PREPARATORYPROBLEMS

Making science together!

Third edition (19-6-3)

Table of contents
Preface ........................................................................................................................................ 3
Fields of advanced difficulty ...................................................................................................... 5
Theoretical problems ............................................................................................................. 10
Problem 1. Butadiene π-electron system .................................................................................. 11
Problem 2. Localization and delocalization in benzene ........................................................... 13
Problem 3. Study of liquid benzene hydrogenation ................................................................. 15
Problem 4. Use of dihydrogen: fuel cells ................................................................................. 16
Problem 5. Hydrogen storage ................................................................................................... 18
Problem 6. Deacidification and desulfurization of natural gas ................................................ 22
Problem 7. Lavoisier’s experiment .......................................................................................... 25
Problem 8. Which wine is it? Blind tasting challenge ............................................................. 26
Problem 9. Nitrophenols: synthesis and physical properties .................................................... 27
Problem 10. French stone flower ............................................................................................. 31
Problem 11. The mineral of winners ........................................................................................ 33
Problem 12. Reaction progress kinetics ................................................................................... 34
Problem 13. Nylon 6 ................................................................................................................ 37
Problem 14. Synthesis of block copolymers followed by size-exclusion chromatography ..... 39
Problem 15. Radical polymerization ........................................................................................ 43
Problem 16. Biodegradable polyesters ..................................................................................... 46
Problem 17. Vitrimers .............................................................................................................. 48
Problem 18. A kinetic study of the Maillard reaction .............................................................. 50
Problem 19. Glycosidases and inhibitors ................................................................................. 53
Problem 20. Fluoro-deoxyglucose and PET imaging .............................................................. 56
Problem 21. Catalysis and stereoselective synthesis of cobalt glycocomplexes ...................... 61
Problem 22. Structural study of copper (II) complexes ........................................................... 62
Problem 23. Synthesis and study of a molecular motor ........................................................... 64
Problem 24. Some steps of a synthesis of cantharidin ............................................................. 67
Problem 25. Study of ricinoleic acid ........................................................................................ 68
Problem 26. Synthesis of oseltamivir ....................................................................................... 70
Problem 27. Formal synthesis of testosterone .......................................................................... 71
Back to 1990: Aqueous solutions of copper salts .................................................................... 73
Practical problems .................................................................................................................. 74
Problem P1: Synthesis of dibenzylideneacetone ...................................................................... 75
Problem P2: Oxidation of (‒)-borneol to (‒)-camphor ............................................................ 77
Problem P3: Aspirin® tablet ..................................................................................................... 79
Problem P4: Illuminated Europe .............................................................................................. 81
Problem P5: Protecting the vineyard ........................................................................................ 83
Problem P6: Equilibrium constant determination .................................................................... 86

51st IChO – Preparatory problems2

Preface
We are happy to provide Preparatory Problems for the 51st International Chemistry Olympiad.
These problems will be an opportunity for students to train for the Olympiad, but also to
discover numerous topics in both modern and traditional chemistry. These problems should be
solved using the topics covered in high school and some topics of advanced difficulty listed
below (six for the theoretical part and two for the practical one).
This booklet contains 27 theoretical and 6 practical problems. Its length should not be seen as
an indication of its difficulty: it merely reflects our commitment to write these problems in a
spirit as similar as possible to the final problems. An additional theoretical task (Back to 1990)
ends the first section. This problem should not be studied as thoroughly as the others, as it is anexcerpt of the tasks proposed to the candidates during the last Olympiad held in France, in 1990.
The official solutions will be sent to the Head Mentors by the end of February 2019, and will
be published on the IChO 2019 website not earlier than the 1st of June 2019.
We will be happy to read and reply to your comments, corrections and questions about the
problems. Please send them to contact-icho2019@laligue.org
Looking forward to seeing you in Paris to enjoy chemistry and to make science together!

The members of the Scientific Committee in charge of the preparatory problems
Didier Bourissou, CNRS, Toulouse
Aurélien Moncomble, Université de Lille
Élise Duboué-Dijon, CNRS, Paris
Clément Guibert, Sorbonne Université, Paris
Baptiste Haddou, Lycée Darius Milhaud, Le Kremlin-Bicêtre
Hakim Lakmini, Lycée Saint Louis, Paris

Acknowledgments
We would like to thank all the authors for their efforts in writing these problems. Their hard
work during numerous months resulted in this booklet that will hopefully be useful for the
young chemists involved in this Olympiad. We are also indebted to the reviewers, including themembers of the steering committee, whose precision and thoroughness significantly improved
these problems.

51st IChO – Preparatory problems

3

Contributing authors
Pierre Aubertin, Lycée Léonard de Vinci, Calais
Laurent Heinrich, Lycée Pierre Corneille, Rouen
Tahar Ayad, Chimie ParisTech, Paris
Lucas Henry, ENS, Paris
Alex Blokhuis, ESPCI, Paris
Claire Kammerer, Univ. Paul Sabatier, Toulouse
Clément Camp, CNRS, Lyon
Dmytro Kandaskalov, Aix Marseille Université
Jean-Marc Campagne, ENSCM, Montpellier
Iuliia Karpenko, Université de Strasbourg
Xavier Cattoën, CNRS, Grenoble
Maxime Lacuve, ENSAM, Paris
Baptiste Chappaz, Collège Les Pyramides, Évry
Hakim Lakmini, Lycée Saint Louis, Paris
Sylvain Clède, Lycée Stanislas, Paris
Julien Lalande, Lycée Henri IV, Paris
Éric Clot, CNRS, Montpellier
Alix Lenormand, Lycée Henri Poincaré, Nancy
Olivier Colin, UVSQ, Versailles
Étienne Mangaud, Univ. Paul Sabatier, Toulouse
Bénédicte Colnet, Mines ParisTech, Paris
Jean-Daniel Marty, Univ. Paul Sabatier, Toulouse
Antton Curutchet, ENS Lyon
Olivier Maury, CNRS, Lyon
Élise Duboué-Dijon, CNRS, Paris
Bastien Mettra, IUT-Lyon1, Villeurbanne
Alain Fruchier, ENSCM, Montpellier
Aurélien Moncomble, Université de Lille
Ludivine Garcia, Lycée Jean Moulin, Béziers
Pierre-Adrien Payard, ENS, Paris
Catherine Gautier, Lycée Algoud Laffemas, Valence Daniel Pla, CNRS, Toulouse
Didier Gigmes, CNRS, Marseille
Romain Ramozzi, Lycée Henri Poincaré, Nancy
Emmanuel Gras, CNRS, Toulouse
Clémence Richaud, Lycée Jules Verne, Limours
Laetitia Guerret, ENS Paris-Saclay, Cachan
Vincent Robert, Université de Strasbourg
Clément Guibert, Sorbonne Université, Paris
Jean-Marie Swiecicki, MIT, Cambridge (USA)Dayana Gulevich, Moscow State University
Guillaume Vives, Sorbonne Université, Paris
Baptiste Haddou, Lycée D. Milhaud, Kremlin-Bicêtre Hanna Zhdanova, Université de Strasbourg

Additional reviewers
Lucile Anthore-Dalion
Quentin Arnoux
Marie AuvraySimon Beaumont
Nicolas Bolik-Coulon
Jérémy Camponovo
Natan Capobianco
Antoine Carof
Guillaume Carret
Fabrice Dalier
Guillaume Didier
Philippe Drabent
Laurence Dupont

51st IChO – Preparatory problems

Olivier Durupthy
Matthieu Emond
Johanna Foret
Mickaël Four
Emma Gendre
Isabelle Girard
Antoine Hoste
Damien Lavergne
Quentin Lefebvre
Christina Legendre
Anne Leleu
Marion Livecchi

Guillaume Mériguet
Mathilde Niocel
Artem Osypenko
Valentin Quint
Emelyne Renaglia
Tristan Ribeyre-Stecki
Clément Robert
Nell Saunders
Laura Scalfi
Freddy Szymczak
Julien Valentin
Dominique Vichard
Vincent Wieczny4

Fields of advanced difficulty
Theoretical
1. Thermodynamics: relation between equilibrium constants and standard reaction Gibbs freeenergy, relation between thermodynamic and electrochemical data.
2. Kinetics: orders of reaction, half-life, rates defined as time derivatives of concentrations,use of integrated rate laws, classic approximations.
3. Basic quantum chemistry: notion of wavefunction, expression of simple molecular orbitals,
electronic energy levels, crystal field theory.
4. Spectroscopy: simple IR spectroscopy (identification of chemical groups only), 1H NMR
spectroscopy (chemical shifts, integrals, couplings and multiplicity).
5. Polymers: block copolymers, polymerization, polydispersity, simple size exclusion
chromatography (SEC).
6. Stereochemistry: stereoisomers in organic and inorganic chemistry, stereoselectivity inorganic synthesis.
Practical
1. Techniques in organic synthesis (drying of a precipitate, recrystallization, TLC).
2. Use of a spectrophotometer (mono-wavelength measurements).

Important notes
Theoretical: the following advanced skills or knowledge WILL NOT appear in the exam set:
 Solid state structures;
 Specific notions about catalysis;
 Specific notions about enzymes;
 Specific carbohydrates chemistry (reactivity at the anomeric position, nomenclature,
representation); Stereochemical aspects associated with the Diels-Alder reaction (supra-supra and endo
approaches); Hückel theory; Calculus (differentiation and integration).
Practical: the following techniques WILL NOT be required during the competition:
 Use of a separatory funnel and extraction using immiscible solvents;
 Use of a rotary evaporator;
 Sublimation;
 Use of a melting point apparatus;
 Use of a pH-meter.

51st IChO – Preparatory problems

5

Physical constants and equations
In this booklet, we assume the activities of all aqueous species to be well approximated by theirrespective concentration in mol L−1. To further simplify formulae and expressions, the standard
concentration c° = 1 mol L−1 is omitted.
Avogadro's constant:
Universal gas constant:Standard pressure:
Atmospheric pressure:
Zero of the Celsius scale:
Faraday constant:
Kilowatt hour:

NA = 6.022∙1023 mol−1
R = 8.314 J mol−1 K−1
p° = 1 bar = 105 Pa
Patm = 1 atm = 1.013 bar = 1.013∙105 Pa
273.15 K
F = 9.6485∙104 C mol−1
1 kWh = 3.6∙106 J

Ideal gas equation:
Gibbs free energy:pV = nRT
G = H – TS
ΔrG° = –RT lnK°ΔrG° = –n F Ecell°
ΔrG = ΔrG° + RT lnQ

Reaction quotient Q for a reactiona A(aq) + b B(aq) = c C(aq) + d D(aq):[C]c [D]dQ=[A]a [B]b[A– ]pH = pKa + log
[AH]
RT
o
E = E – ln𝑄

Henderson–Hasselbalch equation:
Nernst–Peterson equation:
where Q is the reaction quotient of the
reduction half-reaction

Beer–Lambert law:
Clausius-Clapeyron relation:
Arrhenius equation:
Rate laws in integrated form:
Zero order:First order:
Second order:
Half-life for a first order process:Number average molar mass Mn:
Mass average molar mass Mw:
Polydispersity index Ip:

zF

RTat T = 298 K, F ln10 ≈ 0.059 V
A = εlc∆vap 𝐻° 1
P2
1
ln
= −
(
−
)
P1
R
T2
T1
−Ea

k = A e RT

[A] = [A]0 – ktln[A] = ln[A]0 – kt
1
1
=
+ kt
[A] [A]0
ln2
t1/2 =
k∑i 𝑁i 𝑀i
∑i 𝑁i
∑i 𝑁i 𝑀i2𝑀w =∑i 𝑁i 𝑀i
Mw
Ip =
Mn
𝑀n =

The above constants and formulas will be given to the students for the theoretical exam.

51st IChO – Preparatory problems

6Periodic table

1

18

1

H

2

2

13

14

15

16

17

1.008

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

21

25

26

30

22.99

24.31

19

20

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 89103

Rf Db Sg Bh Hs Mt Ds Rg Cn Nh

Fl

Mc Lv

-

-

Ts Og

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

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

51st IChO – Preparatory problems

-

-

-

-

-

-

-

-

-

-

-

7

1H NMR

Chemical shifts of hydrogen (in ppm /TMS)
phenols
alcohols
alkenesaromatics

alkynescarboxylic acids

CH3—NR2

aldehydes

CH3—SiR3

CH3—OR

ketones
CH3—CR3

11.0

10.0

9.0

8.0

7.0

6.0

5.0

4.0

3.0

2.0

1.0

0.0

H-H coupling constants (in Hz)
Hydrogen type|Jab| (Hz)R2CHaHb

4-20

R2HaC—CR2Hb
R2HaC—CR2—CR2Hb
RHaC=CRHb

2-12
if free rotation: 6-8
ax-ax (cyclohexane): 8-12ax-eq or eq-eq (cyclohexane): 2-5
if free rotation: < 0.1
otherwise (rigid): 1-8
cis: 7-12trans: 12-18

R2C=CHaHb

0.5-3

Ha(CO)—CR2Hb

1-3

RHaC=CR—CR2Hb

0.5-2.5

IR spectroscopy table
Vibrational modeσ (cm−1)

Intensity

alcohol O—H (stretching)carboxylic acid O—H (stretching)N—H (stretching)3600-3200
3600-2500
3500-3350

strongstrongstrong≡C—H (stretching)=C—H (stretching)
C—H (stretching)–(CO)—H (stretching)

3300
3100-3000
2950-2840
2900-2800

strong
weak
weak
weak

C≡N (stretching)2250

strong51st IChO – Preparatory problems

8

C≡C (stretching)

2260-2100

variablealdehyde C=O (stretching)anhydride C=O (stretching)ester C=O (stretching)ketone C=O (stretching)amide C=O (stretching)

1740-1720
1840-1800; 1780-1740
1750-1720
1745-1715
1700-1500

strongweak; strong
strongstrongstrongalkene C=C (stretching)aromatic C=C (stretching)1680-1600
1600-1400

weak
weak

CH2 (bending)
CH3 (bending)1480-1440
1465-1440; 1390-1365

mediummediumC—O—C (stretching)
C—OH (stretching)NO2 (stretching)1250-1050 (several)
1200-1020
1600-1500; 1400-1300

strongstrongstrongVisible light
750 nmred

400 nmpurple620 nmorange480 nmblue

590 nmyellow

530 nmgreen

51st IChO – Preparatory problems

9

Theoretical problems

51st IChO – Preparatory problems

10

Problem 1. Butadiene π-electron system
Buta-1,3-diene (simply called butadiene thereafter) is a diene of chemical formula C4H6, which wasisolated for the first time in 1863 by the French chemist E. Caventou and identified in 1886 by the
English chemist H. E. Armstrong. It is a key reagent in the production of synthetic rubber. Over12.7 million tons of butadiene are produced every year. We will study here the properties of its
π-electron system. We will then compare them to those of the hypothetical cyclobutadiene, which has
never been isolated in its free form.
cyclobutadiene
butadiene

1. Give the number of π-electrons of butadiene.
The Molecular Orbitals (MO) Ψi of the π-electron system can be written as a weighted sum(linear combination) of the 2pz atomic orbitals of each carbon atom, φj:4

𝛹i = ∑ 𝑐ij 𝜑j ; 𝑖 = 1 − 4
𝑗=1

We provide below an approximate expression for the MOs together with their associatedenergy. The energy of each MO is expressed as a function of two parameters, α and β, both
negative real numbers. α represents the energy of an electron in an isolated 2pz orbital, and β is
the interaction energy between two neighboring 2pz orbitals.
𝛹1 = 0.3717 𝜑1 + 0.6015 𝜑2 + 0.6015 𝜑3 + 0.3717 𝜑4 ; 𝐸1 = 𝛼 + 1.62 𝛽
𝛹2 = 0.6015 𝜑1 + 0.3717 𝜑2 − 0.3717 𝜑3 − 0.6015 𝜑4 ; 𝐸2 = 𝛼 + 0.62 𝛽
𝛹3 = 0.6015 𝜑1 − 0.3717 𝜑2 − 0.3717 𝜑3 + 0.6015 𝜑4 ; 𝐸3 = 𝛼 − 0.62 𝛽
𝛹4 = 0.3717 𝜑1 − 0.6015 𝜑2 + 0.6015 𝜑3 − 0.3717 𝜑4 ; 𝐸4 = 𝛼 − 1.62 𝛽
2. Draw and fill in the MO diagram of butadiene. Draw schematically each MO and identify
its nature (bonding or anti-bonding).
We consider the formation of the butadiene π-electron system, starting from four carbon atoms,
each bringing an electron in a 2pz orbital of energy α.
3. Calculate the formation energy ΔEf associated with this transformation.
Here, the conjugation energy is defined as the difference between the total π-energy of thestudied compound and that of two non-interacting ethylene molecules. The π-energy of ethylene
is equal to 2(α + β).
4. Calculate the conjugation energy ΔEc of butadiene. Give its sign. Which system is the moststable? Choose the correct answer.
 Butadiene 2 ethylene molecules Both are equally stable
The net charge qj on each carbon atom (i.e., the charge gained or lost by the atom compared toits neutral state) can be calculated in the present case as:

51st IChO – Preparatory problems

11

𝑜𝑐𝑐

𝑞j = 1 − ∑ 𝑛i 𝑐ij 2
𝑖=1

where the sum runs over the occupied MOs, ni is the number of electrons in the ith MO, and cij
is the coefficient of the jth carbon atom in the ith MO.
5. Calculate the net charges q1 and q2 of the butadiene carbon atoms 1 and 2. Deduce thevalues of q3 and q4.
The bond order I is an estimate of the number of π chemical bonds between two atoms. For
instance, a pure single bond would have a bond order I = 0, a pure double bond would
correspond to I = 1, etc. The bond order Irs between two neighboring atoms r and s can beobtained from the MOs as:
𝑜𝑐𝑐

𝐼rs = ∑ 𝑛i 𝑐ir 𝑐is
𝑖=1

where Irs is defined as the sum over the occupied MOs of the product of the number of electrons
in the MO by the coefficient of each of the two atoms r and s in this MO.
6. Calculate for each bond the associated bond order: I12, I23, and I34. Identify the bond(s)that has (have) the strongest double-bond character.
7. Draw alternative Lewis structures of butadiene to reflect the previously obtained results(charges and bond orders).
The MO diagram of the hypothetical cyclobutadiene is provided below. The size of each atomic
orbital is proportional to its coefficient in the considered MO, and its color (grey or white)reflects the sign of the wavefunction.

8. Fill in the MO diagram of cyclobutadiene.
9. Using the provided diagram and considering the symmetry of the molecule, determine themissing coefficients (cij) in the following MO expressions.
𝛹1 = 0.500 𝜑1 + 𝑐12 𝜑2 + 𝑐13 𝜑3 + 𝑐14 𝜑4
𝛹2 = 0.707 𝜑1 + 𝑐22 𝜑2 + 𝑐23 𝜑3 + 𝑐24 𝜑4
𝛹3 = 𝑐31 𝜑1 + 0.707 𝜑2 + 𝑐33 𝜑3 + 𝑐34 𝜑4𝛹4 = 0.500 𝜑1 + 𝑐42 𝜑2 + 𝑐43 𝜑3 + 𝑐44 𝜑4

51st IChO – Preparatory problems

1210. Calculate the formation and conjugation energies, ΔEf’ and ΔEc’, for cyclobutadiene.
Which system is the most stable? Choose the correct answer.
 Cyclobutadiene Two ethylene molecules Both are equally stable
11. Compare the formation energy of cyclobutadiene and that of butadiene. Which compoundis the most stable? Choose the correct answer.
 Butadiene Cyclobutadiene Both are equally stable
We now consider a rectangular deformation of cyclobutadiene, with localization and shorteningof the double bonds and elongation of the simple bonds compared to the square geometry.

12. Choose the correct statement(s) among the following: This deformation stabilizes C=C double bonds.
 This deformation weakens C=C double bonds.
 This deformation does not affect the stability of C=C double bonds.
 This deformation increases the stability due to electronic conjugation.
 This deformation diminishes the stability due to electronic conjugation.
 This deformation does not affect the stability due to electronic
conjugation.
13. Using your previous answers, choose the correct statement among the following. The
π-system after deformation is: More stable than the square cyclobutadiene.
 Less stable than the square cyclobutadiene.
 As stable as the square cyclobutadiene.

Problem 2. Localization and delocalization in benzene
Historically, benzene was first isolated from benjoin (essence of the “Papier d’Arménie”). It wasthen synthesized by the French chemist M. Berthelot in the middle of the 19th century using acetylene
trimerization. In this problem, the objective is to study the electronic properties of this compound, which
is a representative of aromatic molecules. Let us start with benzene by referring to the carbon atoms as
Ci, i = 1 – 6 in a clock-wise manner.

1. Write the reaction from acetylene C2H2 generating benzene.
2. Draw a structure of benzene using three single bonds and three double bonds between
carbon atoms. It is referred to as Kekulé’s benzene.
3. Draw a structure of benzene holding five single and two double bonds. This structure iscalled Dewar’s benzene.

51st IChO – Preparatory problems

13

Let us start with a Kekulé structure K1, holding a double bond between C1 and C2 atoms. A
simple model to describe the  bond between C1 and C2 consists of characterizing thedelocalization of a single electron by an energy t < 0.
4.

Give the energy Eπ of the -system of this bond as a function of t.

5. In K1, double bonds are supposed to be fixed. For this structure K1, calculate the energyof the π-system EK1 as a function of t.
6. Write an analog to K1. It will be called K2.
7. Express the energy EK2 of this structure K2.
Mathematically, the benzene molecule is expressed as a mix between K1 and K2,
K = c1 K1 + c2 K2, where c1 and c2 are real numbers with c12 + c22 = 1 and c1 > 0 and c2 > 0.
This expression stresses that a proper description of benzene cannot be restricted to K1 or K2.
8. On a scheme, show the displacement of the double bond localized between C1 and C2 andthe movement of the other double bonds. These formulae are the resonance structures ofbenzene.
Starting from a localized view K1 or K2, the electronic delocalization over all the carbon atoms
can be accounted for by the introduction of a supplementary energetic term. The energy EK of
K is thus defined as:
EK = c12 EK1 + c22 EK2 + 2 c1 c2 H12
where H12 varies between t and 0, with t < 0. Therefore, EK is a function of c1 and c2.
9.

Express EK as a function of c1 only.

It can be shown that EK is minimal for c1 = 1 / √2. From now on, we assume that c1 = 1 / √2.
10. If H12 = 0, what is the expression of EK? The resonance energy is defined as the differenceE1 = EK(H12 = t) – EK(H12 = 0). Evaluate E1 as a function of t.
11. Specify the sign of E1. Choose the correct statement between: electronic delocalization contributes to stabilize the benzene molecule.
 electronic delocalization contributes to destabilize the benzene molecule.
Alternatively, the  energy of a n carbon atom-system can be evaluated from the occupationsof the molecular orbitals (MOs). C. A. Coulson (C. A. Coulson, Proc. Roy Soc., 1939) showedthat the MOs energies k of a cyclic n carbon atom-system, not necessarily in energy order,
read:
2𝑘π
𝜀k = 2𝑡 cos; 𝑘 ∈ ℕ, 𝑘 ∈ [0 ; 𝑛 – 1]𝑛
12. Draw the MOs diagram of the -system of benzene (n = 6) and calculate the corresponding
energies for each MO.
13. Fill the MOs diagram.
14. Evaluate the -system energy of benzene, EMO, from the filling of the MOs in ascendingorder. Then, calculate the resonance energy E2 = EMO –EK(H12 = 0).

51st IChO – Preparatory problems

1415. Compare E2 and E1.
16. From the previous results, choose one expression for the relation between the standardhydrogenation enthalpy of cyclohexene (ΔrHc°) and that of benzene (ΔrHb°).


ΔrHb° < 3 ΔrHc°
ΔrHb° > 3 ΔrHc°
ΔrHb° = 3 ΔrHc°

Problem 3. Study of liquid benzene hydrogenation
Determination of the standard enthalpy of formation of liquid benzene
1. Write down the balanced chemical equation for the formation of liquid benzene from its
constituent elements in their standard states.
2. Calculate the standard enthalpy of formation of liquid benzene ∆fH°(C6H6(l)) using
standard bond enthalpies, standard enthalpies of dissociation, and the standard enthalpies
of vaporization of benzene and sublimation of graphite.
3. Calculate the standard enthalpy of formation of liquid benzene ∆fH°(C6H6(l)) using Hesslaw and the standard combustion enthalpies.
4. Calculate the difference between the ∆fH°(C6H6(l)) values obtained in the two previous
questions. Choose the correct explanation for this difference.
 The difference is due to experimental errors on the values of standard
enthalpies of combustion reactions.
 The method used at question 2 does not take into account the nature ofbonds in benzene.
 The Hess law is only rigorously applicable with standard enthalpies offormation.
 The method used in question 3 does not take into account the electronic
delocalization.
Successive hydrogenation reactions of liquid benzene study
5. Calculate the enthalpy of reaction for the full hydrogenation of liquid benzene into liquid
cyclohexane.
The different steps of benzene hydrogenation into cyclohexane are given in the scheme 1.

H2

H2

H2

-112.1 kJ mol-1-119.7 kJ mol-1

Scheme 1: benzene hydrogenation

51st IChO – Preparatory problems

15

6. Complete this scheme by calculating the standard enthalpy of hydrogenation of benzene
into cyclohexa-1,3-diene.
The sign of the standard enthalpy of this reaction differs from the sign of the other standard
enthalpies of hydrogenation in scheme 1.
7. What is the main reason for such a difference?
 All the double bounds are not equivalent in benzene: one is stronger thanthe others.
 The breaking of benzene aromaticity.
 The formation of a reaction intermediate (cyclohexa-1,3-diene) with aconstrained geometry.
8. Using only the values given in scheme 1, calculate the resonance energy of cyclohexa-1,3diene and the resonance energy of benzene.
Data:
Standard combustion enthalpies ∆combH° at 298 K in kJ mol‒1CompoundC(graphite)
H2(g)∆combH°‒393.5‒285.6

C6H6(l)‒3268

Standard enthalpy of formation of cyclohexane at 298 K∆fH°(C6H12(l)) = ‒156.4 kJ mol‒1Standard bond enthalpies ∆dH° at 298 K in kJ mol‒1
Bond
C—H
C—C
∆dH°
414.8
346.9

C=C
614.5

Standard enthalpies of dissociation D° at 298 K in kJ mol‒1
Bond
O=O
H—H
D°
498.3
436.0
Standard latent heat at 298 K in kJ mol‒1∆subH°(C(graphite)) = 716.70 kJ mol‒1∆vapH°(C6H6) = 33.90 kJ mol‒1

Problem 4. Use of dihydrogen: fuel cells
In order to generate electricity, the heat produced by fuel combustion (dihydrogen, methanol, etc.)
can be used to evaporate liquid water. The produced steam turns a turbine, which drives a generator. In
such a process, the chemical energy is first converted into thermal energy, then into mechanical energy
and finally into electrical energy. Leaks occur at each conversion step (mainly by heat dissipation),
which decreases the yield of the global process. On the contrary, fuel cells directly convert the chemicalenergy into electrical energy.

The balanced chemical equation for the combustion of one equivalent of fuel A is:
A + x O2(g) = y CO2(g) + z H2O(l)(1)

51st IChO – Preparatory problems

16

∆combH°(A) and ∆combG°(A) are respectively the standard enthalpy of reaction and the standard
Gibbs free energy of reaction associated with reaction (1).
The hydrogen fuel cell
The global reaction in the hydrogen fuel cell is the same as that of H2 combustion. In thisproblem, compounds of the hydrogen fuel cell will be considered in their standard state at
298 K.
1. Write down the redox half-reactions occurring at the anode and the cathode. Write down
the balanced chemical equation for the global reaction, for one equivalent of dihydrogen.
2. Compute the open circuit voltage of such a cell.
3. Compute the theoretical maximum electrical energy recoverable by mole of dihydrogenconsumed.
4. Electric cars consume between 10 and 20 kWh / 100 km. Compute the volume of
dihydrogen necessary to produce an electrical energy of 20 kWh at 1.0 bar.
The thermodynamic eﬃciency of a cell is defined as:∆r 𝐺°
𝛾thermo =∆r 𝐻°
where ∆rG° and ∆rH° are respectively the standard Gibbs free energy of reaction and thestandard enthalpy of reaction associated with the global reaction of the running cell.
5. Calculate the standard enthalpy of the combustion reaction of gaseous dihydrogen∆combH°298K(H2(g)) at 298 K. Deduce the thermodynamic eﬃciency of the dihydrogen fuelcell.
The thermodynamic eﬃciency is smaller than 1 because there is a variation of the entropy ofthe system.
6. Calculate the standard entropy of the dihydrogen combustion reaction ∆combS°298K(H2(g))
at 298 K.
7. Determine if the sign of this standard entropy is consistent with the balanced chemical
equation for the reaction (Yes/No). Justify it by a short calculation using the stoichiometriccoefficients.
The liquid methanol cell
The low energy density of dihydrogen and the necessity of a large pressure for its storage havemotivated the development of batteries using other fuels. In a cell using liquid methanol as afuel, the global reaction is that of the combustion of liquid methanol.
8. Determine the oxidation state of the carbon atom in methanol and in carbon dioxide.
9. Write down the redox half-reactions occurring at the anode and the cathode. Write down
the balanced chemical equation for the global reaction of the running cell for one equivalentof liquid methanol.

51st IChO – Preparatory problems

17

10. Calculate the associated thermodynamic efficiency. Compare it to the efficiency of thedihydrogen fuel cell.
11. Calculate the volume of liquid methanol required to produce 20 kWh. Compare this value
to the previously calculated volume of gaseous dihydrogen.
12. Assuming H2 is an ideal gas, determine the pressure to store the dihydrogen necessary to
produce 20 kWh in the same volume as methanol (question 11).
Data:
Standard enthalpies of formation ∆fH° at 298 K in kJ mol‒1CompoundO2(g)CO2(g)
H2O(g)∆fH°
0.0‒394.0‒241.8

CH3OH(l)‒239.0

Molar heat capacities at constant pressure C°P in J mol‒1 K‒1. They are supposed to be
independent of the temperature.
Compound
H2O(g)
H2O(l)
C°P
33.6
75.3
Standard latent heat of water at 373 K∆vapH°(H2O) = 40.66 kJ mol‒1Standard potentials at 25 °C related to the standard hydrogen electrode (SHE)
E°(O2(g)/H2O(l)) = 1.23 V /SHE
E°(CO2(g)/CH3OH(l)) = 0.03 V /SHE
Liquid methanol densityρmethanol = 0.79 g cm‒3

Problem 5. Hydrogen storage
Dihydrogen is a promising fuel for the future, notably for power production or mobility purposes.
It is an attractive alternative to the use of fossil fuels (hydrocarbons), which release carbon dioxideduring their combustion, thus contributing to global warming. Unfortunately, storing efficiently large
amounts of H2 is not easy. Dihydrogen has a low energy per unit volume at room temperature, is highly
flammable and requires several technological advances to be competitive with fossil fuels. In thisproblem, we investigate the advantages and disadvantages of some hydrogen storing methods.

Storing H2 as a gas
Compressing dihydrogen is one of the methods commonly used to store it. The gas is stored incontainers at a pressure kept between 350 and 700 bars.
1. Calculate the density of an ideal dihydrogen gas at a pressure of 500 bar and at room
temperature (293 K).

51st IChO – Preparatory problems

18

Storing H2 as a liquid
Dihydrogen gas is liquefied and kept in a Dewar flask (a thermally insulated container) usually
under a relatively low pressure (1 to 4 bar). However, the system needs to be kept at very low
temperatures, because the melting point of H2 at a pressure P = 1 atm is Tm = ‒259.2 °C and its
boiling point under the same pressure is Tv = ‒252.78 °C. Its critical point is located at:
Pc = 13.0 bar, Tc = −240.01 °C.
2. At which temperatures can liquid hydrogen be observed?
 16 K 25 K 77 K 293 K
3. Using the Clausius-Clapeyron relation, calculate the pressure needed to liquefy ideal
gaseous dihydrogen at 27.15 K.
Storing dihydrogen as a complex
In 1984, using measurements obtained from neutron diffraction, G. J. Kubas and his
collaborators (G. J. Kubas et al., J. Am. Chem. Soc., 1984) identified a tungsten complex[W(CO)3(P(iPr)3)2(η2-H2)] that possesses a H—H bond with a length of 0.82 Å, close to that of
an isolated H2 molecule (0.74 Å). ((iPr) = iso-propyl). This complex easily dissociates underpartial vacuum or under argon atmosphere, and it can be regenerated in the presence of
dihydrogen.

4. Calculate the mass of the dehydrogenated complex needed to store 1 kg of dihydrogen.
Calculate ρH (the density of hydrogen in the complex, defined as the mass of hydrogenatoms per volume unit of complex).
The next section will study the binding of a H2 molecule to the dehydrogenated complex within
the field of other ligands. The dehydrogenated complex is assumed to be a square-based
pyramid, which the dihydrogen molecule is added to.
Metallic central atom
5. Give the electronic configuration of atomic tungsten. Specify the number of valence
electrons.
6. Fill in the table with the name of each depicted atomic orbital (s, dyz, dz2, d(x2 – y2), dxz, dxy).

51st IChO – Preparatory problems

19

Dihydrogen as a ligand
7. Draw and fill the molecular orbital diagram of dihydrogen.
Kubas complex
Since the complex is considered as a square-based pyramid to which the H2 molecule is added,
we have to take into account the influence of other ligands. The splitting thus obtained is givenin the diagram below.

Figure 1: Simplified diagram of molecular orbitals of the Kubas complex

In order to build the molecular orbital diagram of the Kubas complex, we can study theinteraction of the molecular orbitals of the complex ([W(CO)3(P(iPr)3)2]) —which will be
merely considered as the d orbitals of the metallic central atom— with the H2 molecule orbitals.

Figure 2: Kubas complex and reference axes

8. Give the two planes of symmetry of the Kubas complex (using the axes of figure 2).
9. Indicate for each orbital d of the metallic central atom if they are symmetric orantisymmetric with respect to each of the symmetry planes (using the axes of figure 2).

51st IChO – Preparatory problems

20

Two conformations have been proposed: (1) where H2 is parallel to the phosphine ligands
P(iPr)3, and (2) where H2 is parallel to the CO ligands. Even if steric effects favor conformation(2), conformation (1) is actually more stable.

(1)

(2)

10. Fill in the diagram in figure 1 with electrons.
11. Knowing that only orbitals with the same symmetry interact, enumerate the possible
interactions for each conformation. Which conformation is the most stable one?
Storing hydrogen in form of formic acid
In 2006, a research team of EPFL (Switzerland) (C. Fellay et al., Angew. Chem. Int. Ed., 2008)
proposed to store H2 in form of formic acid. The main idea is to use formic acid as fuel that canbe decomposed on a catalyst made of ruthenium to produce dihydrogen and carbon dioxide
according to the following reaction:
HCOOH(l) → CO2(g) + H2(g)(R1)
12. Calculate ρH (the density of hydrogen at 25 °C defined as the mass of hydrogen atoms pervolume unit of formic acid). Compare this value to those obtained for gaseous dihydrogenat 500 bar and for liquid dihydrogen.
13. Calculate the standard enthalpy and entropy of reaction at 20 °C for reaction (R1).
14. Using the Ellingham approximation (that supposes enthalpy and entropy independent of
temperature), calculate the equilibrium constant at 20 °C for reaction (R1).
Formic acid (2.3 g) is added to a 1 L container with 0.1 g of ruthenium catalyst, under constantatmospheric pressure and at an initial temperature of 25 °C. The container initially contains
dinitrogen.
15. Determine the final composition of the mixture.
Storing hydrogen in metal hydrides
Metal hydrides have also been proposed to store dihydrogen. Compounds with a XxYyHn
formula can store large amounts of hydrogen in a compact way. Moreover, the adsorptiondesorption properties of hydrogen can be tailored by choosing an element X from light elements(Li, Mg, B,…) or other electropositive elements (lanthanides) that have a good affinity withhydride ligands, and an element Y from transition metals that have a low affinity with hydride
ligands. Among the numerous existing metal hydrides, two of them will be studied in theiroperating conditions: LaNi5H6 (300 K, 2 bar) and Mg2NiH4 (550 K, 4 bar).
16. Determine ρH (the density of hydrogen, which is defined as the mass of hydrogen atoms
per volume for these two compounds in their operating conditions).

51st IChO – Preparatory problems

21

The adsorption-desorption equilibrium can be described as a phase change A(g) → A(ads).
Hence, dihydrogen is considered as an ideal gas and the Clausius-Clapeyron relation for a phasetransformation from an ideal gas is a rather good approximation. The latent heat can be
assimilated, in this case, to the adsorption enthalpy. In the following tables, the pressure (MPa)is given as a function of the temperature (K).
LaNi5H6
P (MPa)
T (K)

2.15
370

0.68
333

0.10
285

0.07
278

P (MPa)
T (K)

1.94
667

0.71
625

0.26
588

0.10
555

Mg2NiH4

Table 1: Van’t Hoff plot data (pressure (MPa) as a function of the temperature (K)) of several metal
hydrides (A. Züttel, Naturwissenschaften, 2004)

17. Using table 1, determine the adsorption enthalpies of LaNi5H6 and Mg2NiH4.
Data:
𝑛²𝑎

Van der Waals gas equation: (𝑝 + 𝑉² ) (𝑉 − 𝑛𝑏) = 𝑛R𝑇
Van der Waals coefficients for dihydrogen:a = 0.2476 L2 bar mol‒2b = 0.02661 L mol‒1
Specific latent heat of fusion (at standard pressure):ΔfusH°m = 58.089 kJ kg‒1
Specific latent heat of vaporization (at standard pressure): ΔvapH°m = 448.69 kJ kg‒1
Densities
Gaseous dihydrogen, standard conditions: 0.08988 g L‒1
Liquid dihydrogen, ‒252.78 °C: 70.849 g L‒1
compoundKubas cplx
formic acid
LaNi5H6
conditions-25 °C300 K
ρ
1.94 g cm‒3
1.22 kg L‒18620 kg m‒3

Mg2NiH4
550 K
2643 kg m‒3

Thermodynamic data at normal conditions of temperature and pressure (20 °C, 1 atm)
compoundHCOOH(g) HCOOH(l) CO2(g)
H2(g)
N2(g)
‒1
ΔfH° kJ mol‒378.60‒425.09‒393.510.00
0.00
Sm° J mol‒1 K‒1248.70131.84213.79130.68191.61

Problem 6. Deacidification and desulfurization of naturalgas
95% of dihydrogen is produced by steam reforming from natural gas. The corresponding reaction
is analoguous to the reaction with methane (reaction (1)), which is carried out at about 900 °C inpresence of a catalyst.
cat.

CH4 + H2 O →51st IChO – Preparatory problems

CO + 3 H2(1)22

35% to 40% of the dihydrogen thus obtained is used in ammonia synthesis. However, one sulfur atom
per 1000 nickel atoms is sufficient to poison the nickel-based catalyst. Since acidic gases (H2S and CO2)
contained in natural gas can also damage the pipelines, natural gas must be deacidified and desulfurized.

Steam reforming from natural gas
1. Give the chemical reaction of steam reforming for an alkane CnH2n+2.
2. Calculate the equilibrium constant K° of reaction (1) at 900 °C.
Removal of acidic gases
A common method to remove acidic gases from natural gas is to use an amine solution. Someamine solutions can solubilize all the acidic gases, whereas others are selective due to kinetic
differences between H2S and CO2. This process is modeled below, replacing the hydrocarbons
by N2. The following experiments aim to study deacidification with two different amines:
monoethanolamine (MEA) and methyl-diethanolamine (MDEA), using the apparatus depicted
below.

Flask F1 initially contains 100 mL of a 0.5 mol L‒1 amine solution (n0 = 50 mmol: large
excess).
Flask F2 initially contains 100 mL of a 0.5 mol L‒1 NaOH solution (large excess too).
Step. 1: a gas sample (gas 1) is driven by N2 into a flask containing an amine solution; theoutgoing gas (gas 2) bubbles in a second flask containing a NaOH solution; the final gas (gas3) no longer contains acidic gas.
Step. 2: the liquid contents of each flask are titrated by an HCl solution (cHCl = 1.0 mol L‒1).
Both pH and conductivity are recorded along the titration, so that two curves are obtained foreach experiment (see below).
The sample of gas 1 contains n1 mmol of CO2, n2 mmol of H2S and n3 mmol of CH3SH. The
first experiment is carried out with the primary amine MEA; the second one with the tertiaryamine MDEA.
3. Write down the thermodynamic quantitative (K° >> 1) reactions between the different
gases and (i) the amine solution and (ii) the NaOH solution.
We first study the experiment with MEA. There is no kinetic blockage with this amine.

51st IChO – Preparatory problems

23

4. Determine the amount of each species in the solution (as a function of n1, n2 and n3) in theflask F1 before titration.
5. Which chemical species is/are present in gas 2?
6. Using the curves A1F1 and A1F2, determine (i) n3 and (ii) a relation between n1 and n2.

MDEA reacts with only one of the acid species, the other reaction being kinetically blocked.
7. Determine the amount of the reacting species using curve A2F1.
8. Using curve A2F2, determine if MDEA selectively reacts with CO2 or with H2S. Calculate
the two remaining n1 and n2.

Data at 298 K:‒1

ΔfH° (kJ mol )Sm°(J K‒1 mol‒1)

CO2(g)‒393.5213.8

H2O(g)‒241.8188.8

CH4(g)
‒74.6
186.3

pKa
Amines: MEAH+/MEA; MDEAH+/MDEA
CO2(aq)
H2S
CH3SH

51st IChO – Preparatory problems

C5H12(l)‒178.4260.4CO(g)‒110.5
197.7

CH3CH3(g)
‒84.0
229.2

H2(g)
0.0
130.7

pKa = 9.5
pKa1 = 6.4; pKa2 = 10.3
pKa1 = 7.0; pKa2 = 13.0
pKa = 10.3

24

Problem 7. Lavoisier’s experiment
In 1775, the French chemist A L de Lavoisier, father of modern chemistry, showed by anexperiment that oxygen is one of the constituents of air.

Lavoisier’s experiment (Bussard and Dubois, Leçons élémentaires de chimie, 1897)

The experiment he performed can be summarized as follows: he first introduced 122 g of mercury into a retort, the end of which was inside a cloche(see illustration above) containing 0.80 L of air and placed upside down on a tankcontaining mercury, he then heated the retort in such a way as to keep the mercury boiling for several days,
 after two days, the surface of the mercury began to get covered with red particles, after twelve days, the calcination of mercury seemed to have finished because thethickness of the particle layer was no longer increasing, he then stopped heating, after cooling, he observed the following: only 0.66 L of “air” subsisted under the cloche, this remaining “air” could extinguish a candle or kill a mouse, 2.3 g of red particles had been formed. He called them “rust of mercury”.
The table below shows thermodynamic data at 298 K of some mercury-based compounds anddioxygen.
Compound
ΔfH° (kJ mol‒1)
Sm° (J K‒1 mol‒1)
HgO(s) (red)

‒90

70

HgO(s) (yellow)

‒87

70

Hg2O(s)

‒90

Hg(l)
Hg(g)
O2(g)

51st IChO – Preparatory problems

75
60

175
200

25

1. Standard molar entropy Sm° of mercurous oxide Hg2O has not been experimentally
determined. Choose the value that seems closest to reality:
 0 J K‒1 mol‒1
 100 J K‒1 mol‒1
 200 J K‒1 mol‒1
 300 J K‒1 mol‒1
2. Write down equations for the formation of HgO(s) and Hg2O(s).
3. It is assumed that only the liquid state of mercury reacts, and that either red or yellow HgO
can be formed. Using the value chosen in 1, calculate the equilibrium constants K° at 298 K
for a) red HgO, b) yellow HgO and c) Hg2O.
The red and yellow forms of the mercury (II) oxide have, in particular, very similar standard
potentials and quasi-equal magnetic susceptibilities. However, the yellow form has larger
structural defects than the red form. The red mercury oxide can be obtained by a slow pyrolysis
of Hg(NO3)2, while the yellow oxide can be obtained by precipitation of aqueous mercury (II)ions in an alkaline medium.
4. Write the chemical equations of these processes.
Lavoisier’s experiment is similar to pyrolysis because of the use of heating and the absence ofan aqueous medium, which may explain the formation of red oxide. In the following, we willconsider this one as the only product of the reaction.
5. Calculate the theoretical amount of each species in the final state of Lavoisier’s reaction.
6. Calculate the theoretical mass of mercury (II) oxide in the final state.
7. Choose an explanation to the difference from the mass obtained by Lavoisier.
 Other oxides of the type HgOx (x < 1) are obtained.
 The yield is not maximum.
 Lavoisier measured volumes at T < 25 °C.
 Mercury rust also contains nitride HgxNy.

Problem 8. Which wine is it? Blind tasting challenge
Fermentation of grape juice is a crucial step in the production of wine. During this biochemical
process, sugars accumulated in grapes are converted into ethanol. This process is performed by
microorganisms that are naturally present in the environment, and in particular, on the surface of fruits.
One of the sugars converted by microorganisms is glucose. Action of the microorganisms will not be
considered in the rest of the problem.

1. Write a balanced equation for the transformation of solid glucose (C6H12O6(s)) into liquidethanol (C2H6O(l)) and gaseous carbon dioxide. Does this reaction require the presence of
dioxygen? (Yes/No)
2. Calculate the standard enthalpy, the standard entropy and the standard Gibbs free energy
associated with this reaction at 298 K. Does this reaction generate heat? (Yes/No)
Conversion of glucose into carbon dioxide and water is called cellular respiration.

51st IChO – Preparatory problems

26

3. Write a balanced equation for the transformation of glucose into carbon dioxide and water.
Does this reaction require the presence of dioxygen? (Yes/No)
The concentration of ethanol can vary a lot from one wine to another. Some Riesling winesfrom Germany (named “kabinett”) only contain 7–8% vol of ethanol, while Châteauneuf duPape wines (Rhone Valley, France) usually contain about 14% vol of ethanol (“% vol” means“percent of alcohol by volume” and is defined as the ratio between the volume of ethanol
contained in wine and the total volume of wine, multiplied by 100, at 298 K). It is thus very
important to control the concentration of ethanol in grape juice during fermentation. To
determine the concentration of ethanol in a wine, the following protocol was used: wine X is
diluted 50 times with distilled water. The aqueous solution of wine is added dropwise to a
100 mL aqueous solution of potassium dichromate (5.0·10‒3 mol L‒1) containing sulfuric acid(0.1 mol L‒1). The volume at the equivalence point 𝑉𝑒 is 15 mL.
4. Write a balanced equation for the oxidation reaction of ethanol by dichromate anions.
5. Calculate the equilibrium constant of this reaction. Can it be used to determine theconcentration of ethanol in wine? (Yes/No)
6. Calculate the pH of the solution of potassium dichromate and sulfuric acid before starting
the titration. Here, sulfuric acid can be treated as a strong monoacid.
7. Calculate the pH of the solution of potassium dichromate and sulfuric acid at theequivalence point (sulfuric acid is still considered to be a strong monoacid). Is it possible
to determine the equivalence point using the pH change of the solution? (Yes/No)
8. Calculate the concentration (in % vol) of ethanol contained in wine X. Is this wine a
German Riesling or a French Châteauneuf du Pape?
Data:
Thermodynamic data (at 298 K):
ΔfH° (kJ mol‒1)
S°m (J mol‒1 K‒1)

E° (V)CO2(g)‒393.5
213.6

Cr2O72−/Cr3+
1.33

Glucose(s) Ethanol(l)‒1274‒277.0212.1
160.7
CH3COOH/CH3CH2OH
0.19

Density of ethanol at 293 K: 0.79 g cm‒3

Problem 9. Nitrophenols: synthesis and physical properties
A multicomponent reaction is a reaction where three or more reactants react together to form a
product involving all the reactants. For instance, the Ugi-Smiles coupling has been studied by the French
duo L. El-Kaïm and L. Grimaud in 2005. During the past decade, this coupling has been used for thesynthesis of various heterocyclic compounds using various post-condensations. This reaction involves
an aldehyde, an amine, an isocyanide, and activated phenols, such as nitrophenols.

51st IChO – Preparatory problems

27

In this problem, the synthesis of nitrophenols is examined and some physical properties of the 4nitrophenol are studied.

Synthesis of nitrophenols
In a three-neck reaction flask, sodium nitrate (20.0 g, 235 mmol) is dissolved in water(50.0 mL). After cooling the solution in an ice bath, concentrated sulfuric acid (H2SO4,
14.5 mL) is added in small portions. A solution of phenol (12.5 g in 5.00 mL of water,
133 mmol) is then added slowly under vigorous stirring. The temperature is kept below 20 °C
and the solution is stirred for 2 hours. The liquor is then distillated and a first yellow compound,the 2-nitrophenol, is obtained (46.5 mmol). The residue in the distillation flask is cooled and a
2.00 mol L‒1 solution of sodium hydroxide (NaOH) is used to adjust the pH to 8-9 and charcoal
is added (2.00 g). The mixture is then warmed to reflux for 5 minutes and immediately filtered.
After distillation of 30 mL of water, the concentrated mixture is cooled down in an iced bath.
The obtained crystals are then dissolved and boiled into 50.0 mL of hydrochloric acid (HCl,3.7%) before filtration. The 4-nitrophenol is then obtained (20.0 mmol).
1. A partial scheme for the formation of the 2-nitrophenol starting from the nitronium ion
NO2+ is proposed below. Draw the missing intermediates and products.

2. Give at least two products other than 2-nitrophenol and 4-nitrophenol that could explain
the low yield.
Various characterizations of the 2-nitrophenol and 4-nitrophenol were performed: 1H NMR,
and measurement of their melting point, boiling point and solubility. The results were attributed
anonymously with two labels: A and B.
1

H NMR of A and B:
A (, ppm in CDCl3): 10.6 (large s, 1H), 8.1 (d, J = 8.4 Hz, 1H), 7.6 (dd, J = 8.5, 8.4 Hz, 1H),
7.2 (d, J = 8.4 Hz, 1H), 7.0 (dd, J = 8.5, 8.4 Hz, 1H)
B (, ppm in DMSO-d6): 11.1 (large s, 1H), 8.1 (d, J = 9.1 Hz, 2H), 7.0 (d, J = 9.1 Hz, 2H)

51st IChO – Preparatory problems

28

Properties
A
B

m.p.
44 °C
113-115 °C

b.p.
214 °C--

Solubility in water (298K)
2 g L‒1
15 g L‒1

3. Using the NMR data, determine which product (2-nitrophenol or 4-nitrophenol)
corresponds to A and B. To justify your answer, interpret the NMR chemical shifts of the
products.
4. Which interaction(s) between B and water can explain the higher solubility in comparison
to A? Choose the correct answer(s).
 Intermolecular hydrogen bonds
 Intramolecular hydrogen bond
 Electrostatic interaction
 Van der Waals interactions
 Covalent bond
To check the purity of A and B, a Thin Layer Chromatography (TLC) on silica was performed.
The eluent is a mixture of pentane/diethylether (7:3 in volume). After visualization of the TLC
using a UV light, the retention factor was calculated for the two spots (0.4 and 0.9).
5. Choose the correct statement(s).
A has a ☐ lower ☐ higher retardation factor (Rf) than B on the TLC because:A develops intermolecular hydrogen bonds with the silica.
A develops an intramolecular hydrogen bond.
B develops intermolecular hydrogen bonds with the silica.
B develops an intramolecular hydrogen bond.

Characterization of the 4-nitrophenol
Absorbance. The absorbance (A) versus the wavelength at various pH is given in the figure
below. Absorbance beyond 450 nm is negligible. The two maxima of the absorbance are at
310 nm and 390 nm, respectively.

51st IChO – Preparatory problems

29

6. Which is the color of a solution of 4-nitrophenol in neutral water? Choose the correct
answer.
☐ Blue ☐ Green ☐ Pink ☐ Purple ☐ Red ☐ Yellow
7. Choose the correct answer.
 4-nitrophenol has a longer absorption wavelength than its conjugated basebecause its conjugation is more important.
 4-nitrophenol has a longer absorption wavelength than its conjugated basebecause its conjugation is less important.
 4-nitrophenol has a shorter absorption wavelength than its conjugated basebecause its conjugation is more important.
 4-nitrophenol has a shorter absorption wavelength than its conjugated basebecause its conjugation is less important.
Determination of the pKa. A solution of 10 mL of 4-nitrophenol at c = 1.00·10-4 mol L‒1 wastitrated by a 1.00·10-2 mol L‒1 solution of sodium hydroxide (NaOH). The variation of the pH
as a function of the volume of NaOH is calculated and given in the figure below. Dashed curves
represent the fraction of 4-nitrophenol and 4-nitrophenolate, expressed in percentage on theright side. The pH is indicated as a solid line (scale on the left side).
100

12

10

% of 4-nitrophenol
and 4-nitrophenolate

pH

8

6

4

2
0

0.02

0.04

0.06

0.08

0.1

0
0.12

V (mL)

8. Assign each curve to 4-nitrophenol or 4-nitrophenolate.
9. Estimate the pKa of the 4-nitrophenol.
According to the theoretical curve, the pH jump is expected to be small, which makes theexperimental titration data difficult to analyze.
10. Which alternative method(s) can be used for the titration of 4-nitrophenol? Choose thecorrect answer(s).
 UV-Visible spectroscopy Potentiometry
 NMR
 Conductometry

51st IChO – Preparatory problems

30

Problem 10. French stone flower
Laumontite is a natural zeolite, a hydrated
calcium
aluminosilicate
offormula(CaO)x(A)y(B)z∙yH2O, where A and B are oxides.
It dehydrates in dry air and becomes then very
brittle. Due to this property, it was first called
zéolithe efflorescente (stone flower). But then themineral was named after the French mineralogist
F. Gillet de Laumont who discovered it in 1785.
Laumontite crystallizes into a monoclinic
crystal system of parameters: a = 1.49 nm,
b = 1.37 nm, c = 0.76 nm, α = γ = 90°, β = 112°,
Z = 4. Its density is ρ = 2.17 g cm‒3. After heatingin dry air, the mineral loses 15.3% of its mass, andno further mass change is then observed.

Laumontite from Espira-de-l’Agly deposit,
France (© Christian Berbain)

1. Calculate the stoichiometry y of thewater crystallized in laumontite.
Hint 1: the volume of a monoclinic unit cell is V = abc × sin β.
Hint 2: the mass m of 4 (CaO)x(A)y(B)z∙yH2O in one unit cell is: m = 4 M/NA, where M is themolar mass of the mineral and NA Avogadro’s constant. Also, m = ρV, where ρ is the density
and V the volume of a unit cell.
To determine the composition of this mineral, 0.500 g of laumontite was placed in a crucible
and 2 mL of concentrated hydrochloric acid were then added to it and heated up to 90 °C. The
sample was then washed with distilled water and dried under 120 °C for a few hours. The
insoluble residue was placed in another crucible (m0 = 14.375 g). It was then calcined at atemperature of 900 °C to constant weight. The final mass of the crucible and its content was
found to be m1 = 14.630 g. The residue is a pure binary compound that does not contain chlorine
atoms.
2. Determine the nature of A and B and the values of x and z.
Some samples of laumontite are orange. This coloration is caused by the presence of an
impurity, an element E that partly substitutes calcium, yielding the compound of formula:(EiCa(1‒i)O)x(A)y(B)z∙yH2O
The dissolution of a 0.500 g sample in nitric acid led to the formation of the same precipitate
as before. The filtrate was separated. When a few drops of NH4SCN are added to the filtrate,
the solution turns bright red. The filtrate was then neutralized with an excess of a concentrated
aqueous solution of ammonia (NH3) until complete formation of a precipitate. The latter was
filtered, washed with water and redissolved in 1 mol L‒1 sulfuric acid (H2SO4) followed by theaddition of an excess of zinc powder. The excess of metallic zinc was removed by filtration andthe solution was then transferred to a 100.0 mL volumetric flask and brought up to volume with
distilled water.
A 20.0 mL aliquot was transferred into a titration flask and potentiometrically titrated (using asaturated calomel electrode (SCE) as a reference) by 5.15 mL of a 2.00 mmol L‒1 solution ofCe(SO4)2 in 1 mol L‒1 of H2SO4.
3. Identify the impurity E.
4. Write the equations of the reactions corresponding to the aliquot preparation and titration.
51st IChO – Preparatory problems

31

5. Determine the amount of impurity E (mol. % compared to Ca).
6. Show that the potential at the equivalence point Ee.p. can be expressed as:
𝐸e.p. =

1
(𝐸°(𝐄3+ /𝐄2+ ) + 𝐸°(Ce4+ /Ce3+ ))2

7. According to the following table, determine which compounds would be the two bestindicators in this titration.
Indicator

Eo (V /SCE)

diphenylamine-4-sulfonic acid, sodium salt
5,6-dimethyl,10-phenanthroline
3,3’-dimethoxybenzidine
safranin T
4-ethoxychrysoidine hydrochloride
1,2-benzanthracene

0.60
0.73
0.54
0.00
0.76
1.00

Color
Oxidized form
Reduced form
blue
colorlessyellowred
red
colorlesspurplecolorlessredyellowcolorlesscolorless

Zeolites are widely used as materials in heterogeneous catalysis because of their large specificsurface area, their structural framework and their large number of acid sites. The structured
porous system of zeolites provides a molecular sieve effect. This effect leads to an increase of
the selectivity of some reactions in which the reactants and products have a kinetic diameter(the typical length under which the corresponding molecule will collide with an obstacle)
similar to the pore size of the zeolite. For laumontite, the largest pores present a diameterdmax = 0.604 nm. As a comparison, the kinetic diameter of benzene, 1,4-dimethylbenzene andtoluene is 0.585 nm and that of 1,2-dimethylbenzene is d = 0.680 nm.
Let us study the following reaction:

8. Draw the two main products F and G.
9. This reaction can also be catalyzed by laumontite. Determine which product will mainly
be formed in the pore system of the mineral.
Data at T = 298 K:
Eo(E3+/E2+) = 0.53 V /SCE
Eo(NO3‒/NO2) = 0.56 V /SCE
Eo(Zn2+/Zn) = ‒1.00 V /SCE
Eo(Ce4+/Ce3+) = 1.09 V /SCE
E(SCE) = 0.24 V
Eo(Ox/Red) (V /SCE) = Eo(Ox/Red)(V /SHE) ‒ E(SCE) (V)51st IChO – Preparatory problems

32

Problem 11. The mineral of winners
The mineral pyromorphite (from Greek pyro
‒ fire and morpho ‒ form) has the following
formula: A5(PO4)3B. It was named after its
property to recrystallize after melting. Therefore,
it is also sometimes called mineral of winners. In
France, deposits of this mineral are found in theCentre region.
Pyromorphite crystallizes into a hexagonal
crystal system of parameters: a = b = 0.999 nm,
c = 0.733 nm, α = γ = 90°, β = 120°, Z = 2. Its
density is ρ = 7.111 g cm‒3.
After complete dissolution of 1.000 g ofpyromorphite in concentrated nitric acid, thesolution was neutralized with potassiumhydroxide up to pH ≈ 5. An addition of 1.224 g of
KI was needed to form 1.700 g of a bright yellow
precipitate.

Pyromorphite from Chaillac Mine, Centre,
France (© Didier Descouens)

1. Determine the formula of pyromorphite.
Hint 1: the volume of a hexagonal unit cell is V = abc × sin β.
Hint 2: the mass m of 2 A5(PO4)3B in one unit cell is: m = 2M/NA, where M is the molar massof the mineral and NA Avogadro’s constant. Also, m = ρV, where ρ is the density and V thevolume of a unit cell.
2. Write an equation for a reaction that could occur if the KI was added in excess.
In some cases, A is replaced by the impurity C in a significant proportion. The atomic mass of
A is 3.98 times more than that of C. To determine the amount of the impurity, 1.00 g of themineral was dissolved in HNO3. After addition of Na2SO4 to the solution, a white precipitate
was formed. The precipitate was filtered out and the filtrate was added to an aqueous solutionof ammonia (NH3). Then, C(OH)n was separated and dissolved in a sulfuric acid (H2SO4)
solution. To proceed to the titration of C(+n), this impurity should be pre-oxidized into C(+m).
For this purpose, the solution of C(+n) in H2SO4 was heated in the presence of Ag2S2O8 (Ag+was used as a catalyst). The solution was then transferred to a 100.0 mL volumetric flask andbrought up to volume with distilled water. A 10.0 mL aliquot was then transferred to a titration
flask. Then, 10.0 mL of a 0.100 mol L‒1 acidic solution of Fe(NH4)2(SO4)2 were added. The
obtained mixture was finally titrated by 15.0 mL of an aqueous solution of KMnO4 solution ofconcentration 9.44·10‒3 mol L‒1.
3. Identify the impurity C. Write an equation for each reaction mentioned in the text.
4. Calculate the percentage of C in the studied pyromorphite (w. %).
5. Calculate the equilibrium constant of the titration reaction, for one equivalent of
permanganate ions, at 298 K.
Mn2+ can be added to the solution to indicate the completeness of the C(+n) pre-oxidation
reaction.

51st IChO – Preparatory problems

33

6. Write the equation of the reaction that indicates the completeness of the C(+n) preoxidation reaction. Underline the species that allows the detection of the completeness of
the reaction.
7. Why is Fe(NH4)2(SO4)2 often used in redox titrations instead of FeSO4? Choose the correct
answer: FeSO4 is not stable and get quickly oxidized by the oxygen in the air.
 Fe(NH4)2(SO4)2 is more soluble than FeSO4.
 Fe(NH4)2(SO4)2 is a cheaper reagent than FeSO4.
Data at 298 K:
E°(MnO4‒/Mn2+) = 1.51 V /SHE
E°(Fe3+/Fe2+) = 0.77 V /SHE
E°(C(+m)/C(+n)) = 1.33 V /SHE

Problem 12. Reaction progress kinetics
Kinetic investigations of multistep organic reactions are crucial for fundamental mechanistic studies
and are also necessary for practical applications of organic synthesis. Reaction progress kinetic analysis
is a methodology that makes use of the voluminous data sets that are now readily obtained fromcontinuous monitoring of reactions. The figure below shows a Heck transformation catalyzed by apalladium complex together with the sequence of steps in a typical catalytic cycle where the substrate 1
reacts with the catalyst 4 to form an intermediate 5 (rate constant k1). The reverse reaction is associatedwith a rate constant k‒1. Further reaction of this intermediate with a second substrate 2 delivers the
product 3 and regenerates the catalyst 4 (rate constant k2). The exact nature of the palladium complexes
4 and 5 is unknown.

1. Express the rate r of the reaction as a function of the rate constant k2 and the instantaneousconcentrations of 2 and 5 ([2] and [5], respectively).
2. Express the total concentration in catalyst [4]tot as a function of [4] and [5].
3. Assuming that intermediate 5 is in a steady-state regime, show that the rate r of the reaction
can be written as:𝑟=

𝑘1 𝑘2 [1][2][4]tot
𝑘−1 + 𝑘1 [1] + 𝑘2 [2]Let us define a parameter called [“excess”], which is equal to the difference in the initialconcentrations of the two substrates:[“excess”] = [2]0 – [1]0Hence we can write:[2] = [2]0 ‒ [1]0 + [1] = [“excess”] + [1]51st IChO – Preparatory problems

34

4. Show that the rate can now be written as:𝑟=𝑎where 𝑎 = 𝑘

𝑘1 𝑘2−1 +𝑘2 ["excess"]and 𝑏 = 𝑘["excess"][1] + [1]2
[4]tot
1 + 𝑏[1]𝑘1 + 𝑘2−1 +𝑘2 ["excess"]

For a given set of conditions, the values [4]tot and [“excess”] are constant, thus leaving [1] as
the only variable. Hence, there is a direct relationship between the rate of the reaction r and theinstantaneous concentration of one reactant [1], which can be easily accessed, for instance byabsorbance measurements.
Reaction calorimetry is a technique that can also be used. The heat flowing in or out of thereactor is measured over time, while the temperature is controlled and kept constant. Let us
assume that only one transformation 1 + 2 → 3 is occurring in the reactor.
5. Express the relationship between the heat flow dq(t) at a given time t evaluated during theperiod dt, the volume V of the reactor, the reaction enthalpy rH and the rate r.
By combining the results from these different experimental procedures, it is possible toconstruct reaction progress analysis graphs where the rate r is expressed as a function of theconcentration of 1. The figure below shows experimental results for the relation between therate r of the Heck reaction as a function of substrate concentration, ArX. Two different initialconditions with the same total catalyst concentration and [“excess”] values have been
considered.

Experiment A: [ArX]0 = 0.16 mol L‒1 and [alkene]0 = 0.24 mol L‒1
Experiment B: [ArX]0 = 0.12 mol L‒1 and [alkene]0 = 0.20 mol L‒1

6. For a given concentration [ArX] on the plot, which experiment yielded more product?
(Experiment A / Experiment B)
7. For a given concentration [ArX] on the plot, which is the reaction in which the catalyst hascompleted the more turnovers? (Experiment A / Experiment B)
51st IChO – Preparatory problems

35

In the following questions, choose the correct answer (True / False).
8. Product inhibition would reduce more the rate of the reaction in which more product isformed. (True / False)
9. Catalyst deactivation would reduce the rate of the reaction where the catalyst has done more
turnovers. (True / False)
10. Neither catalyst deactivation nor product inhibition is a factor in the Heck reaction shown.
(True / False)
Experimental results for the reaction progress kinetic analysis in the case of cobalt-catalyzed
epoxide ring-opening are shown below. This reaction tells a different story. For a given epoxide
concentration, a slightly higher rate was observed in Experiment D (lower initial concentration)compared to that shown in Experiment C (higher initial concentration). This result suggests thateither product inhibition or catalyst deactivation influences the reaction.

Experiment C: [epoxide]0 = 1.5 mol L‒1 and [H2O]0 = 2.0 mol L‒1
Experiment D: [epoxide]0 = 1.0 mol L‒1 and [H2O]0 = 1.5 mol L‒1

Let us assume that a new experiment, Experiment E, is performed with the same initialconditions as in Experiment D but with some product of the reaction added right from the
beginning.
11. If the curve obtained for the reaction progress kinetic analysis is identical to that observedfor Experiment D, then catalyst deactivation is the factor responsible for the behavior
shown on the figure above. (True / False)

51st IChO – Preparatory problems

36

Problem 13. Nylon 6
Nylon 6 is a synthetic linear polyamide. Its repeating unit contains six carbon atoms, as shown in
the figure below. This polymer was first synthesized by P. Schlack at IG Farben. Most nylon 6 polymers
tend to be semi-crystalline and are produced in the form of fiber yarns. Nylon is a tough material withgood thermal and chemical resistance.

Nylon 6 can be prepared from ε-caprolactam via a catalyzed anionic ring opening
polymerization. The polymerization can be accelerated by an acylated lactam named I. One ofthe postulated mechanisms for this reaction is shown below.

The first acid base reaction will not be considered in the following study. We further assume
that no reaction other than those listed above occurs.

51st IChO – Preparatory problems

37

1. Determine the relationship between the following concentrations [I]0 (initial concentrationof I), [I], ∑𝑛𝑖=1[A𝑖 ] and ∑𝑛𝑖=1[P𝑖 ].
2. Apply the steady state approximation to all An intermediates.
3. Derive the rate of disappearance of the monomer MH as a function of the reactant
concentrations [I]0, [MNa], [MH], ki and K°.
4. Show that, depending on the rate-limiting step, the partial order with respect to themonomer MH is 0 or 1 and express the conversion τ, fraction of the initial monomer
concentration that has been consumed.
5. In the two limit cases studied in the previous question, draw the conversion of monomer
MH versus time curve.
The monomer conversion τ vs time curve obtained by Macosco et al. is the following:

6. What does the shape of the monomer conversion vs time curve stress out? Choose thecorrect answer(s).
 An inhibition effect of the monomer
 An oscillatory reaction
 A second order reaction
 An autocatalytic process A catalyzed reaction
To explain the experimental kinetic data, a competing mechanism was suggested. This side
reaction decreases the degree of polymerization of nylon:

51st IChO – Preparatory problems

38

The chemical structure of E is:

7. Draw possible structures of B‒, C and D‒.
8. Considering only this mechanism, the disappearance rate of the monomer MH is
proportional to [MH]([MH]0 ‒ [MH]). Plot the disappearance rate of the monomer MH
versus the monomer concentration [MH]. Find the monomer concentration at which therate is the highest.

Problem 14. Synthesis of block copolymers followed bysize-exclusion chromatography
Polymers and polymeric materials offer a wide range of properties. But in practice, different
properties are simultaneously desirable, such as different chemical, thermal or mechanical properties. A
way to achieve such combinations is to combine existing polymers into what is called a block
copolymer. Here, we study a synthetic strategy and a characterization methodology of a block
copolymer composed of polystyrene and polydimethylsiloxane.

Polystyrene (PS) synthesis
1. There are three kinds of polymerization initiators: anionic, cationic and free radicals. Whichof the following is an anionic initiator? Choose the correct answer(s).
 Benzoyl peroxide
 Sulfuric acid
 Azobisisobutyronitrile (AIBN) n-Butyllithium (n-BuLi)
2. Complete the polymerization mechanism for polystyrene synthesis when styrene is inpresence of n-BuLi (block I). Give an example of a reactive species that could be used toget an ester extremity at the termination stage (block II) and the corresponding byproduct(block III).

51st IChO – Preparatory problems

39

In the following we note: M the monomer, A the initiator, and AMi the growing polymer witha degree of polymerization i. The rate constant for the propagation kp is the same for each stageof the chain growth. ka is the rate constant for the initiation stage. We assume that the initiation
reaction is fast and goes to completion.
3. Compare ka and kp.
4. Express the disappearance rate of monomer M as a function of the rate constant kp, thegrowing polymer concentration [AMi], the monomer concentration [M], and i.
We consider that the active species concentration is constant and equal to C.
5. Rewrite the disappearance rate of monomer M obtained in question 4 as a function of C,[M], and kp.
6. Deduce from this equation the half-life time denoted t½ of the polymerization reaction as afunction of kp and C.
7. We now consider different synthesis conditions for styrene polymerization. Fill in withmolecules the mechanism presented below.

8. Is this polystyrene synthesis regioselective? (Yes/No)
Synthesis of Polydimethylsiloxane (PDMS)
We now examine the synthesis of polydimethylsiloxane (figure below), also known as silicone.
These polymers are usually used in sealants, adhesives, lubricants, medicine, cooking utensils,
and thermal insulation.

We will also consider cyclic and short polydimethylsiloxane molecules written Dn with n thenumber of Si atoms. For example, D5 is pictured below.

51st IChO – Preparatory problems

40

9. We consider a reaction medium with D4 and hydroxide ions. Give the structures of A and
B in the mechanism below, leading to the formation of PDMS.

10. During this reaction the synthesis of macrocycles is observed. Draw a mechanism or apattern to explain the formation of such macrocycles.
11. Transfer reactions are also observed. Draw a mechanism showing what a transfer reaction
in such a reaction medium could be.
Size-exclusion chromatography and synthesis of a block copolymer
The copolymer is studied by size-exclusion chromatography (SEC), also called gel permeation
chromatography (GPC). The principles of the method are represented in the figure below: (1)the sample is introduced on a column filled with a microporous packing material (millions ofhighly porous and rigid particles tightly packed together in a column). This material (2) does
not react with polymers. Molecules of various sizes elute from the column at different rates.
The column retains the molecules with low hydrodynamic volume (i.e. the smaller molecules)
for a longer time than the molecules with high hydrodynamic volume. A detector (3), such as arefractive index detector or an infrared absorption detector, detects molecules at the end of thecolumn. For a single species, the intensity of the measured signal (4) is proportional to theconcentration of the molecules detected. The experimental curve (4) can be translated into acurve giving the mass fraction of polymer chains in the sample as a function of their molar
mass. This translation is done using monodisperse polymer standards.
(3)

(1)
(2)

signal intensity(4)

elution time

51st IChO – Preparatory problems

41

PDMS synthesis is performed in THF, starting with D3 and n-BuLi at the concentration [BuLi]0.
Variations of temperature of the reaction medium and/or of [BuLi]0 have consequences on the
product obtained. We introduce the polydispersity index denoted Ip = Mw/Mn, where Mw is theweight average molecular weight and Mn is the number average molecular weight.
12. To better understand the meaning of the polydispersity index, fill in the gaps in thefollowing sentence with the word “low” or “high”:“Mn is more sensitive to molecules of _______ molecular mass while Mw is more sensitive tomolecules of _____ molecular mass. Therefore the more the polymer chains approach uniform chain
length, the more Ip is close to 1.”

Three experiments (V. Bellas et al., Macromolecules, 2000) in different reaction conditions are
performed and the polydispersity index Ip is determined in each case.
(I) 25 °C until 50% conversion is reached. The SEC analysis gives Ip =1.06.
(II) conditions (I) followed by polymerization at ‒20 °C for 8 days.
(III) 25 °C until 100% conversion is reached. The SEC analysis gives Ip =1.3.
The SEC analysis or the three experiments is represented in the figure below.

13. A qualitative analysis of the experimental curves obtained with SEC allows to associate
which reaction conditions lead to the highest Mn. Which of the 3 curves is related to thehighest Mn?
14. Match each curve with reaction conditions (I, II or III with (a), (b) or (c)).
We (finally!) synthesize the block copolymer following the procedure depicted below, andmonitor the reaction by SEC.

51st IChO – Preparatory problems

42

15. Suggest a structure for the polymer that is finally obtained.
16. The final product is then fractionated (a fraction of the polymer chains are separated fromthe sample, according to their length). Associate the SEC experimental curves (1, 2, 3 or
4) measured at different stages of the synthesis (figure below) with the corresponding
molecules (PS, PS-PDMS precursor, unfractionated product, or fractionated product).

Problem 15. Radical polymerization
Radical polymerization is a method of choice for polymer synthesis. Easy to implement andcompatible with a wide range of functional monomers, this process can be carried out under various
experimental conditions, including in the presence of water. Typically, radical polymerization iscomposed of three steps: (i) initiation, (ii) propagation and (iii) termination. The initiation step consistsof the thermal decomposition of an organic compound according to a radical mechanism (first step of
the kinetic scheme below). This leads to a radical species that further initiates radical polymerization.
𝐀𝟐 → 2𝐀∙∙

𝑘d

𝐀 + 𝐌 → 𝐀𝐌 ∙𝐀𝐌𝒏 ∙ + 𝐌 → 𝐀𝐌𝒏+𝟏 ∙𝑘p

51st IChO – Preparatory problems

43

1. Considering a symmetric unimolecular initiator, give the chemical structures of the initiatorand the monomer used for the synthesis of the polymer P1.

At a given temperature, the half-life of the initiator t½ can be determined experimentally byfollowing the evolution of the concentration of the initiator vs time.
2. The table below gives the evolution of A2 concentration over time at 82 °C in
chlorobenzene. Determine graphically the value of t½ for the initiator A2 at 82 °C in
chlorobenzene.
[A2] (mmol L‒1)
1.00 0.81 0.66 0.54 0.44 0.24 0.06
time (h)
0.0
0.3
0.6
0.9
1.2
3.0
6.0
3. Calculate the rate constant for the dissociation of the initiator A2, denoted kd, at 82 °C in
chlorobenzene.
Despite its many advantages, radical polymerization presents some drawbacks mainly related
to the occurrence of irreversible termination reactions (combination, disproportionation,transfer reactions), which limit the possibilities of obtaining polymers with controlled
architectures and compositions.
4. Among the possible termination reactions of P1, write its self-combination reaction.
New techniques such as Reversible-Deactivation Radical Polymerization (RDRP) have beendeveloped to limit the irreversible termination of the propagating radical chains. RDRP
conducted in the presence of a nitroxide, known as Nitroxide Mediated Polymerization (NMP),
consists of using an alkoxyamine as initiator, as shown in the scheme below. Thesealkoxyamines dissociate homolytically under heating to form an alkyl radical that acts as an
initiator and a nitroxyl radical that end-caps reversibly the polymer chain end during thepolymerization process.

51st IChO – Preparatory problems

44

5. Give the chemical structure of the alkoxyamine that will be written ALK1 and is used toobtain the polymer P2.

P2

In the following we write: conv the monomer conversion (% of consumed monomer), m themass of styrene, n the number of moles of initiator, and f the efficiency factor (in the case of
alkoxamine, f = 1).
𝑚

6. Knowing that the number average molar mass Mn can be expressed as 𝑀n = 𝑐𝑜𝑛𝑣 × 𝑓×𝑛 ,give the number of moles and the mass of ALK1 required to obtain 10 g of a polystyrene
sample exhibiting Mn = 20000 g mol‒1 at 100% conversion of styrene.
Thanks to RDRP techniques, access to block copolymers became easier. These copolymers arecomposed of at least two blocks of homopolymer linked together by a covalent bond. They
combine the properties of homopolymers presenting different characteristics (e.g. combinationof a hydrophilic block and a hydrophobic block). For instance, the poly(hydroxyethyl acrylate)b-poly(4-vinylpyridine) diblock copolymer (PHEA-b-P4VP, P4) behaves as a surfactant.
Poly(hydroxyethyl acrylate) is hydrophilic, and poly(4-vinylpyridine) is soluble in water for
pH < 6 (protonation of pyridine) and insoluble in water for pH > 6.

7. In the figure below, assign the state (P4 or P5) of the PHEA-b-P4VP block copolymer inaqueous solution according to the pH values.

A

B

8. Give the expected number of 1H NMR signals with their splitting pattern for the
C(=O)OCH2CH2OH side chains of the copolymer P4 dissolved in deuterated
51st IChO – Preparatory problems

45

tetrahydrofuran. In this solvent, the copolymer P4 is perfectly soluble. (Note: no coupling
will be considered with the terminal OH group).
9. Assign on the following chromatogram obtained by Size Exclusion Chromatography (SEC,
see question 12 in problem 14) which curve corresponds to a polystyrene sample preparedby conventional free radical polymerization (P6) and which one corresponds to polystyrene
prepared by NMP (P7).

Problem 16. Biodegradable polyesters
Replacing conventional polymer materials, sources of pollution, by biodegradable polymers is amajor industrial challenge. The synthesis and characterization of such a biodegradable polymer isstudied here.

1. What is a biodegradable polymer? Choose the correct answer(s).
 A polymer available in native form in nature.
 A polymer made from biobased compounds.
 A polymer transformable by microorganisms into less polluting
molecules.
A polymer functionalized by an ester side-chain is prepared by polymerization of thecorresponding monomer (BED) in the presence of a catalyst. The polymerization is initiated bypentan-1-ol (denoted n-pentOH in the following scheme). Bn corresponds to the benzyl groupi.e. C6H5‒CH2‒.

2. Give chemical function(s) that can explain the biodegradability of this polymer.
The polymer is first characterized by NMR spectroscopy. The obtained 1H NMR spectrum(CDCl3, 300 MHz) is given in the figure below:
51st IChO – Preparatory problems

46

a
O i Ph j

O

b
c

h

d
e

e
f’

0.12

0.65

j

4.40

4.20

4.10

4.30

g

O
O H

f O

k

O

* k’

0.20

i

O

n

4.00

CH2Cl2
k

g,h

f
e

2.0

3.00

3.0

a

4.10

4.0

2.10

114.09

5.0

1.96
0.36
0.59

6.0

56.49

56.87
38.28

145.42

7.0

b,c

d

1.0

3. Give the definition of the number average degree of polymerization Xn.
NMR provides access to the average degree of polymerization Xn,NMR and to the number
averaged molecular weight Mn,NMR. For that, we compare the integrations of the protons at theend of the chains with those of protons of the main chain.
4. By integration of appropriate peaks, determine Xn,NMR.
5. Calculate Mn,NMR from this value knowing that the molecular weight of the monomer unit
and end chain are equal to 278 and 88 g mol‒1, respectively.
6. In general, 1H NMR spectroscopy is an effective method to determine Mn. However, itremains limited for very large polymers. Why? Choose the correct answer(s).
 The peaks at the ends of the chains are not sufficiently resolved comparedto the peaks of the main chain.
 The integration of the different peaks observed may be distorted due to theobserved peak broadening for high mass polymers.
 The number of protons will be too high to be properly analyzed.
The polymer is then analyzed by size exclusion chromatography (SEC, see question 12 inproblem 14). The measured elution volumes can be related to the molar mass of the polymer
obtained by, for example, the prior injection of a range of polystyrene samples of known masses(standard polymers). The obtained value for Mn,SEC is equal to 8950 g mol‒1.
7. Compare Mn,NMR and Mn,SEC. How to explain this difference? Choose the correct
answer(s).
 SEC is not a sufficiently precise analytical method.
 Polystyrene has a larger hydrodynamic volume than that of polyBED.
 Polystyrene has a smaller hydrodynamic volume than that of polyBED.
8. Cite a property of polymers, not accessible by NMR, that SEC can provide.

51st IChO – Preparatory problems

47

Problem 17. Vitrimers
Vitrimers are a new type of material, in which chemical bonds are constantly exchanged. It has
been developed in France by L. Leibler et al. in the early 2010s. These polymers can be easily processedat high temperature, which is used in self-healing materials. Here, we will examine an early example ofvitrimer. The polymer network below was made using A and B under acidic conditions.
Scheme ofa polymer network

A

B

1. Write a structure for the reactive group Q in B, given that the network above is composedof A’ and B’, interlinked by ester bonds (wiggly lines).

A’

B’

Each epoxy group in A reacts with two Q groups from B molecules, to form two esters. A has
two epoxy groups and B has three Q groups. A is added to B in a 3:4 ratio. Of all the reactive
groups (epoxy and Q groups), a fraction ζ reacts (conversion = ζ × 100), such that at 100%
conversion (ζ = 1.0), all epoxy groups in A and all Q groups in B have reacted.
2. Express the average number of ester bonds formed per molecule of A, nEA, as a functionof ζ.
3. Express the average number of ester bonds formed by a molecule B, nEB, as a function ofζ.
The total number of molecules A and B is given by N = NA + NB.
4. Calculate the total number of formed ester bonds, NE, expressed as a function of NA, NB,
nEA, and nEB.
5. Calculate the average number of ester bonds attached to a molecule, 𝑛̅, as a function of ζ(e.g. A’ has four ester bonds attached to it).

51st IChO – Preparatory problems48

6. Calculate the value of ζ for which a crosslinked network starts to form.
The network becomes dynamic when the chemical bonds interchange constantly. In order toreach this state, esterification reactions are essential. The reaction mechanism for acidcatalyzed-transesterification of ethyl acetate (CH3COOEt) with an unknown alcohol V isprovided below.

7. Draw the missing structures U, V, W, X.
If transesterification reactions are rapid enough, a vitrimer can easily be deformed. The
resistance to flow η (viscosity) was measured and 1 / η was shown to follow an Arrhenius law.
Viscosity η is proportional to the time needed for a material to adjust itself, which is inversely
proportional to the transesterification reaction rate constant k: 𝜂 ∝ 1/𝑘.
The plot and table below show that 1 / η follows an Arrhenius law.

η (Pa s)ln η1000 / T (K−1)

6.310·106
15.66
1.8

3.981·108
19.80
2.2

3.162·109
21.87
2.4

7.943·109
22.80
2.5

3.134·1010
24.17
2.7

8. Using the data provided in the table, determine the activation energy EA fortransesterification in kJ mol‒1.
The viscosity η of the material can be controlled by changing the reaction conditions.

51st IChO – Preparatory problems49

9. For the following statements, choose the correct answer:I) When a transesterification catalyst is added to the material
η increases
η remains constantη decreases

II) When the temperature is decreased
η increases
η remains constantη decreases

III) When the pH is lowered from neutral to acidic
η increases
η remains constantη decreases

Problem 18. A kinetic study of the Maillard reaction
The Maillard reaction is a chemical reaction that involves aminoacids and reducing sugars. It leads
to brown molecules (melanoidin polymers) without enzymatic catalysis. Many other products can beformed through the different steps of its complex mechanism. These products can notably enhance taste
and flavors. The Maillard reaction was first described by the French chemist L C Maillard in 1912
(L. C. Maillard, Compt. Rend., 1912). This reaction is observed during roasting and toasting processesof meat, bread, coffee, etc. (S. Martins et al., Trends in Food Science & Technology, 2001). In the firstattempt to describe the Maillard reaction, a reactive model with glycine amino-acid (H2NCH2COOH)
and fructose reducing sugar (C6H12O6) is considered.

O
H2N
OH

Glycine Gly
Fructose Fru

1. Fill in the proposed mechanism for the first step of addition of glycine on fructose that leads
to an imine (acidic conditions).

51st IChO – Preparatory problems

50

In order to describe the formation of melanoidin products, the following mechanism has been
proposed (S. Mundt et al., J. Agric. Food Chem., 2003):
𝐆𝐥𝐲 + 𝐅𝐫𝐮
𝐈

𝑘2
→
slow

𝑘1
→
slow

𝐈

𝐌

(1)
(2)

where Gly stands for glycine, Fru for fructose, I an unknown intermediate and M themelanoidin product.
Previous kinetic studies showed that step (1) exhibits a zero kinetic order for glycine Gly andstep (2) a first kinetic order for the intermediate I. We will now determine the rate constants k1
and k2 as well as the kinetic order (named α) for the fructose during step (1).
We consider the reaction of the intermediate product I with sodium metabisulphite, S(IV). This
reaction is known to be thermodynamically and kinetically favorable and leads to a stableproduct DSH (3,4-dideoxy-4-sulfohexosone), which does not react with other compounds ofthe reaction medium. This reaction strongly favors products and is fast.
𝐈

𝑘3+ 𝐒(IV) →
fast

𝐃𝐒𝐇

(3)

Sodium metabisulphite reacts with Ellman’s reagent (5,5’-dithiobis-(2-nitrobenzoic acid)). It
produces a colored compound, the concentration of which can be spectrophotometricallydetermined at 412 nm. Sodium metabisulphite does not interact with other reactants or products.
The melanoidin product M absorbs at 470 nm and its absorption coefficient in these conditions
is εM = 478 L mol‒1 cm‒1. All other compounds are considered colorless in the range of the UVvisible wavelengths.
1 L of solution was prepared with a sodium acetate/acetic acid buffer (pH = 5.5). All theexperiments were carried out at 55 °C. The length of the cell was l = 1 cm for the
spectrophotometric measurements.
2. Write down the rates of disappearance of Fru and I.
3. Controlling the pH and temperature of the chemical mixture is compelling in this study
case. Choose the correct answer(s).
 Equilibrium constants can depend on pH.
 Equilibrium constants can depend on temperature.
 Rate constants can depend on equilibrium constants.
 Rate constants can depend on pH.
 Rate constants can depend on temperature.
The first experiment (A) was carried out by adding nFru,A,0 = 1 mol of fructose, nGly,A,0 = 0.5 molof glycine and nS(IV),A,0 = 0.02 mol of S(IV) at the same time to a 1 L sodium acetate/acetic acidbuffer. Every 15 hours (roughly), an aliquot of 1 mL was extracted from the reactive medium,put in a cell of l = 1 cm and a drop of Ellman’s reagent was added. The absorbance wasmeasured after stirring. Results obtained from this experiment are shown in figure 1.
4. Choose a wavelength to measure the absorbance of the aliquots of experiment (A).

51st IChO – Preparatory problems

51

5. Demonstrate that the concentration [Fru] of fructose at time t can be deduced from theconcentration [S(IV)] at time t and the initial concentrations [S(IV)]0 and [Fru]0 as:[Fru] = [Fru]0 – [S(IV)]0 + [S(IV)]
6. Using the following graph and the rate equations determined in question 2, determine thekinetic order α of fructose in step (1) of the mechanism. It can be equal to 0, 1 or 2.
7. Determine the kinetic rate constant k1.

Figure 1: [S(IV)] as a function of time for the fructose - glycine - sodium metabisulphite S(IV)reactionA second experiment (B) was carried out by adding nFru,B,0 = 1 mol of fructose andnGly,B,0 = 0.5 mol of glycine without sodium metabisulphite to a 1 L sodium acetate/acetic acidbuffer. A magnetic stirrer was used all along the experiment. Every 6 hours, the absorbance ofan aliquot of 1 mL was directly measured at 470 nm in a cell of l = 1 cm. The obtained data isshown in figure 2. Given the steps (1) and (2), the concentration of the melanoidin product [M]can be written as:
𝑘1[M] = 𝑘1 𝑡 − (1 − e−𝑘2 𝑡 )𝑘2Figure 2: A470 (absorbance at 470 nm) as a function of time for the fructose - glycine reaction

51st IChO – Preparatory problems

52

8. Determine graphically the value ofd𝐴470⁄d𝑡⌋𝑡=80h derivative of A470 absorbance at

470 nm at t = 80 h based on figure 2.
t (h)
15.0
30.5
44.6
62.7
dA470/dt (h‒1) 6.95·10‒3 9.22·10‒3 1.35·10‒2 1.82·10‒280.0

Derivative of A470 as a function of time data for the fructose – glycine reaction
1

9. Assuming that: ln [1 − 𝜀 𝑙𝑘
M

d𝐴470
1

d𝑡

] = −𝑘2 𝑡, determine the rate constant k2.

Problem 19. Glycosidases and inhibitors
The glycosidic bond is one of the most stable bonds among biological polymers. This bond is foundin DNA, in glycoproteins and in polysaccharides. Glycosidases, also called glycoside hydrolases, areenzymes that catalyze this reaction (scheme 1). They are the most abundant and most efficient enzymes
in nature: they can increase the speed of the C—O bond cleavage reaction by a factor of 1017.

Scheme 1: Hydrolysis of the glycosidic bondMechanism of hydrolysis catalyzed by glycosidase
The hydrolysis of the glycosidic bond can occur with inversion (scheme 2) or with retention of
the configuration (scheme 3).

Scheme 2: Mechanism with inversion of configuration

Scheme 3: Mechanism with retention of configuration

51st IChO – Preparatory problems

53

1. Complete the structure of the transition state (scheme 4) for the mechanism with retentionof configuration (step 1), with forming and breaking bonds symbolized by dashed lines (- - -).

Scheme 4: Structure of the transition state of the first step of the mechanism with retention ofconfiguration

Synthesis of a glycosidase inhibitor
Given the essential role of glycosidases, glycosidase inhibitors are of great interest for
researchers: they allow a better understanding of the enzymatic mechanisms of hydrolysis of
the glycosidic bond, but they are also important for therapeutic applications because of the large
number of diseases involving this type of enzymes (diabetes, influenza, cystic fibrosis, etc.).
One way to obtain carbohydrate mimics that could act as inhibitors is to replace the endocyclic
oxygen atom with a nitrogen atom. These inhibitors mimic reaction intermediates or transition
states of glycosidic hydrolysis. In this part, we study the synthesis of a mannoimidazole-typeβ-mannosidase inhibitor.

Scheme 5

In the first step, compound 2 is formed from a natural carbohydrate 1, which reacts in an acidic
environment with methanol.
2. Give the number of the most electrophilic carbon of 1. Draw a tautomeric form in acidic
environment to justify your answer.
3. Compound 2 is C7H14O6. Draw its topological formula.
Compound 2 reacts with sodium hydride NaH and benzyl bromide Bn-Br, to give compound 3.
51st IChO – Preparatory problems

54

4. Chose the correct statement indicating the reactivity of sodium hydride: base acid nucleophile electrophile
5. Chose the correct statement indicating the reaction that occurs in a second step: nucleophilic addition
 nucleophilic substitution
 elimination
 electron transfer
6. Draw the structure of compound 3.
Compound 4 is obtained after hydrolysis of the acetal function of 3 to get a hemiacetal function.

Scheme 6

Compound 4 yields lactone 5. A bicyclic structure is then obtained in 6 steps, not detailed here.
From there, product 7 is obtained after deprotection of the alcohol functions of 6 withdihydrogen in the presence of Pd/C. Three analogs of 7 are then synthesized from 6 in 4 steps(10a, 10b, 10c).
7. Write down the redox half-reaction between 4 and 5.
8. Using the structure given in question 1, draw a scheme explaining the mode of action forthe mimic 7.
A library of potential glycosidase inhibitors has thus been obtained. It is now necessary to know
if the synthesized glycosidase inhibitors are true mimics of the transition state. To do so, we
have to characterize the affinity of each inhibitor with the active site. A kinetic study based ona classical model in enzymatic kinetics, called the Michaelis-Menten model, is presented here.
Michaelis-Menten kinetics
The following mechanism is often used to rationalize the early stages of enzymatic catalysis
processes. In the first step, the enzyme E associates with the substrate S to give an enzymesubstrate complex denoted ES. This equilibrium is fast. The enzyme-substrate complex
dissociates in a second step to give the product P and the regenerated catalyst E.

51st IChO – Preparatory problems

55

9. Demonstrate that the reaction rate r is:
𝑅max [S]𝑘 +𝑘𝑟 = [S]+𝐾
with 𝐾m = − 𝑘 cat and 𝑅max = 𝑘cat [E]totm+and [E]tot the total concentration of enzyme introduced in the solution
10. We consider the two cases where [S] >> Km and [S] << Km. In both cases, determine thereaction rate r and draw the shape of the evolution of r as a function of [S].
kcat is the rate constant of the transformation reaction of the enzyme-substrate complex into the
product. Km is called the Michaelis constant.
11. In the case of k– >> kcat, write the expression of Km.
12. Tick the right boxes to indicate the link between Km and the affinity of the enzyme for thesubstrate:
Km low

Km highHigh affinity
Low affinity

In the presence of a competitive inhibitor, the inhibition constant Ki is the constant of theequilibrium:
EI = E + I
where I is the inhibitor.
The concentration of inhibition ci, a typical concentration for which half of the enzymatic sites
are occupied, can be measured. The following ci values were obtained for the different mimics
studied on Bacteroides thetaiotaomicron β-mannosidase:
Mimicci (mol L‒1)

7
400

10a
57

10b
72

10c
401

11
1000

12
975

13. Write the relation between ci and Ki. Choose the best mimic, i.e. the mimic that exhibits
the highest affinity with the enzyme.

Problem 20. Fluoro-deoxyglucose and PET imaging
Positron Emission Tomography (PET) is a nuclear imaging method allowing in vivo investigationsof the distribution of an isotope emitting positrons (+) upon radioactive decay. Among the + emitters,
18
F has found numerous applications due to its particular properties: (t½ = 109.74 min; decay by +emission; specific activity of 18F = 6.336·1019 Bq mol‒1). 18F can be introduced via a nucleophilic
process using 18F– obtained in water, after desolvation of the fluoride.
N.B.: Bq stands for the Becquerel unit. It is a unit of radioactivity corresponding to one
disintegration per second.
Among the very few radiolabelled molecules commercialized worldwide for PET imaging, 2deoxy-2-[18F]fluoro-glucose ([18F]-FDG) is the most widely used. This problem examines the formationof [18F]-FDG from organic compounds and 18F–.

51st IChO – Preparatory problems

56

Reactivity and stereochemistry at C1
The isomerization from the α to the β form of [18F]-FDG can occur according to two different
mechanisms in protic media.

1. The endocyclic bond cleavage path makes use of the equilibrium between two acyclic forms
that can recyclize to yield both compounds [18F]-α-FDG and [18F]-β-FDG. Draw the two
conformers of the acyclic form that lead to the formation of [18F]-α-FDG and [18F]-β-FDG,
respectively.
2. Choose the correct structural relationship(s) between [18F]-α-FDG and [18F]-β-FDG.
 [18F]-α-FDG and [18F]-β-FDG are enantiomers.
 [18F]-α-FDG and [18F]-β-FDG are epimers.
 [18F]-α-FDG and [18F]-β-FDG are diastereoisomers.
 [18F]-α-FDG and [18F]-β-FDG are atropoisomers.
3. Draw the structure of the missing cyclic intermediate.
Evolution of 18F and molecular consequences of isotopic distribution and decay
18

F is a radioactive isotope that decays by the emission of a positron 01𝛽 + .

4. Write down the nuclear equation for the radioactive decay of 18F.
5. Draw the molecular structure of the hexose arising from the decay of [18F]-FDG.

51st IChO – Preparatory problems

57

6. Establish the equation of the radioactive decay as a function of time and determine thevalue of the radioactive constant (λ, that corresponds to the kinetic constant associated withthe radioactive disintegration reaction).
7. Given that one injection to a human requires 370 MBq for imaging purposes and that thepatient needs to rest for one hour before imaging is processed, calculate the remaining
radioactivity (i) at the imaging processing time, (ii) after 4 h.
2–deoxy–2–[18F]fluoro-glucose by nucleophilic radiofluorination
D-Mannose is the required starting material to achieve the synthesis of 2-deoxy-2-[18F]fluoro-

glucose by nucleophilic radiofluorination.
8. Among the following, choose what type of reaction can be used to convert a compoundfrom the mannose class to a compound of the glucose class with complete stereocontrol: second-order nucleophilic substitution
 first-order nucleophilic substitution
 elimination-addition mechanism

The synthesis of the required peracetylated mannose sulfonate E is described below. E is the
precursor for the radiofluorination reaction.

9. The first step is a peracetylation of D-mannose yielding A. Represent the tetrahedral
intermediate involved in the acetylation of alcohols under the used conditions.
10. The transformation A → B favors the substitution of the acetate in  position of theendocyclic oxygen atom. Draw the carbocationic intermediate involved in thistransformation and represent the electronic effect responsible for the observed selectivity.
51st IChO – Preparatory problems

58

11. In the transformation from B to C, choose the reactivity of ethanol among the following: nucleophile electrophile inert solvent
12. In the same transformation, choose the reactivity of 2,4,6-collidine among the following: nucleophilic catalyst
 base co-solvent
One of the intermediates in the transformation from C to D is C’.

13. Draw the structure of the ionic reaction intermediate between C and C’.
The formation of [18F]-FDG is finally achieved as follows. The compound of general formula
E given above, can be either F, G or H as shown in the following scheme.

14. Identify which of these three compounds F, G or H is the most reactive toward thenucleophilic substitution leading to I.
Crypt-222 has the following structure:

51st IChO – Preparatory problems

59

15. Why is the use of crypt-222 a way to increase fluoride nucleophilicity? Choose the correct
answer: Crypt-222 provides a chelation of potassium ion enhancing the fluoride
nucleophilicity.
 Crypt-222 provides a chelation of fluoride ion enhancing its
nucleophilicity.
 Crypt-222 selectively traps 18F– and enhances the radiofluorination yield.
 Crypt-222 chelates E, promoting the substitution by a fluoride.
C18 solid phase extraction is a purification process that allows for the separation of polar
compounds and non-polar compounds, the latter being first retained on a cartridge when carriedout with water. The non-polar compounds are then eluted with an organic solvent.
16. In the process of [18F]-FDG production, the first wash (or elution) is carried out with aslightly acidic aqueous phase (pH around 3) followed by the second wash with acetonitrile.
Choose the correct statement among the following (a list of pKa values is given at the endof the problem): Crypt-222 is eluted in first place, then [18F]-FDG.
 [18F]-FDG is eluted in first place, then Crypt-222.
 Crypt-222 is retained in the C18 column, meanwhile glucose derivatives
can be eluted.
17. Considering that 18F– is the limiting reagent, both [18F]-FDG and a monosaccharide are
obtained after acid hydrolysis. Draw the structure of this monosaccharide. Why is there noneed to separate this monosaccharide from [18F]-FDG before injection for in vivo imaging?
The whole process from the selected form of E to J takes 30 minutes and presents a 75%
chemical yield.
18. Calculate the minimal molar amount of 18F– required at the beginning of the process toallow one injection to human for imaging purposes.
Data at 298 K:
AcOH/AcO–

pKa values
4.8

ROH/RO–

15.5 – 17
5.23

pyridinium/pyridine
7.43
2,4,6-collidinium/2,4,6-collidine

8.5
monoprotonated Crypt-222/Crypt-222
51st IChO – Preparatory problems

60

Problem 21. Catalysis and stereoselective synthesis ofcobalt glycocomplexes
Stereoselective synthesis of coordination compounds with chiral ligands is a blossoming field. Thisis mainly due to their potential applications in asymmetric catalysis and drug design. One strategy takes
advantage of the sugar scaffold diversity by appending Lewis bases at selected positions around thesugar cycle (F. Cisnetti et al., Dalton Trans., 2007 and F. Bellot et al., Chem. Commun., 2005).

Stereoselective synthesis of cobalt complexes
Glycoligand L and its epimer L’ (on the C3 carbon) were first synthesized. Then, theircorresponding cobalt complexes P and P’ were prepared:
O
N O
N

O O
3

+
O

N

Co(ClO4)2.6H2O+

NH4PF6

P

+

salts

O

L

Below are gathered some results concerning physical and chemical studies on P and P’
complexes.
Crystal structure of P (X-ray diffraction)
Elemental analysis of a neutral salt of P
C: 38.50%; H: 3.71%; N: 4.99%;
Co: 7.00 %; P: 7.35 %.

Counter-ions have been deleted

Visible spectrum of P (*)λmax = 515 nm (ε = 50 L mol‒1 cm‒1)(*) cP = 10‒2 mol L‒1 in an ethanol/acetone(1/1) mixture
Spin state of P
S = 3/2

Scheme of thecoordination sphere

1. Draw L’, the epimer of glycoligand L on C3, using the Cram representation.
2. Thanks to the elemental analysis of a neutral salt of P, give the oxidation number of cobalt
in P. Deduce the electronic configuration of the corersponding free ion.
3. Name the type of transition observed in the visible spectra of P.
4. Draw and fill the d orbital diagram of P consistent with the measured spin state. Name thetype of field created by the ligand L (Low field/High field).
5. Draw the coordination sphere of P’ (as shown in the inset).
SOD-like activity
Cobalt complex C’’ synthesized from glycoligand L’’ was tested for its potential superoxide
dismutase-like (SOD-like) activity as this activity presents a pharmaceutical interest to protect
51st IChO – Preparatory problems

61

against oxidative stress. A modified McCord-Fridovich assay was performed based on the
kinetic competition for the reaction of superoxide with C’’ or ferricytochrome-C.
S1 is the slope of the kinetic plot for the cytcFeIII reduction before the introduction of thecomplex and S2 is the slope recorded after the addition of putative SOD-mimic.

O

N

O

O
O

O
N

O
N

L''

6. Draw the Lewis structure of superoxide radical anion O2•ˉ.
7. Write the redox reaction for the dismutation of superoxide radical that yields dioxygen andhydrogen peroxide.
8. Determine the half maximal inhibitory concentration, IC50, of C’’ that corresponds to theinhibitor concentration needed to inhibit half of the response without inhibitor.

Problem 22. Structural study of copper (II) complexes
Stoichiometry and molecular formula
The stoichiometry of a complex can be determined by various methods. One of them is Job’s
method, also known as the method of continuous variation. This method will be used to findthe formula of the amine aqua copper (II) complex Z.
1. Copper sulfate (CuSO4) in aqueous solution is blue. Write down the formula of thecomplex responsible for the color of the solution and the approximate wavelength λ1 atwhich this complex absorbs.
We consider the reaction between the hexaaqua copper complex and ammonia in water. Its
equilibrium constant is K°. Let us assume that n ammonia ligands replace water in thecoordination sphere of the metal ion. The corresponding reaction equation is then:Cu(H2O)62+ + n NH3 = Cu(H2O)6-n (NH3)n2+ + n H2O

K°

Several solutions were prepared by mixing a copper sulfate solution (c0 = 0.044 mol L‒1), anammonia solution (c0 = 0.044 mol L‒1) and 2.0 g of ammonium nitrate NH4NO3. The
absorbance of each solution was measured at λ1. The blank solution contained NH4NO3 inwater.

51st IChO – Preparatory problems

62

Solution
Cu2+: x mL
NH3: (20-x) mL
ASolution
Cu2+: x mL
NH3: (20-x) mL
A

1
2.50
17.50
0.224
7
4.50
15.50
0.319

2
3.00
17.00
0.262
8
5.00
15.00
0.309

3
3.50
16.50
0.305
9
5.50
14.50
0.295

4
3.75
16.25
0.327
10
6.00
14.00
0.275

5
4.00
16.00
0.329
11
12.0
8.00
0.134

6
4.25
15.75
0.326
12
20.0
0.00
0.080

2. Show that, if the absorbance of ammonia and sulfate ion can be neglected at λ1, thecorrected absorbance A’ can be written:
A’ = A – x/20 A12 = (εZ - εCu) × [Z] × l
where A12 is the absorbance of the 12th solution, εZ the molar absorption coefficient of the amine
aqua copper (II) complex Z, εCu the molar absorption coefficient of free copper ion and [Z] theamine aqua copper (II) complex Z concentration, and l the cuvette path length.
3. Plot the corrected absorbance A’ with respect to x.
4. Assuming that the copper ion is the limiting reagent, determine the corrected absorbance
A’ with respect to x.
5. Assuming that the ligand is the limiting reagent, determine the corrected absorbance A’
with respect to x.
6. Show that the intersection of the two straight lines occurs when xmax = 20/(1+n).
7. Deduce the molecular formula of the amine aqua copper (II) complex Z.
Electronic study of the complexes
8. Assuming a regular octahedral frame of the ligands around the copper center, draw andfill the electronic levels of the d orbitals along an energetic axis.
9. Draw and fill the diagram if a manganese (II) ion is used instead of a copper (II) ion(Mn(H2O)62+ complex). Give the maximum value of the spin for this complex.
10. Draw and fill the diagram if cyano ligands are used instead of water ligands (Mn(CN)64‒
complex). Give the maximum value of the spin for this complex.
11. Using arrows, illustrate the evolution of the electronic levels of the copper (II) complex
when water is replaced by ammonia during the transformation studied in question 1. The
involved ligands will be taken on the z axis.

51st IChO – Preparatory problems

63

Problem 23. Synthesis and study of a molecular motor
In 2016, the Nobel Prize in Chemistry was jointly awarded to J.-P. Sauvage, J. F. Stoddart and
B. L. Feringa for the “design and synthesis of molecular machines”. Within this area, the synthesis ofa new type of molecular motor which allows the conversion of an electron flow into a controlledunidirectional rotary motion was reported in 2008 (G. Vives et al., Tetrahedron, 2008).

The structure of this molecular motor is based on a ruthenium (II) complex featuring a
cyclopentadienyl ligand (the rotating subunit called a rotor, in blue) and a tris(indazolyl)borate ligand(motionless subunit called a stator, in green) functionalized with thioether groups, which provide a tightanchor on gold surfaces. In this problem, the preparation of the ruthenium complex G, the key
intermediate of the synthesis, will first be examined. The synthetic sequence leading totris(indazolyl)borate ligand Q will then be detailed and the redox properties of the whole molecularmotor will finally be addressed.

Synthesis of the intermediate ruthenium complex G

1. Draw the structure of the first reaction intermediate resulting from the action of KOH on
compound A.
2. In these types of reactions, can KOH be used in catalytic amounts? (Yes/No)
3. Draw the structure of a possible reagent D that would lead to the formation of the alcohol
E from C.
In the third step (E → F + F’ + F’’), compound E is treated with HBr in glacial acetic acid togive the corresponding product as a mixture of three regioisomers (and their enantiomers) F,
F’ and F’’ (molecular formula: C36H23Br5).

51st IChO – Preparatory problems

64

4. Select the appropriate type of mechanism involved in the step E → F + F’ + F’’ among thefollowing choices: Electrophilic aromatic substitution
 Nucleophilic aromatic substitution
 Unimolecular nucleophilic substitution SN1
 Bimolecular nucleophilic substitution SN2
5. Give the structure of the reaction intermediate accounting for the formation of these three
regioisomers.
6. Draw the structure of the three isomers F, F’ and F’’.
In the fourth step (F + F’ + F’’ → G), the three regioisomers are reacted with the ruthenium
cluster Ru3(CO)12 to give the ruthenium (II) complex G, the key intermediate in the synthesisof the molecular motor. The evolution of a gas is observed during this reaction.
7. A metallic cluster is a structure involving at least three metal atoms linked via metal-metalbonds. Give the oxidation state of the ruthenium atoms in the cluster Ru3(CO)12.
8. Write the electronic configuration of ruthenium in the free ion corresponding to G.
9. Write a balanced equation for the formation of complex G starting from F and Ru3(CO)12.
Synthesis of the tris(indazolyl)borate ligand

10. Draw the compound which is formed as an intermediate in the first step (H → J).
11. Draw the Lewis structure of the 3-methylbutylnitrite reagent involved in the second step
(J → K).
12. Select appropriate experimental conditions for the third step (K → M).
 NaBH4 in ethanol/water (vol. 50/50)
 LiAlH4 in diethyl ether
 CrO3, H2SO4 in water PTSA in toluene (COCl)2, DMSO, NEt3 in dichloromethane
13. Draw the structure of compound N.

51st IChO – Preparatory problems

65

Alternatively, alcohol M may have been converted into a mesylate by the reaction of M withmethanesulfonyl chloride CH3SO2Cl in the presence of pyridine. However, the efficiency ofthis reaction is limited by the formation of a by-product having the molecular formula
C10H12N2O5S2.
14. Draw the structure of this by-product.
In the fifth step (N → P), the nucleophilic substitution is carried out in the presence of DBU(1,8-diazabicyclo[5.4.0]undec-7-ene).
15. What is the role of DBU in this reaction? Choose the appropriate answer from the following
list: nucleophile electrophile Brønsted base
 Brønsted acid oxidizing agent
 reducing agent
16. Write a balanced equation of the reaction accounting for the formation of potassiumtris(indazolyl)borate Q starting from functionalized indazole P and potassium borohydride.
Redox properties of the molecular motor
The design of the molecular motor involves electroactive groups such as ferrocenes, positioned
at the extremities of four of the rotors arms. These ferrocene groups play an important role inthe control of the rotary motion when the motor is submitted to an electron flow. Cyclic
voltammetry studies of this molecular motor showed that the redox potential of ruthenium ishigher than the redox potential of the iron centers.

17. Give the oxidation state of the iron center in each of the ferrocene groups.
18. The four iron centers should be oxidized selectively. Order the standard potential of an
appropriate oxidant (named “Ox”) regarding to the ones of the ruthenium and iron ions.
19. Fill in the scheme given below with the oxidation state of each metallic center and thecharge of the complex after selective oxidation of the four iron centers.

51st IChO – Preparatory problems

66

Problem 24. Some steps of a synthesis of cantharidin
Cantharidin is a terpenoid that some beetles secrete. Several medicinal uses have been known since
the ancient times, but its isolation by the French chemist P Robiquet in 1810 was a milestone in its
rigorous study. It is now recognized to be a strong poison, especially for horses, and incidentally as a
medication to remove warts. Some steps of the synthesis achieved in 1951 by the Belgium chemist
G Stork are studied in this problem.

1. Draw the structures of A and B.
2. Is A optically active?
 yes
 no
3. Draw the most stable conformation of butadiene.
4. Draw the 3D structure of C (obtained as a single diastereomer).
5. Draw the 3D structure of the transition state yielding product C.
Several steps convert C to D, but they are not studied in this problem.
6. Draw the structure of E and F as a mixture of isomers.

51st IChO – Preparatory problems

67

7. How could we thermodynamically favor the formation of F?
 by heating
 by using anhydrous magnesium sulfate.
 by cooling down
 by using an oxidizing agent
8. Draw the structure of G as a mixture of isomers.

Problem 25. Study of ricinoleic acid
Ricinus seeds contain about 50 to 70% of triglyceride whose fatty acid chains are composed ofnearly 90% of ricinoleic acid. Oleic and linoleic acids are also present in smaller amounts (respectively
around 4 and 3% of fatty acid chains). Fatty acids are carboxylic acids with long aliphatic chains. There
are two main classes of fatty acids: saturated fatty acids and mono- or polyunsaturated fatty acids. Whenone or more double bonds are present, their position on the chain and their configuration are specified
according to the IUPAC rules.

1. Give the general formula of a saturated fatty acid.
Ricinoleic acid is a fatty acid of formula C18H34O3. Its carbon chain is unbranched. It has a
C9=C10 unsaturation and a stereogenic center at C12 with R configuration.
Partial spectroscopic data:
1
H NMR (CDCl3, 300 MHz): the coupling constant measured between the two ethylenic
protons at 5.53 and 5.40 ppm is 7.8 Hz.
IR (σ, cm−1): 1711; 3406.
2. Draw the structure of ricinoleic acid based on the spectroscopic data.
3. Give the number of stereoisomers of ricinoleic acid.
4. Justify the attribution of the stereodescriptor R to C12 by classifying the substituents inorder of their priority.
A total synthesis of racemic ricinoleic acid has been published in 1955 by L. Crombie andA. G. Jacklin according to this scheme. The synthesis of B from A can be done by a
Reformatskii reaction (R-Br with zinc).

51st IChO – Preparatory problems

68Some data about used compounds:A:- mass composition of A: %C = 74%; %H = 12%; %O = 14%- 1H NMR (δ, ppm in CDCl3, 300 MHz): 9.7 (s, 1H), 2.1 (m, 2H), 1.6 (m, 2H), 1.3 (m, 6H), 0.9(t, 3H)B:- general structure:
I-(CH2)n-Cl: %I = 52%; %C = 29%; %Cl = 14%; %H = 5%
5. Give the structures of A and B.
6. Choose the correct sentences.
 the mixture obtained while forming B would rotate light the mixture obtained while forming B would not rotate light B contains one stereogenic carbon
 B contains two stereogenic carbons
 the reaction is stereoselective
 a 50/50 R/S mixture is obtained
7. Choose the reactants that could have been used in this sequence as an alternative to 3,4dihydro-2H-pyran: benzyl bromide PhCH2Br
 ethyl iodide C2H5I
 trimethylsilyl chloride Me3SiCl thionyl chloride SOCl2
 hex-1-ene n-BuCH=CH2
8. Draw the structure of an oxonium involved as a reaction intermediate in the step B → C.
9. Draw the structure of C.
10. Write the chemical equation for the reaction C + Na+, NH2‒.
11. Draw the structure of the transition state involved in the step D → E. The main carbon
chain can be represented as an R group. Give, in the transition state, the geometry of the
carbon atom at which the reaction takes place.
12. Draw the structures of D, E, F, and G.
13. Write the balanced equation of the first step of F → G transformation. During a later stepof this transformation, a gas molecule is produced. Write the formula of this molecule.
14. What are the stereochemical characteristics the step G → H?
 Stereospecific Stereoselective
 Enantiospecific Diastereoselective

51st IChO – Preparatory problems

69

Problem 26. Synthesis of oseltamivir
Oseltamivir is the active ingredient in an antiviral drug (Tamiflu®) used for the prevention andtreatment of influenza A and B. There are several ways to synthesize oseltamivir. The route proposed
below (Karpf-Trussardi synthesis) is based on (‒)-shikimic acid, a stereoisomer of 3,4,5trihydroxycyclohex-1-ene-1-carboxylic acid, that is a precursor for many syntheses in plants. The firststeps are studied in this problem. A series of reactions (not shown) follows the opening of the epoxide
E, and finally lead to the formation of oseltamivir.

1. Give the number of stereoisomers of (‒)-shikimic acid.
2. Give the meaning of the symbol (‒) in (‒)-shikimic acid.
 It is the sign of the specific rotatory power of shikimic acid.
 Shikimic acid is levorotatory.
 Shikimic acid is dextrorotatory.
 Shikimic acid rotates the polarization plane to the right when the observer
faces the source.
 Shikimic acid rotates the polarization plane to the left when the observer
faces the source.
 It gives the absolute configuration of the carboxylic acid function of
shikimic acid.
3. Give the structure of A.
4. Draw the structure of a chlorinated compound involved as a reaction intermediate in theformation of A.
The 1H NMR spectrum of compound A is partly described below (the OH signals have not beenreported). The two hydrogen atoms linked to C6 are diastereotopic and appear as two different
signals.
A (δ, ppm in CDCl3, 400 MHz): 6.78 (1H, m), 4.37 (1H, m), 4.20 (2H, q, 7.3), 4.00 (1H, dt, 7.2
and 5.2), 3.69 (1H, dd, 7.2 and 4.0), 2.70 (1H, dd, 18.4 and 5.2), 2.21 (1H, dd, 18.4 and 5.2),
1.28 (3H, t, 7.3)
5.

Assign all the 1H NMR signals to the corresponding hydrogen atom(s) of A.

51st IChO – Preparatory problems

70

Data about compound X:- Mass composition: %C = 70; %H = 12; %O = 18
- 1H NMR (δ, ppm in CDCl3, 300 MHz): 2.42 (q, 2H), 1.05 (t, 3H)- Partial IR spectroscopic data: intense absorption at 1715 cm-1
6.

Identify X, draw its structure and draw the structure of B.

7.

Draw the structure of a hemiacetal involved as a reaction intermediate in the formation ofB.

8. Indicate the synthetic interest(s) of the transformation A  B:
 Protect the C3 hydroxy group Protect the C3 and C4 hydroxy groups
 Hinder one face of the six-membered ring by a bulky group Prepare the derivatization of the C3 hydroxy group
 Increase the reactivity of the C5 hydroxy group Differentiate chemically the 3 hydroxy groups
9.

Draw the structure of C.

The opening of the ketal C produces a mixture of mesylates (D, D’ and D” in proportion 10/1/1)
among which only one, D, will serve for the next part of the synthesis. D (C15H26O7S) has avibration band around 3300 cm‒1. The by-product D’ is an isomer of D. It also has a vibration
band around 3300 cm‒1 but does not lead to an epoxide under the proposed conditions. D” isnot an isomer of D but has also a vibration band around 3300 cm‒1. Unlike D’, it can undergo
epoxidation under the proposed conditions. The product then obtained, E”, has also a vibration
band around 3300 cm‒1.
10. Draw E with the stereochemistry of all stereochemical centers.
11. Using the given information and a retrosynthetic analysis, draw the structures of D, D’,D”, and E” (including their stereochemistry).
12. Draw F with the stereochemistry of all stereochemical centers.
13. Give the stereochemical descriptors of the stereogenic centers of oseltamivir.

Problem 27. Formal synthesis of testosterone
Testosterone is a hormone, which is a biologically active substance that has signaling properties,
and is produced in a living organism. It is observed in most vertebrates, both in female and maleorganism. Its effects on health are so important that testosterone is included in the World HealthOrganization’s list of essential medicines. Some steps of a formal synthesis of this molecule are studied
in this problem.

51st IChO – Preparatory problems

71

1. Is S optically active? (Yes/No)
2. Draw the structure of A (two diastereomers).
3. Draw the structure of the non-isolated intermediate obtained after the addition of MsCl on
A.
4. Draw the structure of B and B’.
5. Draw the 3D structure of C and the structure of the transition state yielding C (only B isreactive).
6. Draw the structure of D.
7. Circle the most acidic protons of D.
In a mixture of KOH in MeOH, D yields two products. E is the thermodynamic product and E’
the kinetic one. E has three 6-membered rings and one 5-membered ring, while E’ has two 6membered rings and two 5-membered rings.
8. Draw the structure of these two products.

51st IChO – Preparatory problems

72

Back to 1990: Aqueous solutions of copper salts
This problem derives from the 22nd IChO that took place in Paris in 1990. It is not a preparatory
problem, but it is reported here as a reminiscence of the last IChO organized in France.

About the acidity of the hydrated Cu2+ ion and the precipitation of the hydroxide
Consider a 1.00·10−2 mol L‒1 solution of copper (II) nitrate. The pH of the solution is 4.65.
1. Give the equation for the formation of the conjugate base of the hydrated Cu2+ ion.
2. Calculate the pKa of the corresponding acid-base pair.
3. The solubility product of copper (II) hydroxide is Ksp = 1·10‒20. Calculate the pH of
precipitation of Cu(OH)2 in the solution under consideration. Justify your calculation,
showing that the conjugate base of this hydrated Cu2+ ion is present in negligible quantity.
Disproportionation of copper (I) ions
The Cu+ ion is involved in two redox couples:- couple (1): Cu+(aq) + e− = Cu(s)- couple (2): Cu2+(aq) + e− = Cu+(aq)standard potential E1° = 0.52 V
standard potential E2° = 0.16 V

4. Write down the equation for the disproportionation of copper (I) ions and calculate thecorresponding equilibrium constant.
5. Calculate the composition in mol L‒1 of the solution obtained on dissolving 1.00·10−2 molof copper (I) in 1.0 L of water.
6. Name two chemical species other than Cu+ which disproportionate in aqueous solution;
write down the equations of the reactions involved and describe the experimental
conditions under which disproportionation is observed.
We now examine the stability of copper (I) oxide Cu2O in contact with a 1.00·10‒2 mol L‒1solution of Cu2+ ions. The solubility product of copper (I) oxide is: Ksp = [Cu+][OH‒] = 10‒15
7. Calculate the pH at which Cu2O becomes stable.
8. Quote a simple experiment allowing the observation of Cu2O precipitation.
Complex formation involving Cu+ and Cu2+ ions9. The dissociation constant of the complex ion [Cu(NH3)2]+ is KD1 = 1·10‒11. Calculate thestandard electrode potential E3° of the couple: [Cu(NH3)2]+(aq) + e− = Cu(s) + 2 NH3(aq)10. The standard electrode potential of the couple [Cu(NH3)4]2+(aq) + 2 e− = Cu(s) + 4 NH3(aq)is E4° = ‒0.02 V. Calculate the dissociation constant KD2 of the complex ion [Cu(NH3)4]2+.
11. Deduce the standard electrode potential E5° of the couple:[Cu(NH3)4]2+(aq) + e− = [Cu(NH3)2]+ (aq) + 2 NH3(aq)12. Does the disproportionation of the cation [Cu(NH3)2]+ take place? (Yes/No)

51st IChO – Preparatory problems

73

Practical problems

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 anddisposal instructions as regulations are different in each country. Mentors must carefully adapt
the problems accordingly.

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 would fail to respect these rules will not be allowed to enter the lab.

51st IChO – Preparatory problems

74

Problem P1: Synthesis of dibenzylideneacetone
In this task, you will synthesize dibenzylideneacetone (DBA) through an aldol condensation,
starting from acetone and benzaldehyde.

DBA synthesis through aldol condensation

Chemicals
Sodium hydroxide

corrosive

Benzaldehyde

harmful by inhalation

Acetoneflammable

95% Ethanol

flammable

Ethyl Acetateflammable

TLC Eluent
flammable(Cyclohexane/Ethyl
acetate 3:1)

H290-H314; P260-P280-P303 + P361 +P353-P304 + P340 + P310-P305 + P351 +P338
H302 + H312-H315; P264-P270-P280-P301+ P312 + P330-P302 + P352 + P312-P501
H225-H319-H336; P210-P233-P261-P280P303 + P361 + P353-P370 + P378
H225-H319; P210-P233-P280-P303 + P361
+ P353-P337 + P313-P370 + P378
H225-H319-H336; P210-P233-P261-P280P303 + P361 + P353-P370 + P378
H225-H304-H315-H319-H336-H410 ; P210P233-P261-P273- P280-P301 + P310-P331P501-P303 + P361 + P353- P370 + P378

Glassware and equipment
1 Two-neck round-bottom flask, 250 mL
1 Graduated cylinder, 10 mL
1 Graduated cylinder, 100 mL
1 Erlenmeyer flask, 50 mL
1 Erlenmeyer flask, 100 mL
1 Büchner flask, 500 mL
1 Dropping funnel
1 Weighing dish1 Petri dish1 Weighing balance (0.01 g)
1 Condenser
1 Large Büchner funnel
1 Crystallizing dish1 Transfer funnel
1 Thermometer
1 Magnetic stirrer
1 Magnetic rod
1 Laboratory stand
3 TLC sampling vials
1 TLC chamber
1 TLC sheet (with fluorescence indicator)

51st IChO – Preparatory problems

75Filter paper
Pasteur pipettes
SpatulaTLC capillaries
Bossheads and clamps
UV lamp (for TLC)
Laboratory oven (80 °C)
Procedure
1. Clamp the 250 mL two-neck round-bottom flask and add 35 mL of de-ionized water to theflask. Insert the magnetic rod and transfer ca. 3.2 g of sodium hydroxide to the flask. Stirvigorously.
2. Once the dissolution is complete, add 30 mL of 95% ethanol. Let the flask cool down to
20-25 °C using an ice-water bath. Place an ice filled crystallizer beneath the flask and keep
the stirring on until temperature goes down between 20-25 °C.
3. Assemble the condenser on the main neck of the round-bottom flask and the dropping
funnel on the side neck. Be careful that the tap of the dropping funnel is closed.
4. Prepare a solution by mixing 7.6 mL (7.9 g) of benzaldehyde and 2.8 mL (2.2 g) of acetone
in a 50 mL Erlenmeyer flask.
5. Once the mixture is homogenous, transfer it to the dropping funnel. Gently pour half of itin the round-bottom flask. After a few minutes, a yellow blurring appears followed by ayellow puffy precipitate.
6. After 15 minutes of stirring, add the second half of the benzaldehyde-acetone mixture
dropwise.
7. Keep the reaction medium under stirring for 15 minutes.
8. Collect the product by filtration using a Büchner funnel and wash the yellow solid with 3
50 mL portions of (cold) distilled water. Let the solid dry for 5 minutes.
9. Recrystallize the crude product in ethyl acetate (ca. 20-25 mL are needed) in the 100 mL
Erlenmeyer flask.
10. Collect the recrystallized product by filtration. Let the product dry on the Büchner funnel
for 5 minutes.
11. Weigh a Petri dish and record the value. Transfer the recrystallized product to the Petri
dish and let it dry in a laboratory oven (80 °C).
12. Weigh the dried product et calculate the yield.
13. Perform a thin layer chromatography using the recrystallized product and the givenreferences for benzaldehyde and DBA. The eluent is a cyclohexane/ethyl acetate mixture(3:1). Report the Rf values of each compound and check the purity of the recrystallized
DBA.

51st IChO – Preparatory problems

76Problem P2: Oxidation of (‒)-borneol to (‒)-camphor
In this task you will perform the synthesis of (‒)-camphor by oxidation of (‒)-borneol, using
potassium monopersulfate (MPS) and sodium chloride. MPS, commercialized as OxoneTM, has beenchosen as it is both a strong oxidizing reagent and a stable solid quite easy to handle. Furthermore, theproduced sulfate salts are non-toxic.

Borneol oxidation using OxoneTM

Chemicals
Deionized water(‒)-borneol (2.0 g, 13 mmol) Flammable
Sodium chloride (NaCl, 0.2 g)
OxoneTM(potassium Strong oxidizer
monopersulfate, MPS) (4.8 g)
Sodium sulfite
Anhydrous magnesium (orsodium) sulfate
Ethyl acetate (50 mL)
Flammable

H228-H317; P210-P280
H314; P260-P280-P303 + P361 + P353P304 + P340 + P310-P305 + P351 + P338

H225-H319-H336;
P210-P233-P261P280-P303 + P361 + P353-P370 + P378

Starch iodide paper
Glassware and equipment
1 Round-bottom flask (or Erlenmeyer flask), 100 mL
1 Pear-shaped round-bottom flask
1 Magnetic rod
1 Magnetic stirrer
1 Transfer funnel
Filter paper
1 Glass rod
1 Graduated cylinder, 50 mL
1 Graduated cylinder, 10 mL
3 Erlenmeyer flasks, 100 mL
1 Separatory funnel + stopper, 125 mL
1 Sublimation apparatus
1 Laboratory standWeighing dishes
1 Petri dishSpatulaBossheads, ring and clamps
Rotary evaporator
Melting point apparatus

51st IChO – Preparatory problems

77Procedure
1. Clamp the 100 mL round-bottom flask. Add 2.0 g of (‒)-borneol and 10 mL of ethyl
acetate. Insert the magnetic rod and stir to dissolve.
2. With continued stirring, add 4.8 g of OxoneTM to the flask, and then 0.16 g of sodium
chloride and 3 mL of deionized water.
3. Stir vigorously the reaction at room temperature for 50 minutes. Add 0.06 g more of NaCland keep stirring 10 minutes more. Reaction should be complete and excess oxidant has
to be reduced before the extraction of camphor.
4. Add 30 mL of deionized water into the flask and two spatula tips of sodium sulfite. Keepstirring until most of the salts are dissolved. Test aqueous phase with starch iodide paper(dip a glass rod into the aqueous phase and touch the starch paper; a black color reveals thepresence of remaining oxidant). If the test is positive, add another spatula tip of sodium
sulfite and repeat starch iodide paper test, until no color appears.
5. Transfer all the content of the reaction flask in a separatory funnel and separate the phases.
Extract (three times) the aqueous phase with 10 mL of ethyl acetate.
6. Dry the combined organic phases over anhydrous magnesium (or sodium) sulfate. Filter
by gravity into a pre-weighed evaporating round-bottom flask and remove the solvent witha rotary evaporator. Record the mass and the melting point of the crude white solid.
7. Purify the crude solid by sublimation. Record mass and melting point of purified camphor.
8. Calculate the yield of the synthesis.
Note
The following skills will not be asked during the competition:- use a separatory funnel and perform extraction using immiscible solvents;- use a rotary evaporator;- sublimation;- use a melting point apparatus.

51st IChO – Preparatory problems

78

Problem P3: Aspirin® tablet
Acetylsalicylic acid has been used as a drug since ancient Egypt. It has beensynthetized for the first time in 1853 by the French chemist C Gerhardt fromsodium salicylate and acetyl chloride, but the compound thus obtained was
unstable and not pure enough. The preparation method was improved in thefollowing years and in the end F Hoffman, a German chemist and employee ofBayer, succeeded in the total synthesis of the pure compound.
This compound was then marketed under the name Aspirin®, now worldwide
known, by Bayer. The patent and the trademark were deposited 120 years ago, in
1899.
It is used to treat pain, fever, or inflammation, and is as well an antiplatelet drug.
For its 120th anniversary, Aspirin® is still one of the most widely used medications, with anestimated 44 000 tons produced and 120 billion pills consumed each year. Bayer is still responsible for
85% of this production.
It is on the World Health Organization's List of Essential Medicines as one of the safest and mosteffective medicines needed in a health system.

The proposed task aims to determine the amount of aspirin contained in a commercial tablet thanks
to a back titration, using sodium hydroxide solution. A saponification reaction is first performed and theexcess of sodium hydroxide is titrated with hydrochloric acid.
The second part of the task consists of the optimization of TLC eluent in order to monitor thissaponification reaction.

Chemicals
Deionized water
Acetylsalicylic acid
Salicylic acidEluent A (Pure cyclohexane)

flammable H225-H304-H315-H336-H410; P210-P261P273-P301 + P310-P331-P501
Eluent B (Pure ethyl acetate)
flammable H225-H319-H336; P210-P233-P261-P280P303 + P361 + P353-P370 + P378
Eluent C (Mixture 65:30:5 of flammable H225-H304-H315-H319-H336-H410 ; P210 cyclohexane: ethyl acetate:
P233-P261-P273-P280-P301 + P310-P331acetic acid)P501-P303 + P361 + P353- P370 + P378®
Aspirin tablets (500 mg ofacetylsalicylic acid)
Phenolphthalein
indicator
solutionStandardized
0.200 M
hydrochloric acid solution
Sodium hydroxide (pellets) corrosive H290-H314; P260-P280-P303 + P361 + P353(M = 40.00 g mol‒1)
P304 + P340 + P310-P305 + P351 + P338
Acetoneflammable H225-H319-H336; P210-P233-P261-P280P303 + P361 + P353-P370 + P378

51st IChO – Preparatory problems

79

Glassware and equipment
1 Volumetric flask (with stopper), 100 mL
1 Weighing dish1 Spatula
1 Weighing balance (0.1 mg)
1 Transfer funnel
1 Volumetric pipette, 20 mL
1 Volumetric pipette, 10 mL
1 Pipetting bulb
1 Hotplate (with magnetic stirring)
1 Erlenmeyer flask, 100 mL
1 Air condenser
1 Magnetic rod
1 Burette, 25 mL
1 Laboratory stand with burette clamp
Bossheads and clamps
2 Titration flasks, 100 mL
1 Titration flask, 250 mL
1 Stopwatch
2 TLC vials (for sampling)
TLC capillaries
1 TLC chamber
3 TLC sheets (with fluorescence indicator)
Beakers (for transfers)
Procedure for back titration
1. Prepare 100 mL of ca. 0.4 M sodium hydroxide solution, using ca.1.6 g of solid sodiumhydroxide. This solution is called SB.
2. Clamp the 100 mL Erlenmeyer flask to the laboratory stand. Insert the aspirin tablet andadd 20.00 mL of the sodium hydroxide solution SB prepared. Insert the magnetic rod andheat the reaction mixture under reflux with stirring for 15 min.
3. During the reflux, fill the burette with the 0.200 M hydrochloric solution provided.
4. Transfer 10.00 mL of solution SB in the 100 mL titration flask. Add a few drops of thephenolphthalein Indicator Solution. Titrate with the 0.200 M hydrochloric acid solution.
Record the volume V1 and repeat the titration as necessary.
5. After the 15 minutes of reflux, let the Erlenmeyer flask cool down to room temperature.
Transfer the whole content of the Erlenmeyer flask to a 250 mL titration flask and rinse
with deionized water (pour the rinsing water in the titration flask).
6. Add a few drops of the phenolphthalein Indicator Solution. Titrate with the 0.200 M
hydrochloric acid solution. Record the volume V2.
7. Repeat the procedure (1., 2., 5. and 6.) with another aspirin tablet.
8. Calculate the concentration of the sodium hydroxide solution SB.
9. Calculate the amount (in mg) of aspirin in one tablet.

51st IChO – Preparatory problems

80TLC Optimization
10. Prepare TLC samples of acetylsalicylic acid and salicylic acid in acetone.
11. Prepare a TLC sheet by spotting the acetylsalicylic acid and the salicylic acid samples.
12. Let the TLC sheet elute with eluent A.
13. Visualize the TLC sheet with the UV lamp.
14. Repeat the procedure (11., 12., 13.) with eluents B and C.
15. Analyze the TLC sheets and identify which of the eluents is the most appropriate to monitor
aspirin saponification.

Problem P4: Illuminated Europe
In this task you will perform a two-step synthesis of lanthanide complexes. In the first step an acidbase reaction occurs between the 2,6-pyridinedicarboxylic acid and guanidinium carbonate, leading toa salt. Then, this salt reacts with a lanthanide salt (XCl3) to give the lanthanide complex. The scheme
below shows the reactions.
These lanthanide complexes are used in Euro banknotes, as they emit a specific light under UV
irradiation.

Two-step synthesis of the lanthanide complexes

Chemicals
Deionized water
2,6-pyridinedicarboxylic
acid Irritant‒1(H2DPA); M = 167.1 g mol
Guanidinium carbonate (GuaH)2CO3;
M = 180.2 g mol‒1Europium (III) chloride hexahydrate;
M = 366.4 g mol‒1
Lutetium (III) chloride hexahydrate;
M = 389.4 g mol‒1
Terbium (III) chloride hexahydrate;
M = 373.4 g mol‒1

H315-H319-H335; P261-P305 +P351 + P338
H302-H318; P280-P305 + P351 +P338-P313
H315-H319; P305 + P351 + P338
H315-H319; P305 + P351 + P338
H315-H319; P305 + P351 + P338

Glassware and equipment
1 Erlenmeyer flask, 50 mL
1 Graduated cylinder, 25 mL
1 Büchner flask
1 Büchner funnel
51st IChO – Preparatory problems81

1 Crystallizer
1 Magnetic stirrer
1 Magnetic rod
1 Petri dish1 Laboratory stand1 Weighing balance (0.1 mg)
Laboratory oven (80 °C)
Bossheads and clamps
Weighing dishes
UV lamp (365 nm) for banknotes€50 banknote (a copy is provided below)
Filter paper
Spatula
Stopwatch
Procedure
1. Clamp the 50 mL Erlenmeyer flask. Add 0.70 g of 2,6-pyridinedicarboxylic acid and
20 mL of de-ionized water to the flask. Add 0.75 g of guanidinium carbonate and swirl theflask until dissolution of both solids.
2. Insert the magnetic rod and transfer a stoichiometric amount of lanthanide salt into theflask (1 molar equivalent of XCl3 for 3 molar equivalents of 2,6-pyridinedicarboxylic acid).
Stir the flask at room temperature for 1 hour.
3. Cool the mixture in an ice bath for five to ten minutes before filtering.
4. Collect the product by filtration using a Büchner funnel and wash the crystals with small
portions of ice-cold water. Allow the crystals to dry in Büchner funnel for five minutes.
5. Transfer the solid to the pre-weighed Petri dish and let it dry in a laboratory oven (80 °C).
Weigh the complex and calculate the percentage yield.
6. Look at the three complexes under the UV lamp. Record the fluorescence color of eachcomplex.
7. Observe the €50 banknote under the UV lamp. Identify which of the previous complexes
might be used in the ink of the banknote.

€50 banknote under UV irradiation

Note
This problem is dedicated to Europe. No specific knowledge on the lanthanide chemistry nor
the fluorescence properties of such complexes is needed for the competition.
51st IChO – Preparatory problems

82Problem P5: Protecting the vineyard
In order to protect grapes against the mildew, French wine-growers in the area of Bordeaux (southwest of France) developed the so-called “Bordeaux mixture”, which they spread around the vines. The
Bordeaux mixture is composed of copper (II) sulfate CuSO4 and slacked lime Ca(OH)2. The goal of thisproblem is to determine the copper content of the Bordeaux mixture provided.

Vineyard treated with the “Bordeaux mixture”
Picture from Pg1945, under CC BY-SA 3.0 license (Wikipedia page “Bordeaux mixture”)

Chemicals
Bordeaux mixture

H302-H315-H319-H410;
P264-P273-P280-P337 + P313-P391-P501

Standardized
0.001600 M
potassium iodate (KIO3) solution
0.0200 M sodium thiosulfate(Na2S2O3) solution
Potassium iodide (KI)1 M sulfuric acid (H2SO4)
1 M ammonia (NH3) solution

Corrosive

Standardized
0.02000 M
copper (II) sulfate (CuSO4)
solution
Deionized water

H372; P260-P264-P270-P314-P501
H290-H315-H319; P302 + P352-P305 +P351 + P338
H315-H318-H412; P280-P305 + P351 +P338 + P310
H411; P273

Glassware and equipment
1 Laboratory stand with burette clamp
1 Erlenmeyer flask, 250 mL
1 Filter paper
1 Transfer funnel
3 Titration flasks, 250 mL
1 Weighing balance (0.1 mg)
1 Burette, 25 mL
1 Volumetric flask (with stopper), 250 mL
1 Volumetric pipette, 20 mL
3 Graduated pipettes, 5 mL
1 Graduated cylinder, 50 mL

51st IChO – Preparatory problems83

1 Graduated cylinder, 10 mL
3 Weighing dishes
1 Spatula
Aluminum foil
Beakers, 100 mL (for transfers)
1 Spectrophotometer (calibrated at 610 nm)
1 UV-vis plastic absorption cuvette (l = 1.0 cm)
1 Test tube stand7 Test tubes, 15 mL
7 Plastic Pasteur pipettes, 2-3 mL
1 Pipetting bulb
Beakers (for transfers)
Procedure for the iodometric titration of copper
1. Weigh accurately ca. 1 g of Bordeaux mixture (record the mass). Transfer it to the 250 mL
Erlenmeyer flask. Add ca. 50 mL of deionized water and 5 mL of 1 M sulfuric acid. Swirlthe Erlenmeyer flask for 5 minutes (the color of the solution does not change anymore).
2. Using a filter paper and a transfer funnel, transfer the solution into the 250 mL volumetricflask. Carefully rinse the Erlenmeyer flask and the filter paper into the volumetric flask.
Fill the flask with deionized water. Homogenize the solution, which is called SBM.
3. Fill the burette with the 0.0200 M sodium thiosulfate solution.
4. Transfer 20.00 mL of the standardized 0.001600 M potassium iodate solution to a 250 mL
titration flask. Add 2 g of potassium iodide, 25 mL of deionized water and 10 mL of 1 M
sulfuric acid. Swirl until all potassium iodide gets dissolved. Stopper the titration flask andlet it stand in the dark (using aluminum foil or in a cabinet) for 5 minutes.
5. Titrate using the 0.0200 M sodium thiosulfate solution. Record the titration volume V1.
Repeat the titration as needed.
6. Using a 20 mL volumetric pipette, transfer 20 mL of solution SBM into a 250 mL titration
flask. Add 5 mL of 1 M sulfuric acid and 5 g of potassium iodide. Swirl until all potassium
iodide gets dissolved. Stopper the titration flask and let it stand in the dark (usingaluminum foil or in a cabinet) for 5 minutes.
7. Titrate using the 0.0200 M sodium thiosulfate solution. Record the titration volume V2.
Repeat the titration as needed.
Analysis
8. Write down the equations for all the reactions occurring during the standardization of
sodium thiosulfate.
9. Calculate the exact molar concentration of the sodium thiosulfate solution.
10. Write down the equations for all the reactions occurring during the iodometric titration ofcopper (preparation and titration) in solution SBM.
11. Determine the molar concentration of copper in the solution SBM.
12. Calculate the weight percentage of copper %Cu in the Bordeaux mixture.

51st IChO – Preparatory problems

84

Spectrophotometric determination of copper
To confirm the results obtained by iodometry, a spectrophotometric determination of copper as
its ammine complex is performed. A known amount of copper is mixed with an excess ofammonia solution.
Tube #
0.0200 M
copper sulfate
solution
1 M ammonia
solution
Deionized
water
Solution SBM

0

1

2

3

4

5

Bordeaux

0.0 mL0.0 mL

5.0 mL

5.0 mL

5.0 mL0.0 mL0.0 mL 0.0 mL 0.0 mL 0.0 mL 0.0 mL 0.0 mL

5.0 mL

13. Using the calculated concentration for the solution SBM (question 11), fill the previous tablewith volumes that can be used to create a calibration scale for copper.
14. Prepare all these solutions (0 to 5 and “Bordeaux”) in test tubes, using graduated pipettes
for transfers.
15. Record the absorbance value A for each solution, at 610 nm.
16. Plot the absorbance value A versus the molar concentration of copper in each tube (0 to 5).
17. Using this plot, determine the molar concentration of copper in the solution SBM.
18. Calculate the weight percentage of copper %Cu in the Bordeaux mixture. Compare with
the iodometric determination.

51st IChO – Preparatory problems

85Problem P6: Equilibrium constant determination
pH-indicators are used often used in colorimetric titrations of acids and bases. The key point whenchoosing an indicator is to find one having a pKa close to the pH at the equivalence point. Therefore, it
is very important to know accurately the pKa of such acid/base couples.
Fortunately, pH-indicators have very high absorption coefficients (usually ε > 104 cm‒1 L mol‒1).
UV-visible spectroscopy can be used to determine such constants.

Various pH indicators
Picture from TheChimist, under CC BY-SA 3.0 license (Wikipedia page “Potentiel hydrogène”)
The goal of this task is to determine the pKa of bromophenol blue (BPB) using UV-vis spectroscopy.

Chemicals
Bromophenol blue (BPB)
0.2 M hydrochloric acid and 1 M
acetic acid mixed solution (called
HCl/CH3COOH mixture)
1 M sodium acetate
95% Ethanol
flammable

H290

H225-H319; P210-P233-P280-P303 +P361 + P353-P337 + P313-P370 + P378

Deionized water
Glassware and equipment
1 Weighing balance (0.1 mg)
2 Volumetric flasks (with stopper), 250 mL
1 Volumetric flask (with stopper), 100 mL
1 Volumetric pipette, 5 mL
1 Graduated pipette, 10 mL
1 Weighing dish1 Spatula
Beakers, 100 mL (for transfers)
1 Spectrophotometer (calibrated at 590 nm)
1 UV-vis plastic absorption cuvette (l = 1.0 cm)
1 Test tube stand7 Test tubes, 15 mL
7 Plastic Pasteur pipettes, 2-3 mL
1 Pipetting bulb
1 pH-meter with pH-probe (calibrated in the acidic domain)
Beakers (for transfers)
51st IChO – Preparatory problems

86Procedure
1. Weigh ca. 0.100 g of Bromophenol Blue. Transfer it to a 100 mL volumetric flask using
95% ethanol. Dissolve the Bromophenol Blue with 95% ethanol. Homogenize the solution,
which is called S0,BPB.
2. Using a volumetric pipette, transfer 5.00 mL of S0,BPB into a 250 mL volumetric flask. Fillthe flask with the mixture of hydrochloric acid and acetic acid provided. Homogenize thesolution, which is called SA,BPB.
3. Using a volumetric pipette, transfer 5.00 mL of S0,BPB into a 250 mL volumetric flask. Fillthe flask with the 1 M sodium acetate solution. Homogenize the solution, which is called
SB,BPB.
4. For each column in the following table, prepare the solution in a test tube using thevolumes reported in the table. The stock solutions are to be transferred with graduatedpipettes.
Tube #
SA,BPB
SB,BPBpH
Absorbance A(at 590 nm)

1
0.0 mL10.0 mL2
5.0 mL
5.0 mL3
6.0 mL
4.0 mL4
7.0 mL
3.0 mL5
8.0 mL
2.0 mL6
8.5 mL
1.5 mL7
10.0 mL
0.0 mL

5. Using the pH-meter, record the pH of each tube.
6. Record the absorbance value A for each solution (1 to 7) at 590 nm.
Analysis
7. Explain why the analytical concentration of BPB is identical in all tubes. This
concentration will be referred to as 𝑐BPB .
8. Draw the plot of the absorbance A with respect to the pH.
9. Using the plot and assuming hypothesis to be verified, determine the molar absorption
coefficients 𝜀HInd and 𝜀Ind− of the acidic form HInd and the basic form Ind– of BPB as afunction of 𝑐BPB .
10. Derive the equation giving the absorbance A of the solution as a function of the 𝜀HInd,
𝜀Ind− , 𝑐BPB and the molar concentrations [HInd] and [Ind–].
11. Using the Henderson-Hasselbalch equation, derive the equation of the absorbance A for
pH = pKa.
12. Using the plot of A = f(pH), determine the value of the pKa of BPB.
Note
The following skill will not be asked during the competition:- use of a pH-meter.

51st IChO – Preparatory problems

87