USA-4 C-0 C-1

USA-4 C-0 C
Yitian Zhu

IChO
General instructions
Cover sheet

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IChO
USA-4 C-0 G-1

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

G0-1

USA (United States of America)IChO
USA-4 C-0 G-2

G0-2

USA (United States of America)

General Instructions• You are allowed to use pen only to write your answers.
• Your calculator must be non-programmable.
• This examination has 9 problems.
• You can solve the problems in any order.
• You will have 5 hours to solve all problems.
• You can begin working only after the START command is given by the invigilator.
• All results must be written in the appropriate answer boxes with pen on the answer sheets. If you
need scratch paper, use the back of the question sheets. Remember that answers written outside
the answer boxes will not be graded.
• Write relevant calculations in the appropriate boxes when necessary. Work must be shown to receive full credit for correct answers.
• The invigilator will announce a 30-minute warning before the STOP command.
• You must stop working when the STOP command is given. Failure to stop writing will lead to thenullification of your examination.
• The official English version of this examination is available by request for clarification only.
• You are not allowed to leave your seat without permission from the invigilator. If you need any
assistance (broken calculator, need to use a restroom, etc.), raise your hand to get the attention of
the invigilator.
GOOD LUCK!

Problems and Grading Information
Title

Total Score

Percentage

1

Hydrogen at a Metal Surface

24

11

2

Isotope Time Capsule

35

11

3

Lambert–Beer Law?

22

8

4

The Redox Chemistry of Zinc

32

11

5

Mysterious Silicon

60

12

6

The Solid-State Chemistry of Transition Metals

45

13

7

Playing with Non-benzenoid Aromaticity

36

13

8

Dynamic Organic Molecules and Their Chirality

26

11

9

Likes and Dislikes of Capsules

23

10

Total

100

IChO
USA-4 C-0 G-3

G0-3

USA (United States of America)

Physical Constants and Equations
Constants
Speed of light in vacuum𝑐 = 2.99792458 × 108 m s−1

Planck constantℎ = 6.62607015 × 10−34 J s

Elementary charge

𝑒 = 1.602176634 × 10−19 C

Electron mass𝑚e = 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 constant𝑘B = 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 unit𝑢 = 1 Da = 1.66053907 × 10−27 kg

Standard pressure

𝑝 = 1 bar = 105 Pa

Atmospheric pressure𝑝atm = 1.01325 × 105 Pa

Zero degree Celsius

0 ∘ C = 273.15 KÅngstrom

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 = 10−9Part-per-trillion

1 ppt = 10−12pi

𝜋 = 3.141592653589793

The base of the natural logarithm(Euler's number)𝑒 = 2.718281828459045

IChO
USA-4 C-0 G-4

G0-4

USA (United States of America)

Equations
The ideal gas law

Coulomb's law

𝑃 𝑉 = 𝑛𝑅𝑇
, where 𝑃 is the pressure, 𝑉 is the volume, 𝑛 is the amount of substance,
𝑇 is the absolute temperature of ideal gas.
𝑞 𝑞
𝐹 = 𝑘e 1 2 2𝑟, where 𝐹 is the electrostatic force, 𝑘e (≃ 9.0 × 109 N m2 C−2 ) is Coulomb's
constant, 𝑞1 and 𝑞2 are the magnitudes of the charges, and 𝑟 is the distance
between the charges.

The first law of thermodynamics

Δ𝑈 = 𝑞 + 𝑤, where Δ𝑈 is the change in the internal energy, 𝑞 is the heat supplied, 𝑤
is the work done.

Enthalpy 𝐻

𝐻 = 𝑈 + 𝑃𝑉

Entropy based on Boltzmann's principle 𝑆

𝑆 = 𝑘B ln 𝑊
, where 𝑊 is the number of microstates.
𝑞
Δ𝑆 = rev𝑇, where 𝑞rev is the heat for the reversible process.

The change of entropy
Δ𝑆
Gibbs free energy 𝐺

𝐺 = 𝐻 − 𝑇𝑆
Δr 𝐺∘ = −𝑅𝑇 ln 𝐾 = −𝑧𝐹 𝐸 ∘, where 𝐾 is the equilibrium constant, 𝑧 is the number of electrons, 𝐸 ∘ isthe standard electrode potential.

Reaction quotient 𝑄

Δr 𝐺 = Δr 𝐺∘ + 𝑅𝑇 ln 𝑄
For a reaction
𝑎A + 𝑏B ⇌ 𝑐C + 𝑑D𝑐𝑑[C] [D]𝑄=𝑎𝑏[A] [B], where [A] is the concentration of A.

IChO

G0-5

USA-4 C-0 G-5

Heat change Δ𝑞
Nernst equation for redox reaction
Arrhenius equation

USA (United States of America)Δ𝑞 = 𝑛𝑐m Δ𝑇, where 𝑐m is the temperature-independent molar heat capacity.
𝐶
𝑅𝑇
ln ox
𝐸 = 𝐸∘ +𝑧𝐹
𝐶red, where 𝐶ox is the concentration of oxidized substance, 𝐶red is the concentration of reduced substance.
𝐸
𝑘 = 𝐴 exp (− 𝑎 )𝑅𝑇, where 𝑘 is the rate constant, 𝐴 is the pre-exponential factor, 𝐸𝑎 is theactivation energy.
exp(𝑥) = 𝑒𝑥

Lambert–Beer equation

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

Henderson–Hasselbalch
equation

For an equilibrium
HA ⇌ H+ + A−
, where equilibrium constant is 𝐾a ,[A− ]pH = p𝐾a + log ()[HA]𝑐
𝐸 = ℎ𝜈 = ℎ
𝜆, where 𝜈 is the frequency, 𝜆 is the wavelength of the light.

Energy of a photon
The sum of a geometricseriesWhen 𝑥 ≠ 1,

Approximation equationthat can be used to solve
problems

When 𝑥 ≪ 1,
1
≃1+𝑥
1−𝑥

𝑛

1 + 𝑥 + 𝑥2 + ⋯ + 𝑥𝑛 = ∑𝑖=0 𝑥𝑖 =1 − 𝑥𝑛+1
1−𝑥

IChO
USA-4 C-0 G-6

Periodic Table

G0-6

USA (United States of America)IChO
USA-4 C-0 G-7

1

H NMR Chemical ShiftsΔ𝛿 for one alkyl group-substitution: ca. +0.4 ppm

G0-7

USA (United States of America)

USA-4 C-1 C-1

USA-4 C-1 C
Yitian Zhu

IChO
Problem 1
Cover sheet

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Q1-1

IChO
USA-4 C-1 Q-1

USA (United States of America)

Hydrogen at a Metal Surface
11 % of the total
Question

A.1

A.2

B.1

B.2

B.3

B.4

Total

Points

6

4

5

3

3

3

24

Score

Hydrogen is expected to be a future energy source that does not depend on fossil fuels. Here, we willconsider the hydrogen-storage process in a metal, which is related to hydrogen-transport and hydrogenstorage technology.

Part AAs hydrogen is absorbed into the bulk of a metal surface, let us first consider the adsorption process ofhydrogen at the metal surface, H2 (g) → 2H(ad), where the gaseous and adsorbed states of hydrogenare represented as (g) and (ad), respectively. Hydrogen molecules (H2 ) that reach the metal surface(M) dissociate at the surface and are adsorbed as H atoms (Fig. 1). Here, the potential energy of H2 isrepresented by two variables: the interatomic distance, 𝑑, and the height relative to the surface metal
atom, 𝑧. It is assumed that the axis along the two H atoms is parallel to the surface and that the centerof gravity is always on the vertical dotted line in Figure 1.
Figure 2 shows the potential energy contour plot for the dissociation at the surface. The numericalvalues represent the potential energy in units of kJ per mole of H2 . The solid line spacing is 20 kJ mol−1 ,the dashed line spacing is 100 kJ mol−1 , and the spacing between solid and dashed lines is 80 kJ mol−1 .
The zero-point vibration energy is ignored.

IChO
USA-4 C-1 Q-2

Q1-2

USA (United States of America)

Figure 1. Definition of variables. Drawing is not to scale.

Figure 2.

Q1-3

IChO
USA-4 C-1 Q-3

A.1

USA (United States of America)

For each of the following items (i)–(iii), SELECT the closest value from A–G.

6pt(i) The interatomic distance for a gaseous H2 molecule(ii) The interatomic distance between metal atoms (𝑑M in Figure 1)(iii) The distance of adsorbed H atoms from the surface (ℎad in Figure 1)

A.2

A. 0.03 nm

B. 0.07 nm

C. 0.11 nm

E. 0.19 nm

F. 0.23 nm

G. 0.27 nm

D. 0.15 nm

For each of the following items (i)–(ii), SELECT the closest value from A–H.
(i) the energy required for the dissociation of gaseous H2 to gaseous H[H2 (g) → 2H(g)](ii) the energy released during the adsorption of a gaseous H2 [H2 (g) → 2H(ad)]A. 20 kJ mol−1

B. 40 kJ mol−1

C. 60 kJ mol−1

D. 100 kJ mol−1

E. 150 kJ mol−1

F. 200 kJ mol−1

G. 300 kJ mol−1

H. 400 kJ mol−1

4pt

IChO
USA-4 C-1 Q-4

Q1-4

USA (United States of America)

Part B
The adsorbed hydrogen atoms are then either absorbed into the bulk, or recombine and desorb backinto the gas phase, as shown in the reactions (1a) and (1b). H(ab) represents a hydrogen atom absorbed
in the bulk.

The reaction rates per surface site for adsorption, desorption, and absorption are 𝑟1 [s−1 ], 𝑟2 [s−1 ] and𝑟3 [s−1 ] , respectively. They are expressed as:

where 𝑘1 [s−1 Pa−1 ], 𝑘2 [s−1 ] and 𝑘3 [s−1 ] are the reaction rate constants and 𝑃H2 is the pressure of H2 .
Among the sites available on the surface, 𝜃 (0 ≤ 𝜃 ≤ 1) is the fraction occupied by H atoms. It is assumed
that adsorption and desorption are fast compared to absorption (𝑟1 , 𝑟2 ≫ 𝑟3 ) and that 𝜃 remains constant.
B.1

𝑟3 can be expressed as:

EXPRESS 𝐶 using 𝑘1 and 𝑘2 .

5pt

IChO
USA-4 C-1 Q-5

Q1-5

USA (United States of America)

A metal sample with a surface area of 𝑆 = 1.0 × 10−3 m2 was placed in a container (1L = 1.0 × 10−3 m3 )with H2 (𝑃H2 = 1.0 × 102 Pa ). The density of hydrogen-atom adsorption sites on the surface was 𝑁 =1.3 × 1018 m−2 . The surface temperature was kept at 𝑇 = 400 K. As the reaction (1) proceeded, 𝑃H2
decreased at a constant rate of 𝑣 = 4.0 × 10−4 Pa s−1 . Assume that H2 is an ideal gas and that the volumeof the metal sample is negligible.

B.2

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

3pt

B.3

At 𝑇 = 400 K, 𝐶 equals 1.0 × 102 Pa−1 . CALCULATE the value of 𝑘3 at 400 K. (Ifyou did not obtain the answer to B.2, use 𝐴 = 3.6 × 10−7 mol s−1 m−2 .)

3pt

B.4

At a different temperature, 𝐶 = 2.5 × 103 Pa−1 and 𝑘3 = 4.8 × 10−2 s−1 . For 𝑟3 as
a function of 𝑃H2 at this temperature, SELECT the correct curve from (a)–(h).

3pt

IChO
USA-4 C-1 A-1

Hydrogen at a Metal Surface
Part A
A.1 (6 pt)
(i)

(ii)

(iii)

A.2 (4 pt)
(i)

(ii)

A1-1

USA (United States of America)IChO
USA-4 C-1 A-2

Part B
B.1 (5 pt)

𝐶=
B.2 (3 pt)𝐴=

mol s−1 m−2

A1-2

USA (United States of America)IChO
USA-4 C-1 A-3

B.3 (3 pt)

𝑘3 =B.4 (3 pt)

s−1

A1-3

USA (United States of America)

USA-4 C-2 C-1

USA-4 C-2 C
Yitian Zhu

IChO
Problem 2
Cover sheet

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Q2-1

IChO
USA-4 C-2 Q-1

USA (United States of America)

Isotope Time Capsule
11 % of the total
Question

A.1

A.2

A.3

A.4

Total

Points

8

8

10

9

35

Score

Molecular entities that differ only in isotopic composition, such as CH4 and CH3 D, are called isotopologues.
Isotopologues are considered to have the same chemical characteristics. In nature, however, there areslight differences between isotopologues.
Assume that all of the substances shown in this Question are in the gas phase.
Let us consider the following equilibrium:

The entropy, 𝑆, increases as the number of possible microscopic states of a system, 𝑊 , increases:

𝑊 = 1 for 12 C16 O2 and 12 C18 O2 . In contrast, 𝑊 = 2 for a 12 C16 O18 O molecule because the oxygen atoms
are distinguishable in this molecule. As the right-hand side of the equilibrium shown in equation 1 has
two 12 C16 O18 O molecules, 𝑊 = 22 = 4.

IChO
USA-4 C-2 Q-2

A.1

Q2-2

USA (United States of America)

The enthalpy change, Δ𝐻, of equation 3 is positive regardless of the temperature.

8pt

CALCULATE the equilibrium constant, 𝐾, for equation 3 at very low temperature (think of 𝑇 → 0) and again at very high temperature (think of 𝑇 → +∞) .
Assume that the reaction remains unchanged at these temperatures and that
Δ𝐻 converges to a constant value for high temperature.
ΔH of the following process can be explained by molecular vibrations.

At 𝑇 = 0 K, the vibrational energy of a diatomic molecule whose vibration frequency is 𝜈 [s−1 ] is expressedas:In equation 6 , 𝑘 is the force constant and 𝜇 is the reduced mass.
Reduced mass is expressed in terms of the mass of the two atoms in the diatomic molecule, 𝑚1 and 𝑚2 ,according to:

A.2

The wavenumber corresponding to the vibration of H2 is 4161.0 cm−1 . CALCULATE Δ𝐻 of the following equation at 𝑇 = 0 K in units of J mol−1 .

Assume that:• only the vibrational energy contributes to Δ𝐻.
• the 𝑘 values for H2 , HD, and D2 are identical.
• the mass of H is 1 Da and the mass of D is 2 Da.

8pt

IChO
USA-4 C-2 Q-3

Q2-3

USA (United States of America)

The molar ratio of H2 , HD, and D2 depends on the temperature in a system in equilibrium. Here, ΔD2 isdefined as the change of the molar ratio of D2 .

[D2 ]
[D ]
in the sample and 𝑅D∗ 2 to 2 at 𝑇 → +∞. Note that the distribution of isotopes
[H2 ]
[H2 ]
becomes random at 𝑇 → +∞.
Here, 𝑅D2 refers to
A.3

CALCULATE ΔD2 with natural D abundance when the isotopic exchange is inequilibrium at the temperature where 𝐾 (in equation 4) is 0.300. Assume thatthe natural abundance ratios of D and H are 1.5576 × 10–4 and (1 − 1.5576 × 10–4 ),respectively.

10pt

Q2-4

IChO
USA-4 C-2 Q-4

USA (United States of America)

In general, the molar ratio of the doubly substituted isotopologue, which contains two heavy isotope
atoms in one molecule, increases with decreasing temperature. Let us consider the molar ratio of CO2
molecules with molecular weights of 44 and 47, which are described as CO2 [44] and CO2 [47] below. The
quantity Δ47 is defined as:[CO2 [47]][CO2 [47]]∗in the sample and 𝑅47
toat 𝑇 → +∞. The natural abundances of carbon[CO2 [44]][CO2 [44]]and oxygen atoms are shown below; ignore isotopes that are not shown here.

𝑅47 refers to12

natural abundance

16

natural abundanceC

130.988888

0.011112

C

O

17

O

18

0.997621

0.0003790

0.0020000

O

The temperature dependence of Δ47 is determined as follows, where 𝑇 is given as the absolute temperature in units of K:

A.4

The 𝑅47 of fossil plankton obtained from the Antarctic seabed was 4.50865×10–5 .
CALCULATE the approximate temperature using this 𝑅47 . This temperature isinterpreted as the air temperature during the era in which the plankton lived.
Consider only the most common isotopologue of CO2 [47] for the calculation.

9pt

IChO
USA-4 C-2 A-1

Isotope Time Capsule
A.1 (8 pt)

𝑇 →0∶𝐾=, 𝑇 → +∞ ∶ 𝐾 =A2-1

USA (United States of America)IChO
USA-4 C-2 A-2

A.2 (8 pt)

Δ𝐻 =J mol−1

A2-2

USA (United States of America)IChO
USA-4 C-2 A-3

A.3 (10 pt)

ΔD2 =

A2-3

USA (United States of America)IChO
USA-4 C-2 A-4

A.4 (9 pt)

𝑇 =

K

A2-4

USA (United States of America)

USA-4 C-3 C-1

USA-4 C-3 C
Yitian Zhu

IChO
Problem 3
Cover sheet

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Q3-1

IChO
USA-4 C-3 Q-1

USA (United States of America)

Lambert–Beer Law?
8 % of the total
Question

A.1

B.1

B.2

Total

Points

10

6

6

22

Score

In this problem, ignore the absorption of the cell and the solvent. The temperatures of all solutions andgases are kept constant at 25 °C.

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

𝐾=

[A− ][H+ ]
[HA]

(1)

For part A, the optical path length is 𝑙 . Ignore the density change upon dilution. Assume that no chemical
reactions other than equation 1 occur.
A.1

The absorbance of X was 𝐴1 at a wavelength of 𝜆1 . Then, solution X was diluted
to twice its initial volume using hydrochloric acid with pH = 2.500. After dilution,
the absorbance was still 𝐴1 at 𝜆1 . DETERMINE the ratio 𝜀HA /𝜀A− , where 𝜀HA and𝜀A− represent the absorption coefficients of HA and of A− , respectively, at 𝜆1 .

10pt

Q3-2

IChO
USA-4 C-3 Q-2

USA (United States of America)

Part BLet us consider the following equilibrium in the gas phase.
D ⇌ 2M(2)

Pure gas D is filled into a cuboid container that has a transparent movable wall with a cross-section of
𝑆 (see the figure below) at a pressure 𝑃 . Equilibrium is established while the total pressure is kept at 𝑃 .
The absorbance of the gas is 𝐴 = 𝜀(𝑛/𝑉 )𝑙, where 𝜀 is the absorption coefficient, 𝑛 is the amount of gas
in moles, 𝑉 is the volume of gas, and 𝑙 is the optical path length. Assume that all components of the gas
mixture behave as ideal gases.

Use the following definitions if necessary.
Initial state

After equilibrium

D

M

D

M

Partial pressure

𝑃

0

𝑝D

𝑝M

Amount in moles

𝑛0

0

𝑛D

𝑛M

Volume𝑉0

𝑉

B.1

The absorbance of the gas at 𝜆B1 measured from direction 𝑥 was 𝐴B1 both at
the initial state (𝑙 = 𝑙𝑥0 ) and after the equilibrium (𝑙 = 𝑙𝑥 ). DETERMINE the ratio
𝜀D /𝜀M at 𝜆B1 , where 𝜀D and 𝜀M represent the absorption coefficients of D and M,
respectively.

6pt

B.2

The absorbance of the gas at 𝜆B2 measured from direction 𝑦 was 𝐴B2 both at
the initial state (𝑙 = 𝑙𝑦0 ) and after the equilibrium (𝑙 = 𝑙𝑦 ). DETERMINE the ratio
𝜀D /𝜀M at 𝜆B2 .

6pt

IChO
USA-4 C-3 A-1

Lambert–Beer Law?
Part A
A.1 (10 pt)(Continued on the next page)A3-1

USA (United States of America)IChO
USA-4 C-3 A-2

A.1 (cont.)

𝜀HA /𝜀A− =A3-2

USA (United States of America)IChO
USA-4 C-3 A-3Part B
B.1 (6 pt)

𝜀D /𝜀M =A3-3

USA (United States of America)IChO
USA-4 C-3 A-4

B.2 (6 pt)

𝜀D /𝜀M =A3-4

USA (United States of America)

USA-4 C-4 C-1

USA-4 C-4 C
Yitian Zhu

IChO
Problem 4
Cover sheet

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Q4-1

IChO
USA-4 C-4 Q-1

USA (United States of America)

The Redox Chemistry of Zinc
11 % of the total
Question

A.1

A.2

B.1

B.2

B.3

B.4

Total

Points

6

5

4

3

5

9

32

Score

Zinc has long been used in alloys for brass and steel materials. The zinc contained in industrial wastewater is removed by precipitation to detoxify the water, and the precipitate obtained is reduced to recoverand reuse it as metallic zinc.

Part AThe dissolution equilibrium of zinc hydroxide Zn(OH)2 (s) at 25 ∘ C and the relevant equilibrium constants
are given in equations 1–4.

IChO
USA-4 C-4 Q-2

Q4-2

USA (United States of America)

The solubility of zinc, 𝑆, (concentration of zinc in a saturated aqueous solution) is given in equation 5.

A.1

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

6pt

A.2

A saturated aqueous solution of Zn(OH)2 (s) with pH = 7.00 was prepared andfiltered. NaOH was added to this filtrate to increase its pH to 12.00. CALCULATE
the mole percent of zinc that precipitates when increasing the pH from 7.00 to12.00. Ignore the volume and temperature changes.

5ptPart B
Next, the recovered zinc hydroxide is heated to obtain zinc oxide according to the reaction below:

The zinc oxide is then reduced to metallic zinc by reaction with hydrogen:

B.1

In order for reaction (7) to proceed at a constant hydrogen pressure of 1 bar, itis necessary to reduce the partial pressure of the generated water vapor. CALCULATE the upper limit for the partial pressure of water vapor to allow reaction (7) to proceed at 300 ∘ C. The Gibbs formation energies of zinc oxide andwater vapor at 300 ∘ C and 1 bar for all gaseous species are Δ𝐺ZnO (300∘ C) =−2.90 × 102 kJ mol−1 and Δ𝐺H2 O (300∘ C) = −2.20 × 102 kJ mol−1 , respectively.

4pt

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

B.2

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

3pt

IChO
USA-4 C-4 Q-3

Q4-3

USA (United States of America)

Mt. Fuji
B.3

Consider the change of e.m.f. of a zinc–air battery depending on the environment. CALCULATE the e.m.f. at the summit of Mt. Fuji, where the temperature
and altitude are −38 ∘ C (February) and 3776 m, respectively. The atmosphericpressure is represented by equation 9 below at altitude ℎ [m] and temperature
𝑇 [∘ C].

5pt

The molar ratio of oxygen in the atmosphere is 21%. The Gibbs energy changeof reaction (8) is Δ𝐺ZnO (−38∘ C) = −3.26 × 102 kJ mol−1 at −38 ∘ C and 1 bar.
B.4

CALCULATE the Gibbs energy change for reaction (6) at 25 ∘ C. Note that thestandard reduction potentials, 𝐸 ∘ (Zn2+ /Zn) and 𝐸 ∘ (O2 /H2 O) at 25 ∘ C and 1 bar
are given as (10) and (11), respectively.

9pt

IChO
USA-4 C-4 A-1

The Redox Chemistry of Zinc
Part A
A.1 (6 pt)< pH <A4-1

USA (United States of America)IChO
USA-4 C-4 A-2

A.2 (5 pt)

%

A4-2

USA (United States of America)IChO
USA-4 C-4 A-3

Part B
B.1 (4 pt)

bar

𝑝H2 O=
B.2 (3 pt)

g

A4-3

USA (United States of America)IChO
USA-4 C-4 A-4

B.3 (5 pt)

V

A4-4

USA (United States of America)IChO
USA-4 C-4 A-5

B.4 (9 pt)

Δ𝐺∘ =J mol−1

A4-5

USA (United States of America)

USA-4 C-5 C-1

USA-4 C-5 C
Yitian Zhu

IChO
Problem 5
Cover sheet

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Q5-1

IChO
USA-4 C-5 Q-1

USA (United States of America)

Mysterious Silicon
12 % of the total
Question

A.1

A.2

A.3

A.4

B.1

B.2

B.3

Total

Points

9

7

6

10

5

15

8

60

Score

Although silicon is also a group 14 element like carbon, their properties differ significantly.

Part A
Unlike the carbon–carbon triple bond, the silicon–silicon triple bond in a written as R1 –Si ≡ Si–R1 (R: organic substituent), is extremely reactive. For example, it reacts with ethylene to form a cyclic productthat contains a four-membered ring.

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

IChO
USA-4 C-5 Q-2

Q5-2

USA (United States of America)

The 13 C NMR analysis of the corresponding six-membered ring skeletons Si2 C4 shows two signals for B
and one signal for C.
A.1

DRAW the structures of A, B, and C using R1 , R2 , Si, and C. (Only one resonancestructure is needed, if applicable.)

9pt

A.2

CALCULATE the aromatic stabilization energy (ASE) for benzene and compound
C (where R1 = R2 = H) as positive values. The enthalpy change of some hydrogenation reactions of unsaturated systems is shown below (Figure 1).

7pt

Figure 1.

IChO
USA-4 C-5 Q-3

Q5-3

USA (United States of America)

When a xylene solution of C is heated, it undergoes isomerization to give an equilibrium mixture ofcompounds D and E. The mole ratio is D : E = 1 : 40.0 at 50.0 ˚C, and D : E = 1 : 20.0 at 120.0 ˚C.
A.3

CALCULATE Δ𝐻 for the transformation of D to E. Assume that Δ𝐻 does notdepend on temperature.

6pt

The isomerization from C to D and D to E proceeds via transformations of 𝜋-bonds into 𝜎-bonds without
breaking any 𝜎-bonds. A 13 C NMR analysis revealed one signal for the Si2 C4 skeleton of D and two signals
for that of E. The skeleton of D does not contain any three-membered rings, while E has two threemembered rings that share an edge.

A.4

DRAW the structures of D and E using R1 , R2 , Si, and C.

10pt

Part B
Silicon is able to form highly coordinated compounds (> four substituents) with electronegative elements
such as fluorine. Metal fluorides are often used as fluorination reagents and highly coordinated silicon
fluorides also act as fluorination reagents.
The fluorination reaction of CCl4 using Na2 SiF6 was carried out as follows.
Standardization of Na2 SiF6 solution :Preparation• Aqueous solution F: 0.855 g of Na2 SiF6 (188.053 g mol–1 ) dissolved in water (total volume: 200 mL).
• Aqueous solution G: 6.86 g of Ce2 (SO4 )3 (568.424 g mol–1 ) dissolved in water (total volume: 200mL).
Procedure• Precipitation titration of a solution F (50.0 mL) by dropwise adding solution G in the presence ofxylenol orange, which coordinates to Ce3+ , as an indicator. After adding 18.8 mL of solution G,
the color of the solution changes from yellow to magenta. The generated precipitate is a binary
compound that contains Ce3+ , and the only resulting silicon compound is Si(OH)4 .
B.1

WRITE the balanced equation for the reaction of Na2 SiF6 with Ce2 (SO4 )3 .

5pt

Reaction of CCl4 with Na2 SiF6 :(Substance losses, e.g. by evaporation, are negligible during the following steps.)• Na2 SiF6 (𝑥 [g]) was added to CCl4 (500.0 g) and heated to 300˚C in a sealed, pressure-resistant reaction vessel. The unreacted Na2 SiF6 and generated NaCl were removed by filtration. The filtrate wasdiluted to a total volume of 1.00 L with CCl4 (solution H). The 29 Si and 19 F NMR spectra of solution
H showed SiF4 as the only silicon compound. In the 19 F NMR spectrum, in addition to SiF4 , signals
corresponding to CFCl3 , CF2 Cl2 , CF3 Cl, and CF4 were observed (cf. Table 1). The integration ratios in
the 19 F NMR spectrum are proportional to the number of fluorine nuclei.

Q5-4

IChO
USA-4 C-5 Q-4

USA (United States of America)

Table 119

F NMR data

CFCl3

CF2 Cl2

CF3 Cl

CF4Integration ratio

45.0

65.0

18.0

2.0

SiF4 is hydrolyzed to form H2 SiF6 according to the following equation:• Solution H (10 mL) was added to an excess amount of water, which resulted in the complete hydrolysis of SiF4 . After separation, the H2 SiF6 generated from the hydrolysis in the aqueous solution
was neutralized and completely converted to Na2 SiF6 (aqueous solution J).
• The precipitate of unreacted Na2 SiF6 and NaCl, which was removed by filtration in the initial step(underlined above), was completely dissolved in water to give an aqueous solution (solution K; 10.0
L).
• Then, additional precipitation titrations using solution G were carried out. The entire amount ofsolution J required 61.6 mL of titrant to reach the endpoint. Separately, 100 mL of solution K used
44.4 mL of titrant to reach the endpoint. It should be noted here that the coexistence of NaCl or
SiO2 has no effect on the precipitation titration.

B.2

CALCULATE the mass of the NaCl produced in the reaction vessel (underlined
above), and CALCULATE the mass (𝑥 [g]) of the Na2 SiF6 used as a starting material.

15pt

B.3

77.8% of the CCl4 used as a starting material was unreacted. CALCULATE themass of CF3 Cl generated.

8pt

IChO
USA-4 C-5 A-1

A5-1

USA (United States of America)

Mysterious Silicon
Part A
A.1 (9 pt)
A (3 pt)

B (3 pt)

C (3 pt)

A.2 (7 pt)

C6 H6 ∶

kJ mol–1 , C ∶kJ mol–1

IChO
USA-4 C-5 A-2

A.3 (6 pt)

kJ mol–1

Δ𝐻 =A.4 (10 pt)
D (5 pt)

E (5 pt)

A5-2

USA (United States of America)IChO
USA-4 C-5 A-3

Part B
B.1 (5 pt)

B.2 (15 pt)(Continued on the next page)A5-3

USA (United States of America)IChO
USA-4 C-5 A-4

B.2 (cont.)

NaCl ∶

g, Na2 SiF6 ∶

g

A5-4

USA (United States of America)IChO
USA-4 C-5 A-5

B.3 (8 pt)

CF3 Cl ∶g

A5-5

USA (United States of America)

USA-4 C-6 C-1

USA-4 C-6 C
Yitian Zhu

IChO
Problem 6
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Q6-1

IChO
USA-4 C-6 Q-1

USA (United States of America)

The Solid-State Chemistry of Transition Metals
13 % of the total
Question

A.1

A.2

A.3

B.1

B.2

B.3

B.4

C.1

C.2

C.3

Total

Points

6

3

3

6

4

4

4

5

5

5

45

Score

Volcano at Sakurajima island

Part A
Japan is one of the countries with the largest number of volcanos worldwide. When silicate minerals
crystallize from magma, some of the transition-metal ions (Mn+ ) in the magma are incorporated into
the silicate minerals. The Mn+ studied in this problem are coordinated by oxide ions (O2− ) and adopt
a four-coordinate tetrahedral (𝑇d ) geometry in the magma. In the silicate minerals, the ions form sixcoordinate octahedral (𝑂h ) geometry. Both types of complexes exhibit a high-spin electron configuration.
The distribution coefficient of Mn+ between the silicate minerals and magma, 𝐷, can be expressed by:where [M]s and [M]l are the concentrations of Mn+ in the silicate minerals and the magma, respectively.
The table below shows the 𝐷 values of Cr2+ and Mn2+ as examples.

IChO
USA-4 C-6 Q-2

Q6-2

USA (United States of America)

ΔO is the energy separation of the d-orbitals of Mn+ , and CFSEO is the crystal-field stabilization energy ina 𝑂h field. Let ΔT and CFSET be those in a 𝑇d field.
A.1

CALCULATE ΔCFSE, defined as|CFSEO −CFSET | in terms of ΔO for Cr2+ , Mn2+ ,and Co2+ ;assume ΔT = 4/9ΔO .

6pt

A.2

A linear relationship is observed by plotting ln𝐷 against ΔCFSE / ΔO in the Cartesian coordinate system shown below.
CALCULATE the approximate value of 𝐷 for Co2+ .

3pt

Metal oxides MO (M: Ca, Ti, V, Mn, or Co) crystallize in a rock-salt structure wherein the Mn+ adopts an 𝑂hgeometry with a high-spin electron configuration. The lattice enthalpy of these oxides is primarily determined by the Coulombic interactions (due to the radius and charge of the ions) with some contribution
from the CFSE of Mn+ in the 𝑂h field.

A.3

CHOOSE the appropriate set of lattice enthalpies [kJ mol−1 ] from one of the options (a) to (f).

3pt

IChO
USA-4 C-6 Q-3

Q6-3

USA (United States of America)

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

Figure 1.
A can be synthesized by thermal decomposition (pyrolysis) of complex B, which is formed by mixing
metal chlorides in dilute aqueous ammonia solution containing squaric acid (C4 H2 O4 ), a diacid. Upon
pyrolysis in dry air to 200 °C, B loses 29.1% of its mass due to the loss of crystallization water. Anotherweight loss is observed when heating is continued to 700 °C due to the release of CO2 . The total weightloss during the formation of A from B is 63.6%. It should be noted that only water and CO2 are releasedin the pyrolysis reaction.

B.1

WRITE the chemical formula for A and B.

6pt

B.2

CALCULATE 𝑙𝑥 and 𝑙𝑧 using Figure 1.

4pt

B.3

For Cu2+ in the distorted [CuO6 ] octahedron in A of Figure 1, WRITE the namesof the split eg orbitals (d𝑥2 −𝑦2 and d𝑧2 ) on lines (i) and (ii), and PLACE electrons
on the energy diagram in the dotted box on your answer sheet.

4pt

Q6-4

IChO
USA-4 C-6 Q-4

USA (United States of America)

A is an insulator. One hole is generated in the crystal lattice of A when one La3+ is substituted with one
Sr2+ . As a result, the Sr2+ -doped A conducts electricity and shows superconductivity below 38 K. Whenthe substitution described takes place in A, 2.05 × 1027 holes m−3 were generated.

B.4

CALCULATE the percentage of Sr2+ substituted for La3+ based on the mole ratio in the substitution. Note that the valencies of the constituent ions and the
crystal structure are not altered by the substitution.

4pt

Part C−2+

Cu2 (CH3 CO2 )4 is composed of four CH3 CO2 coordinated to two Cu (Figure 2A). Cu2 (CH3 CO2 )4 exhibits−high levels of structural symmetry, with two axes passing through the carbon atoms of the four CH3 CO2
2+
and an axis passing through the two Cu , all of which are oriented orthogonal relative to each other.
−
When a dicarboxylate ligand is used instead of CH3 CO2 , a “cage complex” is formed. The cage complex
2+
L1
Cu4 (L1
L1)4 is composed of planar dicarboxylate L1 (Figure 2B) and Cu (Figure 2C). The structure of thecage complex is determined by the angle θ between the coordination directions of the two carboxylates,
indicated by the arrows in Figure 2B. θ is 0° for L1. Hydrogen atoms are not shown in Figure 2.

Figure 2.

IChO
USA-4 C-6 Q-5

C.1

Q6-5

USA (United States of America)

The θ of the planar dicarboxylate L2 below is fixed to 90˚. If the composition
2+
L2of the cage complex formed from L2 and Cu is Cu𝑛 (L2L2)𝑚 , GIVE the smallest−
integer combination of n and m. Assume that only the CO2 groups of L2 coor2+
dinate to Cu ions.

5pt

Q6-6

IChO
USA-4 C-6 Q-62+

USA (United States of America)−

A zinc complex, Zn4 O(CH3 CO2 )6 , contains four tetrahedral Zn , six CH3 CO2 , and one O2− (Figure 3A). In
Zn4 O(CH3 CO2 )6 , the O2− is located at the origin, and the three axes passing through the carbon atoms−of CH3 CO2 are oriented orthogonal relative to each other. When p-benzenedicarboxylate (Figure 3B,−2+
L3, θ = 180°) is used instead of CH3 CO2 , the Zn clusters are linked to each other to form a crystalline
L3
solid (X) that is called a “porous coordination polymer” (Figure 3C). The composition of X is [Zn4 O(L3L3)3 ]𝑛 ,and it has a cubic crystal structure with nano-sized pores. One pore is represented as a sphere in Figure
2+
3D, and each tetrahedral Zn cluster is represented as a dark gray polyhedron in Figures 3C and 3D.
Hydrogen atoms are not shown in Figure 3.

Figure 3.
C.2

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

5pt

C.3

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

5pt

IChO
USA-4 C-6 A-1

A6-1

USA (United States of America)

The Solid-State Chemistry of Transition Metals
Part A
A.1 (6 pt)

Cr2+ ∶ΔO , Mn2+ ∶ΔO , Co2+ ∶

ΔO

IChO
USA-4 C-6 A-2

A.2 (3 pt)

𝐷∶

A.3 (3 pt)

A6-2

USA (United States of America)

IChO

A6-3

USA-4 C-6 A-3

USA (United States of America)

Part B
B.1 (6 pt)

A∶

,B∶

B.2 (4 pt)

𝑙𝑥 =nm, 𝑙𝑧 =nm

IChO
USA-4 C-6 A-4

B.3 (4 pt), (ii) ∶(i) ∶B.4 (4 pt)

%

A6-4

USA (United States of America)IChO
USA-4 C-6 A-5Part C
C.1 (5 pt)𝑛=,𝑚=

C.2 (5 pt)

𝑎=

cm

A6-5

USA (United States of America)IChO
USA-4 C-6 A-6

C.3 (5 pt)

A6-6

USA (United States of America)

USA-4 C-7 C-1

USA-4 C-7 C
Yitian Zhu

IChO
Problem 7
Cover sheet

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Q7-1

IChO
USA-4 C-7 Q-1

USA (United States of America)

Playing with Non-benzenoid Aromaticity
13 % of the total
Question

A.1

A.2

A.3

B.1

Total

Points

5

2

19

10

36

Score

Prof. Nozoe (1902–1996) started the research field of non-benzenoid aromatic compounds, which are
now ubiquitous in organic chemistry.

Photo courtesy: Tohoku Univ.

Part A
Lineariifolianone is a natural product with a unique structure that was first isolated from Inula linariifolia.
From valencene (1), a one-step conversion gives compound 2. A three-step conversion of 2, via 3, yields
ketone 4. Eremophilene (5) is converted into 6 by performing the same four-step conversion.

IChO
USA-4 C-7 Q-2

Q7-2

USA (United States of America)

Inula linariifolia

A.1

DRAW the structures of 2 and 6 and clearly identify the stereochemistry where
necessary.

5pt

IChO
USA-4 C-7 Q-3

Q7-3

USA (United States of America)

Ketone 4 is ultimately converted into ester 15. Compound 8 (molecular weight: 188) retains all the stereocenters in 7. Compounds 9 and 10 have five stereocenters and no carbon-carbon double bonds. Assume
that H2 18 O is used instead of H2 16 O for the synthesis of 18 O-labelled-lineariifolianones 13 and 14 from 11
and 12, respectively. Compounds 13 and 14 are 18 O-labelled isotopomers. Ignoring isotopic labelling,both 13 and 14 provide the same product 15 with identical stereochemistry.

IChO
USA-4 C-7 Q-4

Q7-4

USA (United States of America)IChO
USA-4 C-7 Q-5

Q7-5

USA (United States of America)

A.2

CHOOSE the appropriate structure for reagent A.

2pt

A.3

DRAW the structures of 8–14 and clearly identify the stereochemistry where
necessary. Also, LABEL the 18 O atoms in compounds 13 and 14 using the notation shown below:

19pt

IChO
USA-4 C-7 Q-6

Q7-6

USA (United States of America)

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

B.1

DRAW the structures of 17–19 and 21. Ignore stereochemistry.

10pt

IChO
USA-4 C-7 A-1

Playing with Non-benzenoid Aromaticity
Part A
A.1 (5 pt)
2 (2 pt)

A.2 (2 pt)

6 (3 pt)

A7-1

USA (United States of America)IChO
USA-4 C-7 A-2

A.3 (19 pt)

A7-2

USA (United States of America)IChO
USA-4 C-7 A-3

Part B
B.1 (10 pt)
17 (2 pt)

18 (2 pt)

19 (3 pt)

21 (3 pt)

A7-3

USA (United States of America)

USA-4 C-8 C-1

USA-4 C-8 C
Yitian Zhu

IChO
Problem 8
Cover sheet

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Q8-1

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USA-4 C-8 Q-1

USA (United States of America)

Dynamic Organic Molecules and Their Chirality
11 % of the total
Question

A.1

A.2

A.3

B.1

B.2

Total

Points

9

3

7

3

4

26

Score

Part A
Polycyclic aromatic hydrocarbons with successive ortho-connections are called [n]carbohelicenes (n represents the number of six-membered rings; see below). [4]Carbohelicene ([4]C) is efficiently preparedaccording to the synthesis shown below. The key photochemical reaction proceeds via the intermediate(Int.) that is then easily oxidized by iodine.

The photochemical reaction proceeds in a manner similar to the following example.

IChO
USA-4 C-8 Q-2

Q8-2

USA (United States of America)

Note: For all of Question 8, please draw alternating single and double bonds in your answers to
the problems as depicted in the examples of carbohelicene. Do not use circles for conjugated 𝜋systems.
A.1

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

9pt

A.2

Attempts to synthesize [5]carbohelicene from the same phosphonium salt andan appropriate starting compound resulted in the formation of only a trace
amount of [5]carbohelicene. Instead, product D whose molecular weight was
2 Da lower than that of [5]carbohelicene was isolated. The 1 H NMR chemical
shifts of D are listed below. DRAW the structure of D.
[D (𝛿, ppm in CS2 , r.t.), 8.85 (2H), 8.23 (2H), 8.07 (2H), 8.01 (2H), 7.97 (2H), 7.91
(2H)]

3pt

[5]- and larger [n]carbohelicenes have helical chirality and interconversion between enantiomers of thesehelicenes is very slow at room temperature. The chirality of [n]carbohelicenes is defined as (M) or (P) as
shown below.

[n]Carbohelicenes with n larger than 4 can be enantiomerically separated by chiral column chromatography, which was developed by Prof. Yoshio Okamoto.

Photo courtesy: The Japan Prize Foundation

IChO
USA-4 C-8 Q-3

Q8-3

USA (United States of America)

Multiple helicenes are molecules that contain two or more helicene-like structures. Considering helicalchirality, several stereoisomers exist in a multiple helicene. For example, compound E contains three[5]carbohelicene-like moieties in one molecule. One of the stereoisomers is described as (P, P, P) andshown below.

A.3

The nickel-mediated trimerization of 1,2-dibromobenzene generates triphenylene. When the same reaction is applied to an enantiomer of F, (P)-F, multiple
helicene G (C66 H36 ) is obtained. Interconversion between stereoisomers does
not occur during the reaction. DRAW one possible stereoisomer of G with thechirality defined as in the example above, including numerical labels. LIST ALL
remaining stereoisomers with location numbers and M and P labels according
to the same numbering. For instance, the other stereoisomers of E would belisted as (1, 2, 3) = (P, M, P), (P, M, M), (P, P, M), (M, M, M), (M, M, P), (M, P, P), and(M, P, M).

7pt

IChO
USA-4 C-8 Q-4

Q8-4

USA (United States of America)

Part B
Sumanene is a bowl-shaped hydrocarbon that was first reported in Japan in 2003. The name "sumanene"comes from a Sanskrit-Hindi word "suman" that means sunflower. The synthesis of sumanene was originally completed by a reaction sequence consisting of a ring-opening and a ring-closing metathesis.

Representative metathesis reactions, catalyzed by a ruthenium catalyst (Ru*), are shown below.

B.1

DRAW the structure of intermediate I (ignore stereochemistry).

3pt

IChO
USA-4 C-8 Q-5

B.2

Q8-5

USA (United States of America)

Starting with the optically active precursor J, the same reaction sequence gives
the optically active sumanene derivative K. The stereocenters in J are unchanged during the metathesis reaction. DRAW the structure of K with appropriate stereochemistry.

4pt

IChO
USA-4 C-8 A-1

A8-1

USA (United States of America)

Dynamic Organic Molecules and Their Chirality
Part A
A.1 (9 pt)
A (3 pt)

A.2 (3 pt)

B (3 pt)

C (3 pt)

IChO
USA-4 C-8 A-2

A.3 (7 pt)

A8-2

USA (United States of America)IChO
USA-4 C-8 A-3

Part B
B.1 (3 pt)

B.2 (4 pt)

A8-3

USA (United States of America)

USA-4 C-9 C-1

USA-4 C-9 C
Yitian Zhu

IChO
Problem 9
Cover sheet

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Q9-1

IChO
USA-4 C-9 Q-1

USA (United States of America)

Likes and Dislikes of Capsule
10 % of the total
Question

A.1

A.2

A.3

A.4

A.5

Total

Points

13

2

2

3

3

23

Score

Good kids don’t do this, but if you unseam a tennis ball, you can disassemble it into two U-shaped pieces.

Based on this idea, compounds 1 and 2 were synthesized as U-shaped molecules with different sizes.
Compound 3 was prepared for comparison to 1. The encapsulation behavior of these compounds wasinvestigated.

IChO
USA-4 C-9 Q-2

Q9-2

USA (United States of America)

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

IChO
USA-4 C-9 Q-3

A.1

Q9-3

USA (United States of America)

DRAW the structures of 4–9; ignore stereochemistry. Use "PMB" as a substituent instead of drawing the whole structure of the p-methoxybenzyl groupshown in the scheme above.

13pt

In the mass spectrum of 1, the ion peak corresponding to its dimer (112 ) was clearly observed, whereas
an ion peak for 3 2 was not observed in the spectrum of 3. In the 1 H NMR spectra of a solution of 1 2 , all
the NH protons were observed to be chemically equivalent. The chemical shift was significantly different
from that of the NH protons of 3. These data indicate that hydrogen bonds are formed between the NH
groups of 1 and atoms X of another molecule of 1 to form the dimeric capsule.

A.2

CIRCLE all the appropriate atom(s) X in compound 1.

2pt

A.3

WRITE the number of hydrogen bonds in the dimeric capsule (112 ).

2pt

IChO
USA-4 C-9 Q-4

Q9-4

USA (United States of America)

The dimeric capsule of 1 (112 ) has an internal space where an appropriate small molecule Z can be encapsulated. This phenomenon is expressed by the following equation:

The equilibrium constant of the encapsulation of Z into 12 is given below:

Encapsulation of a molecule into a capsule could be monitored by NMR spectroscopy. For example, 1 2
in C6 D6 gave different signals in the 1 H NMR spectra before and after addition of CH4 .
Compound 2 also forms a rigid and large dimeric capsule (222 ). The 1 H NMR spectrum of 22 was measured
in C6 D6 , C6 D5 F, and a C6 D6 /C6 D5 F solvent mixture, with all other conditions kept constant. The chemical
shifts for the Ha protons of 2 in the solvents given above are summarized below. No other signals dueto the Ha protons in 2 were observed (except for those listed). Assume that the interior of the capsule isalways filled with the largest possible number of solvent molecules and that each signal corresponds toone species of the filled capsule.

A.4

solvent

𝛿 (ppm) of Ha

C6 D6

4.60

C6 D5 F

4.71

C6 D6 / C6 D5 F

4.60, 4.71, 4.82

DETERMINE the number of C6 D6 and C6 D5 F molecules encapsulated in 2 2 giving
each Ha signal.

3pt

IChO
USA-4 C-9 Q-5

Q9-5

USA (United States of America)

H NMR measurements in C6 D6 revealed that 2 2 can incorporate one molecule of 1-adamantanecarboxylic
acid (AdA), and the association constants (𝐾a ), expressed below, were determined at various temperatures. [solvent@222 ] denotes a species containing one or more solvent molecules.

1

Similarly, 𝐾a values of CH4 and 1 2 (equation 2) at various temperatures in C6 D6 were also determined by
H NMR measurements. The plots of the two association constants (as ln 𝐾a vs 1/T) are shown below.

1

A.5

WRITE the correct letter, A or B, to fill in (1)–(5) in the following paragraph.
No C6 D6 molecule is encapsulated in 1 2 . In line II, the entropy change (Δ𝑆) is( 1 ) and enthalpy change (Δ𝐻) is ( 2 ), indicating that the driving force for theencapsulation in line II is ( 3 ). Therefore, line I corresponds to ( 4 ), and line IIcorresponds to ( 5 ).
A

B

(1)

positive

negative(2)

positive

negative

(3)

Δ𝑆

Δ𝐻

(4)

1 2 and CH4

2 2 and AdA(5)

1 2 and CH4

2 2 and AdA

3pt

IChO
USA-4 C-9 A-1

Likes and Dislikes of Capsule
A.1 (13 pt)
4 (2 pt)

5 (3 pt)

6 (2 pt)

7 (2 pt)

8 (2 pt)

9 (2 pt)

A9-1

USA (United States of America)IChO
USA-4 C-9 A-2

A9-2

USA (United States of America)

A.2 (2 pt)

A.3 (2 pt)
A.4 (3 pt)
𝛿 (ppm) of Ha

numbers of C6 D6numbers of C6 D5 F

4.60 ppm

4.71 ppm

4.82 ppm

A.5 (3 pt)
(1) ∶

(2) ∶

(4) ∶

(5) ∶

(3) ∶