International Chemistry Olympiad

IChO 2013

Moscow, Russia · 8 theory problems with complete solutions

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  1. Problem 1: Clathrate gun

    The only gun that is able to kill all living people in one shot

    On the floors of oceans and seas there are vast reserves of methane in the form of clathrate compounds called methane hydrates. These reserves can be mined and serve as a source of energy or raw materials for organic synthesis. However, scientists are seriously worried about the possibility of spontaneous decomposition of hydrates caused by the raising ocean temperature. It is believed that if a sufficient amount of methane is released into the atmosphere, the oceans will warm up quicker due to the greenhouse effect, further accelerating the decomposition of clathrates. Due to the explosion of the resulting methane-air mixture and/or changes in the composition of the atmosphere, all living creatures may become extinct. This apocalyptic scenario is called a clathrate gun.

    Figure 1

    Upon decomposition of \(1.00\ \mathrm{g}\) of a methane hydrate with a fixed composition at \(25\ ^\circ\mathrm{C}\) and atmospheric (\(101.3\ \mathrm{kPa}\)) pressure, \(205\ \mathrm{cm^3}\) of methane is released.

    1.1 Determine \(n\) (not necessarily integer) in the formula of methane hydrate, \(\ce{CH4}\cdot n\,\ce{H2O}\).

    Real methane hydrate has a non-stoichiometric composition close to \(\ce{CH4 * 6H2O}\). At atmospheric pressure, methane hydrate decomposes at \(-81\ ^\circ\mathrm{C}\). However, under high pressures (e.g. on the ocean floor) it is stable at much higher temperatures.

    Decomposition of methane hydrate produces gaseous methane and solid or liquid water depending on temperature.

    1.2 Write down the equation of decomposition of \(1\ \mathrm{mol}\) of \(\ce{CH4 * 6H2O}\) producing solid water (ice) \(\ce{H2O(s)}\).

    The enthalpy of this process equals \(17.47\ \mathrm{kJ\,mol^{-1}}\). Assume that the enthalpies do not depend on temperature and pressure, the volume change upon decomposition of hydrate is equal to the volume of released methane, and methane is an ideal gas.

    1.3 At what external pressure does decomposition of methane hydrate into methane and ice take place at \(-5\ ^\circ\mathrm{C}\)?

    1.4 What is the minimum possible depth of pure liquid water at which methane hydrates can be stable?

    To answer this question, you should first deduce at which minimum temperature methane hydrate can coexist with liquid water. Choose the correct answer.

    \[ \square\ 272.9\ \mathrm{K} \qquad \square\ 273.15\ \mathrm{K} \qquad \square\ 273.4\ \mathrm{K} \]

    Large methane hydrate stocks on the floor of Baikal lake, the largest freshwater lake in Russia and in the world, have been discovered in July 2009 by the crew of a deep-submergence vehicle «Mir-2». During the ascent from the depth of \(1400\ \mathrm{m}\) methane hydrate samples started to decompose at the depth of \(372\ \mathrm{m}\).

    1.5 Determine the temperature in Baikal lake at the depth of \(372\ \mathrm{m}\). The enthalpy of fusion of ice is equal to \(6.01\ \mathrm{kJ\,mol^{-1}}\).

    Total amount of methane in hydrates on the Earth is no less than \(5\cdot10^{11}\) tons.

    1.6 By how many degrees would the Earth atmosphere heat up, if such amount of methane is burned by reacting with atmospheric oxygen?

    The enthalpy of combustion of methane is \(-889\ \mathrm{kJ\,mol^{-1}}\), the total heat capacity of the Earth’s atmosphere is about \(4\cdot10^{21}\ \mathrm{J\,K^{-1}}\).

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  2. Problem 2: Break down photosynthesis – the Hill reaction

    In the history of photosynthesis research, there were some breakthrough experiments which added much to our knowledge of this very complex process. One of such experiments was performed in 1930s by an English biochemist Robert Hill. In this problem, we consider some of his data together with the data of more recent experiments.

    2.1 In plants, under illumination, carbon dioxide is reduced to carbohydrates (denote as \(\{\ce{CH2O}\}\)) and oxygen is produced. Write the overall equation of photosynthesis in plants.

    Much of the photosynthesis takes place in chloroplasts – organelles found in plant cells and containing chlorophyll – the light-absorbing substance. Hill isolated chloroplasts from the cells by grinding the leaves in the sucrose solutions. The cell-free chloroplasts did not produce oxygen under illumination even in the presence of \(\ce{CO2}\). However, upon adding potassium ferrioxalate \(\ce{K3[Fe(C2O4)3]}\) (with the excess of potassium oxalate) to the chloroplast suspension Hill observed oxygen liberation under illumination even without \(\ce{CO2}\).

    2.2 Hill’s experiment enabled to determine the source of oxygen during photosynthesis. Write the formulas of the oxidant and the reducing agent in the photosynthesis inside the plant cells and in the cell-free chloroplasts (the Hill reaction).

    Hill measured the amount of evolved oxygen using muscle haemoglobin (Hill denoted it \(\ce{Hb}\)) which binds all molecular oxygen in a 1 : 1 ratio to form \(\ce{HbO2}\). The initial concentration of \(\ce{Hb}\) was \(0.6\cdot10^{-4}\ \mathrm{mol\,dm^{-3}}\). Kinetic curves corresponding to different ferrioxalate concentrations are shown in the figure (the upper curve corresponds to \(2.0\cdot10^{-4}\ \mathrm{mol\,dm^{-3}}\)).

    Figure 1

    The fraction of bound haemoglobin \(\ce{HbO2}\) (with respect to the initial amount of \(\ce{Hb}\)) as a function of time. Crosses denote the end of the reaction

    (Figure 2a from the original Hill’s paper: R. Hill. Oxygen produced by isolated chloroplasts. – Proc. R. Soc. B, 1939, v. 127, pp. 192-210.)

    2.3 a. From the figure, estimate the \(\ce{Fe}\) / \(\ce{O2}\) mole ratio at the end of the reaction. Do not take into account the iron from \(\ce{Hb}\).

    b. Write the equation of Hill reaction assuming that it proceeds with a high yield.

    c. Using the table of standard electrode potentials, determine the Gibbs energy of the Hill reaction at \(T=298\ \mathrm{K}\), oxygen pressure \(1\ \mathrm{mm\,Hg}\), \(\mathrm{pH}=8\) and standard concentrations of other species. Is this reaction spontaneous at such conditions?

    Half-reaction \(E^\circ,\ \mathrm{V}\)
    \(\ce{O2 + 4H+ + 4e -> 2H2O}\) \(+1.23\)
    \(\ce{CO2 + 4H+ + 8e ->} \{\ce{CH2O}\} + \ce{H2O}\) \(-0.01\)
    \(\ce{Fe^3+ + e -> Fe^2+}\) \(+0.77\)
    \(\ce{Fe^3+ + 3e -> Fe^0}\) \(-0.04\)
    \(\ce{[Fe(C2O4)3]^3- + e -> [Fe(C2O4)3]^4-}\) \(+0.05\)
    \(\ce{[Fe(C2O4)3]^4- + 2e -> Fe + 3C2O4^2-}\) \(-0.59\)

    Now, the name “Hill reaction” denotes photochemical oxidation of water by any oxidant other than carbon dioxide which is sensitized by plant cells or isolated chloroplasts.

    In another experiment (1952), quinone in an acid solution was used as an oxidant in the Hill reaction initiated by light flashes in the Chlorella algae. Experimental data are shown in the figure. The volume of oxygen (in \(\mathrm{mm^3}\), at temperature \(10\,^\circ\mathrm{C}\) and pressure \(740\ \mathrm{mm\,Hg}\)) per one gram of chlorophyll per one flash was determined as a function of light intensity for natural photosynthesis and for isolated chloroplasts. It was found that the maximum yield of oxygen is the same for natural photosynthesis and the Hill reaction.

    Figure 2

    (Figure 1 from: H. Ehrmantraut, E. Rabinovitch. Kinetics of Hill reaction. – Archives of Biochemistry and Biophysics, 1952, v. 38, pp. 67-84)

    2.4 a. Determine the reaction order of a photochemical Hill reaction with respect to light intensity at low and high intensity. For each case choose one of the following three values: 0, 1 or 2.

    b. How many chlorophyll molecules participate in the formation of one oxygen molecule in the saturation limit of the Hill reaction? (The molecular mass of chlorophyll is about 900 Da).

    The quantum requirement of the light redox reactions is defined as the average number of light photons (not necessarily integer) needed for the transfer of one electron from a reducing agent to an oxidant. The isolated chloroplasts were irradiated during 2 hours by a monochromatic light (wavelength \(672\ \mathrm{nm}\)) with the energy input \(0.503\ \mathrm{mJ\,s^{-1}}\), and the total volume of oxygen formed was \(47.6\ \mathrm{mm^3}\) (under the same conditions as in question 4).

    2.5 Calculate the quantum requirement for the Hill reaction.

    2.6 Try to make conclusions from the above experiments (questions 2.2 – 2.5). Tick the relevant boxes.

    Yes No
    In natural photosynthesis, water oxidation and \(\ce{CO2}\) reduction are separated in space.
    In chloroplasts, \(\ce{O2}\) is produced from \(\ce{CO2}\).
    Oxidation of water in chloroplasts requires light illumination.
    Most of chlorophylls in chloroplasts participate directly in the photochemical \(\ce{O2}\) production.
    In isolated chloroplasts, every absorbed photon causes transfer of one electron.

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  3. Problem 3: Meerwein-Schmidt-Ponndorf-Verley reaction

    In the table for 3.3, A–F are editorial labels identifying the six structures in reading order.

    PROBLEM 3

    Meerwein-Schmidt-Ponndorf-Verley reaction

    Meerwein-Schmidt-Ponndorf-Verley (MSPV) reaction is a useful tool for reduction of carbonyl compounds to alcohols. The reaction is the reduction of carbonyl compounds by low molecular weight alcohols in the presence of alkoxides of aluminium or other metals:

    Figure 1

    The mechanism of the reaction includes coordination of carbonyl compound by aluminium alkoxide, hydride transfer in the inner sphere of the complex and subsequent transalkoxylation. It can be schematically represented as follows (transalkoxylation is shown as a one-step process for brevity):

    Figure 2

    The reaction is reversible and shifting the equilibrium to the desired product requires some excess of the reductant. In some cases (e. g. in the case of reduction of aromatic aldehydes and ketones) the equilibrium constant is so large that the reverse reaction can be neglected.

    The table below contains standard entropies and standard enthalpies of formation of liquid substances at 298 K. The boiling points of the substances at 1 bar are also given.

    Substance \(\Delta_{\mathrm f}H^\circ_{298}\), kJ mol\(^{-1}\) \(S^\circ_{298}\), J mol\(^{-1}\) K\(^{-1}\) \(t_{\mathrm{vap}}\), °C
    Acetone −248.4 200.4 56
    Isopropanol −318.1 180.6 82
    Cyclohexanone −271.2 229.0 156
    Cyclohexanol −348.2 203.4 161

    3.1 a Calculate the minimum isopropanol to cyclohexanone mass ratio which is required to reach a 99 % yield of reaction at 298 K. Assume that:

    i) the reaction mixture eventually gets at equilibrium,

    ii) no products are initially present.

    3.1 b Choose the appropriate way(s) to increase the cyclohexanol yield. Tick the relevant boxes.

    Increase the temperature up to 50 °C using a reflux
    Increase the temperature up to 60 °C, evaporating (distilling) the acetone
    Add some ethanol to the reaction mixture
    Add some ethanal to the reaction mixture

    3.2 Often the rate-limiting step in the MSPV reaction is the hydride transfer or the alcoholysis of the alkoxide after hydride transfer. For these two cases, using the above mechanism (2), derive an expression for the rate of reaction as a function of current concentrations of a carbonyl compound, isopropanol and a catalyst. In both cases determine the rate orders in the reactants and the catalyst. Assume that all reaction steps before the limiting step are fast and reversible. Use equilibrium approximation, if necessary. For brevity use the following notation: A for carbonyl compound, B for isopropanol, C for catalyst. Denote intermediates as you wish.

    MSPV reaction can be used to obtain chiral alcohols, if the chiral catalyst is employed. For instance, Campbell et al. used the catalyst based on the chiral 2,2’-dihydroxy-1,1’-binaphtyl (BINOL), which is synthesized in situ from binaphtol and trimethylaluminium:

    Figure 3

    The chirality of BINOL is due to the sterically hindered rotation around the \(\ce{C-C}\) bond. Though perfectly stable at room temperature, BINOL may racemize when heated.

    3.3 Which of the phenols below can form stable (at room temperature) enantiomers so that they can be used in the same fashion to produce a chiral catalyst? Tick the relevant boxes.

    Substance Can be used Substance Can be used
    A B
    C D
    E F

    A:

    Figure 4

    B:

    Figure 5

    C:

    Figure 6

    D:

    Figure 7

    E:

    Figure 8

    F:

    Figure 9

    Enantiomeric excess, \(ee\), is used to characterize the enantiomeric purity of the substance. This quantity equals to the ratio of the difference of concentrations of enantiomers \(R\) and \(S\) to their sum:

    \[ ee=\frac{[R]-[S]}{[R]+[S]} \]

    Enantiomeric excess of the pure \(R\) isomer is unity, \(ee\) of the racemic mixture is zero.

    3.4 When using the enantiomerically pure \(\ce{(BINOL)Al(OiPr)}\) as a catalyst for reduction of \(\alpha\)-bromoacetophenone, the \(ee\) of the product equals 81 %. What is the \(ee\) of the product if the catalyst \(ee\) equals 50 %? Provide your calculation with an illustration or derivation of the final formula.

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  4. Problem 4: A simple inorganic experiment

    Compound A which contains metal X is a colourless crystalline solid and highly soluble in water. It is used as a reagent in analysis and gives in alkali media a binary compound B containing 6.9 % (mass) of oxygen. Under heating A decomposes with a mass loss of 36.5 %.

    4.1 Determine the metal X and compounds A and B.

    Upon adding some amount of sodium thiosulfate to the solution of A the colour immediately becomes red, then changes to reddish-brown, and after some minutes a dark-brown precipitate C forms (reaction 1). The solution over it is colourless. Being heated on air at \(600\,^\circ\mathrm{C}\), C gives a grey powder X (reaction 2), so as 0.90 g of residue can be obtained from 1.10 g of C. A gas evolved by heating C in vacuum (reaction 3) can be absorbed by calcium hydroxide suspension (reaction 4). Being stored for a long time under saturated solution of barium perchlorate in \(\ce{HClO4}\) (\(c=0.1\ \mathrm{mol\,dm^{-3}}\)), the colour of the precipitate becomes lighter, while the use of magnesium perchlorate does not give such an effect.

    4.2 What is C? Write the equations of the reactions 1 – 4.

    The compound C being stored under the mother liquor (containing an excess of A) changes its colour to yellow due to the transformation into D. If barium ions are added to the suspension of C in the mother liquor, a mixture of D and of a white precipitate is formed.

    4.3 Propose the formula of D, taking into account that it contains 77.5 mass % of X. Give the equation of D formation.

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  5. Problem 5: Simple estimates of graphene properties

    Graphene is a two-dimensional, one atom thick carbon material (Fig. 1 a). Many layers of graphene stack together to form graphite (Fig. 1 b).

    Figure 1

    Fig. 1. (a) The structure of graphene. Spheres are carbon atoms. They are arranged in hexagons. The area of one carbon hexagon is \(5.16\cdot10^{-20}\ \mathrm{m^2}\). (b) Crystal lattice of graphite. Three graphene layers are shown.

    Such atomic structure was long considered to be unstable. However, in 2004 Andrey Geim and Konstantin Novoselov have reported production of the first samples of this unusual material. This groundbreaking invention was awarded by Nobel prize in 2010.

    Experimental studies of graphene are still restricted. Production of massive portions of the new substance still is a challenging synthetic problem. Many properties of graphene were estimated. Usually, there is not enough information for rigorous calculations, so we have to make assumptions and neglect unimportant factors. In this problem, you will estimate the adsorption properties of graphene.

    5.1 Estimate the specific surface of graphene open for adsorption in units \(\mathrm{m^2\,g^{-1}}\). Consider that graphene plane is separated from any other solid or liquid substance.

    The single layer of nitrogen molecules adsorbed on the outer surface of graphite is shown in Fig. 2. Assume that the same arrangement of nitrogen molecules is formed on a graphene surface.

    Figure 2

    Fig. 2. Nitrogen molecules \(\ce{N2}\) (grey circles) on the outer surface of graphite

    5.2 How many grams of nitrogen can be adsorbed on 1 gram of graphene assuming that the graphene layer is placed onto the surface of a solid support? Estimate the volume occupied by these nitrogen molecules after the complete desorption from 1 g of graphene (pressure 1 bar, temperature 298 K).

    Let us consider adsorption as a common chemical equilibrium

    Figure 3

    (\(\ce{A_{gas}}\) are molecules \(\ce{A}\) in the gaseous state, \(\ce{A_{ads}}\) are the same molecules on the surface)

    with the equilibrium constant \(K\):

    \[ K=\frac{n_{\ce{A_{ads}}}\;(\mathrm{mol\,m^{-2}})} {p_{\ce{A_{gas}}}\;(\mathrm{bar})} \]

    (Such an assumption holds if a small number of molecules is adsorbed on the surface.)

    Adsorption properties of graphene can be estimated from the data for adsorption on a regular three-dimensional graphite. The enthalpy of adsorption (\(\Delta H^\circ\) of reaction (1)) of any molecule \(\ce{A}\) on graphene is on average by 10 % less negative compared to that on graphite. On graphite, the adsorbed molecule is bound more strongly due to the interaction with the lower graphene layers in the lattice (Fig. 1 b) and hence the enthalpy of adsorption is more negative. The standard entropies of adsorption on graphene and graphite are assumed to be the same.

    5.3 How many moles, \(n\), of \(\ce{CCl4}\) are adsorbed on 1 g of graphene at \(p(\ce{CCl4})=10^{-4}\ \mathrm{bar}\) if \(2.0\cdot10^{-7}\ \mathrm{mol}\) of \(\ce{CCl4}\) are adsorbed on \(1\ \mathrm{m^2}\) of graphite at \(p(\ce{CCl4})=6.6\cdot10^{-5}\ \mathrm{bar}\)? Assume that graphene is placed onto the surface of a solid support and the interaction of \(\ce{CCl4}\) with the support does not change the enthalpy of adsorption of \(\ce{CCl4}\) on graphene. The temperature in both cases is 293 K. \(\Delta H^\circ\) of adsorption of \(\ce{CCl4}\) on graphite is \(-35.1\ \mathrm{kJ/mol}\).

    The graphene films are expected to be sensitive gas detectors. If \(10^9\) particles of a gas are adsorbed on \(1\ \mathrm{cm^2}\) of a graphene surface this is enough to measure an electrical resistivity change of the graphene layer and to detect the presence of a gas in the environment.

    5.4 Determine the minimal content of ethane, \(\ce{C2H6}\), in the air (in mol. %) at atmospheric pressure (\(T=293\ \mathrm{K}\)) at which a graphene sensor will detect this gas. The known data for the adsorption of alkanes on graphite are shown in Fig 3. Assume that air doesn't affect the adsorption properties of ethane.

    Figure 4
    Figure 5

    Fig. 3
    Thermodynamic properties for adsorption of alkanes on a graphite surface.
    (a) \(\ln K\ [\mathrm{mol\,m^{-2}\,bar^{-1}}]\) as a function of \(\ln M\) (\(M\) - molar mass of alkane in \(\mathrm{g\,mol^{-1}}\));
    (b) \(\Delta H^\circ\) of adsorption as a function of \(\ln M\).
    Linear dependences are assumed in both cases

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  6. Problem 6: Cyclopropanes. So simple. So fancy…

    Cyclopropanes bearing donor and acceptor substituents at the neighbouring C-atoms, for example, A, demonstrate high reactivity behaving similar to 1,3-zwitterion B.

    Figure 1

    Thus, A1 (\(\ce{X = 4-OMe}\)) undergoes the three-membered ring opening in the Lewis acid-catalyzed reaction with 1,3-dimethoxybenzene as a nucleophile giving the product C.

    6.1 Write down structural formula of C.

    A1 participates in cycloadditions, annulations, oligomerizations, and other processes. Thus, [3+2]-cycloaddition between A1 and 4-methoxybenzaldehyde leads to a five-membered ring in D. Complete decarboxylation of D produces E (\(\ce{C18H20O3}\)), the molecule of the latter having a plane of symmetry.

    6.2 Write down structural formulae of D and E indicating their stereochemistry.

    Also, A, can undergo various transformations in the absence of any reaction partners except catalysts. Some transformations typical of A1 are shown in the Scheme below.

    Figure 2

    To determine the structures of F – J, a set of physico-chemical data was obtained (see Table for some results). It was found that:

    a) F and G are structural isomers of A1;

    b) G is formed as the most stable stereoisomer;

    c) H and I are structural isomers;

    d) H is formed as a single diastereomer with \(C_2\) axis of symmetry (the molecule looks the same after rotation through the angle of 180°);

    e) I is formed as a mixture of two diastereomers;

    f) J is naphthalene derivative.

    In the process leading to I, one molecule of A1 demonstrates the described above common reactivity (analogous to that of B). The other molecule of A1 behaves differently. Also, the latter behaviour is demonstrated by cyclopropane A2 (dimethyl 2-(3,4,5-trimethoxyphenyl)cylopropane-1,1-dicarboxylate) when treated with \(\ce{SnCl4}\) in \(\ce{CH3NO2}\) affording K as a mixture of two diastereomers. The major isomer has the centre of symmetry. Similar reactivity is shown by A2 in \(\ce{Sn(OTf)2}\)-catalyzed reaction with G furnishing L.

    6.3 Write down the structural formulae of F – J, and those of L and the major isomer of K.

    Ratio of the number of hydrogen-containing groups

    Non-aromatic \(\ce{CH}\) Non-aromatic \(\ce{CH2}\) Non-aromatic \(\ce{CH3}\) Non-aromatic \(\ce{OH}\) Aromatic Composition
    F 1 1 1+1+1 0 4 in total C 63.62%, H 6.11%
    G 1+1+1 0 2+1 0 4 in total C 63.62%, H 6.11%
    H 1 1 1+1+1 0 4 in total C 63.62%, H 6.11%
    I 1+1+1 1+1 2+1+1+1+1 0 7 in total C 63.62%, H 6.11%
    J 0 0 1+1 1 5 in total C 67.22%, H 5.22%
    K 1+1 1 2+1+1+1 0 1 C 59.24%, H 6.23%
    L 1+1+1+1+1 1 2+2+1+1+1+1 0 5 in total C 61.21%, H 6.18%

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  7. Problem 7: Diverse permanganatometry

    The amount of many reducing agents can be determined by permanganatometric titration in alkaline medium allowing permanganate ion reduction to manganate.

    7.1 Write down the ionic reaction equation for titration of formate with permanganate in an aqueous \(\ce{NaOH}\) solution with a concentration of ~0.5 mol dm\(^{-3}\).

    Titration with permanganate in alkaline medium is often supplemented by addition of a barium salt, which leads to precipitation of manganate as \(\ce{BaMnO4}\).

    7.2 Which side redox processes involving manganate is suppressed by the barium salt? Write down an example of equation of the corresponding reaction.

    A volume of 10.00 cm\(^3\) (\(V_{\mathrm{Mn}}\)) of \(\ce{KMnO4}\) solution with a concentration of 0.0400 mol dm\(^{-3}\) (\(c_{\mathrm{Mn}}\)) was placed in each of flasks A, B, and C and different reactions were conducted in each flask.

    A sample solution containing crotonic acid (CA) \(\ce{CH3-CH=CH-COOH}\), an alkali and barium nitrate (both in an excess) were added to flask A, and the reaction mixture was incubated for 45 min. It is known that crotonic acid loses 10 electrons under the experiment conditions.

    7.3 a Write down the total ionic equation for the reaction.

    A volume of 8.00 cm\(^3\) (\(V_{\mathrm{CN}}\)) of potassium cyanide solution (\(c_{\mathrm{CN}}=0.0100\) mol dm\(^{-3}\)) was then added to the incubated mixture. This resulted in completion of the following reaction:

    Figure 1

    \(\ce{BaMnO4}\) precipitate was then filtered off, and the excess of cyanide in the filtrate was titrated with \(\ce{AgNO3}\) solution (\(c_{\mathrm{Ag}}=0.0050\) mol dm\(^{-3}\)) till detectable precipitation was observed. Note that both \(\ce{CN^-}\) and \(\ce{CNO^-}\) are analogs of halide ions, but \(\ce{CNO^-}\) affords soluble silver salt.

    7.3 b Give the formula for the complex formed when \(\ce{Ag+}\) ions were initially added to the cyanide solution (until the precipitate was formed).

    7.3 c Give the formula of the precipitate formed.

    7.3 d Derive the formula for calculating the amount of substance of crotonic acid in the sample solution. Calculate the mass of crotonic acid (in mg) if 5.40 cm\(^3\) (\(V_{\mathrm{Ag}}\)) of the silver salt solution was consumed for the titration to the endpoint.

    Another sample of crotonic acid and alkali (in an excess) were added to flask B, this mixture lacking barium salt. An excess of \(\ce{KI}\) (instead of cyanide) was added as a reducing agent. The mixture was then acidified and the iodine evolved was titrated with a thiosulfate solution (\(c_{\mathrm{S}}=0.1000\) mol dm\(^{-3}\)). 4.90 cm\(^3\) (\(V_{\mathrm{S1}}\)) of the titrant was used to reach the endpoint.

    7.4 Derive the formula for calculating the amount of substance of crotonic acid in this experiment. Calculate the mass of crotonic acid (in mg).

    A sample containing tin(II) was added to flask C, and the medium was adjusted to weak alkaline. Tin(II) was quantitatively oxidized to \(\ce{Sn(OH)6^2-}\), whereas a precipitate formed as a result of permanganate reduction. The precipitate was isolated, washed off, dried at 250 °C, weighed (the mass of the water-free precipitate (\(m_{\mathrm{prec}}\)), representing a binary compound, was of 28.6 mg), and dissolved in \(\ce{H2SO4}\) in the presence of an excess of potassium iodide. The evolved iodine was titrated with 0.1000 M thiosulfate solution. 2.5 cm\(^3\) (\(V_{\mathrm{S2}}\)) of the latter was consumed to attain the endpoint.

    7.5 a Write down the reaction of precipitation. Confirm it with calculations.

    7.5 b Calculate the mass of tin in the sample (in mg) referred to the metal.

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  8. Problem 8: Unique life of archaea

    Archaea (or archaebacteria) are single-celled microorganisms. They significantly differ from bacteria and eukaryotes at the molecular level.

    Enzymatic reaction of methylamine with water is the major energy source for some archaea. In a particular experiment, an archaea strain was cultivated at pH 7 under anaerobic (oxygen free) conditions with the nutrient medium containing \(\ce{^{13}CH3NH2}\) as the only energy source. After a certain incubation period, the gas over the archaea culture was sampled and analyzed. It was found that the gas contains two substances, A and B, in a molar ratio of 1 : 3 correspondingly (the sample density relative to \(\ce{H2}\) is 12.0).

    8.1 Calculate the volume fractions (in %) of A and B in the mixture.

    8.2 Determine A and B if there are no nitrogen atoms in the gas collected.

    8.3 Write down the equation of enzymatic reaction of methylamine with water described in the above experiment using predominant form of each species.

    Enzymes containing the residue of α-amino acid X, are found in many archaea capable of methylamine utilization. It is known that X:

    • is composed of atoms of 4 elements;
    • contains 18.8 mass % of oxygen;
    • possesses the single individual tRNA and is incorporated into proteins of archaea during translation.

    Amino acid L-lysine (see the structure in scheme below) was identified as the X precursor in archaea. All carbon and nitrogen atoms found in X originate from two starting lysine molecules. Different isotope-labelled L-lysines were introduced into a model system to clarify the biosynthetic pathways of X. The results are summarized in the table.

    Isotope composition of L-lysine Molar mass (rounded to integer) of the X residue \(\ce{[RCH(NH2)CO]}\), bound to tRNA, \((\mathrm{g\,mol^{-1}})\)
    Normal 238
    All carbons \(\ce{^{13}C}\), all nitrogens \(\ce{^{15}N}\) 253
    ε-amino group with \(\ce{^{15}N}\) 239

    8.4 Determine the molecular formula of X.

    X is biosynthesized in archaea according to the following scheme (\(E_1\), \(E_2\), and \(E_3\) – enzymes):

    Figure 1

    At the first step, lysine is transformed into its structural isomer (α-amino acid, C), whereas D contains a peptide bond, and E a formyl group \(\ce{-CHO}\). All reaction coefficients in the above scheme are equal to 1.

    8.5 i) Give the chemical formula of C, D and E.

    ii) From the reaction types given hereunder, choose (tick) only one corresponding to the enzyme \(E_3\) catalyzed reaction.

    • ☐ Oxidative deamination
    • ☐ Decarboxylation
    • ☐ Intermolecular deamination
    • ☐ Hydroxylation
    • ☐ Peptide bond hydrolysis

    X contains the following fragment:

    Figure 2

    R is a massive substituent (\(M>100\ \mathrm{g\,mol^{-1}}\)). The 3rd carbon atom is non-asymmetric. The 4th and 5th carbon atoms are stereogenic centres. All carbon atoms in the cycle are bound with at least one hydrogen atom. Each substituent (H, Me and R) is found only once.

    8.6 Determine the positions of substituents H, Me, and R.

    8.7 Draw structural formulae of C and X with stereochemical details. Mark every stereocenter of X with either \(R\) or \(S\).

    Only one codon is responsible for the incorporation of X residues into proteins in archaea. The nitrogen bases forming this codon contain two exocyclic amino groups and three exocyclic oxygen atoms in total.

    Figure 3

    8.8 Determine the nucleotide composition of the codon by filling in the hereunder table. Write down the number of each nitrogen base in the codon encoding X. Tick only one box in each line.

    The number of bases in the codon

    Nitrogen base 1 2 3 0 or 1 1 or 2
    A
    C
    G
    U

    The fragment of mRNA coding sequence given below contains the codons encoding X residue incorporation into an archaea enzyme:

    \(5'\ldots\mathrm{AAUAGAAUUAGCGGAACAGAGGGUGAC}\ldots3'\)

    8.9 a Using the table of the genetic code, decide how many amino acid residues are incorporated into the enzyme chain due to this fragment translation.

    8.9 b Write down the amino acid sequence translated from this fragment. Note that more than one X residue is found in the fragment.

    Fill in the boxes below with the amino acid abbreviations (from N- to C-terminus).

    Note that the number of boxes is excessive. If there is more than one possibility, write all separated by “/”. If the translation is stopped in a particular position, write “STOP” and leave all the boxes to the right empty.

    Figure 4

    (a) RNA codons for the twenty amino acids

    Second base

    First base U C A G Third base
    U Phe Ser Tyr Cys U
    U Phe Ser Tyr Cys C
    U Leu Ser STOP STOP A
    U Leu Ser STOP Trp G
    C Leu Pro His Arg U
    C Leu Pro His Arg C
    C Leu Pro Gln Arg A
    C Leu Pro Gln Arg G
    A Ile Thr Asn Ser U
    A Ile Thr Asn Ser C
    A Ile Thr Lys Arg A
    A Met (start) Thr Lys Arg G
    G Val Ala Asp Gly U
    G Val Ala Asp Gly C
    G Val Ala Glu Gly A
    G Val Ala Glu Gly G
    Amino Acids Abbreviations:
    Ala = Alanine
    Arg = Arginine
    Asp = Aspartic acid
    Asn = Aaparagine
    Cys = Cysteine
    Glu = Glutamic Acid
    Gln = Glutamine
    Gly = Glycine
    His = Histidine
    Ile = Isoleucine
    Leu = Leucine
    Lys = Lysine
    Met = Methionine
    Phe = Phenylalanine
    Pro = Proline
    Ser = Serine
    Thr = Threonine
    Trp = Tryptophan
    Tyr = Tyrosine
    Val = Valine

    Official English text of the problem.

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