GBR-4 C-0 C-1

GBR-4 C-0 C
Zachary McGuire

IChO
General instructions
Cover sheet

Please return this cover sheet together with all the related question sheets.

IChO
GBR-4 C-0 G-1

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

G0-1

British English (United Kingdom)

IChO

G0-2

GBR-4 C-0 G-2

British English (United Kingdom)

General Instructions• You are only allowed to use pen to write the 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.
• All results must be written in the appropriate answer boxes with pen on the answer sheets. Use
the back of the question sheets if you need rough paper. Remember that answers written outside
the answer boxes will not be graded.
• Write relevant calculations in the appropriate boxes when necessary. Full marks will be given forcorrect answers only when your work is shown.
• 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 on request only for clarification.
• You are not allowed to leave your working place without permission. If you need any assistance(broken calculator, need to go to the toilet, etc.), raise your hand and wait until an invigilator arrives.
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
GBR-4 C-0 G-3

G0-3

British English (United Kingdom)

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 = 9.64853321 × 104 C mol−1

Gas constant

𝑅 = 𝑁A × 𝑘B = 8.3144626 J K−1 mol−1= 8.2057366 × 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 degrees 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
GBR-4 C-0 G-4

G0-4

British English (United Kingdom)

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
GBR-4 C-0 G-5

Heat change Δ𝑞
Nernst equation for a redox reaction
Arrhenius equation

G0-5

British English (United Kingdom)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 the oxidized substance, 𝐶red is the concentration of the 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, and 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
GBR-4 C-0 G-6

The Periodic Table

G0-6

British English (United Kingdom)

IChO
GBR-4 C-0 G-7

1

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

G0-7

British English (United Kingdom)GBR-4 C-1 C-1

GBR-4 C-1 C
Zachary McGuire

IChO
Problem 1
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q1-1

IChO
GBR-4 C-1 Q-1

British English (United Kingdom)

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 anticipated to be a future energy source that does not depend on fossil fuels. Here, wewill consider the hydrogen storage process in a metal, which is related to hydrogen transportation andstorage technology.

Part AAs hydrogen is absorbed into the bulk of a metal via its surface, let us first consider the adsorption
process of hydrogen at the metal surface, H2 (g) → 2H(ad), where the gaseous and adsorbed states ofhydrogen are 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 is represented by two variables: the interatomic
distance, 𝑑, and the height relative to the surface metal atom, 𝑧. It is assumed that the axis along thetwo H atoms is parallel to the surface and that the centre of gravity is always on the vertical dotted line
in Fig. 1.
Fig. 2 shows the potential energy contour plot for the dissociation at the surface. The numerical values
represent the potential energy in units of kJ per mole of H2 . The solid line spacing is 20 kJ mol−1 , thedashed 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
GBR-4 C-1 Q-2

Q1-2

British English (United Kingdom)

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

Fig.2

Q1-3

IChO
GBR-4 C-1 Q-3

A.1

A.2

British English (United Kingdom)

For each of the following items (i)–(iii), select the closest value from A–G.
(i) The interatomic distance for a gaseous H2 molecule(ii) The interatomic distance between metal atoms (𝑑M in Fig. 1)(iii) The distance of adsorbed H atoms from the surface (ℎad in Fig. 1)
A. 0.03 nm

B. 0.07 nm

C. 0.11 nm

E. 0.19 nm

F. 0.23 nm

G. 0.27 nm

6pt

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 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
GBR-4 C-1 Q-4

Q1-4

British English (United Kingdom)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
GBR-4 C-1 Q-5

Q1-5

British English (United Kingdom)

A metal sample with a surface area of 𝑆 = 1.0 × 10−3 m2 was placed in a 1L container (1L = 1.0 × 10−3 m3 )with H2 at pressure 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 volume of the metal sample is negligible.

B.2

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

3pt

B.3

Calculate the value of 𝑘3 at 400 K, where 𝐶 = 1.0 × 102 Pa−1 .
If you 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 𝑇 , 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 plot from (a)–
(h).

3pt

IChO
GBR-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

British English (United Kingdom)

IChO
GBR-4 C-1 A-2

Part B
B.1 (5 pt)

𝐶=
B.2 (3 pt)𝐴=

mol s−1 m−2

A1-2

British English (United Kingdom)

IChO
GBR-4 C-1 A-3

B.3 (3 pt)

𝑘3 =B.4 (3 pt)

s−1

A1-3

British English (United Kingdom)GBR-4 C-2 C-1

GBR-4 C-2 C
Zachary McGuire

IChO
Problem 2
Cover sheet

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Q2-1

IChO
GBR-4 C-2 Q-1

British English (United Kingdom)

Isotope Time Capsule
11 % of the total
Question

A.1

A.2

A.3

A.4

Total

Points

8

8

10

9

35

Score

Molecules that differ only in isotopic composition, such as CH4 and CH3 D, are called isotopologues. Isotopologues are widely considered to have the same chemical characteristics. In fact, however, there is aslight difference.
Assume that all of the substances in this Question are in the gas phase.
Consider the following equilibrium:

The entropy, 𝑆, increases with increasing the number of possible microscopic states of a system, 𝑊 :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 in this molecule are distinguishable. As the right-hand side of the equilibrium shown in eq. 1 has two 12 C16 O18 O molecules, 𝑊 = 22 = 4.

IChO
GBR-4 C-2 Q-2

A.1

Q2-2

British English (United Kingdom)

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

8pt

Calculate the equilibrium constant, 𝐾, for eq. 3 at very low temperatures (as𝑇 → 0) and at very high temperatures (as 𝑇 → +∞).
Assume that the reaction remains unchanged at these temperatures and that
Δ𝐻 converges to a constant value for high temperatures.
The ΔH of the following process can be explained by considering molecular vibrations.

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

A.2

The vibration of H2 is at 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 the 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
GBR-4 C-2 Q-3

Q2-3

British English (United Kingdom)

In a system in equilibrium, the molar ratios among H2 , HD, and D2 depend on the temperature.
We define ΔD2 as the change of the molar ratio of D2 .

Here, 𝑅D2 refers to

[D2 ]
[D ]
in the sample and 𝑅D∗ 2 to 2 at 𝑇 → +∞.
[H2 ]
[H2 ]

It should be noted here that the distribution of isotopes becomes random at 𝑇 → +∞.
A.3

Calculate ΔD2 when the isotopic exchange is equilibrated at the temperature
where 𝐾 in eq. 4 is 0.300.
Assume that the natural abundance of D is 1.5576 × 10–4 and the natural abundance of H is (1 − 1.5576 × 10–4 ).

10pt

Q2-4

IChO
GBR-4 C-2 Q-4

British English (United Kingdom)

A doubly substituted isotopologue contains two heavy isotope atoms in one molecule.
The molar ratio of the doubly substituted isotopologue increases with decreasing temperature.
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:

𝑅47 refers to[CO2 [47]][CO2 [47]]∗in the sample and 𝑅47
toat 𝑇 → +∞.
[CO2 [44]][CO2 [44]]The natural abundances of carbon and oxygen atoms are shown below; ignore isotopes that are notshown here.
12

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 .
Estimate the temperature using this 𝑅47 .
This temperature corresponds to 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
GBR-4 C-2 A-1

Isotope Time Capsule
A.1 (8 pt)

𝑇 →0∶𝐾=, 𝑇 → +∞ ∶ 𝐾 =A2-1

British English (United Kingdom)

IChO
GBR-4 C-2 A-2

A.2 (8 pt)

Δ𝐻 =J mol−1

A2-2

British English (United Kingdom)

IChO
GBR-4 C-2 A-3

A.3 (10 pt)

ΔD2 =

A2-3

British English (United Kingdom)

IChO
GBR-4 C-2 A-4

A.4 (9 pt)

𝑇 =

K

A2-4

British English (United Kingdom)GBR-4 C-3 C-1

GBR-4 C-3 C
Zachary McGuire

IChO
Problem 3
Cover sheet

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Q3-1

IChO
GBR-4 C-3 Q-1

British English (United Kingdom)

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)

In Part A, the optical path length is 𝑙 . Ignore the density change upon dilution. Assume that no chemical
reactions other than eq 1 occur.
A.1

The absorbance of X was 𝐴1 at a wavelength of 𝜆1 .
Solution X was then diluted to twice its initial volume using hydrochloric acidwith a pH = 2.500. After the dilution, the absorbance was still 𝐴1 at 𝜆1 .
Determine the ratio 𝜀HA /𝜀A− , where 𝜀HA and 𝜀A− represent the absorption coefficients (at 𝜆1 ) of HA and A− , respectively.

10pt

Q3-2

IChO
GBR-4 C-3 Q-2

British English (United Kingdom)Part B
Consider the following equilibrium in the gas phase.
D ⇌ 2M(2)

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

Use the following definitions if necessary.
Initial state
D

M

D

M

Partial pressure

𝑃

0

𝑝D

𝑝M

Amount in moles

𝑛0

0

𝑛D

𝑛M

VolumeB.1

After equilibrium

𝑉0

𝑉

The absorbance of the gas at 𝜆B1 measured in direction 𝑥 (i.e. 𝑙 = 𝑙𝑥 ) was 𝐴B1
both at the initial state and after the equilibrium.

6pt

Determine the ratio 𝜀D /𝜀M at 𝜆B1 , where 𝜀D and 𝜀M represent the absorption
coefficients of D and of M, respectively.
B.2

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

6pt

IChO
GBR-4 C-3 A-1

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

British English (United Kingdom)

IChO
GBR-4 C-3 A-2

A.1 (cont.)

𝜀HA /𝜀A− =A3-2

British English (United Kingdom)

IChO
GBR-4 C-3 A-3Part B
B.1 (6 pt)

𝜀D /𝜀M =A3-3

British English (United Kingdom)

IChO
GBR-4 C-3 A-4

B.2 (6 pt)

𝜀D /𝜀M =A3-4

British English (United Kingdom)GBR-4 C-4 C-1

GBR-4 C-4 C
Zachary McGuire

IChO
Problem 4
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q4-1

IChO
GBR-4 C-4 Q-1

British English (United Kingdom)

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 separated by precipitation to purify the water. The precipitate is reduced to recover metallic zinc.

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

IChO
GBR-4 C-4 Q-2

Q4-2

British English (United Kingdom)

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

A.1

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)] once the equilibria in eq. 1–4 areestablished.

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.

5pt

Calculate the molar percentage of zinc that precipitates when the pH is increased from 7.00 to 12.00.
Ignore the volume and temperature changes.

Part 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

It is necessary to reduce the partial pressure of the water vapour formed, so
reaction (7) can proceed at a constant hydrogen pressure of 1 bar.
Calculate the upper limit for the partial pressure of water vapour to allow reaction (7) to proceed at 300 ∘ C.
Gibbs free energies of formation at 300 ∘ C and 1 bar for all gaseous species are:Δ𝐺ZnO (300∘ C) = −2.90 × 102 kJ mol−1
Δ𝐺H2 O (300∘ C) = −2.20 × 102 kJ mol−1

4pt

IChO
GBR-4 C-4 Q-3

Q4-3

British English (United Kingdom)

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

B.2

A zinc–air battery was discharged at 20 mA for 24 hours.

3pt

Calculate the change in mass of the negative electrode (anode) of the battery.

Mt. Fuji
B.3

Consider the change in e.m.f. of a zinc–air battery due to the environment.
Calculate the e.m.f. at the summit of Mt. Fuji, where the temperature andaltitude are −38 ∘ C (February) and 3776 m, respectively.
The atmospheric pressure is represented byat altitude ℎ [m] and temperature 𝑇 [∘ C].
The molar proportion of oxygen in the atmosphere is 21%.
The Gibbs free energy change of reaction (8) at −38 ∘ C and 1 bar is
Δ𝐺ZnO (−38∘ C) = −3.26 × 102 kJ mol−1 .

5pt

IChO
GBR-4 C-4 Q-4

B.4

Q4-4

British English (United Kingdom)

Calculate the Gibbs free energy change for reaction (6) at 25 ∘ C.
The standard reduction potentials at 25 ∘ C and 1 bar, 𝐸 ∘ (Zn2+ /Zn) and𝐸 ∘ (O2 /H2 O) are given as (10) and (11).

9pt

IChO
GBR-4 C-4 A-1

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

A4-1

British English (United Kingdom)

IChO
GBR-4 C-4 A-2

A.2 (5 pt)

%

A4-2

British English (United Kingdom)

IChO
GBR-4 C-4 A-3

Part B
B.1 (4 pt)

bar

𝑝H2 O=
B.2 (3 pt)

g

A4-3

British English (United Kingdom)

IChO
GBR-4 C-4 A-4

B.3 (5 pt)

V

A4-4

British English (United Kingdom)

IChO
GBR-4 C-4 A-5

B.4 (9 pt)

Δ𝐺∘ =J mol−1

A4-5

British English (United Kingdom)GBR-4 C-5 C-1

GBR-4 C-5 C
Zachary McGuire

IChO
Problem 5
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q5-1

IChO
GBR-4 C-5 Q-1

British English (United Kingdom)

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

The group 14 elements carbon and silicon differ significantly in their properties.

Part A
Unlike the carbon–carbon triple bond, the silicon–silicon triple bond, R1 –Si ≡ Si–R1 (R: organic substituent), is extremely reactive. For example, it reacts with ethene to form 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 A with another molecule of R2 –C ≡ C–R2 gives
isomers B and C, both of which have benzene-like cyclic conjugated structures.
These so-called ‘disilabenzenes’ contain a six-membered ring and can be formulated as (R1 –Si)2 (R2 –C)4 .

IChO
GBR-4 C-5 Q-2

Q5-2

British English (United Kingdom)

13

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

Draw the structural formulae of A, B, and C using R1 , R2 , Si, and C, as one of thepossible resonance structures.

9pt

A.2

Calculate the aromatic stabilisation energy (ASE) for benzene and C (in the caseof R1 = R2 = H) as positive values, given the enthalpy change of some hydrogenation reactions of unsaturated systems shown below (Fig. 1).

7pt

Fig. 1IChO
GBR-4 C-5 Q-3

Q5-3

British English (United Kingdom)

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

Calculate Δ𝐻 for the transformation of D to E.
Assume that Δ𝐻 does not depend on temperature.

6pt

The isomerisation from C to D and to E proceeds via transformations of 𝜋-bonds into 𝜎-bonds without
breaking any 𝜎-bonds. 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 structural formulae of D and E using R1 , R2 , Si, and C.

10pt

Part B
Silicon is able to form highly coordinated compounds (more than four substituents) with electronegative elements such as fluorine. Like metal fluorides, highly coordinated silicon fluorides can also act as
fluorination reagents.
The fluorination reaction of CCl4 using Na2 SiF6 was carried out as follows.
• Standardisation 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: 200 mL).
· Precipitation Titration Procedure
Solution F (50.0 mL) was titrated with solution G in the presence of xylenol orange, an indicator which
coordinates to Ce3+ . After adding 18.8 mL of solution G, the colour of the solution changed from yellow
to magenta. The precipitate formed 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 vessel.
The unreacted Na2 SiF6 and NaCl produced were removed by filtration. The filtrate was diluted 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 integrals in the 19 F NMR spectrum are proportional to the number of fluorine nuclei.

Q5-4

IChO
GBR-4 C-5 Q-4

British English (United Kingdom)

Table 119

F NMR data

Integral

CFCl3

CF2 Cl2

CF3 Cl

CF4

45.0

65.0

18.0

2.0

SiF4 is hydrolysed to form H2 SiF6 according to the following eq. 8:

Solution H (10 mL) was added to excess water, which resulted in the complete hydrolysis of SiF4 . After separation, the H2 SiF6 generated from the hydrolysis was neutralised 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 (previously underlined), was completely dissolved in water to give an aqueous solution (solution K; 10.0 L).
Then, additional precipitation titrations using solution G were carried out, and the endpoints of the titrations with G were as follows:·For the entire amount of solution J: 61.6 mL.
·For 100 mL of solution K: 44.4 mL.
The presence of NaCl or SiO2 has no effect on the precipitation titration.
B.2

Calculate the mass of NaCl produced in the reaction vessel (information previously underlined), and calculate the mass (𝑥 [g]) of the Na2 SiF6 used as starting
material.

15pt

B.3

77.8% of the CCl4 used as a starting material did not react.
Calculate the mass of CF3 Cl generated.

8pt

IChO
GBR-4 C-5 A-1

A5-1

British English (United Kingdom)

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
GBR-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

British English (United Kingdom)

IChO
GBR-4 C-5 A-3

Part B
B.1 (5 pt)

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

A5-3

British English (United Kingdom)

IChO
GBR-4 C-5 A-4

B.2 (cont.)

NaCl ∶

g, Na2 SiF6 ∶

g

A5-4

British English (United Kingdom)

IChO
GBR-4 C-5 A-5

B.3 (8 pt)

CF3 Cl ∶g

A5-5

British English (United Kingdom)GBR-4 C-6 C-1

GBR-4 C-6 C
Zachary McGuire

IChO
Problem 6
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q6-1

IChO
GBR-4 C-6 Q-1

British English (United Kingdom)

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 has many volcanoes. When silicate minerals crystallize from magma, some 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− ). They adopt a four-coordinate
tetrahedral (𝑇d ) geometry in the magma and a six-coordinate octahedral (𝑂h ) geometry in the silicate
minerals, both of which exhibit a high-spin electron configuration.
The distribution coefficient of Mn+ between the silicate minerals and magma, 𝐷, can be expressed by:where [M]s and [M]l are the concentrations of Mn+ in the silicate minerals and the magma, respectively.

IChO
GBR-4 C-6 Q-2

Q6-2

British English (United Kingdom)

The table below shows the 𝐷 values of Cr2+ and Mn2+ as examples.

In an 𝑂h field, ΔO is the energy separation of the d-orbitals of Mn+ and CFSEO is the crystal-field stabilization energy.
ΔT and CFSET are the equivalents in a 𝑇d field.
A.1

6pt

ΔCFSE is defined as:
ΔCFSE = |CFSEO − CFSET |Calculate ΔCFSE in terms of ΔO for Cr2+ , Mn2+ , and Co2+ .
Assume ΔT = 4/9ΔO .

A.2

A linear relationship is observed by plotting ln𝐷 against ΔCFSE / ΔO as shown
below.
Estimate 𝐷 for Co2+ .

3pt

IChO
GBR-4 C-6 Q-3

Q6-3

British English (United Kingdom)

Metal oxides, MO (where M is Ca, Ti, V, Mn, or Co), crystallise in a rock-salt structure wherein the Mn+
adopts an 𝑂h geometry with a high-spin electron configuration.
The lattice enthalpy of these oxides is mainly governed by the Coulomb interactions based on the radiusand charge of the ions and some contributions 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
GBR-4 C-6 Q-4

Q6-4

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

Fig. 1
A can be synthesised by thermal decomposition of complex B. B is formed by mixing metal chlorides indilute aqueous ammonia solution containing the diacid, squaric acid C4 H2 O4 .
The thermal decomposition of B in dry air shows a weight loss of 29.1% up to 200 °C due to the loss of
crystallisation water, followed by another weight loss up to 700 °C due to the release of CO2 .
The total weight loss during the formation of A from B is 63.6%. Only water and CO2 are released in thethermal decomposition reaction.
B.1

Write the chemical formulae for A and B.

6pt

B.2

Calculate 𝑙𝑥 and 𝑙𝑧 using Fig. 1.

4pt

B.3

Complete the diagram in your answer sheet for Cu2+ in the distorted [CuO6 ]octahedron in A of Fig. 1:• Write the names of the split eg orbitals (d𝑥2 −𝑦2 and d𝑧2 ) in (i) and (ii).
• Draw the electron configuration in the dotted box.

4pt

Q6-5

IChO
GBR-4 C-6 Q-5British English (United Kingdom)

A is an insulator.
When one La3+ is substituted with one Sr2+ , one hole is generated in the crystal lattice that can conduct
electricity. As a result, the Sr2+ -doped A shows superconductivity below 38 K.
When a substitution reaction took place for A, 2.05 × 1027 holes m−3 were generated.
B.4

Calculate the percentage of La3+ substituted by Sr2+ based on the mole ratio
in the substitution reaction.
Note that the valence of the constituent ions and the crystal structure are notaltered by the substitution reaction.

4pt

Part C−

Cu2 (CH3 CO2 )4 is composed of four CH3 CO2 ions coordinated to two Cu

2+ions (Fig. 2A).

Cu2 (CH3 CO2 )4 exhibits high levels of structural symmetry, with two axes passing through the carbon−
2+
atoms of the four CH3 CO2 and an axis passing through the two Cu , all of which are oriented orthogonal relative to each other.
−

A “cage complex” is formed if a dicarboxylate ligand is used instead of CH3 CO2 .
2+

L1
The cage complex Cu4 (L1L1)4 is composed of planar dicarboxylate L1 (Fig. 2B) and Cu (Fig. 2C). The
angle θ between the coordination directions of the two carboxylates, indicated by the arrows in Fig. 2B,
determines the structure of the cage complex.
For L1 𝜃 = 0∘ .
Note that hydrogen atoms are not shown in Fig. 2.

Fig. 2IChO
GBR-4 C-6 Q-6

C.1

Q6-6

British English (United Kingdom)

The θ of the planar dicarboxylate L2 below is fixed to 90˚.
2+
L2
The cage complex formed from L2 and Cu is Cu𝑛 (L2L2)𝑚 .
Give the smallest integer combination of n and m.
−
2+
Assume that only the CO2 groups of L2 form a coordination bond to Cu ions.

5pt

Q6-7

IChO
GBR-4 C-6 Q-7

British English (United Kingdom)−

2+

A zinc complex, Zn4 O(CH3 CO2 )6 , contains four tetrahedral Zn , six CH3 CO2 , and one O2− (Fig. 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.
−2+

When p-benzenedicarboxylate (Fig. 3B, L3, θ = 180°) is used instead of CH3 CO2 , the Zn clusters arelinked to each other to form a crystalline solid (X) that is called a “porous coordination polymer” (Fig. 3C).
L3
The composition of X is [Zn4 O(L3L3)3 ]𝑛 , and it has a cubic crystal structure with nano-sized pores.
2+

One pore is represented as a sphere in Fig. 3D, and each tetrahedral Zn cluster is represented as adark grey polyhedron in Fig. 3C and 3D. Note that hydrogen atoms are not shown in Fig. 3.

Fig. 3
C.2

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

5pt

Calculate 𝑎 in [cm].
C.3

X contains a considerable number of pores.
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
GBR-4 C-6 A-1

A6-1

British English (United Kingdom)

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

Cr2+ ∶ΔO , Mn2+ ∶ΔO , Co2+ ∶

ΔO

IChO
GBR-4 C-6 A-2

A.2 (3 pt)

𝐷∶

A.3 (3 pt)

A6-2

British English (United Kingdom)

IChO

A6-3

GBR-4 C-6 A-3

British English (United Kingdom)Part B
B.1 (6 pt)

A∶

,B∶

B.2 (4 pt)

𝑙𝑥 =nm, 𝑙𝑧 =nm

IChO
GBR-4 C-6 A-4

B.3 (4 pt), (ii) ∶(i) ∶B.4 (4 pt)

%

A6-4

British English (United Kingdom)

IChO
GBR-4 C-6 A-5Part C
C.1 (5 pt)𝑛=,𝑚=

C.2 (5 pt)

𝑎=

cm

A6-5

British English (United Kingdom)

IChO
GBR-4 C-6 A-6

C.3 (5 pt)

A6-6

British English (United Kingdom)GBR-4 C-7 C-1

GBR-4 C-7 C
Zachary McGuire

IChO
Problem 7
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q7-1

IChO
GBR-4 C-7 Q-1

British English (United Kingdom)

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) founded the research field of non-benzenoid aromatic compounds.

Photo credit: Tohoku Univ.

Part A
Lineariifolianone is a natural product which was isolated from Inula linariifolia. From valencene (1), a
one-step conversion yields 2, before a three-step conversion via 3 yields ketone 4. Eremophilene (5) isconverted into 6 by performing the same four-step conversion.

IChO
GBR-4 C-7 Q-2

Q7-2

British English (United Kingdom)

Inula linariifolia

A.1

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

5pt

IChO
GBR-4 C-7 Q-3

Q7-3

British English (United Kingdom)Then, ketone 4 is converted into ester 15. Compound 8 (molecular weight: 188) retains all the stereocentres in 7. Compounds 9 and 10 have five stereocentres 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. Both 13 and 14 provide the same product 15 withidentical stereochemistry ignoring isotopic labelling.

IChO
GBR-4 C-7 Q-4

Q7-4

British English (United Kingdom)

IChO
GBR-4 C-7 Q-5

Q7-5

British English (United Kingdom)

A.2

Choose the appropriate structure for A.

2pt

A.3

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

19pt

IChO
GBR-4 C-7 Q-6

Q7-6

British English (United Kingdom)Part B
Compound 19 is synthesised as shown below. Linking to non-benzenoid aromaticity, 19 can be used as anactivator for alcohols, and 20 was converted to 22 via ion-pair intermediate 21. Although the formationof 21 was observed by NMR, 21 gradually decomposes to give 18 and 22.

B.1

Draw the structures of 17–19 and 21. Stereochemistry is not required.

10pt

IChO
GBR-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

British English (United Kingdom)

IChO
GBR-4 C-7 A-2

A.3 (19 pt)

A7-2

British English (United Kingdom)

IChO
GBR-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

British English (United Kingdom)GBR-4 C-8 C-1

GBR-4 C-8 C
Zachary McGuire

IChO
Problem 8
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q8-1

IChO
GBR-4 C-8 Q-1

British English (United Kingdom)

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 (here,n represents the number of six-membered rings) (see below).
[4]Carbohelicene ([4]C) is efficiently prepared by a route using a photoreaction as shown below, via anintermediate (Int.) that is readily oxidized by iodine.

The photoreaction proceeds in a manner similar to the following example.

IChO
GBR-4 C-8 Q-2

Q8-2

British English (United Kingdom)

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 affording product D whose molecularweight was 2 Da lower than that of [5]carbohelicene.

3pt

Draw the structure of D.
The 1 H NMR chemical shifts of D are listed below.
[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)]

IChO
GBR-4 C-8 Q-3

Q8-3

British English (United Kingdom)[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 credit: The Japan Prize Foundation

IChO
GBR-4 C-8 Q-4

Q8-4

British English (United Kingdom)

A molecule that contains two or more helicene-like structures is called a multiple-helicene. If helicalchirality is considered there can be several stereoisomers of a multiple-helicene.
For example, compound E contains three [5]carbohelicene-like moieties in one molecule. One of thestereoisomers is described as (P, P, P) as shown below.

IChO
GBR-4 C-8 Q-5

A.3

Q8-5

British English (United Kingdom)

The nickel-mediated trimerisation of 1,2-dibromobenzene generates triphenylene. When the same reaction is applied to an enantiomer of F, (P)-F, multiplehelicene G (C66 H36 ) is obtained.
Given that interconversion between stereoisomers does not occur during thereaction, identify all the possible stereoisomers of G formed in this process,without duplicating stereoisomers.
As a reference, one isomer should be drawn completely with the chirality defined as in the example above, with numerical labels.
The other stereoisomers should be listed with location numbers and M and P
labels according to the same numbering.
For instance, the other stereoisomers of E should be listed as (1, 2, 3) = (P, M, P),(P, M, M), (P, P, M), (M, M, M), (M, M, P), (M, P, P), and (M, P, M).

7pt

IChO
GBR-4 C-8 Q-6

Q8-6

British English (United Kingdom)Part B
Sumanene is a bowl-shaped hydrocarbon that was first reported in Japan in 2003. The name "sumanene"derives from a Sanskrit-Hindi word "suman" that means sunflower. The synthesis of sumanene wasachieved by a reaction sequence that consists of a ring-opening and a ring-closing metathesis.

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

B.1

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

3pt

IChO
GBR-4 C-8 Q-7

B.2

Q8-7

British English (United Kingdom)Starting from the optically active precursor J, the same reaction sequence gives
the optically active sumanene derivative K. The stereocentres in J suffer no inversion during the metathesis reaction. Draw the structure of K with the appropriate stereochemistry.

4pt

IChO
GBR-4 C-8 A-1

A8-1

British English (United Kingdom)

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
GBR-4 C-8 A-2

A.3 (7 pt)

A8-2

British English (United Kingdom)

IChO
GBR-4 C-8 A-3

Part B
B.1 (3 pt)

B.2 (4 pt)

A8-3

British English (United Kingdom)GBR-4 C-9 C-1

GBR-4 C-9 C
Zachary McGuire

IChO
Problem 9
Cover sheet

Please return this cover sheet together with all the related question sheets.

Q9-1

IChO
GBR-4 C-9 Q-1

British English (United Kingdom)

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

If you cut a tennis ball along the seam, you can disassemble it into two U-shaped pieces.

Compounds 1 and 2 are U-shaped molecules with different sizes, inspired by this idea. Compound 3 was
prepared for comparison with 1 and the encapsulation behaviour of these compounds was investigated.

IChO
GBR-4 C-9 Q-2

Q9-2

British English (United Kingdom)

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

IChO
GBR-4 C-9 Q-3

A.1

Q9-3

British English (United Kingdom)

Draw the structures of 4–9; stereochemistry can be neglected. Use the "PMB"abbreviation instead of drawing the whole structure of 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 32 was not observed in the spectrum of 3.
In the 1 H NMR spectrum of a solution of 1 2 , all the NH protons derived from 1 were observed to bechemically equivalent, and their chemical shift was significantly different from that of the NH protons of
3. These data indicate that hydrogen bonds are formed between the NH moieties of 1 and atoms X ofanother molecule of 1 to form the dimeric capsule.
A.2

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

2pt

A.3

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

2pt

IChO
GBR-4 C-9 Q-4

Q9-4

British English (United Kingdom)

The dimeric capsule of 1 (112 ) has an internal space wherein 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 shown below:

Encapsulation of a molecule into a capsule can be monitored by NMR spectroscopy. For example, 1 2 inC6 D6 gave different signals in the 1 H NMR spectra before and after addition of CH4 .
Compound 2 also forms a larger, rigid 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 being kept constant. The
chemical shifts for the Ha proton of 2 in the above solvents are summarized below, and no other signals
from Ha in 2, except for those listed, were observed. Assume that the interior of the capsule is always
filled with the largest possible number of solvent molecules and that each signal corresponds to one
species of 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
GBR-4 C-9 Q-5

Q9-5

British English (United Kingdom)

H NMR measurements in C6 D6 revealed that 2 2 can incorporate one molecule of 1-adamantanecarboxylic
acid (AdA).

1

The association constants (𝐾a ) which are expressed below were determined for various temperatures.
[solvent@222 ] denotes a species containing one or more solvent molecules.

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

No C6 D6 is encapsulated in 1 2 .
A.5

Choose the correct options in gaps (1)–(5) in the following paragraph from A
and B.
In line II, the entropy change (Δ𝑆) is ( 1 ) and enthalpy change (Δ𝐻) is ( 2 ),indicating that the driving force for the encapsulation in line II is ( 3 ). Therefore,line I corresponds to ( 4 ), and line II corresponds to ( 5 ).
A

B

(1)

positive

negative(2)

positive

negative

(3)

Δ𝑆

Δ𝐻

(4)

1 2 and CH4

2 2 and AdA(5)

1 2 and CH4

2 2 and AdA

3pt

IChO
GBR-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

British English (United Kingdom)

IChO
GBR-4 C-9 A-2

A9-2

British English (United Kingdom)

A.2 (2 pt)

A.3 (2 pt)
A.4 (3 pt)
𝛿 (ppm) of Ha

number of C6 D6number of C6 D5 F

4.60 ppm

4.71 ppm

4.82 ppm

A.5 (3 pt)
(1) ∶

(2) ∶

(4) ∶

(5) ∶

(3) ∶