International Chemistry Olympiad

IChO 2024

Riyadh, Saudi Arabia · 9 theory problems with complete solutions

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  1. Problem 1: Ammonia

    The Haber–Bosch process to produce ammonia from nitrogen and hydrogen is one of the most impactful industrial processes in history and estimated to sustain around half of today’s population. The process typically occurs at constant temperature and pressure, 400 °C and 100 bar, respectively. Equilibrium between the reactants and the product is established over Fe-based catalysts in the reactor.

    The standard gas phase thermochemistry data of \(\ce{N2}\), \(\ce{H2}\), and \(\ce{NH3}\) are shown below. Assume the enthalpy and entropy of the reaction are temperature independent.

    \(\ce{N2(g)}\) \(\ce{H2(g)}\) \(\ce{NH3(g)}\)
    \(\Delta_f H^\circ/(\mathrm{kJ\,mol^{-1}})\) 0 0 −45.9
    \(S^\circ/(\mathrm{J\,mol^{-1}\,K^{-1}})\) 191.6 130.7 192.8

    1.1 — 7 pt

    Find the molar nitrogen conversion (in percent) to ammonia under the production conditions from a stoichiometric mixture of nitrogen and hydrogen. Show your calculations for partial credit. If you cannot find a result, use 66% in subsequent calculations.

    One of the reactants in the Haber–Bosch process is hydrogen, mainly produced by steam reforming of methane. The two-step process starts with \(\ce{CH4}\) reacting with water to produce \(\ce{H2}\) and \(\ce{CO}\). In a subsequent step, \(\ce{CO}\) reacts with water to form \(\ce{CO2}\) and additional hydrogen.

    1.2 — 2 pt

    Write the overall reaction equation for the combined process. Give the mass of \(\ce{CO2}\) produced in the reaction per 1.0 kg of \(\ce{H2}\) (\(m_{\ce{CO2}}\)).

    In reality, 7.0 kg of carbon dioxide is produced for each 1.0 kg of hydrogen in steam reforming. The generated ammonia is classified based on \(\ce{CO2}\) management. For "gray" ammonia, the \(\ce{CO2}\) is released in the atmosphere; for "blue" ammonia, it is captured and stored. "Green" ammonia is made from hydrogen produced without carbon emissions.

    1.3 — 3 pt

    Calculate the mass of the carbon dioxide (\(m_{\ce{CO2},\,\mathrm{blue}}\)) that needs to be captured to produce 40,000 kg of “blue” ammonia. Assume that carbon dioxide only comes from a real steam reforming process. Assume further that the ammonia is produced at equilibrium from a stoichiometric mixture of nitrogen and hydrogen.

    Using cheap renewable hydrogen, the prospect of converting it with captured \(\ce{CO2}\) into methanol appears promising. However, in this procedure, there is a competing reaction that produces \(\ce{CO}\):

    Figure 1

    Into a reactor maintained at 250 °C and 50 bar, a mixture containing \(\ce{CO2}:\ce{H2}\) in a molar ratio of \(1:4\) is fed. It has been found that the \(\ce{CO2}\) conversion is 30% and the \(\ce{H2}\) conversion is 18.5% when these two reactions are at equilibrium. No other processes need to be considered.

    1.4 — 5 pt

    Calculate the thermodynamic equilibrium constants (at 250 °C) of the two reactions.

    Heterogeneous copper-containing catalysts are commonly used for \(\ce{CO2}\) hydrogenation. Two specific, Cu-containing catalysts (\(A\) and \(B\)) showed promise for converting \(\ce{CO2}\) to methanol. These catalysts differ not only in their catalytic rates, but also in surface area, Cu content, and Cu dispersion (the ratio of surface Cu to the total Cu in the catalyst).

    Cu dispersion is important, because only Cu species at the surface are active in \(\ce{CO2}\) hydrogenation. Under this assumption, 1 mole of active sites corresponds to 1 mole of surface Cu. The efficiency of these active sites is shown by their turnover frequencies (TOF, moles of product formed per moles of active site in a given time).

    The table below reports the performance (reported as mass of produced methanol per total catalyst mass in a given time) and the specific surface areas of the two catalysts. The Cu content and dispersion of the catalysts was also measured and reported as follows:

    1. At the start, all the copper in the catalysts (oxidized beforehand) is present as Cu(II); that was fully reduced to Cu(0) with hydrogen. No other species in the catalysts reacted with hydrogen besides copper oxides.
    2. \(\ce{N2O}\) was introduced to the reduced samples at room temperature. At this temperature, \(\ce{N2O}\) only oxidizes active surface sites to Cu(I).
    3. The Cu(I) active surface sites were reduced again to Cu(0) with hydrogen.
    Figure 2
    Catalyst performance \((\mathrm{g}_{\mathrm{MeOH}}\,\mathrm{g}_{\mathrm{cat}}^{-1}\,\mathrm{h}^{-1})\) Specific surface area \((\mathrm{m^2}\,\mathrm{g}_{\mathrm{cat}}^{-1})\) \(\ce{H2}\) consumed in step 1 \((\mathrm{mmol}\,\mathrm{g}_{\mathrm{cat}}^{-1})\) \(\ce{H2}\) consumed in step 3 \((\mathrm{mmol}\,\mathrm{g}_{\mathrm{cat}}^{-1})\)
    Catalyst \(A\) 0.80 100 5.0 0.30
    Catalyst \(B\) 0.90 120 4.0 0.70

    1.5 — 3 pt

    Find the methanol production rate per total mass of Cu (in \(\mathrm{g}_{\mathrm{MeOH}}\,\mathrm{g}_{\mathrm{Cu}}^{-1}\,\mathrm{h}^{-1}\)) for catalysts \(A\) and \(B\).

    1.6 — 3 pt

    Find the TOF (in \(\mathrm{h}^{-1}\)) for catalysts \(A\) and \(B\).

    If you cannot find results, use \(\mathit{TOF}_{\mathrm{Catalyst}\ A}=100\ \mathrm{h}^{-1}\) and \(\mathit{TOF}_{\mathrm{Catalyst}\ B}=1\ \mathrm{h}^{-1}\) in subsequent calculations.

    1.7 — 1 pt

    Indicate the correct statement on the answer sheet.

    a. Catalyst \(A\) has more efficient active sites.
    b. Catalyst \(B\) has more efficient active sites.
    c. The two catalysts have active sites with the same efficiency.

    Avogadro constant: \(N_A=6.022\cdot10^{23}\ \mathrm{mol^{-1}}\) Gas constant: \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\)
    Ionic product of water at 298.15 K: \(K_w=10^{-14}\) Zero of the Celsius scale: \(0\,^\circ\mathrm{C}=273.15\ \mathrm{K}\)
    Faraday constant: \(F=96485\ \mathrm{C\,mol^{-1}}\) Planck constant: \(h=6.626\cdot10^{-34}\ \mathrm{J\,s}\)
    Speed of light: \(c=2.998\cdot10^8\ \mathrm{m\,s^{-1}}\) Standard pressure: \(p^\circ=1\cdot10^5\ \mathrm{Pa}=1\ \mathrm{bar}\)
    Ideal gas law: \(pV=nRT\) Beer-Lambert law: \(A=\log\left(\frac{I_0}{I}\right)=\varepsilon\cdot l\cdot c\)
    Enthalpy: \(H=U+pV\) Gibbs energy: \(G=H-TS\)
    \(\Delta_rG^\circ=-RT\ln K=-nFE^\circ_{\mathrm{cell}}\) Henderson-Hasselbalch equation: \(\mathrm{pH}=\mathrm{p}K_a+\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\)
    Arrhenius equation: \(k=A\exp\left(-\frac{E_A}{RT}\right)\) Zeroth order integrated rate law: \([\ce{A}]=[\ce{A}]_0-kt\)
    Surface of a sphere: \(A=4\pi R^2\) First order integrated rate law: \(\ln[\ce{A}]=\ln[\ce{A}]_0-kt\)
    Volume of a sphere: \(V=\frac{4\pi}{3}R^3\) \(\displaystyle\sum_{k=0}^{\infty}ar^k=\frac{a}{1-r}\quad\text{for }\lvert r\rvert<1\)

    Periodic table

    Figure 3

    Official English text of the problem.

    Answers and solution →

  2. Problem 2: Electronic nose

    Resistive gas sensors, especially those using semiconducting metal oxides (SMOX), are becoming increasingly popular. They can detect tiny amounts of impurities, are small and stable, and are easy to make and use. These features make SMOX sensors stand out among the many types of gas sensors that work on different physical and chemical principles.

    The mixed oxide X that has the normal spinel crystal structure, \(\ce{A^{2+}B2^{3+}O4^{2-}}\), is an example of one of these SMOX materials. X can be obtained by decomposition of the crystalline hydrated metal oxalate \(\ce{ZC2O4}\cdot k\ce{H2O}\) of metal Z on air. When heated to \(140\,^\circ\mathrm{C}\), the crystalline hydrate loses 19.7% of its mass. Further heating in air leads to the formation of 2.407 g of black colored X at \(500\,^\circ\mathrm{C}\) and \(3.8\,\mathrm{dm^3}\) of carbon dioxide at 101325 Pa.

    2.1 Determine the formula of X and the value of \(k\) in \(\ce{ZC2O4}\cdot k\ce{H2O}\). 4 pt

    In the spinel structures oxygen atoms form a face centered cubic (FCC) lattice. In normal spinels the \(\ce{A^{2+}}\) cations occupy part of the tetrahedral, and the \(\ce{B^{3+}}\) cations occupy part of the octahedral lattice sites.

    2.2 Calculate the percentage of the tetrahedral sites that is occupied in the case of a normal spinel \(\ce{A^{2+}B2^{3+}O4^{2-}}\). 2 pt

    In addition to the normal spinels, the \(\ce{MM'2O4}\) general formula also describes the inverse and the mixed spinel structures. Here \(\ce{M}\) are the tetrahedrally coordinated, \(\ce{M'}\) the octahedrally coordinated cations, but contrary to normal spinels, particular cations are not restricted to a single lattice site. A characteristic of the inverse spinel structures is that the electron configuration of the \(\ce{M'}\) site cations is \(d^0\), \(d^5\) or \(d^{10}\).

    2.3 Select on the answer sheet two of the given formulas, which correspond to the structure of inverse spinel: \(\ce{Fe3O4}\), \(\ce{NiFe2O4}\), \(\ce{Mn3O4}\), \(\ce{FeCr2O4}\). 1 pt

    Semiconductors are materials whose orbitals contain a small excess or deficit of electrons, leading to negative or positive charges (holes), respectively. These charges can move through the semiconductor and allow it to conduct electricity. Chemical reactions can increase or decrease the charge on a semiconductor, altering its conductivity.

    X is an example of a semiconductor where electricity is carried by positive charges (holes). For a resistive gas sensor based on X, the resistance \(R\) is measured over time \(t\), while passing a gas over the sensor. This allows redox reactions to occur between the gas and the semiconductor surface, which will change the resistance of the sensor depending on the donor/acceptor properties of the gas. Figure 1 shows the two different types of signal obtained this way.

    Figure 1

    Figure 1: plot of resistance against time for two possible signal types, (a) and (b).

    Grey bars = air, white bars = air + analyte gas.

    2.4 Correlate the signal type (a or b) obtained from the X based sensor with the analyte gases (\(\ce{H2S}\), \(\ce{O3}\), \(\ce{NO2}\), \(\ce{NH3}\)) during their detection at low concentrations. 2 pt

    The dependence of the specific conductivity (\(\sigma\)) of X on temperature is described by the Arrhenius law: \(\sigma(T)=\sigma_0\exp(-E_a/(RT))\), where \(E_a\) is the activation energy of conductivity. Three types of conductivity can be distinguished in different temperature ranges:

    Ionization (I): 444–570 K, Impurity (II): 585–765 K, and Intrinsic (III): 800–905 K.

    The sketch below illustrates the conductivity-temperature relation for the three temperature ranges.

    Figure 2

    2.5 Based on the sketch above, order the activation energies associated with the different conductivity types, \(E_a(I)\), \(E_a(II)\), and \(E_a(III)\), from largest to smallest. 3 pt

    Avogadro constant: \(N_A=6.022\cdot10^{23}\,\mathrm{mol^{-1}}\) Gas constant: \(R=8.314\,\mathrm{J\,mol^{-1}\,K^{-1}}\)
    Ionic product of water at 298.15 K: \(K_w=10^{-14}\) Zero of the Celsius scale: \(0\,^\circ\mathrm{C}=273.15\,\mathrm{K}\)
    Faraday constant: \(F=96485\,\mathrm{C\,mol^{-1}}\) Planck constant: \(h=6.626\cdot10^{-34}\,\mathrm{J\,s}\)
    Speed of light: \(c=2.998\cdot10^8\,\mathrm{m\,s^{-1}}\) Standard pressure: \(p^\circ=1\cdot10^5\,\mathrm{Pa}=1\,\mathrm{bar}\)
    Ideal gas law: \(pV=nRT\) Beer-Lambert law: \(A=\log\left(\frac{I_0}{I}\right)=\varepsilon\cdot l\cdot c\)
    Enthalpy: \(H=U+pV\) Gibbs energy: \(G=H-TS\)
    \(\Delta_rG^\circ=-RT\ln K=-nFE^\circ_{\mathrm{cell}}\) Henderson-Hasselbalch equation: \(\mathrm{pH}=\mathrm{p}K_a+\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\)
    Arrhenius equation: \(k=A\exp\left(-\frac{E_A}{RT}\right)\) Zeroth order integrated rate law: \([\ce{A}]=[\ce{A}]_0-kt\)
    Surface of a sphere: \(A=4\pi R^2\) First order integrated rate law: \(\ln[\ce{A}]=\ln[\ce{A}]_0-kt\)
    Volume of a sphere: \(V=\frac{4\pi}{3}R^3\) \(\displaystyle\sum_{k=0}^{\infty}ar^k=\frac{a}{1-r}\quad\text{for }\lvert r\rvert<1\)

    Periodic table

    Figure 3

    Official English text of the problem.

    Answers and solution →

  3. Problem 3: Tyrosinase

    The enzyme tyrosine 3-monooxygenase has an important biological role because it catalyzes the initial and rate-limiting step in the biosynthesis of the neurotransmitter dopamine. In this first step, L-DOPA is produced from the amino acid L-tyrosine (Tyr) in the presence of dissolved oxygen, tetrahydrofolic acid and ferrous ions. In a subsequent step, another enzyme transforms L-DOPA into dopamine.

    Figure 1

    The activity of the tyrosine 3-monooxygenase enzyme is described well by the simplest model of enzyme action, the Michaelis-Menten equation. This gives the rate of the formation (\(r\)) of L-DOPA as a function of the enzyme (E) and substrate Tyr (S) concentrations as follows:

    \[ r=\frac{k[E][S]}{K_{\mathrm M}+[S]} \]

    The following values of enzymatic turnover frequency (\(k=250\ \mathrm{min^{-1}}\)) and Michaelis constant (\(K_{\mathrm M}=0.49\ \mathrm{mmol\,dm^{-3}}\)) can be used in this task when needed.

    Catechol (1,2-dihydroxybenzene) inactivates the enzyme tyrosine 3-monooxygenase irreversibly by turning it into an inactive form. When a carefully purified sample of catechol is mixed with the enzyme with the exclusion of dissolved oxygen, no change is seen at all. So the reaction of catechol with dissolved oxygen, which produces o-quinone, must be important.

    Figure 2

    This process can be studied in the absence of the enzyme. A test experiment in a reactor not in contact with ambient air at \(37\,^\circ\mathrm C\) and at pH 7.5 gave the following kinetic data:

    time (min) catechol concentration (\(\mathrm{mmol\,dm^{-3}}\)) oxygen concentration (\(\mathrm{mmol\,dm^{-3}}\)) o-quinone concentration (\(\mu\mathrm{mol\,dm^{-3}}\))
    0 2.00 0.200 0
    20.0 1.97 0.170 30
    40.0 1.94 0.140 60
    60.0 1.91 0.111 89
    80.0 1.88 0.083 117
    100.0 1.85 0.054 146

    3.1 — 5.0 pt

    Identify the only by-product (which is not one of the molecules shown in the table above) and write the balanced equation of the process with molecular formulas based on the kinetic data.

    3.2 — 2.0 pt

    Calculate the initial rate (\(r_0\)) of o-quinone formation in this experiment with an appropriate unit.

    The rate law of the process was determined by varying the initial concentrations of the reactants. Changing the oxygen concentration did not influence the initial rate of the process. Using different catechol concentrations gave the following initial rates for o-quinone formation:

    catechol concentration (\(\mathrm{mmol\,dm^{-3}}\)) initial rate of o-quinone formation (\(\mu\mathrm{mol\,dm^{-3}\,min^{-1}}\))
    1.0 0.75
    4.0 3.0
    6.0 4.5
    8.0 6.0
    10.0 7.5

    3.3 — 5.0 pt

    Determine the rate law of the reaction between catechol and oxygen producing o-quinone. Calculate the rate constant of this process (\(k_{\mathrm{catechol}}\)) with an appropriate unit.

    In separate experiments, it was confirmed that neither o-quinone, nor the by-product inactivates the enzyme alone, nor even in combination.

    An experiment in which the enzyme is reacted with catechol in the presence of dissolved oxygen did show inactivation, though.

    The experiment was done in an open vessel so that any oxygen consumed in the process could be continuously replenished from air. The initial concentration of the enzyme was \(4.2\ \mathrm{nmol\,dm^{-3}}\), the initial concentration of catechol was \(2.00\ \mathrm{mmol\,dm^{-3}}\) at \(37\,^\circ\mathrm C\) and at pH 7.5.

    Samples were taken regularly from the enzyme-catechol reaction mixture, and the activity of the enzyme was assayed by adding L-tyrosine, tetrahydrofolic acid and ferrous ions. During the enzyme activity measurements, the concentration of added L-tyrosine was \(10.0\ \mathrm{mmol\,dm^{-3}}\) and the concentration of L-DOPA in each assay sample was measured after 2.0 minutes. The following results were obtained:

    Figure 3
    inactivation reaction time \(t\) (min) L-DOPA concentration after 2.0 minutes from the start of the assay \([\mathrm{DOPA}]\) (\(\mu\mathrm{mol\,dm^{-3}}\))
    0 2.0
    20.0 1.9
    40.0 1.7
    60.0 1.3
    70.0 1.00
    80.0 0.72
    90.0 0.38

    3.4 — 4.0 pt

    Derive a formula that allows you to calculate the actual concentration of active enzyme (\([E]\)) from the data of these experiments in terms of \(k\) (\(=250\ \mathrm{min^{-1}}\)), \(K_{\mathrm M}\) (\(=0.49\ \mathrm{mmol\,dm^{-3}}\)), and/or reagent concentrations.

    The change in the active enzyme concentration as a function of time is well described by a parabola (\([E]=[E]_0-k_{\mathrm{obs}}t^2\)), where \(k_{\mathrm{obs}}\) is an observed rate constant (i.e. one that can be written later as a combination of elementary rate constants and initial concentrations), and \([E]_0\) is the initial concentration of the enzyme.

    3.5 — 3.0 pt

    Give the value of \(k_{\mathrm{obs}}\) with an appropriate unit.

    3.6 — 3.0 pt

    Determine the first half-life of the enzyme inactivation reaction. Derive a formula that gives the first half-life as a function of the initial enzyme concentration and of \(k_{\mathrm{obs}}\).

    The first half-lives for enzyme inactivation were determined using other initial catechol concentrations. The dependences of active enzyme concentrations on time were very well described by parabolas in these cases as well.

    catechol concentration (\(\mathrm{mmol\,dm^{-3}}\)) first half-life (min)
    1.0 140
    4.0 35
    6.0 23
    8.0 17.5
    10.0 14

    3.7 — 6.0 pt

    Derive a rate equation of the enzyme inactivation process consistent with all the experimental observations.

    3.8 — 3.0 pt

    What are the reactants of the rate determining step for the enzyme inactivation process? Select the ONLY correct option on the answer sheet.

    A) \(\ce{\text{catechol} + O2 ->} \ldots\)
    B) \(\ce{\text{catechol} + \text{o-quinone} ->} \ldots\)
    C) \(\ce{\text{o-quinone} + O2 ->} \ldots\)
    D) \(\ce{\text{enzyme} + 2 \text{catechol} ->} \ldots\)
    E) \(\ce{\text{enzyme} + O2 ->} \ldots\)
    F) \(\ce{2 \text{catechol} + O2 ->} \ldots\)
    G) \(\ce{\text{enzyme} + \text{catechol} ->} \ldots\)
    H) \(\ce{\text{enzyme} + \text{by-product} ->} \ldots\)

    Official English text of the problem.

    Answers and solution →

  4. Problem 4: Potassium

    Elementary potassium was the first metal synthesized by electrolysis (1807). Humphry Davy used voltaic cells and electrodes immersed into molten \(\ce{KOH}\).

    4.1 Give balanced equations for the processes taking place on the cathode (1) and the anode (2). 2 pt

    It is little known that only a year after Davy, French chemists have produced potassium in a chemical reduction. One of the methods involved a mixture of potassium carbonate and finely powdered carbon. This mixture was heated to very high temperatures in a gun barrel. Carbon monoxide and potassium vapors were leaving the end of the barrel, where the metal could be condensed to a solid.

    4.2 Write down the balanced equation for the reaction giving only these two products. 1 pt

    The chemical production of potassium had a low yield. Depending on the circumstances, a considerable amount of reddish paste was left behind in the gun barrel. Gmelin isolated an interesting compound (X) from this residue in 1825 by recrystallization in water. The exact structure of the byproduct X was determined more than one and a half century later.

    X is a hydrated salt (2 mol water : 1 mol of salt) forming golden yellow crystals. Its anion has an exceptionally symmetric (shown by vibrational spectra) and stable planar cyclic structure.

    The diprotic acid (Y) that can be formed from the salt X has surprisingly (at least to the superficial observer) low \(\mathrm{p}K_a\) values: 0.8 and 2.2.

    4.3 Give the pH where 95% of all dissolved acid Y is in the neutral acid form. 3 pt

    4.4 Calculate the pH of a \(0.01\ \mathrm{mol\,dm^{-3}}\) solution of the salt X in pure water. 4 pt

    The key step in the structure identification of X was the treatment of the anhydrous salt with methyl iodide. The salt reacted with two equivalents of methyl iodide. The somewhat volatile compound that was obtained had a molar mass of \(170.12\ \mathrm{g/mol}\) (measured in a mass spectrometer).

    A number of analogous salts were also identified in the mixture in smaller quantities. The appropriate potassium salts when treated with two equivalent of \(\ce{MeI}\) gave derivatives with molar masses of: \(142.11\ \mathrm{g/mol}\), \(114.10\ \mathrm{g/mol}\), \(198.13\ \mathrm{g/mol}\).

    4.5 Give the empirical formula of Gmelin’s compound X. 4 pt

    4.6 Draw the structure of the free acid Y. 3 pt

    The extraordinary stability of the series of anions in the salts analogous to X can be attributed to electron delocalization.

    4.7 Draw one resonance structure with a significant contribution for the anion in each of the four salts. 4 pt

    Avogadro constant: \(N_A=6.022\cdot10^{23}\ \mathrm{mol^{-1}}\) Gas constant: \(R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\)
    Ionic product of water at 298.15 K: \(K_w=10^{-14}\) Zero of the Celsius scale: \(0^\circ\mathrm{C}=273.15\ \mathrm{K}\)
    Faraday constant: \(F=96485\ \mathrm{C\,mol^{-1}}\) Planck constant: \(h=6.626\cdot10^{-34}\ \mathrm{J\,s}\)
    Speed of light: \(c=2.998\cdot10^8\ \mathrm{m\,s^{-1}}\) Standard pressure: \(p^\circ=1\cdot10^5\ \mathrm{Pa}=1\ \mathrm{bar}\)
    Ideal gas law: \(pV=nRT\) Beer-Lambert law: \(A=\log\left(\frac{I_0}{I}\right)=\varepsilon\cdot l\cdot c\)
    Enthalpy: \(H=U+pV\) Gibbs energy: \(G=H-TS\)
    \(\Delta_rG^\circ=-RT\ln K=-nFE^\circ_{\mathrm{cell}}\) Henderson-Hasselbalch equation: \(\mathrm{pH}=\mathrm{p}K_a+\log\left(\frac{[\ce{A-}]}{[\ce{HA}]}\right)\)
    Arrhenius equation: \(k=A\exp\left(-\frac{E_A}{RT}\right)\) Zeroth order integrated rate law: \([\ce{A}]=[\ce{A}]_0-kt\)
    Surface of a sphere: \(A=4\pi R^2\) First order integrated rate law: \(\ln[\ce{A}]=\ln[\ce{A}]_0-kt\)
    Volume of a sphere: \(V=\frac{4\pi}{3}R^3\) \(\displaystyle\sum_{k=0}^{\infty}ar^k=\frac{a}{1-r}\quad\text{for }\lvert r\rvert<1\)

    Periodic table

    Figure 1

    Official English text of the problem.

    Answers and solution →

  5. Problem 5: Unknown

    Metal containing salt A can be prepared in a simple exchange reaction when the cold, saturated aqueous solutions of the two distinctly colored compounds, B and C are mixed in stoichiometric ratio. 10.00 g B in solution mixed with the solution of 12.86 g C, and immediately cooled to \(2\,^\circ\mathrm{C}\) will give 4.90 g solid A. The yield of A is 32.6 %.

    To determine the composition of A, first iodometry is performed. A known mass of A is added to a titration flask, acidified with sulfuric acid, then \(\ce{KI}\) is added in excess and there is a precipitate formation. After a few minutes, sodium citrate solution is added until the solution is free of precipitate. The citrate ions form a strong complex with a metal ion present in the mixture, reversing completely the reaction leading to the precipitate. The resulting mixture is titrated with sodium thiosulfate solution (Titration I via reaction 1). To the blue titrated solution another portion of sulfuric acid (significantly more than the initial amount) is added to protonate the citrate. The previous solid precipitate is formed once again [reaction 2]. The mixture is titrated with the same thiosulfate solution (titration II via reaction 1).

    The average volumes for 100.0 mg A with \(5.000\cdot10^{-2}\ \mathrm{mol\,dm^{-3}}\) thiosulfate are \(54.12\ \mathrm{cm^3}\) in Titration I and \(5.41\ \mathrm{cm^3}\) in Titration II.

    When the aqueous solution of A is heated, the blue precipitate D can be observed [reaction 3]. Compound E can be crystallized from the liquid above D. The low temperature during the synthesis of A is important to avoid contamination by E.

    5.1 Write balanced equations for reactions [1] and [2]. 4.0 pt

    The thermal decomposition of A was studied in detail. When pure A is slowly heated, already at around \(75\,^\circ\mathrm{C}\) it explodes. When the compound is dispersed in aluminium oxide and the mixture is heated, then the matrix absorbs the excess heat, and the explosion can be avoided. Two decomposition steps can be observed. In the first step (at \(65\,^\circ\mathrm{C}\)) in addition to the 14.1 % mass decrease, a two-component solid residue forms [reaction 4]. The components of this residue can be easily separated, as F is well soluble in water, while G does not dissolve at all. The F:G mass ratio is 1.00 : 2.97. On further heating, F decomposes without a solid residue [reaction 5]. That means pure G is the final solid decomposition product of A. G contains two other elements in addition to 27.0 wt% oxygen. One of the two components of the gas mixture forming in the first decomposition step can be easily quantitated if it is absorbed in acid solution.

    Some hints

    • Compound A decomposes without releasing oxygen.
    • Compound A contains two different metals.
    • Compounds A and B both contain a complex ion.
    • C is a well-known compound for any chemistry student.
    • Compound F does not contain any metal.

    5.2 Give chemical formulae of A—G. You don’t have to show your calculations, but if your compounds are incorrect you may gain partial marks from correct calculations. 28.0 pt

    5.3 Write balanced equations for reactions [3]—[5]. 12.0 pt

    Periodic table

    Figure 1

    Official English text of the problem.

    Answers and solution →

  6. Problem 6: Penicillin

    Penicillins are a group of antibiotics that have played a crucial role in medicine since their discovery.

    Figure 1

    Penicillin VK

    6.1. Mark the stereogenic carbon centers on the structure of Penicillin VK on the answer sheet. Write the corresponding \(R/S\) stereodescriptors next to the centers on the answer sheet. 6 pt

    6.2. Give the total number of possible stereoisomers of Penicillin VK. 2 pt

    One of the many synthetic methods to obtain Penicillin VK starts from the amino acid valine, as shown in the scheme on the next page of this problem.

    Hints:

    • Among other signals, the \({}^{1}\mathrm{H}\)NMR spectrum of compound A contains two signals that each integrate to 3H. Both of these signals are doublets.
    • Compounds A and D are open-chain compounds, while B and E each have a five membered ring (in addition to any rings in the PhthN-group).
    • Brucine is used for chiral resolution.
    Figure 2

    *Please note that in the formation of (+)-C, the word "Separation" means separation.

    **Conc. HCl means concentrated HCl.

    6.3. Choose the suitable reagent for the numbered steps \((1–4)\) on the answer sheet. 8 pt

    a) \(2\,\mathrm{M}\) \(\ce{HCl_{(aq)}}\)

    b) \(\ce{SO3, H2SO4}\)

    c) \(\ce{NaOH_{(aq)}}\)

    d) \(\ce{HCO2H, Ac2O}\)

    e) i) \(\ce{N2H4}\) ii) \(\ce{HCl_{(aq)}, AcOH}\)

    f) i) \(\ce{H2S, Na, CH3OH}\) ii) \(\ce{H2O}\)

    6.4. Draw the structures of A, B, (+)-C, D, E and F on the answer sheet. Show the stereostructure using wedge–dash notation (

    Figure 3

    ) when relevant. 18 pt

    6.5. Select the role of acetone in the step leading to B on the answer sheet. 2 pt

    a) solvent

    b) catalyst

    c) electrophile

    d) nucleophile

    6.6. Select the role of DCC in the step leading from F to Penicillin VK on the answer sheet. 2 pt

    a) oxidizing reagent

    b) catalyst

    c) coupling reagent

    d) radical initiator

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  7. Problem 7: SABIC

    The Saudi Basic Industries Corporation (SABIC) was established in the mid-1970s as part of Saudi Arabia's plan to diversify its economy. Today, SABIC is a global chemical giant, producing 69 million tons of products annually.

    Part A

    SABIC's own unique ethene oligomerization technology (oligomers are polymers with few monomer units) produces linear alpha olefins (1-alkenes), which are used in numerous industrial applications.

    Linear alpha olefins are produced by catalytic oligomerization of ethene, resulting in a range of chain lengths with a distribution. Different catalysts can yield various distributions of products.

    In the geometric or Schulz–Flory distribution, each oligomer fraction is related to the previous fraction by the equation, \(T_{n+1}=\alpha T_n\), where \(T_n\) represents the molar amount of the oligomer with \(n\) ethene units, and \(\alpha\) is a constant called the probability of chain propagation. Ethene, the monomer is not a member of the series (\(n\geq2\)).

    One catalytic oligomerization process can be described by a relatively simple mechanism:

    Figure 1

    Although this mechanism is not a classical chain reaction because the final product is not produced within the cycle, the usual kinetic treatment of chain reactions can be applied to it. The three processes shown (1–3) are all chain propagation steps. R is lengthened by each cycle.

    All three reactions are first order with respect to the chain carriers. For ethene, use an unknown order \(\rho\), to be determined later.

    7.1 Write the rate equations for reactions 1–3 using \(k_1\), \(k_2\), and \(k_3\) as the corresponding rate constants, which are independent of the carbon chain length. 1 pt

    Chain termination step, producing the oligomers, is a first order reaction competing with reaction 1.

    7.2 Write i) the chemical equation and ii) rate equation of termination. Use \(k_t\) as the corresponding rate constant. 2 pt

    7.3 Find the value of \(\rho\) which ensures that the product mixture of olefins shows the required ratio of products in accordance with the Schulz–Flory distribution when the reaction is run in a closed system. Derive a formula for this case that gives \(\alpha\) as the function of rate constants. 4 pt

    Aluminum alkyls are used as co-catalysts in the ethene oligomerization process. It was found that they also catalyzed the Friedel–Crafts alkylation of the aromatic solvents used for the oligomerization.

    In a typical oligomerization reaction, an alkylated aromatic compound Q was isolated from the complex reaction mixture of ethene oligomerization in toluene as solvent. The alkylated aromatic compound Q has the molecular formula \(\ce{C11H16}\) and its \(\ce{^1H}\) NMR spectrum contains the following peaks.

    \(\ce{^1H}\) NMR of Q: \(\delta\) 0.9 (\(t\), 3H); 1.35 (\(m\), 2H); 1.59 (\(m\), 2H); 2.34 (\(s\), 3H), 2.60 (\(t\), 2H); 7.18 (\(d\), 2H); 7.22 (\(d\), 2H).

    7.4 Draw the structure of Q. 3 pt

    Part B

    Vitrimers are adaptive polymers with "reversible" covalent bonds, allowing shape-changing and self-healing under specific conditions. This subtask explores the principles of vitrimer chemistry, an important topic in SABIC's polymer research.

    A recently elucidated fast and reversible reaction involving dioxaborolanes can serve as the chemical background for such an adaptable polymer network. The two dioxaborolanes below were prepared by the reaction of phenylboronic acid with 1,2-butanediol and 3,5-dimethylphenylboronic acid with 1,2-propanediol, respectively.

    Figure 2

    They participate in a model metathesis (exchange) reaction. The two carefully purified compounds were mixed in the absence of solvent and the mixture was kept at \(60\,^\circ\mathrm{C}\) for two hours. Interestingly, the gas chromatograms of the mixture at the start and at the end of the incubation period showed only the peaks below, with equimolar quantities of the components. The peaks in this system separate in a way that correlates with molar mass: compounds with lower mass appear at earlier elution times.

    Figure 3

    No catalyst, water, diol or acid was detectable in the mixture.

    7.5 Draw the compounds associated with peaks I–IV in the chromatogram. 2 pt

    Linking functional units to a polymer chain is known as grafting, and the functional units are referred to as grafting agents. In this case, the dioxaborolane units are linked to the polymer chains by maleimide groups via a radical process, in which one maleimide group is coupling to the polymer chain at a random carbon atom. The scheme below introduces a flexible synthesis route towards the dioxaborolane maleimide grafting agent used in the vitrimers.

    Figure 4

    7.6 Draw the structures of A–C. 3 pt

    7.7 Select on the answer sheet which reagent(s) and/or condition is/are appropriate for linking the dioxaborolane maleimide units to polyethylene. 1 pt

    Figure 5

    The vitrimer was prepared as follows:

    • Polyethylene is grafted with dioxaborolane maleimide. Assume no side reactions between the dioxaborolane maleimide compounds.
    • The reagents are removed from the macromolecules.
    • Propanediol functionalized bis-dioxaborolane is added to enable crosslinking.
    Figure 6
    • The mixture is carefully heated for metathesis (exchange) reaction to take place.

    To determine the extent of grafting and crosslinking in the vitrimer, infrared (IR) spectroscopy was used at room temperature where the metathesis reaction is extremely slow. Assume a homogeneous vitrimer.

    • All of the small molecules are removed from the polymer.
    • The IR spectrum of a thin polymer film is measured.
    • IR spectrum contains absorption peaks that correspond to specific functional groups.
    • The Beer–Lambert law can be used to quantify the concentration of these functional groups.

    The provided data table gives the wavelengths, corresponding molar (related to the specified functional groups) absorption coefficients for relevant characteristic peaks together with the absorbances of the vitrimer sample.

    Vibration wavenumber, with the functional group indicated Molar absorption coefficient, \(\epsilon\) \((\mathrm{dm^3\,mol^{-1}\,\mu m^{-1}})\) Sample absorbance at this wave number, \(A\)
    Imide group band at \(1710\ \mathrm{cm^{-1}}\) 2.67 0.451
    Monosubstituted benzene ring band at \(1600\ \mathrm{cm^{-1}}\) 0.28 0.022
    Disubstituted benzene ring band at \(1520\ \mathrm{cm^{-1}}\) 0.68 0.042
    \(\ce{CH2}\) groups of polyethylene, band at \(1470\ \mathrm{cm^{-1}}\) 0.021 0.904

    7.8 Calculate the grafting density (\(X_g\)) of polyethylene based on the measurements. That is the ratio of the grafted groups over the ethylene units. 3 pt

    7.9 Calculate the percentage of the grafted groups that are not part of a crosslink (\(p_{nc}\)). 4 pt

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  8. Problem 8: Safflower

    In Saudi culture, safflower drink is a traditional remedy used for its calming effects.

    Figure 1

    One ingredient with health benefits is kaempferol. The following scheme shows its total synthesis.

    Figure 2

    Hints:

    • C has a tricyclic system. Also, it has the same number of hydroxyl groups as the compound preceding it.
    • The step C to D is necessary to avoid any unwanted reactivity with DMDO.
    • DMDO is a hydroxylating agent.

    8.1. Draw the structures of compounds A–E. 15 pt

    A unique pentacyclic indole alkaloid, serotobenine found in safflower has antimicrobial properties. Its total synthesis is shown below:

    Figure 3

    Hints:

    • G is a bicyclic compound with one of the rings being a heterocycle.
    • H isomerizes upon heating in the presence of the amine base.
    • K is a product of an oxidative cleavage of the newly formed functional group in J. K has a \({}^{1}\mathrm{H}\) NMR signal at 12 ppm.

    8.2. Draw the structures of compounds F–K. 18 pt

    Carthamin gives safflower its red color.

    Figure 4

    Carthamin

    In the total synthesis of carthamin the assembly was tested on a model substrate L. Note that M resembles carthamin’s core structure.

    Figure 5

    8.3. Draw the structure of compound M including its stereochemistry. (Hint: In methanol-\(d_4\) L has 11 and M has 10 distinct proton signals in their \({}^{1}\mathrm{H}\) NMR spectra). 5 pt

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  9. Problem 9: Porphyrin complexes

    Part A. Metals in petroleum

    Saudi Arabian petroleum, primarily hydrocarbons, contains elements like vanadium in porphyrin complexes, hinting at its biological origins.

    Vanadium complex B is presumably formed in petroleum from chlorophyll according to the following scheme:

    Figure 1
    Step Name
    1) Demetallization
    2) Hydrolysis
    3) Decarbomethoxylation
    4) Reduction
    5) Aromatization
    6) Decarboxylation
    7) Metal chelation

    9.1 Assign the molecular formulae a–e (given in the answer sheet) to the intermediates I–V. 4.0 pt

    Figure 2

    9.2 Draw the structures of intermediate A and vanadium complex B using the pre-drawn framework of porphyrin. (If you need a new template to redraw your structure(s), please ask the supervisor.) 7.0 pt

    Figure 3

    9.3 Indicate the oxidation state of vanadium in B. 1.0 pt

    Part B. Porphyrin non-metal complexes

    Porphyrins are known to form chelate-type complexes not only with metals but also with some non-metals, for example, with boron and phosphorus. Octaethylporphyrin (\(\ce{H2OEP}\)) is often used to model natural porphyrins and to study porphyrin complexes. It has the following structure and can be represented as:

    Figure 4

    To produce dicationic planar boron porphyrin complex \(\ce{[X3]^2+}\), either \(\ce{BCl3}\) or another binary compound X1 can be used as a precursor:

    Figure 5

    9.4 Draw the structures of X1, intermediate complex X2 and the product \(\ce{[X3]^2+}\). 3.0 pt

    Figure 6

    Phosphorus forms porphyrin cationic complexes \(\ce{[Y1]^{n+}}\), \(\ce{[Y2]^{n+}}\), and \(\ce{[Y3]^{n+}}\). The complex \(\ce{[Y3]^{n+}}\) has one less plane of symmetry compared to \(\ce{[Y1]^{n+}}\) and \(\ce{[Y2]^{n+}}\) (ignore ethyl substituents in OEP ligand when considering symmetry). Pyridine (Py) is used here as a basic solvent, “Hal” is one of the halogens: F, Cl, or Br:

    Figure 7

    9.5 Indicate: a) the charge “\(n+\)” of phosphorus porphyrin complexes; b) the number of planes of symmetry \(N(\sigma)\) that complex \(\ce{[Y1]^{n+}}\) has. 2.0 pt

    Figure 8

    9.6 Draw the structures of \(\ce{[Y1]^{n+}}\), \(\ce{[Y2]^{n+}}\), and \(\ce{[Y3]^{n+}}\). 4.0 pt

    Figure 9

    Part C. Porphyrin sandwiches

    Porphyrin sandwich-type complexes form when there are several porphyrin rings aligned relative to each other. Examples are \(\ce{Zr(OEP)2}\), \(\ce{Eu2(OEP)3}\), \(\ce{Bi2(OEP)2Cl2}\) that have at least 3 planes of symmetry each. Note that the hole size in the OEP ligand is ca. \(4.0\ \text{Å}\), and the average metal–N distances in these complexes are \(2.4,\ 2.5,\ 2.3\ \text{Å}\), respectively.

    9.7 Suggest the structures of \(\ce{Zr(OEP)2}\), \(\ce{Eu2(OEP)3}\), and \(\ce{Bi2(OEP)2Cl2}\). You may use a simplified representation of the OEP ligand. 3.0 pt

    Figure 10

    Another type of porphyrin sandwich-type complex is metal–porphyrin dimers with single or multiple metal–metal bonds. For example, \(\ce{[Ru(OEP)]2}\) has a Ru–Ru double bond (considering only d-electrons). Below is the molecular orbital diagram for an eclipsed dimer of the type \(\ce{[M(porphyrin)]2}\) (interaction between \(\mathrm d_{x^2-y^2}\) is not considered as these orbitals lie high in energy):

    Figure 11

    9.8 Draw the \(\sigma(\mathrm d_{z^2}+\mathrm d_{z^2})\), \(\pi(\mathrm d_{xz}+\mathrm d_{xz})\), \(\pi(\mathrm d_{yz}+\mathrm d_{yz})\), and \(\delta(\mathrm d_{xy}+\mathrm d_{xy})\) bonding orbitals according to the specified coordinate system. 4.0 pt

    Figure 12

    9.9 Calculate the metal–metal bond order in the following complexes in eclipsed conformation: \(\ce{[Mo(OEP)]2}\), \(\ce{[Ir(OEP)]2}\), \(\ce{[Re(OEP)]2^{+}}\). 3.0 pt

    Figure 13

    The eclipsed conformation is generally preferred for metal–porphyrin dimers with 7, 8, and 9 d-electrons. Otherwise, these complexes are more stable in a staggered conformation, where the porphyrin rings are rotated relative to each other by \(45^\circ\).

    9.10 Select how the splitting (i.e., the energy between antibonding and bonding orbitals) changes for each type of orbital interaction during the conversion of eclipsed conformation into staggered:

    a) increases;
    b) remains unchanged;
    c) decreases (not to zero);
    d) decreases to zero.

    4.0 pt

    Figure 14

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