PREPARATORY PROBLEMS

52nd International Chemistry Olympiad
Istanbul, Turkey

Preparatory Problems

CHEMISTRY FOR A BETTER TOMORROW

Fifth edition (04.08.2020)

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preface
We are very glad to provide Preparatory Problems for the 52nd International Chemistry
Olympiad, which will be held in 2020 in Istanbul, Turkey. We prepared these problems with
the intention of facilitating the training and preparation of participants. The contents of the
problems have been carefully selected so as to cover a broad range of challenging topics thatcan be encountered in modern as well as classical chemistry. The problems can be solved byapplying the fundamental principles of chemistry covered at high school level along with 6
topics of advanced difficulty for the theoretical section, and 3 topics of advanced difficulty forthe practical section. These advanced topics are listed explicitly under “Topics of Advanced
Difficulty” and their applications are demonstrated in the tasks. We expect the participants to
be familiar with these advanced topics.
The problems listed in this booklet consist of 25 theoretical and 8 practical tasks. The
solutions were sent to the Head Mentor of each country by e-mail by February 10th, 2020 andpublished by June 01st, 2020 on our IChO 2020 website. We welcome any comments,suggestions, corrections, or questions about the problems at icho2020@tubitak.gov.tr.
The International Chemistry Olympiad presents a great opportunity to inspire younger
generations to pursue a career in fundamental sciences and make a positive influence on public
attitudes towards science, and in particular chemistry. We hope you will enjoy solving these
problems and we look forward to seeing you in July in Istanbul, Turkey.

Acknowledgments
I would like to express my deep gratitude to all the authors for their dedication and effort incontributing to the Preparatory Problems as well as the members of the International Steering
Committee for their valuable comments and suggestions. We are also highly appreciative of the
Scientific and Technological Research Council of Turkey (TUBITAK), in collaboration with
the Faculty of Science, Istanbul Technical University (ITU), for facilitating all organizationaltasks before and during IChO 2020.
On behalf of the Scientific Committee,
Dr. Arif DAŞTAN

1

52nd International Chemistry Olympiad, Istanbul, TURKEY

Table of Contents
Authors ....................................................................................................................................... 4
Physical Constants and Equations .............................................................................................. 5
Periodic Table of Elements ........................................................................................................ 71

H NMR Chemical Shifts ........................................................................................................... 8

Typical Coupling Constants ....................................................................................................... 813

C NMR Chemical Shifts .......................................................................................................... 9

IR Absorption Frequency Table ................................................................................................. 9
Fields of Advanced Difficulty .................................................................................................. 11Part I: Theoretical Problems ..................................................................................................... 13
Problem 1. Salvia Species Growing in Turkey: Isolation and Total Synthesis of Abietane
Diterpenoids ............................................................................................................................. 14
Problem 2. Istanbulins and Related Sesquiterpene Natural Products ....................................... 19
Problem 3. Çay, Cha, Chai, Te, Tea, Tee, Thé, Thee, and Earl Grey Tea Flavor: Bergamot .. 23
Problem 4. Early Russian Organic Chemists and Markovnikov’s Rule .................................. 25
Problem 5. Arndt–Eistert Homologation .................................................................................. 29
Problem 6. Atovaquone ............................................................................................................ 32
Problem 7. Which is (±)-Trikentrin A? .................................................................................... 36
Problem 8. Stereoisomers of 1,2,3-Triphenylpropane-1,3-diol ................................................ 40
Problem 9. NMR, Symmetry, and Structural Analysis ............................................................ 41
Problem 10. Woodward–Hoffmann Rules and Pericyclic Reactions ....................................... 44
Problem 11. Benzoporphyrin .................................................................................................... 48
Problem 12. Blue to Green, Turquoise ..................................................................................... 52
Problem 13. Spinel Oxides ....................................................................................................... 58
Problem 14. Platinum Complexes as Anticancer Drugs........................................................... 62
Problem 15. Sodium Compounds from Salt ............................................................................. 66
Problem 16. Thermal Springs of Turkey and Sulfur Chemistry .............................................. 69
Problem 17. Electrochemical Determination of Rutin ............................................................. 73
Problem 18. Particle in a Box Problem: Free Electron Model ................................................. 77
Problem 19. Harmonic Oscillator and Rigid Rotor Models ..................................................... 79
Problem 20. Journey to Different Earth-Like Planets .............................................................. 81
Problem 21. Rate Constant Models and Kinetic Isotope Effect ............................................... 83
Problem 22. Parallel Reaction Kinetics .................................................................................... 86
Problem 23. Reaction Kinetics with Absorbance Measurement .............................................. 882

52nd International Chemistry Olympiad, Istanbul, TURKEY

Problem 24. Acridine Orange / DNA Binding Interactions ..................................................... 90
Problem 25. Spectrophotometric Determination of an Antihistaminic Drug ........................... 94
Part II: Practical Problems ........................................................................................................ 96
Problem P1. Drug Delivery from a Polymeric Hydrogel System .......................................... 100
Problem P2. Determination of the Total Carbon Content of Oltu Stone (Black Amber)
Samples................................................................................................................................... 105
Problem P3. Spectrophotometric Determination of the Equilibrium Constant for the Formationof a Complex .......................................................................................................................... 109
Problem P4. 1-Bromobutane .................................................................................................. 113
Problem P5. Cannizzaro Reaction .......................................................................................... 117
Problem P6. 2,3-Dihydro-5,6-diphenylpyrazine .................................................................... 124
Problem P7. Determination of Rate Constants for n-Butyl Acetate Hydrolysis .................... 127
Problem P8. Activation Energy of Bromide / Bromate Reaction .......................................... 130

3

52nd International Chemistry Olympiad, Istanbul, TURKEY

Authors
ALANYALIOĞLU, Murat, Atatürk University
ARSLAN, Yasin, Burdur Mehmet Akif Ersoy UniversityAYDOĞAN, Abdullah, İstanbul Technical University
BOZKAYA, Uğur, Hacettepe University
BURAT, Ayfer Kalkan, İstanbul Technical University
DAĞ, Ömer, Bilkent UniversityDAŞTAN, Arif, Atatürk University (Chair of Scientific Committee)ELTUĞRAL, Nurettin, Karabük UniversityGÖLCÜ, Ayşegül, İstanbul Technical University
KANBUR, Yasin, Karabük UniversityKILIÇ, Hamdullah, Atatürk UniversityMETİN, Önder, Koç UniversitySARAÇOĞLU, Nurullah, Atatürk UniversityTÜRKMEN, Yunus Emre, Bilkent University
ÜNLÜ, Caner, İstanbul Technical University
YILMAZ, İsmail, İstanbul Technical University

Edited by: SARAÇOĞLU, Nurullah, Atatürk University

4

52nd International Chemistry Olympiad, Istanbul, TURKEY

Physical Constants and Equations
Avogadro's constant, 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 = 9.10938215 × 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 = 1.66053904 × 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∆𝐺 0 = −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:

0

𝑅𝑇

𝐶

𝐸 = 𝐸 + 𝑛𝐹 𝑙𝑛 𝐶 𝑜𝑥

𝑟𝑒𝑑

ℎ𝑐

Energy of a photon:

𝐸= 𝜆Integrated rate law
Zeroth order:First order:[𝐴] = [𝐴]0 − 𝑘𝑡𝑙𝑛[𝐴] = 𝑙𝑛[𝐴]0 − 𝑘𝑡Second order:
1
1
=
+ 𝑘𝑡
[𝐴] [𝐴]0
Arrhenius equation: 𝑘 = 𝐴𝑒 −𝐸𝑎/𝑅𝑇
Equation of linear calibration curve: 𝑦 = 𝑚𝑥 + 𝑛
552nd International Chemistry Olympiad, Istanbul, TURKEY

Standard deviation:2∑𝑁
𝑥=1(𝑥1 − 𝑥̅ )
√
𝑠=
𝑁−1

Lambert–Beer equation: 𝐴 = 𝜀𝑙𝑐

6

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 Chemistry7

52nd International Chemistry Olympiad, Istanbul, TURKEY
1

H NMR Chemical Shifts

Typical Coupling Constants

8

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 insymmetricalalkynesN–Hstretch3500–3300

C–N

stretch1360–1080

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
medium (primary amines have
two bands; secondary have one
band, often very weak)medium-weak9

52nd International Chemistry Olympiad, Istanbul, TURKEY
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

bending

1600

mediumstretchstretch3100–3000
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

10

52nd International Chemistry Olympiad, Istanbul, TURKEY

Fields of Advanced Difficulty
Theoretical
1. Pericyclic reactions (Cycloaddition and electrocyclization reactions).
2. Nucleophilic substitution reactions at sp2 carbon centers.
3. Spectroscopy: Basic 1H and 13C NMR spectroscopy (chemical shifts, signal multiplicity,
intensity and coupling constants); simple IR spectroscopy.
4. Kinetics: Rate constant models and kinetic isotope effect.
5. Basic quantum chemistry: Electronic energy levels, transitions applied to conjugated systems,
vibrational and rotational motions of molecules (formulas provided), and simple theories ofconjugated systems.
6. Inorganic chemistry: Coordination chemistry (crystal structure, crystal field theory, andisomerism) and molecular orbital energy diagrams of homo/heteronuclear diatomic molecules.
Notes:i) The following topics WILL NOT appear in the exam set:








Metal-catalyzed cross-coupling reactions and olefin metathesis reactions.
Use of Microsoft Excel or any related computer software.
Use of derivatives and integrals.
Although a few examples in the preparatory problems are related to biomolecules,
students are not expected to cover any biochemistry or carbohydrate chemistry as
advanced topics.
Inorganic reaction mechanisms.
Molecular orbital diagrams of polyatomic molecules.

ii) Unless important, the reaction conditions such as solvent and temperature have not beenshown on the arrows in the reaction schemes.

Practical
1. Use of a spectrophotometer (mono/double-wavelength measurements).
2. Basic techniques in organic synthesis: recrystallization, thin layer chromatography (TLC),
filtration, and drying of precipitates following the described procedures.
3. Distillation and extraction.

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52nd International Chemistry Olympiad, Istanbul, TURKEY

Notes:
During the practical exam, students WILL NOT be expected to: Determine melting points.
 Use a rotary evaporator.
 Handle and work up moisture-sensitive compounds (using syringes and balloons).
 Perform column chromatography.
 Produce the hydrogel system by polymerization through the experiments.

12

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICALPart I:
Theoretical
Problems

13

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Problem 1. Salvia Species Growing in Turkey: Isolation andTotal Synthesis of Abietane Diterpenoids
The genus Salvia, named after a Latin word, salvare (“healer”), has a variety of species with
important medicinal activities. They have been used for the treatment of colds, flu, andmenstrual disorders in most regions of the world since ancient times. In Turkish folk medicine,
Salvia L. species have also been used as a carminative, diuretic, hemostatic, spasmolitic, andstomachic, and in the treatment of mouth and throat irritations due to their antibacterial andwound healing properties. The genus Salvia includes over 900 species across the world, 58 ofwhich are endemic in Turkey.
Female Turkish scientists Ulubelen & Topçu with co-workers have studied Anatolian Salvia
plants growing in Turkey, and isolated and characterized more than 320 natural products, mostof which are terpenoids, while one third are new diterpenoids.

In one of their studies on Salvia multicaulis Vahl., Ulubelen & Topçu isolated four new aromatic
abietane norditerpenoids (1–4), which showed strong antituberculous activity. In addition to theantibacterial and antifungal activities of the isolated diterpenoids, the plant extracts also showedantioxidant, antiinflammatory, and cholinesterase inhibitory activities. S. multicaulis has
folkloric use in Anatolia, such as an appetizer, for wound healing, against scorpion stings, andin the treatment of respiratory and urinary infections and diabetes.

1452nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Later, a research group in Turkey developed a synthetic route to obtain derivatives of natural
products 1–4. This problem covers the synthesis of related compounds. The following reaction
schemes illustrate the total synthesis of diterpenoids 1 and 5.
1.1. Draw the structure of the products A–M, without any stereochemical detail. Hint: In
second step (𝐀 → 𝐁), combination of lithium bromide and cerium(IV) ammonium nitrate(CAN) is used as a brominating reagent. Compound C is a benzaldehyde derivative and used in
the synthesis step of compound M.
1.2. During the cyclization of H to I-1, another isomeric compound, I-2, with the formula
C18H20O, is also formed. Draw the structure of I-2.

15

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

1.3. The following reaction scheme is related to the synthesis of 6, a desmethyl derivative of thediterpenoids 1 and 2. Draw the structures of products N–Y, without any stereochemical detail.
Hint: Compounds R, S and T exhibit acidic character. The transformation of compound V to
W includes Robinson annulation and a possible deformylation reaction steps.
16

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

1.4. During the transformation of compound V to W (Robinson annulation step), the use of aprecursor of the α,β-unsaturated ketone, such as a β-chloroketone or N,N,N,-trialkyl-3oxobutan-1-aminium halide (as used in the reaction scheme), can be more favorable. Explain.
1.5. Draw possible tautomeric forms of compound V.
1.6. Compound Y can be also obtained via ring-closing (electrocyclization) of the compound
Z. Draw structure of Z.

17

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

1.7. For the transformation of X to Y, which of the following reagents can also be used? (Ignore
SN2' type reactions).
☐
☐
☐
☐
☐

i) PBr3/pyridine; ii) n-Bu3SnH/AIBNi) PBr3/pyridine; ii) Na/t-BuOHi) MnO2; ii) DDQ
i) TsCl/pyridine; ii) LiAlH4i) TsCl/pyridine; ii) DBU

18

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Problem 2. Istanbulins and Related Sesquiterpene Natural
Products

Some elements received their names from different places around the world. In this respect, therecord belongs to the Swedish village of Ytterby, after which four elements were named:ytterbium (Yb), yttrium (Y), erbium (Er), and terbium (Tb). However, elements are not the only
chemical entities that owe their names to such places. Interestingly, a class of natural products,
istanbulins A–E, received their names from the city of Istanbul. The first two members of thisfamily, istanbulins A and B, were first isolated by Prof. Dr. Ayhan Ulubelen and co-workers
from the plant Smyrnium olusatrum in 1971. The isolation of the remaining members,
istanbulins C–E, was reported by Ulubelen and co-workers between 1979 and 1982.

19

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Istanbulins constitute a subclass of a much larger family of natural products called
sesquiterpenes. Two important sesquiterpene natural products with a similar 6-6-5 fused ringsystem are vernolepin (1) and vernomenin (2). Danishefsky and co-workers reported an elegant
total synthesis of these two natural products in 1976 via the utilization of the Diels–Alder (DA)chemistry of the so-called Danishefsky’s diene.
Please note that all formulae depicting chiral molecules in this question refer to racemic
mixtures.

In this context, Danishefsky’s diene (3) and the Rawal–Kozmin diene (4) are two electron-rich
dienes that found widespread use in organic synthesis, and their structures are shown below.

TMS: trimethylsilyl; TBS: tert-butyldimethylsilyl
2.1. Draw the major resonance structures of dienes 3 and 4. Indicate the carbon atoms withhigher electron density on each diene.
2.2. Compounds 3 and 4 have been extensively used as diene components in Diels–Alderreactions. Draw the conformations of 3 and 4 required to be able to enter a DA reaction. Predict
which compound is a more reactive diene in a DA reaction with maleic anhydride (5).

20

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

2.3. When a mixture of Danishefsky’s diene (3) and compound 6 was heated followed bytreatment with acid (TsOH, p-toluenesulfonic acid), compound A was obtained as the major
product.
Draw the structures of all possible Diels–Alder products with the molecular formula of
C12H14O3 that can be obtained from the reaction of 3 and 6. Drawing only one enantiomer of anenantiomeric pair is sufficient.

2.4. Determine the structure of the major product A.
2.5. Diels–Alder adduct A was converted to compound 7 via a sequence of 4 steps as shown
below. Compound B is known to be acidic. Draw the structures of B–D.

2.6. When compound 7 is reacted with 1 equiv of m-CPBA, product E was obtained as a major
product. Circle the functional group that reacts selectively with m-CPBA, and draw thestructure of E.

21

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

2.7. The syntheses of vernolepin (1) and vernomenin (2) were completed as shown in the scheme
below. Draw the structures of compounds F–J. In the final step, compound I is the precursor
of 1.

22

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Problem 3. Çay, Cha, Chai, Te, Tea, Tee, Thé, Thee, and Earl
Grey Tea Flavor: Bergamot

Çay

Thee
Thee
Cha
Thee
Tea Shay Tea
Te
Shay Tejay čaj čaj Tee Shay ChaaÇč e ChaTheečaj Çay čaj Theeečaj Thee
The
Thee
Th
Thee
Tea
čaj
Thee
Te
Te
Thee
Shay
Shay
Thee
Çay
Thee
Te
Cha
čaj Thee Thee Thee Cha
Tee Thea čaj Teh

Te
a

a

y
Ça h
Te

ay
Sh

Çay Teačaj Chai Shay

y
Ça
Te
i
j T
a
e Thee Ch

The

Cha Chai

TečaÇay ShayÇaySha
y

Cha
Teh

Ch

Chinese, Japanese, Korean, Portuguese…..
Russian, Persian…
Turkish, Azerbaijani…
Bosnian, Croatian, Czech, Serbian, Slovak..
Arabic…
Italian, Spanish…
English…
German…
French…
Dutch…
Hindi…
…

y
Ça

Cha
Chai
Çay
čaj
Shay
Te
Tea
Tee
Thé
Thee
Chaay
…..

ee
Th

Tea (in Turkish: çay) is popular throughout Turkey and the Turkish diaspora. Turkish tea
culture also extends from Azerbaijan to some countries in the Balkan Peninsula. Turkey has thehighest per capita tea consumption in the world, i.e. 2.5 kg/person per year, followed by
the United Kingdom (2.1 kg/person per year).

Bergamotene and derivatives (1-4), sesquiterpenes, are analogues of pinnae monoterpenes.
23

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICALFound in bergamot oil, the bergamotenes contribute to the aroma and flavor of Earl Grey tea.

3.1. The following reaction scheme illustrates the synthesis of α-trans-bergamotene (1). Draw
the structures of products AG.
3.2. What is the function of Me3NO reagent in the transformation of A to B?

24

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Preparatory problems: THEORETICAL

Problem 4. Early Russian Organic Chemists and Markovnikov’s
Rule

The last year was devoted to the 150th anniversary of the discovery of Markovnikov's rule,
formulated by Vladimir V. Markovnikov in 1869. Markovnikov was a PhD student of thefamous early Russian scientist Alexander Butlerov. In his PhD thesis in 1869, Markovnikov
discovered the famous rule that exists in almost every textbook on organic chemistry. According
to Markovnikov’s rule, when an unsymmetrical alkene or alkyne reacts with a hydrogen halide(hydrogen chloride, hydrogen bromide, or hydrogen iodide), the hydrogen atom of HX adds tothe carbon atom having the highest number of hydrogen atoms. However, depending on thereagent or substrate, in some cases, opposite results could also be possible, and these kinds ofreactions are called anti-Markovnikov addition. Although Markovnikov's rule was developed
for and is specifically applied to the addition of hydrogen halides to alkenes or alkynes, manyother additions are also described as Markovnikov or anti-Markovnikov depending on theregioselectivity of the addition reaction.

25

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Actually, the rule should be revised as follows: “addition to this kind of double or triple bondproceeds through more stable intermediates”. In some cases, besides electronic effects, stericeffects can also affect the formation of Markovnikov or anti-Markovnikov addition products.
The following problems are mainly related to discoveries described by the student of the more
distinguished organic chemist Alexander Butlerov or his colleagues at Kazan University,
Tatarstan, Russia.
4.1. Draw the structures of major products A-E, including the appropriate stereochemistry(ignore optical isomerism).

4.2. Draw the structures of major products F and G for the following reactions.

Wagner–Meerwein Rearrangement (WMR)
Wagner is another famous scientist who worked at Kazan University contemporaneously withButlerov and Markovnikov. Wagner proposed that bornyl chloride undergoes an internal
rearrangement to form pinene. Meerwein then generalized this type of rearrangement. Thus,
this kind of reaction was named Wagner–Meerwein rearrangement. These reactions take placewhen a carbocation is formed. Generally, a carbocation is rearranged to a more stable26

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

carbocation, if possible, by neighboring group migration. In addition, if the reaction does notproceed through a carbocation or borderline carbocation intermediates, rearrangements do nottake place.
4.3. Considering the formation of intermediates for every reaction, draw the structures ofreagents H and I and major products J–M.

Acid-catalyzed Wagner–Meerwein Rearrangement
The acid-catalyzed reaction of 4,4-dimethylcyclohexa-2,5-dien-1-one resulted in the formationof a compound, the NMR data of which are given below.

27

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICALFor N; 1H NMR (300 MHz, CDCl3): δ = 6.95 (d, J = 8.0 Hz, 1H), 6.61 (d, J = 2.8 Hz, 1H), 6.57
(dd, J = 8.0, 2.8 Hz, 1H), 5.39 (bs, 1H), 2.16 (s, 3H), 2.14 (s, 3H). 13C NMR (100 MHz, CDCl3):
δ 153.4, 137.9, 130.4, 128.6, 116.6, 112.3, 19.8, 18.7.
4.4. Find the structure of product N and propose a plausible mechanism.
4.5. What kind of difference do you expect in the 1H NMR spectrum after a drop of D2O isadded to the solution in the NMR tube?
Zaitsev’s Rule
Zaitsev, who described a rule named after him (Zaitsev's or Saytzeff's or Saytzev's rule), wasanother PhD student of Butlerov’s. Zaitsev's rule is an empirical rule for estimating preferredalkene product(s) in elimination reactions. At Kazan University, the chemist Alexander Zaitsev
studied various elimination reactions and observed a general trend in the resulting alkenes. Moregenerally, Zaitsev's rule stipulates that in an elimination reaction the most substituted productwill be formed. The following problem is mainly related to Zaitsev’s rule.
4.6. Draw the structures of elimination products O–Q and compound R. What is the major
product formed by the thermal reaction of R described in the following scheme?

4.7. Which base(s) can be used to increase the ratio of Q relative to EtONa?
☐
☐
☐
☐
☐
☐
☐

NaOMe
KOMe
i-PrOK
t-BuOK
NH3
DBU
i-Pr2NEt28

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Preparatory problems: THEORETICAL

Problem 5. Arndt–Eistert Homologation
Fritz Georg Arndt (6 July 1885–8 December 1969) was a German chemist who had a great
influence on the development of chemistry in Turkey. He was employed for two decades of his
professional life at Istanbul University in two distinct periods. He discovered the Arndt–Eistert
synthesis with Bernd Eistert. The Arndt–Eistert synthesis is the chemical reaction for onecarbon homologation (i.e. the conversion of RCO2H to RCH2CO2H) of carboxylic acids and iscalled the homologation process. In the Arndt–Eistert homologation, the key step is the Wolff
rearrangement of diazoketones to ketenes, which can be achieved thermally, photochemically,
or by silver (I) catalysis. The reaction is conducted in the presence of nucleophiles such as water,
alcohols, or amines to capture the ketene intermediate to yield carboxylic acids, esters, or
amides, respectively. In this problem, synthesis of indolizidine alkaloids is studied.

5.1. As depicted in the scheme below, the synthesis of indolizidines 167B and coniceine could
be easily and concisely achieved from -unsaturated ester B. The key step (A → B) is theWolff rearrangement. Compound C has a lactam core, which is a bicyclic heterocycle
containing a six-membered ring fused to a saturated five-membered ring, one of the bridging
atoms being nitrogen.
Draw the structures of A–D without any stereochemical detail.

29

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Preparatory problems: THEORETICAL

5.2. In the Arndt–Eistert homologation reaction, an α-diazo ketone can undergo photochemical
Wolff rearrangement to form α-ketocarbene via nitrogen extrusion. This intermediate undergoes
a 1,2-alkyl shift to give the ketene product.
Draw the structures of the α-ketocarbene and ketene intermediates in the second step (A → B).
5.3. Addition of propylmagnesium bromide to compound C, followed by AcOH/NaBH4, is thelast step in the total synthesis of indolizidine 167B.
Draw the structure of an intermediate (C11H20N+) in the fourth step (C → D).
5.4. An alternative synthesis of coniceine is depicted below. Draw the structures of E–J.

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Preparatory problems: THEORETICAL

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Problem 6. Atovaquone
Atovaquone, an approved drug, is used to treat pneumocystosis and malaria. Ketoester 1 andaldehyde 2 are key compounds in the synthetic process of atovaquone.

6.1. The synthesis of key compound ketoester 1 is shown below. A mixture of phthalicanhydride and Et3N is treated with diacid. Gas evolution is observed during this period.
Treatment of the reaction mixture with aq. HCl solution provides formation of acid 3 throughintermediate A with two carboxylic acid groups. Acid 3 is converted to the isomericintermediate B, containing both hemiacetal and ester functionalities, followed by dehydration
to the alkene C, which is then brominated to give D under acidic condition. Dibromide D
undergoes solvolysis in a hot mixture of H2O/AcOH to give tertiary carbocation intermediate
E, which is then trapped with water to give intermediate hemiacetal F. Finally, rearrangementof intermediate hemiacetal F provides key compound 1.
Note: The square brackets denote that the product was not isolated but reacted further without
purification. The conversion of 3 to 1 is a one-pot reaction that involves a series of reactions occurring
one after another in the same vessel without isolation and purification of intermediates.

32

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Preparatory problems: THEORETICAL

Spectroscopic data for intermediates B and C: B: 1H NMR  = 7.86–7.52 (4H), 4.13 (bs, 1H,
exchangeable with D2O), 1.97 (s, 3H). C: 1H NMR  = 7.92–7.58 (4H), 5.24 (m, 2H); 13C NMR
 = 166.8, 151.8, 139.0, 134.4, 130.4, 125.3, 125.1, 120.6, 91.3; MS m/z = 146.0
Draw the structure of intermediates A–F in the synthesis of 1.
6.2. The synthesis of aldehyde 2 starts from cyclohexene by key steps including Friedel–Crafts
acylations, haloform, reduction, and oxidation. Friedel–Crafts acylation of cyclohexene withacetyl chloride yields chlorocyclohexyl methyl ketone J. Reaction of cyclohexene with acetyl
chloride produces an initial carbocation G that undergoes two successive Wagner–Meerwein
hydride migrations to form isomeric carbocations H and I, respectively. Trapping of carbocation
I with chloride ion produces J, the Friedel–Crafts reaction of which with chlorobenzene
provides K. Haloform reaction of methyl ketone K using sodium hypochlorite (NaOCl) gives
the corresponding acid L. Acid L is converted into the aldehyde 2 in a several-step reaction
sequence.
Draw structure of isomeric carbocations GI formed in this reaction.

6.3. Are these carbocations chiral?

G

☐
☐

Yes
No

H

☐
☐

Yes
No

I

☐
☐

Yes
No

6.4. Draw the structure of JL.

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Preparatory problems: THEORETICAL

6.5. Choose all correct statements for L.
☐
☐
☐
☐
☐
☐
☐

L has 4 stereoisomers.
L is a chiral compound.
L is an achiral compound.
L is a meso compound.
L has 2 stereoisomers.
Stereoisomers of L are diastereomers of each other.
Stereoisomers of L are enantiomers of each other.

6.6. Which of the following compound(s) result(s) in the haloform reaction of K?
☐☐☐☐

CH2Cl2
CH3Cl
CHCl3
CCl4

6.7. Which of the following reagents are appropriate to form aldehyde 2 from L?
Choose all correct reactions.

☐

☐

☐

☐

☐

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Preparatory problems: THEORETICAL

☐

☐

35

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Problem 7. Which is (±)-Trikentrin A?
Although the indole skeleton is ubiquitous in nature, annulated indoles at any of the benzenoid
positions are uncommon. The trikentrins and the structurally similar herbindoles represent
fascinating such examples of 6,7-annulated indole or polyalkylated cyclopent[g]indole natural
products. The trikentrins were isolated from the marine sponge Trikentrion flabelliforme andpossess antibacterial activity. Possible structures for trikentrin A are shown in the Figure below.
In this problem, we will find out which of these structures is trikentrin A.

There are several ways to synthesize trikentrin A. Two routes below involve aryne-based andhydrovinylation strategies and both finally lead to the formation of trikentrin A. The first stepfor problems 8.1 and 8.2 is the Bartoli reaction or Bartoli indole synthesis, which is the organic
reaction of ortho-substituted nitroarenes with vinyl Grignard reagents to yield substituted
indoles. In particular, it is the most efficient route to 7-substituted indoles.

36

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL(±)-Trikentrin A: 13C NMR (CDCl3): δ 143.4101.6 (8 signals), 44.815.1 (7 signals).

Aryne-based strategy

7.1. Draw the structures of A–I.
7.2. Draw the structure of the aryne involved as a reaction intermediate in step D → E.

37

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Preparatory problems: THEORETICAL

Hydrovinylation strategy

7.3. Chemical transformation of bromo-nitrobenzene to corresponding 7-vinylindole J includes
in Bartoli reaction followed by the vinylation step with vinylstannane. Draw the structure of J.
7.4. The second step is the Ni(II)-catalyzed asymmetric hydrovinylation of J. The ligands (K1–
K4) used for hydrovinylation are given above.
Note: ee = enantiomeric excess; % ee = % major enantiomer - % minor enantiomer
Choose the correct statement(s):☐☐☐☐Ligand 3 gave the best enantioselectivity.
Ligand 4 gave a racemic mixture.
Each of the ligands K1–K4 is chiral.
Each of the ligands K1–K4 gave excellent yield (>95%) of the product.
38

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Preparatory problems: THEORETICAL

7.5. For the hydrovinylation step, choose the correct statement(s):☐☐☐☐(allyl)2Ni2Br2 or [(allyl)NiBr]2 is a source of vinyl.
In this Ni-allyl complex, each nickel has oxidation number +2.
In this Ni-allyl complex, the electron count of Ni is 18.
This complex has a square planar geometry.

7.6. Draw the structures of L–P. The absolute configuration of the asymmetric center in thehydrovinylation product is S. Hint: In the 13C NMR spectrum of compound M, one carbonyl
carbon signal was observed at δ = 178.3 ppm.

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Preparatory problems: THEORETICAL

Problem 8. Stereoisomers of 1,2,3-Triphenylpropane-1,3-diol

8.1. Draw all possible stereoisomers of 1,2,3-triphenylpropane-1,3-diol.
8.2. List all the achiral compounds.
8.3. List all the chiral compounds.
8.4. Which of the following properties or methods can be used to distinguish between the chiralcompounds from question 8.3? Choose all correct statements.
☐
☐
☐
☐
☐
☐
☐
☐

boiling point
UV spectroscopy
refractive index
melting point
optical rotation
dipole moment
NMR spectroscopy in an achiral environment
IR spectroscopy

40

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Preparatory problems: THEORETICAL

Problem 9. NMR, Symmetry, and Structural Analysis
Naphthalene halides: Key compounds for many applications
Besides benzene, naphthalene is one of the best-known aromatic hydrocarbons. Therefore, thechemistry of naphthalene (1) has been extensively studied and many naphthalene derivatives
have been synthesized. Halogen derivatives of this kind of compound are key for manytransformations. For this reason, nearly all halogenated derivatives of naphthalene are known
in the literature. Both 1H and 13C NMR spectra of symmetric compounds are characteristic, andallow researchers to exclude possible non-symmetrical structures to analyze the correct
structures. Let us consider naphthalene tetrabromide isomers 2.

9.1. Draw the structures of all naphthalene tetrabromide(s) with 3 signals in 13C NMR and one
signal (singlet) in 1H NMR spectra.
9.2. Draw the structures of all naphthalene tetrabromide(s) with 5 signals in 13C NMR spectra.
9.3. Draw the structures of all naphthalene tetrabromide(s) with 6 signals in 13C NMR and adoublet (J = 8–9 Hz) in 1H NMR spectra.
9.4. Draw the structures of all naphthalene tetrabromide(s) with 6 signals in 13C NMR and adoublet (J = 1.5–2.0 Hz) in 1H NMR spectra.
Dynamic NMR: fast transformation between tautomeric forms and identical nuclei in
NMR
Bullvalene (3) is very suitable for degenerate Cope rearrangements. Without counting
enantiomers, the number of possible valence tautomers of a bullvalene with ten distinguishable
positions is 10!/3 = 1,209,600. This arrangement enables all carbon and hydrogen atoms toappear equivalent on the NMR timescale. At sufficiently high temperature, both 1H NMR and13
C NMR spectra of bullvalene show only one signal, average to a rounded peak. However, at−60 °C, as Cope rearrangements do not take place, olefinic and aliphatic protons are observed
separately.
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Preparatory problems: THEORETICAL

9.5. At low temperature, ignoring any Cope rearrangement, how many carbon signals do you
expect from the 13C NMR spectrum of bullvalene?
Label identical carbon atoms with letters a, b, c… on the molecular structure.
9.6 Owing to fast tautomerism, some molecules give clearer spectra due to apparent symmetry.
In light of this information, how many signals do you expect from the 13C NMR spectra of thefollowing compounds?

9.7. In the literature, it has been shown that the tropolone diacetate derivative 4 has fewer signals
than expected in 13C NMR spectroscopy.
Draw reasonable resonance structure(s) and/or transformation(s) responsible for this symmetry.
How many signals do you expect for this molecule in the 13C NMR spectrum?

42

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Preparatory problems: THEORETICAL

Stereochemistry of the epoxidation reaction of bicyclic alkenes.
9.8. Considering the following pieces of information, draw the structures of all possible
stereoisomers formed under the given reaction conditions.
Hint: A and B are isomers with 3 signals and C is an isomer with 4 signals in 13C NMR
spectroscopy.

9.9. Draw the structures of the stereoisomer(s) formed under the given reaction conditions. How
many signals do you expect for the epoxide product(s) in 13C NMR spectra?

4352nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Problem 10. Woodward–Hoffmann Rules and Pericyclic
Reactions
The Woodward–Hoffmann rules (or the pericyclic selection rules), developed by Robert B.
Woodward and Roald Hoffmann, are used to rationalize or predict some stereochemical aspects
and the activation energy of pericyclic reactions. They are for all classes of pericyclic reactions(and their reverse 'retro' processes), such as cycloadditions, sigmatropic shift, electrocyclization,ene, and cheletropic reactions.
Woodward–Hoffmann rules for electrocyclic reactions
System4n
4n+2

Conditions
thermal (∆)
photochemical (hν)
thermalphotochemical

Motionconrotatory (con)disrotatory (dis)disrotatory
conrotatory44

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Preparatory problems: THEORETICAL

10.1. Thermal reaction of compound 1 results in the formation of endiandric acid 2 by a seriesof pericyclic reactions. Show all steps and classify their pericyclic processes.

How many π electrons are involved in the following reactions? Are these reactions thermally or
photochemically allowed according to the Woodward–Hoffmann rules?
10.2.

10.3.

10.4. Domino Diels–Alder reaction of A with succinimide results in the formation of adduct 3.
Draw the structures of A–C.

10.5. The following reaction scheme illustrates the synthesis of endo-isomer of benzenoid
tetracyclic hydrocarbon I starting from o-xylene. Br2-elimination of tetrabromo-o-xylene D45

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

with sodium iodide leads to a reactive intermediate which undergoes a 4 electrocyclization toyield compound F. Draw the structures of intermediates and products D–I.

retro-Diels–Alder Reaction
The retro-Diels–Alder (rDA) reaction is the reverse of the Diels–Alder reaction, i.e., theformation of diene and dienophile from cyclohexene. Generally, an rDA reaction is initiated byheating. In some cases, low temperature is sufficient for this transformation, depending on thenature of the substrate.
10.6. Cyclopentadienes are very useful synthetic intermediates in the fields of organic andcoordination chemistry. Parent (unsubstituted) cyclopentadiene is obtained by the thermal
decomposition of dicyclopentadiene. However, substituted cyclopentadienes are generally
unstable due to the facile migration of the endocyclic double bonds. Consequently, practicaland general methods for the synthesis of substituted cyclopentadienes are limited. In thefollowing reaction scheme, the synthesis of a substituted cyclopentadiene derivative is given.
Besides rDA, some steps also involve the inverse-Diels–Alder reaction, which is a
cycloaddition between an electron-rich dienophile and an electron-poor diene (such as tetrazine4), through the interaction of the HOMO orbital of dienophile and the LUMO orbital of diene.
Draw the structures of the intermediates and products J–N.

46

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10.7. Nucleophilic aromatic substitution reactions constitute an important class of reactions insynthetic organic chemistry. In the following scheme, the reactions of aryl halide 5 proceed via
two different kinds of intermediates in presence of a cyclic 1,3-diene depending on the reaction
conditions and the nature of the substituent on the aromatic ring. Draw the structures of products(O and P), and discuss possible intermediates responsible for the formation of these products.

47

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Preparatory problems: THEORETICAL

Problem 11. Benzoporphyrin
The name "porphyrin" derives from the Greek word porphyra, meaning purple. Porphyrins area group of macrocycle organic compounds, composed of four modified pyrrole subunits. They
have a total of 26 π-electrons, 18 of which form a planar porphyrin ring structure. They are often
described as aromatic. Metal complexes derived from porphyrins occur naturally. One of thebest-known families of porphyrin complexes is heme, the pigment in red blood cells. A
benzoporphyrin is a porphyrin with a benzene ring fused to pyrrole unit(s).
11.1. Benzoporphyrins can be prepared starting from a masked pyrrole derivative E. The
synthesis of E starts with a reaction of cis-1,2-dichloroethene and thiophenol to give A.
Oxidation of A yields B having phenylsulfonyl units. The cis-product B is then converted to its
trans isomer C when treated with a catalytic amount of Br2 under UV light. The Diels–Alderreaction between C and 1,3-cyclohexadiene under thermal conditions gives the product D,which is converted to a pyrrole carboxylic acid ester when reacted with ethyl isocyanoacetate.
Ester then is treated with TFA to give the pyrrole derivative E.
Draw the structures of compounds A–E including stereochemistry when necessary.

11.2. Porphyrins can easily be prepared via a cyclization reaction of pyrrole derivatives withaldehydes. Draw the structure of aldehyde F and determine the oxidation state of zinc incompound H.

48

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11.3. When H is heated under vacuum, it can give a more conjugated product through a retroDiels–Alder reaction.
To complete the structure of I, draw the structures of the dashed circle part of I (all the circles
are identical) and J.

49

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Preparatory problems: THEORETICAL

Ammonia is a major metabolic compound and the importance of its sensitive detection has beenemphasized recently because of its correlation with specific diseases. In normal physiologicalconditions, ammonia can be expelled from slightly alkaline blood and emanated through theskin or exhaled with the breath. Dysfunction in the kidney or liver that converts ammonia tourea can result in an increase in the ammonia concentration in breath or urine. Consequently,
the detection of the ammonia present in breath or urine can be used for the early diagnostics ofliver or stomach diseases. The development of sensor devices for measuring ammonia with asensitivity of 50 ppb–2 ppm and with a fast response time is highly desired.
For that purpose, I was used to prepare a fiber-optic ammonia gas sensor. Exposure of thissensor to ammonia changes the transmittance of the fiber-optic. By using an appropriate
spectrometer, ammonia gas in different concentrations was passed through the sensor and thechange in transmittance was measured. The results of these measurements are listed in the Table
below.
[NH3] (ppm)
0.500
1.00
2.00
4.00
7.00
9.00
11.0
20.0
25.0
30.0

Sensor response(%)–0.2540–0.7590–1.354–1.838–2.255–2.500–2.600–2.947–3.152–3.25650

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11.4. Using the linear region of sensor response data prepare a calibration curve and find thecalibration equation as 𝑦 = 𝑎 + 𝑏𝑥.
11.5. This sensor is then used for the detection of ammonia in human breath. When a kidney
patient’s breath was fed into the sensor, a –3.812% change in the response is observed.
Calculate the ammonia concentration in the patient’s breath.

5152nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Problem 12. Blue to Green, Turquoise
The beauty of the turquoise color of Lake Salda, where blue meets white sands, fascinates those
who see it. Lake Salda, in the southern province of Burdur’s Yeşilova district, has been referred
to as “Turkey’s Maldives” in recent years for its white sandy beaches and turquoise water. In
fact, turquoise is an opaque, blue to green mineral that is a hydrated phosphate of copper andaluminum with the chemical formula of CuAl6(PO4)4(OH)8·4H2O, and is known as a gemstone.
The word turquoise dates back to the 17th century and is derived from the French turquois,
meaning "Turkish" because the mineral was first brought to Europe through Turkey, from minesin the historical Khorasan Province of Persia. Phosphorus, which is also in the structure ofturquoise, is an essential part of life. Without the phosphates in biological molecules such as
ATP, ADP, and DNA, we would not survive. Phosphorus compounds can be found in theminerals in our bones and teeth. With few exceptions, minerals containing phosphorus are in
the maximally oxidized state as inorganic phosphate rocks, which are partially made of apatite,and they are today the chief commercial source of this element. Phosphate products are used as
fertilizers in agriculture. They are also used in animal feeds, as a leavening agent in baking
powder and flour, as an additive to beverages, and in pharmaceuticals. Industrial uses include
water softening, rust proofing, fire proofing, in insecticides and detergents, and for the
manufacture of elemental phosphorus.

Lake Salda
There are three important allotropes of phosphorus: X, Y, and Z. However, another form ofphosphorus, W, also exists (given below). X is a soft, waxy solid. It is exceptionally harmful5252nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

and to a great degree reactive and also displays chemiluminescence. Crystals of X are composedof P4 molecules. Y is obtained by heating X to 250 °C within the sight of daylight. It is
nonpoisonous and odorless. Y does not show chemiluminescence. It exists as a polymeric solid.
Z is produced from X under inert atmosphere. Z is the most stable allotrope of phosphorus andhas a layered structure. W is a form of phosphorus that can be produced by day-long annealingof Y above 550 °C.

The interconvertible forms of all allotropes of phosphorus
12.1. Identify allotropes of phosphorus indicated by X, Y, Z, and W.
12.2. Draw the structure of X, Y, Z allotropes of phosphorus and sketch the geometry of X.
12.3. P4 ignites suddenly in air at around 35 °C to form a phosphorus oxide derivative. Thus, it
is kept under water. When P4 reacts with different amounts of dry halogens, phosphorus
trihalides (PX3) or phosphorus pentahalides (PX5) are obtained. PX5 can also be obtained by thereaction of the halogens with PX3. The phosphorus pentahalides undergo hydrolysis in two steps
to form acid. The phosphoryl halides can be prepared by the hydrolysis of the appropriate
pentahalides in a limited amount of water or by the reaction of the trihalides with oxygen.
Dropping of the oxide derivative of phosphorus into water produces a hissing sound, heat, andacid product. The reaction of P4 with sodium or potassium hydroxide produces phosphine gas
as the major product and potassium or sodium hypophosphite as a by-product. Phosphine burns
in chlorine spontaneously, forming a phosphorus trihalide (PX3) or phosphorus pentahalide(PX5).
Write the formulas of products A–F.

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When phosphorus reacts with excess of halogens, it can form five-coordinated compounds suchas PCl5. Phosphorus mixed pentahalides like PF2Cl3 are prepared by the addition of one halogen
to the phosphorus trihalide of a second halogen.
12.4. Draw the Lewis structures of PCl5 and PF2Cl3 molecules.
12.5. By using VSEPR theory, predict the molecular geometries of PCl5 and PF2Cl3.
12.6. Estimate the polarity of PCl5 and PF2Cl3 molecules.
12.7. Compare the axial P–Cl bond length to the equatorial P–Cl bond length in PCl5.
12.8. Draw the hybridization scheme of the PF2Cl3 molecule and estimate which hybrid orbitals
are used to form the axial and equatorial bonds.
12.9. The synthesis of PH3 from hydrogen with white phosphorus is given below. Calculate
ΔH for the following reaction, using bond energies (single bond energies (BE) (in kJ.mol–1) for
P–P: 213, H–H: 435, P–H: 326).
𝑃4 (𝑔) + 6𝐻2 (𝑔) → 4𝑃𝐻3 (𝑔)Organophosphorus compounds are organic compounds containing phosphorus. Phosphorus canadopt a variety of oxidation states, and organophosphorus compounds are generally classified
based on their derivatives of phosphorus(V) or phosphorus(III), which are the predominantclasses of compounds. Organophosphorus compounds are widely used as nucleophiles andligands. Two major applications are as reagents in the Wittig reaction and as supporting54

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phosphine ligands in homogeneous catalysis. Their nucleophilicity is evidenced by theirreactions with alkyl halides to give phosphonium salts. Phosphines are nucleophilic catalysts inorganic synthesis, e.g., the Rauhut–Currier reaction and Baylis–Hillman reaction.
Triphenylphosphine (PPh3) is a common organophosphorus compound and it is widely used in
the synthesis of organic and organometallic compounds. When a toluene solution of compound
1 and excess of PPh3 are heated to reflux, first compound 2 is formed and then compound 3.

Spectral data of compounds 1–3 are given below (for 1H NMR and 13C NMR data [δ values(relative area)]:
1
H NMR

4.83
singlet

13

224.3
187.2
185.3
184.0
73.3

1

C NMR

IR
MS (m/z)2
7.62–7.41 (m, 15H)
4.19 (m, 4H)
231.0
194.9
189.9
188.9
129.0–134.7 (several peaks)
72.2
2038 cm–1
1958 cm–1
1906 cm–1
684.5

3
7.70–7.32 (m, 30H)
3.49 (s, 4H)
237.1
201.8
193.8
127.7–134.0 (several peaks)
68.80
1944 cm–1
1860 cm–1
919.7

12.10. Identify the structures of 2 and 3.
Hint: The 13C NMR signal of 1 at 224.3 ppm is similar to the chemical shift observed for carbene
carbons; the peaks between 184 and 202 ppm correspond to carbonyls; and the peak at δ 73.3
is typical for CH2CH2 bridges in dioxycarbene complexes.
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12.11. Determine if 2 is more likely to be the facial (fac) or meridional (mer) isomer.
Hint: The three ν(CO) bands with equal intensities are observed in the IR spectrum of compound
2. Protons of the carbene ligand occur as a multiplet in the 1H NMR spectrum.
12.12. Determine if 3 is more likely to be cis or trans isomer.
Hint: The two ν(CO) bands are of approximately equal intensity at 1944 and 1860 cm–1 in the
IR spectrum of compound 3. The 31P NMR spectrum shows a single resonance signal.
Some organophosphorus compounds such as sarin, soman, and VX are often referred as “nerve
gases” despite the fact that they are liquids at room temperature. Each country signing the 1997
Chemical Weapons Convention agreed to ban the development of chemical weapons and todestroy chemical weapons and associated production facilities by 2012. Sarin can be destroyed
by room temperature hydrolysis using aqueous Na2CO3 to give NaF and the sodium salt of anorganophosphate. The hydrolysis of nerve agent VX is more difficult. It reacts slowly withaqueous NaOH at room temperature, and the reaction has to be carried out at 360 K over several
hours.

12.13. Determine the organophosphorus salt formed in the following hydrolysis reaction.

Two chromium complexes containing the ligands CO, PF3, and PCl3 in octahedral geometry are
given below. In an octahedral complex, the molecular orbitals created by coordination can beseen as resulting from the donation of two electrons by each of six σ-donor ligands to the d56

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orbitals on the metal, called σ-bonding. π-bonding (Pi bonding) in octahedral complexes is also
possible when the ligand has p, d or π* molecular orbitals available. Ligands such as CO, CNand phosphines (of formula PR3) are π acceptor, with empty orbitals that can interact with metal
d orbitals in a π fashion. In most cases, the net back π bonding predominates, and electron
density is transferred from the metal to the ligand. π-bonding can affect metal-ligand bondenergy and bond length in carbonyl and phosphine complexes.
Answer the following questions considering the π interaction.
12.14. In which complex is the C–O bond shorter, Cr(CO)5(PF3) or Cr(CO)5(PCl3)?
12.15. In the infrared spectrum of which complex do the C–O stretching bands have higher
energy, Cr(CO)5(PF3) or Cr(CO)5(PCl3)?

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Problem 13. Spinel Oxides
The simple d-block oxides such as Fe3O4 and Co3O4 and many related mixed metal compounds
have important properties. They have structures related to the mineral spinel, MgAl2O4, andmay be given a general formula of AB2O4.
Stoichiometric amounts of two aqua complexes of transition metal (A and B) nitrate salts arethermally reacted to form a spinel AB2O4 crystalline solid that has a face-centered cubic (fcc)
structure with a unit cell composition of 8 AB2O4. Depending on the location of these twocations (A and B), the spinel structures are divided into two categories as normal and inverse
spinels. In a normal spinel, the A2+ ions occupy the tetrahedral holes and the B3+ ions occupy
the octahedral holes, but in the inverse spinel structures, the 2+ ions are replaced by half of the3+ ions in the structure.
Crystalline solid has an ordered structure in which the unit cell repeats along all 3 principal axesof a three-dimensional matter. The smallest group of atoms in the material that constitute thisrepeating pattern is the unit cell of the structure. The unit cell completely reflects the geometry
and structure of the entire crystal, which is built up by repetitive translation of the unit cell along
the principal axes. Face centered cubic (fcc) is one of a common structure type of crystalline
solid. Anions (X) are in the corners and faces of a cube (1/8 from each corners and ½ from eachfaces, because the corners and faces are shared by 8 and 2 unit cells, respectively) in the simplest
fcc structure. The cations (M) occupy the holes among the anions. There are 8 tetrahedral(corners) and 4 octahedral holes (1 at the middle and 3 on the edges, each edge has ¼ octahedralhole) in a fcc structure. Therefore, the unit cell composition is M4X4 with an empirical formulaof MX. However, the unit cell of a spinel structure is constructed by 8 of these fcc units.

29.746 g salt of A was mixed with 58.202 g salt of B in a thermal process to produce 24.724 g
pure product, AB2O4. In the spinel formation process, the metal ion of salt A keeps its oxidation
state but the metal ion B undergoes oxidation. Both salts contain the same number of the water58

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molecule(s), metal ion, and nitrate ion(s). Elemental analysis of the spinel provided thefollowing data: 6.538 g metal A and 11.786 g metal B. Assume the end product is a diamagnetic
solid matter. Considering the information provided above. Answer the following questions.
13.1. Suggest formulas for the salts of A and B.
13.2. Draw the structure of one of the complex ions i) without and ii) with one of the nitrates
being in the coordination sphere as a bidentate ligand and identify if the inversion center ispresent in the complexes. Inversion is a symmetry operation that translates every atom throughthe center to the opposite side.
13.3. Place the metal ions in an appropriate location in the crystal structure and suggest if it isa normal or inverse spinel.
The x-ray diffraction data of AB2O4 provides a unit cell parameter of 8.085 Å, which isconstructed from 8 fcc units and corresponds to a length of the edges of the cube.
13.4. Sketch one of the fcc units of AB2O4 and place the atoms in the unit.
13.5 What is the density of AB2O4? (hint: 1 Å is 1.0 x 10-10 m)
Reacting this spinel with other transition metals (M) produces M doped AB2O4, where M has achoice of occupying the place of either A or B-sides. The side product is AO (mono-oxide ofA).
13.6. M is Mn2+ in compound C and Ni2+ in compound D, suggest the location of Mn2+ andNi2+ ions in the structure of C and D, respectively. Assume splitting energy in Ni2+ and B3+ are
11500 cm-1 and 20800 cm-1 in the octahedral field, respectively, and the pairing energy is 19500
cm-1.
If the doping is in a small quantity or in some cases the doped metal ion behaves like a free ion
in the lattice (it means, electrons of M only feel the surrounding atoms and localized to M andits 1 shell of atoms in the structure). Assume Mn2+ is behaving like a free ion in the lattice andcreating its own localized electronic energy levels.
13.7. Draw the d-orbital splitting and identify if the Mn2+ species are paramagnetic ordiamagnetic.
Magnetic susceptibility could be calculated from the spin only formula:59

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICALµ(spin only) = (n(n+2))1/2,
where n is a number of unpaired electrons. However, some other electronic couplings affect themagnetic moment such that a correction term is needed. The correction term α is related to
ground state (where α = 4 for a non-degenerate and 2 for a degenerate ground state, degeneracyof a ground state can be determined from the electron configurations, such as completely filled
and half-filled set of orbitals creates a non-degenerate and a partially filled set of orbitals create
degenerate states) and λ = 88 cm-1 for Mn2+ and -315 cm-1 for Ni2+), and splitting energy (Δ is5000 cm-1 for Mn2+ and 11500 cm-1 for Ni2+) and the magnetic moment is:µ𝑒𝑓𝑓 = µ(𝑠𝑝𝑖𝑛 𝑜𝑛𝑙𝑦) (1 −

𝑎λ
).
∆

The magnetic susceptibility can be experimentally determined and it is interrelated with themagnetic moment (if we ignore the diamagnetic contributions) with the following formula:
3𝑘𝑋 𝑇

𝑒ℎ

µ𝑒𝑓𝑓 = √𝐿µ 𝑥µ𝑚 2 = 0.7977√𝑋𝑚 𝑇 and 1µ𝐵 = 4𝜋𝑚 = 9.27 × 10−24 𝐽𝑇 −10

𝐵𝑒

Where 𝑘 = Boltzmann constant; 𝐿 = Avogadro number; 𝜇0 = vacuum permeability; 𝑇 =temperature in Kelvin and 𝑋𝑚 (molar magnetic susceptibility)𝑖𝑠 𝑖𝑛 𝑐𝑚3 𝑚𝑜𝑙 −1.
13.8. What is the magnetic susceptibility of the products at 25 oC, if the samples C and D weigh25.433 and 25.471 g, respectively (each obtained from 24.724 g AB2O4)?
13.9. Place all the metal ions (A, B, Mn2+, and Ni2+) into their appropriate locations in the lattice
and fill up the following table. Use t2g for dxy, dxz, and dyz and eg for dx2-y2, dz2 orbitals inoctahedral (Oh) and t2 and e orbitals in tetrahedral (Td) cases. If there is distortion, predict thetype of distortion(s) and show the d-orbital splitting.

Hint: dxz, dxy, dyz orbitals are represented by t2g and t2 in octahedral and tetrahedral geometries,
respectively. Similarly, dx2-y2 and dz2 are represented as eg and e in octahedral and tetrahedral
geometries, respectively.
M

Local
geometry

Electronconfiguration

Degeneracy

Type ofdistortion60

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Problem 14. Platinum Complexes as Anticancer Drugs

Medicinal inorganic chemistry based on metal‐based drugs is broadly defined as the area of
research related to metal ions and metal complexes and their clinical applications. It is a newresearch area that developed from the discovery of the anticancer agent cisplatin. Cisplatin, cisdiamminedichloroplatinum(II), is a yellow powder and an anticancer drug widely used in thetreatment of a variety of tumors, especially those of the testes, ovaries, head, and neck.
The synthesis of cisplatin starts with K2[PtCl4], but has undergone several improvements since
it was published more than 100 years ago. The main problem is the occurrence of impurities
and the formation of the by-product trans-platin. Nowadays, the synthetic routes are mostly
based on a method published in the 1970s by Dhara. In the initial step, K2[PtCl4] is reacted withexcess KI, and the platinum complex is converted into the iodo analogue (A). Subsequently,
NH3 is added to the compound A and compound B is formed by ligand exchange in which two
NH3 ligands are exchanged with two iodo ligands. B is a yellow solid that is filtered, dried, andmixed with the aqueous solution of AgNO3. The insoluble AgI can be filtered off and cisdiamminediaquaplatinum(II) nitrate (C) is formed; then excess KCl is added to the solution of
C to yield cisplatin (D).
The success of the synthesis relies on the strong trans effect of the iodo ligands. The spectator
ligands T that are trans to the leaving group in square-planar complexes influence the rate of
substitution. This phenomenon is called the trans effect. Key point is that a strong σ-donor
ligand or π-acceptor ligand greatly accelerates substitution of a ligand situating in the transposition. Trans effects follow the order given below.
- For a T σ -donor: OH < NH3 < Cl < Br < CN , CH3 < I < SCN < PR3, H- For a T π-acceptor: Br < I < NCS < NO2 < CN < CO, C2H4

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14.1. Write the formulas of A–D.
14.2. Draw molecular structures of A–D.
14.3. Is the complex D polar?
14.4. Sketch the d-orbital splitting of cisplatin complex D in view of Crystal Field Theory andshow the electron distribution diagram.
14.5. Determine magnetic nature of complex A.
The platinum complex binds to DNA and causes cross-linking, which triggers the programmed
cell death (apoptosis). However, the other geometrical isomer of the square planar structuretransplatin, trans-diamminedichloroplatinum(II) (F), is not effective for the treatment of cancer.
Transplatin is synthesized starting from [Pt(NH3)4]2+ to which the first and second Cl– ligands
are added to form transplatin (F) as represented in the scheme below.

14.6. Draw the molecular structures of E and F.
The most important classes of antitumor agents, cisplatin, carboplatin, and oxaliplatin as
platinum(II) diamines are widely used in chemotherapy to treat a wide variety of cancers.

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However, the therapeutic index of these agents is relatively narrow; their use is often plagued
with severe toxicity and the development of resistance, which leads to disease progression.
Recently, oxoplatin, iproplatin, ormaplatin and satraplatin are Pt complexes that have been usedclinically (oxoplatin) or in clinical trials.

14.7. All complexes have the same geometry and oxidation number for the Pt central atom.
Write the oxidation state of Pt and geometry of the complexes.
14.8. Which Pt complex, cisplatin or satraplatin, is kinetically more inert for substitution
reactions?
14.9 Oxaplatin is an isomer of [Pt(NH3)2Cl2(OH)2] complex. Draw all stereoisomers andindicate the chiral one(s).
Platinum complexes (oxoplatin, iproplatin, ormaplatin, and satraplatin) can be considered
prodrugs that are primarily intracellularly activated by biological reducing agents such as thiols,
ascorbic acid, and glutathione (GSH) to kill cancer cells.
In a study, for example, the reduction of cis,trans,cis-[PtCl2(OCOCH3)2(NH3)2] (G, prodrug),
which has a similar structure to satraplatin, by aqueous extract of cancer cells (A2780,
A2780cisR, and HT-29) yields cisplatin (D, drug) and free acetate ion as given below.

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14.10. Draw the molecular structure of G.
14.11. Sketch the d-orbital splitting of the metal ion in G and write the electronic configuration.
14.12. Decide whether G is paramagnetic or diamagnetic.
14.13. The complex G crystallizes into a monoclinic crystal system of parameters: the lengthsof the unit cell: a = 14.9973, b = 8.57220, c = 11.1352 Å, the β angle in the unit cell = 126.7690°,
the number of the molecules in the unit cell (Z) = 4, M = 436.16 g/mol (the complex has one
water molecule in the crystal structure).
Calculate the density (ρ) of the complex.
Hint: the volume of a monoclinic crystal unit cell is 𝑉 = 𝑎 × 𝑏 × 𝑐 × sin 𝛽65

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Problem 15. Sodium Compounds from Salt
The Salt Lake basin in Turkey is of great importance for the conservation of biological diversity
and is classed as a wetland according to international criteria. It is also one of Turkey’s richestlakes for the presence of birds. There are 85 bird species, 129 insect species (4 of which areendemic), 15 mammal species, and 38 endemic plant species. Some 40% of Turkey’s salt needs(as table salt) are supplied from this lake. Salt in the Salt Lake is formed by meteorologicalwaters draining underground and melting the previously formed salt domes and carrying themalong the tectonic lines. Salt production in the Salt Lake is done by evaporation of lake water
under the sun. A pooling system is used in the salt production with solar energy.

Salt Lake
Table salt is one of the most common household chemicals. It is 97% to 99% sodium chloride,
which is an ionic compound with the chemical formula NaCl, representing a 1:1 ratio of sodium
and chloride ions. NaCl is the compound most responsible for the salinity of seawater and of
the extracellular fluid of many multicellular organisms. In its edible form of table salt, it iscommonly used as a condiment and food preservative. A second major application of sodium
chloride is de-icing of roadways in subfreezing weather. Large quantities of sodium chloride
are also used in many industrial processes such as the chloro-alkaline industry and soda-ash
industry as well as in miscellaneous industrial uses: water softening, medicine, agriculture,
firefighting, and cleanser. NaCl is used, directly or indirectly, in the production of many sodium
compounds, which consume most of the world's production. The scheme below shows the
preparation of some sodium compounds starting from NaCl.
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Preparation of some sodium compounds starting from NaCl.
15.1. Write the formulas of products A–G.
Sodium carbonate (Na2CO3, soda ash) is used primarily in the manufacture of glass, which is
produced mostly from natural sources, such as the mineral trona, Na2CO3.NaHCO3.nH2O. It canbe also manufactured mostly from NaCl, CaCO3, and NH3 using a process introduced by the
Belgian chemist Ernest Solvay in 1863. The key step involves the reaction of NH3(g) andCO2(g) in saturated NaCl(aq). Of the possible ionic compounds that could precipitate from sucha mixture (NaCl, NH4Cl, NaHCO3, and NH4HCO3), the least soluble is sodium hydrogencarbonate (sodium bicarbonate, NaHCO3). It is isolated from solution by filtration and thenconverted to sodium carbonate (Na2CO3) by heating. According to this explanation,
15.2. Balance the reactions given below.
𝑁𝑎𝐶𝑙(𝑎𝑞) + 𝐶𝑂2 (𝑔) + 𝑁𝐻3 (𝑔) + 𝐻2 𝑂(𝐼) →∆2𝑁𝑎𝐻𝐶𝑂3 (𝑠) →15.3. Using CaCO3 (limestone), how can you produce the CO2 gas you need to prepare
NaHCO3?
15.4. Write the Lewis structure of CaCO3 with all resonances and show formal charges for each
atom in the structure.

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15.5. Describe the molecular geometry and propose a plausible hybridization scheme for thecentral atom in the ion CO32–.
15.6. Compare the bond lengths of CO3–2, CO, and CO2 in increasing order.
NaCl crystallizes in a face-centered cubic (fcc) structure. The density of NaCl is 2180 kg/m3
and the ionic radius of Na+ is 99 pm.
15.7. How many atoms are there in the unit cell? Which atoms occupy octahedral holes?
15.8. Calculate the length of the unit cell of NaCl and the ionic radius of the Cl– ion (as pm).
15.9. The alkali metals react rapidly with oxygen to produce several different ionic oxides.
Under appropriate conditions, generally by carefully controlling the supply of oxygen, the oxide
M2O can be prepared for each of the alkali metals. Lithium reacts with excess oxygen to give……(A)…. and a small amount of ……(B)……. Sodium reacts with excess oxygen to give
mostly ….(C)…… and a small amount of ……(D)……..Potassium, rubidium, and cesium reactwith excess oxygen to form …….(E)….., ……(F)….., and ……(G)…..
Fill in the blanks above (for A–G) with convenient formulas of metal oxides.
15.10. Draw the Lewis structures of oxide, peroxide, and superoxide ions.
15.11. Draw the molecule orbital energy level diagram of peroxide and superoxide ions andcompare their bond lengths and energies.
When LiClO4, NaClO4, and KClO4 crystallize from an aqueous solution that may or may notcontain water molecules called water of crystallization as part of the solid structures, althoughno simple rule exists for predicting with certainty whether the ions will retain all or part of their
hydration spheres in the solid state, cations with high charge densities tend to retain all or partof their hydration spheres in the solid state. When the cations have low charge densities, thecations tend to lose their hydration spheres; thus, they tend to form anhydrous salts. The ionicradius of Li+, Na+, and K+ is 76 pm, 102 pm, and 138 pm, respectively.
15.12. Calculate the charge densities of the ions in C mm–3.
15.13. Which perchlorate salt is most susceptible to form an anhydrous compound?

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Problem 16. Thermal Springs of Turkey and Sulfur Chemistry
Turkey is one of the 7 countries in the world in terms of thermal source richness with almost
1300 thermal springs throughout Anatolia. There are thermal hotels in many cities such as
Ankara, Bursa, Balıkesir, Yalova, Erzurum, Sivas and Afyonkarahisar. Afyonkarahisar, locatedin the Aegean region, is the most famous city in Turkey for its thermal springs. The thermal
waters of Afyonkarahisar contain over 42 different minerals and many trace elements. The mostconcentrated ones are sulfur, calcium, chloride, sodium, and carbonates. Among these minerals,
sulfur is important as “nature’s beauty mineral” because the human body needs it to manufacture
collagen, which keeps human skin elastic, beautiful, and young looking. Moreover, sulfur isused to minimize the symptoms of many skin diseases including dermatitis, eczema, dandruff,
and warts. People with arthritis may obtain pain relief from taking a soothing bath in thermal
sulfur springs. Mineral water containing sulfur compounds is also shown to decrease cholesteroland blood pressure. Therefore, sulfur chemistry is an important topic. In this question, you will
explore sulfur chemistry by studying its different reactions and compounds.

Hot spring
Sulfur is extracted as the element from underground deposits. It has many allotropes and its
allotropy is complicated, but the most common sulfur allotrope is the puckered rings of S8(orthorhombic sulfur, -form).
16.1. Sketch the molecular structure of S8 and indicate whether the molecule has a horizontal
mirror plane or not.
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Upon the burning of S8 with oxygen, compound A is produced. Further catalytic oxidation ofcompound A yields compound B. The reaction of A and B with water (hydrolysis) yields C andD. Compound D is an oxoacid and a central substance of the chemical industry worldwide.
16.2. Write the formulas of compounds A–D.
16.3. Draw molecular shape of the compounds by giving the name of geometries.
16.4. Write the oxidation state of the sulfur atoms in C and D.
16.5. Give balanced chemical equations for the synthesis of A–D.
Compound A can also be obtained by heating alkaline or alkaline earth sulfide minerals like
CaS in an excess of air.
16.6. Write the balanced chemical equation for the synthesis of A from CaS.
Upon the reaction of D and B, compound E which is a dense and corrosive liquid that is used
as a basic chemical for sulfonation processes is produced.
16.7. Give a balanced chemical equation for the synthesis of E from D.
16.8. Write molecular formula and draw the molecular shape of E.
16.9. Determine the oxidation state of the sulfur atoms in E.
The reaction of S8 with a stoichiometric amount of chlorine gas yields compound F and thefurther reaction of F with excess chlorine gas results in the formation of G, which is used as aprecursor for the synthesis of sulfur dyes and synthetic rubber. The reaction of G with B yields
the compounds H and A. H is a toxic compound used as the chlorinating agent in organic
synthesis.
16.10. Write molecular formulas and draw the molecular shapes of F, G, and H.
16.11. Give balanced chemical equations for the synthesis of compounds F, G, and H.
One of the most common naturally occurring sulfur minerals is pyrite (FeS2: iron(II) disulfide),called fool’s gold because it is a brass-yellow mineral and thus most people suppose that it isgold ore. The treatment of pyrite with hydrochloric acid results in the formation of a colorless,flammable, water-soluble gas with a “rotten egg” odor, compound I. Compound I is dissolved
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in thermal waters for spa applications since it is reported that the therapeutic effects of thermal
water are directly correlated to its sulfur concentration. Compound I is slightly heavier than air
and can be detected by lead(II) acetate paper strip test in which a reaction occurs between
lead(II) acetate and I, producing compound J. Moreover, upon the oxidation of I, compound A
can be yielded.
16.12. Write the molecular formulas of I and J.
16.13. Draw the molecular shape of I and write the name of the shape.
16.14. Give balanced chemical equations for the synthesis of I and J.
The sulfur oxoacids are chemical compounds that contain sulfur, oxygen, and hydrogen atoms.
Sulfur has several oxoacids; one of them is thiosulfuric acid, with the molecular formula
H2S2O3, which can be synthesized by the reaction of sulfite with I. On the other hand, thecontrolled oxidation of sulfur trioxide by MnO2 in acidic solution yields another sulfur oxoacid,called dithionic acid, H2S2O6.
16.15. Give balanced chemical equations for the synthesis of H2S2O3 and H2S2O6.
16.16. Draw the molecular shapes of H2S2O3 and H2S2O6.
On the other hand, the thiosulfate ion (S2O32–) is a very good complexing agent for Ag+ andthus it is used in photography for removing unchanged AgBr from exposed photographic film.
Upon the reaction of sodium thiosulfate ion with AgBr, sodium salt of a coordination compoundwith coordination number 2 is yielded.
16.17. Give a balanced chemical equation for the reaction of AgBr with Na2S2O3.
16.18. Draw the molecular structure of the yielded coordination complex considering its
geometry.
16.19. Write the electron configuration of the silver ion in the coordination compound.
The determination of H2S content in thermal waters is important for spa applications. An
iodometric titration method can be utilized for this purpose. In a typical experiment, 500 mL ofsample is collected from a thermal water source and purged with N2(g) to ensure the transfer ofall H2S gas into 50 mL of 0.010 M NaOH solution in a closed system. After adjusting pH of thesolution to 6.0, 12.5 mL of 0.300 M I2 solution and 1.0 g of KI are added to this solution andthe resultant solution is stored in the dark for 15 minutes after sealing it with parafilm. After71

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adding 1.0 mL of 20 mg/mL starch solution, the resultant solution is titrated with 0.0500 M
Na2S2O3 until the end-point and consumed Na2S2O3 volume is recorded as 95.62 mL.
16.20. Write all balanced equations of this experiment.
16.21. Calculate H2S concentration in the thermal water source in ppm by assuming that there
is no interfering species in the water source and all H2S content of thermal water is swept into
the NaOH solution.

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Problem 17. Electrochemical Determination of Rutin
Flavonoids are a group of natural products with some phenolic groups that are present in many
fruits and vegetables. Flavonoids are commonly used in our daily life due to their antioxidant
and anticarcinogenic properties. Rutin is a flavonoid class substance composed of the flavonol
quercetin and disaccharide rutinose.

Chemical structure of rutin.
It is of very low toxicity for human health and it is known that rutin can supply electrons toreactive free radicals to produce more stable and healthy structures. Rutin is also known as
vitamin P, in which P is due to its permeability. Rutin is an electrochemically active material
and many researchers have extensively studied its electrochemical behavior using different
electrochemical techniques.
Cyclic voltammetry is a useful technique for electrochemical measurement of an analyte, which
is dissolved in a useful electrolyte solution. There are three electrodes in an electrochemical cellsolution; working, counter, and reference electrodes. The potential of working electrode isscanned versus reference electrode because reference electrode has a constant potential value.
Reverse electrochemical reactions of the working electrode occur at the counter electrode.
Therefore, current flows between working and counter electrodes. Reference electrode is used
to adjust the potential of the working electrode at a known value. This technique is applied
based on potentiodynamic application. Potential of the working electrode is scanned versus
reference electrode between two potential values depending on time. Cyclic voltammetry
application results in a graphic (voltammogram) of current versus scanned potential. There are
two important parameters to evaluate a voltammogram; peak potential and peak current. The

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Preparatory problems: THEORETICAL

peak potential and peak current are calculated using x-axis and y-axis of a voltammogram at thepeak maximum, respectively.
The cyclic voltammetry (CV) behavior of rutin at 25 °C has been tested using a glassy carbon
electrode, a saturated calomel electrode (SCE), and a Pt wire as working, reference, and counterelectrodes, respectively. In this study, CV data for 1.0 × 10−4 mol/dm3 rutin solutions at different
pH values have been obtained by scanning the potential between 0.00 and 0.80 V at a scan rateof 100 mV/s. Anodic peak potential (Epa), cathodic peak potential (Epc), anodic peak current(Ipa), and cathodic peak current (Ipc) values supplied from related CVs depending on the pH are
presented in the following Table.
Table. Some CV parameters depending on the pH of a solution containing 1.0×10−4 mol/ dm3
rutin.
pH
1.5
2.0
3.0
4.0
5.0
6.0

Epa/mV
643
609
544
499
441
372

Epc/mV
614
578
514
470
410
344

Ipa/A
0.105
0.118
0.116
0.104
0.093
0.099

Ipc/A–0.104–0.119–0.117–0.104–0.092–0.10017.1. In a three-electrode system, electrochemical oxidation or reduction of an analyte in theelectrochemical cell occurs on the ____________________ because its potential is adjusted
against the _____________________.
Which of the following words fit into the blanks in the above sentence?
a) working electrode / reference electrodeb) counter electrode / working electrodec) reference electrode / working electroded) working electrode / counter electrode17.2. Both anodic and cathodic peak potentials shift to negative potential values by increasing
the pH because the electrochemical reaction of rutin includes ______________.
Which of the following words fits into the blank in the above sentence?
a) Na+b) K+
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Preparatory problems: THEORETICAL

c) H+
d) I–
17.3. Electrochemical oxidation of rutin is _________________ because of the fact that
Ipa/Ipc is about 1 and ΔEp is almost 0.0592/n V.
Which of the following words fits into the blank in the above sentence?
a) irreversibleb) reversiblec) quasi-reversibled) catalyzed

17.4. How long does it take to obtain each CV value?

17.5. Calculate the number of transferred electrons for the electrochemical reaction of rutin
including 2 H+.
17.6. Propose an electrochemical redox mechanism for rutin.
17.7. The SCE reaction is 𝐻𝑔2 𝐶𝑙2 (𝑠) + 2𝑒 − → 2𝐻𝑔(𝑙) + 2𝐶𝑙 − and the SCE contains saturated
KCl solution prepared by dissolving 342 g of KCl in 1.0 L of aqueous solution. How does thepotential of the SCE change (decrease or increase) in the case of 1.0 M KCl?
In order to determine the amount of rutin in a vitamin P tablet, the following procedures havebeen used:i) A 500 mg vitamin P tablet is dissolved in deionized water, pH is adjusted to 2.0, and total
volume is completed to 500 mL in a volumetric flask. A 10 mL part of this solution is placed ina three-electrode cell. CV is obtained with an anodic peak current (Ipa) of 2.26 A.
ii) A solution in the absence of rutin has been prepared at pH 2.0. After placing all electrodes
into this solution, CV has been recorded three times by cleaning the electrode with deionizedwater for each measurement. Then Ipa values have been read as 0.16, 0.11, and 0.18 A,
respectively.

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Preparatory problems: THEORETICALiii) The standard rutin solutions of 1.0, 5.0, 10.0, 20.0, 30.0, and 50.0 mM have been preparedand Ipa values of these solutions have been obtained from related CVs as demonstrated in thefollowing Table.
Table. Ipa values for various rutin standard solutions.
Ipa/A
Crutin/mM
1.0
1.11
5.0
6.43
10.0
12.62
20.0
24.73
30.0
36.20
50.0
58.55
Note that all of the CVs have been obtained by using same working electrode beyond thisexperiment.
17.8. Draw a calibration curve for the rutin determination method.
17.9. Write a mathematical equation for the calibration curve.
17.10. Calculate the rutin amount in the vitamin P tablet as wt %.
17.11. Calculate the calibration sensitivity and limit of detection (LOD) of this method for a
signal to noise ratio (S/N) of 3.0.
Note: Limit of detection:

𝐿𝑂𝐷 =𝑘×𝑠𝑏𝑙𝑎𝑛𝑘𝑚76

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Preparatory problems: THEORETICAL

Problem 18. Particle in a Box Problem: Free Electron Model
The particle in an one-dimensional box model is a crude approximation for conjugatedmolecules. In this model, π electrons are assumed to move freely over the carbon framework of
conjugated bonds. Therefore, this model is also called the free electron model (FEM). The
length of the box may be approximated via 𝐿 = 𝑛𝐶 × 1.40 Å, where L is the box length and 𝑛𝐶
is the number of carbons. Furthermore, the Pauli principle is applied when electrons are filled
to the energy levels. The energy of a particle in an one-dimensional box can be written as
follows:
𝐸𝑛 =

𝑛2 ℎ2
8𝑚𝐿2

where 𝑚 is the mass of the particle, ℎ is the Planck constant, and 𝑛 is a positive integer.
For the 1,3,5,7-octatetraene molecule assuming FME:
18.1. Draw an energy diagram, fill the electrons, and calculate orbital energies.
18.2. Calculate the total π energy of the molecule.
18.3. Determine the wave length of the light (in nm) that require to excite an electron from thehighest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital(LUMO).
For two-dimensional conjugated systems, we may use the particle in a two-dimensional box
model. In this case, the energy can be written as follows:ℎ2 𝑛12 𝑛22
𝐸𝑛1 ,𝑛2 =( + 2)
8𝑚 𝐿21
𝐿2
where 𝐿1 and 𝐿2 are the lengths and 𝑛1 and 𝑛2 are the quantum numbers of the first and second
dimensions, respectively.
Graphene is a sheet of carbon atoms in the form of a two-dimensional hexagonal lattice in which
one atom forms each vertex.

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For a square shaped graphene sheet with 𝐿1 = 𝐿2 = 11 Å:
18.4. The distance between two adjacent carbons in the hexagonal 6-carbon unit isapproximately 1.4 Å. Calculate the number of electrons in a (11 Å × 11 Å) sheet of graphene.
For this problem you may ignore edge electrons. (Area of a regular hexagon with a side of L is
𝐴=

3√3
2

𝐿2 ).

18.5. Calculate the energy of the HOMO.
18.6. Calculate the energy of the LUMO.
18.7. The difference between energies of the LUMO and HOMO is called the band gap (Eg).
Calculate the band gap.
The models for a particle in a one- and two-dimensional box can be extended to a threedimensional rectangular box of dimensions 𝐿1 , 𝐿2 , and 𝐿3 , yielding the following expression
for the allowed energy levels:ℎ2 𝑛12 𝑛22 𝑛32
𝐸𝑛1 ,𝑛2 ,𝑛3 =( + 2 + 2 ),8𝑚 𝐿21
𝐿2
𝐿3
where 𝑛1 , 𝑛2 , and 𝑛3 are the quantum numbers of the first, second, and third dimensions,
respectively. For a particle in a cubic box of length 𝐿:
18.8. Give the expressions for the five different lowest energies.
18.9. Draw a diagram showing all the five energy levels. Indicate degeneracy of each level.

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Preparatory problems: THEORETICAL

Problem 19. Harmonic Oscillator and Rigid Rotor Models
Vibration of a diatomic molecule is reminiscent of two masses on a spring with a potential
energy that is a function of the displacement from equilibrium. Hence, the harmonic oscillator
model is utilized to compute vibrational frequencies. These frequencies are called harmonic
vibrational frequencies. The energy of a harmonic oscillator can be written as follows:1
𝐸𝑛 = ℎ 𝜈 (𝑛 + )2where 𝜈 is the harmonic vibrational frequency, ℎ is the Planck constant, and 𝑛 is a nonnegative
integer. The harmonic vibrational frequency can be calculated as follows:

𝜈=

1 𝑘
√
2𝜋 𝜇

where 𝑘 is the force constant and 𝜇 is the reduced mass:
𝜇=

𝑚1 𝑚2
𝑚1 + 𝑚2

where 𝑚1 and 𝑚2 are the masses of the first and the second atoms, respectively.
For the 12C16O molecule the value of the force constant is 1902.4 N/m. For this problem, theatomic masses of isotopes can be approximated by their mass numbers.
19.1. Calculate the harmonic vibrational frequency of the 12C16O molecule in Hz.
19.2. Express the harmonic vibrational frequency of the 12C16O molecule in cm–1.
19.3. Calculate the zero-point vibrational energy (ZPVE) of the 12C16O molecule in kcal/mol.
19.4. Calculate the harmonic vibrational frequency of the 13C16O molecule in cm–1.
19.5. Calculate the harmonic vibrational frequency of the 12C17O molecule in cm–1.

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Preparatory problems: THEORETICAL

The harmonic oscillator model can readily be extended to polyatomic molecules. In this case,the total vibrational energy of a molecule with nfreq vibrational frequencies can be written as
follows:𝑛𝑓𝑟𝑒𝑞

1= ℎ ∑ 𝜈𝑖 (𝑛𝑖 + )𝑓𝑟𝑒𝑞
2

𝐸𝑛1 𝑛2 …𝑛𝑛

𝑖=1

where 𝜈𝑖 are the harmonic vibrational frequencies, ℎ is the Planck constant, and 𝑛𝑖 arenonnegative integers.
For the water molecule the harmonic vibrational frequencies are 1649, 3832, and 3943 cm –1.
Using the harmonic oscillator model, for the water molecule:19.6. Calculate the ZPVE value (in J and cm–1 units).
19.7. Calculate the first 5 energy levels (in cm–1).
To describe the rotational motion of a diatomic molecule, the rigid rotor model is used. In thismodel the bond length (𝑅) of the diatomic molecule is kept constant during the rotationalmotion. Using the rigid rotor model, the rotational energy of a diatomic molecule can be written
as follows:ℎ2
𝐸𝑙 =𝑙(𝑙 + 1)8 𝜋2𝐼
where 𝐼 is the moment of inertia and 𝑙 is a nonnegative integer. The moment of inertia can be
written as follows:
𝐼 = 𝜇 𝑅2
where 𝜇 is the reduced mass and 𝑅 is the bond length of the diatomic molecule.
In the microwave spectrum of the 12C16O molecule the value of frequency for the lowest energy
transition is 115.270 GHz.
19.8. Calculate the bond length of the 12C16O molecule in Å.
19.9. For the 12C16O molecule predict the frequency of the next two absorptions (selection
rule is ∆𝑙 = ±1).
19.10. For the 12C17O molecule, calculate the frequency of the lowest energy absorption.
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Preparatory problems: THEORETICAL

Problem 20. Journey to Different Earth-Like Planets
In the future, humankind will most likely consume all resources that are necessary for life onearth and will have to relocate to an earth-like planet. Assume that you have started to live on a
new planet where standard pressure condition is 2 bar, standard concentration is 1 mol dm−3,
and all types of gases behave as an ideal gas. On this planet, you are asked to determine
equilibrium conditions for the reaction below:
𝑋𝑌4 (𝑔) ⇌ 𝑋(𝑠) + 2𝑌2 (𝑔)∆𝑟 𝑆° = 80 𝐽𝐾 −1 𝑚𝑜𝑙 −1 at 298 𝐾
20.1. Calculate the change in standard enthalpy of the reaction at 298 K by using thefollowing information:
𝑋4 𝑌8 (𝑠) → 4𝑋(𝑠) + 4𝑌2 (𝑔)∆𝑟 𝐻1° = 123.34 𝑘𝐽 𝑚𝑜𝑙 −1𝑌2 (𝑔) + 𝑋4 𝑌6 (𝑙) → 𝑋4 𝑌8 (𝑠)∆𝑟 𝐻2° = −48.48 𝑘𝐽 𝑚𝑜𝑙 −1

𝑋4 𝑌6 (𝑙) → 2𝑋2 𝑌3 (𝑔)∆𝑟 𝐻3° = 32.84 𝑘𝐽 𝑚𝑜𝑙 −1

𝑋2 𝑌3 (𝑔) + 1⁄2 𝑌2 (𝑔) → 𝑋(𝑠) + 𝑋𝑌4 (𝑔)∆𝑟 𝐻4° = −53.84 𝑘𝐽 𝑚𝑜𝑙 −1

20.2. Calculate ΔrG° of the reaction at 298 K.
20.3. Calculate K° of the reaction at 298 K.
20.4. Assume that ΔrH° of the reaction does not depend on temperature. Find K of thereaction at 50 °C.
20.5. Calculate the percent degree of dissociation for XY4 at 298 K where total pressure is 0.2
bar.
20.6. In order to increase the amounts of products, which one do you choose to increase (ifyou choose both, put a cross next to both of them):☐ pressure☐ temperature of the reaction vessel8152nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: THEORETICAL

Moreover, in this future, the Earth will have a very unstable climate. The surface temperature
could increase or decrease all of a sudden. Suppose that you travelled through time to the era inwhich the Earth’s climate is extremely unstable. Your task in this era is to observe thethermodynamics of phase transitions of water, the most precious substance where all life hasoriginated. Suppose that the temperature suddenly decreased to −20 °C.
One mole of water becomes supercooled liquid water at −20 °C and 1 bar pressure and thenturns into ice at the same temperature (note that the temperature of the surroundings is constantat −20 °C).
By using the following data for water:
The heat of fusion (ΔmH°) of ice at 0 °C and 1 bar is 6020 J mol−1
𝐶𝑝,𝑚 (𝐻2 𝑂(𝑠)) = 37.7𝐽 𝑚𝑜𝑙 −1 𝐾 −1
𝐶𝑝,𝑚 (𝐻2 𝑂(𝑙)) = 75.3 𝐽 𝑚𝑜𝑙 −1 𝐾 −1
During the conversion of supercooled liquid water to ice at −20 °C:
20.7. Calculate the total entropy change in the system.
20.8. Calculate the total entropy change in the surroundings.
20.9. Calculate the total entropy change in the universe.
𝑇𝑞∆S = 𝐶𝑝 𝑙𝑛 𝑇 𝑓𝑖𝑛𝑎𝑙 and ∆S = − 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛𝑇
𝑖𝑛𝑖𝑡𝑖𝑎𝑙82

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Preparatory problems: THEORETICAL

Problem 21. Rate Constant Models and Kinetic Isotope EffectTransition state theory (TST) is a very helpful model to explain the reaction rates of elementary
chemical reactions. The TST assumes a quasi-equilibrium between reactants and transition
state.

Similar to the Arrhenius model, the TST proposes the following temperature-dependent rate
constant expression:
𝑘𝐵 𝑇
∆𝐺 ‡
𝑘𝑇𝑆𝑇 =
exp [−]ℎ
𝑅𝑇
where, 𝑘𝐵 is the Boltzmann constant, ℎ is the Planck constant, and ∆𝐺 ‡ is the activation freeenergy.
The TST rate constant introduces a simple temperature-dependent factor instead of Arrhenius
factor 𝐴. Further, the TST model allows us to better understand the activation energy concept
and build a bridge between the theory and experiment. Moreover, the TST activation free energy
is a temperature dependent parameter instead of Arrhenius temperature independent 𝐸𝑎 .

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For the decomposition of an organic compound obeying first-order reaction kinetics, thefollowing rate constant values are obtained at given temperatures:
10
30
50
70
t (°C)-4-1
k/10 (s ) 1.1408 17.2075 185.5042 1515.7157

21.1. Using the Arrhenius model, calculate the activation energy.
21.2. Calculate the Arrhenius factor 𝐴.
21.3. Calculate the half-life of the organic compound at 75 °C.
21.4. Assume that the rate constants provided obey the TST model, instead of the Arrhenius
model. Then, calculate activation free energy at 30 °C.
21.5. Assume that the rate constant obtained from both Arrhenius and TST models are equal toeach other. Then, derive expressions for activation enthalpy and entropy in terms of theactivation energy and the Arrhenius factor. It is assumed that entropy is constant.
21.6. Using the expressions obtained, calculate the activation enthalpy at 80 °C.
The kinetic isotope effect (KIE) is the change in the reaction rate of a chemical reaction when
one of the atoms in the reactants, generally hydrogen, is replaced by one of its isotopes,
generally deuterium. KIE is generally utilized in organic chemistry in a procedure called as“deuterium labelling” by changing one or more hydrogen(s) with deuterium(s).
In one of the common theoretical approaches to explain KIE, it is assumed that the change inreaction rate is a quantum chemical effect that primarily results from heavier isotopes having
lower vibrational frequencies compared to their lighter counterparts. Hence, one may assume
that the TST model is valid, and the change in the activation free energy arises solely from thechange in the zero-point vibrational energies (ZPVEs). Therefore, we may write the following
equation:𝑘𝐻
exp[(𝑍𝑃𝑉𝐸(𝑅, 𝐻) − 𝑍𝑃𝑉𝐸(𝑇𝑆, 𝐻))/𝑅𝑇]=𝑘𝐷
exp[(𝑍𝑃𝑉𝐸(𝑅, 𝐷) − 𝑍𝑃𝑉𝐸(𝑇𝑆, 𝐷))/𝑅𝑇]
where 𝑘𝐻 and 𝑘𝐷 are the rate constants of reactions including hydrogen and deuterium,
respectively, 𝑍𝑃𝑉𝐸(𝑅, 𝐻) and 𝑍𝑃𝑉𝐸(𝑅, 𝐷) are the ZPVE values of the reactants includinghydrogen and deuterium, respectively, and 𝑍𝑃𝑉𝐸(𝑇𝑆, 𝐻) and 𝑍𝑃𝑉𝐸(𝑇𝑆, 𝐷) are the ZPVE
values of TSs including hydrogen and deuterium, respectively.

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For a thermal decomposition of an organic compound, the difference between ZPVE values of
deuterium including TS (TS-D) and hydrogen including TS (TS-H) is -2.3 kJ/mol. Further, the
ZPVE value of hydrogen including reactant (R-H) is 3.0 kJ/mol higher than that of deuterium
including reactant (R-D).
𝑘

21.7. Calculate the 𝑘𝐻 value at 298.15 K.
𝐷𝑘

21.8. Calculate the 𝑘𝐻 value at 330.0 K.
𝐷

21.9. If the rate constant 𝑘𝐻 is 2.5 × 102 and 𝑘𝐷 is 2.0 × 102 , then what is thetemperature?

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Preparatory problems: THEORETICAL

Problem 22. Parallel Reaction Kinetics
The reaction in which a reactant undergoes two or more independent reactions concurrently iscalled a parallel or competing reaction. Dehydration of ethanol, nitration of phenol, and nitrationof benzene are examples of parallel reactions. The reaction given below is an example of aparallel first-order reaction.
k1
2B
A

k2
C
22.1. For the parallel first-order reactions of A given above, find the concentrations of B and C
as an equation depending on initial A concentration at time t after the start of the reaction. Find
the ratio of B concentration to C concentration.
𝐻𝑖𝑛𝑡: ∫ 𝑒 𝑎𝑥 𝑑𝑥 =

1 𝑎𝑥
𝑒 +𝑐
𝑎

22.2. The effective rate constant (keff) for the decomposition of A can be defined as (k1+k2).
Assume that the effective rate constant satisfies the Arrhenius equation. Write the expression
for the effective activation energy (EA,eff) in terms of k1, k2, Ea,1, and Ea,2 and estimate the EA,eff
for the given values (k1=6.2 min–1, k2=3.2 min–1, Ea,1=35 kJ mol–1, and Ea,2=60 kJ mol–1).
𝐻𝑖𝑛𝑡:

𝑑 𝑎𝑥
𝑒 = 𝑎. 𝑒 𝑎𝑥
𝑑𝑥

Calculate the half-life for the effective rate constant (𝑡1⁄ (𝑒𝑓𝑓)).
2

22.3. If k1 and k2 values of the given parallel first-order reactions of A are 6.2 and 3.2 min–1,
respectively, at 278 K, find the temperature for the production of equimolar concentrations of
B and C. (Ea energies for the formation of B and C are 35 and 60 kJ mol–1, respectively).
22.4. Draw symbolic concentration change curves for [A], [B], and [C] if k1>k2.

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22.5. The reaction given below is an example of parallel-consecutive first-order reactions witha reversible step.
k1

B

A

k4

C

k2
k3

C

The following data are given for this reaction:
k1 = 0.109 min–1, k2 =0.0752 min–1, k3 = 0.0351 min‒1, k4 =0.0310 min–1.
Time (min)
12.9
𝑡 [𝐴]

𝜃𝐴,𝑡 = ∫0 [𝐴] 𝑑𝑡
0

𝜃𝐴,𝑡 (min)6.89

𝜃𝐵,𝑡 (min)
3.79

𝑡 [𝐵]

𝜃𝐵,𝑡 = ∫0 [𝐴] 𝑑𝑡
0

Find the [A], [B], [C] concentrations after 12.9 minutes if [A]0= 5 mol dm–3.

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Preparatory problems: THEORETICAL

Problem 23. Reaction Kinetics with Absorbance Measurement
It is often experimentally convenient to use an analytical method that provides an instrumental
signal that is proportional to concentration, rather than providing an absolute concentration.
Absorbance, fluorescence intensity, and conductance are examples of this type of instrument
response. The requirements are that the reactants and products both give a signal that is directly
proportional to their concentrations and there is an experimentally usable change in the observed
property as the reactants are transformed into the products. In this experiment, absorption
spectroscopy and Beer’s law can be used.
A is the reactant and B is the only product (A → B) and both give an absorbance at selected
wavelength that is directly proportional to their concentrations. In this experiment, ƐA≠ƐB, whereƐ is the molar absorptivity.
For the hydrolysis of A to B in aqueous solution, the absorbance-time data are given in the Table
below. The experimental conditions: pH of the medium is 7.0 and temperature is 25 °C. The
initial A concentration is 4.0 × 10–6 M and measurements are recorded at 400 nm in a 5-cm cell.
t/s
0
20
60
120
160
200
∞

At
0.08400.10900.15150.20100.22550.24400.317023.1. Calculate the molar absorptivities of A and B under these conditions.
23.2. Find the rate constant.
23.3. Find the half-life (t1/2).
23.4. After how many seconds [A] is equal to 1.0 × 10–6 M?
23.5. If k is equal to 0.01029 s–1 at 30 °C, calculate Ea.

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23.6. Assuming that the transition state rate constant is equal to one obtained from theexperimental data, calculate activation free energy.
𝑘𝑇𝑆𝑇 =
whereis the Boltzmann constant,

𝑘𝐵 𝑇 −∆𝐺‡ /𝑅𝑇𝑒ℎis Planck’s constant, and R is the universal gas constant.

The condensation reaction of acid catalyzed ethylene glycol and terephthalic acid is monitored
as a function of extent of the reaction (p: ratio of condensed [A] to [A]0) and the reaction obeys
second-order kinetics. The concentration of each monomer is equal to each other and it is[A]0=4.8 mol dm–3.

Time (h)
0
0.5
1.5
2.5

Extent of Reaction
0
0.636
0.839
0.897

23.7. Find the rate constant.
23.8. Find the half-life of the reaction.
23.9. What will the monomer concentration be after one hour?

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Preparatory problems: THEORETICAL

Problem 24. Acridine Orange / DNA Binding Interactions
Acridine orange (AO) is a fluorescent dye that binds to DNA via an intercalative mode ofbinding. AO can insert itself into the DNA base pairs. Interactions of intercalating agents suchas AO with DNA have been widely studied, and complexation can be followed with
spectrometric titrations by varying the DNA-to-dye ratio. Stock solutions of DNA can be
standardized spectrophotometrically ( = 13,200 mol dm–3 cm–1 at 260 nm for a molar DNA
concentration, CDNA, expressed in base pairs.)
24.1. Give an expression to calculate the pure DNA concentration from an absorbance readingat 260 nm from a UV spectrum of solution containing DNA (quartz cuvette length: 1.0 cm).
The interaction between DNA and AO to form the AO–DNA complex can be expressed by thefollowing reaction:
𝐴𝑂 + 𝐷𝑁𝐴 ⇌ 𝐴𝑂 − 𝐷𝑁𝐴,
whose equilibrium constant is[AO−DNA]K = [AO][DNA](1)

Where [DNA], [AO] and [AO-DNA] are equilibrium concentrations.
24.2. Provide a mass balance expression for the overall AO concentration (CAO) at equilibrium
conditions.
Binding of AO to DNA can be followed by recording fluorescence intensity (F). Both AO and
AO–DNA complex display a maximum emission intensity at em = 520 nm. In dilute solutions,concentration is proportional to F. Therefore, quantitative estimation of complexation can be
determined by using F.
𝐹 = 𝜑𝑖 × 𝐶𝑖
where φi is the fluorescence constant and Ci is the concentration for species i.
24.3. Provide an expression for the overall F in terms of φ and concentrations of AO and of
DNA at equilibrium.
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Preparatory problems: THEORETICAL

Consider that initially there is only AO in the measuring cell displaying an emission at em =520 nm, and finally at equilibrium both AO and AO–DNA complex have emission at the same
wavelength. F − φAO 𝐶AO = ΔF, and φAO−DNA − φAO = Δφ is given.
24.4. Show that ΔF = [AO − DNA]Δφ.
The binding equilibrium constant for AO intercalation to DNA (ignore AO self-aggregation anddimerization) can be determined from the equation:
𝐶AO

1

1

1

= 𝛥φ + 𝛥φ𝐾 [𝐷𝑁𝐴]ΔF(2)

24.5. Derive equation (2) starting from equation (1).
Spectrofluorometric titration is performed by adding increasing amounts of the DNA directly
into the cell containing the AO. Each time DNA is added, the fluorescence intensity at em =520 nm, where only free and bound AO have emission, is recorded.
CAO (mol dm–3)
1.857 × 10–7
1.832 × 10–7
1.800 × 10–7
1.725 × 10–7
1.604 × 10–7

CDNA (mol dm–3)
6.535 × 10–6
1.032 × 10–5
1.521 × 10–5
2.671 × 10–5
4.516 × 10–5

F520 (a.u.)*
162
188
210
240
260

* a.u. is arbitrary unit
24.6. Calculate the equilibrium binding constant for AO–DNA using the data given in the Table
above. Assume that there is no AO self-aggregation or dimerization. Take φAO = 5.00 × 108 mol
dm–3, and [DNA]  CDNA.

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Preparatory problems: THEORETICAL

11,2
y = 2.54E3x + 2.2
R² = 0.9993

11

ln K

10,8
10,6
10,4
10,2
10
9,8
0,00300,0031

0,00320,00330,00340,0035

1/T (K-1)
𝐺°

24.7. Given that 𝐾 = 𝑒 − 𝑅𝑇 and the plot above, calculate the values of rH°, rS°, and rG°
for the complexation of AO with DNA at 25 °C. (Assume that rH° and rS° do not changewith the temperature.)
AO can undergo some self-aggregation (dimerization) at higher concentrations. The
quantitative analysis of the dimerization can be expressed as follows:

Here D represents AO monomer while D2 represents dimeric AO, and kf and kd are the rate
constants for dimer formation and dimer dissociation, respectively. According to that reaction,
AO concentration dependence of the relaxation time,  , which represents the time passed for asystem to reach the new equilibrium when a sudden change is applied, is expressed by the
relationship:1

= 𝑘d + 4𝑘f 𝐶𝐴𝑂

Data for dimerization of AO at 25 °C are given in the Table below.
106 CAO (mol dm–3)
2.50
4.50
8.00
11.0

105 Relaxation time,  (s)
2.32
2.27
2.18
2.11

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Preparatory problems: THEORETICAL

24.8. Calculate the values of kd and kf.
The absorbance spectra of AO derivative at various concentrations (0 to 7.3 × 10–5 mol dm–3)
in water are shown in the Figure below. The spectra indicate that two absorbance peaks exist:
one at 496 nm and the other at 475 nm. The inset gives the ratio of absorbance peaks (A475/A496)
as a function of AO concentration.

24.9. Choose correct statement(s) according to the absorbance spectra of the AO derivative.
☐ The band observed at 496 nm is attributed to the monomeric form.
☐ If there were only the monomeric form, the ratio of absorbance peaks (A475/A496)
would remain constant.
☐ To reduce dimerization, the concentration of AO should be reduced.
24.10. If the initial concentration of AO is 1.0 × 10–5 mol dm–3, then calculate the dimer
fraction at equilibrium.

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Preparatory problems: THEORETICAL

Problem 25. Spectrophotometric Determination of an
Antihistaminic Drug
Spectrophotometric procedures are simple, rapid, and accurate methods that can be utilized forthe determination of drug molecules. The method is based on formation of a complex between
two reagents. Many complexes are colored and absorb in the visible region; thus they can be
determined spectrophotometrically.
An antihistaminic drug, D, acts as an electron donor group and complexes with a π-acceptor, S.
Absorbance of the resulting complexes recorded at relevant maxima (460 nm) with respect todrug concentration shows a linear tendency with good correlation coefficients.
𝐷 + 𝑆 ⇌ 𝐷𝑆
𝐾=

[𝐷𝑆]
[𝐷][𝑆]

where [DS], [D], and [S] represent the equilibrium concentrations of the DS complex, D, and S,
respectively.
𝐶𝐷 = [𝐷] + [𝐷𝑆]where CD is the overall concentration of the drug.
At a wavelength where only the formed DS complex absorbs light, the following expression
holds:
𝐴 = 𝜀𝐷𝑆 𝑙[𝐷𝑆]where l is the length of the measuring cuvette.
The binding equilibrium constant of the complexation can be calculated using the Benesi–
Hildebrand equation, which depends on the experimental conditions where one of thecomponent species should be present in large excess so that its concentration is not altered onformation of complex.
𝐶D
1
1
1
=
+
×
𝐴DS
DS DS K
𝐶S

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Preparatory problems: THEORETICAL

where CS and CD are total concentrations of S and D. ADS is the absorbance of the complex, εDS
is the molar absorptivity of the complex, and K is the equilibrium constant.
7,0E-06

CD/ADS (mol dm-3)

6,0E-06

y =6.20E-11 x 2.23E-7
R² = 0.997

5,0E-06
4,0E-06
3,0E-06
2,0E-06
1,0E-06

0,0E+00
0,0E+00

2,0E+04

4,0E+04

6,0E+04

8,0E+04

1,0E+05

1/ CS (mol-1dm3)

25.1. Considering the Benesi–Hildebrand plot recorded at 25 °C, find the equilibrium constantfor the complex formation and molar absorptivity of the complex.
25.2. The initial equal concentration of D and S is 9 × 10–5 mol dm–3. Calculate the fraction of
the complex formed when equilibrium is reached. Consider there is a 1:1 molar ratio between
D and S in the complexation.
25.3. Calculate the rG° in kJ mol–1 at 25 °C.
The kinetics of complexation of D with S is studied by varying temperature (25, 45, and 60 °C).
The Table gives the rate constant of complexation at different temperatures.
T (°C)
25
45
60

k (sec–1)
0.02000.05040.094425.4. Calculate the activation energy, Ea.
25.5. Given that 𝑘𝑇𝑆𝑇 =

𝑘𝐵 𝑇
ℎ

∆𝐺‡

𝑒 − 𝑅𝑇 , calculate the activation enthalpy (H‡), activation entropy(S‡), and free activation enthalpy (G‡) of the reaction for 25 °C.

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Preparatory problems: PRACTICALPart II:
Practical
Problems

9652nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: PRACTICAL

SafetyParticipants in the Olympiad must be prepared to work in a chemical laboratory and be awareof all relevant rules and safety procedures. The organizers will strictly enforce the safety rules
given in Appendix A of the IChO Regulations during the Olympiad.
The Preparatory Problems are designed to be carried out in properly equipped chemicallaboratories under competent supervision only. For each chemical, the GHS hazard andprecautionary numbers are reported. We did not include specific and detailed safety and disposal
instructions as regulations are different in each country. Mentors must carefully adapt the
problems accordingly.

A Material Safety Data Sheet (MSDS)
A Material Safety Data Sheet (MSDS) is a technical document that contains detailed andcomprehensive information about the hazards of a chemical and how to work safely with thechemical product. You have to know about the hazards with the chemicals in experiments.

Dress code
During the examination, the students will be required to wear:- pants covering their whole legs;- closed and flat shoes;- a lab coat with long sleeves;- if applicable, long hair tied back.
Safety glasses will be supplied and must be carried during the whole examination, even if thestudent wears prescription glasses. Contact lenses are prohibited.
Any student that fails to respect these rules will not be allowed to enter the lab.
The GHS hazard statements (H-phrases) associated with the materials used are indicated in the
problems. Their meanings are as follows.
Definition of GHS hazard statements:
Physical hazards
H224 Extremely flammable liquid and vapour.
H225 Highly flammable liquid and vapour.
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H226
H272
H290
H290

Preparatory problems: PRACTICAL

Flammable liquid and vapour.
May intensify fire; oxidizer.
May be corrosive to metals.
May be corrosive to metals.

Health hazards
H301 Toxic if swallowed.
H302 Harmful if swallowed.
H302 + H312 + H332 Harmful if swallowed, in contact with skin or if inhaled.
H304 May be fatal if swallowed and enters airways.
H311 Toxic in contact with skin.
H312 Harmful in contact with skin.
H312 + H332 Harmful in contact with skin or if inhaled.
H314 Causes severe skin burns and eye damage.
H315 Causes skin irritation.
H317 May cause an allergic skin reaction.
H318 Causes serious eye damage.
H319 Causes serious eye irritation.
H330 Fatal if inhaled.
H334 May cause allergy or asthma symptoms or breathing difficulties if inhaled.
H335 May cause respiratory irritation.
H336 May cause drowsiness or dizziness.
H340 May cause genetic defects.
H350 May cause cancer.
H360FD May damage fertility. May damage the unborn child.
H372 Causes damage to organs through prolonged or repeated exposure.
Environmental hazards
H400 Very toxic to aquatic life.
H410 Very toxic to aquatic life with long lasting effects.
H411 Toxic to aquatic life with long lasting effects.
H412 Harmful to aquatic life with long lasting effects.
Precautionary Statements
P201 Obtain special instructions before use.
P210 Keep away from heat, hot surfaces, sparks, open flames and other ignition sources. No
smoking.
P221 Take any precaution to avoid mixing with combustibles, heavy-metal compounds,
acids and alkalis.
P260 Do not breathe dust or mist.
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Preparatory problems: PRACTICAL

P261 Avoid breathing dust/ fume/ gas/ mist/ vapours/ spray.
P264 Wash skin thoroughly after handling.
P273 Avoid release to the environment.
P280 Wear protective gloves.
P280 "Wear protective gloves/ protective clothing/ eye protection/ faceprotection."
P301 + P310 + P331 IF SWALLOWED: Immediately call a POISON CENTER/doctor. Do
NOT induce vomiting.
P301 + P312 + P330 IF SWALLOWED: Call a POISON CENTER/doctor if you feel unwell.
Rinse mouth.
P301 + P330 + P331 IF SWALLOWED: Rinse mouth. Do NOT induce vomiting.
P302 + P352 IF ON SKIN: Wash with plenty of water.
P302 + P352 + P312 "IF ON SKIN: Wash with plenty of water. Call a POISONCENTER/doctor if you feel unwell."P303 + P361 + P353 "IF ON SKIN (or hair): Take off immediately all contaminatedclothing. Rinse skin with water."
P304 + P340 IF INHALED: Remove victim to fresh air and keep at rest in a position
comfortable for breathing.
P304 + P340 + P310 "IF INHALED: Remove person to fresh air and keep comfortablefor breathing. Immediately call a POISON CENTER/doctor."
P304 + P340 + P312 "IF INHALED: Remove person to fresh air and keep comfortable forbreathing. Call a POISON CENTER/doctor if you feel unwell."
P305 + P351 + P338 IF IN EYES: Rinse cautiously with water for several minutes. Remove
contact lenses, if present and easy to do. Continue rinsing.
P305 + P351 + P338 + P310 IF IN EYES: Rinse cautiously with water for several minutes.
Remove contact lenses, if present and easy to do. Continue rinsing. Immediately call a
POISON CENTER/doctor.
P308 + P310 IF exposed or concerned: immediately call a POISON CENTER or doctor/
physician.
P314 Get medical advice/ attention if you feel unwell.
P337 + P313 If eye irritation persists: Get medical advice/ attention.
P370 + P378 "In case of fire: Use dry sand, dry chemical or alcohol-resistant foam toextinguish."P403 + P233 Store in a well-ventilated place. Keep container tightly closed.
P403 + P235 Store in a well-ventilated place. Keep cool.
Risk and Safety Statements
R 51/53
Toxic to aquatic organisms, may cause long-term adverse effects in the aquatic
environment.
S 61 Avoid release to the environment. Refer to special instructions/ Safety data sheets.
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Preparatory problems: PRACTICAL

Problem P1. Drug Delivery from a Polymeric Hydrogel System

Chemical structure of paracetamol.

Nowadays, studies in the field of medicine aim to reduce the drug dose to a minimum, extend
the dose range, and improve the quality of life by ensuring that the patient is not affected by theside effects of the drug. “Controlled release systems” provide the best response to this goal.
Controlled release is a system that can be prepared in membrane or matrix form. Controlled
release of a drug from systems such as a hydrogel film or sphere allows the drug to be more
effectively introduced into the body.
Paracetamol is a drug active substance with analgesic and antipyretic properties. It is one of themost common drugs used to reduce moderate/mild pain and fever.
In this problem, you will examine the controlled release of paracetamol active substance froma hydrogel polymer system.
Chemicals
Substance

Name

State

Paracetamol

Solid

GHS Hazard
Statements
H302, H315,
H317, H319,
P280, P301 + P312
+ P330, P305 +P351 + P338
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Preparatory problems: PRACTICAL

Phosphate bufferedsaline (PBS)

Phosphate buffered-saline (PBS)
solution (pH: 7.4)
Acrylic acid

Solution

N,N′-Methylenebisacrylamide

Solid(NH4)2S2O8

Ammonium persulfate

Solid

Distilled water

Deionized water

LiquidLiquidH319, P305 +P351 + P338
H226-H302 +H312 + H332H314-H335-H400,
P210, P261, P273,
P280, P303 + P361
+ P353, P305 +P351 + P338
H302, P301 +P312 + P330
H272, H302,
H315, H317,
H319, H334,
H335, P210, P280,
P301 + P312 +P330, P302 +P352, P305 + P351
+ P338Not Hazardous

Glassware and Equipment
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1 Beaker, 100 mL
1 Graduated cylinder, 100 mL
1 Volumetric pipette, 5 mL
1 Volumetrik flask with stopper, 250 mL
5 Volumetric flasks with stoppers, 10 mL
1 Pipette filler bulb
1 Spatula (small)
6 Test tubes with plastic caps
1 Weighing dish1 Petri dish1 Glass rod
1 Wash bottle
1 Magnetic hotplate stirrer
1 Stopwatch
15 Disposable plastic pipette, 3 mL
2 Sheets of millimetric paper
10152nd International Chemistry Olympiad, Istanbul, TURKEY


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

Preparatory problems: PRACTICAL

1 Ruler
1 Felt-tip pen for glassware
1 Photometer2 UV-vis quartz (or quality plastic, <200 nm) absorption cuvettes
1 Balance
1 stirring bar
1 marker
1 thermometer
1 Vortex
1 Hot air gun
Vials
Water bath
Syringe
Plastic Straws

Preparation of pH: 7.4 PBS: Prepare 150 mL of distilled water in a 250 mL flask. Add 1.6 g
NaCl, 0.4 g KCl, 288 mg Na2HPO4, and 48 mg NaH2PO4. Add distilled water until volume is
250 mL and mix to dissolve all salts in the solution.
Preparation of Hydrogel
1. Weigh different amounts of AA monomer (5 mL, 72.9 mmol), MBAA (0.026 g, 0.17 mmol)
as crosslinker, paracetamol (known amount), APS (0.1 g, 0.44 mmol) as initiator in a vial (20mL), add deionized water (5.0 mL).
2. Use a vortex or magnetic stirrer to help dissolve the mixture in the vial.
3. Fill into pipettes with the aid of a syringe (the bottom of the paper pipettes is previously
closed with a hot air gun).
4. Place in a hot water bath at 60 oC for polymerization of these mixtures in pipettes.
5. Remove the polymerized gels from the paper pipettes, and divide them into equal pieces.
Part A. Plotting the Calibration Curve
In this section, you are asked to prepare the standard solutions of paracetamol as seen in Table
1. After reading the absorbance (A) values of each solution by using a UV-Vis
spectrophotometer, fill in the blanks in Table 1.

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Preparatory problems: PRACTICAL

Table 1. Data of paracetamol standards.
Cparacetamol (mg/L)
2
4
6
8
10

Absorbance at 243 nm

Procedure
1. Go to the millimetric paper page.
2. Write the standard paracetamol concentrations on the x-axis and the corresponding
absorbance values on the y-axis. Write the units of the axes.
3. Pass a straight line over the points and determine the equation of this calibration graph. If
you have any problem to obtain the linearity between absorbance and concentration, you canrepeat the experiment till you obtain a linear line.
4. Find the calibration equation.
Part B. Release of Paracetamol from the Polymeric Hydrogel System
Procedure
1. Turn on the magnetic hotplate stirrer and put a 250 mL beaker on it. Add 100 mL of PBS
solution into beaker and adjust temperature of the solution to 37 °C using a thermometer.
2. Stir the solution at 250 rpm.
2. Place the hydrogel sample provided to you into the beaker. Using a glass rod, allow hydrogel
to be in the solution (release solution) using a glass rod.
4. At different time intervals (0, 10, 20, 30, 40, and 50 min), place 2 mL of the release solution
into different test tubes, seal each tube with a plastic cap, and add 2 mL of PBS solution to therelease solution in each case.
5. Read the absorbance at 243 nm for all solutions you collected. Use PBS solution as blank.
6. Fill in the following Table 2.

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Preparatory problems: PRACTICAL

Table 2. Data for the release of paracetamol from the hydrogel system depending on time.
Time (min)
0
10
20
30
40
50

Absorbance (A)Calculations & Analysis
In this section, the release behavior of paracetamol from the hydrogel system will be examined.
Use the absorbance values in Table 2 and the calibration equation.
P1.1. Calculate the cumulative drug release using the following equation:𝑣 𝑋 𝑐𝑖 +𝑣2 ∑ (𝑐𝑖−1 )× 100%𝑚

Cumulative drug release ratio = 1

In this equation:
v1: total volume of PBS solution (100 mL),ci: concentration in the release medium,v2: volume of the measured sample (2 mL),
m: mass of paracetamol in the hydrogel.
Table 3. Experimental data.
Time (min)
0
10
20
30
40
50

Paracetamol concentration

Cumulative drug release (%)

P1.2. Using another sheet of millimetric paper, record the release times (t=0, 10, 20, 30, 40, 50min) on the x-axis and the cumulative drug release values on the y-axis. Pass a straight line
through the points.
P1.3. Calculate the time (in minutes) required to take 20% of paracetamol from this hydrogel
system for an individual person suffering from pain using the graphic of Table 3.
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Preparatory problems: PRACTICAL

Problem P2. Determination of the Total Carbon Content of Oltu
Stone (Black Amber) Samples
Erzurum is a city in the northeastern part of Anatolia and is called the peak of Turkey due to its
high altitude of 1900 m. Erzurum has a unique black stone (Oltu stone, black amber). This stone
has been carved since the 18th century to produce jewelry and different souvenirs. Different
products of Oltu stone such as rings, earrings, necklaces, bracelets, tie pins, smoking pipes,
cigarette holders, and prayer beads are produced by polishing and are sold in Taşhan (Rustem
Pasha Caravanserai) Bazaar in Erzurum, which was established in 1561.

There are around 600 active quarries mining Oltu stone. The beds are about 80 cm in thickness
and form by diastrophism and folding of fossilized trees. Oltu stone is soft when excavated but
begins to harden when it is exposed to the air. It is generally black but can be dark brown, gray,or greenish too. Oltu stone is an important material for the electric and electronics industries
because its graphite-like carbon black nature. Therefore, Oltu stone contains plenty of theelement carbon.
This task aims to quantitatively determine the amount of carbon in Oltu stone samples thanks
to back titration, using dichromate solution. Carbon element in the sample is first oxidized toCO2(g) with dichromate and then excess dichromate is back titrated with a standard iron(II)sulfate solution.
Note: This experiment can also be performed using another carbon-rich material such as coalor pencil graphite instead of Oltu stone.
Caution: Potassium dichromate (K2Cr2O7) is a very strong oxidizer and corrosive. Its contact
with other materials may cause a fire. Therefore, this experiment must be performed in a fume
hood by using glassware, laboratory coat, and glove. After you complete the experiment, waste
solutions and/chemicals should be placed in the waste container.

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Preparatory problems: PRACTICAL

Chemicals
Substance
K2Cr2O7

H2SO4

H3PO4

FeSO4

Name
Potassium
dichromate

State
Solid

GHS Hazard Statement
H340, H350, H360FD, H272, H301,
H312, H314, H317, H330, H334,
H335, H372, H410, P201, P221,
P273, P280, P301 + P330 + P331,
P302 + P352, P304 + P340, P305 +P351 + P338, P308 + P310
Sulfuric acid
Liquid H290, H314, P280, P301 + P330 +P331, P303 + P361 + P353, P305 +P351 + P338 + P310
Phosphoric
Liquid H290, H302, H314, P260, P280,
acid
P301 + P312 + P330, P301 + P330 +P331, P303 + P361 + P353, P305 +P351 + P338 + P310
Iron(II) sulfate Solid H302, H315, H319; P302 + P352,
P305 + P351 + P338, P301 + P312 +P330
Diphenylamine Solid H315, H319, H335
sodiumsulfonate

Glassware and equipment
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2 Volumetric flasks (with stoppers), 1000 mL
1 Volumetric flask (with stopper), 250 mL
2 Volumetric flasks (with stoppers), 100 mL
1 Volumetric flask (with stopper), 10 mL
1 Measuring cylinder, 100 mL
1 Measuring cylinder, 50 mL
1 Erlenmeyer flask, 250 mL
1 Burette, 50 mL
1 two-necked digestion flask
1 Weighing dish1 Spatula
1 Transfer funnel
1 Laboratory stand with burette clamp
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52nd International Chemistry Olympiad, Istanbul, TURKEY

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Preparatory problems: PRACTICAL

1 Weighing balance (0.1 mg)
1 Magnetic hotplate stirrer
1 Magnetic bar
1 Ice-bath or water-bath
1 Thermometer
1 Stopwatch
Beakers (for transfers)
Bossheads and clamps
Water-ice bath
Pasteur pipettes

Reagent and Standard Solutions
Potassium dichromate solution: Potassium dichromate (8.825 g), previously oven dried at about
140 °C for 1 hour, is transferred into a 100 mL flask, dissolved in about 90 mL of deionizedwater, and then diluted to the volume with deionized water (0.3 M K2Cr2O7).
Phosphoric acid-sulfuric acid solution: 1.5 mL of concentrated H2SO4 is added slowly to 5 mLof deionized water under stirring in a 10 mL flask. 1.5 mL of 85% phosphoric acid is also added
to this flask, cooled at room temperature, and diluted to exact volume with deionized water.
Note: Be careful during this step because the process of adding acid to water is exothermic.
Iron(II) sulfate standard solution: Iron(II) sulfate (3.038 g) is transferred into a 100 mL flask
and dissolved in about 80 mL of deionized water. Then 2 mL of concentrated H2SO4 is added
to this FeSO4 solution slowly under stirring. The mixture is cooled to room temperature anddiluted to exact volume (0.2 M FeSO4).
Diphenylamine sodium sulfonate indicator solution: 20 mg of 4-diphenylamine sodium
sulfonate is dissolved in concentrated H2SO4 in a 10 mL flask.
Procedure
1. Weigh about 10 mg of Oltu stone sample using a weighing balance and write down the exact
weight of the sample.
2. Place this Oltu stone sample into a digestion flask over a magnetic hotplate stirrer and thenadd 5 mL of 0.3 M K2Cr2O7 in a fume hood.
3. Add slowly 20 mL of concentrated H2SO4 while cooling the flask in a water-ice bath. Put amagnetic bar and stir at 200 rpm.
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Preparatory problems: PRACTICAL

4. Attach a thermometer to the stopper and heat rapidly until a temperature of 160 °C is reached.
5. Maintain the temperature at 160 ± 2 °C for 10 min.
6. Transfer the solution to a 250 mL Erlenmeyer flask and cool to room temperature with tap
water.
7. Add 8 drops of the phosphoric acid/sulfuric acid mixture and 4 drops of 4-diphenylamine
sodium sulfonate indicator to the solution.
8. Fill the burette with standard 0.2 M FeSO4 solution until the volume line.
9. Titrate the excess K2Cr2O7 with a 0.2 M FeSO4 solution. The color change is from violet todirty gray (end point) and finally green.
10. Repeat steps 1–9 with another Oltu stone sample as necessary.
11. Perform the same experiment using a blank (in the absence of Oltu stone sample)
solution.

Figure P2-1. A titration setup
Calculations & Analysis
P2.1. Write all balanced equations during this experiment.
P2.2. Calculate the mean total carbon content of the Oltu stone sample as weight %.
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Problem P3. Spectrophotometric Determination of theEquilibrium Constant for the Formation of a Complex
In this task, the complexation reaction between I2 and pyridine (pyr) will be followed byspectrophotometry to determine the equilibrium constant (K) of this complexation reaction. I2
and I2.pyr complex can absorb visible region of electromagnetic radiation but pyr cannot
because it is colorless. Analysis of the spectral changes with variation in pyr concentration andwith constant total iodine concentration provides the determination of K of the complexation
reaction.

Caution: All operations should be carried out in a fume hood except for spectrophotometricmeasurements. After you complete the experiment, waste solutions and/or chemicals should beplaced in the waste container.
Chemicals
Substance Name
C5H5N
PyridineState
LiquidC6H12Cyclohexane Liquid

I2

IodineSolid

GHS Hazard Statement
H225, H302 + H312 + H332, H315, H319; P210,
P280, P301 + P312 + P330, P302 + P352 + P312,
P304 + P340 + P312, P305 + P351 + P338
H225, H304, H315, H336, H410, P210, P273, P301
+ P310 + P331, P302 + P352
H312 + H332, H315, H319, H335, H372, H400,
P273, P280, P302 + P352 + P312, P304 + P340 +P312, P305 + P351 + P338, P314

Glassware and equipment






Spectrophotometer2 UV-vis glass, quartz or plastic absorption cuvettes
1 Volumetric flask (with stopper), 50 mL
6 Volumetric flasks (with stoppers), 25 mL
1 Pipette, 1 mL
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Preparatory problems: PRACTICAL

1 Pipette, 10 mL
1 Pipetting bulb

Reagent and Standard Solutions
1. 0.050 M pyr in cyclohexane (50 mL for each student, concentrations must be known
accurately).
2. 0.010 M I2 in cyclohexane (25 mL for each student, concentrations must be known
accurately).
Procedure
1. Pipette the following volumes of stock solutions into six 25-mL volumetric flasks labeled as
F-0, F-1, F-2, F-3, F-4, and F-5; dilute to the mark with cyclohexane; and mix them well.
Flask
F-0
F-1
F-2
F-3
F-4
F-5

Volume of pyr stock
solution/mL
0.0
1.0
2.0
3.0
4.0
5.0

Volume of 𝐼2 stock
solution/mL
1.0
1.0
1.0
1.0
1.0
1.0

2. Use two glass absorption cells with solvent in both the sample and the reference cells andscan the wavelength between 350 and 800 nm to record a baseline.
3. Record all spectra for the samples using this background.
4. Measure the absorbance values at the wavelengths of the two maxima in each spectrum bysubtracting the absorbance of the blank from each.
Depending on the complexation reaction,
𝐾=[𝐼2 . 𝑝𝑦𝑟][𝐼2 ][𝑝𝑦𝑟]Consider a series of solutions in which increments of pyr are added to a constant amount of I2.
Letting I2(0) be the total concentration of I2 (in the forms I2 and I2.pyr), we can write110

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: PRACTICAL[𝐼2 ] = [𝐼2(0) ] − [𝐼2 . 𝑝𝑦𝑟]K expression can be rearranged as follows:[𝐼2 . 𝑝𝑦𝑟]
= 𝐾[𝐼2 ][𝑝𝑦𝑟][𝐼2 . 𝑝𝑦𝑟]
= 𝐾([𝐼2(0) ] − [𝐼2 . 𝑝𝑦𝑟])[𝑝𝑦𝑟]
Based on the last equation, a graph of[𝐼2 .𝑝𝑦𝑟][𝑝𝑦𝑟]vs. [𝐼2 . 𝑝𝑦𝑟] has a slope of –K.

If we know[𝐼2 . 𝑝𝑦𝑟], we can find [𝑝𝑦𝑟] with the mass balance
𝑝𝑦𝑟0 = [𝑡𝑜𝑡𝑎𝑙 𝑝𝑦𝑟] = [𝐼2 . 𝑝𝑦𝑟] + [𝑝𝑦𝑟]To measure[𝐼2 . 𝑝𝑦𝑟], we use absorbance values. Suppose that 𝐼2 and 𝐼2 . 𝑝𝑦𝑟 each has someabsorbance at wavelength λ, but 𝑝𝑦𝑟 has no absorbance at this wavelength. Let us assume that
we measure absorbance values at a path length of 1 cm so that we can omit it when writing
Beer’s law. The absorbance at every wavelengths is the sum of absorbances of 𝐼2 and 𝐼2 . 𝑝𝑦𝑟
𝐴 = 𝜀𝐼2 .𝑝𝑦𝑟 [𝐼2 . 𝑝𝑦𝑟] + 𝜀𝐼2 [𝐼2 ]Substituting [𝐼2 ] = [𝐼2(0) ] − [𝐼2 . 𝑝𝑦𝑟], we can write
𝐴 = 𝜀𝐼2 .𝑝𝑦𝑟 [𝐼2 . 𝑝𝑦𝑟] + 𝜀𝐼2 [𝐼2(0) ] − 𝜀𝐼2 [𝐼2 . 𝑝𝑦𝑟]In the last equation 𝜀𝐼2 [𝐼2(0) ] is A0, the initial absorbance before any pyr is added. Therefore,∆𝐴

𝐴 = [𝐼2 . 𝑝𝑦𝑟](𝜀𝐼2 .𝑝𝑦𝑟 − 𝜀𝐼2 ) + 𝐴0 and [𝐼2 . 𝑝𝑦𝑟] = ∆𝜀where ∆𝜀 = 𝜀𝐼2 .𝑝𝑦𝑟 − 𝜀𝐼2 and ∆𝐴 = 𝐴 − 𝐴0 is the observed absorbance after each addition ofpyr minus the initial absorbance.
Substituting[𝐼2 .𝑝𝑦𝑟][𝑝𝑦𝑟][𝐼2 . 𝑝𝑦𝑟]

from

∆𝐴

𝐴 = ∆𝜀 (𝜀𝐼2 .𝑝𝑦𝑟 − 𝜀𝐼2 ) + 𝐴0

andwritingitin= 𝐾([𝐼2(0) ] − [𝐼2 . 𝑝𝑦𝑟]) gives the following equation:

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52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: PRACTICAL∆𝐴= 𝐾∆𝜀[𝐼2(0) ] − 𝐾∆𝐴[𝑝𝑦𝑟]This equation is known as the Scatchard equation.
Calculations & Analysis∆𝐴

P3.1. Draw a graph of [𝑓𝑟𝑒𝑒 𝑝𝑦𝑟] vs. ∆𝐴 (Scatchard plot).
P3.2. Determine the K value of this complexation reaction from the slope.
P3.3. Determine the ∆𝜀 value using the intercept.
P3.4. Determine the 𝜀𝐼2 value using the absorption band of I2.
P3.5. Calculate the 𝜀𝐼2 .𝑝𝑦𝑟 value.
P3.6. Observe if there is an isosbestic point in this experiment.
P3.7. If yes, explain why an isosbestic point is observed.

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Preparatory problems: PRACTICAL

Problem P4. 1-Bromobutane
Nucleophilic substitution reactions are the most fundamental reactions among the organic
functional group transformations. In these reactions, an electronegative atom or electronwithdrawing group is replaced by another atom or group. Replacement of the hydroxyl groupis subject to special interest since it has a vital role in carbohydrate chemistry. For instance,
sucrose is commonly used in our everyday life since many carbohydrates taste sweet. When we
consume sucrose, excess calories are stored as fat in our bodies. To avoid the storage of fat
because of excessive sucrose consumption, sucrose can be converted to synthetic sweetener,
sucralose, through substitution reactions. On the other hand, sucralose is known diet friendly
and it is much sweeter than sucrose.

Considering the impact of substitution reactions in our everyday life, in this task, you are asked
to synthesize 1-bromobutane starting from 1-butanol by an SN2 reaction. Among several routes,
an aqueous solution of sodium bromide and excess sulfuric acid method is preferred.

Chemicals
Substance NameC4H9OH
1-Butanol

State
Liquid

H2SO4

Sulfuric acid

Aqueoussolution

NaBr

SodiumbromideSolid

GHS Hazard Statement
H226, H302, H315, H318, H335, H336, P210,
P280, P301 + P312 + P330, P302 + P352, P305
+ P351 + P338 + P310
H290, H314, P280, P301 + P330 + P331, P303 +P361 + P353, P305 + P351 + P338 +P310
Not hazardous

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NaOH
CaCl2

Sodiumhydroxide
Calciumchloride

Aqueoussolution
Solid

Preparatory problems: PRACTICAL

H290, H314, P280, P301 + P330 + P331, P303 +P361 + P353, P305 + P351 + P338 + P310
H319, P264, P280, P305 + P351 + P338, P337 +P313

Glassware and equipment

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







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1 Graduated cylinder, 10 mL
1 Graduated cylinder, 50 mL
1 Weighing dish1 Weighing balance (0.01 g)
1 Spatula
2 Laboratory stands
1 Thermometer
1 Magnetic stirrer
Bossheads and clamps
1 Condenser
1 Reflux condenser
1 Round-bottom flask, 100 mL
1 Dropping funnel
1 Distillation head
1 Distillation adapter
1 Erlenmeyer flask, 50 mL
1 Separatory funnel, 100 mL
Ice-water bath

Procedure
1. Add 13.3 g of sodium bromide, 15 mL of water, and 10 mL of 1-butanol to a 100 mL roundbottom flask.
2. Cool the mixture in an ice-water bath.
3. Slowly add 11.5 mL of concentrated sulfuric acid while stirring and cooling the reaction
mixture. (Warning: Be careful in handling concentrated sulfuric acid.)
4. Place the flask on a hot plate, clamp it securely, and fit it with a short reflux condenser. Heat
the resulting mixture to boiling point. Adjust the heat for brisk and steady refluxing.

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Figure P4-1. Refluxing apparatus
5. Reflux for 45 minutes, remove the heater, and let the condenser drain for a few minutes.
6. Remove the condenser, mount the distillation head to the flask, and set up the condenser for
simple distillation through the distillation adapter into the 50 mL Erlenmeyer flask. Clamp jointsof the glassware securely.

Figure P4-2. Distillation apparatus

7. Distill the mixture until clear water droplets appear.
8. Pour the distillate into a separatory funnel, shake with about 10 mL of water, and note thatn-butyl bromide (1-bromobutane) now forms the lower layer. Pink coloration in this layer canbe discharged by adding a little bit of sodium bisulfite and shaking the separatory funnel again(See P5.5 for detailed explanation about using separatory funnel and extraction process).

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9. Drain the lower layer of 1-bromobutane into a clean flask, clean and dry the separatory
funnel, and return the 1-bromobutane to the cleaned separatory funnel.
10. Then add 10 mL of precooled concentrated sulfuric acid to the separatory funnel, shake thefunnel, and wait 5 minutes for separation of the layers. (Warning: Be careful in handlingconcentrated sulfuric acid.)
11. Separate the layers. Then wash 1-bromobutane with 10 mL of 3 M sodium hydroxide
solution to remove traces of acid.
12. Dry the cloudy 1-bromobutane by adding anhydrous calcium chloride pellets (about 1 g)
while stirring until the liquid becomes clear.
13. After 5 minutes, decant or filter the dried liquid into a 25 mL flask.
14. Distill and collect the product boiling in the range of 99–103 °C using a simple distillation
apparatus.
QuestionP4.1. In step 8, the organic layer is treated with a little bit of sodium bisulfite. Explain thereason.
P4.2. Explain the roles of NaBr and H2SO4 in the reaction.
P4.3. Is this reaction possible in the absence of a strong acid?
P4.4. Compare the 1H NMR spectra of 1-butanol and 1-bromobutane in CDCl3 in terms ofnumber of peaks and multiplicities.
P4.5. Would you expect other by-products to be formed in this reaction? What is it?

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Problem P5. Cannizzaro Reaction
Aldehydes are widespread in nature. Most of them are known for their sweet odors. For instance,
vanillin exists in vanilla with its distinct odor. Aldehyde functional group-containing molecules
are often used in perfumes because of their pleasant fragrance. Benzaldehyde provides thedistinctive smell of almonds. Stanislao Cannizzaro discovered in 1853 that benzaldehyde on
treatment with a base gave equimolar quantities of benzoic acid and benzyl alcohol. In this task,
you are asked to prepare p-chlorobenzoic acid and p-chlorobenzyl alcohol through a Cannizzaro
reaction of p-chlorobenzaldehyde.

Cannizzaro is a disproportionation reaction in which an internal oxidation/reduction reaction
occurs. The reaction of an aldehyde with a strong base takes place only if the aldehyde has noalpha hydrogen atoms. The mechanism of the Cannizzaro reaction involves hydride ion (H)transfer, which takes place only in the presence of a strong nucleophile.
Chemicals
Substance
Name
ClC6H5CHO p-Chlorobenzaldehyde

State
Solid

KOH

Potassium hydroxide

Solid(C2H5)2O

Diethyl ether

LiquidHClHydrochloric acid

Aqueoussolution

NaHSO3

Sodium bisulfite

Aqueoussolution

GHS Hazard Statement
H302, H315, H317, H319, H411,
P273, P280, P302 + P352, P305 +P351 + P338
H290, H302, H314, P260, P280,
P301 + P312 + P330, P301 + P330 +P331, P303 + P361 + P353, P305 +P351 + P338 + P310
H224, H302, H336, H412, P210,
P273, P301 + P312 + P330, P403 +P233
H290, H314, H335, P260, P280,
P303 + P361 + P353, P304 + P340 +P310, P305 + P351 + P338 + P310
H302, P301 + P312 + P330

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NaHCO3

Sodium bicarbonate

Aqueoussolution

Na2SO4C2H5OH

Sodium sulfate
Ethanol

Solid
Liquid

CH3OH
CH2Cl2CH3COCH3

Methanol
Methylene chloride
AcetonePetroleum ether

LiquidLiquidLiquidLiquidPreparatory problems: PRACTICAL

Not a hazardous substance or
mixture according to Regulation(EC) No. 1272/2008.
Not hazardous
H225, H319, P210, P305 + P351 +P338
H225, H301 + H311 + H331, H370
H315, H319, H336, H351
H225, H319, H336
H225, H304, H315, H336, H411

Glassware and equipment
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1 Two-neck round bottom flask, 50 mL
1 Magnetic stirrer
1 Graduated cylinder, 50 mL
1 Condenser
1 Separatory funnel, 100 mL
1 Büchner funnel, 250 mL
1 Büchner flask
1 Filter paper
2 Erlenmeyer flasks
1 Funnel
1 TLC development tank
TLC sheets
1 UV lamp
1 Hot plate

Procedure
1. Dissolve 2 g of p-chlorobenzaldehyde in 6 mL of ethanol in a 50 mL two-neck round bottom
flask while stirring with a magnetic rod.
2. Add 6 mL of water to the above solution followed by the addition of 3.2 g of KOH in portions.
3. Mount a condenser above the reaction flask and heat the reaction mixture to 60–70 C.

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Figure P5-1. The reaction apparatus with two-neck round bottom flask.

4. After 45 minutes, remove the flask from the water bath, allow the reaction mixture cool down
to room temperature, transfer into a separatory funnel, and add 30 mL of water.
5. Perform extraction with 7 mL of methylene chloride three times. To do the extraction:

Figure P5-2. Extraction apparatus.
Before adding methylene chloride to the separatory funnel, check the stopcock.
To have sufficient room for extraction, fill the separatory funnel no more than threefourths full.

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










Preparatory problems: PRACTICAL

When the stoppered funnel is shaken to distribute the components between methylene
chloride and water, pressure always develops through volatilization of methylene
chloride from the heat of the hands can increase the pressure.
The funnel is grasped so that the stopper is held in place by one hand and the stopcockby the other. After a brief shake or two, the funnel is held in the inverted position, andthe stopcock is opened cautiously to release pressure. This process is repeated withpressure released as necessary.
When equilibration is judged to be complete, the layers are allowed to separate.
p-chlorobenzyl alcohol is distributed wholly or largely into the bottom methylene
chloride layer, whereas p-chlorobenzoic acid sodium salt, inorganic salts, acids or bases
pass into the upper water layer.
Separate the methylene chloride and aqueous layers by drowning off the bottom
methylene chloride layer.
This process is repeated three times and organic layers are collected in to an Erlenmeyer.
Aqueous layer is drawn of in to another Erlenmeyer.

6. Acidify the aqueous basic solution with concentrated HCl until the solution is acidic (Acidity
of solution can be inspected by Litmus paper). This will cause p-chlorobenzoic acid toprecipitate as a white solid.
7. After cooling down the solution, filter the white solid under vacuum using a Büchner funnel.

Figure P5-3. Vacuum filtration with Büchner funnel and flask.
8. Recrystallize the obtained p-chlorobenzoic acid from ethanol. After air drying, weigh the
product, and calculate the percent yield.To recrystallize p-chlorobenzoic acid from ethanol:
Place the p-chlorobenzoic acid in an Erlenmeyer flask (never use a beaker), add enough ethanol
to cover the crystals, and then heat the flask on a steam. Add ethanol gradually, keeping it at the120

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: PRACTICAL

boiling point, until all of the solute dissolves (Be sure no flames are nearby when working withethanol). Once it has been ascertained that the hot solution is saturated with the p-chlorobenzoic
acid just below the boiling point of ethanol, allow the solution to cool down to room temperatureslowly. With slow cooling, recrystallization should begin immediately. If not, add a seed crystal
or scratch the inside of the Erlenmeyer with a glass rod. Once recrystallization is complete, pchlorobenzoic acid crystals must be filtered using Büchner funnel-Büchner flusk and washedwith ice-cold ethanol.
9. Organic phases from step 5 are gathered together in a separatory funnel and shaken with 15
mL of 40% bisulfite solution. The mixture is then washed with saturated sodium bicarbonate
until a neutral solution is obtained.10. Drying the organic phase with sodium sulfate followedby filtration from filter paper and removal of solvent via simple distillation will afford crude pchlorobenzyl alcohol.
11. p-Chlorobenzyl alcohol is recrystallized from acetone/petroleum ether (1:9).
12. Perform TLC analysis for p-chlorobenzoic acid, p-chlorobenzyl alcohol and starting
material p-chlorobenzaldehyde using CH2Cl2/MeOH (9/1) as the eluent. Report the Rf values.
TCL analysis can be performed as follow:Transfer the TLC eluent (CH2Cl2/MeOH: 4/1, approximately 1 mL) in to the TLC
development tank using a Pasteur pipette.
Insert the TLC plate using tweezers, cover the tank with its cap and let the eluent reachapproximately 0.5 cm bellow the top edge of the plate.
Using tweezers, take the TLC plate out, draw the eluent front line and let the plate airdry.
Place the TLC plate under the UV lamp in a hood. With a pencil, circle all the visualized
spots and calculate the Rf values of p-chlorobenzaldehyde and products designated as
spots A, B, and C on Figure P5-3.

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Figure P5-4. Sample TLC plate and representative Rf values.
13. Measure the melting points of the products and report their purity based on TLC results
and melting points.
QuestionP5.1. If the aldehyde in this reaction has an -hydrogen what kind of a reaction do you expect?
P5.2. Write the products if butanal or pivalaldehyde is used as reactant in this reaction.
P5.3. Tick the bases that can be used instead of KOH in this reaction.
☐
☐
☐
☐

K2CO3
NaOH
NaHCO3Et3N

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P5.4. Which reacts faster in the Cannizzaro reaction if the initial nucleophilic attack is the rate
determining step?
☐☐P5.5. What is the intermediate state in this reaction?
P5.6. Tick the oxidation and reduction products of this reaction.
Oxidation product☐☐

Reduction product
☐
☐

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Preparatory problems: PRACTICAL

Problem P6. 2,3-Dihydro-5,6-diphenylpyrazine
An imine is a functional group or chemical compound containing a carbon–nitrogen doublebond. Some imine compounds can sometimes be referred to as Schiff bases. Imines may findutility in a wide range of contexts, including the development antimicrobial, antiviral andanticancer agents. Imines are also common intermediates in enzymatic reactions and are used
as common ligands in coordination chemistry. They are also used in nanotechnology for water
treatment, encapsulation and functionalized magnetic nanoparticle production.
In this experiment you are asked to synthesize 2,3-dihydro-5,6-diphenylpyrazine (DPP) throughan imine formation reaction, starting from benzil and ethylenediamine.

Chemicals
Substance
C14H10O2
H2NCH2CH2NH2

C2H5OH

Name
Benzil

State
Solid

GHS Hazard Statement
H315, H319, P302 + P352, P305 + P351 +P338
Ethylenediamine Liquid H226, H302 + H332, H311, H314, H317,
H334, H412, P210, P273, P280, P301 +P330 + P331, P302 + P352, P304 + P340,
P305 + P351 + P338, P308 + P310
EthanolLiquid H225, H319, P210, P305 + P351 + P338

Glassware and equipment
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
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

1 Round-bottom flask, 250 mL
1 Stirring bar
1 Pipette, 10 mL
1 Reflux condenser
1 Beaker, 100 mL
1 Crystallization dish, 500 mL
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
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Preparatory problems: PRACTICAL

1 Büchner funnel
1 Büchner flask
1 Filter paper
1 TLC development tank
1 TLC sheets
1 UV lamp
1 Magnetic stirrer with a hot plate
1 Ice-water bath

Procedure
1. Dissolve 10 g of benzil in 30 mL of ethanol (95%) by heating in a 250 mL round bottom
flask. Add 4.5 mL of 68% ethylenediamine (or an equivalent quantity of ethylenediamine indifferent concentrations).
2. While stirring, heat the mixture in a water bath under a reflux condenser for 45 minutes (SeeFigure P4.1 for reflux condenser apparatus).
3. If crystals have not formed in the flask, immediately transfer the hot supersaturated solution
into a 100- or 150-mL beaker. The difficulty of removing the crystals from the flask is thus
avoided.
4. Often crystallization occurs at once when the solution is poured into the beaker, and sufficientheat is evolved to cause the alcohol to boil. Cool to room temperature. Finally, place in an icebath (For detailed recrystallization see P5.8).
5. The loss due to solubility in cold alcohol is negligible. Filter the crystals, and wash themwith a little alcohol. Dry the product using a suction filter (For vacuum filtration apparatus seeP5.7).
6. Weigh the dried product and calculate the percent yield.
7. Determine melting point (highly purified DPP melts at 161.5162.5 C), and reserve a little
amount of product for TLC analysis.
8. Perform TLC analysis using the recrystallized product and reference benzil (See P5.12 for asample procedure of TLC analysis).
9. Report the Rf values of each compound and check the purity of the recrystallized DPP.
QuestionP6.1. What is the product when DPP is oxidized?

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P6.2. Is the oxidation product of the DPP aromatic?
P6.3. Which of the following reactants or methods could be used for the oxidation of DPP?
☐ 2,3-Dichloro-5,6-dicyanobenzoquinone (DDQ)☐ Heating in air☐ Et3N☐ Na2CO3☐ Slight heating under vacuum
P6.4. What are the hybridizations of nitrogen (b) and carbons (a and c) in DPP?

P6.5. Draw the structures of the products when 1,3-propanediamine and 1,4-butanediamine
were used instead of ethylenediamine.

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Problem P7. Determination of Rate Constants for n-Butyl
Acetate Hydrolysis
Hydrolysis, in other words splitting with water, is one of the most important reaction types thatan ester can undergo. The hydrolysis of ester can be catalyzed either by acids or by bases andwhen the catalyst is a base, the reaction yields a carboxylate salt and an alcohol.
In this experiment, you are asked to determine rate constants of the alkaline hydrolysis of nbutyl acetate using sodium hydroxide, which is a typical second-order reaction.

The rate of the reaction can be calculated by the following equation:
Reaction rate = k[CH3COOC4H9][OH − ]
Then
1
1
–
= k × 𝑡,[A] [A]0

where [A] is the concentration of reactant at time t ([A]= [B]), [A]0 is the initial concentrationand k is the second-order constant, which has dimension of concentration–1 time–1 (L mol–1 s–1). In this case, a characteristic plot that will produce a linear function is 1/ [A] vs. time t, witha slope of k.
Chemicals
SubstanceNamen-Butyl acetate

State
Aqueous solution

GHS Hazard Statement
H226, H336, P210

NaOH

Sodium hydroxide

Aqueous solution

HCl

Hydrochloric acid

Aqueous solution

H290, H314, P280, P301 +P330 + P331, P303 + P361
+ P353, P305 + P351 +P338 + P310
H290, H314, H335, P260,
P280, P303 + P361 +P353, P304 + P340 +P310, P305 + P351 + P338
+ P310127

52nd International Chemistry Olympiad, Istanbul, TURKEY

Phenolphthalein

Solution

Preparatory problems: PRACTICAL

H226, H319, P210, P280,
P303 + P361 + P353, P337
+ P313, P370 + P378,
P403 + P235

Glassware and equipment








1 Laboratory stand with burette clamp
5 Volumetric pipettes, 20 mL
5 Pipette pump
5 Titration flasks, 250 mL
10 Volumetric flasks, 250 mL
1 Burette, 50 mL
Stop watch

Procedure

1. Fill the burette with the solution of hydrochloric acid (HCl) (0.02 M).
2. Transfer 60 mL of the solution of n-butyl acetate (0.02 M) into the volumetric flask and 60
mL of the solution of sodium hydroxide (0.02 M) into another volumetric flask at room
temperature. Mix the two solutions in titration flask.
3. Five minutes after mixing, pipette 20 mL of reaction mixture into a titration flask. Add 4
drops of phenolphthalein indicator to the solution.
4. Titrate the sample solution with HCl (0.02 M) until the solution become as colorless. Record
the amount of HCl used. Hint: you can add 6 ml of HCl solution immediately and then carryout the rest of titration with more care.
5. Repeat steps 3 and 4 for 10, 15, 20, and 25 minutes from the moment of mixing. Fill in thetable below.
Hint: You can repeat each step several times to increase the accuracy of data.
Calculations & Analysis:
Fill in the blanks in the following table with the data measured during the experiment.

128

52nd International Chemistry Olympiad, Istanbul, TURKEY

Time (min)
5
10
15
20
25

Preparatory problems: PRACTICAL

VHCl (mL)P7.1. Calculate the concentration of [OH − ] at each time.
1

P7.2. Plot [OH−] vs. time .
P7.3. Calculate the rate constant.
P7.4. Calculate the reaction rate.
P7.5. Calculate the initial half-life for the reaction with initial conditions.

129

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: PRACTICAL

Problem P8. Activation Energy of Bromide / Bromate Reaction
Activation energy is the minimum amount of energy which is required for a chemical reaction
to occur. Activation energy can be defined also as the energy difference between the reactants
and the activated complexes.

In this experiment, you are asked to calculate the activation energy of the following reaction:
𝐾𝐵𝑟𝑂3 + 5𝐾𝐵𝑟 + 3𝐻2 𝑆𝑂4 → 3𝐾2 𝑆𝑂4 + 3𝐵𝑟2 + 3𝐻2 𝑂
In this reaction, the reaction order for KBrO3 and KBr is the same and is observed to be one. By
using Arrhenius equation:
𝐸𝑎𝐸

𝑘 = 𝐴 . 𝑒 −𝑅𝑇 or 𝑙𝑛 𝑘 = 𝑙𝑛 𝐴 – 𝑅𝑇𝑎 products ;For a first rate reaction where A k𝑑[𝐴]− 𝑑𝑡 = 𝑘[𝐴]
𝐴

𝑡
𝑑[𝐴]
−∫
= 𝑘 ∫ 𝑑𝑡
𝐴0 [𝐴]
0[𝐴]0𝑙𝑛=𝑘×𝑡=𝑝[𝐴]

𝑝 = 𝑘 × 𝑡 and it is a constant. Logarithm of this equation:
ln𝑝 = ln𝑘 + ln𝑡
ln𝑘 = ln𝑝 − ln𝑡
If we use ln k  ln p  ln t for Arrhenius equation then;𝐸𝑎
𝑅𝑇𝐸𝑎− ln 𝑡 = (ln 𝐴 − ln 𝑝) −𝑅𝑇
ln 𝑝 − ln 𝑡 = ln 𝐴 −Consider, ln 𝐴 − ln 𝑘 = 𝐾 then;ln 𝑡 =

𝐸𝑎
−𝐾
𝑅𝑇
130

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: PRACTICAL

Completion of the reaction will be observed by following the decoloration of the solution. The
reaction yields Br2, which gives a very rapid reaction with phenol yielding tribromophenol.
When all the phenol is used, the remaining Br2 will decolorate the indicator.
Chemicals
Substance

Name

State

PhenolSolution

KBr

Potassium bromide

Solid

KBrO3

Potassium bromate

Solid

H2SO4

Sulfuric acid

Aqueoussolution

Methyl red

Solution

GHS Hazard
Statement
H301 + H311 + H331,
H314, H341, H373,
P201, P260, P280, P301
+ P310 + P330, P303 +P361 + P353, P305 +P351 + P338 +P310
H319, P280, P305 +P351 + P338, P337 +P313
H271, H301, H350,
P201, P210, P301 +P310 + P330
H290, H314, P280, P301
+ P330 + P331, P303 +P361 + P353, P305 +P351 + P338 +P310

R 51/53 , S 61

Glassware and equipment







3 Glass volumetric pipettes, 10 mL
10 Glass test tubes, 15 mL
Wash bottle
2 Laboratory stands with appropriate clamps
Thermostated water bath
Stop watch131

52nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: PRACTICAL

Procedure
1. Prepare the following solutions in two separate test tubes:
Solution I: 5 mL of 0.01 M phenol, 5 mL of KBr–KBrO3 solution (dissolve 50 mg of KBr and
14 mg of KBrO3 in 5 mL of deionized water), a few drops of methyl red indicator
Solution II: 2.5 mL of 0.5 M H2SO4

2. Place them into a thermostated circulating water bath.
3. When the temperature reaches 25 °C, mix the two solutions together, start your timer, andstop the timer when the red color completely disappears. Record the time.
4. Repeat steps 1–3 for 35 °C, 45 °C, 55 °C, 65 °C.
If you do not have a thermostat, instead of different temperatures, perform the experiment in an
ice bath (or cold water) and at room temperature, then measure the temperature, redesign thetables in step 8.2 and 8.3 accordingly.
Calculations & Analysis
P8.1. Calculate final concentrations of H2SO4, KBr, and KBrO3.
P8.2. Fill in the following table.
T (°C)t (seconds)

25

35

45

55

65

P8.3. Calculate 𝑙𝑛 𝑡 and 1/𝑇 for each step and fill in the table below.
ln t
1/T (K–1)
P8.4. Plot 𝑙𝑛 𝑡 vs. 1/𝑇 and determine the slope of the plot.
P8.5. Calculate Ea.

13252nd International Chemistry Olympiad, Istanbul, TURKEY

Preparatory problems: PRACTICAL

The End133