International Chemistry Olympiad

IChO 2020

Istanbul, Turkey (held remotely) · 9 theory problems with complete solutions

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  1. Problem 1: Two Beauties of Turkey: the Van Cat and the Ankara Cat

    Figure 1

    The most beautiful of cats, the Van cat is a pure breed living only in Lake Van basin. Another endemic cat breed is the Ankara cat. They are called Angora cats. Their most important feature is their two different eye colors.

    Figure 2

    Van cat

    Figure 3

    Ankara cat

    Figure 4

    Nepeta cataria (catnip)

    Just like people, cats can sometimes be stressed and angry. Just as people are made happy by melatonin, the stress of cats can be reduced and they can be made happy thanks to a natural product. Nepetalactone is an organic compound isolated from the plant catnip (Nepeta cataria), which acts as a cat attractant. Nepetalactone is a ten-carbon bicyclic mono- terpenoid compound derived from isoprene with two fused rings: a cyclopentane and a lactone.

    Figure 5

    Cat eCat eating catnip in the garden

    Figure 6

    Cat's dream

    Total synthesis of nepetalactone:

    Figure 7

    1.1 The above scheme describes the total synthesis of nepetalactone. Draw structures of A–G, without stereochemical details.

    Hints:

    • Compound A has strong and sharp band at \(3300\ \mathrm{cm^{-1}}\) in the IR spectrum.
    • A, B, and F are monocyclic, while C, D, E, and G are bicyclic compounds.
    • F has one doublet at ~ \(9.8\ \mathrm{ppm}\) in the \({}^{1}\mathrm{H}\)-NMR spectrum.

    Reactions of nepetalactone:

    Figure 8

    The above scheme includes a few reactions of one of the enantiopure nepetalactone 1 isomers. Three of the reaction products (5, 6, and J) are used as insect repellents in industry.

    1.2 For the relationship between 5 and 6, which of the following is/are true? Tick the box next to the correct answer(s) on your answer sheets.

    ☐ Enantiomers
    ☐ Diastereomers
    ☐ Identical
    ☐ Stereoisomers

    Reaction of 1 with DDQ gives highly conjugated compound H. Also, thermal reaction of compound H with p-quinone gives I with molar mass of \(226.28\ \mathrm{g/mol}\).

    1.3 Draw the structures of H, I, and J indicating stereochemistry.

    Hints:

    • During the formation of I, sequential pericyclic reactions and an oxidation reaction (due to the presence of \(\ce{O2}\)) take place, and a well-known gas forms during the reaction.
    • J has a strong and very broad band between 3300 and \(2500\ \mathrm{cm^{-1}}\) in the IR spectrum.

    Official English text of the problem.

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  2. Problem 2: A Tale of a Reactive Intermediate

    Arynes constitute a special class of reactive intermediates. The first experimental evidence for the structure of an aryne (benzyne) was demonstrated in 1953 via the elegant labeling experiments by John D. Roberts and coworkers.

    In one such experiment, chlorobenzene, whose carbon at position 1 was labeled with radioactive \(\ce{^{14}C}\), was reacted with \(\ce{KNH2}\) in liquid \(\ce{NH3}\) to give nearly equal amounts of isotopic isomers A and B along with the inorganic salt C. This reaction proceeds via the formation of aryne intermediate D.

    Figure 1

    2.1 Draw the structure of A, B and D, and provide the formula of C. Indicate the position(s) of \(\ce{^{14}C}\)-labeled carbon(s) with an asterisk (*) whenever applicable.”

    Analysis of the \(\ce{^{14}C}\)-labeled product(s) was achieved via degradation experiments (the \(\ce{^{14}C}\)- labeled carbons are not shown on the structures). Radioactivities of the intermediates and final products were examined.

    Figure 2

    2.2 Tick the appropriate boxes on the answer sheet for the intermediates and products that you expect to exhibit radioactivity.

    Considering only A: Considering only B:
    ☐ Compound 1 ☐ Compound 1
    ☐ \(\ce{BaCO3}\) (Batch 1) ☐ \(\ce{BaCO3}\) (Batch 1)
    ☐ Compound 2 ☐ Compound 2
    ☐ \(\ce{BaCO3}\) (Batch 2) ☐ \(\ce{BaCO3}\) (Batch 2)

    With the aim of facilitating aryne formation, Kobayashi and co-workers developed a fluoride-induced aryne generation protocol. Using this method, benzene derivative 3 is reacted with furan (4) in the presence of \(\ce{CsF}\), resulting in the formation of E, F, and G.

    Figure 3
    • Combustion analysis of E revealed the following atom content: 75.8 % carbon, 5.8 % hydrogen, and 18.4% oxygen.
    • E does not have a proton that is exchangeable with \(\ce{D2O}\) in \(\ce{^1H}\)-NMR spectroscopy.
    • F is an ionic compound.

    2.3 Determine the structures of E, F, and G (without stereochemical details).

    In the absence of a nucleophile or a trapping agent, arynes can undergo [2+2]-type cyclodimerization or [2+2+2]-type cyclotrimerization reactions under suitable conditions. The aryne derivative that is obtained when 3 is treated with one equivalent of \(\ce{CsF}\) in MeCN can give, in principle, four different dimerization and trimerization products (H – K).

    • H has two planes of symmetry.
    • I is expected to exhibit 21 signals in its \(\ce{^{13}C}\)-NMR spectrum.
    • I and J both exhibit an \(m/z\) value of 318.1 in their mass spectra.

    2.4 Determine the structures of H – K.

    When 5 is reacted with \(\beta\)-ketoester 6 in the presence of 2 equivalents of \(\ce{CsF}\) at 80 °C, L is obtained as the major product. The \(\ce{^1H}\)-NMR and \(\ce{^{13}C}\)-NMR data for L, in \(\ce{CDCl3}\), are as follows:

    • \(\ce{^1H}\)-NMR: \(\delta\) 7.79 (dd, \(J=7.6,\ 1.5\ \mathrm{Hz}\), 1H), 7.47–7.33 (m, 2H), 7.25–7.20 (m, 1H), 3.91 (s, 2H), 3.66 (s, 3H), 2.56 (s, 3H) ppm.
    • \(\ce{^{13}C}\)-NMR: \(\delta\) 201.3, 172.0, 137.1, 134.4, 132.8, 132.1, 130.1, 127.5, 51.9, 40.2, 28.8 ppm.
    Figure 4

    2.5 Determine the structure of L.

    2.6 In the reaction shown in task 2.5, which of the following statement(s) describe(s) the function of \(\ce{CsF}\)?

    • The \(pK_\mathrm{a}\) values of \(\ce{HF}\) and \(\beta\)-ketoester 6 in dimethyl sulfoxide (DMSO) are about 15 and 14, respectively.

    ☐ \(\ce{F-}\) hydrolyzes the trifluoromethanesulfonate (\(\ce{O3SCF3}\)) group of 5

    ☐ \(\ce{F-}\) attacks the \(\ce{-SiMe3}\) group of 5.

    ☐ \(\ce{F-}\) acts as a base to deprotonate 6.

    ☐ \(\ce{F-}\) acts as a nucleophile and attacks the ester group of 6.

    Diazapyrone derivative 8 was shown to be a useful reactant for the construction of a variety of cyclic frameworks. Its preparation from phenylglyoxylic acid (7) and its use in two different reactions are described below.

    • Q and T are gases under ambient conditions.
    • O and P are constitutional isomers.
    • Q does not have any signals in its IR spectrum.
    • Heating 1 mol of R at 85 °C generates 1 mol of reactive intermediate S.
    • Reaction of 8 with two equivalents of S gives U, Q, and T.
    Figure 5

    Note:

    equiv = equivalent

    cat = catalyst

    2.7 Determine the structures of M – U.

    Official English text of the problem.

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  3. Problem 3: (±)-Coerulescine

    A spiro compound is typically an organic compound containing rings linked together by one common atom (spiroatom) as carbon atom with bold in figure below. The spiro[pyrrolidin-3,3′- oxindole] ring system is a structural framework incorporated in several cytostatic alkaloids and unnatural compounds. Coerulescine (1) and horsfiline are the simplest prototype members of this subfamily that can be synthesized by the route shown below.

    Figure 1

    Claisen rearrangement, which is a [3,3]-sigmatropic rearrangement, is a powerful carbon– carbon bond-forming reaction in which an allyl vinyl ether is converted thermally to an unsaturated carbonyl compound as shown in the Scheme below. When compound A is heated, it undergoes Claisen rearrangement to give carbonyl compound B.

    For this entire task, your answers can be given without any stereochemical details.

    Figure 2
    Figure 3

    3.1 Draw the structures of A and B.

    • A is an inseparable mixture of cis/trans isomers.
    • B has IR absorption at \(1726\ \mathrm{cm^{-1}}\).

    3.2 Draw structures for C, D, E, and F.

    • D–F have a bicyclic structure.

    3.3 Choose the correct order of steps for the transformation of F to G.

    ☐ Imine formation, then reduction, then amidation

    ☐ Amidation, then imine formation, then reduction

    ☐ Reduction, then amidation, then imine formation

    3.4 Draw the structures for G and H (both spiro compounds).

    3.5 Draw the structure of the intermediate produced by treatment with n-BuLi in the step H → coerulescine (1).

    Coerulescine (1), on treatment with N-bromosuccinimide (NBS), gives the bromo derivative, which upon heating with sodium methoxide in the presence of cuprous iodide gives horsfiline (I) in 60% yield.

    3.6 Choose the correct structure for compound I consistent with the following selected \(\ce{^1H}\)-NMR data: \(\delta\) 7.05 (d, \(J=1.4\ \mathrm{Hz}\), 1H), 6.78 (d, \(J=8.0\ \mathrm{Hz}\), 1H), 6.72 (dd, \(J=8.0,\ 1.4\ \mathrm{Hz}\), 1H) ppm. Tick the relevant box.

    Figure 4

    3.7 When the allyl ether of 2-naphthol is heated a sigmatropic rearrangement is initiated. Write the structure of major product isolated from this reaction.

    Figure 5

    Official English text of the problem.

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  4. Problem 4: Symmetry Does Matter!

    There are numerous reactions in organic chemistry that proceed through cyclic transition states and these are classified as pericyclic reactions. Woodward–Hoffmann rules, developed by Robert B. Woodward and Roald Hoffmann, are used to rationalize stereochemical aspects and the activation energy of pericyclic reactions.

    Figure 1

    Woodward–Hoffmann rules

    Number of electrons Electrocyclic reactions: Thermal \((\Delta)\) Electrocyclic reactions: Photochemical \((h\nu)\) Cycloadditions: Thermal \((\Delta)\) Cycloadditions: Photochemical \((h\nu)\)
    \(4n\), \((n=1,2,..)\) Conrotatory (con) Disrotatory Disfavored Favored
    \(4n+2\), \((n=1,2,..)\) Disrotatory (dis) Conrotatory Favored Disfavored

    4.1 Fill in the table for reactions (i)–(iii) or products 2–5:

    Reaction Product [? + ?] cycloaddition \(\Delta\) or \(h\nu\)
    i 2
    ii 3
    iii 4
    5

    There are three possible benzotropone isomers. Although two of the benzotropone isomers were isolated, 3,4-benzotropone (1) has not been isolated. Its instability is attributed to the o- quinoidal structure of 1 because it has no sextet electron system in the benzene ring.

    4.2 Draw the structures of stable benzotropone isomers A (with 6 signals in its \(\ce{^{13}C}\)-NMR) and B (with 11 signals in its \(\ce{^{13}C}\)-NMR).

    4.3 When the following tetraene is reacted under photochemical conditions, symmetry-allowed product(s) of three different ring sizes can form according to the Woodward–Hoffmann rules. Tick the correct answer in each row.

    Figure 2
    Figure 3
    Figure 4

    Prof. Dr. Aziz Sancar

    The Nobel Prize in Chemistry 2015 was awarded jointly to the Turkish scientist Aziz Sancar, Swedish scientist Tomas Lindahl, and American scientist Paul Modrich for their “mechanistic studies of DNA repair”. Pyrimidine bases found in DNA may undergo a photochemical [2+2]-cycloaddition reaction (see figure above) with UV light that reaches a person’s skin, causing damage to DNA, which may ultimately lead to skin cancer. The research by Professor Aziz Sancar focused on the DNA repair mechanism for this type of damage.

    Thymine (T) is one of the nucleobases that can undergo such a photochemical reaction with UV light. Let us assume that we have a solution of free thymine that was subjected to UV irradiation.

    Figure 5

    4.4 Considering stereochemistry, draw the structures of all possible products of this reaction between two free thymine (T) molecules. Circle the compound(s) which is/are chiral. A drawing only one enantiomer of an enantiomeric pair is sufficient. Please note that only \(\ce{C=C}\) bonds participate in this reaction.

    A broad range of halogenated derivatives of norbornadiene (N) are known in the literature. Tribromo-norbornadiene (\(\ce{C7H5Br3}\)) has six achiral (meso) isomers. Three of these isomers (6, 7 and 8) are given below.

    Figure 6

    4.5 How many signals do you expect from the \(\ce{^{13}C}\)-NMR spectra of 6, 7, and 8? Fill in the following boxes.

    Figure 7

    4.6 Draw structures of the remaining achiral (meso) tribromo-norbornadiene (\(\ce{C7H5Br3}\)) isomers (C, D, and E) in addition to 6 – 8 over the given figures in the boxes.

    Figure 8

    The NMR spectrum of ether 9 is complex. Two \(\ce{MeO-}\) groups are different as are all the hydrogen atoms on the rings. However, diphenol 10 has a very simple NMR spectrum and there are only three types of protons (marked as a, b, and c). A reasonable average structure responsible for all resonance structures and its symmetry is shown as 11.

    Figure 9

    4.7 How many signals do you expect from the \(\ce{^{13}C}\)- and \(\ce{^{1}H}\)-NMR spectra of 12 and 13?

    Figure 10

    Official English text of the problem.

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  5. Problem 5: Konya, Carrot, Beta-Carotene, Vitamin-A, Immune System, Vision

    Figure 1

    Mevlana (Rumi) was a great mystic and Sufi poet who lived out his days in Konya in the \(13^{\mathrm{th}}\) century. The indirect relevance of Konya to chemistry is that the city provides 65% of the country’s carrot production, from which one of the essential vitamins (vitamin A) is obtained.

    Carrot is an important source of β-carotene, which gives the vegetable its orange color. This molecule is a red-orange pigment naturally found in plants and fruits and is a provitamin A carotenoid. It is converted to vitamin A, which is essential for normal growth and development, the immune system, and vision function.

    Figure 2

    Figure 1. Ball and stick representation of the structure of β-carotene. The gray and white spheres represent the carbon and hydrogen atoms, respectively. The numbered carbon atoms belong to the linear conjugated π-segment of the molecule.

    β-Carotene has an extended polyene chain of 22 carbon atoms. It is a conjugated \(\pi\)-system, having alternating single and double bonds. Its experimental maximum absorption wavelength (\(\lambda_{\max}\)) is 455 nm. We assume that all the bonds between \(C_1\) and \(C_{22}\) are conjugated bonds. There are 22 \(\pi\)-electrons in the molecule (Figure 1).

    To a crude approximation, the electrons in the \(\ce{C}\text{-}2p_z\) orbitals, which are perpendicular to the molecular plane, are assumed to move along the entire molecule, without interacting with each other. They are like independent particles confined in a molecule moving along the x-axis in one dimension. These characteristics of π-electrons make them eligible for being treated by the simplest model called the particle in one-dimensional box model.

    The wave function and the energies of the quantized levels for an electron moving in a one- dimensional box with infinite potential walls are given as follows:

    \[ \Psi_n(x)=\sqrt{\frac{2}{L}}\sin\frac{n\pi x}{L} \]

    here \(n\) is the quantum number, \(n=1,2,3,4,\ldots\infty\), and \(L\) is the box length.

    \[ E_n=\frac{n^2h^2}{8m_eL^2} \]

    In two dimension system, within the framework of independent particle approximation, the wavefunction is expressed as a product of one-dimensional wavefunctions, and the energy is expressed as a sum of one-dimensional energies. The energy levels of the two dimensional rectangular box is given as follows:

    \[ E_{n_xn_y} = \left[\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}\right] \left\{\frac{h^2}{8m_e}\right\} \]

    where \(n_x,n_y\) are the quantum numbers and they are positive integers. \(L_x,L_y\) are the dimensions of the box in the 2D model. They are positive numbers.

    5.1 The β-carotene molecule is orange in color because:

    i) It absorbs in the visible region of the electromagnetic spectrum.

    ii) HOMO → LUMO transition occurs by absorption of IR photon.

    iii) The spacing between the \(22^{\mathrm{nd}}\) and the \(23^{\mathrm{rd}}\) energy levels is equal to the energy of the IR photon at the orange wavelength.

    iv) It absorbs green/blue light and it transmits red/yellow color.

    v) It absorbs in the UV-VIS region since the molecule has no net dipole moment.

    Choose which pair of the above sentences is correct?

    The pairs of sentences are as follows:

    ☐ a) i and ii ☐ b) i and iii ☐ c) i and iv ☐ d) i and v
    ☐ e) ii and iii ☐ f) ii and iv ☐ g) ii and v ☐ h) iii and iv
    ☐ j) iii and v ☐ k) iv and v

    Although it is highly unrealistic, let us assume that the conjugated segment of the molecule is linear and treated with the particle in a one-dimensional box model as shown in figure 2. In this case, the length of the box can be approximated as \(L=1.40\times n_{\mathrm C}\) (in Å), where \(n_{\mathrm C}\) is the number of carbon atoms in the conjugated segment.

    Use this information to answer the questions 5.2–5.6.

    Figure 3

    Figure 2. Schematic representation of the conjugated line segment made up carbon atoms of β-carotene in a one-dimensional box of length L.

    5.2 Calculate the energies (in J) of the lowest two energy levels.

    5.3 Draw the wavefunctions of the lowest two energy levels with proper labelling the x-axis.

    5.4 Sketch the energy level diagram up to \(n=4\) showing the relative spacing.

    5.5 What is the total \(\pi\)-energy (in J) of the molecule?

    5.6 Calculate the wavelength (in nm) at which the transition between the highest occupied and lowest unoccupied energy levels occurs.

    Use the particle in a two-dimensional box model to answer questions 5.7 – 5.8.

    Figure 4

    Figure 3. Schematic representation of the conjugated carbon atoms of β-carotene in a two-dimensional box.

    Assume that the conjugated segment is made up of conjugated bonds that are all-trans to each other. The motion of the π-electrons is studied in the two-dimensional rectangular box with the dimensions \(L_x=26.0\) Å, \(L_y=3.0\) Å (Figure 3).

    5.7 Calculate the energies (in J) of the highest occupied and the lowest unoccupied energy levels and the wavelength (in nm) at which the transition between these energy levels occurs.

    5.8 What should be the \(L_x\) value (in Å) in order for the molecule to absorb light at the experimental \(\lambda_{\max}=455\) nm if \(L_y\) is kept constant at 3.0 Å. (Assume that the quantum numbers for homo and lumo are the same as in the question 5.7.)

    Official English text of the problem.

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  6. Problem 6: Thermodynamics through an Interstellar Journey

    Part 1

    In a hypothetical universe, an unknown amount of diborane participates in the following reaction:

    Figure 1

    Assume that in this hypothetical universe, \(\ce{H3BO3(s)}\) obtained from this reaction was completely sublimed at 300 K. The necessary energy for sublimation was obtained through work released by one cycle of an ideal heat engine in which one mole of monoatomic perfect gas flows through the cycle described in the pressure (\(p\)) – volume (\(V\)) diagram below:

    • A→B; isothermal reversible expansion receiving 250 J by heat transfer (\(q_H\)) at a temperature of 1000 K (\(T_H\)) from a hot source.
    • B→D; reversible adiabatic expansion.
    • D→C; isothermal reversible compression at a temperature of 300 K (\(T_C\)) releasing some amount of heat (\(q_C\)) to a cold sink.
    • C→A; reversible adiabatic compression.
    Figure 2

    After heat transfers, the remaining energy is released as work (\(w\)). Also, \(q_H\) and \(q_C\) are related to \(T_C\) and \(T_H\) as follows:

    \[ \frac{|q_H|}{|q_C|}=\frac{T_H}{T_C} \]

    The efficiency of the cycle can be found by work released by cycle (\(w\)) divided by heat absorbed by cycle (\(q_H\)).

    You are provided with the change in enthalpies of the following reactions at 300 K.

    Figure 3

    6.1 Calculate the molar enthalpy of sublimation (in \(\mathrm{kJ\,mol^{-1}}\)) for \(\ce{H3BO3}\) at 300 K.

    6.2 Calculate the \(\Delta_r U\) (internal energy) in terms of \(\mathrm{kJ\,mol^{-1}}\) at 300 K for the reactions (2) and (4) given above (assume ideal gas behavior for each gaseous species in each reaction).

    6.3 Calculate the amount of overall work produced by a heat engine (\(|w|\)) in terms of J and the amount of overall heat released to the cold sink (\(|q_C|\)) in terms of J.

    6.4 Calculate the efficiency of the heat engine described above.

    6.5 Calculate the entropy change (\(\Delta S\)) for A→B and D→C processes in the heat engine in terms of \(\mathrm{J\,K^{-1}}\).

    6.6 Calculate the Gibbs energy change (\(\Delta G\)) in terms of J for A→B and D→C processes in the heat engine.

    6.7 Calculate the ratio of pressure at point A to the pressure at point B in the cycle (standard pressure: 1 bar).

    6.8 Calculate the amount of \(\ce{H2(g)}\) (in moles) produced according to the reaction given at the beginning of the task for one cycle of the engine.

    Part 2

    Interstellar journeys can be done by using diborane as rocket fuel. Combustion of diborane is shown below:

    Figure 4

    Combustion of diborane is experimented in a closed container (\(100\ \mathrm{dm^3}\)) at different temperatures and the equilibrium amounts were recorded.

    Temperature 8930 K 9005 K
    \(\ce{B2H6(g)}\) 0.38 mol 0.49 mol
    \(\ce{H2O(g)}\) 0.20 mol 0.20 mol

    Partial pressure of \(\ce{O2(g)}\) was stabilized to 1 bar and kept constant at all conditions.

    Assume that in this hypothetical universe; \(\Delta_r S^\circ\) and \(\Delta_r H^\circ\) are independent of temperature, the standard molar entropy (\(S^\circ\)) of \(\ce{B2O3(s)}\) does not change with pressure, all the gas species behave as an ideal gas, and all species remain in the same phase, without any further decomposition before or after reaction, at all temperatures then:

    6.9 Calculate \(K_p\) (pressure based equilibrium constant) at 8930 K and 9005 K.

    6.10 Calculate \(\Delta_r G^\circ\) of the reaction in terms of \(\mathrm{kJ\,mol^{-1}}\) at 8930 K and 9005 K. (If you failed to find \(K_p\), please use \(K_p\) (8930 K) = 2, \(K_p\) (9005 K) = 0.5.)

    6.11 Calculate \(\Delta_r G^\circ\) (in terms of \(\mathrm{kJ\,mol^{-1}}\)), \(\Delta_r H^\circ\) (in terms of \(\mathrm{kJ\,mol^{-1}}\)), and \(\Delta_r S^\circ\) (in terms of \(\mathrm{J\,mol^{-1}\,K^{-1}}\)) of the combustion reaction at 298 K. (If you failed to find \(K_p\), please use \(K_p\)(8930 K) = 2, \(K_p\)(9005 K) = 0.5)

    6.12 Tick the correct answer in the table by determining whether combustion reactions are favored or not at given temperature \(T\) below under standard pressure (1 bar).

    Temperature Favored Unfavored
    298 K ☐ ☐
    8930 K ☐ ☐
    9005 K ☐ ☐
    9100 K ☐ ☐

    6.13 Calculate the \(\Delta_f H\) (\(\mathrm{kJ\,mol^{-1}}\)) and \(S^\circ\) (\(\mathrm{kJ\,mol^{-1}\,K^{-1}}\)) of \(\ce{H2O(g)}\) using the values given in the table below. (\(\Delta_f H\) = enthalpy of formation, \(S^\circ\) = standard entropy)

    (If you fail to find \(\Delta_r H^\circ\) and \(\Delta_r S^\circ\) of the combustion, please use \(\Delta H^\circ=1000\ \mathrm{kJ\,mol^{-1}}\), \(\Delta S^\circ=150\ \mathrm{J\,K^{-1}\,mol^{-1}}\))

    Compound \(\Delta H_f\) (298 K) \(S^\circ\) (298 K)
    \(\ce{B2H6(g)}\) \(36.40\ \mathrm{kJ\,mol^{-1}}\) \(0.23\ \mathrm{kJ\,mol^{-1}\,K^{-1}}\)
    \(\ce{O2(g)}\) \(0.00\ \mathrm{kJ\,mol^{-1}}\) \(0.16\ \mathrm{kJ\,mol^{-1}\,K^{-1}}\)
    \(\ce{B2O3(s)}\) \(-1273\ \mathrm{kJ\,mol^{-1}}\) \(0.05\ \mathrm{kJ\,mol^{-1}\,K^{-1}}\)

    Official English text of the problem.

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  7. Problem 7: Phthalocyanines

    Figure 1
    Figure 2

    Emeritus Professor Özer Bekaroğlu

    The term phthalocyanine (Pc) takes its origin from the Greek “naphtha”, which means rock oil, and “cyanine”, which means dark blue. Turkish scientist Özer Bekaroğlu can be regarded as the pioneer of Pc chemistry in Turkey.

    Metal-free phthalocyanine (1, \(\ce{H2Pc}\)) is a large planar macrocyclic compound with the formula \(\ce{(C8H4N2)4H2}\).

    7.1 How many π-electrons are there in the bold region of the \(\ce{H2Pc}\) molecule in compound 1 shown above?

    Figure 3

    Pcs containing one or two metal ions are called metallo-phthalocyanines (MPcs) and they exhibit different geometries as given above.

    7.2 Complete the table in your answer sheet by determining the coordination number of central ions in 2–5.

    Central ion Copper ion Titanium ion Silicon ion Cerium ion
    Coordination number

    7.3 Complete the table in your answer sheet by determining the oxidation number of each metal (Cu, Ti, and Ce) in 2, 3, and 5.

    Metal in compounds 2 3 5
    Oxidation number

    7.4 Complete the table in your answer sheet by determining the geometry of compounds 2–5.

    Geometry Compound
    Octahedral
    Square prism
    Square pyramidal
    Square planar

    7.5 Complete the table in your answer sheet by determining the magnetic property of compounds 2–5.

    • Use the letter "p" for paramagnetic property and letter "d" for a diamagnetic property.
    Compound Magnetic property
    2
    3
    4
    5

    7.6 Write the ground-state electron configuration of the silicon (Si) ion in the compound 4 and find all the quantum numbers for the 2p electrons in its ground state.

    Electron configuration:
    Quantum numbers for 2p electrons: \(n\) \(l\) \(m_l\) \(m_s\)

    Metal-free phthalocyanine (1, \(\ce{H2Pc}\)) is commonly formed through the cyclotetramerization of phthalonitriles. On the other hand, Pcs having different substituents are called asymmetric, which can be prepared by the statistical cyclization of two different phthalonitriles. This method has no selectivity and the product is a mixture of all possible isomers.

    7.7 Draw the possible products that may occur in the statistical cyclization method by using F and G. If there are any stereoisomers label as cis- or trans-.

    • F and G represent two different symmetrical phthalonitriles.
    • One of the products is \(\mathrm{F}_4\) as given below.
    • Draw other products similar to the format of \(\mathrm{F}_4\).
    Figure 4

    Pcs are used as photosensitizers in the photodynamic therapy (PDT) of cancer due to their strong absorption in the visible spectrum and high molar absorption coefficients. PDT consists of three essential components: photosensitizer, light, and oxygen. None of these is individually toxic, but together they initiate a photochemical reaction resulting in the generation of cytotoxic singlet oxygen (\(\ce{^{1}O2}\)) that can destroy cancer cells.

    Figure 5
    • The multiplicity of an energy level is defined as \(2S+1\)
    • If the two spins are parallel (\(\uparrow\uparrow\)), \(S=1\), and if the two spins are antiparallel (\(\uparrow\downarrow\)), \(S=0\).

    7.8 Draw the molecule orbital (MO) diagram of the lowest energy singlet state of dioxygen (\(\ce{^{1}O2}\)) and calculate bond order.

    • There are no unpaired electrons in that state!

    7.9 If the wavelength of the light needed to excite triplet oxygen into singlet oxygen is 1270 nm, calculate the energy (in kJ/mol) needed for this transition process.

    Official English text of the problem.

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  8. Problem 8: Boron Compounds and Hydrogen Storage

    14% of the total

    Question 8.1 8.2 8.3 8.4 8.5 8.6 8.7 8.8 Total
    Points 3 3 4 4 4 4 6 3 58
    Question 8.9 8.10 8.11 8.12 8.13 8.14 8.15 8.16
    Points 2 3 2 4 4 4 2 2
    Question 8.17
    Points 4

    Problem 8. Boron Compounds and Hydrogen Storage

    Figure 1

    Sodium borohydride (\(\ce{NaBH4}\)) and ammonia borane (\(\ce{BNH6}\)) are the most studied chemical hydrogen storage materials. In this question, you will explore the chemistry of boron and the use of boron compounds as hydrogen storage materials.

    Borax (\(\ce{Na2B4O7 * nH2O}\)) is a boron mineral that is produced by ETI Mining Company in Turkey. \(\ce{NaBH4}\) can be synthesized by the reduction of anhydrous borax with metallic sodium under high-pressure hydrogen gas in the presence of silicon dioxide (silica) at 700 °C, namely the Bayer process. In this process, all hydrogen is stored in \(\ce{NaBH4}\). On the other hand, it has been shown that ammonia borane (\(\ce{BNH6}\)) can be synthesized by the reaction of \(\ce{NaBH4}\) and ammonium sulfate in dry tetrahydrofuran (THF) at 40 °C (Hint: \(\ce{BNH6}\) synthesis must be conducted in a well-ventilated fume-hood because flammable gas is generated as one of the by-products). While \(\ce{NaBH4}\) is an ionic compound, ammonia borane is a Lewis acid–base adduct.

    8.1. Write a balanced chemical equation for the synthesis of \(\ce{NaBH4}\) from anhydrous borax.

    8.2. Write a balanced chemical equation for the synthesis of ammonia borane from \(\ce{NaBH4}\).

    8.3. Draw the molecular geometry of the \(\ce{BH4^-}\) ion and \(\ce{BNH6}\) molecule.

    8.4. Calculate the hydrogen content of \(\ce{NaBH4}\) and \(\ce{BNH6}\) as a percentage by mass (wt%).

    The hydrogen stored in both compounds can be released via hydrolysis reactions in the presence of a suitable catalyst at room temperature. Upon the hydrolysis reactions, 4 and 3 moles of \(\ce{H2}\) gas are released from the hydrolysis of 1 mole of \(\ce{NaBH4}\) and \(\ce{BNH6}\), respectively, along with metaborate anion including B-O bonds.

    8.5. Write the balanced chemical equations for the hydrolysis of \(\ce{NaBH4}\) and \(\ce{BNH6}\).

    One of the simplest stable borates is diboron trioxide (\(\ce{B2O3}\)). Higher borates such as \(\ce{B3O6^{3-}}\) having cyclic structures containing B-O-bonds can be formed. Since \(\ce{B2O3}\) is an acidic compound, it is easily reacted with water to produce boric acid (\(\ce{H3BO3}\)). On the other hand, the high temperature and high-pressure reaction of \(\ce{B2O3}\) with ammonia yields two-dimensional boron nitride, which consists of planar graphite-like sheets of alternating B and N atoms.

    8.6. Write the balanced chemical equations for the synthesis of boric acid and boron nitride.

    8.7. Draw the molecular structures of the \(\ce{B3O6^{3-}}\) ion, boric acid, and a single two-dimensional boron nitride sheet. Hint: show at least 10 B atoms in the boron nitride structure.

    Furthermore, B-H compounds, called boranes, are an important class of boron compounds. The simplest stable borane is diborane (\(\ce{B2H6}\)) and many of the higher boranes can be prepared by the pyrolysis of diborane. Diborane can be synthesized via metathesis of a boron halide and a hydride source.

    8.8. Write a balanced chemical equation for the synthesis of diborane from the reaction of \(\ce{BF3}\) and \(\ce{LiBH4}\). Hint: both products are boron compounds.

    8.9. Draw the molecular geometry of the diborane molecule. Hint: there is no B-B bond in the structure.

    \(\ce{BH3}\) (borane) is an unstable and highly reactive molecule. Therefore, it is not possible to isolate it as \(\ce{BH3}\) under ordinary conditions. However, it can be stabilized via its reaction with carbon monoxide to yield borane carbonyl (\(\ce{BH3CO}\)) compound, which is an adduct of borane. The preparation of \(\ce{BH3CO}\) plays an important role in exploring the chemistry of boranes as it indicates the likely existence of the borane molecule.

    8.10. Sketch the Lewis dot structure of \(\ce{BH3-CO}\) molecule by showing the formal charges.

    8.11. Which of the statements given in your answer sheet is observed in the \(\ce{C-O}\) bond of a \(\ce{CO}\) molecule upon the bond formation between \(\ce{BH3}\) and \(\ce{CO}\)? Tick the correct box.

    • ☐ It gets longer because there will be \(\pi\)-back donation from \(\ce{BH3}\) to \(\ce{CO}\).
    • ☐ It gets longer because \(\ce{CO}\) donates \(\pi\)-bonding electrons to \(\ce{BH3}\).
    • ☐ No or slight change on it because \(\ce{CO}\) donates its mainly non-bonding electrons to \(\ce{BH3}\).
    • ☐ It gets shorter because \(\ce{CO}\) donates \(\pi^*\) anti-bonding electrons to \(\ce{BH3}\).

    Borazine consists of single and double bonded cyclic \(\ce{B-N}\) units and hydrogen atoms attached to these atoms with the molecular formula \(\ce{B3N3H6}\) and is isostructural to benzene. Borazine can be synthesized by using a two-step procedure including the synthesis of symmetrically trisubstituted chlorine derivatives of borazine (\(\ce{B3N3H3Cl3}\)) from the reaction of ammonium chloride and boron trichloride, and then reduction of \(\ce{B3N3H3Cl3}\) with \(\ce{LiBH4}\) in THF.

    8.12. Write the balanced chemical equations for the two-step synthesis of borazine starting from ammonium chloride in THF (tetrahydrofuran). Hint: THF stabilizes one of the products by forming a Lewis acid-base adduct.

    8.13. Draw the molecular structures of borazine and its symmetrically trisubstituted chlorine derivative.

    Catalysts are substances that accelerate the rate of reactions by allowing them to proceed in a lower energy pathway. The catalytic activity of the catalysts is generally determined by the turnover frequency (TOF), which is calculated by dividing the molar amount of the product to the mole of active catalyst and the time (TOF = mole product/(mole catalyst x time)). A typical hydrolysis of \(\ce{BNH6}\) was carried out in 10.0 mL of water by using 100.0 mM \(\ce{BNH6}\) and 5.0 mg of CuPt/C catalyst (CuPt alloy nanoparticles supported on carbon black containing 8.2 wt% Pt atom). 67.25 mL of hydrogen gas was generated in 5 minutes.

    8.14. Assuming the catalytic reaction is performed under standard conditions (1 atm and 273.15 K), calculate the TOF (\(\mathrm{min^{-1}}\)) of the CuPt/C catalyst in terms of only Pt atoms in the hydrolysis of \(\ce{BNH6}\) by considering the volume of the hydrogen gas generated.

    As a result of detailed crystal analysis of a synthesized \(\ce{Cu_{x}Pt_{y}}\) alloy nanoparticles (the subscripts indicate molar percentages of the atoms in the alloy structure), it was determined that the face centered cubic (fcc) unit cell was formed by Pt atoms and the Pt atoms on the face of the fcc unit cell are supposed to be replaced with Cu atoms to form \(\ce{Cu_{x}Pt_{y}}\) displacement alloy nanoparticles. According to this information, answer the following questions.

    8.15. Determine the composition of the alloy nanoparticles by finding \(x\) and \(y\) in the \(\ce{Cu_{x}Pt_{y}}\).

    8.16. Sketch the shape of the described crystal unit cell of \(\ce{Cu_{x}Pt_{y}}\) alloy nanoparticles by showing the position of all atoms on the unit cell.

    8.17. Another alloy has a \(\ce{Cu2Pt1}\) composition. Assume that this alloy also has a fcc unit cell with an edge length of 380 pm, but the Cu and Pt atoms are randomly distributed in the atomic positions. Calculate the density of this alloy in \(\mathrm{g/cm^3}\).

    Official English text of the problem.

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  9. Problem 9: Quantification of Heavy Metal Ions

    | 12% of the total | Points | 5 | 5 | 8 | 16 | 5 | 6 | 25 | 30 | 100 |

    Problem 9. Quantification of Heavy Metal Ions

    For the quantitative analysis of heavy metal ions in a factory’s wastewater pool, the following steps have been applied by an analyzer at 298 K:

    Step 1) 10-mL samples of each were obtained from five different regions in a wastewater pool, mixed in a 100-mL beaker, and then stirred for 5 minutes using a magnetic stirrer.

    Step 2) 10-mL of sample solution was taken from the 100-mL beaker and 142 mg of \(\ce{Na2SO4}\) was added while stirring, followed by transfer to a three-electrode cell as seen in Figure 1a. In this electrochemical cell, \(\ce{Pt}\) wire, \(\ce{Ag/AgCl}\) (3 M \(\ce{KCl}\)), and \(\ce{Pt}\) foil served as the working, reference, and counter electrodes, respectively.

    Step 3) These electrodes were connected to a potentiostat and a constant potential of −0.50 V vs. \(\ce{Ag/AgCl}\) for 14 minutes was applied as seen in Figure 1b (horizontal line). It is assumed that 14 min. is sufficient to complete the expected electrochemical reactions.

    Figure 1

    Figure 1. a) Electrochemical cell design; 1) Working electrode (\(\ce{Pt}\) wire), 2) reference electrode (\(\ce{Ag/AgCl}\), 3M \(\ce{KCl}\)), 3) counter electrode (\(\ce{Pt}\) foil), 4) cell tap, 5) electrochemical cell, 6) 10-mL of sample solution. b) Potential change of working electrode as a function of time. y-axis: potential/V vs \(\ce{Ag/AgCl}\), x-axis: time/min.

    Step 4) The electrodes were rinsed with distilled water, placed into another electrochemical cell including 10-mL of 0.1 M \(\ce{H2SO4}\) solution, and potential was scanned between −0.50 and +0.50 V as seen in Figure 1b (downward sloping line in 2 min.). Current vs. potential data for this step are presented in Figure 2a, which is like an excellent view of Mount Ararat (Ağrı Dağı), the highest mountain in Turkey (Figure 2b).

    Figure 2

    Figure 2. a) Potential scan of the working electrode as a function of current in 0.1 M \(\ce{H2SO4}\) solution after keeping it constant at a potential of −0.50 V in 10-mL of wastewater sample as seen in Figure 1b (horizontal line). y-axis: current/μA, x-axis: potential/V vs \(\ce{Ag/AgCl}\), b) A view of Great and Little Ararat peaks.

    Step 5) Another 10-mL of the sample solution prepared in step 1 was taken and the processes explained in steps 2 and 3 were applied in that order. The electrodes were rinsed with distilled water and placed into 10-mL of 0.1 M \(\ce{H2SO4}\) solution. Then the potential of the working electrode was kept constant at +0.05 V for 14 minutes. It is assumed that 14 min. is sufficient to complete the expected electrochemical reactions.

    Step 6) After step 5 was performed, the solution in the electrochemical cell was placed in a suitable oven to evaporate at 150 °C until dry solid was obtained.

    Step 7) 5-mL of ethylenediaminetetraacetic acid (EDTA, \(\ce{H4Y}\)) (Figure 3) solution was added to the solid obtained in step 6 and it was shaken to dissolve. It is known that 1-mL of EDTA solution is equivalent to 3.85 mg/mL \(\ce{BaCO3}\). Then, pH of the solution was adjusted to 10.0. Excess EDTA was titrated with standard 0.0010 M \(\ce{Ni(NO3)2}\) solution and it was observed that 95.60 mL of \(\ce{Ni(NO3)2}\) solution was consumed up to the endpoint.

    Figure 3

    Figure 3. Chemical structure of EDTA (\(\ce{H4Y}\)).

    • In water saturated with \(\ce{H2S}\), equilibrium concentration of \([\ce{H2S}]\) is 0.1 M.
    • \(K_{\mathrm{sp}}(\ce{NiS})=4.0\times10^{-20}\); \(K_{\mathrm{sp}}(\ce{CuS})=1.0\times10^{-36}\)
    • \(K_{\mathrm{a1}}(\ce{H2S})=9.6\times10^{-8}\); \(K_{\mathrm{a2}}(\ce{H2S})=1.3\times10^{-14}\)
    Reaction \(E^\circ/\mathrm{V}\) (at 298 K)
    \(\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}\) −0.83
    \(\ce{Ni^2+(aq) + 2e- -> Ni(s)}\) −0.24
    \(\ce{2H+(aq) + 2e- -> H2(g)}\) 0.00
    \(\ce{Cu^2+(aq) + 2e- -> Cu(s)}\) +0.34
    \(\ce{Ag+(aq) + e- -> Ag(s)}\) +0.80
    \(\ce{O2(g) + 4H+(aq) + 4e- -> 2H2O(l)}\) +1.23

    9.1. Which of the following can be considered for peak 1 and peak 2 in Figure 2a, respectively? Tick the correct box on the answer sheet.

    • ☐ Peak 1: electrochemical reduction of \(\ce{Ni}\) / Peak 2: electrochemical reduction of \(\ce{Cu}\)
    • ☐ Peak 1: electrochemical reduction of \(\ce{Cu}\) / Peak 2: electrochemical reduction of \(\ce{Ni}\)
    • ☐ Peak 1: electrochemical reduction of \(\ce{Ni}\) / Peak 2: electrochemical oxidation of \(\ce{Cu}\)
    • ☐ Peak 1: electrochemical oxidation of \(\ce{Ni}\) / Peak 2: electrochemical oxidation of \(\ce{Cu}\)
    • ☐ Peak 1: electrochemical oxidation of \(\ce{Cu}\) / Peak 2: electrochemical oxidation of \(\ce{Ni}\)

    9.2. Which of the statement is expected, if the potential is applied as −1.2 V instead of -0.5 V at the first step (horizontal line) in Figure 1b? Tick the correct box on the answer sheet.

    • ☐ \(\ce{NO}\) evolution
    • ☐ \(\ce{NO2}\) evolution
    • ☐ Nitrogen evolution
    • ☐ Oxygen evolution
    • ☐ Hydrogen evolution

    9.3. Calculate the scan rate of data presented in Figure 2a as mV/s at 298 K.

    The potential of the following cell is measured as 0.437 V.

    \[ \ce{Pt},\ce{H2}(0.92\ \mathrm{bar}) \mid \ce{HCl}(1.50\times10^{-2}\ \mathrm{M}),\ce{AgCl}(sat) \mid\ce{Ag} \]

    9.4. Calculate the standard electrode potential value (V) of half-cell of \(\ce{AgCl(s) + e- -> Ag(s) + Cl-(aq)}\) at 298 K. Note: You must show all works.

    9.5. Which of the statements is the main purpose of step 5 in this analysis? Tick the correct box on the answer sheet.

    • ☐ Modification of \(\ce{Pt}\) wire with \(\ce{Ni}\)–\(\ce{Cu}\) alloy film
    • ☐ Modification of \(\ce{Pt}\) wire with \(\ce{Ni}\) film
    • ☐ Electrochemical stripping of both \(\ce{Cu}\) and \(\ce{Ni}\) from \(\ce{Cu}\)–\(\ce{Ni}\)-modified \(\ce{Pt}\) wire to the solution
    • ☐ Electrochemical stripping of \(\ce{Cu}\) from \(\ce{Cu}\)–\(\ce{Ni}\)-modified \(\ce{Pt}\) wire to the solution
    • ☐ Electrochemical stripping of \(\ce{Ni}\) from \(\ce{Cu}\)–\(\ce{Ni}\)-modified \(\ce{Pt}\) wire to the solution

    9.6. Write net ionic equations for the complexation and the back titration reaction of step 7 on the answer sheet

    9.7. Calculate \(\ce{Ni^2+}\) concentration as mg/L in the wastewater of the factory. Note: You must show all works.

    9.8. Calculate the minimum pH value for starting the precipitation of \(\ce{Ni^2+}\) ions in the solution obtained in step 5 by passing \(\ce{H2S}\) gas into the solution until saturation. If you cannot solve question 9.7, use 20 mg/L \(\ce{Ni^2+}\) sample for this question. Note: You must show all works.

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