2 H2(g) + 2 Cl2(g) → HCl(g)𝚫𝐫 𝐇(𝟒) = −𝟗𝟐. 𝟑 𝐤𝐉 𝐦𝐨𝐥−𝟏

6.1. Calculate the molar enthalpy of sublimation (in kJ mol-1) for H3BO3 at 300 K.
6.2. Calculate the ΔrU (internal energy) in terms of kJ mol-1 at 300 K for the reactions (2) and(4) given above (assume ideal gas behavior for each gaseous species in each reaction).
6.3. Calculate the amount of overall work produced by a heat engine (|w|) in terms of J and theamount of overall heat released to the cold sink (|qC|) in terms of J.
6.4. Calculate the efficiency of the heat engine described above.
6.5. Calculate the entropy change (ΔS) for A→B and D→C processes in the heat engine interms of J K −1.
6.6. Calculate the Gibbs energy change (ΔG) in terms of J for A→B and D→C processes in
the heat engine.
6.7. Calculate the ratio of pressure at point A to the pressure at point B in the cycle (standard
pressure: 1 bar).

29

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

6.8. Calculate the amount of H2(g) (in moles) produced according to the reaction given at the
beginning of the task for one cycle of the engine.
Part 2
Interstellar journeys can be done by using diborane as rocket fuel. Combustion of diborane isshown below:
B2H6 (g)+ 3O2 (g) → B2O3 (s) + 3H2O (g)Combustion of diborane is experimented in a 100 L closed container at different temperatures
and the equilibrium amounts were recorded.

B2H6(g)8930 K
0.38 mol

9005 K
0.49 mol

H2O(g)

0.20 mol

0.20 molPartial pressure of O2 (g) was stabilized to 1 bar and kept constant at all conditions. Assume
that in this hypothetical universe; ΔrS° and ΔrH° are independent of temperature, the standard
molar entropy (S°) of B2O3(s) does not change with pressure, all the gas species behave as an
ideal gas, and all species remain in the same phase, without any further decomposition before
or after reaction, at all temperatures then:
6.9. Calculate Kp (pressure based equilibrium constant) at 8930 K and 9005 K.
6.10. Calculate ΔrG° of the reaction in terms of kJ mol-1 at 8930 K and 9005 K. (If you failed
to find Kp, please use Kp (8930 K) =2, Kp (9005 K) = 0.5)
6.11. Calculate ΔrG°(in terms of kJ mol−1 ), ΔrH° (in terms of kJ mol−1 ), and ΔrS°(in terms of J mol−1 K −1)of the combustion reaction at 298 K. (If you failed to find Kp, please
use Kp (8930 K) =2, Kp (9005 K) = 0.5)
6.12. Tick the correct answer in the table by determining whether combustion reactions arefavored or not at given T below under standard pressure (1 bar).

298 K

favored☐Unfavored☐

8930 K☐☐

9005 K☐☐

9100 K☐☐30

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

6.13. Calculate the ΔfH ( kJ mol–1) and S°( kJ mol–1 K–1) of H2O(g) using the values given in
the table below. (ΔfH = enthalpy of formation, S° = standard entropy)(If you fail to find ΔrH° and ΔrS° of the combustion, please use ΔH° = 1000 kJ mol-1, ΔS°= 150
J K-1 mol-1)

B2H6 (g)O2 (g)
B2O3 (s)ΔHf (298 K)
36.40 kJ mol–1
0.00 kJ mol–1–1273 kJ mol–1

S° (298 K)
0.23 kJ mol–1 K–1
0.16 kJ mol–1 K–1
0.05 kJ mol–1 K–1

31

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 7
12% of thetotal

Problems

Question

7.1

7.2

7.3

7.4

7.5

7.6

7.7

7.8

7.9

Points

4

8

6

8

8

14

19

12

6

Total
85

Problem 7. Phthalocyanines

Emeritus Professor ÖzerBekaroğlu
The term phthalocyanine (Pc) takes its origin from the Greek “naphtha”, which means rock oil,and “cyanine”, which means dark blue. Turkish scientist Özer Bekaroğlu can be regarded as thepioneer of Pc chemistry in Turkey.
Metal-free phthalocyanine (1, H2Pc) is a large planar macrocyclic compound with the formula(C8H4N2)4H2.
7.1. How many π-electrons are there in the bold region of the H2Pc molecule in compound 1
shown above?

32

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

Pcs containing one or two metal ions are called metallo-phthalocyanines (MPcs) and they
exhibit different geometries as given above.
7.2. Complete the table in your answer sheet by determining the coordination number of central
ions in 2–5.
Central ion

Copper ion

Titaniumion

Silicon ion

Cerium ion

Coordination number
7.3. Complete the table in your answer sheet by determining the oxidation number of eachmetal (Cu, Ti, and Ce) in 2, 3, and 5.
Metal in compounds
Oxidation number

2

3

5

7.4. Complete the table in your answer sheet by determining the geometry of compounds 2–5.
Geometry

CompoundOctahedral
Square prism
Square pyramidal
Square planar

7.5. Complete the table in your answer sheet by determining the magnetic property ofcompounds 2–5.


Use the letter "p" for paramagnetic property and letter "d" for a diamagnetic property.
CompoundMagnetic property

2
3
4
5

33

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

7.6. Write the ground-state electron configuration of the silicon (Si) ion in the compound 4 andfind all the quantum numbers for the 2p electrons in its ground state.
Electron configuration:
n

l

m1

ms

Quantum numbers for 2p electrons:

Metal-free phthalocyanine (1, H2Pc) is commonly formed through the cyclotetramerization ofphthalonitriles. On the other hand, Pcs having different substituents are called asymmetric,
which can be prepared by the statistical cyclization of two different phthalonitriles. This method
has no selectivity and the product is a mixture of all possible isomers.
7.7. Draw the possible products that may occur in the statistical cyclization method by using F
and G. If there are any stereoisomers label as cis- or trans-.
F and G represent two different symmetrical phthalonitriles.
One of the products is F4 as given below.
Draw other products similar to the format of F4.

Pcs are used as photosensitizers in the photodynamic therapy (PDT) of cancer due to their strong
absorption in the visible spectrum and high molar absorption coefficients. PDT consists of threeessential components: photosensitizer, light, and oxygen. None of these is individually toxic,
but together they initiate a photochemical reaction resulting in the generation of cytotoxic
singlet oxygen (1O2) that can destroy cancer cells.

34

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

The multiplicity of an energy level is defined as 2S+1
If the two spins are parallel (↑↑), S = 1, and if the two spins are antiparallel (↑↓), S = 0.

7.8. Draw the molecule orbital (MO) diagram of the lowest energy singlet state of dioxygen(1O2) and calculate bond order.
 There are no unpaired electrons in that state!

7.9. If the wavelength of the light needed to excite triplet oxygen into singlet oxygen is 1270
nm, calculate the energy (in kJ/mol) needed for this transition process.

35

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 8
14% of the total

Question
PointsQuestion
PointsQuestion
Points

8.1
3
8.9
2
8.17
4

8.2
3
8.10
3

8.3
4
8.11
2

Problems

8.4
4
8.12
4

8.5
4
8.13
4

8.6
4
8.14
4

8.7
6
8.15
2

8.8
3
8.16
2

Total

58

Problem 8. Boron Compounds and Hydrogen Storage

Sodium borohydride (NaBH4 ) and ammonia borane (BNH6 ) are the most studied chemicalhydrogen storage materials. In this question, you will explore the chemistry of boron and theuse of boron compounds as hydrogen storage materials.
Borax (Na2 B4 O7 ∙ nH2 O) is a boron mineral that is produced by ETI Mining Company inTurkey. NaBH4 can be synthesized by the reduction of anhydrous borax with metallic sodium
under high-pressure hydrogen gas in the presence of silicon dioxide (silica) at 700 oC, namely
the Bayer process. In this process, all hydrogen is stored in NaBH4. On the other hand, it has
been shown that ammonia borane (BNH6 ) can be synthesized by the reaction of NaBH4 andammonium sulfate in dry tetrahydrofuran (THF) at 40 C (Hint: BNH6 synthesis must be
conducted in a well-ventilated fume-hood because flammable gas is generated as one of the byproducts). While NaBH4 is an ionic compound, ammonia borane is a Lewis acid–base adduct.
8.1. Write a balanced chemical equation for the synthesis of NaBH4 from anhydrous borax.
8.2. Write a balanced chemical equation for the synthesis of ammonia borane from NaBH4.
8.3. Draw the molecular geometry of the BH4− ion and BNH6 molecule.
8.4. Calculate the hydrogen content of NaBH4 and BNH6 as a percentage by mass (wt%).
36

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

The hydrogen stored in both compounds can be released via hydrolysis reactions in the presenceof a suitable catalyst at room temperature. Upon the hydrolysis reactions, 4 and 3 moles of H2
gas are released from the hydrolysis of 1 mole of NaBH4 and BNH6 , respectively, along withmetaborate anion including B-O bonds.
8.5. Write the balanced chemical equations for the hydrolysis of NaBH4 and BNH6 .
One of the simplest stable borates is diboron trioxide (B2O3). Higher borates such as B3 O3−
6having cyclic structures containing B-O-bonds can be formed. Since B2 O3 is an acidic
compound, it is easily reacted with water to produce boric acid (H3 BO3 ). On the other hand, thehigh temperature and high-pressure reaction of B2O3 with ammonia yields two-dimensionalboron nitride, which consists of planar graphite-like sheets of alternating B and N atoms.
8.6. Write the balanced chemical equations for the synthesis of boric acid and boron nitride.
8.7. Draw the molecular structures of the B3 O3−
6 ion, boric acid, and a single two-dimensionalboron nitride sheet. Hint: show at least 10 B atoms in the boron nitride structure.
Furthermore, B-H compounds, called boranes, are an important class of boron compounds. The
simplest stable borane is diborane (B2 H6 ) and many of the higher boranes can be prepared by
the pyrolysis of diborane. Diborane can be synthesized via metathesis of a boron halide and ahydride source.
8.8. Write a balanced chemical equation for the synthesis of diborane from the reaction of BF3and LiBH4. Hint: both products are boron compounds.
8.9. Draw the molecular geometry of the diborane molecule. Hint: there is no B-B bond in
the structure.
BH3 (borane) is an unstable and highly reactive molecule. Therefore, it is not possible to isolate
it as BH3 under ordinary conditions. However, it can be stabilized via its reaction with carbon
monoxide to yield borane carbonyl (BH3 CO) compound, which is an adduct of borane. The
preparation of BH3 CO plays an important role in exploring the chemistry of boranes as it
indicates the likely existence of the borane molecule.
8.10. Sketch the Lewis dot structure of BH3 − CO molecule by showing the formal charges.

37

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

8.11. Which of the statements given in your answer sheet is observed in the C − O bond of a CO
molecule upon the bond formation between BH3 and CO? Tick the correct box.

☐ It gets longer because there will be -back donation from BH3 to CO.
☐ It gets longer because CO donates -bonding electrons to BH3.
☐ No or slight change on it because CO donates its mainly non-bonding electrons to BH3.
☐ It gets shorter because CO donates * anti-bonding electrons to BH3.

Borazine consists of single and double bonded cyclic B − N units and hydrogen atoms attachedto these atoms with the molecular formula B3 N3 H6 and is isostructural to benzene. Borazine
can be synthesized by using a two-step procedure including the synthesis of symmetricallytrisubstituted chlorine derivatives of borazine (B3 N3 H3 Cl3 ) from the reaction of ammonium
chloride and boron trichloride, and then reduction of B3 N3 H3 Cl3 with LiBH4 in THF.
8.12. Write the balanced chemical equations for the two-step synthesis of borazine starting fromammonium chloride in THF (tetrahydrofuran). Hint: THF stabilizes one of the products byforming a Lewis acid-base adduct.
8.13. Draw the molecular structures of borazine and its symmetrically trisubstituted chlorine
derivative.
Catalysts are substances that accelerate the rate of reactions by allowing them to proceed in alower energy pathway. The catalytic activity of the catalysts is generally determined by theturnover frequency (TOF), which is calculated by dividing the molar amount of the product tothe mole of active catalyst and the time (TOF = mole product/(mole catalyst x time)). A typicalhydrolysis of BNH6 was carried out in 10.0 mL of water by using 100.0 mM BNH6 and 5.0 mg
of CuPt/C catalyst (CuPt alloy nanoparticles supported on carbon black containing 8.2 wt% Ptatom). 67.25 mL of hydrogen gas was generated in 5 minutes.
8.14. Assuming the catalytic reaction is performed under standard conditions (1 atm and 273.15
K), calculate the TOF (𝑚𝑖𝑛−1) of the CuPt/C catalyst in terms of only Pt atoms in thehydrolysis of BNH6 by considering the volume of the hydrogen gas generated.
As a result of detailed crystal analysis of a synthesized Cux Pt y alloy nanoparticles (thesubscripts indicate molar percentages of the atoms in the alloy structure), it was determined thatthe face centered cubic (fcc) unit cell was formed by Pt atoms and the Pt atoms on the face ofthe fcc unit cell are supposed to be replaced with Cu atoms to form Cux Pt y displacement alloy
nanoparticles. According to this information, answer the following questions.
38

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

8.15. Determine the composition of the alloy nanoparticles by finding x and y in the Cux Pt y .
8.16. Sketch the shape of the described crystal unit cell of Cux Pt y alloy nanoparticles byshowing the position of all atoms on the unit cell.
8.17. Another alloy has a Cu2 Pt1 composition. Assume that this alloy also has a fcc unit cellwith an edge length of 380 pm, but the Cu and Pt atoms are randomly distributed in the atomic
positions. Calculate the density of this alloy in g/cm3.

39

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 9
12% of the total

Question
Points

9.1
5

9.2
5

9.3
8

Problems

9.4
16

9.5
5

9.6
6

9.7
25

9.8
30

Total
100

Problem 9. Quantification of Heavy Metal Ions
For the quantitative analysis of heavy metal ions in a factory’s wastewater pool, the following
steps have been applied by an analyzer at 298 K:
Step 1) 10-mL samples of each were obtained from five different regions in a wastewater pool,mixed in a 100-mL beaker, and then stirred for 5 minutes using a magnetic stirrer.
Step 2) 10-mL of sample solution was taken from the 100-mL beaker and 142 mg of Na2SO4
was added while stirring, followed by transfer to a three-electrode cell as seen in Figure 1a. In
this electrochemical cell, Pt wire, Ag/AgCl (3 M KCl), and Pt foil served as the working,reference, and counter electrodes, respectively.
Step 3) These electrodes were connected to a potentiostat and a constant potential of –0.50 V
vs. Ag/AgCl for 14 minutes was applied as seen in Figure 1b (horizontal line). It is assumed
that 14 min. is sufficient to complete the expected electrochemical reactions.
(a)

(b)

Figure 1. a) Electrochemical cell design; 1) Working electrode (Pt wire), 2) reference electrode(Ag/AgCl, 3M KCl), 3) counter electrode (Pt foil), 4) cell tap, 5) electrochemical cell, 6) 10mL of sample solution. b) Potential change of working electrode as a function of time. y-axis:
potential/V vs Ag/AgCl, x-axis: time/min.

40

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

Step 4) The electrodes were rinsed with distilled water, placed into another electrochemical cell
including 10-mL of 0.1 M H2SO4 solution, and potential was scanned between –0.50 and +0.50
V as seen in Figure 1b (downward sloping line in 2 min.). Current vs. potential data for this step
are presented in Figure 2a, which is like an excellent view of Mount Ararat (Ağrı Dağı), thehighest mountain in Turkey (Figure 2b).

(a)

(b)

Figure 2. a) Potential scan of the working electrode as a function of current in 0.1 M H2SO4
solution after keeping it constant at a potential of –0.50 V in 10-mL of wastewater sample asseen in Figure 1b (horizontal line). y-axis: current/A, x-axis: potential/V vs Ag/AgCl, b) Aview of Great and Little Ararat peaks.

Step 5) Another 10-mL of the sample solution prepared in step 1 was taken and the processes
explained in steps 2 and 3 were applied in that order. The electrodes were rinsed with distilledwater and placed into 10-mL of 0.1 M H2SO4 solution. Then the potential of the working
electrode was kept constant at +0.05 V for 14 minutes. It is assumed that 14 min. is sufficient
to complete the expected electrochemical reactions.
Step 6) After step 5 was performed, the solution in the electrochemical cell was placed in asuitable oven to evaporate at 150 °C until dry solid was obtained.
4152nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

Step 7) 5-mL of ethylenediaminetetraacetic acid (EDTA, H4Y) (Figure 3) solution was added
to the solid obtained in step 6 and it was shaken to dissolve. It is known that 1-mL of EDTA
solution is equivalent to 3.85 mg/mL BaCO3. Then, pH of the solution was adjusted to 10.0.
Excess EDTA was titrated with standard 0.0010 M Ni(NO3)2 solution and it was observed that95.60 mL of Ni(NO3)2 solution was consumed up to the endpoint.

Figure 3. Chemical structure of EDTA (H4Y).


In water saturated with H2S, equilibrium concentration of [H2S] is 0.1 M.

Ksp(NiS) = 4.0 × 10–20; Ksp(CuS) = 1.0 × 10–36
Ka1(H2S) = 9.6 × 10–8; Ka2(H2S) = 1.3 × 10–14
Reaction
2𝐻2 𝑂(𝑙) + 2𝑒 − → 𝐻2 (𝑔) + 2𝑂𝐻 − (𝑎𝑞)𝑁𝑖 2+ (𝑎𝑞) + 2𝑒 − → 𝑁𝑖(𝑠)2𝐻 + (𝑎𝑞) + 2𝑒 − → 𝐻2 (𝑔)𝐶𝑢2+ (𝑎𝑞) + 2𝑒 − → 𝐶𝑢(𝑠)𝐴𝑔+ (𝑎𝑞) + 𝑒 − → 𝐴𝑔(𝑠)𝑂2 (𝑔) + 4𝐻 + (𝑎𝑞) + 4𝑒 − → 2𝐻2 𝑂(𝑙)E˚/V (at 298 K)
−0.83
−0.24
0.00
+0.34
+0.80
+1.23

9.1. Which of the following can be considered for peak 1 and peak 2 in Figure 2a, respectively?
Tick the correct box on the answer sheet.
☐ Peak 1: electrochemical reduction of Ni / Peak 2: electrochemical reduction of Cu☐ Peak 1: electrochemical reduction of Cu / Peak 2: electrochemical reduction of Ni☐ Peak 1: electrochemical reduction of Ni / Peak 2: electrochemical oxidation of Cu☐ Peak 1: electrochemical oxidation of Ni / Peak 2: electrochemical oxidation of Cu☐ Peak 1: electrochemical oxidation of Cu / Peak 2: electrochemical oxidation of Ni

4252nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

9.2. Which of the statement is expected, if the potential is applied as –1.2 V instead of -0.5 V at
the first step (horizontal line) in Figure 1b? Tick the correct box on the answer sheet.
☐ NO evolution☐ NO2 evolution☐ Nitrogen evolution☐ Oxygen evolution☐ Hydrogen evolution
9.3. Calculate the scan rate of data presented in Figure 2a as mV/s at 298 K.

The potential of the following cell is measured as 0.437 V.
𝑃𝑡, 𝐻2 (0.92 𝑏𝑎𝑟)|𝐻𝐶𝑙(1.50 × 10−2 𝑀), 𝐴𝑔𝐶𝑙(𝑠𝑎𝑡)|𝐴𝑔
9.4. Calculate the standard electrode potential value (V) of half-cell of 𝐴𝑔𝐶𝑙(𝑠) + 𝑒 − →𝐴𝑔(𝑠) + 𝐶𝑙 − (𝑎𝑞) at 298 K. Note: You must show all works.

9.5. Which of the statements is the main purpose of step 5 in this analysis? Tick the correct box
on the answer sheet.
☐ Modification of Pt wire with Ni–Cu alloy film
☐ Modification of Pt wire with Ni film
☐ Electrochemical stripping of both Cu and Ni from Cu–Ni-modified Pt wire to the solution☐ Electrochemical stripping of Cu from Cu–Ni-modified Pt wire to the solution☐ Electrochemical stripping of Ni from Cu–Ni-modified Pt wire to the solution
9.6. Write net ionic equations for the complexation and the back titration reaction of step 7 on
the answer sheet
9.7. Calculate Ni2+ concentration as mg/L in the wastewater of the factory. Note: You mustshow all works.
9.8. Calculate the minimum pH value for starting the precipitation of Ni2+ ions in the solution
obtained in step 5 by passing H2S gas into the solution until saturation. If you cannot solve
question 9.7, use 20 mg/L Ni2+ sample for this question. Note: You must show all works.

43

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problems

The End44