International Chemistry Olympiad

IChO 2021

Osaka, Japan (held remotely) · 9 theory problems with complete solutions

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  1. Problem 1: Hydrogen at a Metal Surface

    Figure 1

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

    PART A

    As 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, \(\ce{H2(g) -> 2H(ad)}\), where the gaseous and adsorbed states of hydrogen are represented as (g) and (ad), respectively. Hydrogen molecules (\(\ce{H2}\)) that reach the metal surface (M) dissociate at the surface and are adsorbed as \(\ce{H}\) atoms (Fig. 1). Here, the potential energy of \(\ce{H2}\) is represented by two variables: the interatomic distance, \(d\), and the height relative to the surface metal atom, \(z\). It is assumed that the axis along the two \(\ce{H}\) atoms is parallel to the surface and that the center of gravity is always on the vertical dotted line in Fig. 1. Fig. 2 shows the potential energy contour plot for the dissociation at the surface. The numerical values represent the potential energy in units of kJ per mole of \(\ce{H2}\). The solid line spacing is \(20\ \mathrm{kJ\,mol^{-1}}\), the dashed line spacing is \(100\ \mathrm{kJ\,mol^{-1}}\), and the spacing between solid and dashed lines is \(80\ \mathrm{kJ\,mol^{-1}}\). The zero-point vibration energy is ignored.

    Figure 2

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

    Figure 3

    Fig.2

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

    (i) The interatomic distance for a gaseous \(\ce{H2}\) molecule

    (ii) The interatomic distance between metal atoms (\(d_{\mathrm M}\) in Fig. 1)

    (iii) The distance of adsorbed \(\ce{H}\) atoms from the surface (\(h_{\mathrm{AD}}\) in Fig. 1)

    A. \(0.03\ \mathrm{nm}\) B. \(0.07\ \mathrm{nm}\) C. \(0.11\ \mathrm{nm}\) D. \(0.15\ \mathrm{nm}\)
    E. \(0.19\ \mathrm{nm}\) F. \(0.23\ \mathrm{nm}\) G. \(0.27\ \mathrm{nm}\)

    1.2 For each of the following items (i)–(ii), select the closest value from A–H.

    (i) the energy required for the dissociation of gaseous \(\ce{H2}\) to gaseous \(\ce{H}\) [\(\ce{H2(g) -> 2H(g)}\)]

    (ii) the energy released during the adsorption of a gaseous \(\ce{H2}\) [\(\ce{H2(g) -> 2H(ad)}\)]

    A. \(20\ \mathrm{kJ\,mol^{-1}}\) B. \(40\ \mathrm{kJ\,mol^{-1}}\) C. \(60\ \mathrm{kJ\,mol^{-1}}\) D. \(100\ \mathrm{kJ\,mol^{-1}}\)
    E. \(150\ \mathrm{kJ\,mol^{-1}}\) F. \(200\ \mathrm{kJ\,mol^{-1}}\) G. \(300\ \mathrm{kJ\,mol^{-1}}\) H. \(400\ \mathrm{kJ\,mol^{-1}}\)

    PART B

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

    Figure 4
    Figure 5

    The reaction rates per surface site for adsorption, desorption, and absorption are \(r_1[\mathrm{s^{-1}}]\), \(r_2[\mathrm{s^{-1}}]\) and \(r_3[\mathrm{s^{-1}}]\), respectively. They are expressed as:

    \[ r_1=k_1P_{\ce{H2}}(1-\theta)^2 \tag{2} \]

    \[ r_2=k_2\theta^2 \tag{3} \]

    \[ r_3=k_3\theta \tag{4} \]

    where \(k_1\ [\mathrm{s^{-1}\,Pa^{-1}}]\), \(k_2\ [\mathrm{s^{-1}}]\) and \(k_3\ [\mathrm{s^{-1}}]\) are the reaction rate constants and \(P_{\ce{H2}}\) is the pressure of \(\ce{H2}\). Among the sites available on the surface, \(\theta\) (\(0\leq\theta\leq1\)) is the fraction occupied by \(\ce{H}\) atoms. It is assumed that adsorption and desorption are fast compared to absorption (\(r_1,r_2\gg r_3\)) and that \(\theta\) remains constant.

    1.3 \(r_3\) can be expressed as:

    \[ r_3=\frac{k_3}{1+\sqrt{\dfrac{1}{P_{\ce{H2}}C}}} \tag{5} \]

    Express \(C\) using \(k_1\) and \(k_2\).

    A metal sample with a surface area of \(S=1.0\times10^{-3}\ \mathrm{m^2}\) was placed in a container (\(1\mathrm{L}=1.0\times10^{-3}\ \mathrm{m^3}\)) with \(\ce{H2}\) (\(P_{\ce{H2}}=1.0\times10^2\ \mathrm{Pa}\)). The density of hydrogen-atom adsorption sites on the surface was \(N=1.3\times10^{18}\ \mathrm{m^{-2}}\). The surface temperature was kept at \(T=400\ \mathrm K\). As the reaction (1) proceeded, \(P_{\ce{H2}}\) decreased at a constant rate of \(v=4.0\times10^{-4}\ \mathrm{Pa\,s^{-1}}\). Assume that \(\ce{H2}\) is an ideal gas and that the volume of the metal sample is negligible.

    1.4 Calculate the amount of \(\ce{H}\) atoms in moles absorbed per unit area of the surface per unit time, \(A\ [\mathrm{mol\,s^{-1}\,m^{-2}}]\).

    1.5 At \(T=400\ \mathrm K\), \(C\) equals \(1.0\times10^2\ \mathrm{Pa^{-1}}\). Calculate the value of \(k_3\) at \(400\ \mathrm K\). If you did not obtain the answer to 1.4, use \(A=3.6\times10^{-7}\ \mathrm{mol\,s^{-1}\,m^{-2}}\).

    1.6 At a different \(T\), \(C=2.5\times10^3\ \mathrm{Pa^{-1}}\) and \(k_3=4.8\times10^{-2}\ \mathrm{s^{-1}}\) are given. For \(r_3\) as a function of \(P_{\ce{H2}}\) at this temperature, select the correct plot from (a)–(h).

    Figure 6

    Official English text of the problem.

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  2. Problem 2: Isotope Time Capsule

    Figure 1

    Molecular entities that differ only in isotopic composition, such as \(\ce{CH4}\) and \(\ce{CH3D}\), 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:

    Figure 2

    The entropy, \(S\), increases with increasing the number of possible microscopic states of a system, \(W\):

    \[ S=k_{\mathrm B}\ln W \tag{2} \]

    \(W=1\) for \(\ce{^{12}C^{16}O2}\) and \(\ce{^{12}C^{18}O2}\). In contrast, \(W=2\) for a \(\ce{^{12}C^{16}O^{18}O}\) molecule because the oxygen atoms are distinguishable in this molecule. As the right-hand side of the equilibrium shown in eq. 1 has two \(\ce{^{12}C^{16}O^{18}O}\) molecules, \(W=2^2=4\).

    2.1 The enthalpy change, \(\Delta H\), of eq. 3 is positive regardless of the temperature.

    Figure 3

    Calculate the equilibrium constants, \(K\), for eq. 3 at very low (think of \(T\to0\)) and very high (think of \(T\to+\infty\)) temperatures. Assume that the reaction remains unchanged at these temperatures and that \(\Delta H\) converges to a constant value for high temperatures.

    The \(\Delta H\) of the following process can be explained by molecular vibrations.

    Figure 4

    At \(T=0\ \mathrm K\), the vibrational energy of a diatomic molecule whose vibration frequency is \(\nu\ [\mathrm{s^{-1}}]\) is expressed as:

    \[ E=\frac12h\nu \tag{5} \]

    \[ \nu=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}} \tag{6} \]

    Wherein \(k\) is the force constant and \(\mu\) the reduced mass, which is expressed in terms of the mass of the two atoms in the diatomic molecule, \(m_1\) and \(m_2\), according to:

    \[ \mu=\frac{m_1m_2}{m_1+m_2} \tag{7} \]

    2.2 The vibration of \(\ce{H2}\) is at \(4161.0\ \mathrm{cm^{-1}}\) when reported as a wavenumber.

    Calculate the \(\Delta H\) of the following equation at \(T=0\ \mathrm K\) in units of \(\mathrm{J\,mol^{-1}}\).

    Figure 5

    Assume that:

    • only the vibrational energy contributes to the \(\Delta H\).
    • the \(k\) values for \(\ce{H2}\), \(\ce{HD}\), and \(\ce{D2}\) are identical.
    • the mass of \(\ce{H}\) to be \(1\ \mathrm{Da}\) and the mass of \(\ce{D}\) to be \(2\ \mathrm{Da}\).

    The molar ratio of \(\ce{H2}\), \(\ce{HD}\), and \(\ce{D2}\) depends on the temperature in a system in equilibrium. Here, \(\Delta_{\ce{D2}}\) is defined as the change of the molar ratio of \(\ce{D2}\).

    \[ \Delta_{\ce{D2}}=\frac{R_{\ce{D2}}}{R_{\ce{D2}}^*}-1 \tag{9} \]

    Here, \(R_{\ce{D2}}\) refers to \(\dfrac{[\ce{D2}]}{[\ce{H2}]}\) in the sample and \(R_{\ce{D2}}^*\) to \(\dfrac{[\ce{D2}]}{[\ce{H2}]}\) at \(T\to+\infty\). It should be noted here that the distribution of isotopes becomes random at \(T\to+\infty\).

    2.3 Calculate \(\Delta_{\ce{D2}}\) with natural \(\ce{D}\) 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 \(\ce{D}\) and \(\ce{H}\) are \(1.5576\times10^{-4}\) and \(1-1.5576\times10^{-4}\), respectively.

    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 \(\ce{CO2}\) molecules with molecular weights of 44 and 47, which are described as \(\ce{CO2}[44]\) and \(\ce{CO2}[47]\) below. The quantity \(\Delta_{47}\) is defined as:

    \[ \Delta_{47}=\frac{R_{47}}{R_{47}^*}-1 \tag{10} \]

    \(R_{47}\) refers to \(\dfrac{[\ce{CO2}[47]]}{[\ce{CO2}[44]]}\) in the sample and \(R_{47}^*\) to \(\dfrac{[\ce{CO2}[47]]}{[\ce{CO2}[44]]}\) at \(T\to+\infty\). The natural abundances of carbon and oxygen atoms are shown below; ignore isotopes that are not shown here.

    \(\ce{^{12}C}\) \(\ce{^{13}O}\)
    natural abundance 0.988888 0.011112
    \(\ce{^{16}O}\) \(\ce{^{17}O}\) \(\ce{^{18}O}\)
    natural abundance 0.997621 0.0003790 0.0020000

    The temperature dependence of \(\Delta_{47}\) is determined as follows, where \(T\) is given as the absolute temperature in units of K:

    \[ \Delta_{47}=\frac{36.2}{T^2}+2.920\times10^{-4} \tag{11} \]

    2.4 The \(\Delta_{47}\) of fossil plankton obtained from the Antarctic seabed was \(4.50865\times10^{-5}\).

    Estimate the temperature using this \(R_{47}\). This temperature is interpreted as the air temperature during the era in which the plankton lived. Consider only the most common isotopologue of \(\ce{CO2}[47]\) for the calculation.

    Official English text of the problem.

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  3. Problem 3: Lambert–Beer Law?

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

    PART A

    An aqueous solution X was prepared using \(\ce{HA}\) and \(\ce{NaA}\). The concentrations \([\ce{A^-}]\), \([\ce{HA}]\), and \([\ce{H+}]\) in solution X are \(1.00 \times 10^{-2}\ \mathrm{mol\,L^{-1}}\), \(1.00 \times 10^{-3}\ \mathrm{mol\,L^{-1}}\), and \(1.00 \times 10^{-4}\ \mathrm{mol\,L^{-1}}\), respectively, which are correlated via the following acid-base equilibrium:

    Figure 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.

    3.1 The absorbance of X was \(A_1\) at a wavelength of \(\lambda_1\). Then, solution X was diluted to twice its initial volume using hydrochloric acid with pH = 2.500. After the dilution, the absorbance was still \(A_1\) at \(\lambda_1\). Determine the ratio \(\varepsilon_{\ce{HA}}/\varepsilon_{\ce{A^-}}\), where \(\varepsilon_{\ce{HA}}\) and \(\varepsilon_{\ce{A^-}}\) represent the absorption coefficients of \(\ce{HA}\) and of \(\ce{A^-}\), respectively, at \(\lambda_1\).

    PART B

    Let us consider the following equilibrium in the gas phase:

    Figure 2

    Pure gas \(\ce{D}\) 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 \(A=\varepsilon(n/V)l\), where \(\varepsilon\), \(n\), \(V\), 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.

    Figure 3

    Use the following definitions if necessary.

    Initial state After equilibrium
    \(\ce{D}\) \(\ce{M}\) \(\ce{D}\) \(\ce{M}\)
    Partial pressure \(P\) \(0\) \(p_{\mathrm D}\) \(p_{\mathrm M}\)
    Amount in moles \(n_0\) \(0\) \(n_{\mathrm D}\) \(n_{\mathrm M}\)
    Volume \(V_0\) \(V\)

    3.2 The absorbance of the gas at \(\lambda_{\mathrm{B1}}\) measured from direction \(x\) (\(l=l_x\)) was \(A_{\mathrm{B1}}\) both at the initial state and after the equilibrium. Determine the ratio \(\varepsilon_{\mathrm D}/\varepsilon_{\mathrm M}\) at \(\lambda_{\mathrm{B1}}\), where \(\varepsilon_{\mathrm D}\) and \(\varepsilon_{\mathrm M}\) represent the absorption coefficients of \(\ce{D}\) and of \(\ce{M}\), respectively.

    3.3 The absorbance of the gas at \(\lambda_{\mathrm{B2}}\) measured from direction \(y\) was \(A_{\mathrm{B2}}\) both at the initial state (\(l=l_{y0}\)) and after the equilibrium (\(l=l_y\)). Determine the ratio \(\varepsilon_{\mathrm D}/\varepsilon_{\mathrm M}\) at \(\lambda_{\mathrm{B2}}\).

    Official English text of the problem.

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  4. Problem 4: The Redox Chemistry of Zinc

    Figure 1

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

    PART A

    The dissolution equilibrium of zinc hydroxide \(\ce{Zn(OH)2(s)}\) at 25 °C and the relevant equilibrium constants are given in eq. 1–4.

    Figure 2

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

    \[ S=[\ce{Zn^2+(aq)}]+[\ce{Zn(OH)2(aq)}]+[\ce{Zn(OH)4^2-(aq)}] \tag{5} \]

    4.1 When the equilibria in eq. 1–4 are established, calculate the pH range in which \([\ce{Zn(OH)2(aq)}]\) is the greatest among \([\ce{Zn^2+(aq)}]\), \([\ce{Zn(OH)2(aq)}]\) and \([\ce{Zn(OH)4^2-(aq)}]\).

    4.2 A saturated aqueous solution of \(\ce{Zn(OH)2(s)}\) with pH = 7.00 was prepared and filtered. NaOH was added to this filtrate to increase its pH to 12.00. Calculate the molar percentage of zinc that precipitates when increasing the pH 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:

    Figure 3

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

    Figure 4

    4.3 In order for reaction (7) to proceed at a hydrogen pressure kept at 1 bar, it is necessary to reduce the partial pressure of the generated water vapor. Calculate the upper limit for 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 \(\Delta G_{\ce{ZnO}}(300\,^\circ\mathrm{C})=-2.90\times10^2\ \mathrm{kJ\,mol^{-1}}\) and \(\Delta G_{\ce{H2O}}(300\,^\circ\mathrm{C})=-2.20\times10^2\ \mathrm{kJ\,mol^{-1}}\), respectively.

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

    Figure 5

    4.4 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.

    Figure 6

    Mt. Fuji

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

    \[ P\,[\mathrm{bar}] =1.013\times \left(1-\frac{0.0065h}{T+0.0065h+273.15}\right)^{5.257} \tag{9} \]

    at altitude \(h\) [m] and temperature \(T\) [°C]. The molar ratio of oxygen in the atmosphere is 21%. The Gibbs energy change of reaction (8) is \(\Delta G_{\ce{ZnO}}(-38\,^\circ\mathrm{C})=-3.26\times10^2\ \mathrm{kJ\,mol^{-1}}\) at −38 °C and 1 bar.

    4.6 Calculate the Gibbs energy change for reaction (6) at 25 °C. Note that the standard reduction potentials, \(E^\circ(\ce{Zn^2+/Zn})\) and \(E^\circ(\ce{O2/H2O})\) at 25 °C and 1 bar are given as (10) and (11), respectively.

    Figure 7

    Official English text of the problem.

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  5. Problem 5: Mysterious Silicon

    Figure 1

    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 \(\ce{R^{1}-Si#Si-R^{1}}\) (R: organic substituent) is extremely reactive. For example, it reacts with ethylene to form a cyclic product that contains a four-membered ring.

    Figure 2

    When \(\ce{R^{1}-Si#Si-R^{1}}\) is treated with an alkyne (\(\ce{R^{2}-C#C-R^{2}}\)), the four-membered-ring compound A is formed as an initial intermediate. Further reaction of another molecule of \(\ce{R^{2}-C#C-R^{2}}\) with A affords isomers B and C, both of which have benzene-like cyclic conjugated structures, so-called ‘disilabenzenes’ that contain a six-membered ring and can be formulated as \(\ce{(R^{1}-Si)2(R^{2}-C)4}\).

    Figure 3

    The \(\ce{^{13}C}\) NMR analysis of the corresponding six-membered ring skeletons \(\ce{Si2C4}\) shows two signals for B and one signal for C.

    5.1 Draw the structural formulae of A, B, and C using \(\ce{R^{1}}\), \(\ce{R^{2}}\), \(\ce{Si}\), and \(\ce{C}\), with one of the possible resonance structures.

    5.2 Calculate the aromatic stabilization energy (ASE) for benzene and C (in the case of \(\ce{R^{1}}=\ce{R^{2}}=\ce{H}\)) as positive values, considering the enthalpy change in some hydrogenation reactions of unsaturated systems shown below (Fig. 1).

    Figure 4

    Fig. 1

    When a xylene solution of C is heated, it undergoes isomerization to give an equilibrium mixture of compounds D and E. The molar ratio is D : E = 1 : 40.0 at 50.0 °C and D : E = 1 : 20.0 at 120.0 °C.

    5.3 Calculate \(\Delta H\) for the transformation of D to E. Assume that \(\Delta H\) does not depend on temperature.

    The isomerization from C to D and to E proceeds via transformations of \(\pi\)-bonds into \(\sigma\)-bonds without breaking any \(\sigma\)-bonds. A \(\ce{^{13}C}\) NMR analysis revealed one signal for the \(\ce{Si2C4}\) skeleton of D and two signals for that of E. The skeleton of D does not contain any three-membered rings, while E has two three membered rings that share an edge.

    5.4 Draw the structural formulae of D and E using \(\ce{R^{1}}\), \(\ce{R^{2}}\), \(\ce{Si}\), and \(\ce{C}\)

    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 \(\ce{CCl4}\) using \(\ce{Na2SiF6}\) was carried out as follows.

    Standardization of \(\ce{Na2SiF6}\) solution:

    • Preparation

    Aqueous solution F: 0.855 g of \(\ce{Na2SiF6}\) (188.053 g mol\(^{-1}\)) dissolved in water (total volume: 200 mL).

    Aqueous solution G: 6.86 g of \(\ce{Ce2(SO4)3}\) (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 \(\ce{Ce^{3+}}\), 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 \(\ce{Ce^{3+}}\), and the only resulting silicon compound is \(\ce{Si(OH)4}\).

    5.5 Write the balanced equation for the reaction of \(\ce{Na2SiF6}\) with \(\ce{Ce2(SO4)3}\).

    • Reaction of \(\ce{CCl4}\) with \(\ce{Na2SiF6}\):

    (Substance losses by e.g. evaporation are negligible during the following operations.)

    \(\ce{Na2SiF6}\)(\(x\) [g]) was added to \(\ce{CCl4}\) (500.0 g) and heated to 300°C in a sealed pressure-resistant reaction vessel. The unreacted \(\ce{Na2SiF6}\) and generated \(\ce{NaCl}\) were removed by filtration. The filtrate was diluted to a total volume of 1.00 L with \(\ce{CCl4}\) (solution H). The \(\ce{^{29}Si}\) and \(\ce{^{19}F}\) NMR spectra of solution H showed \(\ce{SiF4}\) as the only silicon compound. In the \(\ce{^{19}F}\) NMR spectrum, in addition to \(\ce{SiF4}\), signals corresponding to \(\ce{CFCl3}\), \(\ce{CF2Cl2}\), \(\ce{CF3Cl}\), and \(\ce{CF4}\) were observed (cf. Table 1). The integration ratios in the \(\ce{^{19}F}\) NMR spectrum are proportional to the number of fluorine nuclei.

    Table 1

    \(\ce{^{19}F}\) NMR data \(\ce{CFCl3}\) \(\ce{CF2Cl2}\) \(\ce{CF3Cl}\) \(\ce{CF4}\)
    Integration ratio 45.0 65.0 18.0 2.0

    \(\ce{SiF4}\) is hydrolyzed to form \(\ce{H2SiF6}\) according to the following eq. 8:

    Figure 5

    Solution H (10 mL) was added to an excess amount of water, which resulted in the complete hydrolysis of \(\ce{SiF4}\). After separation, the \(\ce{H2SiF6}\) generated from the hydrolysis in the aqueous solution was neutralized and completely converted to \(\ce{Na2SiF6}\) (aqueous solution J).

    The precipitate of unreacted \(\ce{Na2SiF6}\) and \(\ce{NaCl}\), 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.

    It should be noted here that the coexistence of \(\ce{NaCl}\) or \(\ce{SiO2}\) has no effect on the precipitation titration.

    5.6 Calculate the mass of the \(\ce{NaCl}\) produced in the reaction vessel (information underlined), and calculate the mass (\(x\) [g]) of the \(\ce{Na2SiF6}\) used as a starting material.

    5.7 77.8% of the \(\ce{CCl4}\) used as a starting material was unreacted. Calculate the mass of \(\ce{CF3Cl}\) generated.

    Official English text of the problem.

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  6. Problem 6: The Solid-State Chemistry of Transition Metals

    Figure 1

    Volcano at Sakurajima island

    PART A

    Japan is one of the countries with the highest numbers of volcanos worldwide. When silicate minerals crystallize from magma, a part of the transition-metal ions (\(\ce{M^{n+}}\)) in the magma is incorporated into the silicate minerals. The \(\ce{M^{n+}}\) studied in the problem are coordinated by oxide ions (\(\ce{O^2-}\)) and adopt a four-coordinate tetrahedral (\(T_d\)) geometry in the magma and six-coordinate octahedral (\(O_h\)) geometry in the silicate minerals, both of which exhibit a high-spin electron configuration. The distribution coefficient of \(\ce{M^{n+}}\) between the silicate minerals and magma, \(D\), can be expressed by:

    \[ D=\frac{[\mathrm M]_{\mathrm s}}{[\mathrm M]_{\mathrm l}} \]

    where \([\mathrm M]_{\mathrm s}\) and \([\mathrm M]_{\mathrm l}\) are the concentrations of \(\ce{M^{n+}}\) in the silicate minerals and the magma, respectively. The table below shows the \(D\) values of \(\ce{Cr^2+}\) and \(\ce{Mn^2+}\) as examples.

    \(\ce{Cr^2+}\) \(\ce{Mn^2+}\)
    \(D\) 7.2 1.1

    Let \(\Delta_{\mathrm O}\) and \(\mathrm{CFSE}^{\mathrm O}\) be the energy separation of the d-orbitals of \(\ce{M^{n+}}\) and the crystal-field stabilization energy in a \(O_h\) field, respectively. Let \(\Delta_{\mathrm T}\) and \(\mathrm{CFSE}^{\mathrm T}\) be those in a \(T_d\) field.

    6.1 Calculate \(\left|\mathrm{CFSE}^{\mathrm O}-\mathrm{CFSE}^{\mathrm T}\right|=\Delta\mathrm{CFSE}\) in terms of \(\Delta_{\mathrm O}\) for \(\ce{Cr^2+}\), \(\ce{Mn^2+}\), and \(\ce{Co^2+}\); assume \(\Delta_{\mathrm T}=4/9\Delta_{\mathrm O}\).

    6.2 A linear relationship is observed by plotting \(\ln D\) against \(\Delta\mathrm{CFSE}/\Delta_{\mathrm O}\) in the Cartesian coordinate system shown below. Estimate \(D\) for \(\ce{Co^2+}\).

    Figure 2

    Metal oxides \(\ce{MO}\) (M: Ca, Ti, V, Mn, or Co) crystallize in a rock-salt structure wherein the \(\ce{M^{n+}}\) adopts an \(O_h\) geometry with a high-spin electron configuration. The lattice enthalpy of these oxides is mainly governed by the Coulomb interactions based on the radius and charge of the ions and some contributions from the CFSE of \(\ce{M^{n+}}\) in the \(O_h\) field.

    6.3 Choose the appropriate set of lattice enthalpies [\(\mathrm{kJ\,mol^{-1}}\)] from one of the options (a) to (f).

    \(\ce{CaO}\) \(\ce{TiO}\) \(\ce{VO}\) \(\ce{MnO}\) \(\ce{CoO}\)
    (a) 3460 3878 3913 3810 3916
    (b) 3460 3916 3878 3810 3913
    (c) 3460 3913 3916 3810 3878
    (d) 3810 3878 3913 3460 3916
    (e) 3810 3916 3878 3460 3913
    (f) 3810 3913 3916 3460 3878

    PART B

    A mixed oxide A, which contains \(\ce{La^3+}\) and \(\ce{Cu^2+}\), crystallizes in a tetragonal unit cell shown in Fig.1. In the \(\ce{[CuO6]}\) octahedron, the Cu–O length along the z-axis (\(l_z\)) is longer than that of the x-axis (\(l_x\)), and \(\ce{[CuO6]}\) is distorted from the regular \(O_h\) geometry. This distortion removes the degeneracy of the \(e_g\) orbitals (\(d_{x^2-y^2}\) and \(d_{z^2}\)).

    Figure 3

    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 \(\ce{C4H2O4}\), i.e., a diacid. The pyrolysis behavior of B in dry air shows a weight loss of 29.1% up to \(200\,^\circ\mathrm C\) due to the loss of crystallization water, followed by another weight loss up to \(700\,^\circ\mathrm C\) due to the release of \(\ce{CO2}\). The total weight loss during the formation of A from B is 63.6%. It should be noted that only water and \(\ce{CO2}\) are released in the pyrolysis reaction.

    6.4 Write the chemical formulae for A and B.

    6.5 Calculate \(l_x\) and \(l_z\) using Fig. 1.

    6.6 For \(\ce{Cu^2+}\) in the distorted \(\ce{[CuO6]}\) octahedron in A of Fig. 1, write the names of the split \(e_g\) orbitals (\(d_{x^2-y^2}\) and \(d_{z^2}\)) in (i) and (ii), and draw the electron configuration in the dotted box in your answer sheet.

    A is an insulator. When one \(\ce{La^3+}\) is substituted with one \(\ce{Sr^2+}\), one hole is generated in the crystal lattice that can conduct electricity. As a result, the \(\ce{Sr^2+}\)-doped A shows superconductivity below 38 K. When a substitution reaction took place for A, \(2.05\times10^{27}\) holes \(\mathrm{m^{-3}}\) were generated.

    6.7 Calculate the percentage of \(\ce{Sr^2+}\) substituted for \(\ce{La^3+}\) based on the mole ratio in the substitution reaction. Note that the valences of the constituent ions and the crystal structure are not altered by the substitution reaction

    PART C

    \(\ce{Cu2(CH3CO2)4}\) is composed of four \(\ce{CH3CO2^-}\) coordinated to two \(\ce{Cu^2+}\) (Fig. 2A). \(\ce{Cu2(CH3CO2)4}\) exhibits high levels of structural symmetry, with two axes passing through the carbon atoms of the four \(\ce{CH3CO2^-}\) and an axis passing through the two \(\ce{Cu^2+}\), all of which are oriented orthogonal relative to each other. When a dicarboxylate ligand is used instead of \(\ce{CH3CO2^-}\), a “cage complex” is formed. The cage complex \(\ce{Cu4(L1)4}\) is composed of planar dicarboxylate L1 (Fig. 2B) and \(\ce{Cu^2+}\) (Fig. 2C). The angle \(\theta\) between the coordination directions of the two carboxylates, indicated by the arrows in Fig. 2B, determines the structure of the cage complex. The \(\theta\) is \(0^\circ\) for L1. Note that hydrogen atoms are not shown in Fig. 2.

    Figure 4

    Fig. 2

    6.8 The \(\theta\) of the planar dicarboxylate L2 below is fixed to \(90^\circ\). If the composition of the cage complex formed from L2 and \(\ce{Cu^2+}\) is \(\ce{Cu_{n}(L2)_{m}}\), give the smallest integer combination of \(n\) and \(m\). Assume that only the \(\ce{CO2^-}\) groups of L2 form a coordination bond to \(\ce{Cu^2+}\) ions.

    Figure 5

    A zinc complex, \(\ce{Zn4O(CH3CO2)6}\), contains four tetrahedral \(\ce{Zn^2+}\), six \(\ce{CH3CO2^-}\), and one \(\ce{O^2-}\) (Fig. 3A). In \(\ce{Zn4O(CH3CO2)6}\), the \(\ce{O^2-}\) is located at the origin, and the three axes passing through the carbon atoms of \(\ce{CH3CO2^-}\) are oriented orthogonal relative to each other. When \(p\)-benzenedicarboxylate (Fig. 3B, L3, \(\theta=180^\circ\)) is used instead of \(\ce{CH3CO2^-}\), the \(\ce{Zn^2+}\) clusters are linked to each other to form a crystalline solid (X) that is called a “porous coordination polymer” (Fig. 3C). The composition of X is \(\ce{[Zn4O(L3)3]_{n}}\), and it has a cubic crystal structure with nano-sized pores. One pore is represented as a sphere in Fig. 3D, and each tetrahedral \(\ce{Zn^2+}\) cluster is represented as a dark gray polyhedron in Fig. 3C and 3D. Note that hydrogen atoms are not shown in Fig. 3.

    Figure 6

    Fig. 3

    6.9 X has a cubic unit cell with a side length of \(a\) (Fig. 3C) and a density of \(0.592\,\mathrm{g\,cm^{-3}}\). Calculate \(a\) in [cm].

    6.10 X contains a considerable number of pores, and 1 g of X can accommodate \(3.0\times10^2\) mL of \(\ce{CO2}\) gas in the pores at 1 bar and \(25\,^\circ\mathrm C\). Calculate the average number of \(\ce{CO2}\) molecules per pore.

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  7. Problem 7: Playing with Non-benzenoid Aromaticity

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

    Figure 1

    Photo courtesy: Tohoku Univ.

    PART A

    Lineariifolianone is a natural product with a unique structure, 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) is converted into 6 by performing the same four-step conversion.

    Figure 2

    Inula linariifolia

    Figure 3

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

    Then, ketone 4 is 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 \(\ce{H2^{18}O}\) is used instead of \(\ce{H2^{16}O}\) for the synthesis of \(\ce{^{18}O}\)-labelled-lineariifolianones 13 and 14 from 11 and 12, respectively. Compounds 13 and 14 are \(\ce{^{18}O}\)-labelled isotopomers. Ignoring isotopic labelling, both 13 and 14 provide the same product 15 with identical stereochemistry.

    Figure 4

    7.2 Choose the appropriate structure for A.

    Figure 5

    7.3 Draw the structures of 8–14 and clearly identify the stereochemistry where necessary. Also, indicate the introduced \(\ce{^{18}O}\) atoms for 13 and 14 as shown in the example below.

    Figure 6

    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 the formation of 21 was observed by NMR, 21 gradually decomposes to give 18 and 22.

    Figure 7

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

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  8. Problem 8: Dynamic Organic Molecules and Their Chirality

    PART A

    Polycyclic aromatic hydrocarbons with successive ortho-connections are called [n]carbohelicenes (here, n represents the number of six-membered rings) (see below). [4]Carbohelicene ([4]C) is efficiently prepared by a route using a photoreaction as shown below, via an intermediate (Int.) that is readily oxidized by iodine.

    Figure 1

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

    Figure 2

    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 \(\pi\) systems.

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

    8.2 Attempts to synthesize [5]carbohelicene from the same phosphonium salt and an appropriate starting compound resulted in the formation of only a trace amount of [5]carbohelicene, instead affording product D whose molecular weight was 2 Da lower than that of [5]carbohelicene. The \({}^{1}\mathrm H\) NMR chemical shifts of D are listed below. Draw the structure of D. [D (\(\delta\), ppm in \(\ce{CS2}\), r.t.), 8.85 (2H), 8.23 (2H), 8.07 (2H), 8.01 (2H), 7.97 (2H), 7.91 (2H)]

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

    Figure 3

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

    Figure 4

    Photo courtesy: The Japan Prize Foundation

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

    Figure 5

    8.3 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 (\(\ce{C66H36}\)) is obtained. Given that interconversion between stereoisomers does not occur during the reaction, identify all the possible stereoisomers of G formed in this process, without duplication. 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 \((M,P,M)\).

    Figure 6
    Figure 7

    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 was achieved by a reaction sequence that consists of a ring-opening and a ring-closing metathesis.

    Figure 8

    Representative metathesis reactions catalyzed by a ruthenium catalyst (\(\mathrm{Ru^*}\)) are shown below.

    Figure 9
    Figure 10

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

    Figure 11

    8.5 Starting from the optically active precursor J, the same reaction sequence gives the optically active sumanene derivative K. The stereocenters in J suffer no inversion during the metathesis reaction. Draw the structure of K with the appropriate stereochemistry.

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  9. Problem 9: Likes and Dislikes of Capsule

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

    Figure 1

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

    Figure 2

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

    Figure 3

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

    In the mass spectrum of 1, the ion peak corresponding to its dimer (\(\mathbf{1}_2\)) was clearly observed, whereas an ion peak for \(\mathbf{3}_2\) was not observed in the spectrum of 3. In the \({}^{1}\ce{H}\) NMR spectra of a solution of \(\mathbf{1}_2\), all the NH protons derived from 1 were observed to be chemically 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 of another molecule of 1 to form the dimeric capsule.

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

    9.3 Give the number of the hydrogen bonds in the dimeric capsule (\(\mathbf{1}_2\)).

    The dimeric capsule of 1 (\(\mathbf{1}_2\)) has an internal space wherein an appropriate small molecule Z can be encapsulated. This phenomenon is expressed by the following equation:

    \[ \ce{Z}+\mathbf{1}_2\ce{->}\ce{Z}@\mathbf{1}_2 \tag{1} \]

    The equilibrium constant of the encapsulation of Z into \(\mathbf{1}_2\) is given as below:

    \[ K_a=\frac{[\ce{Z}@\mathbf{1}_2]}{[\ce{Z}][\mathbf{1}_2]} \tag{2} \]

    Encapsulation of a molecule into a capsule could be monitored by NMR spectroscopy. For example, \(\mathbf{1}_2\) in \(\ce{C6D6}\) gave different signals in the \({}^{1}\ce{H}\) NMR spectra before and after addition of \(\ce{CH4}\).

    Compound 2 also forms a rigid and larger dimeric capsule (\(\mathbf{2}_2\)). The \({}^{1}\ce{H}\) NMR spectrum of \(\mathbf{2}_2\) was measured in \(\ce{C6D6}\), \(\ce{C6D5F}\), and a \(\ce{C6D6}/\ce{C6D5F}\) solvent mixture, with all other conditions being kept constant. The chemical shifts for the \(\ce{H}^{a}\) proton of 2 in the above solvents are summarized below, and no other signals from the \(\ce{H}^{a}\) in 2, except for the 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 the filled capsule.

    Figure 4
    solvent \(\delta\) (ppm) of \(\ce{H}^{a}\)
    \(\ce{C6D6}\) 4.60
    \(\ce{C6D5F}\) 4.71
    \(\ce{C6D6}/\ce{C6D5F}\) 4.60, 4.71, 4.82

    9.4 Determine the number of \(\ce{C6D6}\) and \(\ce{C6D5F}\) molecules encapsulated in \(\mathbf{2}_2\) giving each \(\ce{H}^{a}\) signal.

    \({}^{1}\ce{H}\) NMR measurements in \(\ce{C6D6}\) revealed that \(\mathbf{2}_2\) can incorporate one molecule of 1-adamantanecarboxylic acid (AdA), and the association constants (\(K_a\)) which are expressed below were determined for various temperatures. \([\mathrm{solvent}@\mathbf{2}_2]\) denotes a species containing one or more solvent molecules.

    \[ K_a=\frac{[\ce{Z}@\mathbf{2}_2]} {[\ce{Z}][\mathrm{solvent}@\mathbf{2}_2]} \tag{3} \]

    Similarly, the \(K_a\) values of \(\ce{CH4}\) and \(\mathbf{1}_2\) given as eq (2) at various temperatures in \(\ce{C6D6}\) were also determined by \({}^{1}\ce{H}\) NMR measurements. The plots of the two association constants (as \(\ln K_a\) vs \(1/T\)) are shown below.

    Figure 5
    Figure 6

    No \(\ce{C6D6}\) molecule is encapsulated in \(\mathbf{1}_2\). In line II, the entropy change (\(\Delta S\)) is ( 1 ) and enthalpy change (\(\Delta H\)) 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 ).

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

    A B
    ( 1 ) positive negative
    ( 2 ) positive negative
    ( 3 ) \(\Delta S\) \(\Delta H\)
    ( 4 ) \(\mathbf{1}_2\) and \(\ce{CH4}\) \(\mathbf{2}_2\) and AdA
    ( 5 ) \(\mathbf{1}_2\) and \(\ce{CH4}\) \(\mathbf{2}_2\) and AdA

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