NZL-1 C-0 C-1

NZL-1 C-0 C
Abhinav Chawla

IChO
General instructions
Cover sheet

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IChO
NZL-1 C-0 G-1

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

G0-1

NZL English (New Zealand)

IChO

G0-2

NZL-1 C-0 G-2

NZL English (New Zealand)

General Instruction• You are only allowed to use pen to write 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 paper for working. Remember that answers written
outside the answer boxes will not be graded.
• You must write relevant calculations in the appropriate boxes when necessary. Full marks will only
be given for correct answers 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 for clarification.
• You are not allowed to leave your working place without permission. If you need any assistance(e.g., broken calculator, need to visit a restroom), 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
NZL-1 C-0 G-3

G0-3

NZL English (New Zealand)

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
NZL-1 C-0 G-4

G0-4

NZL English (New Zealand)

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 tothe system, 𝑤 is the work done on the system.

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

NZL-1 C-0 G-5

Heat change Δ𝑞
Nernst equation for redox reaction
Arrhenius equation

NZL English (New Zealand)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
NZL-1 C-0 G-6

Periodic Table

G0-6

NZL English (New Zealand)

IChO
NZL-1 C-0 G-7

1

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

G0-7

NZL English (New Zealand)

NZL-1 C-1 C-1

NZL-1 C-1 C
Abhinav Chawla

IChO
Problem 1
Cover sheet

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

Q1-1

IChO
NZL-1 C-1 Q-1NZL English (New Zealand)

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 -storage
technology.

Part AAs hydrogen is absorbed into the bulk of a metal via its surface, let us first consider the adsorption processof hydrogen 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 is represented by two variables: the distance between
the hydrogen atoms, 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 centre of mass is always on the verticaldotted line in Fig. 1.
Fig. 2 shows the potential energy contour plot for the dissociation at the surface. The numerical values
on the graph 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
NZL-1 C-1 Q-2

Q1-2
NZL English (New Zealand)

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

Fig.2 Potential energy contour plot for the dissociation at the surface

Q1-3

IChO
NZL-1 C-1 Q-3

A.1

A.2

NZL English (New Zealand)

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 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
NZL-1 C-1 Q-4

Q1-4
NZL English (New Zealand)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 𝜃remains constant.
B.1

𝑟3 can be expressed as:

Express 𝐶 using 𝑘1 and 𝑘2 .

5pt

IChO
NZL-1 C-1 Q-5

Q1-5
NZL English (New Zealand)

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 number of hydrogen-atom adsorption sites per unit area 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 the amount of H atoms in moles absorbed per unit area of the surface
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.
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 are given. For 𝑟3 as afunction of 𝑃H2 at this temperature, select the correct plot from (a)–(h).

3pt

IChO
NZL-1 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

NZL English (New Zealand)

IChO
NZL-1 C-1 A-2

Part B
B.1 (5 pt)

𝐶=

A1-2

NZL English (New Zealand)

IChO
NZL-1 C-1 A-3

B.2 (3 pt)𝐴=

mol s−1 m−2

B.3 (3 pt)

𝑘3 =B.4 (3 pt)

s−1

A1-3

NZL English (New Zealand)

NZL-1 C-2 C-1

NZL-1 C-2 C
Abhinav Chawla

IChO
Problem 2
Cover sheet

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Q2-1

IChO
NZL-1 C-2 Q-1

NZL English (New Zealand)

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 exists
a slight difference.
Assume that all of the substances shown in this Question are in a gas phase.
Let us 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 are distinguishable in thismolecule. As the right-hand side of the equilibrium shown in eq. 1 has two 12 C16 O18 O molecules,
𝑊 = 22 = 4.

IChO
NZL-1 C-2 Q-2

A.1

Q2-2
NZL English (New Zealand)

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

8pt

Calculate the equilibrium constants, 𝐾, for eq. 3 at very low (think of 𝑇 → 0) andvery high (think of 𝑇 → +∞) temperatures. 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 molecular vibrations.

At 𝑇 = 0 K, the vibrational energy of a diatomic molecule whose vibration frequency is 𝜈 [s−1 ] is expressedas:Wherein 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 when reported as a wavenumber.
Calculate the 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 to be 1 Da and the mass of D to be 2 Da.

8pt

IChO
NZL-1 C-2 Q-3

Q2-3
NZL English (New Zealand)

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 → +∞. It should be noted here that the
[H2 ]
[H2 ]
distribution of isotopes becomes random at 𝑇 → +∞.
Here, 𝑅D2 refers to
A.3

Calculate ΔD2 with natural D abundance when the isotopic exchange is in equilibrium at the temperature where 𝐾 in eq. 4 is 0.300.
Assume that the natural abundance ratios of D and H are 1.5576 × 10–4 and1 − 1.5576 × 10–4 , respectively.

10pt

Q2-4

IChO
NZL-1 C-2 Q-4

NZL English (New Zealand)

In general, the molar ratio of a 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 .
Estimate the temperature using this 𝑅47 . This temperature is interpreted as
the air temperature during the era in which the plankton lived. Consider only
the most common isotopologue of CO2 [47] for the calculation.

9pt

IChO
NZL-1 C-2 A-1

Isotope Time Capsule
A.1 (8 pt)

𝑇 →0∶𝐾=, 𝑇 → +∞ ∶ 𝐾 =A2-1

NZL English (New Zealand)

IChO
NZL-1 C-2 A-2

A.2 (8 pt)

Δ𝐻 =J mol−1

A2-2

NZL English (New Zealand)

IChO
NZL-1 C-2 A-3

A.3 (10 pt)

ΔD2 =

A2-3

NZL English (New Zealand)

IChO
NZL-1 C-2 A-4

A.4 (9 pt)

𝑇 =

K

A2-4

NZL English (New Zealand)

NZL-1 C-3 C-1

NZL-1 C-3 C
Abhinav Chawla

IChO
Problem 3
Cover sheet

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Q3-1

IChO
NZL-1 C-3 Q-1

NZL English (New Zealand)

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 are correlated via the following acid-base equilibrium:
HA ⇌ A− + H+

𝐾=

[A− ][H+ ]
[HA]

(1)

The optical path length is 𝑙 in Part A. 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 .
Then, solution X was diluted to twice its initial volume using hydrochloric acidwith 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 of HA and of A− , respectively, at 𝜆1 .

10pt

Q3-2

IChO
NZL-1 C-3 Q-2

NZL English (New Zealand)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 , and equilibrium is established while the total pressure is kept at
𝑃.
The absorbance of the gas is 𝐴 = 𝜀(𝑛/𝑉 )𝑙, where 𝜀, 𝑛, 𝑉 , and 𝑙 are the absorption coefficient, amount of
the gas in moles, volume of the gas, and optical path length, respectively.
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 (𝑙 = 𝑙𝑥 ) was 𝐴B1
both at the initial state and after the equilibrium.
Determine the ratio 𝜀D /𝜀M at 𝜆B1 , where 𝜀D and 𝜀M represent the absorption
coefficients of D and of 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
NZL-1 C-3 A-1

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

A3-1

NZL English (New Zealand)

IChO
NZL-1 C-3 A-2

A.1 (cont.)

𝜀HA /𝜀A− =A3-2

NZL English (New Zealand)

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

𝜀D /𝜀M =A3-3

NZL English (New Zealand)

IChO
NZL-1 C-3 A-4

B.2 (6 pt)

𝜀D /𝜀M =A3-4

NZL English (New Zealand)

NZL-1 C-4 C-1

NZL-1 C-4 C
Abhinav Chawla

IChO
Problem 4
Cover sheet

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

Q4-1

IChO
NZL-1 C-4 Q-1

NZL English (New Zealand)

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 as alloys for brass and steel materials. The zinc contained in industrial wastewater is separated by precipitation to detoxify the water, and the precipitate is reduced to re-use it asmetallic 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
NZL-1 C-4 Q-2

Q4-2
NZL English (New Zealand)

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

A.1

When the equilibria in eq. 1–4 are established, calculate the pH range
in which [Zn(OH)2 (aq)] is the greatest among [Zn2+ (aq)], [Zn(OH)2 (aq)] and[Zn(OH)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 molar percentage of zinc that precipitates when increasing thepH from 7.00 to 12.00.
Ignore any volume or 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 hydrogen pressure kept at 1 bar, it isnecessary 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.
Here, the Gibbs formation energies of zinc oxide and water vapor at 300 ∘ C
and 1 bar for all gaseous species are Δ𝐺ZnO (300∘ C) = −2.90 × 102 kJ mol−1 andΔ𝐺H2 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 the change in mass of the negative electrode (anode) of the battery.

3pt

IChO
NZL-1 C-4 Q-3

Q4-3
NZL English (New Zealand)

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. of a zinc–air battery at the summit of Mt. Fuji, where thetemperature and altitude are −38 ∘ C (February) and 3776 m, respectively. The
pressure of the atmosphere at a given altitude is represented by

5pt

at altitude ℎ [m] and temperature 𝑇 [∘ C].
The molar ratio of oxygen in the atmosphere is 21%.
The Gibbs energy change of reaction (8) is Δ𝐺ZnO (−38∘ C) = −3.26 × 102 kJ mol−1
at −38 ∘ C and 1 bar.
B.4

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

9pt

IChO
NZL-1 C-4 A-1

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

NZL English (New Zealand)

IChO
NZL-1 C-4 A-2

A.2 (5 pt)

%

A4-2

NZL English (New Zealand)

IChO
NZL-1 C-4 A-3

Part B
B.1 (4 pt)

bar

𝑝H2 O=
B.2 (3 pt)

g

A4-3

NZL English (New Zealand)

IChO
NZL-1 C-4 A-4

B.3 (5 pt)

V

A4-4

NZL English (New Zealand)

IChO
NZL-1 C-4 A-5

B.4 (9 pt)

Δ𝐺∘ =J mol−1

A4-5

NZL English (New Zealand)

NZL-1 C-5 C-1

NZL-1 C-5 C
Abhinav Chawla

IChO
Problem 5
Cover sheet

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Q5-1

IChO
NZL-1 C-5 Q-1

NZL English (New Zealand)

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 compound formulated as
R1 –Si ≡ Si–R1 (R: organic substituent) is extremely reactive.
For example, it reacts with ethylene to form a cyclic product that 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 havebenzene-like cyclic conjugated structures, so-called ‘disilabenzenes’ that contain a six-membered ringIChO
NZL-1 C-5 Q-2

Q5-2
NZL English (New Zealand)and can be formulated as (R1 –Si)2 (R2 –C)4 .

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 structural formulae of A, B, and C using R1 , R2 , Si, and C, with one ofthe possible resonance structures.

9pt

A.2

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

7pt

Fig. 1IChO
NZL-1 C-5 Q-3

Q5-3
NZL English (New Zealand)

When a xylene solution of C is heated, it completely undergoes isomerisation to give a mixture of compounds D and E in equilibrium.
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 isomerization from C to D and 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 three-membered 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 (> four substituents) with electronegative elements
such as fluorine. As metal fluorides are often used as fluorination reagents, highly coordinated silicon
fluorides also act as fluorination reagents.
The fluorination reaction of CCl4 using 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: 200 mL).
· Procedure
Precipitation titration of a solution F (50.0 mL) by dropwise adding solution G in the presence of xylenol
orange, which coordinates to Ce3+ , as an indicator.
After adding 18.8 mL of solution G, the colour of the solution changes from yellow to magenta.
The precipitate formed is a binary compound that contains Ce3+ , and the only resulting silicon compoundis 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 :(Material losses by e.g., evaporation, are negligible during the following operations.)
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 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.

Q5-4

IChO
NZL-1 C-5 Q-4

NZL English (New Zealand)

In the 19 F NMR spectrum, in addition to SiF4 , signals corresponding to CFCl3 , CF2 Cl2 , CF3 Cl, and CF4 wereobserved (see Table 1). The integration ratios in the 19 F NMR spectrum are proportional to the numberof fluorine nuclei.
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 eq. 8:

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 andcompletely 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), 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 solution J (entire amount): 61.6 mL.
·For 100 mL of solution K: 44.4 mL.
Note: the presence of NaCl or SiO2 has no effect on the precipitation titration.
B.2

Calculate the mass of the NaCl produced in the reaction vessel (information
underlined), 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 the massof CF3 Cl generated.

8pt

IChO
NZL-1 C-5 A-1

A5-1

NZL English (New Zealand)

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
NZL-1 C-5 A-2

A.3 (6 pt)

Δ𝐻 =kJ mol–1

A5-2

NZL English (New Zealand)

IChO
NZL-1 C-5 A-3

A.4 (10 pt)
D (5 pt)

E (5 pt)

A5-3

NZL English (New Zealand)

IChO
NZL-1 C-5 A-4

Part B
B.1 (5 pt)

A5-4

NZL English (New Zealand)

IChO
NZL-1 C-5 A-5

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

A5-5

NZL English (New Zealand)

IChO
NZL-1 C-5 A-6

B.2 (cont.)

NaCl ∶

g, Na2 SiF6 ∶

g

A5-6

NZL English (New Zealand)

IChO
NZL-1 C-5 A-7

B.3 (8 pt)

CF3 Cl ∶g

A5-7

NZL English (New Zealand)

NZL-1 C-6 C-1

NZL-1 C-6 C
Abhinav Chawla

IChO
Problem 6
Cover sheet

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Q6-1

IChO
NZL-1 C-6 Q-1

NZL English (New Zealand)

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 a very large number of volcanos. When silicate minerals crystallize from magma, sometransition-metal ions (Mn+ ) in the magma are incorporated into the silicate minerals.
The Mn+ in this problem are coordinated by oxide ions (O2− ) and adopt a four-coordinate tetrahedral(𝑇d ) geometry in the magma and six-coordinate octahedral (𝑂h ) geometry in the silicate minerals, bothof which exhibit a high-spin electron configuration.
The distribution coefficient of Mn+ between the silicate minerals and magma, 𝐷, can be expressed as:[M]s and [M]l are the concentrations of Mn+ in the silicate minerals and the magma, respectively. The
table below gives 𝐷 for Cr2+ and Mn2+ examples.

IChO
NZL-1 C-6 Q-2

Q6-2
NZL English (New Zealand)

ΔO and CFSEO represent the energy separation of the d-orbitals of Mn+ and the crystal-field stabilizationenergy in a 𝑂h field, respectively. ΔT and CFSET are these in a 𝑇d field.
A.1

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

6pt

A.2

A linear relationship is observed when ln𝐷 is plotted against ΔCFSE / ΔO in the
Cartesian coordinate system shown below.
Estimate 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
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
NZL-1 C-6 Q-3

Q6-3
NZL English (New Zealand)Part B
A mixed oxide A, which contains La3+ and Cu2+ , crystallizes in a tetragonal unit cell shown in Fig.1.
In the [CuO6 ] octahedron, the Cu–O length along the z-axis (𝑙𝑧 ) is longer than that of the x-axis (𝑙𝑥 ), and[CuO6 ] is distorted from the regular 𝑂h geometry.
This distortion removes the degeneracy of the eg orbitals (d𝑥2 −𝑦2 and d𝑧2 ).

Fig. 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).
Pyrolysis B in dry air results in a weight loss of 29.1% up to 200 °C due to the loss of waters of crystallization. This is followed by release of CO2 at temperatures up to 700 °C .
The total weight loss during the formation of A from B is 63.6%. It should be noted that only water andCO2 are released in the pyrolysis reaction.
B.1

Write the chemical formulae for A and B.

6pt

B.2

Calculate 𝑙𝑥 and 𝑙𝑧 using Fig. 1.

4pt

B.3

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

4pt

Q6-4

IChO
NZL-1 C-6 Q-4

NZL English (New Zealand)

A is an insulator. When one La3+ is substituted with one Sr2+ , one hole is generated in the crystal latticethat can conduct electricity. As a result, the Sr2+ -doped A shows superconductivity below 38 K.
When a substitution reaction took 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 reaction.
Note that the valences of the constituent ions and the crystal structure are notaltered by the substitution reaction.

4pt

Part C−2+

Cu2 (CH3 CO2 )4 is composed of four CH3 CO2 coordinated to two Cu (Fig. 2A). Cu2 (CH3 CO2 )4 exhibits−high levels of structural symmetry. Two axes pass through the carbon atoms of the four CH3 CO2 and2+an axis passes through the two Cu . All of these 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 (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. The θ is 0° for L1. Note that hydrogen atoms are notshown in Fig. 2.

Fig. 2IChO
NZL-1 C-6 Q-5

C.1

Q6-5
NZL English (New Zealand)

The θ of the planar dicarboxylate L2 below is fixed to 90˚. Assume the compo2+
L2
sition of 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 CO22+groups of L2 coordinate to Cu ions.

5pt

Q6-6

IChO
NZL-1 C-6 Q-6NZL English (New Zealand)

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 areoriented 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. One poreis represented as a sphere in Fig. 3D.
2+Each tetrahedral Zn cluster is represented as dark grey polyhedron in Fig. 3C and 3D. Hydrogen atoms
are not shown in Fig. 3.

Fig. 3
C.2

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

5pt

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
NZL-1 C-6 A-1

A6-1

NZL English (New Zealand)

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

Cr2+ ∶ΔO , Mn2+ ∶ΔO , Co2+ ∶

ΔO

IChO
NZL-1 C-6 A-2

A.2 (3 pt)

𝐷∶

A.3 (3 pt)

A6-2

NZL English (New Zealand)

IChO

A6-3

NZL-1 C-6 A-3

NZL English (New Zealand)Part B
B.1 (6 pt)

A∶

,B∶

B.2 (4 pt)

𝑙𝑥 =nm, 𝑙𝑧 =nm

IChO
NZL-1 C-6 A-4

B.3 (4 pt), (ii) ∶(i) ∶B.4 (4 pt)

%

A6-4

NZL English (New Zealand)

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

C.2 (5 pt)

𝑎=

cm

A6-5

NZL English (New Zealand)

IChO
NZL-1 C-6 A-6

C.3 (5 pt)

A6-6

NZL English (New Zealand)

NZL-1 C-7 C-1

NZL-1 C-7 C
Abhinav Chawla

IChO
Problem 7
Cover sheet

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Q7-1

IChO
NZL-1 C-7 Q-1

NZL English (New Zealand)

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) initiated research into non-benzenoid aromatics, compounds which are now
ubiquitous in organic chemistry.

Photo courtesy: Tohoku Univ.

Part A

Inula linariifolia

IChO
NZL-1 C-7 Q-2

A.1

Q7-2
NZL English (New Zealand)

The natural product lineariifolianone, was isolated from Inula linariifolia. Valencene (1) can be converted in a one-step process to 2. 2 is converted in threestep process via 3 to ketone 4. Eremophilene (5) is converted into 6 using thesame four-step process.
Draw structures of 2 and 6 clearly show stereochemistry where necessary.

5pt

Then, ketone 4 is converted by a multi-step process into ester 15.
Compound 8 (molecular weight: 188) retains all 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. If the isotopic
labelling is ignored, both 13 and 14 provide the same product 15 with identical stereochemistry.

IChO
NZL-1 C-7 Q-3

Q7-3
NZL English (New Zealand)

IChO
NZL-1 C-7 Q-4Q7-4
NZL English (New Zealand)

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
NZL-1 C-7 Q-5Q7-5
NZL English (New Zealand)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 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. Identifying the stereochemistry is notnecessary.

10pt

IChO
NZL-1 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

NZL English (New Zealand)

IChO
NZL-1 C-7 A-2

A.3 (19 pt)

A7-2

NZL English (New Zealand)

IChO
NZL-1 C-7 A-3

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

18 (2 pt)

19 (3 pt)

21 (3 pt)

A7-3

NZL English (New Zealand)

NZL-1 C-8 C-1

NZL-1 C-8 C
Abhinav Chawla

IChO
Problem 8
Cover sheet

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Q8-1

IChO
NZL-1 C-8 Q-1

NZL English (New Zealand)

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 prepared,
using a photoreaction of the type shown below, via an intermediate (Int.) that is readily oxidized byiodine.

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

IChO
NZL-1 C-8 Q-2

Q8-2
NZL English (New Zealand)Note: For all of Question 8, please draw alternating single and double bonds in your answers asshown in the examples of carbohelicene. Do not use circles for conjugated 𝜋 systems.
A.1

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

9pt

A.2

Attempts to synthesize [5]carbohelicene from the phosphonium salt above andan appropriate starting compound resulted in the formation of only a trace
amount of [5]carbohelicene. This reaction gave product D having a molecular weight 2 Da lower than [5]carbohelicene. The 1 H NMR chemical shifts of D
are shown 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)]
Draw the structure of D.

3pt[5]- and larger [n]carbohelicenes have helical chirality and interconversion between enantiomers of thesehelicenes is significantly slow at room temperature. The configuration of [n]carbohelicenes is defined as(M) or (P) as shown below.

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

Photo courtesy: The Japan Prize Foundation

IChO
NZL-1 C-8 Q-3

Q8-3
NZL English (New Zealand)

Multiple helicenes are molecules that contain two or more helicene-like structures. If its helical chirality
is considered, several stereogenic (stereoisomeric) centres 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) as shown below.

A.3

The nickel-mediated trimerization of 1,2-dibromobenzene generates triphenylene as shown below. When the same reaction is applied to an enantiomer ofF, (P)-F, multiple helicene G (C66 H36 ) is obtained.
Assuming that interconversion between stereoisomers does not occur during
the reaction, identify all possible stereoisomers of G formed in this process,without duplication.
One isomer should be drawn completely with the chirality defined as in the example, using numerical labels. 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
NZL-1 C-8 Q-4Q8-4
NZL English (New Zealand)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 catalyzed by a ruthenium catalyst (Ru*) are shown below.

B.1

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

3pt

IChO
NZL-1 C-8 Q-5

B.2

Q8-5
NZL English (New Zealand)Starting from the optically active precursor J, the same reaction sequence gives
the optically active sumanene derivative K. The stereocenters in J are not inverted during the metathesis reaction. Draw the structure of K with appropriate stereochemistry.

4pt

IChO
NZL-1 C-8 A-1

A8-1

NZL English (New Zealand)

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
NZL-1 C-8 A-2

A.3 (7 pt)

A8-2

NZL English (New Zealand)

IChO
NZL-1 C-8 A-3

Part B
B.1 (3 pt)

B.2 (4 pt)

A8-3

NZL English (New Zealand)

NZL-1 C-9 C-1

NZL-1 C-9 C
Abhinav Chawla

IChO
Problem 9
Cover sheet

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

Q9-1

IChO
NZL-1 C-9 Q-1

NZL English (New Zealand)

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

Based on this idea, compounds 1 and 2 were synthesized as U-shaped molecules of different sizes. Compound 3 was prepared as a model of 1. The encapsulation behaviour of these was investigated.

IChO
NZL-1 C-9 Q-2

Q9-2
NZL English (New Zealand)

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

IChO
NZL-1 C-9 Q-3

A.1

Q9-3
NZL English (New Zealand)

Draw the structures of 4–9. It is not necessary to show stereochemistry. Use"PMB" as an abbreviation so that you do not need to draw the structure of pmethoxybenzyl group shown in the scheme above.

13pt

In the mass spectrum of 1, an ion peak corresponding to its dimer (112 ) was clearly observed, whereas
an ion peak for the dimer of 3 (332 ) was not observed in the spectrum of 3. In the 1 H NMR spectra ofa solution of 1 2 , all N-H protons derived from 1 were observed to be chemically equivalent, and theirchemical shift was significantly different from that of the N-H protons of 3. This indicates that hydrogenbonds are formed between the NH groups of 1 and atoms X of another molecule of 1 to form the dimericcapsule.
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
NZL-1 C-9 Q-4Q9-4
NZL English (New Zealand)

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 reaction equation:

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

Encapsulation of a molecule can be monitored by NMR spectroscopy. For example, 1 2 in C6 D6 gavedifferent signals in the 1 H NMR spectra before and after addition of CH4 .
Compound 2 also forms a rigid and larger 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. No other signals
from the Ha in 2, except for those listed, were observed. 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
NZL-1 C-9 Q-5Q9-5
NZL English (New Zealand)

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

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 the encapsulation in line II is (3). Therefore, line I corresponds to (4), and line II corresponds to (5).
A.5

Choose the correct option, from A and B, for each of the gaps (1)–(5) in theparagraph above.
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
NZL-1 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

NZL English (New Zealand)

IChO
NZL-1 C-9 A-2

A9-2

NZL English (New Zealand)

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) ∶