International Chemistry Olympiad

IChO 2016

Tbilisi, Georgia · 8 theory problems with complete solutions

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  1. Problem 1: Nitrogen fluorides

    Nitrogen trifluoride is a surprisingly stable compound that was first prepared in the melt electrolysis of a mixture of ammonium fluoride and hydrogen fluoride.

    1.1 On which electrode does nitrogen trifluoride form? Write a balanced chemical equation for the electrode half reaction for the formation of \(\ce{NF3}\).

    Interestingly the related fluoroamine (\(\ce{NH2F}\)) and difluoroamine (\(\ce{NHF2}\)) are very unstable materials; decomposition of either pure substance can even be explosive. This is dangerous as they are formed in the electrolysis as side products.

    1.2 Which of \(\ce{NF3}\), \(\ce{NHF2}\) or \(\ce{NH2F}\) compound is expected to condense at the lowest temperature?

    The \(\ce{N-F}\) bond lengths in these molecules were determined to be 136, 140 and 142 pm. The change in the bond lengths can be explained with a simple electrostatic model taking into account the partial charges on the atoms.

    1.3 Assign the \(\ce{N-F}\) bond lengths (136, 140, 142 pm) to the molecules.

    When \(\ce{NHF2}\) is bubbled through a solution of \(\ce{KF}\) in \(\ce{HF}\), a binary nitrogen – fluorine compound can be obtained as a mixture of two geometric isomers.

    1.4 Write a balanced chemical equation for the formation of the binary nitrogen-fluorine compound.

    Tetrafluoroammonium ion (\(\ce{NF4+}\)) and its corresponding salt can form from \(\ce{NF3}\) with elementary fluorine in the presence of an appropriate reagent.

    1.5 Propose a suitable reagent and write a balanced chemical equation for the reaction.

    \(\ce{NF4+}\) ions form stable salts with a number of anions. These are very sensitive to humidity, because \(\ce{NF4+}\) ion hydrolyzes forming \(\ce{NF3}\) and \(\ce{O2}\). Interestingly nitrogen trifluoride always forms quantitatively, while the quantity of oxygen is often less than expected due to side reactions.

    1.6 Write a balanced chemical equation for the hydrolysis of \(\ce{NF4+}\) Write a balanced chemical equation for a possible side reaction that can decrease the theoretically expected \(\ce{O2}:\ce{NF3}\) mole ratio.

    Tetrafluoroammonium salts were investigated for use as solid rocket fuels, because \(\ce{NF3}\) and \(\ce{F2}\) are released from them on heating. One of them has a fluorine content of 65.6 \(m/m\%\), all of which is converted into \(\ce{NF3}\) and \(\ce{F2}\) upon decomposition. During the decomposition 2.5 times as many moles of \(\ce{F2}\) are formed as of \(\ce{NF3}\).

    1.7 Determine the formula of the salt in question.

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  2. Problem 2: Copper(I) oxide

    One of the first materials used in solid state electronics was red copper(I) oxide. Interest is renewed nowadays because it could be a non-toxic and cheap component of solar cells.

    Figure 1
    Figure 2

    The two figures above depict the cubic unit cell of the \(\ce{Cu2O}\) crystal. The lattice constant of the structure is 427.0 pm.

    2.1.1 Which of the atoms (A or B) is copper?

    Which basic structure (primitive cubic, face centered cubic, body centered cubic, diamond) is formed by the A atoms and which structure is formed by the B atoms?

    What are the coordination numbers of the atoms?

    2.1.2 Calculate the smallest \(\ce{O-O}\), \(\ce{Cu-O}\) and \(\ce{Cu-Cu}\) distances in the structure?

    2.1.3 What is the density of pure copper(I) oxide?

    A common defect in this crystal is some copper atoms missing with the oxygen lattice unchanged. The composition of one such crystal sample was studied, and 0.2% of all copper atoms were found to be in oxidation state +2.

    2,2 What percentage of normal copper sites are empty in the crystal sample? What is \(x\) in the empirical formula \(\ce{Cu_{2-x}O}\) of the crystal?

    Copper(I) oxide is insoluble in water. It is stable in dry air, but humidity in the air catalyses a transformation (Reaction 1).When copper(I) oxide is dissolved in dilutesulfuric acid, a blue solution containing a precipitate is formed without evolution of a gas (Reaction 2). When hot, concentrated sulfuric acid is used, no precipitate remains, but an odorous gas forms (Reaction 3).The same gas forms when the precipitate from reaction 2 is dissolved in hot concentrated sulfuric acid.

    2.3 Write balanced chemical equations for reactions (1-3).

    Copper (I) oxide can be produced in a number of ways. Heating copper in air is a common method in the synthesis of semiconductor \(\ce{Cu2O}\). In a pure oxygen atmosphere, the three species containing copper (\(\ce{Cu(s)}\), \(\ce{Cu2O(s)}\) or \(\ce{CuO(s)}\)) can potentially interconvert.

    Suppose that the \(\Delta_\mathrm{f}H^\circ\) and \(S^\circ\) data given for \(10^5\ \mathrm{Pa}\) are independent of temperature:

    Compound \(\Delta_\mathrm{f}H^\circ/\mathrm{kJ\,mol^{-1}}\) \(S^\circ/\mathrm{J\,mol^{-1}\,K^{-1}}\)
    \(\ce{Cu(s)}\) 0 65
    \(\ce{O2(g)}\) 0 244
    \(\ce{CuO(s)}\) −156 103
    \(\ce{Cu2O(s)}\) −170 180

    2.4. Determine the temperature ranges, if any, of thermodynamic stability of copper and its oxides between 500 and 1500 K in a \(10^5\ \mathrm{Pa}\) oxygen atmosphere.

    Important data are given for 298 K. Use this temperature in the following calculations:

    \[ K_\mathrm{sp}(\ce{Cu(OH)2})=2\times10^{-19} \]

    \[ \ce{Cu2O(s) + H2O(l) + 2e- -> 2Cu(s) + 2OH-(aq)} \qquad E^\circ=-0.360\ \mathrm{V} \]

    \[ \ce{Cu^{2+}(aq) + e- -> Cu+(aq)} \qquad E^\circ=+0.159\ \mathrm{V} \]

    \[ \ce{Cu^{2+}(aq) + 2e- -> Cu(s)} \qquad E^\circ=+0.337\ \mathrm{V} \]

    One possibility for producing \(\ce{Cu2O}\) is the anodic oxidation of copper. Electrolysis of an aqueous basic solution (e.g. \(\ce{NaOH}\)) with a copper anode and platinum cathode can lead to formation of copper(I) oxide on the anode.

    2.5 Write the half reaction equations for the electrode processes during the anodic production of \(\ce{Cu2O}\) in \(\ce{NaOH}\) solution with a platinum cathode and copper anode.

    Electrolytic reduction of copper(II) ions in solution is another possibility.

    2.6.1 Write the half reaction equation of the cathode process giving \(\ce{Cu2O}\) in acidic medium.

    Let us use \(0.100\ \mathrm{mol\,dm^{-3}}\) \(\ce{Cu^{2+}}\) solution and carry out electrolysis with platinum electrodes.

    2.6.2 What is the maximum pH of this solution at which the concentration of copper(II) can be maintained at \(0.100\ \mathrm{mol\,dm^{-3}}\)?

    If the pH is too low, reduction to metallic copper is preferred to the formation of copper(I) oxide.

    2.6.3 What is the minimum pH at which the cathodic production of \(\ce{Cu2O}\) in a \(\ce{Cu^{2+}}\) solution (\(c=0.100\ \mathrm{mol\,dm^{-3}}\)) is still possible?

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  3. Problem 3: Determination of iodine in iodized salt

    Iodine deficiency is of special concern in Georgia because it occupies a region where iodine is scarce in soil and water. Iodine deficiency can be effectively and inexpensively prevented if salt for human consumption is fortified with small amounts of iodine. Methods for analyzing salt for iodine content are thus important. Current regulations in Georgia are that iodized salt must contain between 25 - 55 ppm iodine (1 ppm = 1 mg iodine/kg salt).

    Most salt is iodized by fortification with potassium iodate (\(\ce{KIO3}\)). Iodate content can be determined in salt samples using iodometric titration. In a typical procedure, 10.000 g of an iodized salt sample is dissolved in 100 \(\mathrm{cm^3}\) aqueous \(\ce{HCl}\) (\(c=1.0\ \mathrm{mol\,dm^{-3}}\)) to which 1.0 g \(\ce{KI}\) has been added. The solution is then titrated with aqueous sodium thiosulfate solution (\(c=0.00235\ \mathrm{mol\,dm^{-3}}\)) to a starch endpoint; this requires 7.50 \(\mathrm{cm^3}\) of titrant.

    3.1.1 Write a balanced net ionic equation for the reaction when iodate reacts with excess iodide in acidic solution.

    3.1.2 Write a balanced net ionic equation for the reaction taking place during the titration with thiosulfate.

    3.1.3 Calculate the iodization level, in ppm, of this salt sample.

    A less common agent for iodizing salt is potassium iodide, which cannot be easily measured by iodometric titration.

    One possible method for analyzing iodide in the presence of chloride is potentiometric titration. However, this method is not very precise in the presence of large amounts of chloride. In this method, a silver wire is immersed in the solution (containing iodide and chloride) to be analyzed and silver ion is gradually added to the solution. The potential of the silver wire is measured relative to a reference electrode consisting of a silver wire in a solution of \(\ce{AgNO3}\) (\(c=1.000\ \mathrm{mol\,dm^{-3}}\)). The measured potentials are negative and the absolute values of these potentials are reported. The solution to be analysed has a volume of 1.000 \(\mathrm{dm^3}\) (which you may assume does not change as silver ion is added) and \(t=25.0\ ^\circ\mathrm{C}\).

    The results of this experiment are governed by three equilibria: the solubility of \(\ce{AgI(s)}\) [\(K_{\mathrm{spI}}\)] and \(\ce{AgCl(s)}\) [\(K_{\mathrm{spCl}}\)] and the formation of \(\ce{AgCl2^-(aq)}\) [\(K_{\mathrm f}\)]. (Iodide also forms complex ions with silver but this may be neglected at the very low concentrations of iodide present in this experiment).

    Figure 1

    Below are shown the results of two experiments measuring the observed potential as a function of added number of moles of silver ion. Experiment A (solid circles) was carried out with 1.000 \(\mathrm{dm^3}\) of iodide solution with concentration \(1.00\times10^{-5}\ \mathrm{mol\,dm^{-3}}\) and containing no chloride ion. Experiment B (open circles) was done using 1.000 \(\mathrm{dm^3}\) of solution with a concentration of iodide \(1.00\times10^{-5}\ \mathrm{mol\,dm^{-3}}\) and that of chloride equal to \(1.00\times10^{-1}\ \mathrm{mol\,dm^{-3}}\).

    Figure 2
    \(\mu\mathrm{mol}\ \ce{Ag+}\) added \(\lvert E\rvert,\ \mathrm{V}\), exp. A \(\lvert E\rvert,\ \mathrm{V}\), exp. B
    1.00 0.637 0.637
    3.00 0.631 0.631
    5.00 0.622 0.622
    7.00 0.609 0.610
    9.00 0.581 0.584
    10.0 0.468 0.558
    11.0 0.355 0.531
    12.0 0.337 0.517
    13.0 0.327 0.517
    15.0 0.313 0.517

    3.2.1 Select an appropriate data from the experiments and use it to calculate the solubility product of \(\ce{AgI}\), (\(K_{\mathrm{spI}}\).).

    3.2.2 Select an appropriate data from the experiments and use it to calculate the solubility product of \(\ce{AgCl}\), (\(K_{\mathrm{spCl}}\).).

    3.2.3 Select an appropriate data from the experiments and use it to calculate \(K_{\mathrm f}\). You may need to use values of \(K_{\mathrm{spI}}\) or \(K_{\mathrm{spCl}}\) to do this calculation. If you were unable to carry out the calculations in 3.2.1 or 3.2.2 you may use the arbitrary values of \(K_{\mathrm{spI}}=1.00\times10^{-15}\) and \(K_{\mathrm{spCl}}=1.00\times10^{-9}\) without penalty.

    An analytical method that is more practical, because it is not sensitive to the presence of chloride, uses the Sandell-Kolthoff reaction. This is the reaction of \(\ce{H3AsO3}\) with \(\ce{Ce(IV)}\) to give \(\ce{Ce(III)}\) in acidic solution, which is strongly catalyzed by iodide ion.

    3.3.1 Write balanced net ionic equations for the reaction of cerium(IV) with \(\ce{H3AsO3}\) in acidic solution, as well as reactions of cerium(IV) with a species containing the element iodine and \(\ce{H3AsO3}\) with a species containing the element iodine, that could reasonably account for the catalysis of the net reaction by iodide.

    The reaction of \(\ce{Ce(IV)}\) with \(\ce{H3AsO3}\) can be monitored by measuring the absorbance at 405 nm, as \(\ce{Ce(IV)}\) is orange and absorbs significantly at 405 nm, while the other reactants and products are colorless and do not absorb appreciably. Three runs were carried out, all in \(0.50\ \mathrm{mol\,dm^{-3}}\) \(\ce{H2SO4}\) at 25.0°C using the following initial concentrations:

    Run \([\ce{H3AsO3}]_0\), \((\mathrm{mol\,dm^{-3}})\) \([\ce{Ce(IV)}]_0\), \((\mathrm{mol\,dm^{-3}})\) \([\ce{I^-}]_0\), \((\mathrm{mol\,dm^{-3}})\)
    1 0.01250 0.00120 \(1.43\cdot10^{-6}\)
    2 0.00625 0.00060 \(1.43\cdot10^{-6}\)
    3 0.01250 0.00120 \(7.16\cdot10^{-7}\)

    An analyst initiated the reactions by mixing the reagents in a cuvette. After a short variable delay absorbance measurements were started, with the first measurement recorded at \(t=0\ \mathrm{s}\). The data obtained are shown below:

    Figure 3
    \(t,\ \mathrm{s}\) \(A_{405}\), Run 1 \(A_{405}\), Run 2 \(A_{405}\), Run 3
    0 0.621 0.287 0.818
    20 0.348 0.149 0.608
    40 0.198 0.083 0.455
    60 0.113 0.046 0.340
    80 0.064 0.025 0.254
    100 0.037 0.014 0.191

    Under these conditions (\(0.5\ \mathrm{mol\,dm^{-3}}\) \(\ce{H2SO4}\), 25.0°C), the rate law for the reaction can be written as

    \[ \text{Rate}=k[\ce{H3AsO3}]^m[\ce{Ce(IV)}]^n[\ce{I^-}]^p \]

    where \(m\), \(n\) and \(p\) are integers.

    3.3.2. Determine the values of \(m\), \(n\), and \(p\) and calculate the value of \(k\) (be sure to specify its units).

    A 1.000 g sample of iodized salt is dissolved in water to give 10.00 \(\mathrm{cm^3}\) of solution. A 0.0500 \(\mathrm{cm^3}\) aliquot of this solution is added to a mixture of 1.000 \(\mathrm{cm^3}\) \(0.025\ \mathrm{mol\,dm^{-3}}\) \(\ce{H3AsO3}\) in \(0.5\ \mathrm{mol\,dm^{-3}}\) \(\ce{H2SO4}\) and 0.800 \(\mathrm{cm^3}\) \(0.5\ \mathrm{mol\,dm^{-3}}\) \(\ce{H2SO4}\). To this mixture is added 0.200 \(\mathrm{cm^3}\) \(0.0120\ \mathrm{mol\,dm^{-3}}\) \(\ce{Ce(NH4)2(NO3)6}\) in \(0.5\ \mathrm{mol\,dm^{-3}}\) \(\ce{H2SO4}\) and the absorbance at 405 nm is measured as a function of time at 25.0°C:

    Figure 4
    \(t,\ \mathrm{s}\) \(A_{405}\)
    0 0.756
    20 0.531
    40 0.373
    60 0.262
    80 0.185
    100 0.129

    3.3.3. Calculate the iodization level, in ppm, of this salt sample.

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  4. Problem 4: Kinetics: water treatment and degradation of pharmaceutical products

    Application of kinetic studies in water treatment

    Industrial waste is a major cause of water pollution and kinetic studies are carried out in a laboratory to design effluent treatment. 1,4-dioxane, more commonly known as dioxane (\(\ce{C4H8O2}\)), an industrial solvent and by-product, is a significant water contaminant. It can be oxidised to hazard free chemicals using oxidants such as peroxodisulfate, ozone or hydrogen peroxide.

    The data obtained in the kinetic study of oxidation of dioxane with potassium peroxodisulfate (\(\ce{K2S2O8}\)) as oxidant and \(\ce{AgNO3}\) as catalyst at \(T=303.15\ \mathrm{K}\) are given below. The reaction was monitored by the estimation of unreacted peroxodisulfate. The concentration of \(\ce{AgNO3}\) used in this study was \(1.00\cdot10^{-3}\ \mathrm{mmol\,dm^{-3}}\).

    Trial Dioxane / \(\mathrm{mmol\,dm^{-3}}\) \(\ce{K2S2O8}\) / \(\mathrm{mmol\,dm^{-3}}\) Initial rate / \(\mathrm{mmol\,dm^{-3}\,min^{-1}}\)
    1 0.0100 2.50 \(1.661\cdot10^{-2}\)
    2 0.0100 5.10 \(3.380\cdot10^{-2}\)
    3 0.00500 13.8 \(9.200\cdot10^{-2}\)
    4 0.0110 13.8 \(9.201\cdot10^{-2}\)

    In many countries the accepted maximum level of dioxane in drinking water is specified as \(0.35\ \mathrm{\mu g\,dm^{-3}}\).

    A water sample contains an initial dioxane concentration of \(40.00\ \mathrm{\mu g\,dm^{-3}}\). Assume that 1 mol dioxane requires 1 mol of peroxodisulfate for oxidation. The concentration of \(\ce{AgNO3}\) used in this study was \(1.00\times10^{-3}\ \mathrm{mmol\,dm^{-3}}\).

    4.1.1. Calculate the time in minutes the oxidation process has to continue in order to reach the accepted level of dioxane at 303.15 K if the initial concentration of \(\ce{K2S2O8}\) is \(5.0\times10^{-6}\ \mathrm{mol\,dm^{-3}}\). Assume that the rate law obtained from the data above is valid under these conditions.

    Various mechanisms have been proposed for the peroxodisulfate oxidation of dioxane. Misra and Ghosh (1963) proposed the following mechanism:

    Figure 1

    4.1.2 Assuming Ag(III) to be in steady state, deduce the rate equation for the oxidation of dioxane.

    4.1.3 Which of the following is/are correct?

    A) The rate equation based on the mechanism given in 4.1.2, at very high concentrations of dioxane, is consistent with the experimental data in 4.1.1.

    B) The rate equation based on the mechanism given in 4.1.2, at very low concentrations of dioxane, is consistent with the experimental data in 4.1.1.

    C) The units of the observed rate constant are \(\mathrm{dm^3\,mol^{-1}\,s^{-1}}\) at very high concentrations of dioxane.

    D) The units of the observed rate constant are \(\mathrm{dm^3\,mol^{-1}\,s^{-1}}\) at very low concentrations of dioxane.

    Degradation of pharmaceutical products – a kinetic overview

    Kinetic studies are important in deciding the shelf life of a pharmaceutical product. Several chemical reactions can affect the shelf life of pharmaceutical products and the rates of these reactions depend on conditions such as pH, temperature, humidity.

    Lysine acetylsalicylate (LAS) is prescribed as a pain killer and anti-inflammatory drug under the brand name Aspegic. LAS on hydrolysis forms lysine salicylate and acetic acid.

    Figure 2

    Hydrolysis of LAS can proceed via three different pathways (a) acid catalysed, (b) uncatalysed and (c) base catalysed.

    If [LAS] denotes the concentration of LAS at time \(t\), the overall rate of the hydrolysis reaction can be written as

    \[ -\frac{d[\mathrm{LAS}]}{dt} = k_{\mathrm H}[\mathrm{LAS}][\ce{H+}] +k_{\mathrm O}[\mathrm{LAS}] +k_{\mathrm{OH}}[\mathrm{LAS}][\ce{OH-}] \]

    where \(k_{\mathrm H}\), \(k_{\mathrm O}\) and \(k_{\mathrm{OH}}\) are the rate constants of the acid catalysed, uncatalysed and base catalysed pathways of hydrolysis, respectively. The overall rate constant is defined as:

    \[ -\frac{d[\mathrm{LAS}]}{dt} =k_{\mathrm{obs}}[\mathrm{LAS}] \]

    4.2.1 Write an expression for \(k_{\mathrm{obs}}\) in terms of \(k_{\mathrm H}\), \(k_{\mathrm O}\), \(k_{\mathrm{OH}}\) and \([\ce{H+}]\).

    Hydrolysis of LAS was carried out at 298.15 K at various pH values (0.50 to 13.0). A very low initial concentration of LAS ensured that the pH did not change during the course of the reaction.

    The following graph shows the pH dependence of the hydrolysis of LAS.

    Figure 3

    4.2.2 Which of the following is/are correct?

    A) \(k_{\mathrm{obs}}\approx k_0\) at pH = 12

    B) \(k_{\mathrm{obs}}\approx k_0\) at pH = 5.0

    C) The rate of the reaction increases when the pH is changed from 0.50 to 1.0.

    D) The rate of the reaction increases when the pH is changed from 10 to 12.

    4.2.3 Using the diagram and the data given below, calculate \(k_{\mathrm H}\), \(k_0\) and \(k_{\mathrm{OH}}\). Make sure to specify the units.

    pH 1.300 5.300 12.180
    \(\log(k_{\mathrm{obs}}/\mathrm{min^{-1}})\) −3.886 −4.000 −1.726

    Acetylsalicylic acid, more commonly known as aspirin is a medicine often used for reducing fever, pain and inflammation. Like LAS, the hydrolysis of aspirin can also take different pathways depending on the pH. The pH rate profile of aspirin hydrolysis at 333.15 K is given below:

    Figure 4

    The following are possible reactions for the hydrolysis of aspirin. Depending on the pH, one or more of these reactions will predominate.

    Figure 5

    4.3.1 Using the pH-rate profile diagram and the reactions given above, state which of the following statements is/are correct. (\(\mathrm{p}K_{\mathrm a}\) of aspirin = 3.57 at 333.15 K)

    a) In the region C-D, reaction IV is predominant

    b) In the region C-D, reaction V is predominant

    c) In the region D-E reaction VI is predominant

    d) In the region A-B, reaction II is predominant

    The plot of \(k_{\mathrm{obs}}\) vs pH for the hydrolysis of aspirin has been confirmed to show a minimum at a particular pH. At 290.15 K the following rate constants for reactions I, II and III were determined:

    \(k_{\mathrm H}\) (reaction I) \(k_0\) (reaction II) \(k_{\mathrm{OH}}\) (reaction III)
    \(1.99\ \mathrm{dm^3\,mol^{-1}\,day^{-1}}\) \(2.29\times10^{-4}\ \mathrm{day^{-1}}\) \(3.18\times10^9\ \mathrm{dm^3\,mol^{-1}\,day^{-1}}\)

    The ionic product of water at 290.15 K can be taken as \(1.0\times10^{-14}\).

    4.3.2 Assuming that only reactions I, II and III occur, calculate the value of the pH at the minimum.

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  5. Problem 5: Egyptian blue and Chinese blue

    5500 years ago in ancient Egypt people learned for the first time how to synthesize a blue pigment. Now we know this pigment as Egyptian blue. About 2000 years later in ancient China another pigment was widely used, which is now referred to as Chinese blue. The two pigments are similar in structure, but have different elemental compositions.

    Figure 1

    Ushabti figurines from Egyptian pharaoh tomb covered with Egyptian blue
    and a Chinese blue soap dinspenser sold at Alibaba

    The ancient method of preparation for these pigments can easily by reproduced in a modern laboratory.

    When considering the amounts, assume that all of the compounds in this task are pure, and the yields are quantitative.

    To make Egyptian blue, one should heat 10.0 g of mineral A with 21.7 g of \(\ce{SiO2}\) and 9.05 g of mineral B at 800 – 900°C for a prolonged time. A volume of 16.7 \(\mathrm{dm^3}\) of a mixture of two gaseous products are released (the volume is measured at 850°C and \(1.013\times10^5\) Pa (1.013 bar) pressure. In result, 34.0 g of the pigment was obtained. No other products are formed. As the gas mixture is cooled, one component of the mixture condenses. As the remaining gas is further cooled to 0°C, the gaseous volume reduces to 3.04 \(\mathrm{dm^3}\).

    5.1.1 Calculate the mass of the gaseous mixture formed upon heating of A with B and \(\ce{SiO2}\).

    5.1.2 Determine the quantitative composition of this gas mixture.

    When 10.0 g of mineral A is heated with 21.7 g of \(\ce{SiO2}\) in the absence of B, it forms 8.34 \(\mathrm{dm^3}\) of gaseous products at a temperature of 850°C and pressure of \(1.013\cdot10^5\) Pa (= 1.013 bar). Mineral A contains only one metal.

    5.1.3 Calculate the molar mass and determine the formula of mineral B. Hint: it is an ionic solid insoluble in water and containing no water of crystallization.

    In order to obtain Chinese blue, one should take 17.8 g of mineral C instead of mineral B (keeping the amounts of mineral A and \(\ce{SiO2}\) same as for Egyptian blue), and run the reaction at higher temperatures. Besides the pigment, the same gaseous products in the same quantities are formed as in the preparation of Egyptian blue.

    5.1.4 Determine the formula of mineral C.

    5.1.5 Determine the formulae of Egyptian blue and Chinese blue.

    5.1.6 Determine the formula of mineral A.

    Elemental analysis of some samples of Chinese blue shows traces of sulfur. This led to a conclusion that those were synthesized using another common mineral instead of C.

    5.2.1 Suggest a formula for the mineral used in place of C.

    5.2.2 Could the temperature of synthesis of Chinese blue be decreased if this mineral is used instead of C?

    If during the synthesis of Chinese blue we take a smaller amount of silica than in the process above, we will obtain a purple pigment: Chinese violet. It was used, in particular, for coloring the famous Terracotta army soldiers.

    Figure 2

    Terracotta army from Xian, China and reconstruction of its original coloring.

    5.3 Write down the formula of a binary compound that forms under the conditions required for Chinese violet and is responsible for the change of the color.

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  6. Problem 6: Synthesis of galantamine

    Although there is currently no known cure for Alzheimer’s disease, there are medications available to manage the neurodegenerative disorder. Among these are acetylcholinesterase inhibitors, of which galantamine 1 is an example. This molecule can be isolated from the Caucasian snowdrop, a plant native to Georgia; however, the large amounts needed for therapy require a synthetic route. Shown below is the route used to prepare galantamine industrially.

    Figure 1

    Notes about the synthesis:

    • \(\ce{^1H}\) NMR of A indicates 2 aromatic protons in a para arrangement.
    • C is labile in aqueous conditions, so it is not isolated, but rather reacted immediately with \(\ce{NaBH4}\) to convert it to D.

    6.1.1 Suggest structures for A, B, C, D, F, and G. None of the reactions except for the final transformation with L-selectride are stereoselective. Therefore, stereochemistry does not need to be indicated in your answers.

    6.1.2 Give the formula for a possible reagent, X, to convert compound D to E.

    The optical rotation of the material obtained by resolution was \(-400^\circ\ \mathrm{cm^2\,g^{-1}}\), while that of the enantiomerically pure compound is \(-415^\circ\ \mathrm{cm^2\,g^{-1}}\) when measured under the same conditions. You may assume that the only optical impurity is the other enantiomer.

    One way of describing optical purity is enantiomeric excess (\(ee\)). It is defined as the difference in the percentages of the enantiomers in a mixture. For example in a mixture of 70% \(R\) and 30% \(S\), the \(ee\) is 40%.

    6.2.1 What is the enantiomeric excess of the resolved compound as prepared by the industrial route?

    L-selectride is a commercial reagent that performs the final reaction stereoselectively.

    6.2.2 Assign the labelled stereocentres \((\alpha,\beta,\gamma)\) in (−)-1 as \(R\) or \(S\).

    Figure 2

    6.2.3 Give the formula for a reagent that carries out the same reaction as L-selectride, converting H to 1. You need not worry about stereoselectivity.

    An alternative route to galantamine occurs with the seven-membered ring being the last ring to form.

    Figure 3

    6.3.1 Give the formula for compound Y to carry out the first step of the route.

    6.3.2 Suggest structures for J and K.

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  7. Problem 7: Synthesis of dolasetron mesylate

    This question looks at the synthesis of dolasetron mesylate, Z (shown right), a drug sold under the tradename Anzemet and used to treat post-operative nausea and vomiting.

    Figure 1

    The synthesis begins as shown below.

    Figure 2

    First cyclic compound A is made, which contains \(\ce{C}\), \(\ce{H}\), and \(\ce{O}\) only. Compound G is achiral and can be prepared directly from D using ozone under reductive conditions, or via stereoisomers E1 and E2 using \(\ce{OsO4}\), or via stereoisomers F1 and F2 using the peracid shown.

    7.1 Determine the empirical formula of G from the percentage masses given.

    7.2 Give the structures of A, B, C, D, E1, E2, F1, F2 and G.

    Compound G is used in the next stage of the synthesis, under buffered conditions, to form H (as a mixture of two achiral diastereoisomers). Reduction of H with \(\ce{NaBH4}\) gives alcohol I (as a mixture of four achiral diastereoisomers). I reacts with acidified dihydropyran to form J (as a mixture of even more diastereoisomers). J is then treated first with t-butoxide base, then refluxed with acid before finally extracting under weakly basic conditions to form K as a mix of two diastereomers, K1 (major product) and K2 (minor product). These could be separated, and K1 was used in the final stages of the synthesis.

    Figure 3

    7.3.1 Give the structures of H, I, and J. There is no need to show the different diastereoisomers formed.

    7.3.2 Give the structures of diastereoisomers K1, and K2.

    In the final stage of the synthesis, L and M react to form intermediate N. N then reacts with K1 to form, after extraction, the neutral amine which gives the target compound upon protonation with \(\ce{CH3SO3H}\).

    Figure 4

    7.4 Give the structure of N.

    Official English text of the problem.

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  8. Problem 8: Sugar derivatives from D-glucose

    An exotic, but biologically relevant sugar analogue can be prepared from D-glucose in the following manner. Heating a mixture of D-glucose and acetone with a few drops of concentrated acid results in the formation of a diacetonide A. Then A can be hydrolyzed selectively to B.

    Figure 1

    8.1.1 Which of the following sentences is true? (Tick the relevant boxes.)

    • ☐ A is an α isomer.
    • ☐ A is neither α nor β.
    • ☐ A is a β isomer.
    • ☐ A is a mixture of α and β isomers.

    Give your chosen answers to the questions on the answer sheet.

    8.1.2 Which of the following sentences is true?

    • ☐ We can get product A only if we use α-D-glucose as starting material.
    • ☐ We can get product A only if we use β-D-glucose as starting material.
    • ☐ We can get product A either from α- or from β-D glucose as starting material.

    8.1.3 Which one of these reagents can be utilized as X for the selective hydrolysis of A?

    • ☐ 50% acetic acid
    • ☐ concentrated \(\ce{H2SO4}\)
    • ☐ \(\ce{HCl}\) in water (\(6\ \mathrm{mol\,dm^{-3}}\))
    • ☐ \(\ce{NaOH}\) aqueous solution (\(1\ \mathrm{mol\,dm^{-3}}\))
    • ☐ \(\ce{HCl}\) in acetic acid (\(6\ \mathrm{mol\,dm^{-3}}\))

    8.1.4 Which is the stereochemically correct structure for compound B?

    Figure 2

    B is treated with sodium metaperiodate to get C. C is then reacted with an aqueous solution of \(\ce{NaCN}\), then heated with 10% \(\ce{NaOH}\) solution to get a mixture of two diastereomeric compounds D₁ and D₂. These compounds can be separated by column chromatography.

    Figure 3

    Reaction of D₁ with \(\ce{LiAlH4}\) followed by heating with 1M \(\ce{HCl}\) solution gives sugar F that is the hydrolysis product of the most abundant natural polysaccharide.

    Figure 4

    8.2.1 Draw the structures of C, D₁, D₂, E and F including stereochemical information. Show F as the more stable 6-membered ring containing isomer using the ring skeleton. Indicate with a wavy line if absolute chirality around a carbon is not known.

    8.2.2 The reaction sequence from glucose to F does not seem to be useful. In some cases, however, this is the most economical way to produce F. In which case?

    • ☐ \(\ce{^{13}C}\) labelling at carbon 6 of F
    • ☐ \(\ce{^{13}C}\) labelling at carbon 5 of F
    • ☐ \(\ce{^{13}C}\) labelling at carbon 1 of F
    • ☐ \(\ce{^{15}O}\) labelling at glycosidic OH of F
    • ☐ synthesis of an uncommon isomer of F

    Neutralization of D₂ with \(\ce{HCl}\) followed by heating in toluene results in dehydration and formation of G, which has a tricyclic structure in water-free solvents. Boiling G in \(\ce{HCl}\) solution (\(c = 1\ \mathrm{mol\,dm^{-3}}\)) gives H (\(\ce{C6H10O7}\)), which is a natural sugar derivative containing a 6 membered ring. H is a building block of heparin, an anticoagulant polysaccharide produced by our bodies.

    Figure 5

    8.3.1 Draw the structure of G including the stereochemistry.

    Draw H as the more stable 6-membered ring containing isomer using the ring skeleton. Indicate with a wavy line if absolute chirality around a carbon is not known.

    8.3.2 How are the rings of G fused together?

    • ☐ both junctions cis,
    • ☐ one cis and one trans junction,
    • ☐ both junctions trans.

    8.3.3 Which of the following is true for H? (You can choose more than one option.)

    • ☐ H is a reducing sugar (reacts with Fehling’s reagent)
    • ☐ H is an aldaric acid (dicarboxylic derivative of an aldose)
    • ☐ H is an aldonic acid (α-carboxyl derivative of an aldose)
    • ☐ H is a uronic acid (ω-carboxyl derivative of an aldose)
    • ☐ H is a ketose
    • ☐ H is a lactone
    • ☐ H is a D sugar
    • ☐ H is achiral
    • ☐ H is a meso compound

    Official English text of the problem.

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