International Chemistry Olympiad 2021 Japan
53rd IChO2021 Japan
25th July – 2nd August, 2021
https://www.icho2021.org

Educational Version
Published in April, 2022

The preparation of the problems for the 53rd International Chemistry Olympiad (ChO2021) started
with the first meeting of the Science Committee on April 24th (Sunday), 2018. Since then, the
Science Committee had actively worked for over three years, and the fruits of the activity weresummarized as the IChO2021 Official English version, which is available on the IChO2021 website.
The problems of IChO must satisfy the following criteria:- The problems must be of high standard and quality, for which highly talented students from allover the world should use the best of their creative ability.
- The knowledge required to solve the problems must be open to all students. Only the secondary
education-level (as international baccalaureate) of knowledge is allowed to use for the problems.
- If any knowledge beyond the secondary education-level is necessary to solve the problems, it
must be announced preliminarily, and at least two problems regarding this knowledge must be
included in the preparatory problems.
- The number of advanced fields of knowledge must not exceed six, and each of them must be in
the range that all students can master in a two- or three-hour lecture.
- The total length of the problems must not exceed 25,000 characters, so that the students haveenough time to read all the problems during the examination time.
- The problems never ask any knowledge itself.
- The solutions must be logically derived from the problems and must be represented only byequations and chemical formulae.
The problems prepared under these criteria should be creative, intellectually stimulative, andeducationally valuable. Since the members of the Science Committee are the professors who
actively engage in research and education in various fields of chemistry, they worked hard to
prepare high-quality problems with their sincere professionalism. The Science Committee “finalized”
the problems (February version) along with the preparatory problems on February 1st, 2021.
International Jury is, however, the only body that can finalize the problems for the competition. In
defense of the problems from comments at the International Jury, the Science Committee had tofurther shorten the length of the problems, and had to simplify the problems to avoid ambiguity in
grading. For example, questions based on similar ideas and questions on physical chemistry in the
problems of organic chemistry were deleted to shorten the total length of the problems. In addition,a question that is very sensitive to significant figures and a question that is based on the linearityof data were deleted or simplified to avoid ambiguity in grading. After these intensive trimmings,
the problems became slim to be suitable for the competition. However, at the same time, somenarratives and tastes that fascinated us in the February version had been lost.
In IChO2021, the Official English version was greatly admired in the International Jury. The Science
Committee is proud of this praise. However, it should be noted that some essences in the February
version, which is another most fruitful works achieved by the Science Committee, are missing in1

the Official English version. Thus, we decided to prepare this Educational version, where the highly
polished Official English version is made up with the missing contents of February version.
The Educational version reflects the entire works made by the Science Committee. We sincerely
hope that the Educational version is widely utilized in chemical education as well as the Official
English version.
Vice-Chair of IChO2021 Science Committee
Nobuhiro Kihara
April, 20222

I am very pleased to publish the Educational Version of the 53rd IChO Problems 2021. This
Educational Version has been edited by imparting the following information to the official IChO
Problems/Solutions (https://www.icho2021.org/problems/icho2021/).
1) Questions that were unfortunately deleted before the publication of the official IChO Problems
because of the IChO exam length regulation. (after yellow highlight)2) Questions that were unfortunately deleted before the publication of the official IChO Problems
because of potential difficulty in marking by following the IChO marking regulation. (after yellow
highlight)3) Tutorial comments and useful literature for further study (Written in pale green)
I sincerely hope that you can touch the passion of the Science Committee members through thisEducational Version and that this Version is valuable for young people who participated in IChO,
high school students who are eager to participate in IChO, and their mentors.
I would like to express my deepest gratitude to all the Theoretical Problem Subcommittee members
for their great efforts in creating both preparatory and IChO2021 problems. I would also like to thankthe reviewers for their valuable comments and suggestions.
All the best for all those who have been involved in IChO,
Hideki Yorimitsu
On behalf of 53rd IChO 2021 Theoretical Problem Subcommittee, Science Committee

3

The 53rd International Chemistry Olympiad 2021, Japan
Science Committee ChairNISHIHARA, Hiroshi (Professor, Tokyo University of Science)
Science Committee Vice-Chair
KIHARA, Nobuhiro (Professor, Kanagawa University)
Theoretical Problem Subcommittee Chair
YORIMITSU, Hideki (Professor, Kyoto University)
Science Committee Secretary
HIROI, Takashi (Researcher, National Institute for Materials Science)
Science Committee Assistant
SAITO, Hayate (Assistant Professor, Kyoto University)
Physical Chemistry Branch（Question 1–4）
MATSUMOTO, Yoshiyasu（Fellow, Toyota Physical and Chemical Research Institute）＊Chair
HASEGAWA, Takeshi（Professor, Kyoto University）
SATO, Hirofumi（Professor, Kyoto University）
OKUYAMA, Hiroshi（Associate Professor, Kyoto University）
HIROI, Takashi（Researcher, National Institute for Materials Science）
Inorganic/Analytical Chemistry Branch（Question 5 and 6）
NISHI, Naoya（Associate Professor, Kyoto University）＊ChairUCHIDA, Sayaka（Associate Professor, University of Tokyo）
SASAMORI, Takahiro（Professor, University of Tsukuba）
HORIKE, Satoshi（Associate Professor, Kyoto University）
HOSOKAWA, Saburo（Associate Professor, Kyoto Institute of Technology）
HIROI, Takashi（Researcher, National Institute for Materials Science）
Organic Chemistry Branch（Question 7–9）
SHIMOKAWA, Jun（Associate Professor, Kyoto University）＊ChairTSUBAKI, Kazunori（Professor, Kyoto Prefecture University）
YAMAGUCHI, Hiroyasu（Professor, Osaka University）
SHINTANI, Ryo（Professor, Osaka University）
TANAKA, Takayuki（Associate Professor, Kyoto University）
SAITO, Hayate（Assistant Professor, Kyoto University）

4

Physical Constants and Equations
Constants
Speed of light in vacuum, c = 2.99792458 × 108 m s–1
Planck constant, h = 6.62607015 × 10–34 J s
Elementary charge, e = 1.602176634 × 10–19 C
Electron mass, me = 9.10938370 × 10–31 kg
Electric constant (permittivity of vacuum), ε0 = 8.85418781 × 10–12 F m–1
Avogadro constant, NA = 6.02214076 × 1023 mol–1
Boltzmann constant, kB = 1.380649 × 10–23 J K–1
Faraday constant, F = NA × e = 9.64853321233100184 × 104 C mol–1
Gas constant, R = NA × kB = 8.31446261815324 J K–1 mol–1= 8.2057366081 × 10–2 L atm K–1 mol–1
Unified atomic mass unit, u = 1 Da = 1.66053907 × 10–27 kg
Standard pressure, p = 1 bar = 105 Pa
Atmospheric pressure, patm = 1.01325 × 105 Pa
Zero degree Celsius, 0 °C = 273.15 KÅngström, 1 Å = 10–10 m
Picometer, 1 pm = 10–12 m
Electronvolt, 1 eV = 1.602176634 × 10–19 J
Part-per-million, 1 ppm = 10–6
Part-per-billion, 1 ppb = 10–9
Part-per-trillion, 1 ppt = 10–12pi, π = 3.141592653589793
The base of the natural logarithm (Euler's number), e = 2.718281828459045

5

Equations
The ideal gas law:

PV = nRT, where P is the pressure, V is the volume,n is the amount of substance, T is theabsolute temperature of ideal gas.

The first law of thermodynamics:

ΔU = q + w, where ΔU is the change in the internalenergy, q is the heat supplied, w is thework done.

Enthalpy H:

H = U + PV

Entropy based on Boltzmann's principle S:

S = kB lnW, where W is the number of microstates.

The change of entropy ΔS:∆𝑆 =Gibbs free energy G:

G = H – TS
ΔrG 0 = –RT lnK = –zFE 0, where K is the equilibrium constant, z is
the number of electrons, E 0 is the standard𝑞!"#𝑇, where qrev is the heat for the reversible
process.

electrode potential.
Reaction quotient Q:

ΔrG = ΔrG 0 + RT lnQ
For a reaction
aA+bB⇌cC+dD[𝐶]$ [𝐷]%𝑄=[𝐴]& [𝐵]', where [A] is the concentration of A.

Heat change Δq:Δq = ncmΔT, where cm is the temperature-independent
molar heat capacity.

Nernst equation for redox reaction:

𝑅𝑇
𝐶)*
ln Q
R
𝑧𝐹
𝐶+,, where Cox is the concentration of oxidized
substance, Cred is the concentration of
reduced substance.

Arrhenius equation:

𝐸 = 𝐸( +𝑘 = 𝐴 exp Q−

6

𝐸&
R
𝑅𝑇

, where k is the rate constant, A is the preexponential factor, Ea is the activationenergy.
exp (x) = e x
Lambert–Beer equation:

A = εlc, where A is the absorbance, ε is the molar
absorption coefficient, l is the optical path
length, c is the concentration of thesolution.

Henderson−Hasselbalch equation:

For an equilibrium
HA ⇌ H+ + A–
, where equilibrium constant is Ka,[A. ]pH = p𝐾& + log ^_[HA]

Energy of a photon:𝑐
𝜆, where ν is the frequency, λ is the
wavelength of the light.

The sum of a geometric series:

When x ≠ 1,

𝐸 = ℎ𝜈 = ℎ

0

1 − 𝑥 034
1 + 𝑥 + 𝑥 +. . . +𝑥 = j 𝑥 =
1−𝑥
/

0

1

12(

Approximation equation that can be used
to solve problems:

When 𝑥 ≪ 1,
1
~1+𝑥
1−𝑥

7

1

44.955908

39

Calcium

40.078

Potassium

39.0983

H NMR Chemical Shifts

8RaRadium[226]

Fr

Francium[223]

Actinoids

Ac-Lr :

89-103Lanthanoids

La-Lu :

57-71

88

87

Lanthanoids

Barium137.327

56

Ba

55

Cs

Caesium

57-71

La-Lu

87.62

85.4678132.905452

88.90584

Strontium

Rubidium

90

Thorium

232.0377

Actinium

[227]

Th

89

Ac

Cerium140.116Lanthanum

Ce

58

[267]

Rutherfordium

Rf

104

178.49HafniumHf

72

91.224

Zirconium

Zr

40

47.867

Titanium

138.90547

La

57

Actinoids

Ac-Lr

89-103

Yttrium

Y

38

Sr

37

Rb

ScandiumTi

231.03588

Protactinium

Pa91

140.90766

Praseodymium

Pr

59

[268]

DubniumDb

105

180.94788

TantalumTa

73

92.90637

NiobiumNb

41

50.9415

VanadiumV

238.02891

UraniumU92

144.242

Neodymium

Nd

60

[271]

Seaborgium

Sg

106

183.84Tungsten

W

74

95.95

MolybdenumMo

42

51.9961

Chromium

Cr

[237]

Neptunium

Np

93

[145]

Promethium

Pm

61

[272]

BohriumBh

107

186.207

Rhenium

Re

75

[99]

TechnetiumTc

4354.938044

Manganese

Mn

26

[239]

PlutoniumPu

94

150.36

Samarium

Sm

62

[277]

HassiumHs

108

190.23

OsmiumOs

76

101.07

RutheniumRu

44

55.845

Iron

Fe

27

[243]

AmericiumAm95

151.964Europium

Eu

63

[276]

Meitnerium

Mt

109

192.217

Iridium

Ir

77

102.90550

RhodiumRh

45

58.933194

Cobalt

Co

28

[247]

CuriumCm

96

157.25Gadolinium

Gd

64

[281]

Darmstadtium

Ds

110

195.084

PlatinumPt

78

106.42PalladiumPd

46

58.6934

Nickel

Ni

29

[247]

BerkeliumBk

97

158.92535

Terbium

Tb

65

[280]

Roentgenium

Rg

111

196.966569

Gold

Au

79

107.8682

SilverAg

47

63.546

Copper

Cu

30

[252]

Californium

Cf98

162.500

Dysprosium

Dy

66

[285]

Copernicium

Cn

112

200.592

MercuryHg

80

112.414

Cadmium

Cd

48

65.38

Zinc

Zn

[252]

EinsteiniumEs99

164.93033

Holmium

Ho

67

[278]

NihoniumNh113

204.384

ThalliumTl

81

114.818

IndiumIn

49

69.723

Gallium

Ga

31

Sc

20

Ca

19

K

Aluminium

26.981539

24.306

13

Magnesium

25

Boron

10.814

Sodiumatomic weight [in parenthesis for the radioactive element]

name

B

22.989769[278]

Nihonium

Symbol

Al
24

113

Nh

5

13

12

23

12

Mg

6.968

22

11

11

Beryllium

9.012183

Lithium

21

10

Na

4

Be

3

Li

atomic number

9

[257]

Fermium

Fm

100

167.259

Erbium

Er

68

[289]

Flerovium

Fl

114

207.2

Lead

Pb

82

118.710

Tin

Sn

50

72.630GermaniumGe

32

28.085

SiliconSi

1412.0106

Carbon

C

6

14

[258]

Mendelevium

Md

101

168.93422

Thulium

Tm

69

[289]

Moscovium

Mc115

208.98040

BismuthBi

83

121.760

Antimony

Sb

51

74.921595

Arsenic

As

33

30.973762

PhosphorusP1514.00686

Nitrogen

N

7

15

[259]

NobeliumNo102173.045

Ytterbium

Yb

70

[293]

Livermorium

Lv

116

[210]

PoloniumPo

84127.60TelluriumTe52

78.971

SeleniumSe

34

32.068SulfurS

16

15.9994

Oxygen

O

8

16

[262]

Lawrencium

Lr

103

174.9668

Lutetium

Lu

71

[293]

Tennessine

Ts

117

[210]

AstatineAt85

126.90447

IodineI

53

79.904

BromineBr35

35.452ChlorineCl

17

18.998403

Fluorine

F

9

17

18

[294]

Oganesson

Og

118

[222]

Radon

Rn

86

131.293

Xenon

Xe

54

83.798KryptonKr

36

39.948

Argon

Ar

18

20.1797

Neon

Ne

10

Helium8

4.002602

Key:

7

1.00798

6

Hydrogen

5

2

4

He

3

H

2

1

1

Periodic tableΔδ for one alkyl group-substitution: ca. +0.4 ppm

9

13

O

H

O

12

-COOH
-SO3H

11
10

-CHO

9
8

Ar H

-CONH2

7

6

H

Ar OH

5

4

RO CH3

Ar NH23

R2N CH3

2

Ar CH3CH3

RCO CH3

R OH

R NH2

1

R CH3

0
δ (ppm)

R3Si CH3

Question 1: Hydrogen at a Metal Surface

Hydrogen is expected to be a future energy source that does not depend on fossil fuels. Here, wewill consider the hydrogen-storage process in metal, which is related to hydrogen-transport and-storage technology.
Part AAs hydrogen is absorbed into the bulk of a metal via its surface, let us first consider the adsorption
process of hydrogen at the metal surface, H2 (g) → 2H (ad), where the gaseous and adsorbed
states of hydrogen are represented as (g) and (ad), respectively. Hydrogen molecules (H2) thatreach the metal (M) dissociate at the surface and are adsorbed as H atoms (Fig. 1). Here, thepotential energy of H2 is represented by two variables: the interatomic distance d and the height
relative to the surface metal atom, z. It is assumed that the axis along the two H atoms is parallelto the surface and that the center of gravity is always on the vertical dotted line in Fig. 1. Fig. 2
shows the potential energy contour plot for the dissociation at the surface. The numerical values
represent the potential energy in units of kJ per mole of H2. The solid line spacing is 20 kJ mol−1,
the dashed line spacing is 100 kJ mol−1, and the spacing between solid and dashed lines is 80 kJ
mol−1. The zero-point vibration energy is ignored.

10A.1 For each of the following items (i)-(iii), select the closest values from A–G.
(i) The interatomic distance for a gaseous H2 molecule.
(ii) The interatomic distance between metal atoms (dM in Fig. 1).
(iii) The distance of adsorbed H atoms from the surface (had in Fig. 1).
A. 0.03 nm, B. 0.07 nm, C. 0.11 nm, D. 0.15 nm, E. 0.19 nm, F. 0.23 nm, G. 0.27 nm(i) B, (ii) F, (iii) A
A.2

For each of the following items (i)-(ii), select the closest value from A–H.
(i) The energy required for the dissociation of gaseous H2 molecule to gaseous H [H2 (g) →2H (g)].
(ii) The energy released during the adsorption of a gaseous H2 [H2 (g) → 2H (ad)]A. 20 kJ mol−1, B. 40 kJ mol−1, C. 60 kJ mol−1, D. 100 kJ mol−1,
E. 150 kJ mol−1, F. 200 kJ mol−1, G. 300 kJ mol−1, H. 400 kJ mol−1(i) H, (ii) D

Fig. 1 Definition of variables. Drawing is not in scale.

11Fig. 2 Potential energy for the adsorption process of hydrogen at the metal surface as a functionof d and z.
In Part A, dissociative adsorption process of hydrogen molecules on a metal surface is consideredusing two-dimensional potential by taking the intramolecular distance as a variable in addition to
the distance between the molecule and the surface. In general, there are as many variables as thenumber of degrees of freedom (six) of the reactant (in this case, hydrogen molecule), and the sixdimensional potential can be considered by considering all of them.

12Part B
The adsorbed hydrogen atoms are then either absorbed into the bulk, or recombine and desorb
back into the gas phase, as shown in the reactions (1a) and (1b). H(ab) represents a hydrogen
atom absorbed in the bulk.
(1a)
(1b)
The reaction rates per site for adsorption, desorption, and absorption are 𝑟𝑟1 [s−1], 𝑟𝑟2 [s−1], and 𝑟𝑟3[s−1], respectively. They are expressed as:

𝑟𝑟1 = 𝑘𝑘1 𝑃𝑃H2 (1 − 𝜃𝜃)2
𝑟𝑟2 = 𝑘𝑘2 𝜃𝜃 2
𝑟𝑟3 = 𝑘𝑘3 𝜃𝜃

(2)
(3)
(4)

where 𝑘𝑘1 [s−1 Pa−1], 𝑘𝑘2 [s−1], and 𝑘𝑘3 [s−1] are the reaction rate constants and 𝑃𝑃H2 [Pa] is thepressure of H2. Among the sites available on the surface, 𝜃𝜃 (0 ≤ 𝜃𝜃 ≤ 1) is the fraction occupied by

H atoms. It is assumed that adsorption and desorption are fast compared to absorption (𝑟𝑟1 , 𝑟𝑟2 ≫𝑟𝑟3 ) and that 𝜃𝜃 remains constant.

13

B.1 𝑟𝑟3 can be expressed as:

𝑟𝑟3 =Express C using 𝑘𝑘1 and 𝑘𝑘2.
From 𝑟𝑟1 , 𝑟𝑟2 ≫ 𝑟𝑟3 and 𝑟𝑟1 = 𝑟𝑟2 + 𝑟𝑟3,

From 𝑟𝑟3 = 𝑘𝑘3 𝜃𝜃:

Thus,

1
1 +�
𝑃𝑃H2 C(5)

𝑟𝑟1 = 𝑟𝑟2

Then,
Solve for 𝜃𝜃:

𝑘𝑘3

𝑘𝑘1 𝑃𝑃H2 (1 − 𝜃𝜃)2 = 𝑘𝑘2 𝜃𝜃 2
𝜃𝜃 =

𝑟𝑟3 =

1

𝑘𝑘
1+� 2
𝑃𝑃H2 𝑘𝑘1
𝑘𝑘3

𝑘𝑘
1+� 2
𝑃𝑃H2 𝑘𝑘1
𝐶𝐶 =

𝑘𝑘1
𝑘𝑘2

This is a question regarding the equilibrium steady state between the second-order adsorption
process (2) and the second-order desorption process (3). The expression of 𝜃𝜃 is different fromthat of the well-known first-order Langmuir equation.

14

A metal sample with a surface area of S = 1.0×10−3 m2 was placed in a container (1 L = 1.0×10−3m3) with H2 (𝑃𝑃H2 = 1.0×102 Pa). The density of hydrogen-atom adsorption sites on the surface was
N = 1.3×1018 m−2. The surface temperature was kept at T = 400 K. As the reaction (1)proceeded, 𝑃𝑃H2 decreased at a constant rate of v = 4.0×10−4 Pa s−1. Assume that H2 is an idealgas, and that the volume of the metal sample is negligible.

B.2 Calculate the amount of H atoms in moles absorbed per unit area of the surface per unit time,
A [mol s−1 m−2].

The change in the amount of hydrogen atoms per unit time in the gas phase is 𝐴𝐴 × 𝑆𝑆. Thus,

𝐴𝐴 × 𝑆𝑆 =

2𝑣𝑣𝑉𝑉
𝑅𝑅𝑅𝑅

= 2 × 4.0 × 10−4 ×
Therefore,

10−3

8.31×400= 2.4 × 10−10 mol s−1
𝐴𝐴 = 2.4 × 10−7 mol s −1 m−2

The reduction rate of hydrogen molecules in the container is obtained from the pressure change,
which is associated with the rate of absorption into the metal. Note that 1 mol of hydrogen molecule
corresponds to 2 mol of hydrogen atom.

15

B.3 At T = 400 K, C equals 1.0×102 Pa−1. Calculate the value of 𝑘𝑘3 at 400 K.
The relationship between 𝑟𝑟3 and 𝐴𝐴 is:

𝐴𝐴 = 𝑟𝑟3 ×

Thus,

𝑟𝑟3 = 𝐴𝐴 ×

𝑁𝑁
𝑁𝑁𝐴𝐴

𝑁𝑁𝐴𝐴
= 1.1 × 10−1 s −1
𝑁𝑁

Solution 1:
𝑟𝑟3 =
Thus,

𝑘𝑘3

1
1 + �𝑃𝑃 𝐶𝐶
H2

=

𝑘𝑘3

1
1 + �10000

=

𝑘𝑘3
1.01

𝑘𝑘3 = 1.01 × 𝑟𝑟3 = 1.1 × 10−1s −1
Solution 2:
Under the conditions (𝑃𝑃H2 𝐶𝐶 ≫ 1), it follows that:
𝑟𝑟3 =

𝑘𝑘3

1+�

1

𝑃𝑃H2 𝐶𝐶

≅

𝑘𝑘3
1

= 𝑘𝑘3

𝑘𝑘3 = 𝑟𝑟3 = 1.1 × 10−1 s−1
Because 𝑟𝑟3 and 𝐴𝐴 are directly related, one can find the value of 𝑟𝑟3 . 𝑘𝑘3 and 𝑟𝑟3 are related in theequation of B.1. In the present case, one can find 𝑘𝑘3 from 𝑟𝑟3 ≅ 𝑘𝑘3 (𝑃𝑃H2 𝐶𝐶 ≫ 1). In this situation,
the adsorption sites on the surface are almost saturated (𝜃𝜃 ≅ 1). In general, the reaction rate (𝑟𝑟3 ,etc.) depends on the pressure and 𝜃𝜃, but the reaction rate constant (𝑘𝑘3, etc.) depends only on thetemperature.

16B.4 At different T, C = 2.5×103 Pa−1 and 𝑘𝑘3 = 4.8×10−2 s−1 are given. For 𝑟𝑟3 as a function of 𝑃𝑃H2
at this temperature, select the correct plot from (a)–(h).

The figures show the region in which 𝑃𝑃H2 𝐶𝐶 ≪ 1. Therefore,
𝑘𝑘3

𝑟𝑟3 =
≅

1 + �𝑃𝑃

𝑘𝑘3

�𝑃𝑃

1

H2 𝐶𝐶

1

H2 𝐶𝐶

= 𝑘𝑘3 �𝑃𝑃H2 𝐶𝐶 = 2.4�𝑃𝑃H2

Thus, (b)Because the temperature is different, 𝑘𝑘3 and 𝐶𝐶 are also different from the previous question. The

horizontal axis of the graph represents the region of 𝑃𝑃H2 𝐶𝐶 ≪ 1. This corresponds to the region

where 𝜃𝜃 is small, and it can be seen from the equation of B.1 that 𝑟𝑟3 depends on the square rootof 𝑃𝑃H2 . In this situation, the adsorption sites on the surface are almost vacant (𝜃𝜃 ≪ 1), but thepredetermined conditions 𝑟𝑟1 , 𝑟𝑟2 ≫ 𝑟𝑟3 are maintained. However, when the pressure decreases toabout 𝑃𝑃H2 = 1 × 10−8 Pa (the line is excluded in the graph), this condition does not hold and theequation of B.1 is no longer valid.

17

Question 2: Isotope Time Capsule

Molecular entities that differ only in isotopic composition, such as CH4 and CH3D, are calledisotopologues. Isotopologues are considered to have the same chemical characteristics. In nature,
however, there exists a slight difference.
Let us consider the following equilibrium:
12

16

C O2 +

12

18

12

16

18

C O2 ⇄ 2 C O O

𝐾=#C 16O 18O$

2

#C 16O2 $#C 18O2 $(1)

The entropy, S, increases with increasing the number of possible combinations, W:
𝑆 = 𝑘! ln 𝑊
W = 1 for

12

16

C O2 and

12

18

(2)

C O2. In contrast, W = 2 for a

12

16

18

C O O molecule because the oxygen

atoms are distinguishable. As the right-hand side of the equilibrium shown in eq. 1 has two
12

C16O18O molecules, W = 22 = 4.

The following question A.1 is not included in the official IChO Problems 2021 because of the IChO
exam length regulation.
A.1

The equilibrium of the oxygen isotope is accomplished between CO2 and H2O over a long
period. A separation factor, 𝛼, of the oxygen atom for this equilibrium is defined as follows.
18

𝛼 = 18

O/ 16O in CO2
O/ 16O in H2 O=

(2#C 18O2 $+[C 16O 18O])/(2#C 16O2 $+#C 16O 18O$)
18
16[H2 O] / [H2 O](3)When T→+∞, 𝛼 is expressed without using [C16O18O] as follows.
a𝛼=#C 18O2 $ / #C 16O2 $
18

c

16

#H2 O$ / #H2 O$
Using eq. 1, calculate the values of a−d.

18

b

d

(4)

[C #$O #%O]&= 4 → [C #$O #%O] = 2[C #$O& ]#/& [C #%O& ]#/&
[C #$O& ][C #%O& ]Therefore,##

2[C #%O& ] + [C #$O #%O] 2[C #%O& ] + 2[C #$O& ]& [C #%O& ]&=##2[C #$O& ] + [C #$O #%O]
2[C #$O& ] + 2[C #$O& ]& [C #%O& ]&
#

=

#

#

[C #%O& ]& 4[C #$O& ]& + [C #%O& ]& 5###

[C #$O& ]& 4[C #$O& ]& + [C #%O& ]& 5#=

[C #%O& ]&#[C #$O& ]&Substitute this relationship into eq. 3:##(2[C #%O& ] + [C #$O #%O])/(2[C #$O& ] + [C #$O #%O]) [C #%O& ]& / [C #$O& ]&𝛼==[H& #%O] / [H& #$O]
[H& #%O] / [H& #$O]
Therefore, a = b = 1/2, c = d = 1.
A.2

The enthalpy change, ΔH, of eq. 5 is positive regardless of the temperature.
H2 + DI ⇄ HD + HI

(5)

Calculate the equilibrium constant, K, for eq. 5 at very low (think of T→0) and very high (thinkof T→+∞) temperatures. Assume that the reaction remains unchanged at these temperatures
and that ΔH converges to a constant value for low / high temperatures.
ΔG =ΔH − TΔS = −RT ln K → ln K = −ΔH /RT +ΔS /R(K = exp(−ΔH / RT ) exp(ΔS / R ))T→0: As ΔH > 0, ln K converges to −∞ and therefore K = 0
T→+∞: ln K =ΔS /RGiven that ΔS per 1 mole is 𝑁( 𝑘! ln 𝑊 = 𝑅 ln 2, K = 2
The followings are the other examples of K for T→+∞.
H2O + HDS ⇄ HDO + H2S

K→1

NH3 + HD ⇄ NH2D + H2

K → 3/2

14

K→1

N15NO + 14NO ⇄ 14N2O + 15NO

The molecules whose W is not 1 are; HDS: W = 2, HDO: W = 2, HD: W = 2, NH2D: W = 3.
Note that two nitrogen atoms in 14N15NO are not equivalent, and therefore K = 1 for T→+∞.

19

The following question A.3 is not included in the official IChO Problems 2021 because of potential
difficulty in marking by following the IChO marking regulation.
By using the temperature dependence of the equilibrium constant between isotopologues insubstances, we can estimate the temperature at which the substances existed. The temperaturedependence of the equilibrium constant can be determined experimentally. As an example, let us
estimate the temperature at which the CaCO3 was produced in water using the isotope ratio ofoxygen, 18O / 16O. A separation factor of the oxygen atom for the equilibrium between CaCO3 andH2O is defined as follows. [Geochim. Cosmochim. Acta 61, 3461 (1997)]
18

𝛼=

16

O/ O in CaCO3

18

16

O/ O in H2 O

(6)

The temperature dependence of 𝛼 in 10 − 40 °C is determined as follows where T is given as
absolute temperature (unit: K).
ln 𝛼 =

A.3

18.03− 3.242×10-2
T

(7)

18

O / 16O of CaCO3 produced at a certain temperature was 2.051741×10−3. By assuming that18

O / 16O of H2O is 1.988557×10−3 regardless of the temperature, estimate the temperature

at which the CaCO3 was produced.
α = 2.051741×10−3/ 1.988557×10−3 = 1.031774
T = 18.03 / (ln α + 3.242×10−2) = 283.0 K
For the estimation of the temperature using 18O / 16O, we have to assume 18O / 16O of H2O. However,18

O / 16O of H2O varies with time and place. To solve this problem, the molecule which contains tworare isotopes, called doubly-substituted isotopologue, can be used.
The ΔH of eq. 8 originates from molecular vibration.
2HD ⇄ H2 + D2

(8)

At T = 0 K, the vibrational energy of a diatomic molecule whose vibration frequency is 𝜈 s-1 isexpressed as:
1
𝐸 = ℎ𝜈
2

(9)

1 𝑘
B
2𝜋 𝜇

(10)

𝜈=

Wherein 𝑘 is the force constant and 𝜇 the reduced mass, which is expressed in terms of the massof the two atoms in the diatomic molecule, 𝑚# and 𝑚& , according to:
𝜇=

𝑚# 𝑚&
𝑚# + 𝑚&

20

(11)

A.4

The vibrational frequency of H2 is 4161.0 cm−1 in units of wavenumbers. Calculate the ΔHof the following equation at T = 0 K in units of J mol−1.
2HD → H2 + D2

(12)Assume that:Ø only the vibrational energy contributes to the ΔH.
Ø the 𝑘 values for H2, HD, and D2 are identical.
Ø the masses of H and D are 1.0078 Da and 2.0141 Da, respectively.
𝜇)! = 0.5039 Da, 𝜇)* = 0.6717 Da, 𝜇)! = 1.0071 Da
Using 𝜈 =

#
&+

,

-"

.

.#

-"

"!

#!

O- , ."# = O- ! and . ! = O- ! are obtained.
"!

"#

The frequency of the H2 vibration is 4161.0 cm−1 in units of wavenumbers.
Therefore, the frequencies of the molecular vibration for HD and D2 are calculated to be3604.0 cm−1 and 2943.4 cm−1, respectively.
The difference of the zero-point energies of eq. 12 is calculated to be𝜈P = (4161.0 + 2943.4) / 2 − 3604.0 = −51.8 cm−1.
𝐸 = 𝑁/ ℎ𝜈 = 𝑁/ ℎ𝑐𝜈P (𝜈P: frequency in wavenumbers), and therefore E =ΔH = −620 J mol−1.
As the temperature increases, the equilibrium constant K of eq. 8 converges to 1/4 because thecontribution of entropy to K overwhelms the contribution of enthalpy. Therefore, the abundance
ratio among H2, HD, and D2 depends on the temperature. Although the amount of this change isquite small, the ratio of change is observable for D2, the amount of which is inherently very small.
Let us use this dependence on the temperature in a system in equilibrium. Δ*& is defined as theratio of change of the molar ratio of D2.
ΔD =

RD2
−1
RD2*

(13)

Here, RD2 refers to [D2] / [H2] in the sample and RD2* to [D2] / [H2] at T→+∞. It should be noted herethat the distribution of isotopes becomes random at T→+∞.
A.5

Calculate ΔD2 with natural D abundance when the isotopic exchange is in equilibrium at thetemperature where K in eq. 8 is 0.300. Assume that the natural abundance ratios of D and H
are 1.5576×10−4 and 1 − 1.5576×10−4, respectively.

Solution 1:
Let the sum of the concentrations of H2, HD, and D2 be C.
・T→+∞ (K = 1/4)
[H& ]0 = (1 − 1.5576 × 1012 )& 𝐶 = 9.9969×10−1 C[D& ]0 = (1.5576 × 1012 )& 𝐶 = 2.4261×10−8 C

21

・K = 0.300
Let the amount of change in the molar ratio be x.
[H& ][D& ] ([H& ]0 /𝐶 + 𝑥)([D& ]0 /𝐶 + 𝑥)
=
=𝐾
[HD]&
([HD]0 /𝐶 − 2𝑥)&
Solve the equation for x when K = 0.300:(1 − 4𝐾)𝑥 & + V[HD]0[HD]&0[H& ]0 [D& ]0[H& ]0 [D& ]0
+
+ 4𝐾
−
𝐾
W𝑥 + VW = 0,𝐶𝐶
𝐶
𝐶&
𝐶&

x = 4.8504×10−9.
From this value, we obtain [H& ] = 9.9969×10−1 C and [D& ] = 2.9112×10−8 C.
Δ*& =

𝑅*&
2.9112 × 101% /9.9969 × 101#−1=
− 1 = 0.200∗
𝑅*&
2.4261 × 101% /9.9969 × 101#

Solution 2:
By using an appropriate approximation, we can obtain the answer without calculating theconcentration of each species. The use of this approximation is rationalized considering that thevalues of [H& ], [HD], and [D& ] differ by 4 orders of magnitudes while the required significant digits
for K are 3.
[H& ][D& ] ([H& ]0 + 𝑦)([D& ]0 + 𝑦) [H& ]0 [D& ]
=
≃
[HD]&
([HD]0 − 𝑦)&[HD]0 &
Δ*& =[D& ]/[H& ][D& ][H& ]0 [D& ]/[HD]&0
𝑅*&
0.300
−
1
=
−
1
≃
−
1
=
−1≃
− 1 ≃ 0.200∗&[D& ]0[D& ]0 /[H& ]0
𝑅*&[H& ]0 [D& ]0 /[HD]00.250

The problem raised in A.3 is that 18O / 16O of H2O varies with time and place despite the absolutevalue of the abundance ratio is required for the estimation of temperature. In contrast, Solution 2
above shows that the abundance ratio of D does not affect the numerical value of Δ*& (if theabundance ratio of D is significantly small). This is the reason why we use the ratio of change ofthe molar ratio such as Δ*& and Δ24 (next question).
In general, as described in the example above, the molar ratio of the doubly-substituted
isotopologue that contains two heavy isotope atoms in one molecule increases as the temperature
decreases. Let us consider the molar ratio of CO2 molecules with molecular weights of 44 and 47,
which are described as CO2[44] and CO2[47] below. The quantity Δ47 is defined as:
Δ47 =

R47
−1
R47*

(14)

R47 refers to [CO2[47]] / [CO2[44]] in the sample and R47* to [CO2[47]] / [CO2[44]] at T→+∞. The
natural abundances of carbon and oxygen atoms are shown below; ignore isotopes that are notshown here.

22

12

natural abundance
16

natural abundance
A.6

C

13

0.988888

0.011112

C

O

17

O

18

0.997621

0.000379

0.002000O

List all possible isotopologues of CO2[47], indicating the isotopes of carbon and oxygen
explicitly (for example: 12C16O2). Indicate which is the most common isotopologue among
these.

13

C16O18O, 12C17O18O, 13C17O2

Most common isotopologue: 13C16O18O
The temperature dependence of Δ47 is determined as follows, where T is given as the absolutetemperature in units of K: [Geochim. Cosmochim. Acta 61, 3461 (1997)]
Δ47 =

A.7

36.2
2

T

+ 2.920×10–4(15)

The R47 of fossil plankton obtained from the Antarctic seabed was 4.50865×10−5. Estimate
the temperature using this R47. This temperature is interpreted as the air temperature during
the era in which the plankton lived. Consider only the most common isotopologue of CO2[47]for the calculation. [Paleoceanography 30, 1305 (2015)]

The molar ratio of 13C16O18O in the case where all the isotopes are distributed randomly is 0.011112
×0.002000×0.997621×2 [O is indistinguishable] = 4.43423×10−5.
The molar ratio of 12C16O2 in the case where all the isotopes are distributed randomly is 0.988888
×0.9976212=9.84188×10−1.
R47* = 4.43423×10−5 / 9.84188×10−1 = 4.50547×10−5
Δ47 = 7.06×10−4
T = 296 K

One of the technical problems for this method is the requirement of high-accuracy measurement of
R47. This is the reason why the use of doubly-substituted isotopologue has not been popular yet.

23

Question 3: Lambert–Beer Law?
In this problem, ignore the absorption of the cell and the solvent. The temperatures of all solutions
and gases are kept constant at 25 °C.
The following Part A is not included in the official IChO Problems 2021 because of the IChO exam
length regulation.
Part A
A.1

The absorption value of an aqueous solution of substances A and B, which do not interact
with each other, was 1.00 at the wavelength of λ1. This solution was diluted twice using purewater. Choose one of the correct absorption values of this diluted solution at λ1.
□ Less than 0.50

A.2

■0.50

□ More than 0.50

The absorbance values of hydrochloric acid with pH = 3 containing a small amount of methyl
orange (pKa = 3.5), which is used as a pH indicator, at wavelengths of 400 nm, 465 nm, and
500 nm were A400, A465, and A500, respectively. This solution was diluted 10 times using purewater. Choose one of the correct absorption values of this diluted solution at wavelengths of
400 nm, 465 nm, and 500 nm. Refer to the absorption spectrum of methyl orange (Fig. 1) if
necessary. The Isosbestic points of methyl orange exist at wavelengths of 352 nm and 465
nm.

Fig. 1 Absorption spectra of methyl orange solution with pH = 3.0 and 4.0. The concentration of
methyl orange for both solutions is the same. [RSC Adv., 10, 11311 (2020)]

24

At 400 nm:□ Less than A400/10 (< A400/10)□ A400/10■ More than A400/10 (> A400/10)■ A465/10□ More than A465/10 (> A465/10)□ A500/10□ More than A500/10 (> A500/10)

At 465 nm:□ Less than A465/10 (< A465/10)
At 500 nm:■ Less than A500/10 (< A500/10)

Note that the pH after dilution is 4 and therefore the composition of protonated and deprotonatedspecies changes by the dilution.
The message of Part A is that Lambert−Beer law holds only when the chemicals do not show any
chemical change (such as acid−base equilibrium). Through the use of this fact, we made aseemingly mysterious situation in Part B; absorbance does not change even if the solution was
diluted.

25

Part BAn aqueous solution X was prepared using AH and NaA. The concentrations [A−], [AH], and [H+] insolution X are 1.00×10−2 mol L−1, 1.00×10−3 mol L−1, and 1.00×10−4 mol L−1, respectively, which arecorrelated via the following acid−base equilibrium:
AH ⇄ A− + H+

𝐾𝐾 =[A− ][H+ ]
[AH]

(1)

The optical path length was l in Part B. Ignore the density change upon dilution. Assume that no
chemical reactions other than eq 1 occur.
B.1

The absorbance of X was A2 at a wavelength of λ2. Then, solution X was diluted to twice its
initial volume using hydrochloric acid with pH = 2.500. The absorbance of the diluted
solution was also A2 at λ2. Calculate the ratio of εAH : εA, where εAH and εA represent theabsorption coefficients of AH and of A−, respectively, at λ2.

𝐾𝐾 =[A− ][H+ ] (1.00 × 10−2 )(1.00 × 10−4 )== 1.00 × 10−3 mol L−1[AH]
1.00 × 10−3

Concentrations before the dilution:[AH] = 1.00×10−3 mol L−1[A−] = 1.00×10−2 mol L−1[H+] = 1.00×10−4 mol L−1

Concentrations just after the dilution (nominal initial concentrations before the equilibrium):[AH] = 5.00×10−4 mol L−1[A−] = 5.00×10−3 mol L−1[H+] = (1.00×10−4 + 3.16×10−3 ) / 2 = 1.63×10−3 mol L−1（pH = 2.500 → [H+] = 3.16×10−3 mol L−1）
Equilibrium after the dilution:[A− ][H+ ] (5.00 × 10−3 − 𝑥𝑥)(1.63 × 10−3 − 𝑥𝑥)== 1.00 × 10−3 mol L−1[AH](5.00 × 10−4 + 𝑥𝑥)
Solve the equation for x: x = 1.19 ×10−3 mol L−1→ [A−] = 3.81 ×10−3 mol L−1, [AH] = 1.69 ×10−3 mol L−1
Therefore,
1.00 × 10−2 𝜀𝜀A + 1.00 × 10−3 𝜀𝜀AH = 3.81 × 10−3𝜀𝜀A + 1.69 × 10−3 𝜀𝜀AH

By solving this equation: εAH = 9.0 εA26

In the case where εAH > 9.0 εA, absorbance at λ2 increases by dilution! We would like to emphasizethat the concentration of chemicals fulfills conservation law even under such a seemingly
mysterious situation.
The following question B.2 is not included in the official IChO Problems 2021 because of the IChO
exam length regulation.
B.2

X was diluted to four times its initial volume using hydrochloric acid with a certain pH value.
The absorbance of this diluted solution was also A2 at λ2. Calculate the pH value of thehydrochloric acid used for the dilution.

Concentrations before the dilution:[A− ][H+ ] (1.00 × 10−2 )(1.00 × 10−4 )== 1.00 × 10−3 mol L−1[AH]
1.00 × 10−3
Concentrations after the dilution:[A− ][H+ ] (2.50 × 10−3 − 𝑥𝑥)[H+ ]== 1.00 × 10−3 mol L−1[AH](2.50 × 10−4 + 𝑥𝑥)
As the absorbance is not changed by the dilution, the following relationship holds:
1.00 × 10−2 𝜀𝜀A + 1.00 × 10−3𝜀𝜀AH = (2.50 × 10−3 − 𝑥𝑥)𝜀𝜀A + (2.50 × 10−4 + 𝑥𝑥)𝜀𝜀AH

Solve the equation for x using εAH = 9.0 εA: x = 1.8×10−3 mol L−1→ [A−] = 7.2×10−4 mol L−1, [AH] = 2.0×10−3 mol L−1

Using these values provides the concentration of protons after the dilution: [H+] = 2.8×10−3 mol L−1
Let the concentration of the hydrochloric acid used for the dilution be C:[(3C + 1.00×10−4) / 4] − x = 2.8×10−3 mol L−1
Solve the equation for C: C = 6.1×10−3 mol L−1 → pH = 2.21

27

Part CLet us consider the following equilibrium in the gas phase.
D ⇄ 2M(2)

This gas is filled with a cuboid container that has a transparent movable wall with a cross−sectionof S as shown in Fig. 2. The partial pressures of D and M are pD and pM, respectively. Lambert–
Beer law also holds for the absorption value of gas: the absorption value A of the gas is the
product of the absorption coefficient ε, the number density of the gas n/V (n is the mole number ofthe gas and V is the volume of the gas), and the optical path length of l: A = ε (n/V) l. Assume thatthe temperature of the gas was kept at a constant value and the gas is an ideal gas.

Fig. 2 A container with a transparent movable wall and the direction of absorption measurements.

28

C.1

The container is filled with pure gas D at the pressure P0 and the volume V0, and theequilibrium is achieved under constant pressure. The absorption values of the gas at the
wavelength of λ3a measured from the direction of A are A3a both just after the filling (onlygas D exists) and after the equilibrium is achieved. After this, the pressure was set to be P1
and the equilibrium is achieved. Calculate the absorption values of the gas at the
wavelength of λ3a measured from the direction of A.
Hint: First, determine the ratio 𝜀𝜀D /𝜀𝜀M at λ3a, where 𝜀𝜀D and 𝜀𝜀M represent the absorption
coefficients of D and of M, respectively.

Using the Lambert–Beer law, the absorbance of D and M at an arbitrary volume V are expressed
as follows. Here, let the optical path length be 𝑙𝑙A . The ideal gas law is used.
𝑛𝑛D
𝑝𝑝D
𝐴𝐴D = 𝜀𝜀D � � 𝑙𝑙A = 𝜀𝜀D𝑙𝑙
𝑉𝑉
𝑅𝑅𝑅𝑅 A

𝑛𝑛M
𝑝𝑝M
𝐴𝐴M = 𝜀𝜀M � � 𝑙𝑙A = 𝜀𝜀M𝑙𝑙
𝑉𝑉
𝑅𝑅𝑅𝑅 AFirst, determine the ratio 𝜀𝜀D /𝜀𝜀M at λ3a. There are two ways of solutions.
Solution 1:

Let the initial number of moles of D be 𝑛𝑛0 . The absorbance at the initial state is:
The absorbance after equilibrium is:

𝐴𝐴3a = 𝐴𝐴D =𝐴𝐴3a = 𝐴𝐴D + 𝐴𝐴M =𝜀𝜀D 𝑛𝑛0𝑙𝑙
𝑉𝑉0 A

𝜀𝜀D 𝑛𝑛D + 𝜀𝜀M 𝑛𝑛M
𝑙𝑙A𝑉𝑉

From the ideal gas law, the following relationship is obtained:
𝑛𝑛0 𝑃𝑃0 𝑛𝑛D + 𝑛𝑛M==
𝑉𝑉0 𝑅𝑅𝑅𝑅𝑉𝑉

From these equations, the following relationship is obtained:
𝐴𝐴3a =

𝜀𝜀D 𝑛𝑛0𝜀𝜀D (𝑛𝑛D + 𝑛𝑛M )𝜀𝜀D 𝑛𝑛D + 𝜀𝜀M 𝑛𝑛M
𝑙𝑙 =
𝑙𝑙A =
𝑙𝑙A
𝑉𝑉0 A
𝑉𝑉
𝑉𝑉
𝜀𝜀D 𝑛𝑛M = 𝜀𝜀M 𝑛𝑛M

0 = (𝜀𝜀M − 𝜀𝜀D )𝑛𝑛M

As 𝑛𝑛M > 0 after the equilibrium, 𝜺𝜺𝐃𝐃 (𝝀𝝀𝟑𝟑𝐚𝐚 ) = 𝜺𝜺𝐌𝐌 (𝝀𝝀𝟑𝟑𝐚𝐚 ) holds.
Solution 2:

The absorbance at the initial state is:
𝐴𝐴3a = 𝐴𝐴D = 𝜀𝜀D
29

𝑃𝑃0
𝑙𝑙
𝑅𝑅𝑅𝑅 AThe absorbance after equilibrium is:
𝐴𝐴3a = 𝐴𝐴D + 𝐴𝐴M =𝜀𝜀D 𝑝𝑝D + 𝜀𝜀M 𝑝𝑝M
𝑙𝑙A
𝑅𝑅𝑅𝑅

From these equations, the following relationship is obtained:
𝐴𝐴3a = 𝜀𝜀DUsing the fact that 𝑝𝑝D = 𝑃𝑃0 − 𝑝𝑝M,

𝑃𝑃0
𝜀𝜀D 𝑝𝑝D + 𝜀𝜀M 𝑝𝑝M
𝑙𝑙A =
𝑙𝑙A
𝑅𝑅𝑅𝑅
𝑅𝑅𝑅𝑅

𝜀𝜀D 𝑃𝑃0 = 𝜀𝜀D (𝑃𝑃0 − 𝑝𝑝M ) + 𝜀𝜀M 𝑝𝑝M = 𝜀𝜀D 𝑃𝑃0 + (𝜀𝜀M − 𝜀𝜀D )𝑝𝑝M
0 = (𝜀𝜀M − 𝜀𝜀D )𝑝𝑝M

As 𝑝𝑝M > 0 after the equilibrium, 𝜺𝜺𝐃𝐃 (𝝀𝝀𝟑𝟑𝐚𝐚 ) = 𝜺𝜺𝐌𝐌 (𝝀𝝀𝟑𝟑𝐚𝐚 ) holds.
Lambert−Beer law at the pressure at P1 is as follows.
𝐴𝐴(𝑝𝑝 = 𝑃𝑃1 ) = �𝜀𝜀D
C.2

𝑝𝑝D 𝑝𝑝M
𝑝𝑝D
𝑝𝑝M
𝑃𝑃1
𝑃𝑃0
𝑷𝑷𝟏𝟏
𝑃𝑃1
+ 𝜀𝜀M � 𝑙𝑙A = 𝜀𝜀D � + � 𝑙𝑙A = 𝜀𝜀D𝑙𝑙 = 𝜀𝜀D𝑙𝑙 ⋅ = 𝑨𝑨𝟑𝟑𝐚𝐚
𝑅𝑅𝑅𝑅
𝑅𝑅𝑅𝑅
𝑅𝑅𝑅𝑅 𝑅𝑅𝑅𝑅
𝑅𝑅𝑅𝑅 A
𝑅𝑅𝑅𝑅 A 𝑃𝑃0
𝑷𝑷𝟎𝟎

The container is filled with the gas D at the pressure P0 and the volume V0, and theequilibrium is achieved under constant pressure. The absorption values of the gas at the
wavelength of λ3b measured from the direction of B are A3b both just after the filling andafter the equilibrium is achieved. After this, the pressure was set to be P2 and the equilibriumis achieved. Calculate the absorption values of the gas at the wavelength of λ3b measured
from the direction of B.
Hint: First, determine the ratio 𝜀𝜀D /𝜀𝜀M at λ3b.

Using the Lambert–Beer law, the absorbance of D and M at an arbitrary volume V are expressed
as follows. Here, let the optical path length be 𝑙𝑙B . Note that 𝑉𝑉 = 𝑙𝑙B 𝑆𝑆.
𝑛𝑛D
𝑛𝑛D
𝐴𝐴D = 𝜀𝜀D � � 𝑙𝑙B = 𝜀𝜀D
𝑉𝑉
𝑆𝑆

𝑛𝑛M
𝑛𝑛M
𝐴𝐴M = 𝜀𝜀M � � 𝑙𝑙B = 𝜀𝜀M
𝑉𝑉𝑆𝑆First, determine the ratio 𝜀𝜀D /𝜀𝜀M at λ3b. There are two ways of solutions.
Solution 1:

Let the initial number of moles of D be 𝑛𝑛0 . The absorbance at the initial state (optical path length:
𝑙𝑙B0) is:𝐴𝐴3b = 𝐴𝐴D =𝜀𝜀D 𝑛𝑛0𝑙𝑙
𝑉𝑉0 B0

The absorbance after equilibrium (optical path length: 𝑙𝑙Bf ) is:30

𝐴𝐴3b = 𝐴𝐴D + 𝐴𝐴M =𝑉𝑉 𝜀𝜀D 𝑛𝑛D + 𝜀𝜀M 𝑛𝑛M
𝜀𝜀D 𝑛𝑛D + 𝜀𝜀M 𝑛𝑛M
𝜀𝜀D 𝑛𝑛D + 𝜀𝜀M 𝑛𝑛M
𝑙𝑙Bf =
𝑙𝑙B0 =
𝑙𝑙B0
𝑉𝑉
𝑉𝑉
𝑉𝑉0
𝑉𝑉0

From these equations, the following relationship is obtained:
𝐴𝐴3b =

Using the fact that 𝑛𝑛M = 2(𝑛𝑛0 − 𝑛𝑛D ),𝜀𝜀D 𝑛𝑛0𝜀𝜀D 𝑛𝑛D + 𝜀𝜀M 𝑛𝑛M
𝑙𝑙B0 =
𝑙𝑙B0
𝑉𝑉0
𝑉𝑉0

𝜀𝜀D 𝑛𝑛0 = 𝜀𝜀D 𝑛𝑛D + 2𝜀𝜀M (𝑛𝑛0 − 𝑛𝑛D ) = 2𝜀𝜀M 𝑛𝑛0 + (𝜀𝜀D − 2𝜀𝜀M )𝑛𝑛D
(𝜀𝜀D − 2𝜀𝜀M )𝑛𝑛0 = (𝜀𝜀D − 2𝜀𝜀M )𝑛𝑛D

As 𝑛𝑛0 > 𝑛𝑛D after the equilibrium, the equality 𝜺𝜺𝐃𝐃 (𝝀𝝀𝟑𝟑𝐛𝐛 ) = 𝟐𝟐𝜺𝜺𝐌𝐌 (𝝀𝝀𝟑𝟑𝐛𝐛 ) holds.
Solution 2:

Let the initial number of moles of D be 𝑛𝑛0 . The absorbance at the initial state is:
The absorbance after equilibrium is:

𝐴𝐴3b = 𝐴𝐴D = 𝜀𝜀D

𝐴𝐴3b = 𝐴𝐴D + 𝐴𝐴M = 𝜀𝜀D

𝑛𝑛0
𝑆𝑆

𝑛𝑛D
𝑛𝑛M
+ 𝜀𝜀M
𝑆𝑆
𝑆𝑆

From these equations, the following relationship is obtained:
𝐴𝐴3b = 𝜀𝜀DUsing the fact that 𝑛𝑛M = 2(𝑛𝑛0 − 𝑛𝑛D ) ,

𝑛𝑛0
𝑛𝑛D
𝑛𝑛M
= 𝜀𝜀D
+ 𝜀𝜀M
𝑆𝑆
𝑆𝑆
𝑆𝑆

𝜀𝜀D 𝑛𝑛0 = 𝜀𝜀D 𝑛𝑛D + 2𝜀𝜀M (𝑛𝑛0 − 𝑛𝑛D ) = 2𝜀𝜀M 𝑛𝑛0 + (𝜀𝜀D − 2𝜀𝜀M )𝑛𝑛D
(𝜀𝜀D − 2𝜀𝜀M )𝑛𝑛0 = (𝜀𝜀D − 2𝜀𝜀M )𝑛𝑛D

As 𝑛𝑛0 > 𝑛𝑛D after the equilibrium, the equality 𝜺𝜺𝐃𝐃 (𝝀𝝀𝟑𝟑𝐛𝐛 ) = 𝟐𝟐𝜺𝜺𝐌𝐌 (𝝀𝝀𝟑𝟑𝐛𝐛 ) holds.
Lambert−Beer law at the pressure at P2 is as follows.
𝐴𝐴(𝑝𝑝 = 𝑃𝑃2 ) = 𝜀𝜀D

𝑛𝑛D
𝑛𝑛M
𝑛𝑛D + 𝑛𝑛M /2
𝑛𝑛0
+ 𝜀𝜀M
= 𝜀𝜀D
= 𝜀𝜀D
= 𝑨𝑨𝟑𝟑𝐛𝐛
𝑆𝑆
𝑆𝑆
𝑆𝑆
𝑆𝑆

These questions in Part C can be solved without calculating anything by considering what
happens inside the container.
The result of C.1 shows that the absorption value of the gas at the wavelength of λ3a measured
from the direction of A is proportional to the pressure of the container. This is because the number
density of particles in the optical path is proportional to the pressure (in the case of an ideal gas),
and the absorbance measured from the direction of A is determined only by the number densityof particles in the optical path regardless of D or M since 𝜀𝜀D (𝜆𝜆3a ) = 𝜀𝜀M (𝜆𝜆3a ) holds. 𝜀𝜀D (𝜆𝜆3a ) =𝜀𝜀M (𝜆𝜆3a ) is obtained from the fact that the absorption value remains constant by the observation of31

the absorption value under the condition of the constant pressure and optical path length after
filling the gas D, which means that the absorption values do not change whether the particle in the
optical path is D or M.
The result of C.2 shows that the absorption value of the gas at the wavelength of λ3b measured
from the direction of B is independent of the pressure of the container. The absorption value
measured from the direction of B is proportional to the number density of particles in the optical
path multiplied by the length in the direction of B, which is proportional to the number of particles
in the optical path. Because 𝜀𝜀D (𝜆𝜆3b ) = 2𝜀𝜀M (𝜆𝜆3b ), the absorbance measured from the direction of B

is determined by the number of “monomers” M in the optical path if we count “dimer” D as “2monomers”. In this interpretation, the number of “monomers” does not depend on the pressure
and therefore the absorbance becomes constant. 𝜀𝜀D (𝜆𝜆3b ) = 2𝜀𝜀M (𝜆𝜆3b ) is obtained from the factthat the absorption value remains constant by the observation of the absorption value under thecondition of the constant number of monomers (count “dimer” D as “2 monomers”) after filling thegas D, which means that the absorption values do not change when the particle in the optical pathchanges from D to M. This result is based on only the conservation law of particles and therefore
the assumption of an ideal gas is not required.
Please check the reference figure shown on the next page.

32

33

Problem 4: Redox Chemistry of Zinc

Zinc has long been used as alloys for brass and steel materials. The zinc contained in industrial
wastewater is separated by precipitation to detoxify the water, and the obtained precipitate is
reduced to recover and reuse it as metallic zinc.
Part AThe dissolution equilibrium of zinc hydroxide Zn(OH)2 (s) at 25 °C and the relevant equilibrium
constants are given in eq. 1-4.
Zn(OH)2 (s) ⇄ Zn2+ (aq) + 2OH− (aq),Zn(OH)2 (s) ⇄ Zn(OH)2 (aq),Zn(OH)2 (s) + 2OH− (aq) ⇄ Zn(OH)42− (aq),
H2O (l) ⇄ H+ (aq) + OH− (aq),Ksp = 1.74 × 10−17(1)

K1 = 2.62 × 10−6

(2)

Kw = 1.00 × 10−14(4)

K2 = 6.47 × 10−2

(3)

The solubility, S, of zinc (concentration of zinc in a saturated aqueous solution) is given in eq. 5.
S = [Zn2+ (aq)] + [Zn(OH)2 (aq)] + [Zn(OH)42− (aq)]
A.1

(5)

When the equilibria in eq. 1-4 are established, calculate the pH range in which [Zn(OH)2(aq)] is the greatest among [Zn2+ (aq)], [Zn(OH)2 (aq)] and [Zn(OH)42− (aq)].

Solution 1:
𝐾𝐾𝑠𝑠𝑠𝑠From [Zn(OH)2 (aq)] > [Zn2+ (aq)]: [OH− ]2 < 𝐾𝐾1
Solve this for [OH− ]: 8.4 < pH

− 2
From [Zn(OH)2 (aq)] > [Zn(OH)2−
4 ]: 𝐾𝐾1 > 𝐾𝐾2 [OH ]Solve this for [OH− ]: pH < 11.8Thus,

8.4 < pH < 11.8
34

Solution 2:
From (1):log[Zn2+ ][OH− ]2 = log 𝐾𝐾splog[Zn2+ ] = log 𝐾𝐾sp − 2log[OH− ]From (2):log[Zn(OH)2 (aq)] = log 𝐾𝐾1[Zn(OH)2 (aq)] > [Zn2+ (aq)]:log 𝐾𝐾sp − 2log[OH− ] < log 𝐾𝐾1

log 𝐾𝐾sp − 2(−14 + pH) < log 𝐾𝐾1
From (3):− 2log[Zn(OH)2−
4 ]/[OH ] = log 𝐾𝐾2−
log[Zn(OH)2−
4 ] = 2log[OH ] + log 𝐾𝐾2[Zn(OH)2 (aq)] > [Zn(OH)2−4 ]:log 𝐾𝐾1 > 2log[OH− ] + log 𝐾𝐾2

log 𝐾𝐾1 > 2(−14 + pH) + log 𝐾𝐾2
Thus,

8.4 < pH < 11.8

Fig. pH dependences of each species

35

A.2

A saturated aqueous solution of Zn(OH)2 (s) with pH = 7.00 was prepared and filtered. NaOH
was added to this filtrate to increase its pH to 12.00. Calculate the molar percentage of zincthat precipitates when increasing the pH from 7.00 to 12.00. Ignore the volume andtemperature changes.

For pH = 12.00:
2−log[Zn(OH)4 ] = −29.19 + 2 pH = −5.19

log[Zn(OH)2 (aq)] = −5.58

log�Zn2+ � = 11.24 − 2 pH = −12.76 (negligible)

Thus,

For pH = 7.00:

𝑆𝑆 = 9.0865 × 10−6 mol L−1

log�Zn2+ � = 11.24 − 2 pH = −2.76

log[Zn(OH)2 (aq)] = −5.582−log[Zn(OH)4 ] = −29.19 + 2 pH = −15.19 (negligible)

Thus,

𝑆𝑆 = 1.7404 × 10−3 mol L−1
The percentage of zinc precipitated is:

1.7404 − 0.0090865= 0.9948
1.7404= 99.5%In Part A, we consider from the viewpoint of the equilibrium constant that the saturated solubility ofzinc hydroxide, which is a poorly soluble salt, increases as the acidity or basicity increases. This isbecause the salt dissolves as a zinc ion when it is acidic and as a complex ion when it is basic.
Since the saturated solubility changes according to the exponential function with respect to pH, it
was shown that most of the zinc present in the aqueous solution can be recovered as a precipitateby appropriately changing the pH.

36

Part B
Next, the recovered zinc hydroxide is heated to obtain zinc oxide according to the reaction below:
Zn(OH)2 (s) → ZnO (s) + H2O (l)(6)

The zinc oxide is then reduced to metallic zinc by reaction with hydrogen:ZnO (s) + H2 (g) → Zn (s) + H2O (g)
B.1

(7)

In order for reaction (7) to proceed at a hydrogen pressure kept at 1 bar, it is necessary to
reduce the partial pressure of the generated water vapor. Calculate the upper limit for thepartial pressure of water vapor to allow reaction (7) to proceed at 300 °C. Here, the Gibbs
formation energies of the zinc oxide and water vapor at 300 °C and 1 bar are ΔGZnO (300 °C)= −2.90×102 kJ mol−1 and ΔGH2O (300 °C) = −2.20×102 kJ mol−1, respectively.
1

Zn + 2 O2 → ZnO, ∆𝐺𝐺ZnO (300 ℃) = −2.90 × 102 kJ mol−1
1

H2 + O2 → H2 O, ∆𝐺𝐺H2 O (300 ℃) = −2.20 × 102 kJ mol−1

Thus,

2

ZnO + H2 → Zn + H2 O, ∆𝐺𝐺 = ∆𝐺𝐺H2O (300 ℃) − ∆𝐺𝐺ZnO (300 ℃)
ln 𝐾𝐾 = ln

𝑝𝑝H2O
∆𝐺𝐺
=−
𝑝𝑝H2
𝑅𝑅𝑅𝑅

From 𝑇𝑇 = 573.15 K,∆𝐺𝐺 = −∆𝐺𝐺1 + ∆𝐺𝐺2 = 7.0 × 101 kJ mol−1

𝑝𝑝H2O = 4.14 × 10−7 bar = 4.1 × 10−7 bar

The reduction reaction of zinc oxide with hydrogen does not proceed in the standard state because
the standard Gibbs energy change is positive. However, the Gibbs energy change becomes
negative by lowering the partial pressure of the product (water vapor), and thus, the reaction
proceeds. Thus, the Gibbs energy change can be increased or decreased by controlling the partialpressure in most cases (when the product or reactant contains a gas).

37

Metallic zinc is used as a negative electrode (anode) material for metal-air batteries. The electrodeconsists of Zn and ZnO. It uses the following redox reaction to generate electricity with theelectromotive force (e.m.f.) at 25 °C and pressure of 1 bar, E°.
1

E° = 1.65 V

Zn (s) + 2 O2 (g) → ZnO (s),
B.2

(8)

A zinc–air battery was discharged at 20 mA for 24 hours. Calculate the change in mass ofthe negative electrode (anode) of the battery.

The reaction Zn + 2OH− → ZnO + H2O + 2e− occurs at the negative electrode and consumes 2
mol electrons per mol Zn oxidized.
Thus, the weight change is:𝑊𝑊 =0.02 A × 24 × 60 × 60
× 16
2 𝐹𝐹
= 0.14 g

We consider the reaction that occurs at the electrode of zinc-air battery when it is discharged, andcalculate the amount of zinc oxide deposited from Faraday's law.
B.3

Consider the change of e.m.f. of a zinc–air battery depending on the environment. Calculate
the e.m.f. at the summit of Mt. Fuji, where the temperature and altitude are −38 °C (February)and 3776 m, respectively. The atmospheric pressure is represented by
0.0065h

5.257

P [bar]=1.013× �1 − T + 0.0065h + 273.15�

(9)

at altitude h [m] and temperature T [°C]. The molar ratio of oxygen in the atmosphere is 21 %.
The Gibbs energy change of reaction (8) is ΔGZnO (−38 °C) = −3.26×102 kJ mol−1 (1 bar).

Mt. Fuji38Solution 1:
From (9), the air pressure at 3776 m and 𝑇𝑇 = −38 ℃ is

𝑃𝑃 = 0.6011 bar

From the oxygen content of 21%, the partial pressure of oxygen is:
From the Nernst equation (𝑇𝑇 = −38 ℃):
𝐸𝐸(−38 ℃) − 𝐸𝐸 ○ (−38 ℃) = −𝐸𝐸 ○ (−38 ℃) = −Thus,∆𝐺𝐺 ○ (−38 ℃)
2 𝐹𝐹

=

𝑅𝑅𝑅𝑅
2𝐹𝐹

ln

326000
2 𝐹𝐹

1

�𝑃𝑃O2

𝑃𝑃O2 = 0.126 bar= −0.01048 V = −0.01 V= 1.6894 V = 1.69 V

𝑬𝑬(−𝟑𝟑𝟑𝟑 ℃) = 𝟏𝟏. 𝟔𝟔𝟔𝟔 𝐕𝐕

Solution 2:

From (9), the air pressure at 3776 m and 𝑇𝑇 = −38 ℃ is

𝑃𝑃 = 0.6011 bar

From the oxygen content of 21%, the partial pressure of oxygen is:1

𝑃𝑃O2 = 0.126 bar∆𝐺𝐺(−38 ℃) = ∆𝐺𝐺 ○ (−38 ℃) − 𝑅𝑅𝑅𝑅 ln 𝑃𝑃O2 = −3.24 × 102 kJ mol−1
Thus,

2

𝐸𝐸(−38 ℃) = −(° is used for 1 bar)

∆𝐺𝐺(−38 ℃)
2 𝐹𝐹= 1.68 V

We consider how the electromotive force of a zinc-air battery depends on temperature andpressure. The Nernst equation is used in order to study the dependence of the electromotive force
on the pressure and temperature (Solution 1). It can also be obtained from Gibbs energy (Solution2). The Nernst equation is the same with the pressure dependence of Gibbs energy, both of which
are ascribed to the pressure dependence of entropy.

39

B.4

Calculate the Gibbs energy change for reaction (6) at 25 °C. Note that the standard reduction
potentials, E° (Zn2+/Zn) and E° (O2/ H2O) at 25 °C and 1 bar are given as (10) and (11),respectively.
Zn2+ + 2e− → Zn,

E° (Zn2+/Zn) = −0.77 V(10)

O2 + 4H+ + 4e− → 2H2O,

E° (O2/ H2O) = 1.23 V(11)

From (10):Zn2+ + 2 e− → Zn, 𝐸𝐸 ○ = −0.77 V∆𝐺𝐺 ○ = −2𝐹𝐹 × −0.77 = 148.61 kJ mol−1 (1′ )From (8):1

Zn + O2 → ZnO, 𝐸𝐸 ○ = 1.65 V
2∆𝐺𝐺 = −2𝐹𝐹 × 1.65 = −318.45 kJ mol−1 (2′ )○From (11) :O2 + 4H+ + 4e− → 2H2 O, 𝐸𝐸 ○ = 1.23 V∆𝐺𝐺 ○ = −4 × 𝐹𝐹 × 1.23 = −474.71 kJ mol−1 (3′ )From (1):Zn2+ + 2 OH− → Zn(OH)2 (s), 𝐾𝐾sp = 1.74 × 10−17−1∆𝐺𝐺 ○ = −𝑅𝑅𝑅𝑅 ln𝐾𝐾sp= −95.612 kJ mol−1 (4′ )From (4):H+ + OH− → H2 O, 𝐾𝐾w = 1 × 10−14∆𝐺𝐺 ○ = −𝑅𝑅𝑅𝑅ln𝐾𝐾w−1 = −79.912 kJ mol−1 (5′ )From(1′ )×2+(2′ )×2−(3′ )−(4′ )×2+(5′ )×4
2:

Zn(OH)2 (s) → ZnO + H2 O,∆𝐺𝐺 ○ = 3.0~3.4 kJ mol−1(distributed depending on the handling of figures)

The Gibbs energy change of the decomposition reaction of zinc hydroxide is calculated by
combining those of the four elementary reactions. The electromotive force, solubility product,equilibrium constant, etc. are given for each elementary reaction, but it should be understood that
all of these can be associated with the Gibbs energy change of each elementary reaction.

40

Question 5: Mysterious Silicon

Although silicon is also a group-14 element like carbon, their properties differ significantly.
Part A
Unlike the carbon–carbon triple bond, the silicon–silicon triple bond in a compound formulated as
R1–SiºSi–R1 (R: organic substituent) is extremely reactive. For example, it reacts with ethylene toform a cyclic product that contains a four-membered ring.

R1 Si

Si

R1 +

H2C

CH2

Si
R1

Si
R1

When R1–SiºSi–R1 is treated with an alkyne (R2–CºC–R2), the four-membered-ring compound A
is formed as an initial intermediate. Further reaction of another molecule of R2–CºC–R2 with A
affords isomers B and C, both of which have benzene-like cyclic conjugated structures, so-called‘disilabenzenes’ that contain a six-membered ring and can be formulated as (R1–Si)2(R2–C)4. The
13

C NMR analysis of the corresponding six-membered ring skeletons Si2C4 shows two signals for

B and one signal for C.

R1 Si

A.1

Si

R1 +

R2 C

C

R2

A

R2 C

C

R2

Draw the structural formulae of A, B, and C using R1, R2, Si, and C.

41

B

+

C

R1
R1

R1
Si

Si

C

C

R2

R2
A

R1

R2

Si
Si

C
C

C
C

R1

R1
R2

R1

R2

R2

R2
B

Si
Si

C
C

C
C
R2

R2

R2

R2

R2

Si
C

C

C

C
Si

R2

R2

R1
C

Intermediate A is a four-membered ring compound, which is formed by the addition of an alkyne
R2CºCR2 to R1SiºSiR1. In consideration of the [2+2] cycloaddition reaction of R1SiºSiR1 with
ethylene giving the corresponding Si=Si-C-C- four-membered ring, compound A is assumed to bea 1,2-disilacyclobutadiene derivative with -Si=Si-C=C- skeleton. When another molecule ofacetylene is inserted into the Si-C moiety of intermediate A, the corresponding 1,2-disilabenzene
is formed. On the other hand, when acetylene is inserted into the Si-Si moiety of A, thecorresponding 1,4-disilabenzene is formed. It should be helpful information that the product isnoted as "disilabenzene Si2C4," and the structural formulas of H-substituted disilabenzenes areshown in Figure 1. Based on the information that 13C NMR spectra of compounds B and C showthat two signals and one signal, respectively, compound B is the 1,2-disilabenzene and compound
C is the 1,4-disilabenzene.

42Calculate the aromatic stabilization energy [kJ mol–1] for benzene and C (in the case of R1 =A.2

R2 = H) as positive values, using Fig. 1.
H2C

CH2

＋

H2

H 3C

CH3

ΔH = ‒135 kJ·mol‒1 （1）

H2Si

CH2

＋

H2

H3Si

CH3

ΔH = ‒213 kJ·mol‒1 （2）

H2Si

SiH2

＋

H2

H3Si

SiH3

ΔH = ‒206 kJ·mol‒1 （3）

＋

3 H2

HSi

ΔH = ‒173 kJ·mol‒1 （4）

SiH

H2Si
＋

ΔH = ‒326 kJ·mol‒1 （5）

3 H2

SiH
HSi

HSi

SiH

SiH2

SiH2
＋

3 H2

H2Si

＋

3 H2

H2Si

ΔH = ‒368 kJ·mol‒1 （6）SiH2ΔH = ‒389 kJ·mol‒1 （7）Fig. 1 Enthalpy change in some hydrogenation reactions of unsaturated systems.
The aromatic stabilization energy (ASE) can be calculated as the difference of the sum of the heatof hydrogenation of each double bond and the heat of hydrogenation of the aromatic compound.
ASE for benzene: 135 × 3 – 173 = 232 kJ mol–1
ASE for 1,4-disilabenzene (C=C + 2 Si=C): (135 + 213 × 2) – 389 = 172 kJ mol–1

43

A.3

When a xylene solution of C is heated, it undergoes isomerization to give an equilibrium
mixture of compounds D and E. The molar ratio is D : E = 1 : 40.0 at 50.0 °C and D : E = 1 :20.0 at 120.0 °C. Calculate ΔH [kJ mol–1] for the transformation of D to E. Assume that ΔH
does not depend on temperature.

T = 50 °C = 323.15 K (1/RT = 0.3722): KDE = 40
T = 120 °C = 393.15 K (1/RT = 0.3059): KDE = 20
According to ln KDE = –(ΔH°/RT) + ΔS°/R:
ΔH° = –(ln 40–ln 20) / (0.3722–0.3059) = –10.5 kJ mol–1
A.4

The isomerization from C to D and to E proceeds via transformations of π-bonds into
σ-bonds without breaking any σ-bonds. A 13C NMR analysis revealed one signal for the
Si2C4 skeleton of D and two signals for that of E. The skeleton of D does not contain any
three-membered rings, while E has two three-membered rings that share an edge. Draw thestructural formulae of D and E using R1, R2, Si, and C.

Based on the signals observed in the 13C NMR spectra, the following structures could besuggested for D and E:
1 13C signal

2 13C signals

R1
R2

R2

R1

Si
C

C

C

C
Si

R2

R2

R2

R2

DW1 R1

R1

Si
C

C
C

C
Si

R2

R2

R2

R2

PS1 R1

R1

Si
C

C
C

C
Si
R1

BZ1

potential structures for D

R2

R2

R2

R2

R1

Si
C

C
C

C
Si

R2

R2

R2

R2

DW2 R1

Si
C

C
C

C

PS2

Si

R2

R2

R1

potential structures for E(DW: Dewar benzene type, PS: Prismane type, BZ: Benzvalene type)
In addition, considering that D has no three-membered ring in its skeleton while E has two
three-membered rings that share an edge, the structures of D and E can be determined to be
those shown below:
R1
R2

R2
D

R1

Si
C

C

C

C
Si
R1

R2

R2

R2

R2
E

Si
C

C

C

C
Si

R2

R2

R1

44

Part B.
Silicon is able to form highly coordinated compounds (> four substituents) with electronegative
elements such as fluorine. As metal fluorides are often used as fluorination reagents, highly
coordinated silicon fluorides also act as fluorination reagents.
The fluorination reaction of CCl4 using Na2SiF6 was carried out as follows.
Quantification of Na2SiF6:·Preparation
Aqueous solution F: 0.855 g of Na2SiF6 dissolved in water (total volume: 200 mL).
Aqueous solution G: 6.86 g of Ce2(SO4)3 dissolved in water (total volume: 200 mL).
·Procedure
Precipitation titration of a solution F (50.0 mL) by dropwise adding solution G in the presence ofxylenol orange, which coordinates to Ce3+, as an indicator. After adding 18.8 mL of solution G, thecolor of the solution changes from yellow to magenta. The generated precipitate is a binary
compound that contains Ce3+, and the only resulting silicon compound is Si(OH)4.
Reaction of CCl4 with Na2SiF6:(Substance losses by e.g. evaporation are negligible during the following operations.)
Na2SiF6 (x [g]) was added to CCl4 (500.0 g) and heated to 300 ºC in a sealed pressure-resistantreaction vessel. (a)The unreacted Na2SiF6 and generated NaCl were removed by filtration. The
filtrate was diluted to a total volume of 1.00 L with CCl4 (solution H). The 29Si and 19F NMR spectraof solution H showed SiF4 as the only silicon compound. In the 19F NMR spectrum, in addition to
SiF4, signals corresponding to CFCl3, CF2Cl2, CF3Cl, and CF4 were observed (cf. Table 1). The
integration ratios in the 19F NMR spectrum are proportional to the number of moles of the fluorine
nuclei.
SiF4 is hydrolyzed to form H2SiF6 according to equation 8:
3 SiF4 + 2 H2O → SiO2 + 2 H2SiF6(8)

Solution H (10 mL) was added to an excess amount of water, which resulted in the completehydrolysis of SiF4. After separation, the H2SiF6 generated from the hydrolysis in the aqueous
solution was neutralized and completely converted to Na2SiF6 (aqueous solution J).
The precipitate of unreacted Na2SiF6 and NaCl, which was removed by filtration in the initial step(underline (a)), was completely dissolved in water to give an aqueous solution (solution K; 10.0 L).
Then, additional precipitation titrations using solution G were carried out, and the endpoints of thetitrations with G were as follows:·For solution J (entire amount): 61.6 mL.
·For 100 mL of solution K: 44.4 mL.

45

It should be noted here that the coexistence of NaCl or SiO2 has no effect on the precipitation
titration.
Table 119

F NMR data

CFCl3

CF2Cl2

CF3Cl

CF4

Chemical shift (δ)

0.0

−8.0

−28.6

−62.3

Integration ratio

45.0

65.0

18.0

2.0

B.1

Write the reaction formulae for the reaction of Na2SiF6 with Ce2(SO4)3.

Na2SiF6 = 188.0 [g·mol–1]. The concentration of Na2SiF6 in solution F is 0.855 g / 200 mL = 4.548 ×
10–3 mol / 0.2 L = 2.274 × 10–2 mol·L–1 (F)
Ce2(SO4)3 = 568.4 [g·mol–1]. The concentration of Ce2(SO4)3 in solution G is 6.860 g / 200 mL =1.207 × 10–2 mol / 0.2 L = 6.034 × 10–2 [mol·L–1] (G)The concentration of Ce3+ ions in G is 6.034 × 10–2 [mol·L–1] × 2 = 1.207 × 10–1 [mol·L–1]
In 50.0 mL of solution F: 2.274 × 10–2 × (50.0/1000) [mol] = 1.137 × 10–3 [mol] of Na2SiF6 wasincluded.
In 18.8 mL of solution G: 6.034 × 10–2 × (18.8/1000) [mol] = 1.134 × 10–3 [mol] of Ce2(SO4)3 wasincluded.
Accordingly, Na2SiF6 should react with Ce2(SO4)3 in a 1:1 ratio (SiF62– reacts with Ce3+ in a 1:2
ratio)
Na2SiF6 + Ce2(SO4)3 + 4 H2O → 2 CeF3 + Si(OH)4 + Na2SO4 + 2 H2SO4
B.2

Calculate the weights [g] of NaCl produced and of Na2SiF6 used (x [g]) as a starting
material.

As Na2SiF6 reacts with Ce2(SO4)3 in a 1:1 ratio, the amount of SiF62– in the aqueous solution J is
6.034 × 10–2 [mol·L–1] × (61.6 × 10–3 [L]) = 3.717 × 10–3 [mol]. Considering the equation
3 SiF4 + 2 H2O → SiO2 + 2 H2SiF6,
the ratio of the consumed SiF4 and the generated H2SiF6 should be 3:2, and therefore 10 mL of thediluted solution H contains 3.717 × 10–3 [mol] × (3/2) = 5.576 × 10–3 [mol] of SiF4. Thus, 1.00 L of
diluted solution H contains 5.576 × 10–3 [mol] × (1000/10) = 0.5576 [mol] of SiF4.
The amount of fluorine atoms that replace the chlorine atoms of CCl4 should be twice the amountof SiF4 formed during the reaction. Thus, 2 × 0.5576 = 1.115 [mol] of F– should replace Cl– to resultin the formation of NaCl.
1.115 [mol] × 58.4 [g·mol–1]= 65.12 [g] of NaCl was formed.
Answer: 65.1 [g]46As Na2SiF6 reacts with Ce2(SO4)3 in a 1:1 ratio, the amount of SiF62– in 100 mL of aqueous
solution K is 44.4 × 10–3 [L]× (6.034 × 10–2 [mol·L–1]) = 2.679 × 10–3 [mol]. Accordingly, the residual
amount of Na2SiF6 in 10.0 L of aqueous solution K is 2.679 × 10–3 [mol] × 100 = 0.2679 [mol].
In total, the amount of Na2SiF6 used as a starting material is (0.2679 [mol] + 0.5576 [mol]) × 188.0[g·mol–1] = 155.2 [g]Answer: 155 [g]B.3

77.8% of the CCl4 used as a starting material was unreacted. Calculate the weight [g] of
CF3Cl generated.

500.0 [g] = 3.25 [mol] of CCl4 was initially used as a starting material. Thus, the amount of the
products
that contain at least one F atom is 3.25 [mol] × 0.222 = 0.721 [mol].
The ratio of integrals in the 19F NMR spectrum is CFCl3 : CF2Cl2 : CF3Cl : CF4 = 45 : 65 : 18 : 2.0.
Thus, themole ratio of these compounds should be CFCl3 : CF2Cl2 : CF3Cl : CF4 = 45 : 32.5 : 6 : 0.5 = 90 :65 : 12 : 1.
Accordingly, the amount of CF3Cl (104.46 [g·mol–1]) is 0.721 [mol] × (12 / (90 + 65 + 12 + 1)) =0.0515 [mol] = 0.0515 [mol] × 104.46 [g·mol–1] = 5.38 [g]Answer: 5.38 [g]Subject Intention:
1. To learn the process of the structural identification of the unknown products based on the
spectral features and the reaction patterns. To realize that an unfamiliar benzene derivative
where its carbon atoms are replaced by silicon atoms, exhibits the considerable aromaticstabilization.
2. To realize that silicon compounds are one of unique main group elements that can have up to
six bondings, and learn the process of quantitative understanding the phenomenon of such an
unfamiliar fluorination reaction using highly coordinated silicon fluoride in a step-by-stepmanner based on the precipitation titration.

47

Problem 6. The Solid-State Chemistry of Transition Metals

Volcano at Sakurajima island
Part A
Japan is one of the countries with the highest numbers of volcanos. When silicate minerals
crystallize from magma, a part of the transition-metal ions (Mn+) in the magma is incorporated into
the silicate minerals. The Mn+ studied in the problem are coordinated by oxide ions (O2–) and adopt
a four-coordinate tetrahedral (Td) geometry in the magma and six-coordinate octahedral (Oh)geometry in the silicate minerals, both of which exhibit a high-spin electron configuration. The
distribution coefficient of Mn+ between the silicate minerals and magma, Kp, can be expressed by:
Kp =

[M]s
[M]l

(1)

where [M]s and [M]l are the concentrations of Mn+ in the silicate minerals and the magma,
respectively. Table 1 shows the Kp values of Cr2+, Mn2+, and Fe2+.
Fe2+ in Table 1 is not included in the official IChO Problems 2021 because of potential difficulty inmarking by following the IChO marking regulation.
Table 1

Kp

Cr2+

Mn2+

Fe2+

7.2

1.1

2.0

Let ΔO and ΔOM be the energy separation of the d-orbitals of Mn+ and the crystal-field stabilizationenergy (CFSE) in a Oh field, and ΔT and ΔTM those in a Td field.

A.1 Calculate |ΔOM－ΔTM| = ΔA with units of ΔO for Cr2+, Mn2+, and Fe2+; assume ΔT = 4/9ΔO.

48

As explained above, the d-orbitals of a six-coordinate octahedral complex split into two groups eg(dx2-y2, dz2) and t2g (dxy, dyz, dzx) with energy separation of ΔO. The energies of the eg and t2g orbitals
relative to the barycenter are +0.60ΔO and –0.40ΔO, respectively. Likewise, the d-orbitals of a fourcoordinate tetrahedral complex split into two groups t2 (dxy, dyz, dzx) and e (dx2-y2, dz2) with energy
separation of ΔT. The energies of the t2 and e orbitals relative to the barycenter are +0.40ΔT and –0.60ΔT, respectively. Therefore, in the case of high-spin electron configuration,
ΔOM and ΔTM for Cr2+ (3d4: t2g3eg1 or e2t22) are –0.60ΔO and –0.40ΔT (= –0.18ΔO), respectively.
ΔOM and ΔTM for Mn2+ (3d5: t2g3eg2 or e2t23) are both zero.
ΔOM and ΔTM for Fe2+ (3d6: t2g4eg2 or e3t23) are –0.40ΔO and –0.60ΔT (= –0.27ΔO), respectively.
According to these values, |ΔOM－ΔTM| (= ΔA) for each metal ions are,
Cr2+: |–0.60ΔO – (–0.18ΔO)| = 0.42 ΔO
Mn2+: 0Fe2+: |–0.40ΔO – (–0.27ΔO)| = 0.13 ΔO

A.2

A linear relationship can be obtained by plotting
ln Kp against ΔA / ΔO on the graph at right. Estimate
Kp for Co2+ by using this graph.

Co2+ (3d7): |–0.80ΔO – (–0.53ΔO)| = 0.27 ΔO
1. By using the graph and least squares method; y ≃ 4.45x + 0.100
Therefore, Kp = exp(4.45 × 0.27 + 0.100) = 3.67 ≃ 3.7
2. By calculation; The coordinates of Mn2+ and Cr2+ are(0, 0.095) and (0.42, 1.97), respectively. The linear
regression line calculated from these coordinates would be,y ≃ 4.46x + 0.100
Therefore, Kp = exp(4.46 × 0.27 + 0.100) = 3.68 ≃ 3.7
3. If the participants use Fe2+ (0.13, 0.69) instead of Mn2+,
the regression line would be,y ≃ 4.57x + 0.100
Then, Kp = exp(4.57 × 0.27 + 0.100) = 3.79 ≃ 3.8

49

A.3 Metal oxides of MO (M: Ca, Ti, V, Mn, or Co) crystallize in a rock-salt structure wherein the
Mn+ adopts an Oh geometry with a high-spin electron configuration. The lattice enthalpy ofthese oxides is mainly governed by the Coulomb interactions based on the radius and chargeof the ions and some contributions from the CFSE of Mn+ in the Oh field. Choose theappropriate set of lattice enthalpies [kJ mol–1] from one of the options (a) to (f).

CaO

TiO

VO

MnO

CoO

(a)

3460

3878

3913

3810

3916

(b)

3460

3916

3878

3810

3913

(c)

3460

3913

3916

3810

3878

(d)

3810

3878

3913

3460

3916

(e)

3810

3916

3878

3460

3913

(f)

3810

3913

3916

3460

3878

One way to reach the correct answer: The lattice enthalpy is determined by the Coulombinteractions, which are proportional to the product of the valences of the constituent ions andinversely proportional to the sum of the ionic radii. As the target compounds are oxides of divalent
metal ions, we should consider the ionic radii of the metal ions. The radii of divalent metal ions
within the same period decrease with increasing atomic number. Let us compare the latticeenthalpies of CaO and MnO with no contribution from the CFSE: The ionic radius of Mn2+ is smaller
than that of Ca2+, and therefore, the lattice enthalpy is higher for MnO. So the participants shouldchoose (a), (b), or (c). Then, let us compare the lattice enthalpies of TiO (d2) and VO (d3): The ionicradius of V2+ is smaller than that of Ti2+, and the CFSE is higher for VO than for TiO. Accordingly,
the lattice enthalpy is also higher for VO. Based on these observations, the answer should be (a)or (c). Finally, let us compare the lattice enthalpies of TiO(d2) and CoO(d7): The ionic radius of Co2+
is smaller than that of Ti2+, while their CFSEs are comparable. Thus, the lattice enthalpy is higher
for CoO. Thus, the correct answer is (a).

In Part A, we wish to introduce the structural diversity of d-metal complexes. We also show thatthe states of d-metal ions in crystals can be understood using the crystal field theory usually applied
to d-metal complexes, which are molecules.

50

Part B
A mixed oxide A, which contains La3+ and Cu2+,
crystallizes in a tetragonal unit cell shown in Fig.1. In the[CuO6] octahedron, the Cu–O length along the 𝑧-axis (lz)is longer than that of the 𝑥-axis (lx), and [CuO6] is distorted
from the regular Oh geometry. This distortion removes thedegeneracy of the eg orbitals (d𝑥2−𝑦2 and d𝑧2).

A can be synthesized by thermal decomposition(pyrolysis) of complex B, which is formed by mixing metal
chlorides in dilute aqueous ammonia solution containing
squaric acid (C4H2O4), i.e., a diacid. The pyrolysis
behavior of B in dry air shows a weight loss of 29.1% up
to 200 °C due to loss of crystallization water, followed byanother weight loss up to 700 °C due to the release of

Fig. 1CO2. The total weight loss prior to the formation of A is
63.6%. It should be noted that only water and CO2 are released in the pyrolysis reaction.

B.1

Write the chemical formulae of A and B.

The unit cell shown in Fig. 1 contains four La3+, two Cu2+, and eight O2– ions. Therefore, A:
La2CuO4. As the formula weight of La2CuO4 is 405.3, that of B should be 1113.5 considering thefollowing equation: 405.3 ÷ (1 – 0.636). Given that the weight loss due to crystallization water is
29.1%, the number of molecules of crystallization water is 18.00 considering the following equation:(1113.5 × 0.291) ÷ 18 (18 H2O; M = 324). Complex B is a trinuclear complex that consists of two
La3+ and Cu2+ ion. Considering that the synthetic solution is basic, the squaric acid is deprotonatedand coordinates to the metal ion as C4O42–. The number of squaric acid molecules is 4.00 basedon the following equation: (1113.5 – 138.9 × 2 – 63.5 – 324) ÷ 112 (C4O42–; M = 112). B:
La2Cu(C4O4)4(H2O)18 (La2CuC16O34H36)

B.2

Calculate lx and lz using Fig. 1.

lx: 0.3833 ÷ 2 = 0.1917 » 0.192 nm
lz: (1.3313 – 0.2520 × 2) ÷ 4 = 0.2068 » 0.207 nm

51

B.3

For Cu2+ in the distorted [CuO6] octahedron in A of Fig. 1, write the names of the split eg
orbitals (d𝑥2−𝑦2 and d𝑧2) in (i) and (ii), and draw the electron configuration in the dotted box
in your answer sheet.

(i) d𝑥2−𝑦2,B.4(ii) d𝑧2

A is an insulator. When one La3+ is substituted with one Sr2+, one hole is generated in the
crystal lattice that can conduct electricity. As a result, the Sr2+-doped A shows
superconductivity below 38 K. When a substitution reaction took place for A, 2.05 × 1027
holes m–3 were generated. Calculate the percentage of Sr2+ substituted for La3+ based on
the mole ratio in the substitution reaction. Note that the valences of the constituent ions andthe crystal structure are not altered by the substitution reaction.

The corresponding reaction equation, where the amount of Sr is x%, is:
La2CuO4 + (2x/100)Sr2+ → [La{2×(1－x/100)}Sr(2x/100)CuO4](2x/100)－ + (2x/100)La3+
The charge of [La{2×(1－x/100)}Sr(2x/100)CuO4] is negative, and the amount of doped holes is (2x/100)
h+．The volume of the unit cell is 0.38332 × 1.3313 = 0.1956 nm3. The unit cell contains four La3+,
two Cu2+, and eight O2－ ions, i.e., the unit contains two La2CuO4. Accordingly, the number of holes
per unit cell is (4x/100).
Since the concentration of holes is the number of holes divided by the unit cell volume, the following
equation is satisfied: (4x/100) / (0.1956 × 10－27) = 2.05 × 1027. x = 10%

In Part B, we wish to introduce the structures, related reactions, and functions of mixed transitionmetal oxides with La2CuO4 as an example. We also present the structural diversity based on JahnTeller distortion of Cu2+ (d9 configuration), and the functions related with elemental substitution.

52

Part C
[Cu2(CH3CO2)4] is composed of four CH3CO2− coordinated to two Cu2+ (Fig. 2A). [Cu2(CH3CO2)4]
exhibits high levels of structural symmetry, with two axes passing through the carbon atoms of thefour CH3CO2− and an axis passing through the two Cu2+, all of which are oriented orthogonal
relative to each other. When a dicarboxylate ligand is used instead of CH3CO2−, a “cage complex”
is formed. The cage complex [Cu4(L1)4] is composed of planar dicarboxylate L1 (Fig. 2B) and Cu2+
(Fig. 2C). The angle θ between the coordination directions of the two carboxylates, indicated by
the arrows in Fig. 2B, determines the structure of the cage complex. The θ is 0° for L1. Note thathydrogen atoms are not shown in Fig. 2.

Fig. 2

C.1

The θ of the planar dicarboxylate L2 below is fixed to 90°. If the composition of the cagecomplex formed from L2 and Cu2+ is [Cun(L2)m], give the smallest integer combination of nand m. Assume that only the CO2− groups of L2 form a coordination bond to Cu2+ ions.

53n = m = 12

A zinc complex, [Zn4O(CH3CO2)6], contains four tetrahedral Zn2+, six CH3CO2−, and one O2− (Fig.
3A). In [Zn4O(CH3CO2)6], the O2− is located at the origin, and the three axes passing through thecarbon atoms of CH3CO2 − are oriented orthogonal relative to each other. When pbenzenedicarboxylate (Fig. 3B, L3, θ = 180°) is used instead of CH3CO2−, the Zn2+ clusters arelinked to each other to form a crystalline solid (X) that is called a “porous coordination polymer”(Fig. 3C). The composition of X is [Zn4O(L3)3]𝑛, and it has a cubic crystal structure with nano-sizedpores. One pore is represented as a sphere in Fig. 3D, and each tetrahedral Zn2+ cluster isrepresented as a dark gray polyhedron in Fig. 3C and 3D. Note that hydrogen atoms are not shownin Fig. 3.

Fig. 354

C.2

X has a cubic unit cell with a side length of 𝑎 (Fig. 3C) and a density of 0.592 g cm−3.
Calculate 𝑎 in [cm].

The molecular weight of Zn4O(L3)3 is 770.
According to Fig. 3C, there are eight Zn4O(L3)3 units in the unit cell. Therefore, the molecular weight
per unit cell can be calculated as 770×8 = 6160.
The weight of the unit cell is 6160÷𝑁A = 1.02×10−20 g
Let the length of the side of the unit cell be 𝑎 [cm], then,(1.02×10−20) g / 𝑎3 [cm3] = 0.592 g cm−3.
𝑎 = 2.6×10−7 cm

C.3

X contains a considerable number of pores, and 1 g of X can accommodate 3.0×102 mL of
CO2 gas in the pores at 1 bar and 25 °C. Calculate the average number of CO2 molecules
per pore.

There is one pore per Zn4O(L3)3 unit.
Number of pores [mol] per 1 [g] of X: 1 [g] / 770 = 0.00130.
Based on the ideal gas equation, 3.0×102 [mL] of adsorbed CO2 corresponds to:(1×105 [Pa] × 3.0×10−4 [m3]) / (8.31×298 [K]) = 0.0121 [mol] of CO2.
Therefore, 0.0121÷0.00130 = 9.3 molecules of CO2 per pore.

55

The following question C.4 is not included in the official IChO Problems 2021 because of the IChO
exam length regulation.
Let's try to describe the structure of the complex geometrically here, as shown in Fig. 4. As shownin Figs. 4a and 4b, [Cu2(CO2)4] is represented by a square panel with four carbons connected. The
structure of [Cu2(CO2)4] is then described by connecting the square panels using straight lines
connecting the carbons of the two CO2－ groups of the dicarboxylate (Fig. 4C).

Fig. 4

C.4

Following the notation in Fig. 4, the structures of the various coordination polymers consistingof [Cu2(CO2)4] can be represented by connecting the vertices of the square panels withstraight lines. For the structures of the four coordination polymers described in this way (Fig.
5, right), select the most suitable dicarboxylate (Fig. 5, left) to form each of them, andconnect the lines to be 1:1. Note that only the CO2－ group is assumed to coordinate with the
Cu2+ ion, and all dicarboxylates are represented by a straight line connecting the carbons of
the two CO2－ group. The dotted line on the right side of Fig. 5 indicates that the samestructure continues.

56

Fig. 5In Part C, we would like the examinee to imagine and understand to construct more diverse
structures of coordination complexes architectures by changing the organic ligands.

57

Question 7：Playing with Non-benzenoid Aromaticity
Professor Nozoe (1902–1996) opened the research field of non-benzenoid aromatic compounds,
which are now ubiquitous in organic chemistry.

Photo courtesy: Department of Chemistry, Graduate School of Science, Tohoku University.
Part A
Lineariifolianone is a natural product with a unique structure, which was isolated from Inulalinariifolia. From valencene (1), a one-step conversion yields 2, before a three-step conversion via
3 yields ketone 4. Eremophilene (5) is converted into 6 by performing the same four-step
conversion.

Inula linariifolia
Further reading:
Erik J. Sorensen, et al., “Synthesis of (+)-Lineariifolianone and Related CyclopropenoneContaining Sesquiterpenoids” The Journal of Organic Chemistry 2019, Vol. 84, Iss. 9, 5524–5534.
1
58

H

N
NaIO4
OsO4

2

THF, H2O

C14H22O

valencene (1)OH
Si N Si

OH
Si I
CH3CN
80 °C

H

OH

CH3Li

2

THF
–78 °C

N
NaIO4
OsO4

THF
–78 °COLi
3

H
Si N Si

(ii)

H
O
4

O
OH

Si I

(i) CH3Li
THF–78 °C

CH3CN
80 °C

THF, H2O

OH
6

THF–78 °C

eremophilene (5)A.1

Draw the structure of 2 and 6 and clearly identify the stereochemistry where necessary.

6

2

H

H

O

O

2
59

Then, ketone 4 is converted into ester 15. Compound 8 (molecular weight: 188) retains all thestereocenters in 7. Compounds 9 and 10 have five stereocenters and no carbon-carbon double
bonds. Assume that H218O is used instead of H216O for the synthesis of18

O-labelledlineariifolianones 13 and 14 from 11 and 12, respectively. Compounds 13 and 14 are18

O-labelled

isotopomers. Ignoring isotopic labelling, both 13 and 14 provide the same product 15 with identicalstereochemistry.
K
Si N Si(KHMDS)A

H

4O

THF, –78 °C

O

Li
N

H
CF3
S
O O

7
O
Cl

8

OH
O
(mCPBA)
9

NaHCO3

MW: 188

CH2Cl2, 0 °C

9

C14H20OO
S OH
O
H218O, THF

Si CF2Br
NaI
THF, 80 °C

10+

C14H20O

11

13
C15H2218On16O3-n(n = 0—3)

C15H20F2OO
S OHO
H218O, THF

Si CF2Br
NaI
THF, 80 °C
10

12

14
C15H2218On16O3-n

C15H20F2O(n = 0—3)

N

OH
OH

N
16O-13/16O-14(C15H2216O3)

H
CH3OH

O
OCH3
15

3
60

A.2

Choose the appropriate structure for A.

IO
F3C S OHOIIO
F3C S NH2
O

O
CF3
O S
N
CF3
S
O O

III

O
CF3
O S
IV
HN
CF3
S
O O

III

A.3

Draw the structures of 8–14 and clearly identify the stereochemistry where
necessary. Also, indicate the introduced 18O atoms for 13 and 14 as shown in
the example on the right.

8
H

9

10
O

O
H

H

12

11

O

O
H

H

F

13

H 18
O

14
O

H
H

F
F

F

18
O

H

O 18 H
O
H

4
61

18
O

18
O
O

Part B
Compound 19 is synthesized as shown below. In relation to non-benzenoid aromaticity, 19 can beused as an activator for alcohols, and 20 was converted to 22 via ion-pair intermediate 21. Although
the formation of 21 was observed by NMR, 21 gradually decomposes to give 18 and 22.

Br2

O

17

CH3COOH

C15H12Br2O

16

Et3N

Cl

18

CH2Cl2

O
S

Cl

19

C15H10O

C15H10Cl2

OH

Cl
+ 19

20

1H NMR (CD CN, ppm)
3

+

21

–HCl

22

20: δ 7.4–7.2 (5H), 3.7 (2H), 2.8 (2H), 2.2 (1H)
21: δ 8.5–7.3 (15H), 5.5 (2H), 3.4 (2H)

B.1

Draw the structures of 17–19 and 21. Identifying the stereochemistry is not necessary.

18

17

O
O

Br

O
Br

or

Br Br

19

21
Cl

Cl

Cl

O

5
62

18

Further reading:
Tristan H. Lambert, et al., “Aromatic Cation Activation of Alcohols: Conversion to Alkyl Chlorides
Using Dichlorodiphenylcyclopropene” Journal of the American Chemical Society 2009, Vol. 131,
Iss. 39, 13930–13931.

663

Question 8：Dynamic Organic Molecules and Their Chirality
Part A
Polycyclic aromatic hydrocarbons with successive ortho-connections are called [n]carbohelicenes(here, n represents the number of six-membered rings). [4]Carbohelicene is efficiently prepared bya route using a photoreaction as shown below, via an intermediate that is readily oxidized by iodine.

PPh3 BrNaOEt

A
C11H8O

+

B

EtOH
Δ

C

B

hν

hν

C

I2

[4]carbohelicene
C18H12

C18H14The photoreaction proceeds in a manner similar to the following example.
hν

A.1

hν

Draw the structures of A–C. Stereoisomers should be distinguished.

A

B

C

O

64

A.2

Attempts to synthesize [5]carbohelicene in the same way resulted in the formation of only a
trace amount of [5]carbohelicene, instead affording an unexpected product D whosemolecular weight was 2 Da lower than that of [5]carbohelicene. The 1H NMR chemical shiftsof D are listed below. Draw the structure of D.
D (δ, ppm in CS2, 20 ℃): 8.85 (2H), 8.23 (2H), 8.07 (2H), 8.01 (2H), 7.97 (2H), 7.91 (2H)

hν
I2

hν
I2

(–2H)

(–2H)
[5]carbohelicene

D

C22H14

benzo[ghi]perylene
C22H12

The following questions A.3 and A.4 are not included in the official IChO Problems 2021 becauseof potential difficulty in marking by following the IChO marking regulation.
[6]Carbohelicene, which is prepared by the same photoreaction, can be enantiomerically separated
on a chiral column, a process developed by Prof. Yoshio Okamoto. Here, the chirality of[n]carbohelicenes are defined as (M) or (P) as shown below.

ki
ki
(M )

(P )

65

A.3

Enantiomerically pure [6]carbohelicene gradually racemizes upon heating with an helicalinversion rate constant ki. The activation barrier in this case can be determined by plotting
the enantiomeric excess as a function of time as shown in the figure below. Calculate theracemization rate constant, kr, and the activation barrier, Ea, using the enantiomeric excess
at 60 min. Consider the relationship kr = 2ki. The rate of the racemization process is
independent of the concentration.

First, kr is calculated according to Figure 1. Given that the racemization is a unimolecular firstorder reaction,ln (ee) = –krt + y, wherein y is a constant. By recalling kr = 2ki and using the ee values at t = 0 s and 3600 s,
at T = 460 K and 478 K, ki are calculated to be 2.927 ×10–5 s–1 and 1.273 ×10–4 s–1, respectively.
Second, according to the Arrhenius equation,ln ki = –Ea/(RT) + ln A, a plot of ln ki versus 1/T gives-8.5
0.00205-9

a straight line shown on the right, whose gradient can-9.5

be used to determine –Ea/R as –17954.

-10

–1

0.00210.00215–1Given R = 8.31 J·K ·mol , Ea is calculated to be
149.27 kJ mol–1 = 1.5×102 kJ mol–1

-10.5
-11

66

y = –17954x + 28.592

0.0022

A.4

The racemization kinetics of [5]carbohelicene was estimated to be Ea = 98.5 kJ mol–1 with afrequency factor in the Arrhenius equation (preexponential factor) of A = 2.43╳1012 s–1.
Based on these parameters, estimate the half life of the enantiomeric excess (i.e., the timeat which 50%ee is observed) of [5]carbohelicene at 300 K.

ki is calculated to be 1.748 x 10–5 s–1 by using the Arrhenius equation ln ki = –Ea/(RT) + ln A with Ea= 98.5 kJ mol–1, A = 2.43 x 1012 s–1, R = 8.31 J·K–1·mol–1, and T = 300 K.
kr = 2ki = 3.496 x 10–5 s–1 = 2.098 x 10–3 min–1
With ln (ee) = –krt + y, the half life = – (ln 0.5)/kr = 330 min

A message from A.3 and A.4 is that the dynamic behavior of [n]carbohelicenes can be quantitatively
analyzed. The racemization activation energy is reasonably higher for [6]carbohelicene than thatfor [5]carbohelicene. Because the chiral separation experiment is routinely conducted around room
temperature, [6]carbohelicene can be fully separated while [5]carbohelicene cannot due to the rateof racemization.

67

The following question A.5 was modified and is different from the corresponding question in theofficial IChO Problems 2021 because of potential difficulty in marking by following the IChO marking
regulation.
Multiple helicenes are molecules that contain two or more helicene-like structures. If its planarchirality is considered, several stereoisomers exist in a multiple helicene. For example, compound
E (C46H26) contains three [5]carbohelicene-like moieties in one molecule. One of the stereoisomers
is described as (P, P, P) as shown below.

A.5

The nickel-mediated trimerization of 1,2-dibromobenzene generates triphenylene. When thesame reaction is applied to a racemic mixture of F, multiple helicene G (C66H36) is obtained.
Draw all the possible stereoisomers of G. As a reference, one isomer should be drawn
completely with the chirality defined as in the example above; The other isomers should belisted using M and P. For instance, the other stereoisomers of E can be listed as (1, 2, 3) =(P, M, P), (P, M, M), (P, P, M), (M, M, M), (M, M, P), (M, P, P), and (M, P, M).
Ni(cod)2
Br

N

3

N

cod =

Br

Ni(cod)2
Br
3

N

N

G
(C66H36)

Br

F

68

example

①

P

④

③

⑤
②

P

P

P

⑥
P
P

（①, ②, ③, ④, ⑤, ⑥）(P, P, P, P, P, P)(①, ②, ③)
First, consider all the possible isomers withrespect to ①〜③.
Second, for each case, consider all the possible(P, P, P)(P, P, P, P, P, P)(P, P, P, M, P, P)(P, P, P, M, M, P)(P, P, P, M, M, M)

1
2
3
4

(M, P, P)(M, P, P, P, P, P)(M, P, P, M, P, P)(M, P, P, P, M, P)(M, P, P, P, P, M)(M, P, P, M, M, P)(M, P, P, P, M, M)(M, P, P, M, P, M)(M, P, P, M, M, M)

5
6
6
7
8
9
9
10

(M, M, P)(M, M, P, P, P, P)(M, M, P, M, P, P)(M, M, P, P, M, P)(M, M, P, P, P, M)(M, M, P, M, M, P)(M, M, P, P, M, M)(M, M, P, M, P, M)(M, M, P, M, M, M)

ent-10
ent-9
ent-8
ent-9
ent-6
ent-6
ent-7
ent-5

(M, M, M)(M, M, M, P, P, P)(M, M, M, M, P, P)(M, M, M, M, M, P)(M, M, M, M, M, M)

ent-4
ent-3
ent-2
ent-1

isomers with respect to ④〜⑥.
You then can list up all the possible cases.
Carefully check the same compounds.
ent- means the enantiomer of each compound.

（①, ②, ③, ④, ⑤, ⑥） compound

Finally, you find 10 diastereomers and
20 stereoisomers including enantiomers.

69

Part B
Sumanene is a hydrocarbon that was first reported in Japan in 2003. It was named after Suman
that means sunflower in both Hindi and Sanskrit and has a bowl-shaped structure as shown below.

The synthesis of sumanene was achieved by a reaction sequence that consists of a ring-opening
and a ring-closing metathesis. Representative metathesis reactions catalyzed by a ruthenium
catalyst are shown below.
Ring-opening metathesis
Phcatalyst

M

Ph

Ph

polymerization

Mn

M = RuCl2[P(C6H11)3]2

Ring-closing metathesisRM

Ph
catalyst

R

R

+
R

O

Synthesis of sumaneneMPhcatalyst

Cl

CN

Cl

CN
O

I
(C21H18)
H

70

B.1

Draw the structure of intermediate I (its stereochemistry is not required).

I

B.2

Starting from the optically active precursor J, the same reaction sequence gives the optically
active sumanene derivative K with retention of the stereochemistry. Draw the structure of K
with the appropriate stereochemistry.
O

Me

Cl

M

Ph
catalyst

CN

Cl

CN
O
K
(C24H18)

Me
Me
J

Me

Me
K

Me

Further reading:Hidehiro Sakurai, Taro, Daiko, Toshikazu Hirao, “A Synthesis of Sumanene, a Fullerene
Fragment” Science 2003, Vol. 301, Iss. 5641, 1878.

71

Question 9：Likes and Dislikes of Capsule
Part A
Good kids don’t do that, but if you unseam a tennis ball, you can disassemble it into two U-shaped
pieces.

Based on this idea, compounds 1 and 2 were synthesized as U-shaped molecules with different
sizes. Compound 3 was prepared as a comparison of 1 and the encapsulation behavior of thesecompounds was investigated.

72

The synthetic route to 2 is shown below. The elemental analysis of compound 9 revealed: C:
40.49%, H: 1.70%, and O: 17.98% by mass.

73

A.1

Draw the structures of 4–9; the stereochemistry can be neglected. Use "PMB" as a
substituent instead of drawing the whole structure of p-methoxybenzyl group shown in thescheme above.

4

5

6

7

8

9

In the mass spectrum of 1, the ion peak corresponding to its dimer (12) was clearly observed,
whereas an ion peak for 32 was not observed in the spectrum of 3. In the 1H NMR spectra of asolution of 12, all the NH protons derived from 1 were observed to be chemically equivalent, andtheir chemical shift was significantly different from that of the NH protons of 3. These data indicatethat hydrogen bonds are formed between the NH moieties of 1 and atoms X of another moleculeof 1 to form the dimeric capsule.
A.2

Circle all the appropriate atom(s) X in 1.

A.3

Give the number of the hydrogen bonds in the dimeric capsule (12).
8The eight NH protons in 12 are observed equally, thus the number of hydrogen bonds in 12 are
obviously 8.

74

The following questions A.4 and A.5 are not included in the official IChO Problems 2021 becauseof the IChO exam length regulation.
The dimeric capsule of compound 1 has an internal space and can encapsulate the appropriate
small molecules. The 1H NMR spectrum of methane and compound 1 (2.00 x 10–2 mmol) in C6D6
(1.00 mL) was measured. The assignment of characteristic signals and chemical shifts and integral
values are shown below. Here, methane inside and outside the capsule was observed separately
in the 1H NMR analysis.
Attribution of protons

Chemical shifts (ppm)

integrals(a) NH protons of the methane-encapsulated capsule

9.25

2.50

(b) NH protons of the empty capsule

9.20

1.76

(c) Protons of methane outside the capsule

0.23

3.20

(d) Protons of methane inside the capsule

–0.91

1.25

Under the above conditions, all compounds 1 form the dimeric capsule, and the volume of thesolution does not change by adding 1 and methane.
A.4

Determine the concentrations of the methane inclusion capsule and the empty capsule,
respectively.

The concentration of compound 1 is 2.00 × 10–2 mmol/1.00 mL, therefore, the sum ofconcentration of the empty and inclusion capsules is 1.00 × 10–2 mmol/1.00 mL. In the
NMR analysis, the integration ratio of the empty capsule to the inclusion capsule is 1.76 to
2.50. Thus,[inclusion capsule] = 1.00 x 10–2 x (2.50/(1.76 + 2.50)) = 0.587 x 10–2 mol/L[empty capsule] = 1.00 x 10–2 x (1.76/(1.76 + 2.50)) = 0.413 x 10–2 mol/L
A.5

Determine the association constant (Ka) for methane and the dimeric capsule. Note thatthe association constant (Ka) for incorporated molecule Z and capsule 12 is given byequations 1 and 2.
Z + 12 → Z@12

(1)

[Z@12 ]
[Z][12 ]

(2)

Ka =

75

In the NMR spectrum, the integral ratio of NH (8H) and internal methane (equivalent to 4H)
is 2.5:1.25, indicating that methane incorporated into the capsule is a single molecule, i.e.,a 1:1 complex is formed. Therefore, the concentration of the inclusion capsule and theconcentration of methane inside the capsule are equal (0.587 x 10–2 mol/L), and theconcentration of methane outside the capsule is (3.20/1.25) x 0.587 x 10–2 mol/L = 1.50 x
10–2 mol/L from the corresponding integral ratio.
Therefore, the association constant Ka is calculated as follows:
Ka = [0.587 x 10–2]/([0.413 x 10–2] x [1.50 x 10–2]) = 94.8 M–1
Compound 2 also forms a rigid and larger dimeric capsule (22). The 1H NMR spectrum of 22 wasmeasured in C6D6, C6D5F, and a C6D6/C6D5F solvent mixture, with all other conditions being kept
constant. The chemical shifts for the Ha proton of 2 in the above solvents are summarized below.
Assume that the interior of the capsule is always filled with the largest possible number of solvent
molecules and that each signal corresponds to one species.

solvent

Chemical shifts (ppm)

integrals(a) C6D6

4.60

2.10

(b) C6D5F

4.71

5.21

(c) mixed solvents of C6D6 and C6D5F

4.60, 4.71, 4.82

2.48

A.6

Determine the number of C6D6 and C6D5F molecules encapsulated in 22 giving each Ha
signal.

Ha

C6D6

C6D5F

4.60

( 2 ) molecule(s)( 0 ) molecule(s)4.71( 0 ) molecule(s)( 2 ) molecule(s)4.82( 1 ) molecule(s)( 1 ) molecule(s)The observation of the third signal in the mixed solvent indicates that two solvent molecules are
initially encapsulated in the capsule. Thus, the signal at 4.82 ppm is ascribed to a capsulecontaining one molecule of C6D6 and one molecule of C6D5F.

76

1

H NMR measurements in C6D6 revealed that 22 can incorporate one molecule of 1-

adamantanecarboxylic acid (AdA), and the association constants (Ka) which are expressed below
were determined for various temperatures.

Ka =[Z@22 ]
[Z][solvent@22 ](3)

Similarly, the Ka values of CH4 and 12 given as eq (2) at various temperatures in C6D6 were also
determined by 1H NMR measurements. The plots of the two association constants (as ln Ka vs 1/T)
are shown below.

A.7

Choose the correct options in gaps ( 1 )–( 5 ) in the following paragraph from A and B.

No C6D6 molecule is encapsulated in 12. In line II, the entropy change (ΔS) is ( 1 ) and enthalpy
change (ΔH) is ( 2 ), indicating that the driving force for the encapsulation in line II is ( 3 ). Therefore,line I corresponds to ( 4 ), and line II corresponds to ( 5 ).
A

B

(1)

positive

negative(2)

positive

negative

(3)

ΔS

ΔH

(4)

12 and CH4

22 and AdA(5)12 and CH4

22 and AdA( 1 ): A( 2 ): A( 3 ): A( 4 ): A( 5 ): B

77

Transforming ΔG = –RTlnKa = ΔH–TΔS gives lnKa = –ΔH/R x (1/T) + ΔS/R
(If ΔH is negative, the slope is positive; if the y-intercept is negative, ΔS is negative.)

The relationship between 12 and CH4 is that the entropy change is unfavorable (ΔS < 0) giventhat the two components become one component. Nevertheless, the encapsulation of CH4
occurs (ΔG < 0), indicating that the enthalpy change is exothermic and favorable (ΔH < 0).
This result indicates a plot with a positive slope and a negative y-intercept for 12 and CH4.
Therefore, the plot with a negative slope (ΔH > 0) and positive y-intercept (ΔS > 0) is 22 and AdA.
This is due to the release of the two molecules of C6D6 that were initially encapsulated and theencapsulation of one molecule of AdA.

78

The following Part B is not included in the official IChO Problems 2021 because of the IChO exam
length regulation.
Part B
Four plate-shape molecules 10 quantitatively form an octahedral capsule with large internal spaceby coordination bonds with six palladium clips. This capsule can be used as a "small flask" fororganic reactions.

When compounds 11 and 12 were added to an aqueous solution of the octahedral capsule and theresulting mixture was stirred, one molecule of 11 and one molecule of 12 were incorporated into
the octahedral capsule. However, no new compounds were observed when the solution was heated.
Compounds 13 and 12 were also encapsulated into the capsule, and compound 14 was obtained
upon heating of the solution. In the absence of the capsule, no reaction occurred even after heating.
Compounds 15 and 12 were also encapsulated and were quantitatively converted, upon heating,to compound 16, a product of the [4+2] cycloaddition reaction. Heating a solution of 15 and 12 in
the absence of the capsule gave compound 17, a structural isomer of 16.
Note that the notation “11&12@capusule” indicates that compounds 11 and 12 are encapsulated
in the capsule.

79

O
+

O

11

capsuleNheat

11&12@capsuleno reaction

12
O

n-C3H7+ 12

capsule13&12@capsule

heat

H
H

N
O

n-C3H7n-C3H7n-C3H7

13

14
n-C3H7
+ 12

heat

+

12

capsule

+

12

heat

no reaction

n-C3H713
15&12@capsuleheat[4+2] cycloaddition

16

15

15

B.1

17
(an isomerof 16)

Compound 16 has stereoisomers, but the reaction in the capsule provided only 16. Contrarily,
there are no stereoisomers in compound 17. Draw the structures of 16 and 17 with theirstereochemistry clearly shown.

16

17

80

The results of the reactions of 12 with 11 and with 13 tell that the [4+2] cycloaddition reaction (aDiels-Alder reaction) does not proceed when the total volume of the encapsulated molecules issmall enough to allow them to move within the capsule and that the reaction occurs when theencapsulation is so tight to fix the two molecules in close proximity then to induce this intrinsicallyslow dearomatizing bimolecular reaction. Notably, the cycloaddition reaction between 13 and 12
took place on the sterically more accessible benzene ring. The propyl side chains of 13 and thecyclohexyl group of 12 are oriented in the same direction, which would originate from the tightest
packing of the two molecules 13 and 12 in the capsule and led to the selective formation of 14.
The reaction between 15 and 12 in the capsule should follow the rules above and proceeded fromtight encapsulation to form 16. The formation of compound 17 is a well-known reaction, which
occurs at the 9 and 10 positions of anthracene to allow two benzene rings to remain. Products 16
and 17 are identifiable by considering the information about stereoisomerism of each product.
Further reading:
Makoto Fujita, et al., “Naphthalene Diels–Alder in a Self-Assembled Molecular Flask” Journal of
the American Chemical Society 2010, Vol. 132, Iss. 9, 2866–2867.
Makoto Fujita, et al., “Functional Molecular Flasks: New Properties and Reactions within Discrete,
Self-Assembled Hosts” Angewandte Chemie International Edition 2009, Vol. 48, Iss. 19, 3418–
3438.

81