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

Preparatory Problems

Table of Contents
Preface
Contributing Authors
Fields of Advanced Difficulty
Physical Constants and Equations
Constants
Equations
Periodic Table of Elements
1
H NMR Chemical Shifts
SafetyTheoretical Problems
Problem 1.
Revision of SI unit
Problem 2.
Does water boil or evaporate?
Problem 3.
Molecules meet water and metals
Problem 4.
Synthesis of diamonds
Problem 5.
Count the number of states
Problem 6.
The path of chemical reactions
Problem 7.
Molecular vibrations and infrared spectroscopy
Problem 8.
Quantum chemistry of aromatic molecules
Problem 9.
Protic ionic liquids
Problem 10. The Yamada universal indicator
Problem 11. Silver electroplating
Problem 12. How does CO2 in the atmosphere affect the pH value of seawater?
Problem 13. How to produce sulfuric acid and dilute it without explosion
Problem 14. Hydrolysis of C vs Si and the electronegativity of N vs Cl
Problem 15. Sulfur in hot springs and volcanoes
Problem 16. Identification of unknown compounds and allotropes
Problem 17. Metal oxides
Problem 18. Coordination chemistry and its application to solid-state catalysts
Problem 19. Acids and bases
Problem 20. Semiconductors
Problem 21. Carbenes and non-benzenoid aromatic compounds
Problem 22. Nazarov cyclization
Problem 23. Tea party
Problem 24. E-Z chemistry
Problem 25. Fischer indole synthesis
Problem 26. Planar chirality
Problem 27. Cyclobutadiene
Problem 28. Onion-like complexes
Problem 29. Hydrogen-bonded capsules
Problem 30. Synthesis and structural analysis of polymers
Problem 31. Total synthesis of tetrodotoxin

1
2
3
4
5
7
8
9
11
13
15
18
23
27
33
35
37
42
44
46
50
51
56
57
59
63
66
68
71
74
77
81
83
85
88
91
93
96
102

Appendix (Practical Tasks)
Task 1. Analysis of the saponification rate using a pH meter
108
Task 2. Simultaneous acid–base titration
113
Task 3. Synthesis and analysis of a cobalt(III) oxalate complex
116
Task 4. Hinokitine: synthesis of a deep-red-colored natural product120
Task 5. Functionalization of a seven-membered ring: synthesis of tropolone tosylate 123iTask 6. Hydrolysis of polyethylene terephthalate: A small experiment for us, but a giant
leap toward a more sustainable society
125
Task 7. Separation of blue and red components from a green mixture
127ii

Preface
We are very pleased to introduce the Preparatory Problems for the 53rd International
Chemistry Olympiad. These problems cover a wide range of challenging and important
topics in modern chemistry. We hope that both students and their mentors will enjoy
solving these problems and prepare for the Olympics. The problems include 6 topics ofadvanced difficulty for the Theoretical part and 3 topics of advanced difficulty for thePractical part, in addition to the subjects normally covered in high school chemistrycourses. These topics are explicitly listed under "Fields of Advanced Difficulty" andtheir applications are shown in the preparatory problems consisting of 31 theoreticalproblems and 6 practical tasks. The solutions will be emailed to the head mentor of eachcountry by February, 2021 and will be published online in July, 2021. We welcome any
comments, corrections and questions about the problems via email to:preparatory@icho2021.org.
The International Chemistry Olympiad is a great opportunity for young people from allover the world to deepen their understanding of the wonders of chemistry, and inspire
each other. At the same time, it is a wonderful opportunity to make friends around theworld, and enjoy the history and culture of the host country. COVID-19 is widespreadall over the world and the situation is very severe, but we hope that we can meet you in
Osaka, Japan in July.
Acknowledgement
We would like to express our deepest gratitude to all the authors for their great efforts
in creating both preparatory and competition problems. We would also like to thank thereviewers for their valuable comments and suggestions.

Appendix
The preparatory problems are published to help the students and the mentors prepare forthe usual real IChO including the theoretical problems and the practical tasks. However,
because of the pandemic of COVID-19, the Organizing Committee finally decided tohold the IChO2021 Japan as the remote IChO to ensure the safety of the participants.
Since the practical tasks will not be conducted in the remote IChO2021, the practicaltasks in the preparatory problems are out of use for the IChO2021. It is not necessary
for the students who want to participate the IChO2021 to study and/or examine thepractical tasks and the advanced skills included in the preparatory problems.
Instead of the deletion of the practical tasks from the preparatory problems, however,
we moved them to the Appendix part. Even though the practical tasks will not be
conducted in IChO2021, the importance of laboratory experiments does not change inthe chemistry. We hope to have any opportunity where the practical tasks prepared forthe IChO2021 Japan are fully utilized. The practical tasks included in the Appendix partwill help such an event.

1

Contributing Authors
Theoretical Problems
HASEGAWA, Takeshi
HIROI, Takashi
HORIKE, Satoshi
HOSOKAWA, Saburo
MATSUMOTO, Yoshiyasu
NISHI, Naoya
OKUYAMA, Hiroshi
SAITO, Hayate
SASAMORI, Takahiro
SATO, Hirofumi
SHIMOKAWA, Jun
SHINTANI, Ryo
TANAKA, Takayuki
TSUBAKI, Kazunori
UCHIDA, Sayaka
YAMAGUCHI, Hiroyasu

Kyoto University
National Institute for Materials Science
Kyoto University
Kyoto University
Toyota Physical and Chemical Research InstituteKyoto University
Kyoto University
Kyoto University
University of Tsukuba
Kyoto University
Kyoto University
Osaka University
Kyoto University
Kyoto Prefectural University
The University of Tokyo
Osaka University

Practical Tasks
FUKUDA, Takamitsu
KOMINAMI, Hiroshi
MATSUO, Tsukasa
NAYA, Shin-ichi
NOMA, Naoki
SUDO, Atsushi
SUENAGA, Yusaku
YAMAGIWA, Yoshiro

Osaka University
Kindai University
Kindai University
Kindai University
Kindai University
Kindai University
Kindai University
Kindai University

The Chair of Scientific Committee
NISHIHARA, Hiroshi
Tokyo University of Science
The Chair of Theoretical Problem Committee
YORIMITSU, Hideki
Kyoto University
The Chair of Practical Task Committee
KURODA, Takayoshi
Kindai University

2

Fields of Advanced Difficulty: Theoretical subject
Structure and characterization of inorganic compound:solid state structure, unit cell, crystal field theory, a concept of hole.
Quantum mechanical behavior of molecules:
molecular vibration, how to read potential energy surface, definition of entropy basedon Boltzmann's principle.
Thermodynamics and kinetics:
relationship between electrode potential and Gibbs free energy, adsorption isotherm.
Stereochemistry:
conformation of fused bi- or tri-cyclic alkanes, chirality, stereospecific reaction,
stereoselectivity by steric hindrance, axial attack in cyclohexane system.
Number of signals in NMR (1H, 13C, and heteronuclear NMR) and chemical shift in 1H NMR.
Reactive intermediate and species:
carbenoid, non-benzenoid aromatics, organic main group-metal compounds,
heteroatom–heteroatom bonds.
Notes
Students are not expected to cover the following advanced topics:
Pericyclic reaction other than Diels-Alder reaction, Conservation of orbital symmetry, Polymer
chemistry, Phase equilibrium, Catalytic reaction on surface, Normal modes of molecularvibration, Molecular orbital method, Distribution function, Slater rule.
The practical tasks in the Appendix part include the following advanced skills, although thepractical examination will not be conducted in the IChO2021 Japan:
Preparation technique:vacuum filtration following the described procedure.
Purification:
recrystallization and column chromatography following the described procedure.

3

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

4

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:

𝐸 = 𝐸( +

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

5

Arrhenius equation:

Lambert–Beer equation:

𝐸&
R
𝑅𝑇, where k is the rate constant, A is the preexponential factor, Ea is the activationenergy.
exp (x) = e x
𝑘 = 𝐴 exp Q−

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:

𝐸 = ℎ𝜈 = ℎ

The sum of a geometric series:

When x ≠ 1,𝑐
𝜆, where ν is the frequency, λ is the
wavelength of the light.
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

6

44.955908

39

Calcium

40.078

Potassium

39.0983

7

Ra

Radium[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

13

O

H

O

Δδ for one alkyl group-substitution: ca. +0.4 ppm
8

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
b (ppm)

R3Si CH3

1

H NMR Chemical Shifts

Safety
Participants in the Olympiad must be prepared to work in a chemical laboratory and be aware of all
relevant rules and safety procedures. The organizers will strictly enforce the safety rules given in
Appendix A of the IChO Regulations during the Olympiad.
The Preparatory Problems are designed to be carried out in properly equipped chemical laboratories
under competent supervision only. We did not include specific and detailed safety and disposal
instructions as regulations are different in each country. Mentors must carefully adapt the problems
accordingly.
The GHS hazard statements (H-phrases) associated with the materials used are indicated in the
problems. Their meanings are as follows:
Physical Hazards
H225 Highly flammable liquid and vapor.
H226 Flammable liquid and vapor.
H227 Flammable liquid.
H228 Flammable solid.
H271 May cause fire or explosion; strong oxidizer.
H272 May intensify fire; oxidizer.
H290 May be corrosive to metals.
Health Hazards
H301 Toxic if swallowed.
H302 Harmful if swallowed.
H303 Maybe harmful if swallowed.
H304 May be fatal if swallowed and enters airways.
H305 May be harmful if swallowed and enters airways.
H311 Toxic in contact with skin.
H312 Harmful in contact with skin.
H313 May be harmful in contact with skin.
H314 Causes severe skin burns and eye damage.
H315 Causes skin irritation.
H316 Causes mild skin irritation.
H317 May cause an allergic skin reaction.
H318 Causes serious eye damage.
H319 Causes serious eye irritation.
H320 Causes eye irritation.
H331 Toxic if inhaled.
H332 Harmful if inhaled.
H333 May be harmful if inhaled.
H334 May cause allergy or asthma symptoms or breathing difficulties if inhaled.
H335 May cause respiratory irritation.
H336 May cause drowsiness or dizziness.
H351 Suspected of causing cancer.
H361 Suspected of damaging fertility or the unborn child.
H361d Suspected of damaging the unborn child.
H361f Suspected of damaging fertility.
H362 May cause harm to breast-fed children.
H370 Causes damage to organs.
9H371 May cause damage to organs.
H372 Causes damage to organs through prolonged or repeated exposure.
H373 May cause damage to organs through prolonged or repeated exposure.
Environmental Hazards
H400 Very toxic to aquatic life.
H401 Toxic to aquatic life.
H402 Harmful to aquatic life.
H410 Very toxic to aquatic life with long lasting effects.
H411 Toxic to aquatic life with long lasting effects.
H412 Harmful to aquatic life with long lasting effects.
H413 May cause long-lasting harmful effects to aquatic life.
Chemicals or substrates with GHS hazard statement H300, H310, H330, H340, H350, and H360 are
not allowed to be used in IChO and according activities.

10Problem 1. Revision of SI unit
The definitions of the SI base units were fully revised on May 20, 2019. Let us examine some of thedefinitions related to chemistry.
Part 1.
Before the new definition, the Avogadro number, NA, was defined as ‘the number of atoms contained
in 0.012 kg of the carbon isotope 12C’. In other words, a collection of NA atoms of 12C has a mass of
0.012 kg.
To determine NA experimentally, 0.012 kg of 12C must be accurately measured, which requires aweight ‘standard’. Traditionally, the kilogram prototype has served this purpose.
In practice, an international research consortium consisting of eight institutes including a Japanese one
made a single crystal of silicon (Si) composed solely of 28Si, from which a true sphere was cut for thedetermination of NA. The single crystal of Si has a diamond-type crystal system with the cubic unit
cell as shown in Figure 1.

Figure 1. Single crystal of Si.
1-1. Determine the number of atoms of Si, n, involved in the unit cell.
Let us consider the density of the true sphere of the single crystal of Si. The length of each side of theunit cell is a [m]. The atomic mass of 28Si is m = 28.09 u, where u is defined in the former fashion as:u=

1 ´ 10 -3 [kg]NAwhich has the same meaning as the atomic mass unit Dalton (Da).
1-2. Write an equation for the density, d [kg m-3], using n, m, and a.
The radius, r, of the true sphere of Si was measured to be r = 4.69 cm. Using the kilogram prototype,the mass of the sphere was determined to be w = 1.00 kg. The lattice length of the unit cell was
determined to be a = 5.43 Å by an X-ray diffraction analysis.
1-3.Calculate the Avogadro number, NA, using the measured parameters with three significant figures.
Part 2.
In the new definition, NA is defined with a finite number of digits (Table 1). In this situation, acollection of NA atoms of 12C does not exactly have a mass of 0.012 kg.
The new definition enables calculating mass using the equation found in Q1-3. In fact, mass can now
be determined without using the kilogram prototype if the volume is accurately measured. The
kilogram prototype itself is not easy to access, and replicas are also not easily available worldwide. In
addition, the aging of the prototype cannot be ignored. Determining mass without using the prototype
has therefore a significant merit. As a result, seven fundamentally important constants have been
determined using the newest measurement data based on the consensus of multiple research institutes.
11

Three of these constants are presented in Table 1.
Table 1. Three re-defined constants
Constant
Notation
Newly defined value
Avogadro constantNA
6.02214076 × 1023
Planck constanth6.62607015 × 10-34
Velocity of lightc
2.99792458 × 108

Unit
mol-1
Js
m s-1

Here, the product of the Avogadro and Planck constants is theoretically also a constant.
2-1. Based on these facts, select all the appropriate options for the determination of mass.
To determine mass:□ N A alone is sufficient□both N A and h are required□h alone is sufficient

Let us define mass by considering the units of the constants. The Avogadro constant has the unit
mol-1. Since this unit has no apparent unit of mass, discussion of mass based on this constant would be
difficult. On the other hand, the unit of the Planck constant, J s, can be rewritten as kg m2 s-1, which
seems more suitable for obtaining a definition of mass. To use this approach, we must understand thedefinitions of 1 s of time and 1 m of distance.
In the new definitions of the SI units, 1 s is based on the experimental observation that the energy
difference of the hyperfine energy splitting of 133Cs in a magnetic field (details are omitted here) isexpressed in units of Hz as:
Dn Cs = 9192631770 Hz

This highly accurate piece of information can be rewritten using the definition of Hz = s-1 as:
Dn Cs = 9192631770 (1 s)-1

We then obtain the following definition:
1s=

9192631770
Dn Cs

This means that 1 s is 9192631770 times longer than the time of the Dn Cs transition, which is the new
definition of 1 s.
In a similar manner, 1 m of distance is defined by using the velocity of light, c = 299792458 m s-1. Inother words, 1 m is defined as the distance that light travels in 1 s divided by 299792458. The definitionof 1 s can be substituted into this definition of 1 m to afford:
1m=

9192631770 cc= 30.663319
299792458 Dn Cs
Dn Cs, which is the new definition of 1 m.
2-2. Write a new definition for 1 kg of mass using these new definitions with eight significant digits.
On the other hand, mass can also be discussed in terms of energy. The mass of a particle at rest, m, isequivalent to the energy having a frequency of ν (i.e., the rest energy):
hn = mc 2
In this definition, a photon can also be recognized to have mass.
2-3. Calculate the apparent mass of a photon, m [kg], emitted from a laser pointer with a wavelengthof 635 nm.

12Problem 2. Does water boil or evaporate?
When an ideal gas is isothermally expanded from pressure 𝑝 [bar] to 𝑝′ [bar] (𝑝 > 𝑝′), the entropychange DS of the gas is
56(1).
∆𝑆 [J K .4 mol.4 ] = −𝑅 ln r s
5

The enthalpy change when liquid water evaporates to water vapor at 𝑝( [bar] is ∆𝐻#○ [J mol.4 ], andthe entropy change is ∆𝑆#○ [J K .4 mol.4 ]. Let us suppose these values do not depend on temperature,and the gas is an ideal gas. In this case, when liquid water evaporates to water vapor at 𝑝4 [bar], theenthalpy change is ∆𝐻# [J mol.4 ] = ∆𝐻#○ , and the entropy change is ∆𝑆# [J K .4 mol.4 ] = (a) .
1.

Write the appropriate formula for blank (a).

For the following questions, assume that ∆𝐻#○ = 4.070 × 108 J mol.4 and that, ∆𝑆#○ = 1.091 ×
10/ J K .4 mol.4 at 𝑝( = 1 bar.
2.

Calculate the boiling point of water at 1 bar.

3.

Calculate the saturated vapor pressure of water at 87 °C.

4.

Calculate the Gibbs energy change per mole when water evaporates at 87 °C at a humidity of
50%. The humidity represents the ratio of the actual partial pressure of water vapor relative to thesaturated vapor pressure of water.

A container with a piston like that shown in Figure 1 is filled with 0.10 mol of Ar and 1.00 mol ofwater (liquid and vapor). The temperature of the entire container is 87 °C and the total pressure is 1 bar.
It is assumed that the volume of the liquid or cup is negligible with respect to that of the gas, and thatthe temperature of water and Ar is always maintained at 87 °C. In this initial situation, the partialpressure of Ar is (b) bar, the partial pressure of water vapor is (c) bar, the volume of the gas
is (d) L, and the number of moles of liquid water is (e) mol.

Figure 1. A container with a piston containing Ar gas, water vapor and liquid water in a cup.
The piston is then pulled back quickly to fix the gas volume to 15.8 L. At the moment when the piston
is pulled, the partial pressure of Ar decreases to(f)bar and the partial pressure of water vapor
decreases to (g) bar. In this situation, the water will boil, because the total pressure of the gas islower than the saturated vapor pressure of water at 87 °C. The partial pressure of water vapor increases
due to the boiling until the boiling eventually stops. When the boiling stops, the number of moles of13liquid water is(h)mol. After that, evaporation proceeds until equilibrium is reached. At
equilibrium, the number of moles of liquid water is (i) mol.
5.

Calculate the appropriate numerical values for blanks (b) - (i).

14

Problem 3. Molecules meet water and metals
I. Dissolution of gases in water
For gases that are slightly soluble in water, there is a proportional relationship between the partialpressure, P, and the mole fraction, x, of the gas molecules dissolved in water (Henry's law):
𝑃 = 𝑘9 𝑥
Here, kH is the Henry coefficient. The Henry coefficients of nitrogen, oxygen, and argon at 25 ∘ C forwater are 8.57×104, 4.42×104, and 4.02×104 bar, respectively. In the following questions, thecompression of the water by the pressurized gas can be ignored.
Part 1. Decompression sickness
When a diver spends prolonged time at high depths, the high water pressure causes nitrogen moleculesin the body to dissolve in the blood. If the diver rises too quickly to the surface, this dissolved nitrogen
will form bubbles that damage the blood vessels and tissues. With the decompression sickness
phenomenon in mind, answer the following questions regarding the dissolution of gas in water in aclosed container.
1-1. Calculate the pressure [bar] underwater at a depth of 20.0 m. Here, the density of water is 1.00 g
cm-3, the specific gravity of seawater is 1.02, the gravity acting on a mass of 1 kg is 9.81 N, andthe atmospheric pressure is 1.01 bar.
1-2. At a temperature of 25 ∘ C, 5.00 L of water, which is approximately equivalent to the blood volumeof an adult, is added to a container with a piston, and the container is filled with nitrogen. Then,the pressure of the nitrogen is adjusted to the value calculated in Q1-1 by pushing down the piston,and sufficient time is allowed to pass for the gas–liquid equilibrium to be established. Calculate
the amount of nitrogen [mol] dissolved in water.
1-3. The piston is then moved to lower the pressure to 0.800 bar, which is approximately equal to thepartial pressure of nitrogen in the atmosphere, and sufficient time is allowed for the vapor–liquid
equilibrium to be established. Calculate the volume [L] of nitrogen at 25 ∘ C that is released into
the gas phase from the nitrogen dissolved in water in Q1-2.
Part 2. Sparkling beverages
The dissolution of high-pressure gas in water in a closed container is also relevant to sparkling
beverages.
2-1. A container (V = 570 mL) is filled with water (𝑉; = 500 mL) and pressurized with CO2 gas (𝑃( =50 atm), before it is allowed to stand at 10 ∘ C until the vapor–liquid equilibrium is established.
Calculate the pressure of CO2 [atm] in the container and the amount of CO2 [mol] dissolved in
the water. The Henry coefficient of CO2 for water at 10 ∘ C is 𝑘9 = 0.104 × 104 atm, and we willconsider that the reaction of CO2 in water can be ignored.
2-2. Carbon dioxide (N L) dissolved in water (1 L) is called N GV (Gas Volume), where N is thevolume of this gas at 0 °C and 1 atm. Calculate how many GV the aqueous solution obtained in
Q2-1 corresponds to.
2-3. When this container is heated to 50 ∘ C, calculate the pressure of the gas [atm] in the container andthe amount of CO2 [mol] dissolved in water. Here, the Henry constant of CO2 at 50 ∘ C is 0.283 ×
104 atm.
15

Part 3. Removal of dissolved oxygen
3-1. A container (2.0 L) filled with water (1.0 L) has been left exposed to air for a long time at 1.0 bar
and 25 ∘ C. Calculate the mole fraction and mass of oxygen [mg] dissolved in the water. Here,assume that the atmosphere contains 21% oxygen.
3-2. The atmosphere in contact with this water is replaced with 1.0 bar argon and sealed, and the
mixture is stirred sufficiently for the gas–liquid equilibrium to be established at 25 °C. Calculate
the mole fraction of oxygen dissolved in the water. Here, the evaporation of water can be ignored
during the gas replacement.
3-3. Calculate how many times the above operation must be repeated to reduce the mole fraction ofdissolved oxygen to 1 ppt (= 10-12) or less.
Part 4. Complex formation in water
Consider a mixture of two different gases (A and B) for which Henry's law holds. When gas molecules
A and B do not interact with each other in either the gas phase or in water, the amount of each that will
be dissolved in water is determined only by its partial pressure and does not depend on the partialpressure of the other gas. However, let us now consider the case in which A and B form a complex AB
when they are dissolved in water:
A (l) + B (l) = AB (l)Let xi (i = A, B, AB) be the mole fraction of the dissolved species, and let KAB be the equilibrium
constant for the above equation:
𝐾<= = 𝑥<= /𝑥< 𝑥= .
Suppose that water (Vl =1.00 L) and a gas mixture are filled into a container (V = 2.00 L) at 25∘ C. The
Henry coefficients of A and B are 1.0×104, and 2.0×104 bar, respectively, and KAB= 500. Answer thefollowing questions:4-1. The initial partial pressures of each gas, 𝑃<( and 𝑃=( , are both 2 bar. Calculate the pressure of eachgas and the mole fractions of the dissolved gas molecules after the equilibrium is established.
4-2. Calculate the percentage of the mole fraction of dissolved gas B relative to that found in Q4-1
when the initial partial pressure of gas A is increased to 10 bar.
II. Adsorption of molecules on a metal surface
Let us consider the adsorption of gas molecules on a metal surface under the following assumptions:(1) There are specific sites on the metal surface where the gas molecules are adsorbed; (2) when a gas
molecule occupies an adsorption site, the site is blocked so that other molecules cannot be adsorbed at
the occupied site; (3) the probability of the adsorption of a molecule onto an empty site does not depend
on whether or not a molecule is adsorbed at the neighboring sites; i.e., molecules adsorbed on the metal
surface do not interact with each other. Here, we define parameters relevant to the adsorption: 𝑆( [molm-2] is the density of adsorption sites; 𝜃 = 𝑎/𝑆( is the coverage, where 𝑎 [mol m-2] is the number ofadsorbed molecules per unit surface area; 𝑆 [mol m-2] is the number of empty sites per unit area.
Answer the following questions:
Part 5. Adsorption isotherm
5-1. The surface is exposed to a gas at a pressure P. Let the reaction rate constants of adsorption anddesorption be 𝑘& and 𝑘% , respectively. Write an expression for the rates of adsorption anddesorption in terms of 𝑃, 𝑘& , 𝑘% , 𝑎, 𝑎𝑛𝑑 𝑆.
5-2.Adsorption–desorption equilibrium is achieved when the adsorption and desorption rates are equal.
16

Let the equilibrium constant be 𝐾 = 𝑘& /𝑘% . Write an expression for the coverage θ at equilibriumin terms of K and P.
5-3. Express P in terms of 𝜃 and 𝐾 for the case of extremely low coverage (𝜃 ≪ 1).
5-4. Metal particles (𝑚 = 100 g) are placed in a container (𝑉 = 1.00 m3) and a gas A is introduced into
the container at an initial pressure of 𝑃( = 100 Pa. The gas pressure decreases and then reaches aconstant value. Calculate the pressure of the gas [Pa] in the container and the amount of gas
molecules [mol] adsorbed on the surface of the metal particles. Here, assume that the surface areaof the metal particles per unit mass is 𝐴 = 10.0 m2 g-1; 𝑆( = 1.66 × 10-5 mol m-2; 𝐾 = 10.0 Pa-1;
the temperature of the system is constant at T = 300 K and the volume of the metal particles isnegligible.
Part 6. Adsorption of mixed gas
6-1. Consider the case in which two gases (A and B) are adsorbed on the metal surface. Assume thatthese gases do not interact or react with each other (neither in the gas phase nor on the surface
after adsorption), and that an adsorption site can be occupied by either A or B. Express thecoverage of A and B (𝜃< and 𝜃= ) in terms of the partial pressures of the gases (𝑃< and 𝑃= ) as wellas the adsorption and desorption equilibrium constants (𝐾< and 𝐾= ).
6-2. When gases A and B are filled into the same container as in Q5-4, the partial pressures of eachgas after reaching equilibrium are measured to be 𝑃< = 70.0 Pa and 𝑃= = 30.0 Pa. Calculate thecoverage of each gas adsorbed on the surface of the metal particles. Let the adsorption anddesorption equilibrium constants of gases A and B be 𝐾< = 10.0 Pa-1 and 𝐾= = 20.0 Pa-1,
respectively.
6-3. Under the conditions where 𝐾< 𝑃< ≪ 1 and 𝐾= 𝑃= ≪ 1 , how can the coverage 𝜃< and 𝜃= beapproximated? Write approximate expressions in terms of 𝑃< , 𝑃= , 𝐾< , and 𝐾= .
6-4. Gases A and B are introduced into the container described in Q5-4 and the system reaches
adsorption–desorption equilibrium. The partial pressure of gas A is found to be 70.0 Pa, which isthe same as in Q6-2, whereas the partial pressure of gas B is 60.0 Pa, which is twice that in Q6-2.
Calculate the percentage of coverage of gas A compared to that under the conditions of Q6-2.
III. Note
Comparing Part 4 and 6, we understand the following difference between gas dissolution in liquid andadsorption on metal surface. Under conditions where Henry's law holds, the dissolution of molecules
into a liquid depends only on the partial pressure of the gas of interest. The same applies to adsorption
onto a metal surface when the coverage is sufficiently low. However, even under conditions where
Henry's law holds, if there is interaction and association between the molecules of the different species
in the liquid, the amount of each species dissolved will depend not only on the partial pressure of thegas, but also on the partial pressure of the molecules of the other species. In contrast, as the coverageincreases in the case of adsorption, the amount of adsorption for a given species will depend on thepartial pressure of the molecules of the other species even if there is no interaction between themolecules of the different species. This is due to the finite number of sites where the molecules can beadsorbed on the metal surface.

17

Problem 4. Synthesis of diamonds
Diamond, one of the allotropes of carbon, is used not only for jewelry, but also in various tools thattake advantage of its extremely high hardness. Natural diamond is considered to be produced underhigh-pressure and high-temperature conditions in the bowels of the earth. Currently, several techniques
to synthetically generate diamonds are available, and massive amounts of artificial diamonds are usedindustrially.
Let us examine the stability of diamonds at ambient pressure and temperature. The enthalpies ofcombustion (𝛥> 𝐻∘ ) for 1 mole of graphite, diamond, and C60, as well as the standard entropy changes(Δ𝑆 ∘ ) for the conversion of 1 mole of graphite to diamond or C60 under ambient conditions (1 bar,298.15 K) are summarized in Table 1. For graphite and diamond, 1 mole represents 12.01 g.
Table 1. Thermodynamic data for carbon allotropes
𝛥> 𝐻∘ / kJ mol−1
Graphite−393.5
Diamond−395.3
C60−25965

Δ𝑆 ∘ / J K−1 mol−1
(0)
−3.25
1.54

1.

Calculate the change in enthalpy, entropy, and Gibbs free energy for the transformation of 1 moleof graphite into diamond and into C60 under ambient conditions. Then, rank graphite, diamond,and C60 in order of decreasing stability in terms of Gibbs free energy.

2.

Under ambient conditions, diamond does not transform into graphite, which is energetically morestable than diamond. Explain why this transformation does not occur.

The enthalpies of combustion for graphite and diamond are comparable, and therefore, it is difficult todetermine the change in enthalpy from graphite to diamond precisely via the calculation method usedin Q1. To overcome this difficulty, the temperature dependence of the electromotive force of thefollowing battery was measured:
Pt, C(graphite) + CaC2 + CaF2 || CaF2 || CaF2 + CaC2 + C(diamond), Pt(1)

This is a solid-state battery in which diamond is used as the cathode (positive electrode) and graphite
is used as the anode (negative electrode). On the cathode, the diamond is reduced and CaC2 is produced.
On the anode, CaC2 is oxidized and graphite is produced.
3.

Write the half-reactions occurring on the cathode and the anode.

4.

The electromotive force of this battery over the temperature range 940–1260 K can be expressed
as a function of the absolute temperature T according to:
𝐸 = 11.4 + 0.0481𝑇 [mV](2)

Calculate the change in enthalpy and entropy for the conversion of 1 mole of graphite to diamondin this temperature range.

18One of the techniques to synthesize diamond is the direct conversion of graphite into diamond by
applying high pressure to graphite. The mechanism underlying this conversion is known as thepuckering mechanism.
Graphite is composed of stacked layers of graphene. For each layer, there are three possible relative
positionings in which the graphene layer can be placed. From here on, these geometries are referred toas A, B, and C. The crystal structures of graphite can exhibit one of three stacking patterns:
AAAAAA… (1H where H stands for hexagonal), ABABAB… (2H), and ABCABC… (3R where R
stands for rhombohedral). The most stable structure is 2H, and typical graphite also contains a small
amount of the 3R structure.
5.

One of the geometries of graphene is shown below (left, set as A). Draw the other two geometries(B and C, you do not have to distinguish these two).

6.

Assign the following three crystal structures as the 1H, 2H, and 3R structures.

Diamond (cubic crystal) is generated from the 3R structure by displacing each carbon atom in a
direction perpendicular to the plane. A crystal structure similar to that of diamond is obtained bydisplacing each carbon atom of the 1H structure in a direction perpendicular to the plane. The crystal
structure generated from the 1H structure is called lonsdaleite, and is a hexagonal crystal. It has beentheoretically predicted that pure lonsdaleite would be harder than diamond.

19

7.

Assign the following crystal structures as diamond and lonsdaleite.

8.

Under ambient conditions, the distance between covalently bonded carbon atoms in graphite is
1.42 Å, while the distance between the graphene layers of graphite is 3.35 Å; the distance between
covalently bonded carbon atoms in diamond is 1.54 Å. Calculate the density of graphite anddiamond under ambient conditions. The unit cell of diamond is shown below:

In the puckering mechanism, there are two energy barriers: The barrier for the transformation from the
2H structure to the 3R structure (Step 1), and the barrier for the transformation of the carbon atoms
from a planar to a tetrahedral configuration via puckering (Step 2).
9.

The activation energy in Step 2 will be small if there is no volume change between graphite anddiamond. Calculate the distance between the graphene layers of graphite for which there would
be no volume change. Assume that the distances between the covalently bonded carbon atoms ingraphite and diamond do not change on account of the applied pressure.

[Note] The direct conversion method described above requires high temperature (3000–4000 K) to
increase the reaction rate in addition to high pressure (~ 10 GPa). A new conversion method, in which
the graphite is first melted with metals to break the covalent bonds among the carbon atoms before
new covalent bonds are generated under high-pressure conditions where diamond isthermodynamically stable, has been developed.
In addition to the direct conversion method, diamond can also be synthesized by a chemical vapor
deposition (CVD) method. In this technique, a thin layer of diamond is grown on a substrate exposed
to a gas containing hydrocarbon radicals prepared by heating or electrical discharge.
As an example, let us consider a diamond synthesis using methane and hydrogen molecules as sources.
The reaction rates for the synthesis of diamond and graphite on the substrate, 𝑣% and 𝑣? , can be
expressed in terms of the concentration of active hydrocarbons (such as CH3 radicals) near the surface
20

of the substrate, 𝐶@ , and the rate constants, 𝑘% and 𝑘? , according to:𝑣% = 𝑘% 𝐶@(3)𝑣? = 𝑘? 𝐶@(4)

At the same time, diamond and graphite on the substrate are removed in the form of gaseous species(typically hydrocarbon radicals) due to reaction with hydrogen atoms near the surface. This process iscalled etching. The reaction rates for the etching of diamond and graphite from the substrate, 𝑣%6 and𝑣?6 , are expressed in terms of the concentration of hydrogen atoms near the surface of the substrate, 𝐶9 ,and the rate constants, 𝑘%6 and 𝑘?6 , according to:𝑣%6 = 𝑘%6 𝐶9(5)𝑣?6 = 𝑘?6 𝐶9(6)

10. Write an expression for the concentration of hydrogen atoms near the surface of the substrate interms of 𝐶@ and the appropriate rate constants when the amount of graphite on the substrateremains constant.
11. Write an expression for the net growth rate of diamond on the substrate in terms of 𝑣% and theappropriate rate constants under the condition where the amount of graphite on the substrateremains constant. (Based on this result, the condition that the reaction constants must satisfy canbe obtained.)
To determine the activation energy of this process experimentally, the growth rate of diamond on thesubstrate is used. The measured growth rates of the thicknesses of the diamond layer on the diamondsubstrate for different substrate temperatures and gas compositions are shown in Table 2. In the casewhere only hydrogen gas was used, the substrate became thinner due to etching (eq. 5).
Table 2. Growth rates of diamond prepared by the chemical vapor deposition method
Substrate temperature
900 °C
1100 °C−1
CH4 13% + H2 87%
30.34 μm h63.43 μm h−1−1
H2 100%−1.58 μm h−5.52 μm h−1
Although the elementary steps of the synthesis and etching of diamond are not known, it is reasonable
to define the activation energy for the total reaction, because the reaction rates for the synthesis andetching on the substrate follow the Arrhenius equation.
12. Calculate the activation energy of the synthesis of diamond when the composition of the source
gas is CH4 (13%) + H2 (87%). Assume that the effect of etching by hydrogen gas is negligible
compared to the synthesis of diamond.
13. Calculate the pre-exponential factor of the Arrhenius equation and the activation energy for theetching of diamond by hydrogen atoms.

2114. When the experiment is conducted using a CH4 (1%) + H2 (99%) source gas mixture at 1000 °C,
the thickness of the substrate does not change. Assuming that the etching of diamond by hydrogenatoms and the synthesis of diamond occur independently, i.e., that the pre-exponential factor of
the etching and the activation energies of the etching and diamond synthesis are the same as thevalues obtained in the previous questions, calculate the temperature of the substrate where thegrowth rate of diamond is 0.50 μm h−1 using the CH4 (1%) + H2 (99%) source gas. By controlling
the reaction rate of the synthesis of diamond and etching appropriately as shown in this problem,a very thin diamond coating can be achieved.

22

Problem 5. Count the number of states
Boltzmann defined entropy as a measure of the number of possible microscopic states of a system, W,
as:
𝑆 = 𝑘= ln𝑊
Here, 𝑘= is the Boltzmann constant. Hereafter, it is assumed that neither energy nor particles areexchanged between the system of interest and the surrounding environment, and that the system is inequilibrium at a constant temperature. In addition, the gas atoms and molecules (henceforth referred
to as ‘molecules’) are assumed to behave as an ideal gas.
Now, let us explore the entropy of gas molecules using this definition in Questions 1–7.
Assume that ten molecules are arranged in two chambers separated by a wall, A and B, with fourmolecules in A and six in B. We describe this microscopic state as (4,6).

Figure 1. Example of a microscopic state in which ten molecules are distributed in two chambers.
The number of cases in which 10 molecules result in a (4,6) arrangement is calculated as:𝑊(4,6) = 4( 𝐶8 = 10!/6! 4! = 210
For (5,5):𝑊(5,5) = 4( 𝐶A = 10!/5! 5! = 252
Thus, W(5,5) is larger than W(4,6). In this system, the number of cases represents the number of states
for a specific arrangement of gas molecules in the two chambers.
1.

Calculate W(40,60) and W(50,50) with 2 significant digits using the following approximate
formula for a factorial:𝑛 0
𝑛! ≃ √2𝜋𝑛 r s𝑒Removing the boundary wall that separates the two chambers in Figure 1 allows the molecules to movethroughout the entire container. Assume that it is possible to measure the arrangement of the molecules
instantly after the removal of the wall. If this measurement is performed a sufficiently high number oftimes, the probability that a state with an arrangement (n,m) will be found can be calculated.
2.

Calculate the arrangement (𝑛∗ , 𝑁 − 𝑛∗ ) with the highest probability of occurrence for N = 10 and
100, as well as the corresponding probability of occurrence for each state. Moreover, calculate
the probability of observing a state in which 𝑛∗ is in the range 𝑛∗ – 0.02 N ≤ 𝑛∗ ≤ 𝑛∗ + 0.02 N for
N = 10 and N = 100. Here both N and n* are integers.

Let us consider the space inside chamber A, whose volume V1 is composed of small subsections DV.
As a molecule in the chamber is located in one subsection at any given time, the total number ofpossible locations of the molecule in the chamber WA is given by 𝑊< = 𝑉4 /Δ𝑉. Likewise, the numberof possible locations for a molecule that can access the whole chamber, whose volume is V2, can be23written as 𝑊<= = 𝑉/ /Δ𝑉. Thus:

𝑊<= 𝑉/ /Δ𝑉 𝑉/==
𝑊<
𝑉4 /Δ𝑉 𝑉4
Now suppose that chamber A is filled with n moles of gas molecules, while no molecules are present
in chamber B (State 1), i.e., chamber B is a vacuum. Then, the boundary wall is removed so that thegas in chamber A expands spontaneously throughout the entire container (State 2) at a constanttemperature.
3.

Express the change in entropy Δ𝑆 = 𝑆/ − 𝑆4 in terms of the gas constants R, V1, and V2.

4.

Calculate the difference in entropy for an isothermal expansion of the two-chamber systemdescribed above when chamber A is initially filled with oxygen molecules (0.30 mol). Use 𝑉4 =0.10 m3 and 𝑉/ = 0.40 m3.

As shown in Figure 2, let State 1 be the case in which chambers A and B, whose volumes are identical,
are each filled with a different kind of gas molecules. Suppose that these gases are mixed isothermallyin a chamber with the same volume as chambers A and B (State 2).

Figure 2. Isothermal mixing of different kinds of gas molecules in chambers with the same volumes.
5.

Calculate the difference in entropy associated with the mixing process.

As shown in Figure 3, two kinds of gas molecules, a (na moles) and β (nβ moles), are separately filled
into chambers A (volume VA) and B (volume VB), respectively, at a temperature T and a pressure P(State 1). Let State 2 represent the complete mixture of the gas molecules after removing the boundary
wall.

24

Figure 3. Mixing of different kinds of gas molecules accompanied by expansion.
6.

Express the entropy change ΔS = S2−S1 from State 1 to State 2 in terms of R, na , and nβ.

7.

Instead of being filled with molecules of gas β as in Q6, assume that chamber B is filled withmolecules of gas a (nβ moles) (State 1), before the boundary wall is removed (State 2). Calculate
the entropy difference from State 1 to State 2.

In Questions 8 and 9, we will explore the entropy associated with the orientation of molecules in a
crystal.
According to the third law of thermodynamics, the entropy of a pure and perfect crystal approaches
zero as the absolute temperature approaches 0 K. However, in a real molecular crystal, the moleculesin the crystal may not be completely aligned at low temperatures. Thus, even when the absolutetemperature approaches 0 K, the crystal can retain a finite nonzero entropy value. This is called residual
entropy. For example, because carbon monoxide (CO) is a heterogeneous diatomic molecule, CO
molecules have a definite orientation in the CO crystal. Figure 4A shows that the orientation of somemolecules is disordered. Because a CO molecule can exhibit two different orientations in the crystal,
the residual entropy per mole if the molecular orientations are completely random in the crystal wouldbe:
S = kB ln 2C! = R ln 2 = 5.7 J K-1 mol-1

Figure 4. Schematic representation of disorder in the molecular orientation
in a CO crystal. A: Some CO molecules are oriented in the opposite direction.
B: An ideal crystal without any disorder in the molecular orientation.
8.

Every methane (CH4) molecule in a crystal is surrounded by four methane molecules in atetragonal fashion. Calculate the molar residual entropy of isotopically labeled methane, H3CD,
when the molecules are completely randomly oriented.
25

Figure 5 shows a molecular arrangement of ice (solid H2O); a central water molecule is coordinatedby four water molecules in a tetragonal fashion.

Figure 5. Arrangement of water molecules in an ice crystal.
All the water molecules are arranged in the crystal according to the so-called ‘ice rules’:(1) Each hydrogen atom must be located between two adjacent oxygen atoms.
(2) Two of the four hydrogen atoms that surround each oxygen atom must be positioned closer to thatoxygen than to the neighboring oxygen atoms, while the other two hydrogens should be locatedcloser to one of the neighboring oxygen atoms.
9. Now, let us estimate the molar residual entropy of ice using the following procedures.
9-1. There are two stable sites for a hydrogen atom between two adjacent oxygen atoms. Calculate thenumber of possible configurations of hydrogen atoms in a crystal composed of 1 mole of water(NA molecules) without any constraints of the ice rules.
9-2. Calculate the number of possible configurations for four hydrogen atoms around the central
oxygen atom in Figure 5.
9-3. Some of the configurations calculated in Q9-2 such as (H3O)+, in which three protons are arranged
closest to one oxygen atom, violate the ice rules. List all chemical species that break the ice rules
and calculate the number of configurations for each species. Then, calculate the number ofconfigurations that satisfy the ice rules.
9-4. Based on these considerations, calculate the molar residual entropy when the orientation of watermolecules is completely random in ice.

26

Problem 6. The path of chemical reactions
The structure of a molecule and its changes can be understood in terms of the potential energy.
Consider the motion of the two atoms that make up a diatomic molecule AB. The velocity of the atoms
A and B, whose masses are mA and mB, are denoted as vA and vB. The kinetic energy of the system isgiven by1
1
(1)
𝑇 = 𝑚< 𝑣</ + 𝑚= 𝑣=/
22Each atom, A and B, has three degrees of freedom, and there are six degrees of freedom in total.
Because the motion of the molecule AB consists of the part of the translation and rotation of the whole
molecule and that of the relative motion of the two atoms, the above equation can be rewritten as
1
1(2)
𝑇 = 𝑀4 𝑣4/ + 𝜇(𝑣< − 𝑣= )/22where µ is the effective mass defined as
1
1 .4
(3)
𝜇=Q
+
R
𝑚< 𝑚=
1.

Calculate 𝑀4 and 𝑣4 in Equation (2). 𝑀4 and 𝑣4 are quantities with dimensions of mass andvelocity, respectively.

The geometry of a diatomic molecule is determined only by the relative position (coordinates) of thetwo atoms, i.e., the distance between them, R, which corresponds to the vibrational motion of themolecule. The potential energy E as a function of R near the equilibrium structure can be approximatedby1(4)
𝐸(𝑅) − 𝐸( = 𝑘(𝑅 − 𝑅( )/2where E0 is the reference energy and R0 is the equilibrium internuclear distance. Under normal
conditions, the molecule adopts a stable structure with the lowest energy. In this equation, k is aphysical quantity corresponding to the spring constant that increases with increasingly stronger bonds.
This quadratic representation is called the harmonic oscillator approximation, and k is linked to thefrequency 𝜈 via the effective mass.
𝜈=
2.

1 𝑘
“
2𝜋 𝜇

(5)

Suppose that diatomic molecule XY forms either single or double bonds. What relationship canbe expected for each k (kS and kD) and their corresponding equilibrium internuclear distances (RS0and RD0 )? Choose only one answer.

□ kS > kD and RS0 > RD0□ kS = kD and RS0 > RD0□ kS < kD and RS0 > RD0□ kS > kD and RS0 = RD0□ kS = kD and RS0 = RD0□ kS < kD and RS0 = RD0

27□ kS > kD and RS0 < RD0□ kS = kD and RS0 < RD0□ kS < kD and RS0 < RD0Q2. Suppose that atoms X and Y form either single or double bonds. What relationship can be expected for each k (k S and k D ) and their corresponding equilibrium
inter-nuclear distances (R0S and R0D )? Choose one that is correct.

For diatomic molecules in reality, bonds break when R becomes sufficiently large; then atoms noS> k D and R0S > R0D
k S > k D and R0S = R0D
k S > k D and R0S < R0D
longerk interact
with each otherand the energy remains
unchanged. The following equation, the soSD= k D and
R0S > R0D approximation,k S = k D and R20Sis= often
R0D
k S = insteadk D and R0Sof<R
0called kMorsepotentialusedthequadratic equation described above
S
D
S
D
S
D
S
D
S
D
S
D
k
<
k
and
R
>
R
k
<
k
and
R
=
R
k
<
k
and
R
<
R
0
0
0
0
0
0
to better describe the potential energy of a molecule.
/(6)
𝐸(𝑅) − 𝐸( = 𝐷" ”1 − 𝑒 .&(E.E") •For a diatomic molecule in reality, when R bewhere De, a, and R0 are constants.

atoms no longer interact with each other, making
and Y form either single or double bonds. What relathe energy unchanged. The following equationeach k (k S and k D ) and their corresponding equilibrium
is often used instead of the quadratic equationnd R0D )? Choose one that is correct.
described above to better describe the potential

E(R) − E0

comes large enough, the bond breaks and theS
DS
Dof aRmolecule
potential
approximak Senergy> k D andkS >k D and R
0 = R0 (Morse
0 < R0k Stion).
= k D and R0S = R0D
k S < k D and R0S = R0D

k S = k D and R0S < R0D
S

D

S

R

D

!k < k and R0"2< R0
E(R) − E0 = De 1 − e−a(R−R0 ) ,(4)

1 potential
Figure 1.Figure
Morsey, when R be3.e , a Draw
and R0 in the figure below.
where Dand R0 D
aree constants.
reaks and theratic equation

E(R) − E0

wing equation

E(R) − E0

other, makingQ3. Show De and R0 in the figure below.

the potential

ial approxima-

−R0 )

"2

,

(4)

R

Figure 1

4.

Choose all correct explanations for the harmonic and Morse potential approximation energyRcurves.

figure below. Q4. Choose all correct explanations for the harmonic and Morse potential ap-□ For the harmonic oscillator approximation (Eq. 4) and the Morse potential
approximation (Eq. 6), the equilibrium internuclear distance is the same.
□ In the region where R is long, the Morse potential approximation is lower in energy
than the harmonic oscillator approximation.
□ In the region where R is short, the Morse potential approximation and the harmonic
oscillator approximation coincide.
□ As a increases in the Morse potential approximation, the width of the well becomesnarrower.

proximation energy curves.

R
For diatomic molecules,
the energy E is represented as a one-dimensional curve that depends only on
R(Figure
1),
while
for
generalpolyatomic molecules, it is a multi-dimensional surface, i.e., a point on
xplanations for the harmonic and Morse potential apsuch an energy surface corresponds to the structure of the molecule. Figure 2 shows a bird's-eye viewof the energy surface, wherein the direction of the x-axis is the Morse oscillator and the direction of yaxis is the harmonic oscillator.

28

Figure 2. Energy surface in two dimensions
5.

4 the horizontal axis and the y axis
Sketch the contour map of this energy surface with the x axis as
as the vertical axis.
5
4
3

y
2
1
0

0

1

2

x

3

4

5

The molecular geometry of a triatomic molecule can be determined by specifying three parametersthat characterize the structure, such as both internuclear distances and the bond angle. A contour mapof the potential energy surface of a typical triatomic molecular reaction A+BC → AB+C is shown in
Figure 3.
5.0

d by specifying three parameters

4.0

the structure, such as the inter-

3.0

and the bond angle. A contour

tial energy surface of a typicalR (A-B)ometry of a triatomic molecule

2.0
1.0

ar reaction is shown in the fig-

B+ C

5.0
4.0
R (A-B)0/

𝐸(𝑥, 𝑦) = 5.0”1 − 𝑒 ./.((H.4.() • + 20.0(𝑦 − 2.0)/

3.0
2.0
1.0

0

1.0

0

2.0 3.0 4.0 5.0
R (B-C)
1.0 2.0 3.0 4.0
R (B-C)

5.0

Figure 3

Figure 3. Contour map of the reaction A+BC → AB+C

with a diatomic molecule BC, producing a new diatomic molecule
AB and a29

suming that the three atoms are always on the same line, two degrees of freedom

letely determine the arrangement of the three atoms in the reaction process.

Here, A collides with a diatomic molecule BC, producing a new diatomic molecule AB and monatomic
C. Assuming that the three atoms are always on the same line, two degrees of freedom allow us tocompletely determine the arrangement of the three atoms in the reaction process. In Figure 3, thevertical axis represents the internuclear distance of AB while the horizontal axis represents theinternuclear distance of BC.
Thus, the structural change in the reaction can be interpreted as a pathway on the potential energy
plane. Fukui has defined this pathway as the intrinsic reaction coordinate (IRC).
6.

As a reaction path, consider a path that always minimizes the potential energy in any direction
other than the reaction path. Sketch this path in the figure below.

5.0

R (A-B)

4.0
3.0
2.0
1.0
0

1.0

2.0 3.0 4.0
R (B-C)30

5.0

In the actual reaction, the nuclei, which do not participate directly in the recombination of atoms, also
oscillate and do not stop, which is the so-called ‘bobsleigh effect’. This corresponds to a motion in the
direction perpendicular to the reaction coordinates.
7.

Suppose that BC is vibrating before A collides, and that AB is also vibrating after the collision.
Sketch a path of this reaction in the figure below.

5.0

R (A-B)

4.0
3.0
2.0
1.0
0

1.0

2.0 3.0 4.0
R (B-C)31

5.0

Let's generalize further to consider reactions in more complex molecules.
CH3CH2F → CH2=CH2 + HF
As a result of this dissociation reaction, all the atoms change their position, bond length, and angles.
Let’s focus on the distances shown in Figure 4.
RCC: The distance between the carbon atoms
ember 29, 2020
6
RHF: The distance between Hα and fluorine (F)RCH: The average of the four hydrogen (H) -carbon distancess generalize furtherR to :consider
reactions in
CM The distance between the center of mass of HF and CH2CH2 moieties.

e complex molecules.
r 29, 2020

CH3 CH2 F−→ CH2 =CH2 +HF
neralize further to consider reactions in

Hα

6

RHF F

mplex
molecules.
a resultof this dissociation reaction, all thems change their positions, as well as their bond
H3 CH2 F−→ CH2 =CH2 +HF
ths and angles. Therefore, we focus on the

Hα

RHF F RCM

ult of shown
this dissociation
ancesin Figure 4.reaction, all theange their positions, as well as their bond
: The distance between the carbon atomsnd angles. Therefore, we focus on the

H
H

C

RR
CM
CC

C

H
H

Figure4. RCC, RHF, RCH, and RCM in theFigure
reactionin Figure4.
: shown
The distance
betweeH(F)
4 CH3CH2F → CH2=CH2 + HF
α –fluorine

C R
C
CC
H
H
Figure5theshows
howatoms
the structurehedistance ofbetweencarbon: Average
the fourhydrogen(H)-carbon of the molecule changes while the reaction progresses. The verticalH
H
axis of the graph is the change in the interatomic distance ∆R from the beginning of the reaction,

distances while the horizontal axis is the reaction
0.4coordinates, quantifying the progress
A of the reaction, whiche distance betwee
Hα –fluorine (F)Figure 4

proceeds from left to right.

: The distance between the midpoint of theverage of the four hydrogen (H) -carbon
0.2
two carbons (M) and F
tances
0.4

A

∆R/Å

re 5 shows how the structure of the molecule
he distance between the midpoint of the0.0
ges with the reaction. The vertical axis of the 0.2
o carbons (M) and F
h is the change in the interatomic distance
-0.2
shows
howbeginning
the structureof reaction,
the moleculefrom theof theand the
0.0
with
the
reaction.
The verticalaxis of thezontalaxis
is the reaction
coordinates,
quanti-C∆R/Å

distance-0.4
gthe
the change
progressinofthe
theinteratomic
reaction, which
proceeds-0.2
the tobeginningleftright. of the reaction, and theal axis is the reaction coordinates, quantiQ8. Answer which of the four disprogress of the reaction, which proceeds
tances in the graph, A to D , repto right.
resents the above four distances.

B

-0.4

B
C

D
Reaction Coordinate
Figure 5
D
Reaction Coordinate

RCC, RHF, RCH, and RCM during the. Answer which of the four dis- Figure 5. The change of
Figure5 + HF
reaction CH3CH2F → CH2=CH
2

nces in the graph, A to D , rep-

8. fourMatch
A to D in the graph and the four distances mentioned above (RCC, RHF, RCH, and RCM).
ents the above
distances.
32

Problem 7. Molecular vibrations and infrared spectroscopy
Consider the molecular vibration of hydrogen fluoride (HF). When HF diluted with a large excess ofargon (Ar) is sprayed onto a cold cesium iodide (CsI) substrate at 12 K, the gas mixture condenses togenerate a thin film on the substrate, which is known as the ‘Ar matrix’. In this matrix, the rotation of
HF can be ignored, and vibrational interactions between Ar and HF are also negligible. As a result, themolecular vibration of an isolated HF molecule can be measured independently using infrared (IR)
spectroscopy, which quantitatively reveals molecular vibrations. Since CsI does not absorb IR
irradiation, the spectrum of HF in the Ar matrix film can be readily measured using IR spectroscopy,
and a strong IR absorption peak appears at 3953.8 cm-1 in the spectrum.
The frequency of the molecular vibration of HF, n HF , is given by:
n HF =

1
2p

k

µHF .

Here, k denotes the spring constant of the chemical bond between H and F, which increases withincreasing bond order. The reduced mass, µHF , is defined by:
µHF =

mH mF
mH + mF ,where mH and mF are the atomic masses of H and F, respectively.
1.

Calculate the wavenumber of the absorption peak of deuterium fluoride (DF) in an IR spectrum
recorded using the Ar matrix technique. Let the spring constant of D–F be the same as that of H–
F. The atomic masses of F, H, and D are 18.998, 1.0079, and 2.0141, respectively.

The quantized energy levels of molecular vibration, Ev , are defined by:1öæ
Ev = ç v + ÷ hn2øè

( v = 0, 1, 2, ! ).

Here, v is the quantum number of vibration.
The molecular vibrational energy is distributed over the quantized energy levels, and the distribution
ratio is proportional to the following factor:æ E ö
exp ç - v ÷è kBT ø .

This factor is known as the Boltzmann factor. Here, kB and T are the Boltzmann constant and absolutetemperature, respectively. Although this factor is proportional to the distribution ratio, it is not equal
to the probability for each level. Therefore, when all the Boltzmann factors are summed up, the sum
is not unity, but instead yields qvib , i.e., the ‘distribution function’. The actual distribution ratio at eachlevel can be determined using the distribution function as the normalization constant.
2.

Write the equation for the distribution function at a temperature, T, as the summation of ageometric series using the vibrational temperature, q , which is defined as q = hn kB .

3.

Calculate the vibrational temperature of HF.

When the substrate temperature is raised to 26 K, the HF molecules begin to move in the Ar matrix,
which yields a molecular associate of two HF molecules, (HF)2, via hydrogen bonding as observed via
a hydrogen-bonding-specific absorption peak in the IR spectrum. The hydrogen bond is generated fromthe overlapping wave functions of the electron of a lone pair and of a hydrogen atom. The molecule33

donating the hydrogen is called the ‘donor’, whereas the molecule receiving the hydrogen on the lonepair is called the ‘acceptor’. The hydrogen bond between two HF molecules is most stabilized when
the lone pair of the acceptor and the H–F bond of the donor are aligned linearly. In this situation, thehydrogen atom is shared by the two fluorine atoms.
4.

Based on this description, draw a chemical scheme for the molecular associate (HF)2. Use a dottedline to represent the hydrogen bond, and show apparently the lone pairs of the accepter fluorine
atom.

5.

In the IR spectrum, a new absorption peak corresponding to the hydrogen-bonded acceptor isencountered. Does this new peak appear at a lower or higher wavenumber relative to the IR peakof the isolated HF molecule? Choose the correct shift with the most appropriate reason. Consider
the fact that the H–F vibration becomes an F–H–F vibration with a different apparent mass, andthat the bond order of the H–F bond is reduced when the hydrogen atom is shared by the two
fluorine atoms.
(1) k becomes smaller and μ becomes smaller; a prediction of the peak shift is difficult.
(2) k becomes smaller and μ becomes larger, which results in a shift to a lower wavenumber.
(3) k becomes larger and μ becomes smaller, which results in a shift to a higher wavenumber.
(4) k becomes larger and μ becomes larger; a prediction of the peak shift is difficult.

Active IR absorption modes are those with a vibrational energy transition of Dv = +1; such transitions
are observed between various energy levels. At low temperature, however, only the v = 1 ¬ v = 0transition needs to be considered. Here, the leftward arrow indicates an upward energy transition. Thisapproximation is allowed because the occupation number at v = 1, g1, is much lower than that at v = 0,g 0 . As a result, higher energy transitions such as v = 2 ¬ v = 1 can be ignored.
6.

Calculate the ratios g1 g0 at temperatures of 12 K and 25 °C for an absorption peak at 3953.8 cm1. The occupation number for a given level is proportional to the distribution ratio at that level.

34

Problem 8. Quantum chemistry of aromatic molecules
The behavior of π-electrons is of paramount importance for understanding the reactivity andphotophysical properties of aromatic compounds. In the electronic theory of organic chemistry, theelectronic structure is considered to be a superposition of several ‘resonating’ electronic states.
1.

Following the example below, draw all possible resonance structures for the compounds listed
below. It should be noted that the positions of the atoms (nuclei) must remain unchanged. As only
the skeletons of the structures are shown, add bonds as appropriate according to the chemical
formulae. Do not include resonance structures with separated charges or unpaired electrons.

(a)

(b)

C10H8

C14H10(c)(d)

C14H102.

C8H6

In real benzene, all carbon–carbon bonds have the same bond length (1.399 Å), which means thatthe two cyclohexatriene resonance structures are extremes, and their superposition represents thereal molecule. With this in mind, we will consider the bond lengths R1, R2, and R3 in anthracene.
Assuming that RS is the typical length of a carbon–carbon single bond (1.53 Å) and RD that of atypical C=C double bond (1.34 Å), arrange RS, RD, R1, R2, and R3 in order of length.

R1

R2

35

R3

Following Questions are optional:
The properties of molecules are governed by their wavefunctions. Although resonance structures areuseful for discussing various physical properties, it is difficult to use them to predict reactivity. The
frontier orbital theory, proposed by Fukui and co-workers, is suitable for discussing reactivity basedon molecular orbitals. For example, the six molecular orbitals of benzene are shown in the figure below.

The π-orbitals consist of p-orbitals perpendicular to the molecular plane of each atom (left panel); thewhite and black parts of each correspond to positive and negative values of the wavefunction. The
right panel shows the same information more simply.
3.

Give the number of orbitals occupied by π-electrons in benzene, naphthalene, and pyrrole.

4.

Based on the following facts, choose the right pyrrole HOMO (highest occupied molecular orbital)from the options (a)-(c) below. The positive and negative values of wave functions are represented
with white and black in the upper panel. In the lower panel, the circles on the atoms are drawn
with a size proportional to the electron density.
- The reaction readily takes place at the position with the largest electron density in the HOMO.
- The electrophilic substitution reaction of pyrrole occurs faster at the 2 position than at the 3
position.

5.

Arrange these orbitals in order of their energy, from lowest energy to highest energy. The diagramof benzene can be used as a guide.

36

Problem 9. Protic ionic liquids
Water undergoes autoprotolysis according to:(1)
2 H2O ⇄ H3O+ + OH−
The product of the concentration of dissociated ions is constant and known as the autoprotolysis
equilibrium constant.
[H3O+][OH−] = Kw = 1.00 × 10−14 （25 °C）(2)
The negative decadic logarithm of the concentration of H3O+ is known as the pH value, which is used
as a scale of the acidity of an aqueous solution:
pH = −log10[H3O+](3)
When the concentration of the cations and the anions generated by autoprotolysis is the same, thesolution is said to be neutral. In the case of water at 25 °C, the neutral condition, i.e., [H3O+] = [OH−],
holds when pH = 7.
1.

Autoprotolysis also occurs in liquids other than water. For example, the autoprotolysis equilibrium
constant of methanol is 10−16.7 at 25 °C.

1-1. Write the chemical equation for the autoprotolysis of methanol.
1-2. Calculate the concentration of the cation of methanol under neutral conditions at 25 °C.
Substances composed of ionic bonds, such as sodium chloride, usually exist in solid form at room
temperature. In 1914, ethylammonium nitrate was reported as a unique salt that exists as a liquid evenat room temperature. Such salts are known as ionic liquids, and these have attracted considerable
attention recently.

1-Ethyl-3-methylimidazolium bis(trifluoromethanesulfonyl)imide (hereafter denoted as CM+/B−) is acommonly used ionic liquid.

1-Ethylimidazolium bis(trifluoromethanesulfonyl)imide (hereafter denoted as CH+/B−), which has asimilar structure to CM+/B− except that the methyl group on the imidazole ring is replaced by a proton,
is also an ionic liquid.

CH+/B− shows an equilibrium that is similar to autoprotolysis:(4)
CH+/B− ⇄ C + HB
Ks = [C][HB](5)+ −To determine the Ks of CH /B , electrochemical measurements have been conducted (Figure 1). Using
CH+/B− as a solvent, a 0.35 mol L−1 bis(trifluoromethanesulfonyl)imide solution (henceforth: HB
solution) and a 3.2 mol L−1 1-ethylimidazole solution (henceforth: C solution) were prepared.

37

Figure 1. Schematic illustration of the experimental setup.
7.3 mL of the HB solution was titrated with the C solution, and the electromotive force was measured
during the titration using an electrode that responds to HB. The obtained electric voltage, Ei, can be
expressed by:
Ei = E° + RT/F ln[HB](6)
The resulting electric voltages as a function of the amount of the C solution added, Vi, are shown in
Figure 2.

Figure 2. Electric voltages during the titration of the HB solution
as a function of the amount of the C solution added (Vi).
The total concentration of protons, CH, and the total concentration of imidazole, Cim, can be expressedby:CH = [HB] + [CH+](7)+
Cim = [C] + [CH ](8)
In the following problems, assume that the temperature of the solutions is 25 °C.
2.

First, let us consider the case where there is an excess of HB. At the point of neutralization, theobserved electric voltage changes drastically. This indicates that Ks is very small, and therefore
almost all C added prior to the point of neutralization is protonated ([C] << [CH+]).

38

2-1. Consider the case where the initial concentration of HB solution is [HB]0 (= 0.35 mol L−1), theconcentration of C solution is [C]0 (= 3.2 mol L−1), the concentration of CH+ originating from thesolvent is [CH+]sol, and the volume of HB solution before the titration is V0 (= 7.3 mL). Write an
expression for CH and Cim in terms of V0, Vi, [HB]0, [C]0, and [CH+]sol. Assume that [CH+]sol does
not change during the titration.
2-2. When 0.50 mL of the C solution was added, Ei was 0.796 V. Calculate the value of E°. Note thatthe concentration of HB can be approximated as [HB] = CH − Cim.
To analyze the obtained result, Figure 2 is replotted with (V0+Vi)exp(±EiF/RT) as the vertical axis
(Figure 3).

Figure 3. Replotted graphs of Figure 2.
(Left) Before the point of neutralization. (Right) After the point of neutralization.
3.

When the C solution is added beyond the point of neutralization, there is an excess of C. In thiscase, almost all the protons bind to C, and there are few protons that exist in the form of HB.

3-1. Write an expression for [HB] beyond the point of neutralization in terms of CH, Cim, and Ks.
Assume that CH = [HB] + [CH+] ≃ [CH+], which holds when there is an excess of C.
3-2. The slope of (V0+Vi)exp(−EiF/RT) as a function of Vi is 0.0271 beyond the point of neutralization.
Calculate Ks.
Let us consider how to control the reaction rate of the following chemical reaction.

(S1)

An amide bond is produced by the reaction between the amine and the active ester. In the following
problems, assume that all of the produced N-hydroxysuccinimide (pKa = 6.0 in water) exists in the39

deprotonated form (NHS−). Ignore the degradation reaction of the active ester. The temperature of thesolutions is 25 °C.
First, let us consider reaction S1 in a buffered aqueous solution. In an aqueous solution, the acid–baseequilibrium of R1–NH2 is expressed by:(9)
R1–NH3+ + H2O ⇄ R1–NH2 + H3O+
Ka = [R1–NH2] [H3O+] / [R1–NH3+] = 10−9.27(10)
The reaction rate of S1 is expressed as a second-order reaction:d[–NH–CO–]/dt = k [R1–NH2][R2–CO–NHS](11)
The variation of [–NH–CO–] over time is obtained by integrating eq. 11, which gives:[–NH–CO–] = ([R1–NH2]0 + [R1–NH3+]0)(1 − exp(−k[R2–CO–NHS] fa t) )(12)
1
[R –NH2]0 and [R1–NH3+]0 represent the initial concentrations of R1–NH2 and R1–NH3+, respectively,
while fa represents the ratio of the deprotonated form of the amine, which is defined by:
fa = [R1–NH2] / ([R1–NH2] + [R1–NH3+])(13)
When the reaction occurs in a buffered aqueous solution, fa remains constant during the reaction
because the pH of the solution is constant.
4.

The reaction progress in a mixture of 20 μmol of R1–NH2 in 10 mL of the buffer solution and 200
μmol of R2–CO–NHS in 10 mL of the buffer solution is monitored via the absorption of NHS− inthe ultraviolet region. In this case, the concentration of R2–CO–NHS is regarded constant during
the reaction, because it is significantly higher than that of R1–NH2 (pseudo-first-order reaction).
Reaction S1 proceeds only for the deprotonated amine, R1–NH2. The protonated amine,
R1–NH3+, does not react with the active ester.

4-1. Calculate the time at which half of the added R1–NH2 will have reacted when k′ = k[R2–CO–
NHS] = 5.0 × 10−2 s−1 and pH = 9.27.
4-2. In a differently buffered solution, the time at which one-third of the added R1–NH2 has reacted is
500 s when k′ = 5.0 × 10−2 s−1. Calculate the pH of the buffer solution in this case.
Then, let us consider reaction S1 in the ionic liquid CM+/B−. In CM+/B−, the acid–base equilibrium of
R1–NH2 is expressed as follows:(14)
R1–NH3+ + B− ⇄ R1–NH2 + HB
KaA = [R1–NH2] [HB] / [R1–NH3+] = 10−16.40(15)
The reaction rate of S1 is expressed as a second-order reaction:d[–NH–CO–]/dt = kI [R1–NH2][R2–CO–NHS](16)
5.

As shown in Q4, to control the reaction rate of S1, the ratio of [R1–NH2] should be set to theappropriate value by adjusting the proton concentration in the solution. However, the ionic liquid
CM+/B− does not contain a proton. To solve this problem, let us use the ionic liquid CH+/B−, which
was considered in Q2 and Q3, as a proton source for CM+/B−. The acid–base equilibrium of CH+
in CM+/B− can be expressed by:(17)
CH+ + B− ⇄ C + HB
C+
−14.90Ka = [C] [HB] / [CH ] = 10(18)

5-1. Calculate the [HB] of a 6.00 × 10-3 mol L−1 CH+/B− solution dissolved in the ionic liquid CM+/B−.
5-2. A R1–NH2 solution is prepared using the solution from Q5-1 as the solvent. The [HB] of thesolution of 100 μmol R1–NH2 in 10 mL of the solvent is 5.39 × 10-17 mol L−1. Calculate theconcentrations of CH+, C, R1–NH3+, and R1–NH2.
40

As shown in Q5-2, most of the protons added as CH+ are transferred to R1–NH2 and exist in the formof R1–NH3+. Similarly, the protons produced in reaction S1 also bind to R1–NH2. As the resulting
chemical species, R1–NH3+, does not react with the active ester, the reaction rate decreases as thereaction proceeds.
5-3. Reaction S1 is started once 100 μmol of R2–CO–NHS have been dissolved in the solution fromQ5-2. Calculate the concentrations of HB, CH+, C, R1–NH3+, and R1–NH2 when the concentrationof R2–CO–NHS become 8.00 × 10−3 mol L−1.
5-4. To avoid a decrease in the reaction rate, the concentration of HB should remain constant. When
CH+/B− and C are dissolved together, the concentration of HB is determined by the ratio of theconcentrations of C and CH+. Express [HB] in terms of [C], [CH+], and KaC.
A buffer solution is prepared by adding the appropriate amount of C to 0.10 mol L−1 of CH+/B− solution
using CM+/B− as the solvent. The reaction progress of a mixture of 20 μmol R1–NH2 in 10 mL of thebuffer solution and 200 μmol of R2–CO–NHS in 10 mL of the buffer solution is monitored. The
variation in [–NH–CO–] over time can be expressed by:[–NH–CO–] = ([R1–NH2]0 + [R1–NH3+]0)(1 − exp(−kI’ fa t) )(19)
5-5. The time at which one third of the added R1–NH2 will have reacted is 2000 s when kI′ = kI[R2–
CO–NHS] = 8.0 × 10−4 s−1. Calculate the concentration of C in the buffer solution.
[Note] The reaction control technique explained above has already been applied to the condensationof the two polymers shown in Figure 4. As a result, high-strength gels with homogeneous network
structures were prepared from both aqueous solutions and ionic liquids (for an example, see: T. Sakai
et al., Macromolecules 2008, 41, 5379).

Figure 4. Conceptual illustration of the controlled synthesis of high-strength gels.

41

Problem 10. The Yamada universal indicator
In this problem, assume that the optical path length of the cell is 1.0 cm. Ignore the absorption of thecell and the solvent.
1.

The acid dissociation constant of bromothymol blue (BTB) is 10–7.10. The absorbance values of
200 μmol L–1 BTB aqueous solutions with different pH values (2.00, 12.00) were measured, andthe results are shown in Table 1.

Table 1. Absorbance values of 200 μmol L–1 BTB solutions
Wavelength
420 nm
501 nmpH = 2.00
1.74
0.70
pH = 12.00
0.60
0.70

520 nm0.36
1.06

1-1. Calculate the molar absorption coefficients of the protonated/deprotonated forms of BTB at 420
nm, 501 nm, and 520 nm.
1-2. At a certain pH value, the absorbance values of an aqueous solution containing a small amount of
BTB at 420 nm and 520 nm were 1.13 and 0.52, respectively. Calculate the pH of this aqueous
solution.
1-3. The aqueous solution used in Q1-2 was diluted 5 times using hydrochloric acid with pH = 1.
Calculate the absorbance value of the diluted solution at 501 nm.
1-4. The aqueous solution used in Q1-2 was diluted 5 times using aqueous sodium hydroxide with pH= 12. Calculate the absorbance value of the diluted solution at 501 nm.
To estimate the pH of a solution using indicators, appropriate indicators that show a clear color change
in the target pH region should be used. Universal indicators have been developed as easy-to-use
indicators. Universal indicators can be prepared by mixing several kinds of indicators so that theresulting mixture shows color changes over a wide pH region. As a representative universal indicator,
let us consider the Yamada universal indicator, which was developed by a Japanese researcher,
Shinobu Yamada, in 1933.
The components of the Yamada universal indicator, its color under various pH conditions, the molar
absorption coefficients of each indicator, and the acid dissociation constants of each indicator are
summarized in Tables 2, 3, 4, and 5, respectively.
Table 2. Components of the Yamada universal indicator dissolved in 100 mL ethanol
Indicators
Molecular weight
Weight [mg]Thymol blue (TB)
466.6
5.00
Methyl red (MR)
269.3
12.5
Bromothymol blue (BTB)
624.4
50.0
Phenolphthalein (PP)
318.3
100.0
Table 3. Color of the Yamada universal indicator under various pH conditions
pH
–3
3–6
6–8
8 – 11
11 –
Color
Red
Yellow
GreenBlue
Purple42

Table 4. Molar absorption coefficients of each indicator (unit: mol–1 L cm–1)
Wavelength [nm] 400
450
500
550
600
TB, H2A
5500
2500
6000
18000
1500
TB, HA–
6000
7000
3000
500
0
TB, A2–
3500
1500
3500
10000
17500
–
MR, HA
1000
10000
35000
35000
1500
MR, A2–
15500
16000
4000
500
0
–
BTB, HA
7500
9000
3500
500
0
BTB, A2–
5000
1000
3500
10000
20000
–
PP, HA00
0
0
0
PP, A2–
1000
1000
6000
21000
1000
Table 5. Acid dissociation constants of each indicator(In this problem, ignore all acid dissociation constants except for these.)
Indicator
TB
MR
BTB
pKa
pKa1 = 1.7, pKa2 = 8.9 5.1
7.1
2.

650
0
0
2500
0
0
0
10000
0
0

PP
9.7

The Yamada universal indicator (1.0 mL) was added to aqueous solutions (29.0 mL) with various
pH values, and the absorbance of these solutions was measured. The measured absorbance values
and the colors of the solutions are shown in Table 6. Calculate the pH of solutions (1), (2), and(3) after the addition of the Yamada universal indicator up to one decimal place.
Note that under each pH condition, only one of the four indicators exhibits a drastic change in theratio of its protonated/deprotonated forms by the slight change of its pH value.

Table 6. Measured absorbance values and colors of the solutions
Wavelength [nm]
Color
400
450
500
550
pH = 1.0
Red
0.236
0.407
0.655
0.609
(1)
Yellow 0.323
0.455
0.460
0.352
(2)
Green
0.435
0.440
0.176
0.121
(3)
Blue
0.466
0.360
0.645
1.979
pH = 12.0 Purple 0.490
0.384
0.793
2.498

43

600
0.028
0.018
0.179
0.673
0.701

650
0.000
0.002
0.089
0.275
0.276

Problem 11. Silver electroplating
Silver electroplating, in which silver is deposited on a metal surface, has been used to decorate objects
and to increase their electrical conductivity. In silver electroplating, usually an aqueous alkaline
solution containing Ag 3 and CN . is used. The metal to be plated is used as the cathode, and silver ionsare deposited on the metal surface by reductive deposition. Pure silver is used as the anode, from whichsilver ions are supplied to the solution by oxidative dissolution. The half reaction and standard
electrode potential 𝐸° of the reductive deposition/oxidative dissolution of Ag 3 are as follows:
Ag 3 + e. ⇌ Ag(s)𝐸° = 0.799 V

In the solution, Ag 3 forms several kinds of complexes with CN . and OH . . The complex formation
reactions and formation constants are as follows:
Ag 3 + CN . + OH . ⇌ Ag(OH)(CN).
Ag 3 + 2CN . ⇌ Ag(CN).
/Ag 3 + 3CN . ⇌ Ag(CN)/.
K
Ag 3 + 4CN . ⇌ Ag(CN)K.
8

𝐾I9@J = 104K./(
𝛽/ = 10/(.8(𝛽K = 10/4.8(
𝛽8 = 10/(.L(M<N(@J)$ O

where the overall formation constants 𝛽1 are defined as 𝛽1 = P[<N%][@J#$]# S.

Answer the following questions. Unless otherwise noted, [CN . ] = [OH . ] = 1.00 M, and the total
concentration of Ag-related dissolved species is 1.00 mM. Use the absolute temperature 𝑇 = 298.2 K.
1.

Write the five Ag-related dissolved species in this solution in order of concentration. Calculate
the concentration of the species with the highest concentration.
>

>

>

>

2.

Write the half reaction in which the Ag-related dissolved species with the highest concentrationin Q1 is reduced to Ag(s). Calculate the 𝐸° of the half reaction.

3.

If [CN . ] varies from 1.00 M, the relative order of the species in Q1 will also vary. Calculate theconcentration range of [CN . ] within which the relative order in Q1 holds.

4.

Calculate the equilibrium potential of the cathode, assuming that the cathode surface is covered
with Ag(s).

5.

The reductive deposition reaction of silver competes with the hydrogen evolution reaction (HER).
The potentials at which the HER occurs depend on the pH of the solution; the higher the pH, themore negative the potentials for the HER. By comparing the equilibrium potential calculated in
Q4 with that for the HER, calculate the lower pH limit above which the reductive deposition
reaction of silver is more thermodynamically favorable than the HER.

44

6.

Because HCN(g) is highly toxic, the evolution of this gas should be minimized, even when using
a fume hood. Calculate the pH requirements for the solution to keep the equilibrium partialpressure of HCN(g) below 10 ppm (= 10−5.0 bar). Use the following equilibrium equations for the
calculation.
𝐾9 = 104.(
𝐾& = 10.T./

HCN(g) ⇌ HCN(aq)
HCN(aq) ⇌ H 3 + CN .

45

Problem 12. How does CO2 in the atmosphere affect the pH value of seawater?
Let us consider the chemical equilibrium of carbonic acid in ocean acidification, which is caused byan increase in the concentration of carbon dioxide in the atmosphere. Use the following equilibrium
equations and constants:
CO/ (g)⇌ CO/
CO/ + H/ O ⇌ H/ COK
H/ COK⇌ H 3 + HCOK .
HCOK .
⇌ H 3 + COK /.
H/ O⇌ H 3 + OH .

𝐾NUV = 10.4.8W
𝐾XY- = 10./.Z(
𝐾U4 = 10.K.WA
𝐾U/ = 10.4(.KK
𝐾[ = 10.48.((

where (g) denotes a species in the gaseous state, and all species other than CO/ (g) and H/ O are
dissolved in an aqueous solution. Let the current value of the molar fraction of carbon dioxide in theatmosphere, 𝑥@I& , be 410 ppm (= 410 × 10−6 bar).
1.

When pure water is left exposed to the atmosphere, it absorbs carbon dioxide from the atmosphere,
which changes its pH. Calculate the pH of water in equilibrium with the atmosphere using thefollowing charge balance equation:[H 3 ] = [OH . ] + [HCOK . ] + 2ŸCOK /.

2.

To the water in equilibrium with the atmosphere in Q1, a strong base or strong acid was added
such that the pH changed by ±1 (ΔpH = ±1). The solution was then left exposed to the atmosphereagain. What is the change in the amount of absorbed carbon dioxide relative to that before theaddition? Choose the pH change for each scenario from the following Options. Assume that there
is no change in the liquid volume due to the acid/base addition.
(a) Strong base (ΔpH = +1)(b) Strong acid (ΔpH = −1)
Options(1) The amount of absorbed carbon dioxide increases to more than double the initial value.
(2) The amount of absorbed carbon dioxide decreases to less than half the initial value.
(3) The magnitude of the increase or decrease is less than described in Options (1) and (2).

3.

Currently, 𝑥@I& continues to rise. A report has predicted that 𝑥@I& will exceed 800 ppm by 2070
(cf. IPCC RCP 8.5). Calculate the pH of water in equilibrium with the atmosphere if 𝑥@I& weretwice its current value (𝑥@I& = 2 × 410 = 820 ppm) and evaluate the pH shift relative to the pH
found in Q1 where the current 𝑥@I& value was used.

As 𝑥@I& rises, the oceans, like pure water, become more acidic. The current pH of the oceans is 8.05
on average, although it varies spatially and temporally. Compared to the pre-industrialization value inthe 18th century, the current pH of the oceans has decreased by 0.1. If the pH decreases a further 0.2
to 7.85 in the future, the decrease in ŸCOK /. will result in CaCOK (aragonite), which is a componentof the shells of corals and phytoplankton, being dissolved at low temperatures, causing severe damage
to the marine biota. Let us estimate how much a rise in 𝑥@I& would reduce pH.
46

4.

First, let us estimate the ocean acidification due to the rise in 𝑥@I& , ignoring chemical equilibria
other than those of carbonic acid. A variety of ions are dissolved in the ocean. The difference inthe concentration of the cations and anions originating from strong electrolytes, 𝐶@.< , appears as
a constant in the charge balance equation.
[H 3 ] + 𝐶@.< = [OH . ] + [HCOK . ] + 2ŸCOK /.
Calculate the pH shift when 𝑥@I& increases from 410 to 820 ppm, with 𝐶@.< = 0.72 mM.

5.

In real seawater, chemical equilibria other than those of carbonic acid affect this pH shift. Let us
calculate the pH shift taking into account the following ion-pair formation constant for MgCOK(dissolved species):
Mg /3 + COK /. ⇌ MgCOK

𝐾\ = 10/.LL

The charge balance equation can be written as:
2[Mg /3 ] + [H 3 ] + 𝐶@.< = [OH . ] + [HCOK . ] + 2ŸCOK /.
where 𝐶@.< = −105.0 mM. Use 𝐶]N = 53.0 mM as the total concentration of Mg-related species.
Calculate the pH when 𝑥@I& changes from 410 to 820 ppm. It may be helpful to use anapproximation, (1 − 𝑥).4 ~ 1 + 𝑥, for x ≪ 1. Can corals survive?

Let us consider the meaning of the pH shift evaluated in Q3 and Q4 using a virtual weak acid HA
which dissociates as follows:
HA ⇌ H 3 + A.
with the acid dissociation constant 𝐾U = 10.K.(( (note that this constant is different from that ofcarbonic acid). A solution is prepared using a given concentration of HA, 𝐶9< , and a given
concentration of its conjugate base salt NaA, 𝐶JU< . The pH of the solution depends on theseconcentrations. The dependence of the pH on p𝐶9< (= −log4( 𝐶9< ) and p𝐶JU< (= −log4( 𝐶JU< ) isshown as a contour map below. A doubling in 𝐶9< corresponds to a –0.30 shift of p𝐶9< to the left inthis contour map.

47

Fig. 1. Contour map of pH for an HA/NaA solution.
6.

In the situation described in Q3, i.e., pure water in equilibrium with the atmosphere, 𝐶JU< isideally zero and can be neglected. In the above contour map, this situation corresponds to either
Region A, B, or C, where p𝐶JU< is large and the following inequalities for p𝐶9< hold:(Region A)(Region B)(Region C)

p𝐶9< < p𝐾U^_
p𝐾U < p𝐶9< < /'^_'/< p𝐶9<

Derive analytical solutions for the pH as a function of p𝐶9< using the charge balance equation:[H 3 ] = [OH . ] + [A. ]combined with the above inequalities for each Region. Because doubling 𝐶9< corresponds to a–0.30 shift in p𝐶9< as described above, the pH shift can be determined from the derived analyticalsolutions. Calculate the pH shift when 𝐶9< is doubled for Regions A, B, and C, respectively. The
situation in Q3 corresponds to Region. Fill in this blank with either A, B, or C.

48

7.

In the solution with pH = 8.05 in Q4, in which we ignored any chemical equilibria other than thatof carbonic acid, HCOK . is the main component among the carbonic-acid-related dissolved
species. This solution can be regarded as an aqueous solution of NaHCOK , which corresponds toRegion D with a large 𝐶JU< , i.e., a small p𝐶JU< , in the contour map. In Region D, the Henderson–
Hasselbalch equation holds, in which the equilibrium equation 𝐾U =.][9% ][<$ ][9<]is approximatedusing [HA] = 𝐶9< and [A = 𝐶JU< . Calculate the pH shift using the Henderson–Hasselbalch
equation when 𝐶9< doubles in Region D.

49

Problem 13. How to produce sulfuric acid and dilute it without explosion
Japan is a country rich in sulfur resources. Sulfuric acid can be synthesized from sulfide-based mineral
resources such as copper sulfide, iron sulfide, and lead sulfide.
1.

When a mixture of chalcopyrite (CuFeS2) and silica sand (SiO2) is melted at high temperature in
the presence of oxygen in a blast furnace, sulfur dioxide is generated along with Cu2S and FeSiO3(Reaction 1). The Cu2S obtained in Reaction 1 reacts with oxygen at high temperature to form
copper and sulfur dioxide (Reaction 2). Write the reaction formulae for Reaction 1 and Reaction
2.

2.

Calculate how many kg of copper and how many liters of sulfur dioxide (at 25 °C, 1.00 bar) will
be obtained from 100 kg of pure chalcopyrite by the process described in Q1. Assume that all thereactions proceed completely and that all gases are ideal gases.

3.

Sulfur dioxide is oxidized to sulfur trioxide by air using a catalyst such as vanadium(V) oxide.
This oxidation proceeds exothermically, and the reaction reaches an equilibrium state as it is areversible reaction. Which of the following conditions is most appropriate to increase theproduction ratio of sulfur trioxide? Choose one option from (1) to (4).
(1) Low temperature / Low pressure(2) Low temperature / High pressure(3) High temperature / Low pressure(4) High temperature / High pressure

4.

A mixture of sulfur dioxide (7.00 mol) and oxygen (3.00 mol) is placed in a 50.0 L heat- andpressure-resistant container together with a catalyst. When the container is heated to 350 °C, thepressure increases to 8.60 bar. Calculate how many moles of sulfur trioxide are generated in thisreaction.

5.

The absorption of sulfur trioxide by concentrated sulfuric acid produces fuming sulfuric acid.
Subsequent dilution of the fuming sulfuric acid with dilute sulfuric acid can afford the desiredconcentration of sulfuric acid. Calculate how many kg of 60% sulfuric acid are needed to dilute20.0 kg of 32% fuming sulfuric acid (32% by weight SO3, 68% H2SO4) to obtain 98% sulfuric
acid. Assume that the volatilization of the components during the dilution is negligible.

6.

8.0 kg of 98% sulfuric acid is diluted with water to give 60% sulfuric acid. Calculate the heat ofdilution Q [J] released by this dilution process. Note that when 1 mol of H2SO4 is diluted with n
mol of H2O, the heat of dilution Q0 [J] is expressed by the following formula:7.47 × 108 × 𝑛
𝑄( [𝐽] =𝑛 + 1.80
The amount of heat required to raise the temperature of 1 g of water by 1 °C is 4.18 J (1 cal).
Calculate the increase in temperature that the released heat of dilution Q [J] would cause in 10 kgof water (0 °C).

50

Problem 14. Hydrolysis of C vs Si and the electronegativity of N vs Cl
1.

Answer the following questions based on the following values of bond-dissociation energies [kJmol–1]. Assume that the reaction-enthalpy change can be estimated by considering only the bonddissociation energies of the starting materials and the reaction products.
C–Cl: 335, C–O: 392, O–H: 467, H–Cl: 426, Si–Cl: 451, Si–O: 500, C–H: 413, Si–H: 318.

1-1. Carbon tetrachloride is a stable compound in water, while silicon tetrachloride undergoes facilehydrolysis in water. For reactions (1) and (2) below, which are the initial steps of hydrolysis,
calculate ΔH [kJ mol–1], which can be estimated based on the bond dissociation energies.
(1) CCl4 + H2O → CCl3OH + HCl(2) SiCl4 + H2O → SiCl3OH + HCl
In Q1-1, it cannot be anticipated whether the reaction would proceed smoothly based on only theenthalpy change of the bond-dissociation energies between the reaction substrates and the reaction
products. The difference in reactivity between carbon tetrachloride and silicon tetrachloride describedin Q1-1 can be explained by the one-step reaction in the carbon case and the multi-step reaction in thesilicon case.
‡
Cl
(1)

H 2O

C

+
Cl

Cl

O

Cl

Cl

Cl

H

C

H Cl

HO

Cl

+ HCl

C
Cl

Cl

Cl

transition state

Cl
(2)

H 2O

(2-1)
Si

+
Cl

Cl

Cl
ΔH2-1 = 14.0 kJ mol–1

Cle

H
O

Cla

Si

H Cle

Cl

(2-2)

Cle

HO

+ HCl

Si
Cl

Cl

intermediate A

Cle

Cl
(3)

HCl

Si

+
Cl

Cl

Cl

Cla SiH Cle

Cla

ΔH3 = 50.0 kJ mol–1

Cle

Equation (1) is a one-step reaction that requires a very large activation energy because its transition
state involves a penta-coordinated carbon species. Equation (2) is a two-step reaction through anintermediate, which is a penta-coordinated silicon species. For example, as shown in equation (3), apenta-coordinated silicon compound known as a silicate should be generated by the reaction of silicontetrachloride with hydrogen chloride. The molecular structure of the penta-coordinated siliconcompound exhibits a trigonal bipyramidal geometry with a silicon atom at the center and a bonding
atom at each vertex of the trigonal bipyramidal skeleton.
51apical Cla
equatorial Cle
apicalSiCla

Cle equatorial
Cle equatorial

geometry of pentachlorosilicate SiCl5–,
penta-coordinated silicon species

In this intermediate, the bond dissociation energies of the atoms located at the two vertices above andbelow silicon (apical positions) differ from those of the atoms located at the three vertices that form
the triangle centered on silicon (equatorial positions). Assume that the bond-dissociation energies ofthe equatorial Si–Cl bonds (BE(Si–Cle)) are all equal, and that the bond-dissociation energies of theapical Si–Cl bonds (BE(Si–Cla)) are also equal. The bond-dissociation energy between the pentacoordinated silicon and the oxygen atom of the apical H2O molecule is 400 kJ mol–1, and the changes
in the bond-dissociation energies of the O–H and Cl–H bonds upon coordination of the H2O and HClmolecules to the silicon atom to form the corresponding silicates are negligible.
1-2. Provided that ΔH2-1 in Eq.(2-1) is 14.0 kJ mol–1 and ΔH3 in Eq.(3) is 50.0 kJ mol–1, calculate thebond-dissociation energies BE(Si–Cla) and BE(Si–Cle) between the silicon atom and chlorine
atom at the apical and equatorial positions, respectively.

52

1-3. As described in Q1-2, it is important that silicon can adopt, in contrast to carbon, a pentacoordinate geometry in a meta-stable intermediate. This is due to the longer bond lengths of siliconrelative to those of carbon; accordingly, there is a spatial margin around the central silicon atom,
resulting in low repulsion between the five coordinated ligands. As can be found in Q1-2, although
the bond-dissociation energy of each bond in the penta-coordinated silicon compound is smaller
than that in the tetracoordinated silicon compound, the five weak bonds in the penta-coordinatedsilicon compounds should collectively yield sufficient stabilizing energy. Thus, the pentacoordinated silicon compound exhibits stability that is comparable to that of the corresponding
tetracoordinated silicon compound. Based on the aforementioned considerations, choose thegraphs that most appropriately represent the energy changes during the progress of reactions of(1) and (2), respectively.
(i)energy(ii)

energyCl
OH + HCl

Cl

Cl

Cl
Cl

Cl

E

Cl

E

Cl

Cl + H2O

Cl

Cl

E

Cl + H2O

reaction coordinate

(iii)

(iv)

energy

Cl

Cl
E

Cl

Cl

OH + HCl

Cl

Cl
Cl

E

Cl

E

Cl + H2O

Cl

reaction coordinate

reaction coordinate

(v)

Cl
Cl

Cl

E

E

OH + HClCl

Cl
Cl + H2O

Cl

reaction coordinate

Note that the vertical axes in figures (i)–(v) all have the same scale.

53

E
Cl

Cl

Cl + H2O

Cl

energyOH + HClCl

reaction coordinate

energyE
Cl

OH + HCl

For certain elements A and B, the Pauling electronegativities χA and χB are defined according to thefollowing equation:(χA – χB)2 = DAB – (DAA + DBB) / 2
DAB: Bond dissociation energy of an A–B bond [eV]
DAA: Bond dissociation energy of an A–A bond [eV]
DBB: Bond dissociation energy of a B–B bond [eV]The Pauling electronegativities of hydrogen, nitrogen, and chlorine are as follows: χH = 2.10, χN = 3.04,
χCl = 3.16. The bond-dissociation energies [kJ mol–1] are as follows: N≡N: 945, O=O: 498, N–N: 158,
H–H: 436, H–Cl: 426, Cl–O: 206, O–H: 467.
2-1. Based on these values, calculate the bond-dissociation energies DNH, DNCl, and DClCl for the N–
H, N–Cl, and Cl–Cl bonds, respectively.
2-2. Write the molecular formulae that can apply to blanks A to D in the following sentence, andcalculate the numerical value that applies to blank Q.
The Pauling electronegativities of nitrogen and chlorine are χN = 3.04 and χCl = 3.16, respectively, andthus, the polarity of the N–Cl bond can be considered to be N+–Cl–. For example, in the hydrolysis
reaction of monochloroamine (H2NCl), it would be expected that nitrogen-containing product A andchlorine-containing product B will be produced based on the N+–Cl– polarization. In the actual reaction,
however, nitrogen-containing product C and chlorine-containing product D are produced, as shown in
the equation below. This result cannot be interpreted in terms of the Pauling electronegativities. The
reaction between monochloroamine and water can be written as follows, and the enthalpy change inthis reaction can be calculated as ΔH = Q [kJ] based on the values of the bond-dissociation energies,
suggesting that this reaction should proceed endothermically.
(eq.) H2NCl + H2O → C + D

ΔH = Q [kJ]

Allred and Rochow defined electronegativity using the equation below. The Allred–Rochow
electronegativity (χAR) depends on the values of the effective nuclear charge for the valence electrons(ZVEeff) and the covalent radius of the atom (r [pm]), based on the idea that electronegativity should bedetermined by the strength of the electric field on the atomic surface.
𝜒`E = 3590 ^𝑍 ab "cc – 0.35
_ + 0.744
𝑟/

The effective nuclear charge (Zeff) can be estimated according to Slater’s rules as follows:(i) Write the electron configuration for the atom using the following pattern, in which electrons inthe s and p orbitals should be combined into the same group, while those in the d orbitals should
be independent from those in the s and p orbitals. For example, [1s], [2s, 2p], [3s, 3p], [3d].
(ii) Identify the electron of interest and ignore all electrons in higher groups (larger “n” in [ns,np]
group.). These do not shield electrons in lower groups.
(iii) Regarding the shielding experienced by an s- or p- electron, electrons within the same groupcontribute 0.35 to the shielding, except for 1s electrons, which contribute 0.30 to the shielding.
(iv) For the electrons in an [ns, np] group, the electrons in the n-1 group contribute 0.85 to theshielding.
(v) For the electrons in an [ns, np] group, electrons in the n-2 or lower groups contribute 1.00 to the54

shielding.
(vi) For the electrons in an [nd] or [nf] group, electrons within same group contribute 0.35 to theshielding. All electrons in higher groups of [nd] and [nf] orbitals are ignored.
(vii) For the electrons in an [nd] or [nf] group, the electrons in the lower groups contribute 1.00 to theshielding.
For example, according to Slater’s rule, in the case of silicon (14Si), which exhibits an electron
configuration of (1s)2(2s)2(2p)6(3s)2(3p)2, the effective nuclear charge for the valence electrons ZVEeff(Si) can be estimated as 4.15 [= 14–(0.35×3)–(0.85×8)–(1.00×2)].
2-3. The covalent radii of nitrogen and chlorine are r(N) = 70.0 [pm] and r(Cl) = 99.0 [pm],respectively. Using the effective nuclear charge ZVEeff estimated with Slater’s rules, calculate theAllred–Rochow electronegativities χAR(N) and χAR(Cl) of nitrogen and chlorine.
The electronegativity values based on the Allred–Rochow definition can explain the products in thehydrolysis reaction of monochloroamine. Thus, ‘electronegativity’ is a concept that was qualitative innature and whose numerical values vary depending on the definition.

55

Problem 15. Sulfur in hot springs and volcanoes
Japan is known as a volcanic country, and the hot springs that are widely distributed throughout thecountry are deeply beloved. Hot springs exhibit different characteristics regarding e.g. temperature,acidity (pH), color, and smell, and are classified according to the chemical components they contain.
Interestingly, many hot springs contain sulfur components.
1.

Water from a certain hot spring (density: 1.00 g cm–3) contains both thiosulfate ions and hydrogensulfide. An excess of cadmium acetate is added to 100 mL of this hot spring water, and theresulting precipitate (9.50 mg) is separated by filtration. A small amount of starch is added as anindicator to the filtrate, and an aqueous I2/KI solution (effective I2 concentration: 2.50 ×10–2 molL–1) is slowly added dropwise. After the addition of 7.30 mL of the aqueous I2/KI solution to theanalyte, the coloration due to iodine came to appear. Calculate the total sulfur (S) content [mgkg–1] in the initial hot spring water. Assume that sulfur is only derived from the thiosulfate ionsand hydrogen sulfide and that no other species are involved in these reactions..

2.

Hot springs that contain more than 2 mg of sulfur (S) derived from H2S, HS–, S2–, and S2O32– in
1.00 kg of hot spring water are known as ‘sulfur springs’. In a water sample from a hot spring, thedensity is 1.00 g cm–3, the pH is 8.000, [S2–] is 1.00 × 10–8 mol L–1, and 5.26 × 10–5 mol ofthiosulfate ion is present in 1.00 kg of the water. Calculate the total sulfur content [mg kg–1] of
this hot spring.
The first and second acid dissociation constants of hydrogen sulfide are K1 = 9.50 × 10–8 and K2= 1.30 × 10–14, respectively.

3.

To investigate the content of hydrogen sulfide and sulfur dioxide contained in volcanic gas, 10.0
L of volcanic gas are sampled. The gas is absorbed into 500 mL of an aqueous I2/KI basic solution(effective I2 concentration: 2.00×10–2 mol L–1) under basic conditions. The generated elemental
sulfur is removed by filtration, and the volume of the filtrate is adjusted with water to give a total
volume of 1.00 L. Two 100 mL aliquots of this solution are placed in two beakers, which arereferred to as beakers A and B. Barium chloride is added to the solution in beaker A until nofurther precipitate appears, and 41.5 mg of white precipitate is isolated by filtration. The solution
in beaker B is titrated with 2.00 × 10–2 mol L–1 sodium thiosulfate solution using starch as anindicator. After dropping 73.0 mL of the titrant into the solution, the coloration due to iodine
disappears. Calculate the concentrations of hydrogen sulfide and sulfur dioxide in the volcanicgas [mol L–1]. Assume that the only sulfur-containing components in this volcanic gas arehydrogen sulfide and sulfur dioxide, and that during the series of experiments described above,
the hydrogen sulfide does not react with the sulfur dioxide directly, since they are present under
dilute conditions. Additionally, assume that the only substances that react during this series ofexperiments are hydrogen sulfide, sulfur dioxide, and the related substances generated from them.

56Problem 16. Identification of unknown compounds and allotropes
Part 1. Identification of inorganic compounds
A is a crystalline solid of a binary compound. It contains a metal ion Xn+ and is grayish-white incolor. A slowly reacts with water at room temperature to form B , which is scarcely soluble inwater; the generation of a gas is observed during this reaction. Dehydration reaction occurs uponheating B at approximately 350 °C to form the white compound C , which contains 39.7 weight%of oxygen atoms. Heating A generates gas of an amount of 7.6 weight% based on A , and theelemental metal X is formed. (i) Heating a mass M [g] of X under an N2 atmosphere provides a yellowgreen metal nitride. (ii) This metal nitride reacts with a large amount of water to generate D . The
amount of D corresponds to the gas of volume 120 mL at 1 bar and 25 °C.
1-1. Write the chemical formulae of compounds A , B , C , and D .
1-2. Write the chemical reactions corresponding to the underlined parts (i) and (ii).
1-3. Calculate the weight M [g].
Part 2. Allotropes
The term “allotrope” refers to one or more physical forms of a chemical element that occur in the same
physical state. Allotropes generally exhibit different physical properties and may also differ in theirchemical reactivity. For example, graphite and diamond have long been known as allotropes of carbon,
and more recently, fullerenes and carbon nanotubes have been discovered as further carbon allotropes.
One of the allotropes of phosphorus is a solid composed of P4 molecules, which is called white
phosphorus. It has a low melting point of 44 °C and is highly toxic, so care must be taken in its storage.
2-1. The solution 13C NMR spectrum of fullerene C60 (Figure 1) exhibits E peak(s). X-ray diffraction
measurements have revealed F different covalent carbon–carbon bond length(s) in the crystal
structure of fullerene C60. Write the appropriate integer values for E and F .

Figure 1. Structure of fullerene C60.
2-2. Fullerene C60 forms a crystal via van der Waals forces. The crystal structure of fullerene C60 atroom temperature is composed of a cubic unit cell with dimensions of 1.41 nm on each side. The
molecules are located at all vertices and at the centers of all faces of the cubic unit cell. Calculate
the density [g cm–3] of the crystal of fullerene C60.
2-3. The P4 molecule has high structural symmetry, and all the P–P–P bond angles are 60°. Draw thesteric structure of P4.

57

Dioxygen and ozone are the allotropes of oxygen. Ozone acts as a strong oxidizing agent. For example,(iii) the oxidation reaction of potassium iodide by ozone in aqueous solution is accompanied with thegeneration of dioxygen, and the reaction is used for the quantitative determination of ozone. Ozone
also exists in the stratosphere and acts as a filter that absorbs harmful ultraviolet light from sunlight.
Dioxygen in the stratosphere is irradiated with ultraviolet light of wavelengths < 240 nm, where adioxygen molecule is split into two oxygen atoms that react with another dioxygen molecule to produce
ozone. On the other hand, ozone decomposes into dioxygen molecule and an oxygen atom when
exposed to ultraviolet light with a wavelength of 240–315 nm. When these formation anddecomposition reactions of ozone are in equilibrium, the concentration of ozone in the stratosphere ismaintained at a constant level, blocking short-wave ultraviolet rays that are harmful to humans on the
ground. In contrast, (iv) atmospheric nitric oxide reacts with ozone, and (v) nitrogen dioxide reacts
with oxygen atoms, resulting in (vi) a reaction that converts ozone into dioxygen. This creates an ozone
hole that can allow harmful ultraviolet radiation from the sun to reach the surface on the earth.
2-4. Write the reaction formula for the underlined part (iii).
2-5. N [mol] of ozone is bubbled through 15 mL of an aqueous solution of potassium iodide with aconcentration of 1.0 mol L–1; we assume that the ozone reacts completely. The solution is thentitrated with an aqueous solution of sodium thiosulfate with a concentration of 0.20 mol L–1; thetitration requires 3.5 mL of the sodium thiosulfate solution. Calculate N [mol].
2-6. Write the reaction formulae for the underlined parts (iv)–(vi).

58

Problem 17. Metal oxides
1.

In the crystal structures of most metal oxides, the oxide ions (O2‒) form a close-packed structure,
wherein the metal ions occupy the voids created by O2‒. Figure 1(a) shows the cubic close-packed
structure created by O2‒. Note that the ionic radius of O2‒ is 0.140 nm. Answer the following
questions.

(a)

(c)

(b)

Figure 1. (a) Cubic close-packed structure formed by O2‒ in metaloxides, (b) octahedral void, (c) tetrahedral void. ● O2‒ ions, ● voids.
1-1. Considering O2‒ as solid spheres in contact with each other, calculate the percentage volume ofthe voids in a cubic close-packed structure.
1-2. With reference to Figure 1(b) and (c), calculate the maximum radii of the spheres that can beplaced in the octahedral and tetrahedral voids of a cubic close-packed structure composed of O2‒,
respectively.
1-3. Write the chemical formula for a ternary metal oxide AxByOz wherein one metal ion (A) occupies
1/8 of the tetrahedral voids in a cubic close-packed structure composed of O2‒, and a second metalion (B) occupies 1/2 of the octahedral voids.
1-4. Regular arrays of atoms, ions, or molecules in solids are represented by a repeating unit known as
the unit cell. In SrTiO3, which adopts a perovskite structure, the ionic radii of Sr2+ and O2‒ arecomparable, and a cubic close-packed structure is formed by Sr2+ and O2‒. The Ti4+ ions occupy
all octahedral voids formed by O2‒. Based on this information, draw the unit cell of SrTiO3 withreference to Figure 2, assuming Ti4+ (●) is located at the center of the cube. Draw Sr2+ as ● andO2‒ as ×.

59

Figure 2. Unit cell of SrTiO3 showing only Ti4+ at the center.
Most divalent metal oxides (MO) crystallize in a rock salt structure wherein M2+ occupies theoctahedral voids created by O2– (Figure 3).

Figure 3. Unit cell of CaO. Blue and red spheres
represent Ca2+ and O2–, respectively. (● Ca2+, ● O2–)
The thermodynamic properties of metal oxides can be discussed in terms of the standard molar
enthalpy change (ΔHLexp) associated with the formation of gaseous ions from the ionic solid:MO(s) → M2+(g) + O2–(g) ΔHLexp(1)

ΔHLexp is known as the lattice enthalpy; although it cannot be measured directly, it can be determinedfrom enthalpy data (Table 1) using a Born–Haber cycle, which is a closed path of steps that includes
lattice formation as one stage.
Table 1. Enthalpy changes of reactions involved in the Born–Haber cycle
Steps
Enthalpy change [kJ mol–1]
Sublimation of Ca(s): Ca(s) → Ca(g)+193
Ionization of Ca(g) to Ca+(g): Ca(g) → Ca+(g) + e–(g)
+590
+
2+
+
2+
–
Ionization of Ca (g) to Ca (g): Ca (g) → Ca (g) + e (g)+1010Dissociation of O2: O2(g) → 2 O(g)+498––Electron gain by O(g): O(g) + e (g) → O (g)–141Electron gain by O–(g): O–(g) + e–(g) → O2–(g)+878Formation of CaO(s): Ca(s) + ½ O2(g) → CaO(s)–635
2-1. Calculate the ΔHLexp [kJ mol–1] of CaO using the enthalpy data in Table 1 and the Born–Haber
cycle.

60

The lattice enthalpy of an ionic crystal can be calculated theoretically (ΔHLcalc) from the sum of the
Coulomb energies between the constituent ions as point charges. The Coulomb energy E between ionsis expressed by the following equation:
1 𝑍4 𝑍/ 𝑒 /𝐸=
4𝜋𝜀( 𝑟(2)

where ε0 is the permittivity of the vacuum, e the elementary charge, Z1 and Z2 the charge numbers of
the ions, and r the distance between the ions. E is attractive (i.e., negative) between an anion and acation, while E is repulsive (i.e., positive) between two anions or two cations.
Ca2+ is surrounded by six O2– in an octahedral arrangement (Figure 4).

Figure 4. A unit cell of a rock salt structure to calculate the Madelung constant.
For a distance d between Ca2+ and O2– in the rock salt structure depicted in Figure 4, the Coulomb
energy (E1) between a Ca2+ and six O2– is formulated as follows:(3)

In the structure depicted in Figure 4, twelve Ca2+ are located at a distance of √2d from a given Ca2+.
The Coulomb energy (E2) between the Ca2+ and the twelve surrounding Ca2+ is formulated as follows:(4)

Similarly, for any given Ca2+, there are 8 O2–, 6 Ca2+, 24 O2–, and 24 Ca2+ at distances of √3d, √4d,
√5d, and √6d, respectively. The overall Coulomb energy Eall, obtained by the summation of En
converges to the following equation:
𝑁< 𝑍 3 𝑍 . 𝑒 /𝐸Udd =
𝑀
4𝜋𝜀( 𝑑(5)where Z+ and Z– are the charge numbers of the cation and anion, respectively; NA is Avogadro’s
constant, and M is the Madelung constant, which is 1.748 for the rock salt structure. The theoreticallattice enthalpy ΔHLcalc is calculated using the Born–Mayer equation below, which takes into accountthe electron–electron and nucleus–nucleus repulsion of nearby ions:61

Δ𝐻e>Ud> =𝑁< |𝑍 3 𝑍 . |𝑒 /𝑑∗Q1 − R 𝑀
4𝜋𝜀( 𝑑𝑑(6)where d* is a constant (typically: 0.0345 nm).
2-2. Fill in the blanks (3) and (4) with the appropriate equations.
2-3. Calculate ΔHLcalc [kJ mol–1] for CaO using equation (6); use d = 0.240 nm.

62Problem 18. Coordination chemistry and its application to solid-state catalysts
Transition-metal complexes are formed by the coordination of ligands (e.g., H2O and NH3) to transition
metals or their ions. Most transition-metal complexes exhibit beautiful colors that cannot be explainedby the classical theory of chemical bonding. Therefore, crystal-field theory has been developed; therepulsion between the d-electrons and the lone pairs of the ligands as negative point charges splits theenergy of the five degenerate d-orbitals. Transition-metal complexes exhibit color because this energy
separation and the corresponding transitions coincide with the energy of visible light.
Consider an octahedral transition-metal complex with the metal ion at the origin of a three-dimensional
coordinate system and a total of six ligands as negative point charges equidistant from the origin on
the x, y, and z axes (Figure 1a). The energies of the five degenerate d-orbitals are split into two groups
called the eg and t2g orbitals, which are oriented along and between the axes, respectively.
z(a)(-1, -1, 1)
L(1, 1, 1)
L

(b)

L
L

L

Mn+

L

Mn+

y

L
L

x

L

L

(-1, 1, -1)(1, -1, -1)

Figure 1. Arrangement of the metal ion (Mn+) and ligands (L) in (a) octahedral and (b) tetrahedral
complexes. Blue solid lines represent the coordination bonds. The metal ion is located at the origin (0,0, 0) of the three-dimensional coordinate system. The coordinates of the four ligands in the tetrahedral
complex are indicated for clarity.
1.

Classify the five d-orbitals into eg and t2g orbitals.

Let the energy separation between the eg and t2g orbitals (i.e., the crystal-field splitting parameter) be
ΔO, where the subscript O signifies an octahedral arrangement. The barycenter of the energies of theeg and t2g orbitals is equal to the energy of the five degenerate d-orbitals.
2.

Write the energies of the eg and t2g orbitals relative to the barycenter in units of ΔO.

The energetically split d-orbitals, caused by the repulsion between the transition metal ion and theligands, accommodate electrons according to Hund’s rule. For example, Ti3+ with one d-electron (d1)forms an octahedral complex [Ti(H2O)6]3+; this complex is energetically more stable than Ti3+ without
ligands. This additional stability relative to the barycenter is known as the ‘crystal field stabilizationenergy’ (CFSE). The first through third d-electrons (d1 to d3) are exclusively accommodated in the t2g
orbitals. However, there are two possible ground-state electron configurations for d4 to d7. Namely,
Hund’s rule applies when the CFSE is small, resulting in a ‘high-spin’ ground state with a large numberof unpaired electrons. On the other hand, Hund’s rule does not apply when the CFSE is large, resulting
in a ‘low-spin’ ground state with a smaller number of unpaired electrons. In general, the CFSE depends
on the type of ligands (cf. spectrochemical series).
3.

Draw the high-spin electron configurations of octahedral complexes for d4-d7 by referring to theelectron configuration of d3 shown below, and calculate each CFSE in units of ΔO.
63

eg

Δo
t2g

While six-coordinate octahedral complexes are the most common among the transition-metal
complexes, four-coordinate complexes are also very frequently encountered. There are two types offour-coordinate complexes, tetrahedral and square planar complexes. Of these two, tetrahedral
complexes have a lower steric repulsion between the ligands, and similarly to octahedral complexes,
the energy splitting of their d-orbitals can be considered using crystal-field theory. In the tetrahedral
crystal field (Figure 1b), the energy of the five degenerate d-orbitals is split into two sets of orbitals, e
and t2 orbitals.
4.

Draw the energy separation between the e and t2 orbitals with ΔT as the crystal-field splitting
parameter, where the subscript T signifies a tetrahedral arrangement. Classify the five d-orbitals
into e and t2 orbitals. Write the energies of e and t2 orbitals relative to the barycenter in units ofΔT .

When an aqueous solution containing the octahedral complex [Ti(H2O)6]3+ is irradiated with visible
light, a d-electron is excited from the t2g to the eg orbital, which is accompanied by the absorption ofgreen light (λ = 500 nm). Consequently, the aqueous solution takes on a red-purple color, which is thecomplementary color of green.
5.

Calculate the energy separation ΔO (kJ mol–1) between the eg and t2g orbitals in [Ti(H2O)6]3+.

Transition-metal complexes with carbon monoxide (CO) as a ligand are called metal carbonyls.
Representative examples are the octahedral Cr(CO)6 and tetrahedral Ni(CO)4 complexes. The total
number of electrons provided by the metal and ligands in a transition-metal complex is called theeffective atomic number (EAN), and the 18-electron rule states that a complex is stable when the EAN
is equal to the atomic number of noble gases. For instance, Cr and the CO ligands in Cr(CO)6 supply
6 and 2 × 6 = 12 electrons, respectively; similarly, Ni and the CO ligands in Ni(CO)4 supply 10 and 2
× 4 = 8 electrons, respectively, satisfying the 18-electron rule for each metal carbonyl. In addition tomononuclear metal carbonyls with one metal atom, there are also multinuclear metal carbonyls withmultiple metal atoms and metal–metal bonds.
6.

Draw the molecular structure of Mn2(CO)10 and Ir4(CO)12 and show that the EAN per metal atom
is 18.

64

The coordination bonds formed between carbon monoxide (CO) and metals can be used to estimate
the size of metal nanoparticles loaded on catalyst support such as Al2O3. Figure 2a shows a schematic
illustration of a platinum catalyst supported on alumina (Pt/Al2O3), which is used as an automotive
catalyst.

Figure 2. Schematic illustrations of (a) the Pt/Al2O3 catalyst and (b) the CO molecules
adsorbed on the Pt surface. The Pt metal particles in (a) are the same size.
1.0 g of Pt/Al2O3 catalyst that contains 0.010 g of Pt metal is prepared. Upon exposure to CO gas, 0.55
mL (1.0 bar, 273 K) of CO was adsorbed per 1.0 g of catalyst. The Pt metal particles on Al2O3 arespherical, and their surfaces can be regarded as being covered with Pt atoms without gaps. As shownin Figure 2b, one CO molecule coordinates to one Pt atom exposed on the surface. The density of the
Pt metal is 21.45 g cm‒3, and the area occupied by each Pt atom is 8.0 × 10‒2 nm2.
7.

Answer the following questions.

7-1. Calculate the ratio [%] of Pt atoms exposed on the surface to the total number of Pt atoms in thespherical Pt metal particles.
7-2. Calculate the number of CO molecules adsorbed per 1.0 g of Pt metal dispersed on the Al2O3
support.
7-3. Calculate the surface area [m2] per 1.0 g of Pt metal.
7-4. Calculate the diameter of the Pt metal particles.

65Problem 19. Acids and bases
In 1923, Brønsted and Lowry defined an acid–base reaction as the transfer of a proton (H+) from theproton donor (acid) to the proton acceptor (base).
HF + H2O ⇌ H3O+ + F–
H2O + NH3 ⇌ NH4+ + OH–

(1)
(2)

Hydrogen fluoride (HF) is an example of a Brønsted acid, ammonia (NH3) is an example of a Brønsted
base, while water (H2O) is amphoteric, i.e., it can behave as both a Brønsted acid and a Brønsted base.
When acid–base reactions are in chemical equilibrium, the reactions are indicated by two one-sided
arrows as in (1) and (2). When HF donates a proton to H2O, F– and H3O+ are formed as the conjugatebase and conjugate acid, respectively. Brønsted acidity can be evaluated quantitatively using theacidity constant, or qualitatively using the (a) stability of the conjugated base.
1.

Arrange the following Brønsted acids in order of increasing acidity by referring to the underlined
part (a).
HClO HClO2 HClO3 HClO4

The Brønsted acidity of hydrogen halides (HX, X = F, Cl, Br, I) can be evaluated by the standard
enthalpy change (ΔHºHX) of the protonation reaction of their conjugate bases (X–):
X–(g) + H+(g) → HX(g)ΔHºHX (3)This reaction can be divided into three steps:
Electron loss from X–: X–(g) → X(g) + e– (g)Electron gain by H+: H+ (g) + e– (g) → H(g)Combination of H and X: H(g) + X(g) → HX(g)

ΔHº1
ΔHº2
ΔHº3

(4)
(5)
(6)

Therefore, ΔHºHX is calculated as ΔHºHX = ΔHº1 + ΔHº2 + ΔHº3.
2.

Calculate the absolute value of ΔHºHF (|ΔHºHF| in kJ mol–1) by drawing a thermodynamic cycle
for the protonation reaction of F–. Use the following values if necessary:
H–F bond dissociation enthalpy: 567 kJ mol–1; ionization enthalpy of hydrogen: 1312 kJ mol–1;
enthalpy of electron gain by fluorine: –328 kJ mol–1.

3.

Calculate the absolute value of ΔHºHI (|ΔHºHI| in kJ mol–1) by drawing a thermodynamic cycle
for the protonation reaction of I–. Use the following values if necessary:
H–I bond dissociation enthalpy: 298 kJ mol–1; enthalpy of electron gain by iodine: –295 kJ mol–1.

4.

According to Q2 and Q3, |ΔHºHI| (a) |ΔHºHF|; therefore, the Brønsted acidity of hydrogen halides
is arranged in the following order, HF (b) HCl (b) HBr (b) HI. Choose which inequality sign,‘>’ or ‘<’, should be placed in blanks a and b.

Metal ions are hydrated in aqueous solutions, and generally, the first hydration shell of metal ionsconsists of six water molecules. When the oxygen atom of a water molecule is strongly coordinated toa metal ion, the coordinated water acts as a proton donor as shown in (7).
[M(OH2)6]n+(aq) + H2O(l) ⇌ [M(OH2)5(OH)](n-1)+(aq) + H3O+(aq) (7)665.

Arrange the following aqua acids in order of increasing acidity.
[Na(H2O)6]+ [Ca(H2O)6]2+ [Fe(H2O)6]3+ [Zn(H2O)6]2+

Lewis introduced a more general concept of acids and bases, which includes reactions that do not
involve protons. Lewis defined an acid–base reaction as the transfer of an electron pair from theelectron-pair donor (Lewis base) to the electron-pair acceptor (Lewis acid). Note that the Brønstedacid HA is a complex formed by the Lewis acid H+ with the Lewis base A–. All Brønsted acids andBrønsted bases exhibit Lewis acidity and Lewis basicity, respectively.
6.

Identify the Lewis acids and bases in each of the following reactions.
(i) BF3 + NH3 → BF3NH3(ii) Ni + 4 CO → [Ni(CO)4](iii) CO2 + OH– → HCO3–(iv) SiF4 + 2 F– → SiF62–

7.

Phosphoric acid (H3PO4) and boric acid (H3BO3) are both acids with three hydrogens. Whilephosphoric acid reacts as a triprotic Brønsted acid, boric acid reacts as a monoprotic acid. Draw
the Lewis structure of the conjugate base of boric acid by considering the reaction with water.

67

Problem 20. Semiconductors
In molecules that exist as gases or liquids, the molecular orbitals are formed by the overlap between
the atomic orbitals of individual atoms. In contrast, in the electronic state of a solid composed of an
infinite number of atoms, many atomic orbitals are overlapping; therefore, many molecular orbitals
with close energy levels are formed. As the gap between their energy levels is very narrow, theseorbitals are for practical reasons represented as an energy band. Insulators have a wide energy gap
between the valence band (VB), which is completely filled with electrons, and the conduction band
(CB), which does not contain electrons (Figure 1).

Insulator

Semiconductor

EnergyEnergyConduction
bande–Thermal energyh+Valenceband
Figure 1. Schematic view of the electronic band structure in solid-state materials.
In insulators, the electrons are bound within the VB. On the other hand, in semiconductors, someelectrons (e‒) in the VB can be transferred to the CB by thermal energy, as the energy gap between the
VB and CB is relatively small. An electron transition in a semiconductor creates a hole in the VB; thishole behaves like a positively charged particle. The hole is commonly referred to as h+. The electrons
and holes formed in the semiconductor can move, which results in electrical conductivity.
1.

With reference to Figure 1, illustrate the band structure of a metallic conductor, which showselectrical conductivity even when the temperature is lowered.

Semiconductors are classified as either intrinsic or extrinsic semiconductors. An intrinsic
semiconductor is composed of a pure material, and the same number of electrons and holes areproduced by thermal energy and other factors. The electrical conductivity is not very high. On theother hand, there are two types of extrinsic semiconductors, n-type and p-type, whose electricalconductivity is generally higher than that of intrinsic semiconductors. In n-type semiconductors, small
amounts of electrons are added to a pure material (electron doping), while in p-type semiconductors,
small amounts of holes are added to a pure material (hole doping). For example, adding a small amountof P to pure Si, a typical intrinsic semiconductor, results in the doping of an extra electron (Figure 2).
Accordingly, P-doped Si exhibits n-type semiconductor properties.

68

(a)

(b)
Si

Si

Si

Si

Si

Si

Si

Si

Si

Si

P

Si

Si

Si

Si

Si

Si

Si

Figure 2. Schematic view for electron states of (a) Si and (b) P-doped Si.
2.

Indicate which elements in the following list could be used to produce p-type semiconductor
properties in Si.
Ga, Ge, As, B

TiO2, a representative oxide semiconductor, is capable of inducing the water-splitting reaction at room
temperature using light energy as a driving force. The water-splitting reaction, which generates
hydrogen (H2) from water, is expected to be a significant source of renewable energy in the near future.
The discovery of the following phenomena is one reason for the development of photocatalysts suchas TiO2. (1) The electrolysis of water occurs when a voltage is applied. In contrast, (2) when TiO2 andplatinum electrodes are arranged as shown in Figure 3 and the TiO2 side is exposed to light irradiation,oxygen and hydrogen are generated from each electrode. This phenomenon is called the Honda–
Fujishima effect and was discovered in 1972. Electrons and holes are produced by irradiating TiO2
with light, which leads to the reduction and oxidation of water, respectively.
Current

V

TiO2

Pt

Figure 3. Schematic illustration of the photoelectrochemical water-splitting reaction.
3-1. When TiO2 is subjected to a thermal treatment under an H2 flow, it exhibits n-type semiconductor
properties. Explain the reasons for this phenomenon.
3-2. Regarding the underlined part in (1), calculate the voltage at which water can bethermodynamically electrolyzed. The standard Gibbs energy change for the splitting of 1 mole ofwater is 237 kJ.

69

3-3. Write the half-reaction equation including the hole (h+) for the gas-evolution reaction on TiO2
described in the underlined part in (2). When TiO2 is exposed to light irradiation, a current isobserved in the direction shown in Figure 3.
3-4. Choose the option below that best represents the maximum percentage of the photon number insunlight that is available for water-splitting reactions on TiO2. The band gap of TiO2 between the
VB maximum and the CB minimum is 3.0 eV. Figure 4 shows the solar spectrum, and thecalculation of the photon number here does not take into account sunlight below 300 nm and above
800 nm.
(a) <1.0 % (b) 1–10 % (c) 10–20 % (d) 20–30 %Figure 4. Photon number vs wavelength in sunlight.
3-5. The ratio of the number of carriers (electrons or holes) consumed in the water-splitting reaction to
the number of irradiated photons is called quantum efficiency. When TiO2 is irradiated with 1.00
W of light with a wavelength of 350 nm, 5.00 μmol of hydrogen are produced in 100 seconds.
Calculate the quantum efficiency [%].

70

Problem 21. Carbenes and non-benzenoid aromatic compounds
The following problems describe the highly unusual and interesting reactivity of carbenes.
Si CF2Brn-Bu4N Br (cat.)1+

A

toluene, 110 °C

Si Br

1H NMR 7.23 ppm (10H), 2.95 ppm (2H)
19F NMR –134.5 ppm (triplet, J = 8.5 Hz)

Under the conditions described above, olefin 1 is converted into A, which does not contain a double
bond.
1.

Draw the structure of A.

Under similar reaction conditions, tautomer of ketone 2 is converted into B, which is subsequently
transformed into C in the presence of a fluoride ion.
Si CF2BrO

n-Bu4N Br (cat.)n-Bu4N F (cat.)B

C

C13H16F2OSi
2

2.

Draw the structure of B.

3.

Choose all possible structures for C.
F

F
OH

a

OH

b

F

c
OH
F

OH
d

F

F
OH

e

F

F

f

F
OH

Silicon exhibits an extremely high affinity toward oxygen and especially fluorine. An application that
capitalizes on this reactivity is the formation of benzyne species.

71

4.

A four-step transformation converts 3 into 4. Draw the structures of D–F.
Si
OH

N
H

Si

Br

D

BuLi

E

C9H13BrOSi(‒BuBr)C9H13LiOSi

3
O
CF3
O S
O
CF3
S
O O

O
n-Bu4N F

FO

C10H13F3O3SSi

4

In another four steps, 4 is converted into 6, which bears a cycloheptatrienone moiety. The structure of
G contains one more cyclic moiety than the structure of 4.
5.

Draw the structure of G (while the stereochemistry is important for the reaction, it is not the focusof this question).
Cl

CHCl3
NaOH

G
C11H8Cl2O

O

165 °C

O

Cl

PhNO2

4

5
LiAlH4

H2SO4, H2OO

C11H9ClO

66.

HDraw the structure of H.

72

Subsequently, 6 is converted into I by exploiting the unique aromaticity of cycloheptatriene cations.
When I is treated with the Grignard reagent MeMgI, a mixture of J and K is obtained. Treating I withcatalytic amount of HCl in CCl4 furnishes isomer I’, which is again converted into a mixture of J andK upon treatment with Lewis-acidic Grignard reagents.

LiAlH4

NaH
CH3I

MeMgI

6

7.

I

J

C12H12O

C12H12+K
C12H12

Draw the structures of I, J, K.

The ring-opening of the epoxide moiety on the six-membered ring occurs so that both the introduced
functional group and the hydroxy leaving group adopt axial positions. Epoxides 7 and 8 both reactwith trimethylsilyl iodide, but afford different products.

Si I
O

L

H
7
Si I
O

M

H
8

8.

Draw the structures of L and M.

73

Problem 22. Nazarov cyclization
The formation of five-membered rings is a very important class of organic transformations. Answer
the following questions regarding one of those interesting reactions, the so-called ‘Nazarov
cyclization’.

pentadienyl cation

cyclic allylic cation

In general, 4π-electrocyclization reactions proceed via one of two cyclization modes. Conrotatory
cyclization occurs under thermal transformation conditions, while disrotatory cyclization occurs
upon photoirradiating the substrate.
Δ
H H

conrotatoryhν

H H

disrotatory

74

1.

Titanium tetrachloride (TiCl4) is a Lewis acid that can activate the carbonyl group in 1. The
resulting pentadienyl cation intermediate is transformed via a conrotatory Nazarov cyclization.
The resulting cation intermediate is then trapped with chloride to give A. Choose all possible
structures of A.

TiCl4
A
O

(a)

1

(b)

(c)

(d)

(e)

H

H

H

H

H

Cl O

Cl O

Cl O

Cl O

Cl O

(f)

(g)

(h)

(i)

(j)

H

H

H

H

H

Cl O

Cl O

Cl O

Cl O

Cl O

75

2.

Compound 2 is transformed into Taiwaniaquinol (3) using a Nazarov cyclization as a key step.
Draw the structures of B, C, D, E.

OMe

O

Br
N

O

O
n-BuLi
B
–BuBr

MeO
2
AcO OAc
I OAc
OC
C21H32O3

O O
S
Si O
CF3

O
Dess-Martinoxidationhydrolysis

C21H30O3

OH
E(1.1 equiv)DOH

2 stepsMeO

C21H30O3

OH
Taiwaniaquinol B (3)Optional problem:
The frontier molecular orbitals of the allyl cation, which consist of three carbon atoms and one π orbital
shown below. The lowest unoccupied molecular orbital (LUMO) and the highest occupied molecularorbital (HOMO) are labeled in this figure. Draw the molecular orbitals of the pentadienyl cation using
the same style as given in the example and label the LUMO and HOMO orbitals.
φ3allyl cation

φ2

LUMO

φ1

HOMO

76

Problem 23. Tea party
In Japan, the cultivation of tea leaves and the green-tea culture started in the early 1200s. This culture
has grown and developed ever since into the current form of the traditional ‘tea ceremony’, during
which matcha is prepared and drunk. Moreover, regular green tea for daily consumption has becomea staple of Japanese diet. Catechins, which are responsible for the characteristic taste of green tea, havebeen associated with a variety of health benefits. Tea leaves contain several catechins such as
epicatechin, epigallocatechin, epicatechin gallate, and epigallocatechin gallate. Green tea also contains
caffeine, which is bitter, and theanine, which adds a savory (umami) flavor.
X
OH

X
OH
O 2

HO

3

O

OH

OH

O
OH

OH

OH
X = H: (–)-epicatechin
X = OH: (–)-epigallocatechin

1.

HO

O

OH

OH
OH
X = H: (–)-epicatechin gallate
X = OH: (–)-epigallocatechin gallate

Assign the absolute configuration of the stereocenters at the 2- and 3-positions of (–)-epicatechinusing the R/S nomenclature.

77

The following scheme shows a synthetic route to (–)-gallocatechin, which is an epimer of (–)epigallocatechin.
OH

OH
A

OH
HO

MeO

OH

OBn
B

OH

O

OBn
MeO

OH
O

O
(Bn = CH2Ph)
O

+

NH

LiAlH4OBn–
ClCrO3O

HO

OEt
pyridineCD

refluxC32H30O5

BnO
OBn

OBn
OBn

OH
OBn

i Bu2AlH

BnO

HO

OBn

OBn
O

OH

OH
OH
E

C50H43IO7

2.

HO

O

OH
OH

OH
(–)-gallocatechin

From the following conditions, choose the best for A.
□ MeMgBr, THF□ NaOMe, THF□ NaBH4, MeOH
□ HCl, MeOH

78

I

P(n Bu)3
O
Me2N
N
N
NMe2
O

3.

From the following conditions, choose the best for B.
□ PhCH2OH, NaH□ PhCH2Cl, NaH□ PhCHO, HCl□ PhCO2H, HCl

4.

Draw the structures of C–E and clearly identify the stereochemistry where necessary.

Caffeine, which is responsible for the bitter taste of green tea, can be synthesized as follows.
O
OO

NaOEt, EtOH, reflux

+
MeHN

NHMe

HO

HCO2H, Δ

F

CN

MeN
O

C6H9N3O2

N
Me

N
H
O

HNO3, H2SO4, reflux

G

Zn, CH3CO2H

I
H

MeN
O

C7H8N4O5

C7H8N4O2

5.

Draw the structures of F–H.

6.

From the following conditions, choose the best for I.
□ 1) NaH, 2) MeI□ 1) MeMgBr, 2) H2O□ NaOMe, MeOH
□ NMe3, H2O

79

CHO

Me
N

N
N
Me
caffeine(+)-Theanine, which is responsible for the savory (umami) taste of green tea, can be synthesized from(+)-glutamic acid as follows.

O
H 2N

O

A
OH

H 2N

CO2H
(+)-glutamic acid

J

OMe

O
PhCH2O C Cl, Na2CO3
H2O/dioxane

CO2H

1. EtNH2, THF, Δ
2. H2, cat. Pd/C, THF(+)-theanine
C7H14N2O3

C14H17NO6
A

is the same as in question 2

7. Draw the structures of J and (+)-theanine and clearly identify the stereochemistry where necessary.

80Problem 24. E-Z chemistry
Alkenes can undergo trans/cis isomerization upon photoirradiation as shown below, and theisomerization process can be accelerated by adding a sensitizer that is excited upon absorption of lightand subsequently transfers this energy to the alkenes.
R2

hν
R1

R1
trans

R2
cis

This photo-induced trans/cis isomerization also occurs in certain cycloalkenes, which containconnected substituents R1 and R2. In general, trans-cycloalkenes that contain a 7- to 9-membered ringare highly reactive due to their intrinsic ring strain. Their two faces of the π-plane can be distinguishedby the reactivity, as one face is blocked by the carbon chain within the ring, while the other unblocked
face can engage selectively in chemical reactions.
Moreover, asymmetric syntheses can be achieved by exposing cycloalkenes to irradiation in thepresence of a chiral sensitizer. For instance, as shown in the scheme below, photoirradiation of
cycloheptene (1) at low temperature in the presence of chiral sensitizer 2 generates 3, which undergoes
a Diels–Alder reaction with 1,3-diphenylisobenzofuran (4) to enantioselectively afford optically active
5. Under the same photoirradiation conditions, the reaction of 3 with OsO4 (6) results in the formationof optically active 7.
Ph
O
4
hν
Chiral sensitizer 2–70 °CPh

Ph
O–70 °C

Ph
5

3
1

OsO4 (6)7–70 °C
Me Ph
RO2C

CO2R

RO2C

CO2R

Chiral sensitizer 2:

Me
R=
Me

1.

Draw the structures of 3 and 7 and clearly identify the stereochemistry where necessary.

81The naturally occurring compound Coraxeniolide is a bicyclic compound that contains fused 9- and 6membered rings. A synthetic route to Coraxeniolide precursor 15 is shown below, starting from 8 via
cycloalkene derivative 12.
O
S Cl
O

Me
OH

O

OH

NaBH4
CH3COOH

O

pyridine

9(C10H16O3)

Me
8

HO

Me
10

OSiMe2t Bu

NaH

11(C17H22O4S)

12(C10H14O)

OMe

Me

NaOt Bu;then HCHO2.

Ph3P CH214(C13H18O3)

O

O

OMe
cat. Ph3CClO4

13

15(C14H20O2)

A nucleophilic addition to a carbonyl group does not occur easily when the carbonyl group isconnected to a bulky substituent. When the two faces of the carbonyl group can be differentiatedby the bicyclic structure as in 8, the addition of the nucleophile usually occurs selectively fromthe outer face of the core structure. Based on these general considerations, choose the mostsuitable structure for 9 from A–D.
OH

O

Me
A

OH

OH

O

OH

OH

Me

HO

B

Me
C

OH

O

HO

O

Me
D

3.

The nucleophilicity of the hydroxy group in alcohols generally increases with decreasing size ofthe associated alkyl group. Draw the structure of 11 and clearly identify the stereochemistry where
necessary.

4.

Compound 12 is a chiral compound although it does not possess a carbon stereocenter. Draw thestructure of 12 and clearly identify the stereochemistry where necessary.

5.

Draw the structures of 14 and 15 and clearly identify the stereochemistry where necessary.

82

Problem 25. Fischer indole synthesis
The reaction of phenylhydrazone 1 and carbonyl compound 2 under acid catalysis to give indole 3 via
4–8 is known as the Fischer indole synthesis.
R1
N
H
1

R2

+

NH2

R2
N
H
3

O
2

R1

R1
R2

N
H
4

R1

H+ catalyst

R1

R2

N

N
H
5

R1
R2

NH

R1
R2

NH
NH2

N
H

6

7

R2

NH2

N
8

When bicyclic ketone 2a is used as the carbonyl compound, indole 9 is obtained selectively via
intermediate 5a, wherein a carbon–carbon double bond is formed between C2 and C3 in order to avoid
the formation of a highly strained carbon–carbon double bond between bridge-head C1 and C2.

N
H
1

+

NH2

H+ catalyst

3

O

2 1

N
H
9

2a

N
H
5a

1.

Draw the structure of compound 10 in the following reaction.

N
H

2.

NH

NH2

MeCO2H10(C8H8O)+PhN
H

In the following reaction, tricyclic 12 is produced instead of an indole due to the substituent at the
3-position of 11. Draw the structure of 12.
Me
N
H

NH2

3

MeCO2H

+
O

HO
11

83

12
(C11H13NO)

3.

Draw the structures of 14 and 15 in the following scheme and clearly identify the stereochemistrywhere necessary.
Ts
N

O
N
H

MeCO2H

NH2 +

K2CO3
MeOH

14
(C26H26N2O4S)

15(C27H30N2O5S)

Me
O

O
S
O

O
13 ( Ts =contains atetrahydrofuran ring

Me )Compound 15 described above is a precursor for the naturally occurring Aspidophylline A, and 13 canbe synthesized from 16 as shown below.
O

Ph
Ph

O

O

Ph

O
Cu2O, H2O

17
(C23H19NO5)

N

O

N

O

O
16
Br
HC(OMe)3
cat. TsOH
MeOH

MeO
MeO

18
(C16H19NO3)

MeO

19(C21H28INO6S)

Pd-catalyst
base

MeO
MeO

NHTs
I
Me
Cs2CO3O

Ts
N

indiumNH4Cl

FeCl3•6H2O
Me

MeO

O

Ts
N

O
Ts
N

LiN(SiMe3)2;
then

I

21
(C22H27NO5S)

O

*

Me
MeO

Me

O
20

O

O
13

4.

Draw the structures of 17 and 18 and clearly identify the relative stereochemistry where necessary.

5.

Draw the structure of 19 and clearly identify the stereochemistry where necessary.

6.

In the conversion of 20 to 21, the reaction occurs selectively at the sterically less hindered side,and the stereochemistry with asterisk (*) of 13 is set in this step. Draw the structure of 21 andclearly identify the stereochemistry where necessary.
84

Problem 26. Planar chirality
A cyclophane consists of aromatic units (typically benzene rings) and aliphatic chains that form
bridges between two non-adjacent positions of the aromatic rings. As shown below, if the bridge is
ethylene (-CH2-CH2-) and the substitution pattern is para, it is called [2.2]paracyclophane. For somekinds of [2.2]paracyclophanes, enantiomers can be isolated depending on the substituent
configurations on the aromatic rings given that these cannot rotate.
Brominating [2.2]paracyclophane affords mono-brominated A, which exhibits planar chirality, i.e.,(Rp)- and (Sp)-isomers. Here, as shown in the dotted square, the definition of (Rp) is that the orthosubstituent is oriented in clockwise direction when traced from the other aromatic unit through the
ethylene bridge. Further bromination of A furnished dibrominated products B-E.

Br2
FeBr3

Br

[2.2]paracyclophane(Rp)-A

Br
Br2
FeBr3

Br

Br

Br
Br

Br
B

1.

Rp

Br
(Sp)-ABr
C

Choose from B-E all products that are optically active.

85

Br
D

E

Bromination of [2.2]paracyclophane with an excess of bromine in the presence of FeBr3 results indibromination on each benzene ring to generate tetrabrominated products F (F') and G (G'). Note that
F and F' are a pair of enantiomers, while G and G' are identical. By adding two equivalents of n-BuLi
to a solution of F or G at low temperature, one of two bromo-substituents in each benzene unit isexchanged to lithium, and the subsequent addition of an excess of DMF leads to the formation ofseveral products.
Br

Br

Br
Br
Br

Br2
FeBr3

Br
Br

[2.2]paracyclophane

Br

Br
F

G

+

Br

Br

Br

Br

Br

Br

Br
F’
i) n-BuLi (2 eq)i) n-BuLi (2 eq)ii) DMF (excess)ii) DMF (excess)

H

2.

G’

I

Assuming that the reaction proceeds completely, draw all the possible stereoisomers for H and I.

86

A racemate of tetrabrominated F (rac-F) is transformed into J by applying the reaction conditions
shown below. Here J is a racemic mixture of (Rp)-J and (Sp)-J. The racemic mixture reacts with
optically active (-)-camphanic chloride to give a diastereomeric mixture of (Rp)-K and (Sp)-K, which
can be separated by column chromatography. (Rp)-K and (Sp)-K can be hydrolyzed to give opticallypure (Rp)-J and (Sp)-J. Here, the definition of (Rp) is the same as in Q1. When there are several kindsof substituents on the benzene ring, the definition obeys the Cahn-Ingold-Prelog priority rule.
Br
Br

B(OMe)3n-BuLi (1 eq)

H2O2 aq.

NaOH aq.

Br(Rp)-J+ (Sp)-JBr
rac-F(Rp)-J

KOH(Rp)-K

MeOH

COCl
O
O

+
pyridineRpO(Sp)-J

3.

KOH(Sp)-K

MeOH

R* =

O
O

C

Draw the structures of (Rp)-J, (Sp)-J, (Rp)-K, and (Sp)-K. You can abbreviate the acyl moiety
derived from (-)-camphanic chloride as R*.

The photodimerization of anthracene affords product whose planar chirality is similar to that of
cyclophanes. As shown below, anthracene undergoes photodimerization under irradiation withultraviolet light.
hν (365 nm)
hν’ (254 nm)

4.

Draw all the stereoisomers of photodimerized products of 2-substituted anthracene L. Amongthese, draw lines between each pair of enantiomers.

COOH

hν
M
hν’

L

87

Problem 27. Cyclobutadiene
In contrast to benzene, cyclobutadiene adopts a rectangular conformation in order to avoid 4π
antiaromaticity by bond alternation. Even though it is difficult to isolate pristine cyclobutadiene underambient conditions, its tricarbonyliron complex is isolable and relatively stable toward acids and bases.
Moreover, this complex readily engages in electrophilic substitutions, i.e., it exhibits reactivity that issimilar to that of benzene.

6π aromatic

4π antiaromatic(unfavorable)
OC
Fe2(CO)9CO
COFe(CO)3 FeΔ

1.

stableDraw the structures of A and B, which are obtained from cyclobutadiene under the conditions
shown below. The description of the iron complex is allowed to be the same as that of substrate.

Me
CH3COCl
AlCl3

Me
+
N

Cl
–

H Cl
(Vilsmeier reagent)(CO)3 FeA

B
then H2O

2.

The iron moiety can be removed by treatment with cerium ammonium nitrate, Ce(NH4)2(NO3)6.
Although [2+2]-cycloaddition reactions are generally not allowed, the thus generated
cyclobutadiene is so reactive that it can undergo [2+2]-cycloaddition-like reactions with alkenes
and alkynes. Draw the structure of C shown below. Its stereochemistry is not required.
not allowed+(CO)3 Fe

Ce(NH4)2(NO3)6
+

CO2Me

88

C

3.

A part of a synthetic route toward natural product asteriscanolide is shown below.
O

CO2Me

Fe2(CO)9hν(CO)3 Fe CO2Me

LiAlH4BF3·OEt2(CO)3 Fe Me

O
Δ
D

F

E

Me2NCH2NMe2
H3PO4
CH3CO2H

MeI

G(C11H13NO3Fe)

HHO(CO)3 Fe MeNaH

Ce(NH4)2(NO3)6
J

O
I
N Mes
Mes N
Cl Ru
Cl
Ph
PCy3
H2C CH2(cat.)

Cope
rearrangementKL

3-1. Draw the structure of the aromatic compound obtained by heating a mixture of D and dimethyl
acetylenedicarboxylate.
3-2. Draw the structures of G and salt H. The latter is an intermediate of the reaction from G to I.
3-3. Draw the structure of tetracyclic compound J (its stereochemistry is not required).

893-4. The ring opening metathesis of J furnishes K, and the subsequent Cope rearrangement leads tothe formation of L. Draw the structures of tricyclic compounds K and L (their stereochemistry is
not required). Examples of Cope rearrangement and ring-opening metathesis are shown below.
Cope rearrangement:
Δ

Ring-opening metathesis :+H2C CH2

N Mes
Mes N
Cl Ru
Cl
Ph
PCy3
(cat.)

90

Problem 28. Onion-like complexes
As shown below, cyclic molecules composed of alternating benzene and alkyne moieties can form an
onion-like concentric complex if a suitable pair of ring-sizes is chosen.

1.

For n = 6, calculate the length of the outer perimeter of the ring-shaped molecule. A list of relevantbond-length values is shown below. Consider the benzene ring as a regular hexagon. In addition,
calculate the diameter under the assumption that the ring is a true circle.

2.

For n = 9, calculate the length of the outer perimeter and the diameter under the same assumptions
as in Q1.

3.

Ring 1 (n = 6) and ring 2 (n = 9) can form a concentric complex in chloroform. Calculate thewidth of the interspace between the two rings (ignore the volume of the atoms).

4.

From A-C shown below, choose all the possible configurations of the benzene rings in thecomplex of 1 and 2 when viewed vertically relative to the ring plane. Here, a benzene ring isregarded as a circle with a diameter of 0.74 nm.

5.

Similar to the case of the encapsulation of 1 by 2, C60 can be encapsulated by ring 1. Estimate thediameter of C60 under the same assumptions for 1 and 2 regarding the interspace.

91

Cycloparaphenylenes (CPPs) are ring-shaped molecules that consist exclusively of benzene moietiesthat are connected in a para fashion. Depending on the number of constituent benzene rings (n), CPPs
are usually denominated [n]CPP. There are combinations of [n]CPPs that can form onion-like
concentric complexes.

6.

A free-energy change of –18.00 kJ mol-1 was estimated for the encapsulation of [6]CPP by[11]CPP in deuterated dichloroethane at 323.15 K. Calculate the concentration of free [6]CPP(which does not form any complexes) when the same volume of solutions of [6]CPP and [11]CPP
in deuterated dichloroethane (0.800 mmol L–1 each) is mixed at 323.15 K. In this case, the resulting
mixture is homogenous.

7.

Two equivalents of [13]CPP in solution were added in two portions to a solution containing anequimolar amount of five kinds of [n]CPPs (n = 8 – 12). Shown below is the 1H NMR spectralchange. No signals derived from CPPs were observed outside the region shown. Upon adding thefirst equivalent of [13]CPP, the formation of a precipitate was observed. Based on the information
given, choose all correct statements among (a) – (e).

(a) [13]CPP encapsulates [8]CPP to form a precipitate.
(b) [13]CPP quantitatively forms complexes with all other [n]CPPs (n = 8 – 12) and generates aprecipitate.
(c) The new peak that emerges at approximately 7.64 ppm can be ascribed to the complex of[13]CPP and [8]CPP.
(d) The new peak that emerges at approximately 7.64 ppm can be ascribed to [13]CPP.
(e) In the presence of [13]CPP, the peak associated with [8]CPP is substantially downfield shifted.

92

Problem 29. Hydrogen-bonded capsules
Calix[4]arene is a bowl-shaped molecule that is synthesized as shown below.

1.

Draw the structure of intermediate A.

2.

Choose the most suitable conditions for the transformation from C to D.
(a) tBuLi, then B(OMe)3, then AcOH(b) tBuLi, then B(OMe)3, then H2O2/NaOH aq.
(c) BBr3, then H2O2/NaOH aq.
(d) PMe3, then H2SO4(e) PMe3, then H2O2 aq.

93

3.

Two molecules of D form an aggregate via hydrogen-bonds upon addition of DBU (a kind of
weakly nucleophilic but strong organic base; see the structure below) to a chloroform solution of
D. Among (a) – (d), choose the most probable structure for the dominant hydrogen-bond networkof dimers of D. Refer to the graph showing the ratio of dimer (%) as a function of the amount of
DBU added to a solution of D (0.25 mmol) in chloroform.

Dimer ratio (%)

100
80
60
40
20
0
0

0.5

1

1.5

2

DBU added (mmol)

The dimeric capsule of D (host) described in Q3 can encapsulate other molecules (guests). In solution,
the encapsulation and release of the guest(s) are reversible. The association constant K for this process
is given by𝐾 =[𝑠𝑜𝑙𝑣𝑒𝑛𝑡][𝑔𝑢𝑒𝑠𝑡@ℎ𝑜𝑠𝑡][𝑠𝑜𝑙𝑣𝑒𝑛𝑡@ℎ𝑜𝑠𝑡][𝑔𝑢𝑒𝑠𝑡]

Here, the encapsulation of guest A by host B is expressed as A@B. The relative association constantKrel can be calculated by measuring the ratio of free and encapsulated guests in the coexisting state.
Thus, Krel is given by
𝐾+,d =

𝐾/[𝑠𝑜𝑙𝑣𝑒𝑛𝑡][𝑔𝑢𝑒𝑠𝑡2@ℎ𝑜𝑠𝑡] [𝑠𝑜𝑙𝑣𝑒𝑛𝑡][𝑔𝑢𝑒𝑠𝑡1@ℎ𝑜𝑠𝑡] [𝑔𝑢𝑒𝑠𝑡2@ℎ𝑜𝑠𝑡][𝑔𝑢𝑒𝑠𝑡1]=/=
𝐾4[𝑠𝑜𝑙𝑣𝑒𝑛𝑡@ℎ𝑜𝑠𝑡][𝑔𝑢𝑒𝑠𝑡2] [𝑠𝑜𝑙𝑣𝑒𝑛𝑡@ℎ𝑜𝑠𝑡][𝑔𝑢𝑒𝑠𝑡1] [𝑔𝑢𝑒𝑠𝑡1@ℎ𝑜𝑠𝑡][𝑔𝑢𝑒𝑠𝑡2]

94Below, the Krel values of several guest molecules in chloroform relative to benzene (guest1) are shown.

In the presence of a sufficient amount of DBU, D dimerizes almost quantitatively. In the following
questions, free D can be ignored.
4.

Equimolar amounts of pyrazine and 1,4-dioxane are added to a chloroform solution of D in thepresence of excess DBU. After the system reaches the equilibrium state, a 75:25 ratio of free toencapsulated 1,4-dioxane was found. Estimate the ratio of pyrazine encapsulated by D.

5.

A chloroform solution of dimers of D encapsulating benzene is exposed to the following
conditions (a) and (b). For each set of conditions, choose from (1) – (3) what happens to thecapsule structure and the benzene molecule.
(a) Addition of acetyl chloride and pyridine.
(b) Dilution with chloroform.
(1) The capsule dissociates and benzene is released.
(2) Both capsule and benzene remain intact.
(3) The capsule remains intact, but benzene is released.

6.

The same solution used in Q5 is treated with K2CO3 and CH2BrCl (cf. the reaction B→C in thescheme on the first page of this problem). Subsequently, pyrazine, which is more prone to be
encapsulated than benzene, is added. However, an exchange between the encapsulated benzene
and pyrazine is not observed. Explain this observation.

95

Problem 30. Synthesis and structural analysis of polymers
Polymers that contain carbon-carbon based main chains are often synthesized by chain polymerizationof vinyl monomers (Eq. 1).
R1

R1
CH2

C

H 2C

C

(Eq. 1)

R2

R2

n

When the main chain of a polymer adopts a zigzag shape, polymers in which all substituents R1 (orR2) are located on the same side are called ‘isotactic’, while those in which the substituents are locatedon alternating sides of the main chain are called ’syndiotactic’. Polymers that contain R1 = -CH3 andR2 = -COOCH3 are called poly methyl methacrylate (PMMA). Isotactic PMMA is obtained underconditions A in the scheme below, while syndiotactic PMMA is obtained under conditions B.
CH3
CH2

C
C

O

OCH3

n

Poly(methyl methacrylate) (PMMA)
Isotactic PMMA

R1

Conditions A

R2 R1

R2 R1

CH3

n CH2

R2 R1

R2

R1 : -CH3C

R2 : -COOCH3

C

O

Syndiotactic PMMA

OCH3

Conditions B
R1

R2 R2

R1 R1

R2 R2

Conditions A

Conditions B
CH3

CH3
H 3C

C
C

H 3C

H 3C

CH3
OLi + H3C

O
CH

R1

Si

C

OLi

H 3C

C

OLi +

H 3C

CH3

96

CH

Al
H 2C

O

CH3

CH2

CH3
CH3

CH2 CH3

The 1H NMR spectra of isotactic and syndiotactic PMMA are shown in Figure 1.

Figure 1. 100 MHz 1H NMR spectra of two types of PMMA in CD3NO2 at 100 °C
1.

In Figure 1, the intensity ratio of the peaks A, B, and C is 3:2:3. Assign the protons in PMMA tothe corresponding peaks.

2.

In Figure 1, which 1H NMR spectrum belongs to isotactic PMMA? Choose (a) or (b).

97

Among the copolymers that consist of two types of monomers (M1 and M2), those in which thesequences of M1 and M2 are not ordered are called ‘random copolymers’, while those in which the two
types of monomers are arranged alternately are called ‘alternating copolymers’. When the monomers
M1 and M2 are copolymerized, the composition and arrangement of the polymer are determined by the(relative) reactivity of M1 and M2. When the polymerization proceeds via a radical reaction, two active
species at the growth end of the polymer chain and two monomer species reacting with the active
species can be considered. There are four types of monomer-addition reactions to the growth end of
the polymer as shown in Eqs. 3.1 – 3.4. Here, it is assumed that the reaction between the growth endof polymer and the monomers M1 and M2 depends on the radical species (M1· and M2·) at the growth
end of the polymer. When the concentrations of the radically active species of M1 or M2 at the polymer
growth end and monomers M1 and M2 are expressed by [M1·], [M2·], [M1], and [M2], respectively, thereaction rates of reactions 3.1 – 3.4 are expressed as follows:
Reaction ratek11

M1 • + M 1

k12

M1 • + M 2

k21

M2 • + M 1

k22

M2 • + M 2

M1 •k11 [M1•] [M1]

(Eq. 3.1)M2 •k12 [M1•] [M2]

(Eq. 3.2)

M1 •k21 [M2•] [M1]

(Eq. 3.3)M2 •k22 [M2•] [M2]

(Eq. 3.4)

These reactions take place at the reactive chain end terminating in either monomer (M1· and M2·) with
their reaction rate constants k. The consumption rates of M1 and M2 are expressed according to Eqs.
4.1 and 4.2:––

d [M1]dt
d [M2]
dt= k11 [M1•] [M1] + k21 [M2•] [M1]

(Eq. 4.1)= k12 [M1•] [M2] + k22 [M2•] [M2]

(Eq. 4.2)

In a steady state, where the radical concentration is constant during polymerization, the rate at which
the active species M1· at the polymer growth end changes to M2· and the rate at which M2· changes to
M1· are equal. The ratio of the consumption rates of M1 and M2 under such conditions are shown in
Eq. 5 when k11 / k12 = r1 and k22 / k21 = r2. The reactivity ratio (r1 or r2) for each propagating chain end
is defined as the ratio of the rate constant for addition of a monomer of the species already at the chain
end to the rate constant for addition of the other monomer. These ratios are a measure of the tendency
for one of the two monomers to show a preference for insertion into a growing chain in which the last
inserted unit was the same, rather than the other monomer.
d [M1]
d [M2]=[M1] r1 [M1] + [M2]•
[M2] [M1] + r2 [M2]98

(Eq. 5)

3.

What kind of polymers can be obtained when r1 and r2 adopt the values shown in (1), (2), and (3)?
Among (a – g), select those that most appropriately depict the sequence and the ratio of themonomer-unit content in the polymer synthesized by mixing two different monomers in a 1:1 ratio.
When the values of fraction of M1 unit in the polymer (0 – 1.0) and the initial concentration ratioof M1 to M2 ([M1] / ([M1] + [M2]); 0 – 1.0) are plotted on the vertical and horizontal axes,
respectively, select among (A – F) those sequences that most accurately reflect the characteristicsof the polymer obtained in each system.
(1)
(2)
(3)

r1 = 1.0,
r1 = 2.0,
r1 = 0,r2 = 1.0
r2 = 0.50
r2 = 0

Sequence and ratio of the monomer-unit content in the polymer:a-------------M1-M2-M1-M2-M1-M2-M1-M2-M1-M2-----------b-------------M1-M1-M1-M1-M1-M1-M1-M1-M1-M1-----------

c -------------M2-M2-M2-M2-M2-M2-M2-M2-M2-M2----------d-------------M1-M1-M2-M1-M2-M1-M1-M1-M2-M1-----------

e -------------M2-M2-M1-M2-M1-M2-M2-M2-M1-M1-----------

4.

f-------------M1-M1-M1-M1-M1-M2-M2-M2-M2-M2-----------g-------------M1-M2-M2-M1-M1-M2-M1-M2-M2-M1-----------

What kind of polymer can be obtained when r1 = 0.50 and r2 = 0.50? When the vertical axis is thefraction of M1 in the polymer (0 to 1) and the horizontal axis is the initial concentration ratio of
M1 to M2 (0 to 1), select a suitable graph showing the characteristics of the obtained polymer.

99

Graphs for Q3 and Q4

100

5.

A copolymer that contains two types of monomers (M1 and M2) with a monomer reactivity ratioof r1 = 0.40 and r2 = 0.10 is synthesized using a radical polymerization initiator. In order tosynthesize a copolymer with an M1 content of 50 mol% at the initial stage of polymerization, howmuch M1 must be mixed with M2? Your answer should be expressed in terms of mol% of M1
relative to the two monomers.

6.

The configuration of the obtained polymer is different when a) NaOH (anionic polymerization) orb) BF3 in the presence of a small amount of water (cationic polymerization) is added as an initiator
to optically active propylene oxide. Among (A – D), select the appropriate polymer structure for
1 and 2. In addition, select among (E – H) the site where the initiator reacts during the synthesisof 1 and 2.

101

Problem 31. Total synthesis of tetrodotoxin
The ‘Fugu’ fish, which inflates and threatens when angry, has a delicious taste and is marketed as adelicacy in Japan.

However, fugu fish are also very dangerous as they contain highly toxic substances such as
tetrodotoxin (TTX). The structure of TTX has been determined, and its asymmetric total synthesis has
been accomplished.
O

-

HO

+

H 2N

O

H
N

OH O

N
H
OH

HO

Tetrodotoxin (TTX)

102

OH

1.

Compound 7 was synthesized from 1 via the synthetic route shown below. Draw the structures of
2 and 6 and clearly identify the stereochemistry where necessary. Compound 7 is the product of[3,3]-sigmatropic rearrangement from a compound with 3-hetero-atomic-1,5-diene structure asshown in the following scheme.
Z

X

Z

Y

X

Y

X = O, Y = NH

Scheme. [3,3]-Sigmatropic rearrangement reaction103

2. To synthesize 11, a hydroxy group is introduced into 7 in a site-specific and stereoselective manner.
Draw the structures of 9 and 10 and clearly identify the stereochemistry where necessary.

104

3.

More hydroxy groups and oxygen-based functional groups required for the synthesis of TTX are
introduced by epoxidation of the ring olefin, isomerization to an allyl alcohol, oxidation of theallylic position with SeO2, and another epoxidation. Draw the structures of 12, 13, 15, 17-19 in
the reaction scheme below and clearly identify the stereochemistry where necessary.

105

4.

During the synthesis of 24 from 20, a regioselective ring opening of epoxide is carried out in theconversion from 22 to 23. The total synthesis of TTX was achieved by hydrolyzing all protecting
groups under acidic conditions after the introduction of guanidine. Draw the structures of alphaketo acid derivative 21 and cyclization product 23 and clearly identify the stereochemistry where
necessary.

O

O
O

H

O

COCCl3

KMnO4
NaIO4

21

NaHCO3(C37H66Cl3NO12Si3)NH

H2O2 aq.

OAc
TESO

O

H

NaHCO3

COCCl3
NH
OH

TESO

O

CH3OH
COONa

OTES

OTES

OTES

OTES

(C37H66Cl3NO9Si3)(C34H63Cl3NNaO10Si3)

20

22

O
AcO
Cyclization

H+231) TESOTf(C28H50Cl3NO10Si2)

2) (CH3CO)2O
Pyridine

O

O
O

Cl3C

N O
H

O
O

OTES
TESH

(C36H66Cl3NO11Si3)

24

-

OH
HO

+

HN

H

H

O

N
N O
H

O
O
OH

H

H
OH

(C11H17N3O8)

TTX

106

TES

OH

Appendix
Practical Tasks

107

Task 1. Analysis of the saponification rate using a pH meter
Purpose
Although the hydrolysis of esters is very slow in itself, it is significantly accelerated by the presenceof acids or bases, which act as catalysts. A typical ester hydrolysis reaction is the saponification of fats
and oils, which produces soap. In the saponification reaction using sodium hydroxide, hydroxide ionsare consumed because the organic acids produced react with sodium hydroxide. In the saponificationof ethyl acetate using sodium hydroxide, the reaction rate, v, is proportional to the product of theconcentration of ethyl acetate, [CHK COOC/ HA ], and the concentration of hydroxide ions, [OH . ] ,according to:𝑣 = 𝑘/ [CHK COOC/ HA ] [OH . ]where k2 is the rate constant of the second-order reaction.
When ethyl acetate is present in excess relative to sodium hydroxide, the concentration of ethyl acetate
can be considered constant during the reaction. The reaction under such conditions is known as a
pseudo-first-order reaction. By defining the reaction rate constant 𝑘46 as:
𝑘46 = 𝑘/ [CHK COOC/ HA ],the reaction rate, v, can be expressed as:𝑣 = 𝑘46 [OH . ]Therefore, the reaction rate constants 𝑘46 and 𝑘/ can be obtained by examining the change in thehydroxide ion concentration over time. In this experiment, the concentration of the hydroxide ion
during the saponification of ethyl acetate is monitored using a pH meter, and the reaction rate constantis determined by analyzing the results.
Chemicals
Substance
C4H8O2Name
Ethyl acetateStateliquidGHS Codes
H225, H320, H332,
H335, H336
H315, H318, H371
Not applicable

NaOH
Sodium hydroxide aqueous solution
liquid
C6H4(COOH)(COOK) Potassium hydrogen phthalate aqueous
liquidsolution (pH = 4 standard solution)
KH2PO4
Potassium dihydrogen phosphate aqueous liquid H320
solution (pH = 7 standard solution)
Na2HPO4
Disodium hydrogen phosphate aqueous
liquid H320
solution (pH = 7 standard solution)
Sodium tetraborate aqueous solution
liquid H360
Na2B4O7・10H2O*(pH = 9 standard solution)
NaHCO3
Sodium hydrogen carbonate aqueous
liquid Not applicable
solution (pH = 10 standard solution)
Na2CO3
Sodium carbonate aqueous solution
liquid H318, H332, H335,(pH = 10 standard solution)
H336*If students use the pH = 9 standard solution, it should be prepared by the mentor. Otherwise, a pH =10 standard solution should be used.

108

Glassware and equipment
Volumetric flasks (2 pcs; 100 mL), Erlenmeyer flask with stopper (2 pcs), volumetric pipette (2 pcs;
25 mL), volumetric pipette (1 pc; 2 mL), pipette (1 pc), stir bar (1 pc), stopwatch (1 unit), pH meter (1unit), water bath (1 unit), thermometer (1 unit), stirrer (1 unit), precision electronic balance (1 unit)[measurement accuracy: 0.1 mg].
Experimental procedure
Preparation of an aqueous solution of ethyl acetate
1. Add ion-exchanged water (ca. 50 mL) to a volumetric flask (100 mL) and weigh it on the precisionelectronic balance.
2. Add approximately 90 mg of ethyl acetate to the ion-exchanged water in the volumetric flask, andimmediately close the flask with a stopper.
3. Record the mass of ethyl acetate based on the mass increase.
4. Add ion-exchanged water to the marked line to increase the volume to 100 mL and mix it.
5. Using a volumetric pipette, transfer 25 mL of the aqueous solution of ethyl acetate into an
Erlenmeyer flask. Place the stir bar, close the flask with a stopper, and place it in a water bath at
30 °C to keep the temperature of the liquid constant.
Preparation of an aqueous solution of NaOH
1. Add ion-exchanged water (ca. 50 mL) to a volumetric flask (100 mL).
2. Using a volumetric pipette, transfer 2 mL of a 0.05 mol L-1 aqueous solution of NaOH to thevolumetric flask, and then add ion-exchanged water to the marked line to increase the volume to
100 mL.
3. Using a volumetric pipette, transfer 25 mL of the aqueous solution of NaOH into an Erlenmeyer
flask, close it with a stopper, and place the flask in a water bath at 30 °C to keep the temperatureof the liquid constant.
Calibration of the pH meter
Calibrate the pH meter using three different pH standard solutions.
Saponification
1. Begin stirring the aqueous solution of ethyl acetate in the Erlenmeyer flask in the water bath.
2. Quickly add the aqueous solution of NaOH (25 mL), close the flask with a stopper and starttiming with a stopwatch.
3. After 3 minutes, insert the thermometer and the electrodes of the pH meter.
4. Measure the temperature and the pH value of the solution every 5 minutes for 60 minutes.

109

Discussion
Relationship between the pH value and the reaction rate-[I9$ ]The transformation and definite integration of 𝑣 = −= 𝑘46 [OH . ] from t = 0 to t = t gives:-f− ln[OH . ]
= 𝑘46 𝑡[OH . ](

where [OH . ]( is the initial concentration of sodium hydroxide, and [OH . ] is the concentration of
sodium hydroxide at time t. Using the common logarithm, the equation will be:[OH . ]
1− log=
× 𝑘46 𝑡.
[OH . ](
2.303
Furthermore, this equation can be transformed to give:− log[OH . ] + log[OH . ]( =1× 𝑘46 𝑡2.303

(a)Since [OH . ] = pH − 13.833 at 30 °C, equation (a) can be rewritten as:

or:

𝑘46
(pH)( − pH =×𝑡
2.303
pH = −

𝑘46
× 𝑡 + (pH)( .
2.303

Here, (pH)( and pH are the pH value of the solution before the reaction and at time t, respectively.

110

Analysis
1. Calculate the concentration of the prepared aqueous solution of ethyl acetate (C1) and theconcentration of the diluted aqueous solution of NaOH (C2).
C1 [mol L-1]

C2 [mol L-1]

2.

Write the equation for the saponification of ethyl acetate with NaOH.

3.

Record the temperature and the pH value of the solution.
Time(minutes)

Temperature
(°C)

Time
(minutes)pH

111

Temperature
(°C)

pH

4.

Plot the time on the horizontal axis, the pH value on the vertical axis, and determine 𝑘46 from theslope. Indicate the units.

5.

Determine 𝑘/ . Given that the concentrations of the two aqueous solutions are low, it can beassumed that the volume after mixing is equal to the sum of the volumes of the two solutions.

6.

The time when the concentration of the substrate is half of its initial concentration is called thehalf-life (time of half decay), which is represented using the symbol 𝑡( . Determine 𝑡( .
&

112

&

Task 2. Simultaneous acid–base titration
Introduction
Sodium carbonate, which is a raw material for glass, is produced by absorbing ammonia into an
aqueous solution of sodium chloride and subsequent thermal decomposition of the sodium hydrogencarbonate obtained after passing carbon dioxide. Sodium carbonate can also be obtained by reacting
carbon dioxide with an aqueous solution of sodium hydroxide, which is an important chemical for the
production of soap, paper, and fibers.
In this experiment, a differential titration (Warder method) is used to simultaneously determine theconcentrations of two electrolytes with different basicity. Specifically, the mass of the electrolytes ina mixed aqueous solution containing sodium hydroxide (pKa > 13) and sodium carbonate (pKa1 = 6.35,
pKa2 = 10.33) will be determined using two pH indicators.
Chemicals
Substance
HCl

Name
Hydrochloric acid

State
Aqueous solution

C20H14O4

Phenolphthalein

Ethanol solution

C14H14N3NaO3S Methyl orange
NaOH
Sodium hydroxide
Na2CO3
Sodium carbonate

Aqueous solution
Aqueous solution
Aqueous solution

GHS Codes
H301, H314, H318, H330, H331,
H334, H370, H372, H400
H225, H320, H335, H336, H341,
H351, H372, H373
H314, H318, H370, H402
H332, H318, H335, H336

Glassware and Equipment- 1 Volumetric pipette, 20 mL with pipette filler- 1 Burette, 25 mL- 2 Erlenmeyer flasks with stopper (HCl standard solution, unknown alkali solution), 300 mL- 1 Plastic container (for solution-state chemical waste), 1 L- 2 Pipettes and pipetting bulbs (for indicators a and b)- 5 Titration flasks, 100 mL- 1 Erlenmeyer flask (for transferring the HCl standard solution to the burette), 50 mL- 1 Transfer funnel- 1 Laboratory stand with burette clamp
Chemicals- 0.1 mol L-1 HCl standard aqueous solution- Unknown alkali solution containing NaOH and Na2CO3- Indicator a: Phenolphthalein ethanolic solution- Indicator b: Methyl orange aqueous solution- Deionized water

113

Experimental procedure(1) Transfer 20.00 mL of the unknown alkali aqueous solution into a 100 mL titration flask using a
20 mL volumetric pipette. Add indicator a, which will turn the solution reddish purple. Titrate
the unknown alkali solution by adding the HCl standard solution in the burette. When the alkali
solution turns pale pink, you have reached the first equivalence point. Record the volume (Va) in
your notebook.
(2) Add indicator b to the now pale-pink aqueous solution. The color of the solution will turn yellow.
Continue the titration using the HCl standard solution until the aqueous solution turns lightorange, where you have reached the second equivalence point. Record the volume (Vb; volume
from the first equivalence point to the second equivalence point).
Repeat procedures (1) and (2) if necessary.
Results
Titration Step (1)No.

Vfinal (mL)

Vinitial (mL)

V (mL)

Vinitial (mL)

V (mL)

1
2
3
4
5
Accepted volume Va
Titration Step (2)No.
Vfinal (mL)
1
2
3
4
5
Accepted volume Vb

Questions
1. Write the two chemical equations for the neutralization reactions during titration step (1).

2.

Write the chemical equation for the neutralization reaction during titration step (2).

114

3.

From the results of the above titration operation, calculate the weights (g) of NaOH and Na2CO3
contained in 1 L of the unknown alkali aqueous solution.
Weight (g)NaOH
Na2CO3Calculation:

4. Sketch the pH titration curve expected for the unknown alkali solution.

115

Task 3. Synthesis and analysis of a cobalt(III) oxalate complex
Cobalt is a silver-white ferromagnetic metal that belongs to the elements of the iron group. As it is lesseasily oxidized than iron as well as less corroded by acids and bases, alloys of cobalt with nickel,
chromium, molybdenum, and other metals are used in equipment that requires wear resistance, suchas cutting tools and high-grade materials such as scissors. Cobalt is also one of the trace biologicalessential elements because it is the central metal ion of vitamin B12. The most common oxidation
states of cobalt are +2 and +3; cobalt ions in these oxidation states form metal complexes with a hexacoordinated octahedral structure. In this problem, a cobalt(III) oxalate complex in which oxalate ionsare coordinated to a cobalt ion is synthesized. Cobalt(II) chloride is oxidized with hydrogen peroxide
and then treated with oxalic acid to obtain a green cobalt(III) oxalate complex. Below, three possible
formulae for this complex are shown:
-

3O
O

O
O

O

O

Co
O

O

OH2

Co

O

O

OH2
O

O

OH2
O
Co

O

O
O

O

O
O

O

O

OH2
OH2

O
Cl

O

O
K3[Co(ox)3]•3H2O

K[Co(ox)2(H2O)2]•3H2O[Co(ox)Cl(H2O)3]•3H2O

2O

O= oxalate (ox2-)OO

The number of oxalate ligands coordinated to the cobalt(III) oxalate complex synthesized in thisexperiment can be determined by titration with an aqueous potassium permanganate solution after
decomposing the metal complex with an acid. The cobalt(III) oxalate complex is photosensitive andgradually decomposes from a green to a yellowish-brown product when exposed to visible orultraviolet light. Thus, care must be taken when handling the sample in solution. In this experiment,
this effect can be ignored if the measurement is performed immediately after the solution is prepared.

116Chemicals
Substance[CoCl2]·6H2O
H2SO4 (aq), 6 M
K2C2O4·H2O
Na2C2O4
H2O2 (aq), 30 %C2H5OHKMnO4 (aq), ca. 0.02 M
CName
Cobalt(II) chloride
hexahydrate
Sulfuric acid
Potassium oxalatemonohydrate
Sodium oxalate
Hydrogen peroxide
solution
Ethanol
Potassium
permanganate
Activated carbon

State
Solid
AqueoussolutionSolid

GHS Codes
H315, H319, H301, H334, H341,
H351, H361, H335, H317
H314, H318, H330, H402, H370,
H372
H301, H312

Solid
Aqueoussolution
LiquidH319
H272, H314, H318, H302, H312,
H331, H351, H401, H370, H372
H225, H320

Aqueoussolution
PowderH272, H314, H318, H302, H341,
H361, H335, H400, H410, H372
NA

Glassware and equipment- Erlenmeyer flask, 100 mL (2), 50 mL (1), 25 mL (3)- Valves for Pasteur pipettes and rubber pipettes- Hot plate- Graduated cylinder, 25 mL- Bath- Ice bath
- Conical funnel and filter paper- Glass filter for vacuum filtration- Vacuum filtration set (Stand, flask-fixing clamp, aspirator, filtration flask or bottle, rubber adapter
used for filtration)- Burette (25 mL) and stand- Small funnel to transfer the solution to the burette117Procedure
A. Synthesis of the cobalt(III) oxalate complex
1. Add 2.4 g of potassium oxalate monohydrate and 5 mL of water to a 100 mL Erlenmeyer flask,
and heat the mixture to 70 °C in order to dissolve the solid.
2. In another 100 mL Erlenmeyer flask, add 1 g of cobalt(II) chloride hexahydrate followed by 3 mLof water to dissolve it. Then, add ca. 0.02 g of activated carbon powder followed by 1.0 mL ofthe 30% hydrogen peroxide solution.
3. Add the solution from step 1 to the solution from step 2 and heat the mixture while stirring in awater bath at 70 °C; the mixture will begin to foam. After further heating and stirring for about 15
minutes, the foaming will subside, and the red color of the solution will gradually turn to a darkgreen.
4. Remove the activated carbon via vacuum filtration and wash it with a small amount of water toobtain a dark green solution. Transfer the filtrate to a 50 mL Erlenmeyer flask.
5. Add 10 mL of ethanol to the filtrate. A green precipitate will form. To aid the precipitation, cool
the flask in an ice bath for 0.5 h.
6. Separate the precipitate using vacuum filtration and wash it with a small amount of a water–
ethanol (50/50, v/v) solution.
7. Allow the crystals to air-dry or dry them between two sheets of filter paper.
8. Weigh a clean empty vial (sample container) and place the dry crystals in the container todetermine the weight of the resulting cobalt oxalate complex crystals.
B. Analysis of the cobalt oxalate complex
B-1.Standardization of the potassium permanganate aqueous solution (ca. 0.02 M)
Transfer the aqueous potassium permanganate solution (ca. 0.02 M) to a 25 mL burette.
Accurately weigh out ca. 50 mg of sodium oxalate into a 100 mL Erlenmeyer flask, and then add
20 mL of water and 5 mL of 6 M sulfuric acid. Warm the flask in a hot water bath toapproximately 80 °C. Titrate this oxalate solution with the aqueous potassium permanganate
solution. When the mixture turns light pink and the color is maintained for about 1 minute, theend point of the titration has been reached. Record the amount of potassium permanganate
aqueous solution required for the titration and determine its molar concentration.
B-2. Analysis of the cobalt oxalate complex
1. Accurately weigh out ca. 20 mg of the cobalt oxalate complex synthesized in A and place it in a
100 mL Erlenmeyer flask. To this Erlenmeyer flask, add 20 mL of water and 5 mL of 6 M sulfuric
acid. Warm the mixed solution in the Erlenmeyer flask in a hot water bath maintained at ca. 80 °C.
2. Titrate the solution in the warm Erlenmeyer flask with the aqueous solution of potassium
permanganate standardized in B-1. The end point is determined in the same way as above. Record
the amount of the aqueous potassium permanganate solution required for the titration.

118

Questions
1. Write the values of the following data obtained in the experiment.
1-1. Weight of the crystals of the complex obtained in A.

g

1-2 Weight of sodium oxalate used in B-1.

g

1-3 Volume of the aqueous potassium permanganate solution (accepted) used in the titration in B-1.
mL
1-4 Concentration of the aqueous potassium permanganate solution in B-1.
mol L-1
1-5 Weight of cobalt oxalate complex analyzed in B-2.

g

1-6 Volume of the potassium permanganate aqueous solution (accepted) used in the titration in B-2.
mL
2.

Write the chemical equation of the reaction used in B-1.

3.

Find the ratio (% by weight) of the oxalate ion relative to the cobalt oxalate complex. Then,assuming that the synthesized cobalt oxalate complex has one of the following compositions,determine the composition ratio of the cobalt ion to the oxalate ligand.
- tris complex: K3[Co(C2O4)3]·3H2O- bis complex: K[Co(C2O4)2(H2O)2]·3H2O- mono complex: [Co(C2O4)Cl(H2O)3]·3H2O
Weight% of oxalate ion
Cobalt ion to oxalate ligand composition ratio

4.

%
:

.

Calculate the yield of the complex based on the raw cobalt material.
Yield of complex

119%

Task 4. Hinokitine: synthesis of a deep-red-colored natural productNon-benzenoid aromatic compounds based on seven-membered ring systems have been widely
investigated for many years both with respect to fundamental and applied chemistry. One suchcompound, 4-isopropyltropolone, is known as ‘hinokitiol’, as it had originally been isolated from anessential oil of the Chamaecyparis obtuse var. formosana (Taiwan Hinoki) tree. Hinokitiol is used ine.g. cosmetics, sunscreens, and oral care products. Recently, it has also been reported that hinokitiol
can restore iron transport into, within, and/or out of cells, and thus may have potential in treating some
genetic diseases.
O
OH

Hinokitiol

The iron complex of hinokitiol, which is known as ‘hinokitine’, is found in the heartwood of Taiwan
Hinoki, Thuja dolabrata, and Thujopsis dolabrata. In this task, you will synthesize hinokitine startingfrom hinokitiol and iron(III) nitrate nonahydrate.

Chemicals
Substance
C10H12O2
Fe(NO3)3·9H2O
C2H5OH
CH3COOC2H5
SiO2

Name
Hinokitiol
Iron(III) nitratenonahydrate
Ethanol
Ethyl acetate
Silica gel
Sea sand

State
Solid
Solid

GHS Codes
H302, H361
H272, H315, H319, H402, H412

LiquidLiquidSolid
Solid

H225, H320
H225, H320, H332, H335, H336
H351, H402

120

Glassware and equipment- 2 Vials (20 mL)- 1 Spatula- 1 Weighing balance (0.001 g)- 2 Weighing papers- 1 Magnetic stirrer- 1 Magnetic stir bar (small)- 4 Pasteur pipettes- 1 Filter funnel (small)- 1 Filter paper- 1 Filtering flask- 1 Aspirator- 1 Laboratory stand- 1 Bosshead and clamp- 1 Column tube with stopcock (inner diameter: 12 mm)- 1 Beaker (100 mL)- 2 Erlenmeyer flasks (100 mL)- 2 Funnels- Absorbent cotton- 1 Glass rod- 1 Eggplant flask- 1 Rotary evaporator
Procedure
Synthesis of hinokitine
1. Charge a 20 mL vial (vial A) with hinokitiol (50 mg), iron(III) nitrate nonahydrate (25 mg),ethanol (0.8 mL), and a small magnetic stir bar.
2. Vigorously stir the reaction mixture at room temperature for 50 minutes.
3. Collect the product by vacuum filtration using a small filter funnel. Transfer all the contents ofvial A to the filter funnel.
4. Wash the product with a small amount of ethanol using a Pasteur pipette.
5. Let the product dry on the filter funnel by continuing suction for more than 10 minutes.
6. Transfer the product to a 20 mL vial (vial B).
Purification of hinokitine by column chromatography on silica gel
Packing the silica gel column
1. Loosely pack a small plug of cotton at the bottom of the column tube using a glass rod.
2. Clamp the column tube to a laboratory stand. Add sea sand to fill the curved part of the column
tube.
3. Add silica gel (5 g) and ethyl acetate (30 mL) as the eluent to a 100 mL beaker.
4. Carefully pour the silica gel slurry into the column tube using a funnel.
5. Rinse off any silica gel adhered to the side of the column tube with a small amount of the eluent.
6. Tap on the side of the column tube to help the silica gel to settle uniformly.

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Loading the sample onto the column
7. Dissolve the product in vial B by adding a small amount of ethyl acetate.
8. Open the stopcock of the column tube and reduce the eluent level to the top of the silica gel. Be
sure that the silica gel is always covered with the eluent.
9. Close the stopcock and load the solution in vial B onto the top of the silica gel using a Pasteur
pipette.
10. Open the stopcock and reduce the solution level to the top of the silica gel.
11. Close the stopcock, rinse vial B with a small amount of the eluent and load the washings onto the
top of silica gel using the Pasteur pipette.
12. Open the stopcock and reduce the solution level to the top of the silica gel.
13. Repeat steps 11 and 12 one or two more times.
14. Close the stopcock and add a small amount of the eluent to the column using a Pasteur pipette.
Open the stopcock and add more eluent to the column.
Eluting and collecting the sample
15. Collect the colored fraction in a 100 mL Erlenmeyer flask. Add more eluent to the column ifnecessary.
16. Weigh a 100 mL eggplant flask and record the value.
17. Transfer the solution in the Erlenmeyer flask to the eggplant flask using a funnel.
18. Evaporate the solvent in the flask using a rotary evaporator.
19. Submit the flask for the evaluation.
Using the above procedures, hinokitine should be obtained as a deep red solid in more than 80% yield.
Questions
1. Calculate the theoretical yield of hinokitine.

2.

Write the side product other than hinokitine.

3.

Draw the structures of hinokitine. Explain how many isomers, including enantiomers, arepossible for hinokitine.

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Task 5. Functionalization of a seven-membered ring: synthesis of tropolone tosylate
Tropolone is a derivative of tropone (cyclohepta-2,4,6-trienone) wherein a hydroxy group is bound tothe carbon atom adjacent to the ketone group on the conjugated seven-membered ring. Tropone,
tropolone, and their derivatives constitute an important class of organic cyclic compounds that havearomatic character due to the contribution of the tropylium ion structure with six π-electrons. The
natural product hinokitiol (4-isopropyltropolone) is a typical tropolone-type compound derived froman essential oil of the Hinoki tree family, and shows effective antibacterial and antimicrobial properties.

In this task, you will synthesize tropolone tosylate by the reaction of tropolone with 4dimethylaminopyridine (DMAP) followed by tosyl chloride (TsCl). Tropolone tosylate iscommercially available and useful for the construction of fused ring systems such as ‘azulene’
compounds.

Chemicals
Substance
C7H6O2(CH3)2NC5H4N
C2H5OH
CH3C6H4SO2Cl

Name
Tropolone
4-Dimethylamino-pyridine (DMAP)
Ethanol
Tosyl Chloride (TsCl)

Glassware and equipment- 2 Vials (20 mL)- 1 Spatula- 1 Weighing balance (0.001 g)- 2 Weighing papers- 1 Magnetic stirrer- 1 Magnetic stir bar (small)- 2 Pasteur pipettes- 1 Filter funnel (small)- 1 Filter paper- 1 Filtering flask- 1 Aspirator
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State
Solid
Solid
LiquidSolid

GHS Codes
H315, H319, H301, H371
H225, H320
H314, H315, H318

Procedure: Synthesis of tropolone tosylate
1. Charge a 20 mL vial (vial A) with tropolone (125 mg), ethanol (1.0 mL), and a small magnetic
stir bar.
2. Slowly add 4-dimethylaminopyridine (DMAP; 126 mg) to vial A while stirring the reaction
mixture.
3. After the addition of DMAP, stir the reaction mixture at room temperature for another 10 minutes.
4. Slowly add tosyl chloride (TsCl; 196 mg) while stirring the reaction mixture.
5. After the addition of TsCl, stir the reaction mixture at room temperature for another 50 minutes.
6. Collect the product by vacuum filtration using a small filter funnel. Transfer all the contents ofvial A to the filter funnel.
7. Wash the product with a small amount of ethanol using a Pasteur pipette.
8. Let the product dry on the filter funnel by continuing suction for more than 10 minutes.
9. Weigh a 20 mL vial (vial B) and record the value.
10. Transfer the product to vial B for submission.
Using the above procedures, almost pure tropolone tosylate should be obtained as a light-brown solidin about 30% yield, which can be confirmed by 1H NMR spectroscopy.
Questions
1. Draw the structure of the reaction intermediate resulting from the reaction between tropolone and4-dimethylaminopyridine (DMAP). Explain the role of DMAP in this reaction.
2. Describe a procedure for the recovery of DMAP after the tosylation of tropolone.
3. What tosylation product(s) would be obtained from the above reaction if hinokitiol was usedinstead of tropolone? Draw the structure(s) of the tosylation product(s) of hinokitiol.

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Task 6. Hydrolysis of polyethylene terephthalate: A small experiment for us, but a giant leap
toward a more sustainable society
Poly(ethylene terephthalate) (PET), the raw material for PET bottles, is one of the most common
plastics. In recent years, growing concern about the environmental impact of plastic waste has
prompted the development of efficient technologies to recycle plastics. Chemical recycling, which isbased on decomposing plastic into its raw monomer materials and reusing the recycled monomers forthe production of new plastics, is an important technology because it allows the production of recycledplastics with high purity. This task explores the chemical recycling of plastics through an experiment
involving the hydrolysis of PET.

Materials- A strip of PET cut from a PET bottle, 1.00 g- Sodium hydroxide, 3.0 g- Ethanol, 50 mL
Substance(C10H8O4)n
NaOH
C2H6O

Name
Poly(ethylene terephthalate)
Sodium Hydroxide
Ethanol

State
Solid
Solid
LiquidGHS Codes
H314, H318, H370
H225, H320

Glassware and equipment- Laboratory stand with clamps- Magnetic stirrer and a stir bar- Oil bath with a temperature controller- Round-bottom flask (100 mL): A wide-mouthed flask is preferred.
- Round-bottom flask (20 mL)- Reflux condenser and rubber tubes for supplying water as the coolant- Graduated cylinder (100 mL)- Beaker (100 mL)- Buchner funnel and filter paper- Vacuum flask (100 mL) and rubber adapter to connect the Buchner funnel- Aspirator for vacuum filtration. An alternative such as a dry pump can be used.
- Pipette (10 mL)- Petri dish- Tweezers to handle the PET strip- Weighing balance

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Experimental procedure
Hydrolysis of PET
1. Heat the oil bath to 100 °C.
2. Add 5.0 g of sodium hydroxide and 30 mL of ethanolic to a 100 mL round-bottom flask in orderto prepare an ethanol solution of sodium hydroxide.
3. Weigh a strip of PET and place it in the flask.
4. Connect the reflux condenser to the flask and run water. Immerse the bottom of the flask in theoil bath to start heating and refluxing. After 30 minutes, lift the flask from the oil bath.
5. Remove the residual PET from the flask using tweezers and wash it with a small amount of
ethanol. Place the PET strip on a filter paper, dry it under air, and then weigh it.
6. Filter the suspension remaining in the flask using vacuum filtration. Wash the solid on the filter
paper with a small amount of ethanol and dry it by continuing suction to obtain crude crystals ofcompound A shown in the reaction scheme.
7. Transfer the crude crystals of A on the filter paper to a petri dish and weigh them.
Recrystallization of A
1. Place the crude crystals of A in a 20 mL round-bottom flask and add 2 mL of water. Dissolve
the crude crystals by heating the mixture to 100 °C using an oil bath. If necessary, add 0.5 mLof water to dissolve all the crude crystals.
2. Allow the solution to cool to room temperature. Collect the precipitated crystals of A via
vacuum filtration, wash them with a small amount of ethanol and dry them with continuing
suction.
3. Transfer the crystals of A to a petri dish and weigh them.
Questions
1. Draw the structural formulas of A and B.
2. Calculate the yield of A based on the amount of PET used.
3. Calculate the yield of A based on the amount of PET reacted.
4. Draw a reaction mechanism for the cleavage of ester bonds in the reaction of PET and NaOH.
5. This reaction is irreversible. Describe the reasons for this.
6. Which would be more efficient: The hydrolysis of PET conducted in this experiment or thehydrolysis of the polyamide shown below under the same conditions? The reasons for youranswer should be also described.

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Task 7. Separation of blue and red components from a green mixture
In 1903, Tswett reported the first example of chromatography: Leaves were ground and developed on
filter paper to separate pigments such as chlorophyll, carotenes, and xanthophylls. The term
chromatography is derived from the Greek chroma, which means ‘color’, and graphein, which means‘to write’. While chromatography was initially applied to colored compounds, nowadays, it is also
applied to colorless compounds.
Chromatography is a very powerful method to separate the individual constituents of compound
mixtures. In this task, chromatography is compared to recrystallization, which is another popular
separation/purification method.
For that purpose, we will use a mixture of guaiazulene and 4-(phenylazo)phenol (henceforth denotedas the ‘mixture’) in this task. Guaiazulene is a natural pigment that is found in some essential oils.
Given the special electronic structure of the azulene skeleton (vide infra), guaiazulene exhibits a deep
blue color, which stands in stark contrast to other isomeric naphthalene derivatives. The second
component of our mixture, 4-(phenylazo)phenol, is a product of one of the simplest diazo coupling
reactions. Conjugated azo compounds represent an important class of chromophores, and various azo
dyes have found applications in e.g. the textile and food industry.
In this task, considering the historic origins of ‘chromatography’, we will separate the two individualcolored compounds from the ‘mixture’.

N N

OH

4-(phenylazo)phenol

azuleneguaiazulene

Chemicals
Substance
C15H18C6H5N=NC6H4OH
C2H5OH
C6H14
CH3COOC2H5
SiO2

Name
Guaiazulene
4-(Phenylazo)phenol
Ethanol
HexaneEthyl acetate
Silica gel
Sea sand

State
Solid
Solid
LiquidLiquidGHS Codes
H302
H302, H315, H319
H225, H320
H225, H315, H319, H361, H335
H336, H304, H401, H372
Liquid H225, H320, H332, H335, H336
Solid H351, H402
Solid

SamplesA ‘mixture’ for the recrystallization; ca. 1:5 (w/w) mixture of guaiazulene and 4-(phenylazo)phenol.
A ‘mixture’ for the column chromatography; ca. 1:1 (w/w) mixture of guaiazulene and 4-(phenylazo)phenol.
Solutions
A solution for the recrystallization; 1:1 (v/v) mixture of ethanol and hexane.
A solution to charge the sample; 1:1 (v/v) mixture of hexane and ethyl acetate.
Second eluent; 4:1 (v/v) mixture of hexane and ethyl acetate.

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Glassware and equipment- 2 Short test tubes- 1 Test tube rack
- 5 Pasteur pipettes- 3 Spatulas- 1 Weighing balance (accuracy: 0.001 g)- 1 Hot water bath
- 1 Ice bath
- 1 Graduated cylinder (50 mL)- 1 Graduated cylinder (5 mL)- 1 Filter funnel (small)- 1 Filter paper- 1 Filtering flask- 1 Aspirator- 1 Petri dish- 1 Laboratory stand- 1 Clamp- 1 Glass column with stopcock (inner diameter: 12 mm)- 1 Vial tube (4 mL)- 4 Erlenmeyer flasks (100 mL)- 1 Erlenmeyer flask (50 mL)- 3 Funnels- Cotton wool
- 1 Glass rod- 2 Round-bottom flasks (100 mL)- 1 Rotary evaporator
Procedure
Recrystallization of the ‘mixture’
1. Weigh out a test tube and place ca. 0.5 g of the ‘mixture’ for the recrystallization into it; weighthe test tube again.
2. Heat the water bath to 80 °C.
3. Add 1.0 mL of the 1:1 (v/v) mixture of ethanol and hexane to the test tube. Clamp the test tube
and heat it in the water bath until all of the ‘mixture’ has dissolved.
4. Allow the green solution to cool to room temperature. After crystals have formed, further cool thetest tube with iced water.
5. Place the filter funnel on the filtering flask and insert the filter paper. Connect the filtering flask
to the aspirator and start the suction. Wet the filter paper with a small amount of ethanol and press
it to the filter funnel using a spatula.
6. Add the 1:1 mixture of ethanol and hexane to another test tube and cool it with iced water.
7. Transfer all of the precipitate onto the filter funnel with the cold ethanol-hexane mixture. Wash
the crystals with a small amount of the cold ethanol-hexane mixture and dry them using continued
suction.
8. Weigh the empty petri dish and transfer the product to the petri dish; weigh the petri dish again.

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Separation of the ‘mixture’ by column chromatography on silica gel
Packing the silica gel column
1. Use a glass rod to loosely insert a small plug of cotton wool into the small glass tube just before
the stopcock inside the column.
2. Clamp the column onto a laboratory stand. Add sea sand to fill the curved part on the bottom of
the column.
3. Place silica gel (20 g) in an Erlenmeyer flask (50 mL), add hexane (30 mL) as the eluent, and stir
the mixture to make a slurry.
4. Carefully pour the slurry into the column using a funnel.
5. Rinse off any silica gel adhered to the side of the column with a small amount of hexane.
6. Use something soft to gently tap on the side of the column to help the silica gel to settle uniformly.
Loading the sample onto the column
7. Place ca. 0.2 g of the ‘mixture’ for the column chromatography in a vial tube and weigh it.
Dissolve the ‘mixture’ by adding a small amount of the 1:1 (v/v) mixture of hexane and ethyl
acetate.
8. Place an Erlenmeyer flask under the column. Open the stopcock of the column and lower theeluent level to reach the top of the silica gel. Be sure that the silica gel is always covered by theeluent.
9. Close the stopcock and gently load the green solution in the vial onto the top of the silica gel usinga Pasteur pipette.
10. Open the stopcock and lower the solvent level to reach the top of the silica gel.
11. Close the stopcock, rinse the vial with a small amount of the 1:1 hexane-ethyl acetate mixture,
and add the washings onto the top of silica gel using the Pasteur pipette.
12. Open the stopcock and lower the solvent level to reach the top of the silica gel.
13. Repeat steps 11 and 12 one or two more times.
14. Close the stopcock, place sea sand on top of the silica gel, and add hexane to the column.
Eluting and collecting the sample
15. Open the stopcock; after collecting the first colorless fraction, change to the next Erlenmeyer
flask to collect the first colored fraction. Add more hexane to the column if necessary.
16. After collecting the first colored fraction, change the Erlenmeyer flask to collect the subsequentcolorless eluate. Add the 4:1 (v/v) mixture of hexane and ethyl acetate as the second eluent to thecolumn and continue collecting.
17. After collecting the colorless eluate, change the Erlenmeyer flask to collect the second colored
fraction. Add more of the 4:1 hexane-ethyl acetate mixture to the column if necessary. After
collecting the second colored eluate, close the stopcock.
18. Weigh the empty round-bottom flasks.
19. Transfer each colored fraction from the Erlenmeyer flasks to individual round-bottom flasks
using a funnel (use a clean funnel for each fraction).
20. Remove the solvent(s) from each flask using a rotary evaporator.
21. Submit the flask for evaluation.

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Questions
1. Name the constituent obtained from recrystallization.
2. Name the constituent in each colored fraction obtained from column chromatography.
3. Explain why the recovery from recrystallization is almost never quantitative.
4. Chromatography is based on the interaction of the eluite (eluted molecules) with the stationary
phase (in this case: Silica gel). Which molecule engages in stronger interactions with silica gel,guaiazulene or 4-(phenylazo)phenol? Name the functional group and the type of interaction thatcharacterizes the affinity.

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53rd IChO2021 Preparatory Problem version 2.5
Edited by Nobuhiro Kihara, Kanagawa University
version 2.0: Issued at 1st February, 2021.
version 2.1: Issued at 15th February, 2021.
version 2.2: Issued at 15th March, 2021.
version 2.3: Issued at 29th March, 2021.
version 2.4: Issued at 7th April, 2021.
version 2.5: Issued at 8th June, 2021.
final version (3.0): Issued at 12th July, 2021.
Copyright © 2021 by The Organizing Committee of 53rd IChO 2021 Japan