Candidate: AAA-1

THEORETICAL
EXAM

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2019-07-26

51st IChO – Theoretical Exam1

Candidate: AAA-1

General instructions
This theoretical exam booklet contains 60 pages.

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You have 5 hours to complete the exam.

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

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

Use only the pen and calculator provided.



The official English version of the exam booklet is available upon request and serves forclarification only.



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and then make a new empty answer box next to it.

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After the Stop command has been given, place your exam booklet back in your exam envelope,then wait at your seat. The exam supervisor will come to seal the envelope in front of you andcollect it.

GOOD LUCK!

51st IChO – Theoretical Exam

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Candidate: AAA-1

Table of Contents

This theoretical exam is composed of 9 independent problems, as follows. Their relative weight isindicated in parenthesis.

Problem T1: Infinite well and butadiene

(6%)

p. 8

Problem T2: Hydrogen production by water-splitting

(7%)

p. 13

Problem T3: About silver chloride

(5%)

p. 19

Problem T4: From black powder to the discovery of iodine

(7%)

p. 24

Problem T5: Complexes for the formation of nanomachines

(8%)

p. 30

Problem T6: Characterization of a block-copolymer

(8%)

p. 39

Problem T7: Ring motion in a [2]catenane

(6%)

p. 47

Problem T8: Identification and synthesis of inositols

(6%)

p. 52

Problem T9: Synthesis of levobupivacaine

(7%)

p. 57

51st IChO – Theoretical Exam

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Candidate: AAA-1
Physical constants and equations
In these tasks, we assume the activities of all aqueous species to be well approximated by theirrespective concentration in mol L−1. To further simplify formulas 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:
Watt:
Kilowatt hour:
Planck constant:
Speed of light in vacuum:
Elementary charge:
Electron-volt
Electrical power:
Power efficiency:
Planck-Einstein relation:
Ideal gas equation:
Gibbs free energy:

Reaction quotient Q for a reactiona A(aq) + b B(aq) = c C(aq) + d D(aq):Henderson−Hasselbalch equation:
Nernst–Peterson equation:
where Q is the reaction quotientof the reduction half-reaction
Beer–Lambert law:
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:

51st IChO – Theoretical Exam

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 W = 1 J s−1
1 kWh = 3.6∙106 J
h = 6.6261∙10−34 J s
c = 2.998∙108 m s−1
e = 1.6022∙10−19 C
1 eV = 1.6022∙10−19 J
P = ΔE × I
η = Pobtained/Papplied
E = hc/λ = hνpV = nRT
G = H − TS
ΔrG° = −RT lnK°ΔrG° = −n F Ecell°
ΔrG = ΔrG° + RT lnQ[C]c [D]d𝑄=[A]a [B]b[A− ]pH = pKa + log
[AH]
RT
o
E = E − zF ln𝑄
RTat T = 298 K, F ln10 ≈ 0.059 V
A = εlc[A] = [A]0 − ktln[A] = ln[A]0 − kt
1/[A] = 1/[A]0 + kt
ln2
𝑘∑i 𝑁i 𝑀i𝑀n =∑i 𝑁i
∑i 𝑁i 𝑀i2𝑀w =∑i 𝑁i 𝑀i
Mw
Ip =
Mn
4

Candidate: AAA-1

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

51st IChO – Theoretical Exam

238.0

-

-

-

-

-

-

-

-

-

-

-

5

Candidate: AAA-1
1

H NMR

Chemical shifts of hydrogen (in ppm / TMS)

phenols:
alcohols:
alkenes:

CH3—CR3:

alkynes:amines::R—CH2—OR’

amide NH—COR:
CH3—NR2::carboxylic acids

CH3—SiR3::aldehydes:ketones
benzylic CHn—C6H5:

aromatics:
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
eq = equatorial, ax = axial

51st IChO – Theoretical Exam

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Candidate: AAA-1
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)C≡C (stretching)

2250
2260-2100

strongvariablealdehyde 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
1200-1020
1600-1500; 1400-1300

strongstrongstrong51st IChO – Theoretical Exam

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Candidate: AAA-1

Problem
T1
6%

Question

1

2

3

4

5

6

7

8

9

10

11

Total

Points

3

4

4

2

3

2

2

4.5

2.5

3

3

33

Score

Problem T1: Infinite well and butadiene
The buta-1,3-diene molecule is often written CH2=CH–CH=CH2, with alternating single and double
bonds. Nevertheless, its chemical reactivity is not consistent with this description and the π electrons
are better described by a distribution along the three bonds:

This system can be modeled as a 1D box (i.e. infinite well) where the electrons are free. The energy of𝑛2 ℎ 2

an electron in an infinite well of length L is: 𝐸𝑛 = 8𝑚 𝐿2, where n is a non-zero positive integer.
𝑒1.

Two different models are studied. Sketch at least the three lowest-energy levels En for eachmodel in the respective diagrams, showing how the relative energy levels differ within andbetween models.

Model 1 (« localized »): The π electrons are Model 2 (« delocalized »): The π electrons are
localized on the extremal bonds and evolve in two delocalized on the whole molecule and evolve ina single infinite potential well of length 3d.
separate infinite potential wells of length d.

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2.

Place the π electrons for model 1 in the previous diagrams and express the total energy of the πsystem in model 1, as a function of h, me and d.

E(1) =

3.

Place the π electrons for model 2 in the previous diagrams and express the total energy of the πsystem in model 2, as a function of h, me and d.

𝐸(2) =

The conjugation energy is the total energy of the actual π system, minus the sum of the energies of
ethylene molecules involving the same number of electrons.
4.

Express the conjugation energy ΔEc of butadiene, as a function of h, me and d.

∆𝐸c =Models 1 and 2 are too simplistic. A new model will be detailed in the following.
5.

Draw three other resonance structures of butadiene using Lewis notation.

To take into account the size of carbon atoms, model 2 is now modified into model 3, as follows:- the new length of the well is L and is located between the abscissa 0 and L;- the carbon atoms are located at the abscissas L/8; 3L/8; 5L/8 and 7L/8.
For each level n, the π wavefunction is:2𝑛𝜋𝑥
𝜓n (𝑥) = √ sin ()
𝐿
𝐿

51st IChO – Theoretical Exam

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Candidate: AAA-1
and the π electron density for a system with N π electrons is:
𝑁/2

𝜌(𝑥) = 2 ∑|𝜓𝑖 (𝑥)|2
𝑖=1

The four π wavefunctions, which correspond to the molecular orbitals of the π system, are depicted
below (arbitrary order).

6.

Sort the energies of the four π wavefunctions (EA, EB, EC and ED).

<

<

<

7.

Give the labels (A, B, C or D) of the orbitals that are filled with electrons in butadiene.

8.

Within model 3, give the values of the π wavefunctions 𝜓n for occupied levels at positions 0, L/4
and L/2, for n = 1 and n = 2, as a function of L.

𝜓1 (0) =51st IChO – Theoretical Exam

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Candidate: AAA-1

𝐿𝜓1 ( ) =4𝐿
𝜓1 ( ) =2

𝜓2 (0) =𝐿𝜓2 ( ) =4𝐿
𝜓2 ( ) =2

9.

Within model 3, give the value of the π electron density at positions 0, L/4 and L/2.

𝜌(0) =

𝐿
𝜌( ) =
4

51st IChO – Theoretical Exam

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Candidate: AAA-1

𝐿𝜌( ) =
2

10. Draw the π electron density between 0 and L.

11. Sort the following CC bonds (B1, B2, …, B5) by increasing length, using the symbols = or <:
B1:
B2 :
B3 :
B4 :
B5 :

C1C2 in the butadiene molecule
C2C3 in the butadiene molecule
C3C4 in the butadiene molecule
CC in the ethane molecule
CC in the ethene molecule

51st IChO – Theoretical Exam

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Problem
T2
7%

Question

1

2

3

4

5

6

7

8

9

10

Total

Points

1

4

2

3

3

6

4

1

8

2

34

Score

Problem T2: Hydrogen production by water-splitting
Data:
CompoundH2(g)H2O(l)

H2O(g)O2(g)∆fH° (kJ mol−1)0−285.8−241.80Sm° (J mol−1 K−1)

130.6

69.9

188.7

205.2

Molecular hydrogen (H2) can be used as an alternative to carbon dioxide-emitting fuels. Hence,lowering the cost and the environmental impact of its production is a major challenge. In this field,
water-splitting is a promising candidate technology.
1. Write down the balanced equation of liquid water splitting reaction using a stoichiometriccoefficient of 1 for water.

2.

Using only the provided thermodynamic data, justify numerically whether this reaction isthermodynamically favorable at 298 K.

Calculations:

Reaction thermodynamically favorable?
 Yes

 No

Water splitting can be performed electrochemically using two electrodes in an acidic water bath,connected by a generator (Fig. 1). Gas bubbles are formed at both electrodes.

51st IChO – Theoretical Exam

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Candidate: AAA-1

Fig. 1 – Water-splitting electrochemical cell.
3. Write down the balanced net electrochemical half reactions occurring at each electrode.

On electrode (1):On electrode (2):4.

Using only the provided thermodynamic data (or question 2), derive the condition on the applied
voltage Eapplied between electrodes, compared to value Eth (to determine), for the process to be
thermodynamically favorable at 298 K, when all reactants and products are in their standard state.
Tick the right condition and give the numerical value with 3 decimal places.

Calculation:

Eapplied = Eth
 Eapplied > Eth
 Eapplied < Eth

, where Eth ……….. V(give the result with 3 decimal places)
If you could not calculate Eth, the value 1.200 V
can be used in the rest of the problem.

Experimentally, a higher voltage is needed to observe water splitting. For a given Pt cathode, theminimum voltage necessary to observe water splitting, Emin, depends on the nature of the anode, as
displayed in the table below:
Anode
IrOx
NiOx
CoOx
Fe2O3

51st IChO – Theoretical Exam

Emin (V)
1.6
1.7
1.7
1.9

14

Candidate: AAA-1
The difference between Emin and Eth is responsible for losses in the device.
5.

Give the expression of the device power efficiency elec (fraction of the power used for water
splitting) as a function of Eth and Emin. Assuming an identical current value I, calculate thewater electrolysis power efficiency when a Pt cathode and a Fe2O3 anode are used. Give the mostefficient anode.

elec =

Power efficiency when a Pt and a Fe2O3 electrodes are used:elec =%

Most efficient anode:
If you could not calculate elec, the value elec = 75%
can be used in the rest of the problem.

V vs SHE

An alternative to water electrolysis is direct photocatalytic water-splitting. It uses a semiconductor thatcan be activated by absorbing light.

Name of semiconductor
Potential of equivalent cathode : Minimal light energy to activate
Potential of equivalent anodeFig. 2 – Activation condition and equivalent electrode potentials of different semiconductors. Dashedlines correspond to water oxidation and reduction potentials. SHE = Standard Hydrogen Electrode51st IChO – Theoretical Exam

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Candidate: AAA-1

Fig. 3 – Left axis: Spectral distribution of the solar photon flux . The photon flux is the number ofphotons per unit area per unit time arriving on the semiconductor. Right axis and dashed line:
cumulative photon flux (i.e. fraction of the photon flux with smaller wavelength).
6.

Estimate the fraction of the solar photon flux that can activate the following semiconductors:
TiO2, CdS, Si. State explicitly the equations and units used for the computation.

Explanation / calculation:

51st IChO – Theoretical Exam

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Candidate: AAA-1

Approximate
fraction
TiO2

%

CdS

%

Si

%

The activation of the semi-conductor results in a modification of the surface potentials, so that it canbe seen as two electrodes of different potentials.
7.

Using the data in Fig 2, choose the semiconductor(s) in the following list that, once activated, canplay both roles of anode and cathode for water-splitting reaction.

ZrO2

 ZnO

 TiO2

 WO3

CdS

 Fe2O3

 CdSe Si

8.

Give the semiconductor that, used as both cathode and anode, is expected to be the most efficientfor water splitting upon a given solar shining.

The evolution of H2 and O2 when a semiconductor is irradiated by simulated solar light at T = 25 °C at
patm was recently studied. Using an incident power light of P = 1.0 kW m−2 and a photoelectrode with a
S = 16 mm2 surface, the production of V = 0.37 cm3 of H2(g) was measured after t = 1 hour ofreaction.
9.

Calculate the power efficiency direct of the conversion.

Calculation:

51st IChO – Theoretical Exam

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Candidate: AAA-1

direct =%If you could not calculate direct, the value direct = 10%
can be used in the rest of the problem.

Two modes of converting solar energy to hydrogen can thus be compared: direct photocatalysis, andindirect photo-electrolysis combining a photovoltaic panel with an electrolyzer. The efficiency ofphotovoltaic panels on the market is around panels = 20%.
10. Compare the power efficiencies of the two modes, direct and indirect, using Fe2O3 and Pt
electrodes for the electrolysis.
Calculation: direct > indirect direct ≈ indirect

51st IChO – Theoretical Exam direct < indirect18

Candidate: AAA-1

Problem
T3
5%

Question

1

2

3

4

5

6

7

8

9

10

11

12

Total

Points

1

3

3

3

4

2

7

2

2

3

4

6

40

Score

Problem T3: About silver chloride
Data at 298 K:
pKs1(AgCl) = 9.7; pKs2(Ag2CrO4) = 12
Formation constant of the complex [Ag(NH3)n]+:  n = 107.2
Potentials against the standard hydrogen electrode:Standard potential of Ag+/Ag(s): E°(Ag+/Ag(s)) = 0.80 V
Apparent potential of O2(aq)/HO−(aq) (in seawater): E'(O2(aq)/HO−(aq)) = 0.75 V
Part A: Quotes from a chemistry lesson by Louis Joseph Gay-Lussac
The following quotes from a chemistry lesson by Louis Joseph Gay-Lussac (French chemist andphysicist, 1778–1850) deal with some properties of silver chloride.
Quote A: “I will now talk about silver chloride, a milk-white solid. It is easily obtained by pouringhydrochloric acid into an aqueous solution of silver nitrate.”
Quote B: “This salt has no taste since it is insoluble.”
Quote C: “This compound is completely insoluble in alcohol and even in acids, except inconcentrated hydrochloric acid which dissolves it readily.”
Quote D: “On the other hand, silver chloride is highly soluble in aqueous solution of ammonia.”
Quote E: “Then, we can make silver chloride appear again by adding an acid which reacts withammonia.”
Quote F: “If you take a bowl made of silver to evaporate salty seawater, you will get impure sodium
chloride, mixed with a milk-white solid.”
1. Quote A: Write the balanced chemical equation of AgCl(s) synthesis.

2.

Quote B: Calculate the solubility s of AgCl(s) in water at 298 K in mol L−1.

Calculation:s=

51st IChO – Theoretical Exam

mol L−119Candidate: AAA-1
3.

Quote C: In a highly concentrated solution of chloride ions, a well-defined complex of
stoichiometry 1:2 is formed. On the following qualitative axis (with pCl increasing from left toright), place in each domain the silver-containing species that is predominant (or exists, forsolids). pCl values at frontiers are not expected.

Quote D: When ammonia is added to silver chloride, a well-defined complex of stoichiometry n isformed.
4.

Write the balanced equation corresponding to the synthesis of the complex [Ag(NH3)n]+ fromsilver chloride and calculate the corresponding equilibrium constant.

Equation:

Calculation:K=

If you could not calculate K, the following value
can be used in the rest of the problem: K = 10‒3
5.

Ammonia is added to 0.1 mol of silver chloride in 1 L of water until the last grain of solid
disappears. At this moment, [NH3] = 1.78 mol L−1. Determine the stoichiometry of the complex
neglecting dilution effects.

Calculation:n=
51st IChO – Theoretical Exam

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Candidate: AAA-1
6.

Write the balanced chemical equation corresponding to quote E.

7.

Assuming that seawater is slightly basic and rich in dioxygen, and that silver metal can reduce
dioxygen in such conditions, write a balanced chemical equation corresponding to the formationof the solid mentioned in quote F. A stoichiometric coefficient of 1 will be chosen for dioxygen.
Calculate its equilibrium constant at 298 K.

Equation:

Calculation:

K=

Part B: The Mohr method
The Mohr method is based on the colorimetric titration of Cl− by Ag+ in the presence of potassium
chromate (2K+, CrO42−). Three drops (~ 0.5 mL) of a K2CrO4 solution at about 7.76∙10−3 mol L−1 areadded to V0 = 20.00 mL of a sodium chloride solution of unknown concentration CCl. This solution isthen titrated by silver nitrate (Ag+, NO3−) at CAg = 0.050 mol L−1, which immediately leads to theformation of solid A. A red precipitate (solid B) appears at VAg = 4.30 mL.
8.

Write the balanced equations of the two reactions occurring during the experiment. Calculate thecorresponding equilibrium constants.

51st IChO – Theoretical Exam

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Candidate: AAA-1

K°1 =

K°2 =

9.

Identify the solids.

Solid A:
Solid B:

10. Calculate the unknown concentration CCl of chloride ions in the sodium chloride solution.
Calculation:CCl =mol L–1

If you could not calculate CCl, the value CCl = 0.010 mol L−1
can be used in the rest of the problem.
11. Calculate the minimal volume VAg(min) for which AgCl(s) precipitates.
Calculation:

VAg(min) =51st IChO – Theoretical ExammL
22

Candidate: AAA-1
12. Calculate the residual concentration [Cl−]res of chloride ions when silver chromate begins toprecipitate. Justify why CrO42− is a good titration endpoint indicator by comparing two values.

Calculation:[Cl‒]res =mol L‒1

CrO42− is a good titration endpoint indicator because:

51st IChO – Theoretical Exam

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Candidate: AAA-1

Problem
T4
7%

Question

1

2

3

4

5

6

7

8

Total

Points

6

9

8

5

6

2

2

12

50

Score

Problem T4: From gunpowder to the discovery of iodine
In the 19th century, the French entrepreneur B. Courtois specialized in the production of nitrate A(MA(NO3)m), used for gunpowder. Initially imported from Asia, A was later produced from nitrate B(MB(NO3)n) using exchange reaction with compound C, obtained from algae.
1. Find the formulas of nitrates A and B knowing that they are anhydrous salts of alkaline or
alkaline-earth metal (MA and MB). One of the nitrates contains no more than 1 w% of non-metallic
impurities while the other contains 9 ± 3 w% of impurities. The content of metals MA and MB in
the samples is 38.4 w% and 22.4 w% respectively. Support your answer with calculations.

A:
51st IChO – Theoretical Examand B:
24

Candidate: AAA-1
To obtain A, 262.2 g of solid compound C were added to the solution containing 442.8 g of B. B is
known to be in excess. As a result, 190.0 g of white precipitate D were formed and removed byfiltration. The filtrate was evaporated, and the obtained solid mixture E was heated until the mass ofthe sample (containing only nitrites, NO2−) was constant. The only gaseous product was dioxygen:60.48 L at 0 °C at 1 atm (dioxygen can be considered as an ideal gas).
2.

Calculate the composition (in w%) of mixture E considering that it contained only compounds A
and B and no other impurities, and that C was taken in pure anhydrous state.

w% of A:
51st IChO – Theoretical Examand ofB:

25

Candidate: AAA-1
3.

Determine the formulas of compounds C and D and write the balanced reaction equation
between B and C.

C:

and D:

Reaction between B and C:

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26

Candidate: AAA-1
In 1811, when working with algae ashes, Courtois observed that copper vessels were worn out fasterthan usual. While he was studying this phenomenon, his cat entered the laboratory and spilled thesolution of concentrated sulfuric acid on the dry algae ashes: violet vapors instantly came out of thevessel (1, sulfuric acid is the oxidizing agent): iodine (I2) had just been discovered! Iodine was thecause of the copper corrosion (2). However, because of the medicinal applications of iodine, Courtois
opened a new manufacture to produce it by reaction of algae with chlorine (3).
Nowadays, iodine is prepared from the set of reactants (NO3−, I−, H+) (4) or (IO3−, I−, H+) (5).
4.

Write balanced equations for reactions 1–5.

1
2
3
4
5

The solubility of iodine is very low in water but significantly increases when iodide ions are added.
Together they form ions such as triiodide, I3−:
I−(aq) + I2(aq) = I3−(aq)(6)
Equilibrium (6) can be studied through the extraction of I2 with dichloromethane. Indeed, I− and I3− donot dissolve in organic solvents but I2 does and, when extracted, it is 15 times more concentrated indichloromethane than in water.
The following experiment was performed. To prepare the initial solution, a few crystals of solid iodine
were dissolved in 50.0 mL of an aqueous solution of potassium iodide (0.1112 g). Then, 50.0 mL ofdichloromethane were added, and the mixture was vigorously shaken until equilibration. After phaseseparation, each phase was titrated by 16.20 mL (organic phase) and by 8.00 mL (aqueous phase) of
the standard aqueous solution of sodium thiosulphate pentahydrate (14.9080 g in 1.000 L of solution)in the presence of starch. The process is schematically represented below:

51st IChO – Theoretical Exam27

Candidate: AAA-1

5.

Find the correspondence between the stages on the scheme (1–9) and the schematic pictures
representing them (a–i).

StagesPicture

1
2
3
4
5
6
7
8
9
6.

Write balanced equations for the two possible chemical reactions in the aqueous phase during thetitration involving iodine species and sodium thiosulphate.

7.

Calculate the mass of iodine used to prepare the initial solution.

51st IChO – Theoretical Exam28

Candidate: AAA-1

m(I2) =
8.

g

Calculate the equilibrium constant K° for equilibrium of reaction (6).

K° =51st IChO – Theoretical Exam

29

Candidate: AAA-1

Problem
T5
8%

Question

1

2

3

4

5

6

7

8

9

10

11

12

Total

Points

3

4

4

2

5

5

4

3

5

2

2

2

41

Score

Problem T5: Azobenzene – β-cyclodextrin complexes forthe formation of nanomachines
Nanomachines are molecular assemblies that enable the transformation of an energy source into a
nano-movement for applications such as drug delivery. Numerous nanomachines make use of theisomerization of azo compounds (R–N=N−R’) upon irradiation.
1. Draw the stereoisomers of azobenzene (H5C6–N=N–C6H5) and draw a line between the two
carbon atoms that are the furthest apart. Compare these two distances (dtrans and dcis).

transComparison:cis
dtransdcis

Fig. 1 – Possible reactants for the synthesis of M.
2.

M can be synthesized in two steps from simple reactants (Fig. 1). Choose among the suggested
reactants (N to Q) the ones that can provide M with very high regioselectivity. Sodium nitrite(NaNO2) in cold aqueous hydrochloric acid is used as reagent for the first step of the synthesis.

51st IChO – Theoretical Exam30

Candidate: AAA-1

Reactants:

andDetermination of the association constant Kt
β-cyclodextrin (C, Fig. 2) is a cyclic heptamer of glucose, which can form inclusion complexes
with azo compounds. In tasks 3 to 6, we will determine by spectroscopy the association constant Kt,
corresponding to the formation of the inclusion complex CMtrans as depicted in Fig. 2.

Fig. 2 – Formation of the CMtrans inclusion complex.
Several solutions are prepared by mixing C and Mtrans in different proportions to reach initialconcentrations [C]0 and [Mtrans]0. While [Mtrans]0 is identical for all solutions, [C]0 varies. We follow,
at a fixed wavelength, the evolution of the difference in absorbance A between the absorbance ofeach solution and the pure Mtrans solution. We note the molar absorption coefficients of CMtrans andMtrans, εCMtrans and εMtrans, respectively. L is the path length of the beam through the sample. The
absorbance of C (εC) is negligible.
3.

Demonstrate that ΔA=  ·[CMtrans] and express α in terms of known constant(s).

Demonstration:

=
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Candidate: AAA-1
4.

Demonstrate that, when C is in large excess with respect to Mtrans (i.e. [C]0 >> [Mtrans]0), theconcentration of C may be considered as constant, [C] ≃ [C]0.

Demonstration:

5.

Demonstrate that, when C is in large excess with respect to Mtrans (i.e. [C]0 >> [Mtrans]0),∆𝐴 = 𝛼 ∙𝛽∙[𝐂]0and express β in terms of constant(s) and initial concentration(s).
1+𝐾t ∙[𝐂]0

Demonstration:β=

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Candidate: AAA-1
6.

Determine Kt using the following experimental curve (Fig. 3).

Fig. 3 – Evolution of 1/∆A as a function of 1/[C]0.

Calculations:Kt =51st IChO – Theoretical Exam

33

Candidate: AAA-1
Determination of the association constant Kc
In tasks 7 to 9, we will determine by kinetic studies the association constant Kc, corresponding to theformation of the inclusion complex with Mcis, CMcis. A sample containing only Mtrans is irradiated,thus producing a known amount of Mcis, [Mcis]0. Mcis (free or within the inclusion complex) thenthermally isomerizes into Mtrans. In the absence of C, the isomerization follows a first order kinetics
with a rate constant k1. All complexation equilibria are faster than the isomerization processes. The
kinetic scheme corresponding to this experiment is provided in Fig. 4.

Fig. 4 – Kinetic scheme for the isomerization of Mcis in the presence of C.
The rate of disappearance r for the total amount of Mcis (free and complexed) is defined asr = k1[Mcis] + k2[CMcis]
Experimentally, r follows an apparent first order kinetic law with an apparent rate constant kobs:r = kobs([Mcis] + [CMcis])
7.

Demonstrate that 𝑘obs =𝛾+𝛿∙𝑘2 [𝐂]and express γ and δ in terms of known constant(s).
1+𝐾c [𝐂]

Demonstration:

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Candidate: AAA-1

γ =

8.

and

δ=

Choose in which condition(s) the half-life t1/2 corresponding to kobs can be expressed as
ln2

𝑡1/2 = 𝛾 (1 + 𝐾c [𝐂]0 ) given that [C]0 >> [Mcis]0. Mathematically justify your answer.
Very slow isomerization of Mcis within cyclodextrin
Very slow isomerization of free Mcis
CMcis very stable
CMtrans very stable

Demonstration:

9.

Assuming the condition(s) in task 8 satisfied, determine Kc by a linear regression using the data
below. You may use a calculator or plot a graph.
[C]0 (mol L−1)0
1.0·10−4
5.0·10−4
1.0·10−3

51st IChO – Theoretical Exam

t1/2 (s)
3.0
3.2
3.6
4.1

[C]0 (mol L−1)
3.0·10−3
5.0·10−3
7.5·10−3
1.0·10−2

t1/2 (s)
5.9
7.7
9.9
12.6

35

Candidate: AAA-1

Equation of the linear regression:

Kc =51st IChO – Theoretical Exam

36

Candidate: AAA-1
Formation of nanomachines

Fig. 5 – Cleavage of an azobenzene ̶ cyclodextrin inclusion complex induced by a light-triggered
isomerization, which allows delivery of a dye (grey circles).
Another azobenzene compound (for which Kc << Kt), initially in the trans form, is covalently graftedon silica (Fig. 5). The silica pores are filled with a dye (rhodamine B, grey circles in Fig. 5). Upon
addition of C, an inclusion complex is formed, which blocks the pores and prevents the release of thedye.
10. Choose the most appropriate condition (one choice only) so that the pores are initially blocked in
the presence of C, and the dye can be released upon irradiation.






Kt >> 1
Kt >> 1 and Kc << 1
Kt / Kc << 1
Kt >> 1 and Kc >> 1
Kc << 1This azobenzene-silica powder loaded with a dye is placed in the corner of a cuvette (Fig. 6) so thatthe powder cannot move into solution. The powder is irradiated at a wavelength λ1 to trigger the
release of the dye from the pores (Fig. 5). To monitor this release by absorbance spectroscopy wemeasure the absorbance of the solution at wavelength λ 2.

Fig. 6 – Left: experimental setup used to monitor the release of the dye; right: absorption spectra oftrans-azobenzene (full line), cis-azobenzene (dotted line) and rhodamine B (dashed line).
11. Determine λ1.

λ1 =

nm

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Candidate: AAA-1
12. Determine λ2.

λ2 =

nm

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Candidate: AAA-1

Problem
T6
8%

Question

1

2

3

4

5

6

7

8

9

Total

Points

4

4

5

3

10

2

9

6

5

48

Score

Problem T6: Characterization of a block-copolymer
Block-copolymers, obtained by linking different polymers (blocks), have unique properties, such as
the ability to self-assemble. In this problem, the synthesis and characterization of such amacromolecule are studied.

Study of the first block

In this first part, we will study the water soluble homopolymer 1 (α-methoxy-ωaminopolyethyleneglycol).
The 1H NMR spectrum of 1 (DMSO-d6, 60 °C, 500 MHz) includes the following signals:
Indexδ (ppm)Peak Area

a

2.7*

0.6

b

3.3

0.9

c

3.4

0.6

d

~ 3.5

133.7

Table 1, *in the presence of D2O, the signal at 2.7 ppm disappears.
1. Match the 1H NMR signals (a, b, c, d) from Table 1 with each of the corresponding protons.

2.

Express the average degree of polymerization n as a function of the area AOC2H4 of the NMR peakof the repeating unit and the area AOCH3 of the NMR peak of the methyl end group. Calculate n.

51st IChO – Theoretical Exam

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Candidate: AAA-1

n=
If you could not calculate n, the value n = 100
can be used in the rest of the problem.

Study of a diblock-copolymer
The synthesis of the second block of the copolymer is performed through the reaction of 1 with 2 (ε(benzyloxycarbonyl)-lysine N-carboxyanhydride). This yields the block-copolymer 3.

3.

Draw the reaction intermediate that is formed in the first step of the addition of 1 to 2. The second
step of the mechanism leads to the formation of a gas molecule, G. Draw its structure.

G:

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Candidate: AAA-1
4.

Infrared (IR) measurements are performed to characterize the compounds. Match the three IR
spectra with compounds 1, 2 and 3.

Compound: 1 2 3Compound: 1 2 3Compound: 1 2 3

5.

The 1H NMR spectrum of copolymer 3 (in DMSO-d6, at 60 °C, 500 MHz) is reported in Fig. 1.
Using some or all of the NMR signals, the areas of which are reported in Table 2, calculate its
number average molar mass Mn, considering n from question 2. For your calculations, draw acircle around the group(s) of atoms you used and give their corresponding symbol(s) (α, β…).
Table 2
Peak

Area

α

22.4

ß

119

γ

23.8

δ

47.6

ε

622

Fig. 1 – signals marked with * correspond to the solvent and water.

51st IChO – Theoretical Exam

41

Candidate: AAA-1

Mn =kg mol–1

Provide your answer with two decimal places.

51st IChO – Theoretical Exam

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Candidate: AAA-1
This reaction of 1 with 2 yielded the copolymers 3a after 20 h, 3b after 25 h and 3c after 30 h ofreaction at 40 °C. Results of size-exclusion chromatography (SEC) experiments are presented in Fig.
2.

Fig. 2 – SEC chromatograms of 3a, 3b and 3c as a function of the elution volume, Ve.
6.

3a:
3b:
3c:

Match the signals in Fig. 2 with the copolymers 3a, 3b and 3c.

X
X
X

Y
Y
Y

Z
Z
Z

In order to calibrate the chromatogram, a mixture of standard polymers of known masses (3, 30, 130,700 and 7000 kg mol−1) has been studied (Fig. 3).
The log value of the molar mass is a linear function of the elution volume, Ve.

Fig. 3 – SEC chromatogram of the mixture of standards.
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Candidate: AAA-1
7.

Based on the SEC curves in Fig. 2 and 3, determine Ve of the polymer that corresponds to curve
X and use it to estimate the degree of polymerization m of its second block. Detail yourcalculation; you may use a calculator or plot a graph.

Ve =

mL

m=

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Candidate: AAA-1

Triblock copolymer synthesis
For biological applications, involving the formation of micelles, a triblock copolymer 9 can besynthesized through the introduction of a middle block, B, using monomer 5.

8.

Draw the structures of 5, 7 and 8.

5 (no other products than 6:A-B are obtained)

7 (a gas is formed in the final step)

8

9.

Amphiphilic block copolymers, such as 9: A-B-C, can be used for medical applications, as they
self-assemble into micelles in water (pH = 7), which can be used as drug carriers. Assign eachblock of the copolymer to a property. Draw a scheme of the micelle with only 4 polymer chains.

A: hydrophobic hydrophilicB: hydrophobic hydrophilic

51st IChO – Theoretical Exam

45

Candidate: AAA-1

C: hydrophobic

51st IChO – Theoretical Exam hydrophilic46

Candidate: AAA-1

Problem T7: Ring motion in a [2]catenane
Problem
T7
6%

Question

1

2

3

4

5

6

7

8

9

10

11

Total

Points

4

12

2

2

2

5

5

8

4

5

5

54

Score

In 2016, the Nobel Prize in Chemistry was awarded to J.-P. Sauvage, Sir J. F. Stoddart and
B. L. Feringa "for the design and synthesis of molecular machines". An example of these is[2]catenane, a molecule consisting of two interlocked rings. In this system, one macrocycle contains asingle phenanthroline (bidentate) ligand and the second contains two ligands: a phenanthroline and aterpyridine (tridentate) ligand. A copper ion is coordinated by one ligand from each macrocycle.
Depending on the oxidation state of the copper (+I or +II), two configurations are obtained (Fig. 1).

Fig. 1 – Multi-stability of a ring in a [2]catenane.
The synthesis of the macrocycle is the following:

1. Draw the structure of B.
B

51st IChO – Theoretical Exam

47

Candidate: AAA-1
2.

Draw the structures of E, F and G.

E

F

G

3.

Out of the following the reaction conditions, choose which one(s) can produce E from D: H+, H2O
 OH−, H2O
 NaBH4, CH3OH
 H2, Pd/C, THF
4.

In the synthetic strategy, MsCl is used to obtain:

 a leaving group a protecting group a deactivating group a directing group5.

G is obtained by the reaction between F and LiBr in acetone. This reaction is: electrophilic aromatic substitution
 nucleophilic aromatic substitution
 SN1
 SN2
51st IChO – Theoretical Exam48

Candidate: AAA-1
6.

Draw the transition state of the rate-determining step of the reaction F → G, showing the 3D
geometry. Depict only one reaction center. The main carbon chain can be represented as an R
group.

Transition state:

The synthesis of [2]catenane L uses the template effect of a copper complex:

7.

Write the full electronic configuration of Cu(0) in its ground state. Give the oxidation state of Cu
in complex J and write the electronic configuration of Cu in the free ion corresponding to J.

Electronic configuration of Cu(0):Oxidation state of Cu in J:

Electronic configuration of Cu in J:

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49

Candidate: AAA-1
8.

Select the geometry of the copper ion in L. Assuming an ideal geometry of the ligands around thecopper center, draw the electronic levels of the d orbitals subject to the crystal field. Fill theorbital diagram. Give the maximum value of the spin (S) for this complex.

The geometry of Cu in L is: Octahedral Tetrahedral
 Square planar Trigonal bipyramid
Splitting and filling of d orbitals:S=

9.

Out of the following compounds, choose the one(s) that can remove the copper ion in L to obtainthe free [2]catenane: CH3CN NH4PF6
 KCN
 trenIn [2]catenane L, the copper ion can exist in two oxidation states (+I) or (+II), and each of them
exhibits a different coordination sphere (tetra- or penta-coordinated, respectively).

Fig. 2 – [2]catenane L states

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Candidate: AAA-1
The stability of Cu(I) complexes can be inferred by comparing their electronic structures to that of anoble gas.
10. Fill in the blanks with a number or a tick:

The CuIN4 complex has … electrons in the coordination sphere of the metal.
The CuIN5 complex has … electrons in the coordination sphere of the metal.
The CuIN4 complex is  more /  less stable than the CuIN5 complex.
11. Fill in the solid boxes with the designation of the involved complexes in Fig. 2 and complete thesequence to achieve electrochemical control of the system using the following notation for thedashed boxes:(rotation); + e‒ ; ‒ e‒.

51st IChO – Theoretical Exam

51

Candidate: AAA-1

Problem
T8
6%

Question

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Total

Points

2

6

2

2

11

2

4

3

4

2

6

8

2

6

4

64

Score

Problem T8: Identification and synthesis of inositols
In this problem, we define “3D structure” and “perspective formula” as indicated for β-glucose in thefollowing figure.

Inositols are cyclohexane-1,2,3,4,5,6-hexols. Some of these 6-membered carbocycles, in particular
myo-inositol, are involved in a number of biological processes.

Structure of myo-inositol
1. Draw the structural formula of inositols, without stereochemical details.

This family of molecules contains 9 different stereoisomers, including enantiomers.
2.

Draw all 3D structures of the stereoisomers that are optically active.

The structure of a specific inositol, called myo-inositol, is studied here. Only one of its chairconformers is predominant and its structure can be deduced from its 1H NMR spectrum. The spectrum
below was obtained at 600 MHz in D2O. No other signal from that compound was observed in the
spectrum. The integration is indicated on the spectrum below each signal.
51st IChO – Theoretical Exam

52

Candidate: AAA-1

3.

Give the molecular formula of the predominant compound derived from myo-inositol in thissample that is consistent with the number of protons observed in the 1H NMR spectrum.

4.

Based on the number and integrations of the proton signals, give the number of symmetry plane(s)that exist(s) in this molecule.

5.

Complete the following perspective drawing of the most stable conformation of myo-inositol.
Then label each hydrogen with the corresponding letter (a, b, c or d) according to the NMR
spectrum above. Proton a must be on carbon a on the following representation. Draw its 3D
structure.

3D structure:

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53

Candidate: AAA-1

Synthesis of inositols
For medicinal applications, it is useful to synthesize some inositol phosphates on a large scale. We will
study the synthesis of inositol 2 from bromodiol 1.

6.

Choose the correct structural relationship(s) between 2 and 3.

 enantiomers
 epimers
 diastereomers
 atropoisomers
Inositol 2 can be obtained from compound 1 in 7 steps.

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54

Candidate: AAA-1
7.

Draw the 3D structure of 4.

4

8.

The reaction leading to 5 occurs on the double bond with the highest electron density. Consider
below the structure of 1-bromo-1,3-cyclohexadiene, which is a substructure of 4. Circle thedouble bond with the highest electron density. Represent all the electronic effects due to thebromine.

9.

Draw the 3D structure of the major diastereomer 5.

5

10. Give the total number of stereoisomers of 5 possibly obtained by this synthesis, starting fromenantiopure compound 1.

11. For the step 5 → 6, another product with the same molecular formula, denoted 6’, can beproduced. Draw the 3D structures of 6 and 6’.

6

51st IChO – Theoretical Exam

6’

55

Candidate: AAA-1
12. Draw the 3D structures of major diastereomers 8 and 9.

89

13. Select the right set(s) of conditions A to obtain 2.

 H2, Pd/C
 K2CO3, HF
 HCOOH, H2O
 BF3·OEt2
14. If the bromine is not present in compound 1, in addition to 2, another stereoisomer would beobtained. Considering that the stereoselectivity of the reactions that take place in the synthesis
remains unchanged and that the following steps involve the same number of equivalents as for 2,
draw the 3D structure of this stereoisomer and give its relationship with 2.

 enantiomers
 epimers
 diastereoisomers
 atropoisomers
15. During the synthesis of 2 from 1, choose the removal step(s) of protecting or directing groups.

 1→4
 4→5
 5→6
 6→7
 7→8
 8→9
 9→2

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56

Candidate: AAA-1

Problem
T9
7%

Question

1

2

3

4

5

6

7

8

9

10

11

12

13

Total

Points

2

2

4

3

2

17

1

1

2

4

2

2

2

44

Score

Problem T9: Synthesis of levobupivacainePart I.
The local anesthetic bupivacaine (marketed as Marcaine) is on the World Health Organization List of
Essential Medicines. Although the drug is currently used as a racemic mixture, it was demonstrated
that one enantiomer of bupivacaine, levobupivacaine, is less cardiotoxic and, therefore, safer than the
racemate. Levobupivacaine can be synthesized from the natural amino acid L-lysine.

1. Assign the absolute configuration of the stereogenic center in L-lysine hydrochloride and justify
your answer by classifying the substituents in order of their priority.
Configuration:

Priority 1 > 2 > 3 > 4:

R
S

2.

The prefix L in L-lysine refers to relative configuration. Choose all correct statements:






All natural L-amino acids are levorotatory.
Natural L-amino acids can be levorotatory or dextrorotatory.
All natural L-amino acids are (S).
All natural L-amino acids are (R).

Often, we want only one of the amino groups in L-lysine to react. A Cu2+ salt with excess aqueoushydroxide can selectively mask the reactivity of one of the amino groups. After the complex is formed,
only the non-complexed NH2 group is available to react.
3.

Considering that L-lysine acts as a bidentate ligand and that two L-lysines coordinate to one Cu2+
ion in the presence of aqueous hydroxide, draw the structure of the intermediate complex.

Complex

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57

Candidate: AAA-1
Fortunately, in the synthesis of levobupivacaine shown below, the same amino group reacts even
without Cu2+ salt.

From this point on, you can use the abbreviations proposed in the scheme above.
4.

Draw the structure of compound A, including the appropriate stereochemistry.

A

5.

Transformation of L-lysine into A is (choose proper answer(s)):

 an enantioselective reaction.
 an enantiospecific reaction.
 a regioselective reaction.
51st IChO – Theoretical Exam58

Candidate: AAA-1
6.

Draw the structures of compounds B–F, including the appropriate stereochemistry.

B C14H20N2O4

C C16H21NO6D

E C29H34N2O6S

F C21H28N2O4S

7.

What is the role of DCC in the transformation C → D?

 Protecting group for the amino group.
 Protecting group for the hydroxy group.
 Activating agent for the amide bond formation.
8.

TsCl is used in the synthesis to enable:






Nucleophilic substitution of an amino group.
Electrophilic substitution of an amino group.
Nucleophilic substitution of a hydroxy group.
Electrophilic substitution of a hydroxy group.

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59

Candidate: AAA-1
9.

Mark all possible reagents which could be used as reagent H:

 diluted HCl
 K2CO3
 diluted KMnO4
 SOCl2

 Zn/HCl H2SO4
 diluted NaOH
 PCl5

10. Draw the structure of levobupivacaine, including the appropriate stereochemistry.
Levobupivacaine C18H28N2O

Part II.
The synthesis of levobupivacaine requires the use of enantiomerically pure L-lysine. A common
method to confirm the enantiomeric purity of aminoacids is their transformation into amides using
Mosher's acid (see the structure of the (S) isomer below).

11. Draw the structure of the amide formed when the α-amino group of L-lysine is derivatized with(S)-Mosher's acid. Clearly show the stereochemistry of each chiral center.

12. How many products will be formed from racemic lysine and (S)-Mosher's acid (consider that
only the α-amino group of lysine is derivatized)?
Two diastereoisomers.
Four diastereoisomers.
A racemic mixture of two enantiomers.
Four compounds: two enantiomers and two diastereoisomers.

13. Choose the method(s) which can be used to quantitatively determine the enantiomeric purity oflysine after its derivatization with (S)-Mosher's acid:

NMR spectroscopy.
Liquid chromatography.
Mass spectrometry.
UV-vis spectroscopy.

51st IChO – Theoretical Exam

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