International Chemistry Olympiad

IChO 2017

Nakhon Pathom, Thailand · 11 theory problems with complete solutions

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  1. Problem 1: Production of propene using heterogeneous catalysts

    Propene or propylene is one of the most valuable chemicals for the petrochemical industry in Thailand and around the world. One good example of the commercial use of propene is for the production of polypropylene (PP).

    Part A.

    Propene can be synthesized via a direct dehydrogenation of propane in the presence of a heterogeneous catalyst. However, such a reaction is not economically feasible due to the nature of the reaction itself. Provide a concise explanation to each of the questions below. Additional information: \(H_{\mathrm{bond}}(\ce{C=C})=1.77H_{\mathrm{bond}}(\ce{C-C})\), \(H_{\mathrm{bond}}(\ce{H-H})=1.05H_{\mathrm{bond}}(\ce{C-H})\), and \(H_{\mathrm{bond}}(\ce{C-H})=1.19H_{\mathrm{bond}}(\ce{C-C})\), where \(H_{\mathrm{bond}}\) refers to average bond enthalpy of the indicated chemical bond.

    1.1 What is the enthalpy change of the direct dehydrogenation of propane? Show your calculation and express your answer in terms of \(H_{\mathrm{bond}}(\ce{C-C})\).

    1.2 It is difficult to increase the amount of propene by increasing pressure at constant temperature. Which law or principle can best explain this phenomenon? Select your answer by marking “✓” in one of the open circles.

    • ☐ Boyle’s law
    • ☐ Charles’ law
    • ☐ Dalton’s law
    • ☐ Raoult’s law
    • ☐ Le Chatelier’s principle

    1.3 Initially, the system is in equilibrium. Consistent with question 1.1, what is/are correct set(s) of signs for the following thermodynamic variables of the system for the direct dehydrogenation of propane? Select your answer(s) by marking „✓” in any of the open square(s).

    \(\Delta H\) \(\Delta S\) \(\Delta G\) \(T^{*}\)
    ☐ − + + lower
    ☐ − + − higher
    ☐ − − + lower
    ☐ − − − higher
    ☐ + + + lower
    ☐ + + − higher
    ☐ + − + lower
    ☐ + − − higher

    ☐ None of the above is correct

    \(^*\) Relative to the initial temperature at the same partial pressure.

    Part B.

    A better reaction to produce large quantity of propene is the oxidative dehydrogenation (ODH) using solid catalysts, such as vanadium oxides, under molecular oxygen gas. Although this type of reaction is still under intense research development, its promise toward the production of propene at an industrial scale eclipses that of the direct dehydrogenation.

    1.4 The overall rate of propane consumption in the reaction is

    \[ r_{\ce{C3H6}} = \frac{1}{ \dfrac{p^{0}}{k_{\mathrm{red}}p_{\ce{C3H6}}} + \dfrac{p^{0}}{k_{\mathrm{ox}}p_{\ce{O2}}} } \]

    where \(k_{\mathrm{red}}\) and \(k_{\mathrm{ox}}\) are the rate constants for the reduction of metal oxide catalyst by propane and for the oxidation of the catalyst by molecular oxygen, respectively, and \(p^{o}\) is the standard pressure of 1 bar. Some experiments found that the rate of oxidation of the catalyst is 100,000 times faster than that of the propane oxidation. The experimental

    \[ r_{\ce{C3H8}} = k_{\mathrm{obs}}\frac{p_{\ce{C3H6}}}{p^{0}} \]

    where \(k_{\mathrm{obs}}\) is the observed rate constant (\(0.062\ \mathrm{mol\,s^{-1}}\)). If the reactor containing the catalyst is continuously passed through with propane and oxygen at a total pressure of 1 bar, determine the value of \(k_{\mathrm{red}}\) and \(k_{\mathrm{ox}}\) when the partial pressure of propane is 0.10 bar. Assume that the partial pressure of propene is negligible.

    Part C.

    The metal oxide catalyst contains oxygen atoms on its surface that serve as active sites for the ODH. Denoting red* as a reduced site and \(\ce{O(s)}\) as an oxygen atom on the surface of the catalyst, one of the proposed mechanisms for the ODH in the presence of the catalyst can be written as follows:

    Figure 1

    Given

    \[ \beta= \frac{\text{number of reduced sites}} {\text{total number of active sites}} \]

    The rate laws for the above 3 steps are:

    \[ r_1=k_1p_{\ce{C2H8}}(1-\beta) \]

    \[ r_2=k_2p_{\ce{C3H6}}(1-\beta) \]

    and

    \[ r_3=k_3p_{\ce{O2}}\beta \]

    1.5 Assuming that the amount of oxygen atoms on the surface stays constant at any time of reaction, calculate \(\beta\) as a function of

    \[ k_1,\ k_2,\ k_3,\ p_{\ce{C2H8}},\ p_{\ce{C3H6}},\ \text{and }p_{\ce{O2}}. \]

    Official English text of the problem.

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  2. Problem 2: Kinetic isotope effect (KIE) and zero-point vibrational energy (ZPE)

    Calculation of ZPE and KIE

    Kinetic isotope effect (KIE) is a phenomenon associated with a change in rate constant of the reaction when one of the atoms is replaced by its isotope. KIE can be used to confirm whether a particular bond to hydrogen is broken in the reaction. Harmonic oscillator model is used to estimate the difference in the rate between \(\ce{C-H}\) and \(\ce{C-D}\) bond activation (D – deuterium).

    The vibrational frequency (\(\nu\)) represented by harmonic oscillator model is

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

    where \(k\) is the force constant and \(\mu\) is the reduced mass.

    The vibrational energies of the molecule are given by

    \[ E_n=\left(n+\frac{1}{2}\right)h\nu \]

    where \(n\) is vibrational quantum number with possible values of 0, 1, 2, ... The energy of the lowest vibrational energy level (\(E_n\) at \(n=0\)) is called zero-point vibrational energy (ZPE).

    2.1 Calculate the reduced mass of \(\ce{C-H}\) (\(\mu_{CH}\)) and \(\ce{C-D}\) (\(\mu_{CD}\)) in atomic mass unit.

    Assume that the mass of deuterium is twice that of hydrogen.

    Note: If a student is unable to calculate the values for \(\mu_{CH}\) and \(\mu_{CD}\) he/she can use \(\mu_{CH}=1.008\) and \(\mu_{CD}=2.016\) for the subsequent parts of the question. The given values are not necessarily be close to the correct values.

    2.2 Given that the force constant (\(k\)) for \(\ce{C-H}\) stretching is the same as that for the \(\ce{C-D}\) stretching and the \(\ce{C-H}\) stretching frequency is \(2900\ \mathrm{cm^{-1}}\), find the corresponding \(\ce{C-D}\) stretching frequency (in \(\mathrm{cm^{-1}}\)).

    2.3 According to the \(\ce{C-H}\) and \(\ce{C-D}\) stretching frequencies in question 2.2, calculate the zero-point vibrational energies (ZPE) of \(\ce{C-H}\) and \(\ce{C-D}\) stretching in \(\mathrm{kJ\,mol^{-1}}\).

    Note: If a student is unable to calculate the values for ZPE in 2.3, he/she can use use \(ZPE_{CH}=7.23\ \mathrm{kJ\,mol^{-1}}\) and \(ZPE_{CD}=2.15\ \mathrm{kJ\,mol^{-1}}\) for the subsequent parts of the question. The given values are not necessarily be close to the correct values.

    Kinetic isotope effect (KIE)

    Due to the difference in zero-point vibrational energies, a protonated compound and its corresponding deuterated compounds are expected to react at different rates.

    For the \(\ce{C-H}\) and \(\ce{C-D}\) bond dissociation reactions, the energies of both transition states and both products are identical. Then, the isotope effect is controlled by the difference in the ZPE's of the \(\ce{C-H}\) and \(\ce{C-D}\) bonds.

    2.4 Calculate the difference in the bond dissociation energy (BDE) between \(\ce{C-D}\) bond and \(\ce{C-H}\) bond (\(BDE_{CD}-BDE_{CH}\)) in \(\mathrm{kJ\,mol^{-1}}\).

    2.5 Assume that the activation energy (\(E_a\)) for the \(\ce{C-H}/\ce{C-D}\) bond cleavage is approximately equal to the bond dissociation energy and the Arrhenius factor is the same for both \(\ce{C-H}\) and \(\ce{C-D}\) bond cleavage. Find the relative rate constant for the \(\ce{C-H}/\ce{C-D}\) bond cleavage (\(k_{CH}/k_{CD}\)) at \(25\,^\circ\mathrm{C}\).

    Using KIE to study reaction mechanism

    The oxidation of nondeuterated and deuterated diphenylmethanol using an excess of chromic acid was studied.

    Figure 1

    2.6 Let \(C_0\) be the initial concentration of either nondeuterated diphenylmethanol or deuterated diphenylmethanol and \(C_t\) its concentration at time \(t\). The experiment led to two plots (Figure 2a and Figure 2b), from which the first-order rate constant can be determined.

    Figure 2

    Figure 2a

    Figure 3

    Figure 2b

    Which plot should be for the oxidation of nondeuterated diphenylmethanol and which one is for the oxidation of deuterated diphenylmethanol?

    For each statement, select your answer by marking “✓” in one of the square.

    The oxidation of nondeuterated diphenylmethanol:

    ☐ Figure 2a
    ☐ Figure 2b

    The oxidation of deuterated

    ☐ Figure 2a
    ☐ Figure 2b

    2.7 Determine \(k_{CH}\), \(k_{CD}\) (in \(\mathrm{min^{-1}}\)), and the \(k_{CH}/k_{CD}\) of this reaction from the plots in question 2.6.

    2.8 The mechanism has been proposed as follows:

    Figure 4

    According to the information in 2.6 and 2.7, which step should be the rate determining step?

    Select your answer by marking “✓” in one of the corresponding squares:

    ☐ Step (1)
    ☐ Step (2)
    ☐ Step (3)

    Official English text of the problem.

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  3. Problem 3: Thermodynamics of chemical reactions

    Part A.

    Methanol is produced commercially by using a mixture of carbon monoxide and hydrogen over zinc oxide/copper oxide catalyst:

    \[ \ce{CO(g) + 2 H2(g) -> CH3OH(g)}. \]

    The standard enthalpy of formation (\(\Delta H_f^\circ\)) and the absolute entropy (\(S^\circ\)) for each of the three gases at room temperature (298 K) and at a standard pressure of 1 bar are given as follows.

    Gas \(\Delta H_f^\circ\) (kJ mol\(^{-1}\)) \(S^\circ\) (J K\(^{-1}\) mol\(^{-1}\))
    \(\ce{CO(g)}\) −111 198
    \(\ce{H2(g)}\) 0 131
    \(\ce{CH3OH(g)}\) −201 240

    3.1 Calculate \(\Delta H^\circ\), \(\Delta S^\circ\), \(\Delta G^\circ\), and \(K_p\) for the reaction at 298 K.

    (If you are unable to calculate \(K_p\) at 298 K in problem 3.1, use \(K_p=9\times10^5\) later on.)

    3.2 A commercial reactor is operated at a temperature of 600 K. Calculate the value of \(K_p\) at this temperature, assuming that \(\Delta H^\circ\) and \(\Delta S^\circ\) are independent on temperature.

    (If you are unable to calculate \(K_p\) at 600 K in problem 3.2, use \(K_p=1.0\times10^{-2}\) later on.)

    3.3 Production of methanol in industry is based on flowing of the gas comprising 2.00 moles of \(\ce{H2}\) for each mole of \(\ce{CO}\) into the reactor. The mole fraction of methanol in the exhaust gas from the reactor was found to be 0.18. Assuming that equilibrium is established, what is the total pressure in the reactor at a high temperature of 600 K?

    Part B.

    3.4 Consider the following closed system at 300 K. The system comprises 2 compartments, separated by a closed valve, which has negligible volume (see next figure). At the same pressure \(P\), compartment A and compartment B contain 0.100 mol argon gas and 0.200 mol nitrogen gas, respectively. The volumes of the two compartments, \(V_A\) and \(V_B\), are selected so that the gases behave as ideal gases.

    Figure 1

    After opening the valve slowly, the system is allowed to reach equilibrium. It is assumed that the two gases form an ideal gas mixture. Calculate the change in Gibbs free energy at 300 K, \(\Delta G\).

    If you are unable to calculate \(K_p\) at 298 K in problem 3.1, use \(K_p=9\times10^5\) later on.

    Official English text of the problem.

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  4. Problem 4: Electrochemistry

    Part A. Galvanic cell

    The experiment is performed at 30.00°C. The electrochemical cell is composed of a hydrogen half-cell \([\ce{Pt(s)}\mid\ce{H2(g)}\mid\ce{H+(aq)}]\) containing a metal platinum electrode immersed in a buffer solution under a pressure of hydrogen gas. This hydrogen half-cell is connected to a half-cell of a metal (M) strip dipped in an unknown concentration of \(\ce{M^{2+}(aq)}\) solution. The two half-cells are connected via a salt bridge as shown in Figure 1.

    Note: The standard reduction potentials are given in Table 1.

    Figure 1

    Figure 1 The galvanic cell

    Table 1. Standard reduction potential (range 298-308 K)

    Half-reaction \(E^\circ\) (V)-
    \(\ce{Ba^{2+}(aq) + 2e- -> Ba(s)}\) -2.912
    \(\ce{Sr^{2+}(aq) + 2e- -> Sr(s)}\) 2.899
    \(\ce{Ca^{2+}(aq) + 2e- -> Ca(s)}\) -2.868
    \(\ce{Er^{2+}(aq) + 2e- -> Er(s)}\) -2.000
    \(\ce{Ti^{2+}(aq) + 2e- -> Ti(s)}\) -1.630
    \(\ce{Mn^{2+}(aq) + 2e- -> Mn(s)}\) -1.185
    \(\ce{V^{2+}(aq) + 2e- ->}\ ;\ce{V(s)}\) -1.175
    \(\ce{Cr^{2+}(aq) + 2e- -> Cr(s)}\) -0.913
    \(\ce{Fe^{2+}(aq) + 2e- -> Fe(s)}\) -0.447
    \(\ce{Cd^{2+}(aq) + 2e- -> Cd(s)}\) -0.403
    \(\ce{Co^{2+}(aq) + 2e- -> Co(s)}\) -0.280
    \(\ce{Ni^{2+}(aq) + 2e- -> Ni(s)}\) -0.257
    \(\ce{Sn^{2+}(aq) + 2e- -> Sn(s)}\) -0.138
    \(\ce{Pb^{2+}(aq) + 2e- -> Pb(s)}\) -0.126
    \(\ce{2H+(aq) + 2e- -> H2(g)}\) 0.000
    \(\ce{Sn^{4+}(aq) + 2e- -> Sn^{2+}(aq)}\) +0.151
    \(\ce{Cu^{2+}(aq) + e- -> Cu+(aq)}\) +0.153
    \(\ce{Ge^{2+}(aq) + 2e- -> Ge(s)}\) +0.240
    \(\ce{VO^{2+}(aq) + 2H+(aq) + e- -> V^{3+}(aq) + H2O(l)}\) +0.337
    \(\ce{Cu^{2+}(aq) + 2e- -> Cu(s)}\) +0.340
    \(\ce{Tc^{2+}(aq) + 2e- -> Tc(s)}\) +0.400
    \(\ce{Ru^{2+}(aq) + 2e- -> Ru(s)}\) +0.455
    \(\ce{I2(s) + 2e- -> 2I-(aq)}\) +0.535
    \(\ce{UO2^{2+}(aq) + 4H+(aq) + 2e- -> U^{4+}(aq) + 2H2O(l)}\) +0.612
    \(\ce{PtCl4^{2-}(aq) + 2e- -> Pt(s) + 4Cl-(aq)}\) +0.755
    \(\ce{Fe^{3+}(aq) + e- -> Fe^{2+}(aq)}\) +0.770
    \(\ce{Hg2^{2+}(aq) + 2e- -> 2Hg(l)}\) +0.797
    \(\ce{Hg^{2+}(aq) + 2e- -> Hg(l)}\) +0.851
    \(\ce{2Hg^{2+}(aq) + 2e- -> Hg2^{2+}(aq)}\) +0.920
    \(\ce{Pt^{2+}(aq) + 2e- -> Pt(s)}\) +1.180
    \(\ce{MnO2(s) + 4H+(aq) + 2e- -> Mn^{2+}(aq) + 2H2O(l)}\) +1.224
    \(\ce{Cr2O7^{2-}(aq) + 14H+(aq) + 6e- -> 2Cr^{3+}(aq) + 7H2O(l)}\) +1.360
    \(\ce{Co^{3+}(aq) + e- -> Co^{2+}(aq)}\) +1.920
    \(\ce{S2O8^{2-}(aq) + 2e- -> 2SO4^{2-}(aq)}\) +2.010

    4.1 If the reaction quotient (\(Q\)) of the whole galvanic cell is equal to \(2.18\times10^{-4}\) at 30.00 °C, the electromotive force is +0.450 V. Calculate the value of standard reduction potential (\(E^\circ\)) and identify the metal “M”.

    Note: \(\Delta G=\Delta G^\circ+RT\ln Q\)

    4.2 Write the balanced equation of the spontaneous redox reaction of the galvanic cell.

    4.3 The unknown concentration of \(\ce{M^{2+}(aq)}\) solution in the cell (Figure 1) can be analyzed by a iodometric titration. A 25.00 \(\mathrm{cm^3}\) aliquot of \(\ce{M^{2+}(aq)}\) solution is added into a conical flask and an excess of \(\ce{KI}\) is added. A volume of 25.05 \(\mathrm{cm^3}\) of a sodium thiosulfate solution with a concentration of 0.800 \(\mathrm{mol\,dm^{-3}}\) is required to reach the equivalent point. Write all the redox reactions associated with this titration and calculate the concentration of \(\ce{M^{2+}(aq)}\) solution.

    Note: If a student cannot find the answer, the student can use 0.950 \(\mathrm{mol\,dm^{-3}}\) as the concentration of \(\ce{M^{2+}}\) for further calculations.

    4.4 In Figure 1, if the hydrogen half-cell is under 0.360 bar hydrogen gas and the platinum electrode is immersed in a 500 \(\mathrm{cm^3}\) buffer solution containing 0.050 mol lactic acid (\(\ce{HC3H5O3}\)) and 0.025 mol sodium lactate (\(\ce{C3H5O3Na}\)), the electromotive force of the galvanic cell measured is +0.534 V. Calculate the pH of the buffer solution and the dissociation constant (\(K_a\)) of lactic acid at 30.00 °C.

    Note: If a student cannot find the answer, the student can use 3.46 as the buffer pH for further calculations.

    Official English text of the problem.

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  5. Problem 5: Phosphate and silicate in soil

    Distribution and mobility of phosphorus in soil are usually studied by sequential extraction. Sequential extraction is performed by the use of acid or alkaline reagents to fractionate inorganic phosphorus in soil. Soil sample was extracted and analyzed as follows:

    Part A. Determination of total phosphate (\(\ce{PO4^3-}\)) and silicate (\(\ce{SiO4^4-}\))

    A 5.00 gram of soil sample is digested to give a final volume of 50.0 \(\mathrm{cm^3}\) digesting solution which dissolves total phosphorus and silicon. The extract is analyzed for the total concentrations of phosphorus and silicon. The concentrations of phosphorus and silicon are found to be 5.16 \(\mathrm{mg\,dm^{-3}}\) and 5.35 \(\mathrm{mg\,dm^{-3}}\), respectively.

    5.1.1 Determine the mass of \(\ce{PO4^3-}\) in mg per 1.00 g of soil.

    5.1.2 Determine the mass of \(\ce{PO4^3-}\) in mg per 1.00 g of soil.

    Part B. Determination of available \(\ce{PO4^3-}\) in acid extract

    Phosphate can be analyzed by using molybdenum blue method. One mole of phosphate is converted into one mole of molybdenum blue compound. This method is used for determination of phosphate in the acid extract. Absorbance (\(A\)) and transmittance (\(T\)) are recorded at 800 nm. The molar absorptivity of the molybdenum blue compound is 6720 \(\mathrm{dm^3\,mol^{-1}\,cm^{-1}}\) and all measurement is carried out in a 1.00-cm cuvette.

    Transmittance and absorbance are given by the following equations:

    \[ T = I/I_{\mathrm{o}} \]

    \[ A = \log(I_{\mathrm{o}}/I) \]

    where \(I\) is the intensity of the transmitted light and \(I_{\mathrm{o}}\) is the intensity of the incident light.

    5.2.1 When the sample containing high concentration of phosphate is analyzed, a reference solution of \(7.5\times10^{-5}\ \mathrm{mol\,dm^{-3}}\) of molybdenum blue compound is used for adjusting zero absorbance. The transmittance of the sample solution is then measured to be 0.55. Calculate the concentration of phosphate (\(\mathrm{mol\,dm^{-3}}\)) in the sample solution.

    Part C. Determination of \(\ce{PO4^3-}\) and \(\ce{SiO4^4-}\) in alkaline extract

    Both phosphate and silicate ions can react with molybdate in alkaline solution, producing the yellow molybdophosphate and molybdatosilicate. Further reduction with ascorbic acid produces intense color molybdenum blue compounds. Both complexes exhibit maximum absorption at 800 nm. Addition of tartaric acid helps preventing interference from silicate in the determination of phosphate.

    Two series of phosphate standard are treated with and without tartaric acid whereas a series of silicate standard is not treated with tartaric acid. Linear equations obtained from those calibration curves are as follows:

    Conditions Linear equations
    Phosphate with and without tartaric acid \(y = 6720x_1\)
    Silicate without tartaric acid \(y = 868x_2\)

    \(y\) is absorbance at 800 nm,

    \(x_1\) is concentration of phosphate as \(\mathrm{mol\,dm^{-3}}\),

    \(x_2\) is concentration of silicate as \(\mathrm{mol\,dm^{-3}}\)

    Absorbance at 800 nm of the alkaline fraction of the soil extract after treated with and without tartaric acid are 0.267 and 0.510, respectively.

    5.3.1 Calculate the phosphate concentration in the alkaline soil extract in \(\mathrm{mol\,dm^{-3}}\) and calculate the corresponding phosphorus in \(\mathrm{mg\,dm^{-3}}\).

    5.3.2 Calculate the silicate concentration from the soil sample in t the alkaline fraction in \(\mathrm{mol\,dm^{-3}}\) and calculate the corresponding silicon in \(\mathrm{mg\,dm^{-3}}\).

    Part D. Preconcentration of ammonium phosphomolybdate

    A 100 \(\mathrm{cm^3}\) of aqueous sample of ammonium phosphomolybdate (\(\ce{(NH4)3PMo12O40}\)) compound is extracted with 5.0 \(\mathrm{cm^3}\) of an organic solvent. The organic-water partition coefficient (\(K_{\mathrm{ow}}\)) is defined as the ratio of concentration of the compound in the organic phase (\(c_{\mathrm{o}}\)) to that in the water phase (\(c_{\mathrm{w}}\)). \(K_{\mathrm{ow}}\) of the ammonium phosphomolybdate is 5.0.

    The molar absorptivity of ammonium phosphomolybdate in the organic phase is 5000 \(\mathrm{dm^3\,mol^{-1}\,cm^{-1}}\).

    5.4 If the absorbance in the organic phase is 0.200, calculate the total mass of phosphorus (in mg unit) in the original aqueous sample solution. The optical pathlength of the cuvette is 1.00 cm.

    Official English text of the problem.

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  6. Problem 6: Iron

    Iron (\(\ce{Fe}\)) is the fourth most abundant element in the Earth’s crust and has been used for more than 5,000 years.

    Part A.

    Pure iron is easily oxidized, which limits its utilization. Element X is one of the alloying elements that is added to improve the oxidation resistance property of iron.

    6.1 Below is some information about the element X:

    (1) In first ionization, an electron with quantum numbers \(n_1=4-l_1\) is removed.

    (2) In second ionization, an electron with quantum numbers \(n_2=5-l_2\) is removed.

    (3) The atomic mass of X is lower than that of \(\ce{Fe}\).

    What is the element X?

    (Answer by writing the proper symbol according to the periodic table.)

    6.2 Both \(\ce{Fe}\) and X crystallize in the body centered cubic structure. Approximating the \(\ce{Fe}\) atoms as hard-spheres, the volume taken up by the \(\ce{Fe}\) atoms inside the unit cell is \(1.59\times10^{-23}\ \mathrm{cm^3}\). The volume of the unit cell of X is \(0.0252\ \mathrm{nm^3}\). A complete substitutional solid solution usually occurs when

    \[ \Delta R=\left(\frac{|R_X-R_{\ce{Fe}}|}{R_{\ce{Fe}}}\right)\times100 \]

    is less than or equal to 15, where \(R_X\) and \(R_{\ce{Fe}}\) are the atomic radii of X and \(\ce{Fe}\), respectively. Can X and \(\ce{Fe}\) form a complete substitutional solid solution?

    Tick the relevant box ☐ YES　 ☐ NO.

    Show your calculation. The volume of sphere is \(4/3\pi r^3\).

    Part B.

    Iron in natural water is in the form of \(\ce{Fe(HCO3)2}\), which ionizes to \(\ce{Fe^2+}\) and \(\ce{HCO3^-}\). To remove iron from water, \(\ce{Fe(HCO3)2}\) is oxidized to an insoluble complex \(\ce{Fe(OH)3}\), which can be filtered out of the water.

    6.3 \(\ce{Fe^2+}\) can be oxidized by \(\ce{KMnO4}\) in a basic solution to yield \(\ce{Fe(OH)3}\) and \(\ce{MnO2}\) precipitates. Write the balanced ionic equation for this reaction in a basic solution.

    6.4 Under this condition, \(\ce{HCO3^-}\) ions are converted to \(\ce{CO3^2-}\). Write the balanced ionic equation for this reaction in a basic solution.

    6.5 A covalent compound A which contains more than 2 atoms and, a potential oxidizing agent, can be prepared by the reaction between diatomic halogen molecule (\(\ce{X2}\)) and \(\ce{NaQO2}\)

    Figure 1

    where \(x\), \(y\) and \(z\) are the coefficients for the balanced equation. Among the binary compounds between hydrogen and halogen, \(\ce{HX}\) has the lowest boiling point. Identify X and if A has an unpaired electron, draw a Lewis structure of compound A with zero formal charge on all atoms.

    6.6 Compound D is an unstable oxidizing agent that can be used to remove \(\ce{Fe(HCO3)2}\) from natural water. It consists of elements G, Z and hydrogen and the oxidation number of Z is +1. In this compound, hydrogen is connected to the element having the higher electronegativity among them. Below is some information about the elements G and Z:

    (1) G exists in its normal state as a diatomic molecule, \(\ce{G2}\).

    (2) Z has one proton fewer than that of element E. E exists as a gas under standard conditions. \(\ce{Z2}\) is a volatile solid.

    (3) The compound \(\ce{EG3}\) has a pyramidal shape.

    Identify the elements G and Z and draw a molecular structure of compound D.

    Part C.

    \(\ce{^{59}Fe}\) is a radiopharmaceutical isotope which is used in the study of iron metabolism in the spleen. This isotope decays to \(\ce{^{59}Co}\) as follows:

    Figure 2

    6.7 What are a and b in equation (1)? (Mark ✓ in the appropriate boxes.)

    proton neutron beta positron alpha gamma

    6.8 Consider equation (1), if the \(\ce{^{59}Fe}\) isotope is left for 178 days which is \(n\) times of its half-life (\(t_{1/2}\)), the mole ratio of \(\ce{^{59}Co}\) to \(\ce{^{59}Fe}\) is 15 : 1. If \(n\) is an integer, what is the half-life of \(\ce{^{59}Fe}\) in day(s)? Show your calculation.

    Official English text of the problem.

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  7. Problem 7: Chemical Structure Puzzles

    Titanium complexes have been investigated for their antitumor activity. Many factors including isomerism and sizes have shown to affect the potency of the complexes. This question deals with the synthesis and characterization of some titanium complexes.

    7.1 A reaction of 2 equivalents of 2-tert-butylphenol, 2 equivalents of formaldehyde, and \(N,N'\)-dimethylethylene-1,2-diamine under acidic conditions at \(75\,^\circ\mathrm{C}\) affords three major products with the same chemical formula of \(\ce{C26H40N2O2}\), as shown in the equation below. Draw the structure of each product.

    Figure 1

    7.2 If 2,4-di-tert-butylphenol is used as a substrate instead of 2-tert-butylphenol using the same stoichiometry as that in 7.1, only one product X was obtained. Draw the structure of X.

    A reaction between X from 7.2 and \(\ce{Ti(O^{i}Pr)4}\) [\(^{i}\mathrm{Pr}\) = isopropyl] in diethyl ether under an inert atmosphere resulted in the six-coordinate Ti complex Y, as a yellow crystalline solid and isopropanol at room temperature.

    Figure 2

    UV-VIS spectra of X, \(\ce{Ti(O^{i}Pr)4}\), and Y reveal that only the product Y has an absorption at \(\lambda=370\ \mathrm{nm}\). By varying the volumes of X and \(\ce{Ti(O^{i}Pr)}\), each with the concentration of \(0.50\ \mathrm{mol\,dm^{-3}}\), and using benzene as the solvent, the absorbance data at \(\lambda=370\ \mathrm{nm}\) are given below:

    Volume of X (\(\mathrm{cm^3}\)) Volume of \(\ce{Ti(O^{i}Pr)4}\) (\(\mathrm{cm^3}\)) Volume of benzene (\(\mathrm{cm^3}\)) Absorbance
    0 1.20 1.80 0.05
    0.20 1.00 1.80 0.25
    0.30 0.90 1.80 0.38
    0.50 0.70 1.80 0.59
    0.78 0.42 1.80 0.48
    0.90 0.30 1.80 0.38
    1.10 0.10 1.80 0.17
    1.20 0 1.80 0.02

    7.3 Fill in appropriate values in the table provided below.

    \(\displaystyle\frac{\text{mol of X}}{\text{mol of X}+\text{mol of }\ce{Ti(O^{i}Pr)4}}\) Absorbance
    0.05
    0.25
    0.38
    0.59
    0.48
    0.38
    0.17
    0.02

    Next figure: Plot a graph showing a relationship between

    \[ \frac{\text{mole of X}} {\text{mole of X}+\text{mole of }\ce{Ti(O^{i}Pr)4}} \]

    and absorbance in the space provided below.

    The value of

    \[ \frac{\text{mole of X}} {\text{mole of X}+\text{mole of }\ce{Ti(O^{i}Pr)4}} \]

    which maximizes the amount of the product Y represents the stoichiometry of X in the chemical formula of Y. Based on the graph above, show what is the molar ratio Ti : X in the complex Y.

    7.4 The titanium complex Y is six-coordinated. The IR spectrum of Y does not contain a broad absorption band in the range of \(3200–3600\ \mathrm{cm^{-1}}\). Y exists as three diastereomers. Ignoring stereochemistry at N atoms, draw clearly the structures of all three diastereomers.

    Note that you do not need to draw the complete structure of the ligand. Only identify donor atoms that involve in coordination with titanium and the ligand framework between the donor atoms can be drawn as follows:

    For example:

    Figure 3

    (2,2'-bipyridine)

    can be drawn as:

    Figure 4

    **If you did not get a structure of X from 7.2, use the following ligand symbol to represent X (A and Z are donor atoms):

    Figure 5

    7.5 Under certain conditions, the reaction shown in equation 1 affords only one diastereomer of Y. Given that structures of Y are "fixed" (no intramolecular movement), the \(\ce{^1H}\) NMR spectrum of Y in \(\ce{CDCl3}\) shows four singlet resonances at \(\delta\) 1.25, 1.30, 1.66, and 1.72 corresponding to the tert-butyl groups. Draw a structure of the only possible diastereomer of Y.

    (You do not need to draw the complete structure of the ligand. Only identify donor atoms that involve in coordination and the ligand framework between the donor atoms can be drawn as shown in 7.4).

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  8. Problem 8: Silica Surface

    Silica exists in various forms like amorphous and crystalline. Silica can be synthesized via sol- gel process by using silicon alkoxides like tetramethoxysilane (TMOS) and tetraethoxysilane (TEOS) as the details below:

    a. Hydrolysis

    Figure 1

    b. Water condensation

    Figure 2

    c. Alcohol condensation

    Figure 3

    In bulk silica, all silicon atoms are tetrahedrally bonded to four oxygen atoms giving three-dimensional solid network. The silicon environments found inside silica is presented below:

    Figure 4

    8.1 Three silicon atom environments (similar to the example above) are commonly observed at the silica surface. The three structures of the silicon environments must be drawn in the provided boxes.

    Figure 5

    Silica can be used as an effective metal ion adsorbent in water. The proposed structure for metal-silica complex is as follows:

    Figure 6

    8.2 After \(\ce{Cu^2+}\) is being adsorbed, the color of silica changes from white to pale blue. The visible spectrum shows a broad absorption band (with a shoulder) at \(\lambda_{\max}=550\ \mathrm{nm}\).

    If \(\ce{Cu^2+}\) can bind with silica and adopt the structure similar to II, draw the splitting diagram of the \(d\)- orbitals of \(\ce{Cu^2+}\) ion including the label of the \(d\) orbitals in the complex and specify the corresponding electronic transition(s) for the visible absorption.

    8.3 If the first row transition metal ions form complexes with silica analogous to \(\ce{Cu^2+}\), which metal ion(s) do(es) have the analogous to electronic transition(s) to \(\ce{Cu^2+}\)?

    The metal ion(s) must be in +2 or +3 oxidation state. Please note that the silanol groups (\(\ce{Si-OH}\)) and water are weak field ligands.

    However, silica is randomly bonded to various types of metal ion. To increase the selectivity, modification of silica surface has been performed by grafting with various organic molecules like 3-aminopropyltrimethoxysilane and 3-mercaptopropyltrimethoxy-silane.

    Figure 7
    Figure 8

    8.4 If \(\ce{Hg^2+}\) is only bonding to sulfur binding sites in silica-SH, the symmetric complex of \(\ce{[Hg(silica-SH)2]^2+}\) is formed. Draw the structure of \(\ce{[Hg(silica-SH)2]^2+}\), specify the direction of the bond axes, and draw the corresponding \(d\)-orbital splitting. (You may use \(\ce{R-SH}\) instead of drawing the whole structure of silica-SH.).

    The structure:  \(d\)-orbital splitting diagram :

    8.5 Specify whether the following statements are true or false:

    a) \(d\)-\(d\) transition is found in \(\ce{[(Hg(silica-SH)_{x})]^2+}\)

    □ True  □ False

    b) The \(\ce{[Cu(silica-NH2)_{x}]^2+}\) in a similar geometry, is expected to have a color similar to other copper(II) amine complexes.

    □ True  □ False

    c) In the visible absorption spectra, \(\lambda_{\max}\) of \(\ce{[Cu(silica-NH2)_{x}]^2+}\) is greater than that of \(\ce{[Cu(silica-OH)_{x}]^2+}\).

    □ True  □ False

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  9. Problem 9: Into the Unknown

    Organic compound A is chiral and contains only three elements with the molecular weight (MW) of 149 (rounded to an integer).

    \({}^{1}\mathrm{H}\) NMR spectrum of compound A shows among others, three types of aromatic protons, and its \({}^{13}\mathrm{C}\) NMR spectrum shows eight signals, of which four signals are in the range of 120-140 ppm.

    Compound A can be prepared by treating a carbonyl compound with methylamine followed by \(\ce{NaBH3CN}\).

    9.1 Write all possible structural formulae of compound A. No stereochemistry is required, and do not include stereoisomers.

    9.2 One of the position isomers of compound A (structure A1, A2 or A3) can be synthesized from compound B or C and D as shown in the diagram below. Write down the structural formulae of compounds B-F, and the position isomer of compound A.

    9.3 Compound A is the \((R)\)-form of one of structures A1–A3. It can be prepared from vicinal diols X and Y as shown in the diagram below. Both diols are structural isomers, and each structure contains one carbon less than that of compound A.

    Write down the structural formulae of compounds G-N, X, Y and the \((R)\)-form of compound A. You must show stereochemistry of all compounds.

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  10. Problem 10: Total Synthesis of Alkaloids

    Alkaloids are a class of nitrogen-containing natural products. Their tructural complexity and potent biological activities has drawn attentions. Two representative examples of alkaloids – sauristolactam and pancratistatin are highlighted in following questions.

    Part A

    Sauristolactam possesses excellent cytotoxicity against various cancer cell lines. It could be prepared by following synthetic sequence on the page. (\(\ce{^1H}\)-NMR spectra were recorded in \(\ce{CDCl3}\) at 300 MHz.)

    10.1 Draw the structures of A-G in the sequence.

    Part B

    Pancratistatin, isolated from a Hawaiian native plant, spider lily, exhibits potent in vitro and in vivo inhibitory activity of cancer cell growth in addition to its excellent antiviral activity.

    Figure 1

    Pancratistatin

    Pancratistatin could be successfully synthesized via intermediates X1 and X2. The synthesis of these intermediates are shown in the attached scheme.

    10.2 Draw the structures of A and B.

    10.3 Intermediate X1 (a single enantiomer with the stereochemistry shown) is labeled with deuterium with configuration as indicated below, propose the 3-D chair structure of compound E and the structure of compound F, with stereochemistry. Is Y a proton (\(\ce{^1H}\)) or a deuterium (\(\ce{^2H}\))?

    Syntetic sequence:

    Figure 2

    Scheme of the synthesis of Pancratistatin and structures of compound A and B:

    Figure 3
    Figure 4

    Intermediate X1

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  11. Problem 11: Twist & Chirality

    trans-Cyclooctene has a chiral plane and a high barrier for racemization. The double bond of trans-cyclooctene is twisted, as a result, the molecule displays unusual reactivity in cycloaddition reactions.

    In 2011, Fox and coworkers developed a photochemical synthesis towards a variety of trans- cyclooctene derivatives. The process is non-stereocontrolled and the synthetic scheme is as follow.

    Figure 1

    11.1 Draw all possible stereoisomers of compound 3 that could be obtained from the reduction of compound 2. Not necessary to assign \(R,S\) configuration.

    11.2 i) If one of the stereoisomers of compound 3 is converted to compound 4, how many stereoisomeric form(s) of compound 4 will be obtained?

    ii) The number of possible stereoisomeric form(s) of compound 4 = \(\boxed{\phantom{00}}\)

    iii) If there are more than one stereoisomer, is it possible to separate the obtained stereoisomers of compound 4 by achiral chromatography? Tick the relevant circle.

    ○ Yes  ○ No

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