International Chemistry Olympiad

IChO 1999

Bangkok, Thailand · 6 theory problems with complete solutions

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  1. Problem 1: Combustion calorimetry and molecular state of compound Q

    A compound Q (molar mass \(122.0\ \mathrm{g\,mol^{-1}}\)) consists of carbon, hydrogen and oxygen.

    PART A

    The standard enthalpy of formation of \(\ce{CO2(g)}\) and \(\ce{H2O(l)}\) at \(25.00\ ^\circ\mathrm C\) are −393.51 and −285.83 \(\mathrm{kJ\,mol^{-1}}\), respectively. The gas constant, \(R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}\).

    (Relative atomic masses : H = 1.0; C = 12.0; O = 16.0)

    A sample of solid Q that weighs 0.6000 g, is combusted in an excess of oxygen in a bomb calorimeter, which initially contains 710.0 g of water at \(25.000\ ^\circ\mathrm C\). After the reaction is completed, the temperature is observed to be \(27.250\ ^\circ\mathrm C\), and 1.5144 g of \(\ce{CO2(g)}\) and 0.2656 g of \(\ce{H2O(l)}\) are produced.

    1.1 Determine the molecular formula and write a balanced equation with correct state of matters for the combustion of Q.

    If the specific heat of water is \(4.184\ \mathrm{J\,g^{-1}\,K^{-1}}\) and the internal energy change of the reaction \((\Delta U^0)\) −3079 \(\mathrm{kJ\,mol^{-1}}\).

    1.2 Calculate the heat capacity of the calorimeter (excluding the water).

    1.3 Calculate the standard enthalpy of formation \((\Delta H_{\mathrm f}^{0})\) of Q.

    PART B

    The following data refer to the distribution of Q between benzene and water at \(6\ ^\circ\mathrm C\), \(C_{\mathrm B}\) and \(C_{\mathrm W}\) being equilibrium concentrations of the species of Q in the benzene and water layers, respectively:

    Assume that there is only one species of Q in benzene independent of concentration and temperature.

    Concentration \((\mathrm{mol\,dm^{-3}})\)

    \(C_{\mathrm B}\) \(C_{\mathrm W}\)
    0.0118 0.00281
    0.0478 0.00566
    0.0981 0.00812
    0.156 0.0102

    1.4 Show by calculation whether Q is monomer or dimer in benzene. Assume that Q is a monomer in water.

    The freezing point depression, for an ideal dilute solution, is given by

    \[ T_{\mathrm f}^{0}-T_{\mathrm f} =\frac{R(T_{\mathrm f}^{0})^{2}X_{\mathrm s}}{\Delta H_{\mathrm f}} \]

    where \(T_{\mathrm f}\) is the freezing point of the solution, \(T_{\mathrm f}^{0}\) the freezing point of solvent, \(\Delta H_{\mathrm f}\) the heat of fusion of the solvent, and \(X_{\mathrm s}\) the mole fraction of solute. The molar mass of benzene is \(78.0\ \mathrm{g\,mol^{-1}}\). At 1 atm pure benzene freezes at \(5.40\ ^\circ\mathrm C\). The heat of fusion of benzene is \(9.89\ \mathrm{kJ\,mol^{-1}}\).

    1.5 Calculate the freezing point \((T_{\mathrm f})\) of a solution containing 0.244 g of Q in 5.85 g of benzene at 1 atm.

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  2. Problem 2: A diprotic acid and an acid-base indicator

    PART A

    A diprotic acid , \(\ce{H2A}\) , undergoes the following dissociation reactions :

    Figure 1

    A 20.00 \(\mathrm{cm^3}\) aliquot of a solution containing a mixture of \(\ce{Na2A}\) and \(\ce{NaHA}\) is titrated with 0.300 M hydrochloric acid. The progress of the titration is followed with a glass electrode pH meter. Two points on the titration curve are as follows :

    \(\mathrm{cm^3}\) \(\ce{HCl}\) added \(pH\)
    1.00 10.33
    10.00 8.34

    2.1 On adding 1.00 \(\mathrm{cm^3}\) of \(\ce{HCl}\), which species reacts first and what would be the product?

    2.2 What is the amount (mmol) of the product formed in (2.1)?

    2.3 Write down the main equilibrium of the product from (2.1) reacting with the solvent?

    2.4 What are the amounts (mmol) of \(\ce{Na2A}\) and \(\ce{NaHA}\) initially present?

    2.5 Calculate the total volume of \(\ce{HCl}\) required to reach the second equivalence point.

    PART B

    Solutions I, II and III contain a \(pH\) indicator \(\ce{HIn}\) (\(K_{\mathrm{In}}=4.19\times10^{-4}\)) and other reagents as indicated in the table. The absorbance values at 400 nm of the solutions measured in the same cell, are also given in the table. \(K_a\) of \(\ce{CH3COOH}\) is \(1.75\times10^{-5}\).

    Table:

    Solution I Solution II Solution III
    Total concentration of indicator \(\ce{HIn}\) \(1.00\times10^{-5}\ \mathrm{M}\) \(1.00\times10^{-5}\ \mathrm{M}\) \(1.00\times10^{-5}\ \mathrm{M}\)
    Other reagents \(1.00\ \mathrm{M}\) \(\ce{HCl}\) \(0.100\ \mathrm{M}\) \(\ce{NaOH}\) \(1.00\ \mathrm{M}\) \(\ce{CH3COOH}\)
    Absorbance at 400 0.000 0.300 ?

    2.6 Calculate the absorbance at 400 nm of solution III.

    2.7 Apart from \(\ce{H2O}\), \(\ce{H+}\) and \(\ce{OH-}\), what are all the chemical species present in the solution resulting from mixing solution II and solution III at 1 : 1 volume ratio?

    2.8 What is the absorbance at 400 nm of the solution in (2.7)?

    2.9 What is the transmittance at 400 nm of the solution in (2.7)?

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  3. Problem 3: The thorium decay series

    One of naturally occurring radioactive decay series begins with \(\ce{^{232}_{90}Th}\) and ends with a stable \(\ce{^{208}_{82}Pb}\).

    3.1 How many beta (\(\beta^-\)) decays are there in this series? Show by calculation.

    3.2 How much energy in MeV is released in the complete chain?

    3.3 Calculate the rate of production of energy (power) in watts (\(1\ \mathrm{W}=\mathrm{J\,s^{-1}}\)) produced by 1.00 kilogram of \(\ce{^{232}Th}\) (\(t_{½}=1.40\times10^{10}\) years).

    3.4 \(\ce{^{228}Th}\) is a member of the thorium series. What volume in \(\mathrm{cm^3}\) of helium at \(0\ ^\circ\mathrm{C}\) and 1 atm collected when 1.00 gram of \(\ce{^{228}Th}\) (\(t_{1/2}=1.91\) years) is stored in a container for 20.0 years. The half-lives of all intermediate nuclides are short compared to the half-life of \(\ce{^{228}Th}\).

    3.5 One member of thorium series, after isolation, is found to contain \(1.50\times10^{10}\) atoms of the nuclide and decays at the rate of 3440 disintegrations per minute. What is the half-life in years?

    The necessary atomic masses are :

    \[ \ce{^{4}_{2}He}=4.00260\ \mathrm{u},\quad \ce{^{208}_{82}Pb}=207.97664\ \mathrm{u},\quad \ce{^{232}_{90}Th}=232.03805\ \mathrm{u}; \quad\text{and }1\mathrm{u}=931.5\ \mathrm{MeV} \]

    \[ 1\ \mathrm{MeV}=1.602\times10^{-13}\ \mathrm{J} \]

    \[ N_A=6.022\times10^{23}\ \mathrm{mol^{-1}} \]

    The molar volume of an ideal gas at \(0\ ^\circ\mathrm{C}\) and 1 atm is \(22.4\ \mathrm{dm^3\,mol^{-1}}\).

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  4. Problem 4: Iron and chromium complexes of ligand L

    Ligand L can form complexes with many transition metals. L is synthesized by heating a mixture of a bipyridine, glacial acetic acid and hydrogen peroxide to 70 – 80 °C for 3 hrs. The final product L crystallizes out as fine needles and its molecular mass is 188. An analogous reaction with pyridine is ;

    Figure 1

    Complexes of L with \(\ce{Fe}\) and \(\ce{Cr}\) have the formulae of \(\ce{FeL_m(ClO4)_n * 3H2O}\) (A) and \(\ce{CrL_xCl_y(ClO4)_z * H2O}\) (B), respectively. Their elemental analyses and physical properties are given in Tables 4a and 4b. The relationship of colour and wavelength is given in Table 4c.

    Table 4a: Elemental analyses.

    Complex Elemental analyses , (wt. %)
    A Fe 5.740, C 37.030, H 3.090 , Cl 10.940, N 8.640
    B Cr 8.440, C 38.930, H 2.920, Cl 17.250, N 9.080

    Use the following data:

    Atomic number: Cr = 24, Fe = 26

    Relative atomic mass: H = 1, C = 12, N = 14, O = 16, Cl = 35.45, Cr = 52, Fe = 55.8

    Table 4b: Physical properties

    Complex Magnetic moment , \(\mu\) B.M. Colour
    A 6.13 Yellow
    B Not measured Purple

    Table 4c Relationship of wavelength to colour.

    Wavelength (nm) and colour absorbed Complementary colour
    400 (violet) Yellow Green
    450 (blue) Yellow
    490 (blue green) Orange
    500 (green) Red
    570 (yellow green) Violet
    580 (yellow) Blue
    600 (orange) Blue green
    650 (red) Green

    4.1 Write down the molecular formula of L.

    4.2 If L is a bidentate chelating ligand, draw the structure of the bipyridine used. Also draw the structure of L.

    4.3 Does the ligand L have any charge, i. e. net charge?

    4.4 Draw the structure when one molecule of L binds to metal ion (M).

    4.5 From the data in Table 4a, determine the empirical formula of A. What are the values of m and n in \(\ce{FeL_m(ClO4)_n * 3H2O}\)? Write the complete formula of A in the usual IUPAC notation. What is the ratio of cation to anion when A dissolves in water?

    4.6 What is the oxidation number of Fe in A? How many d-electrons are present in Fe ion in the complex? Write the high spin and the low spin configurations that may exist for this complex. Which configuration, high or low spin, is the correct one? What is the best evidence to support your answer?

    4.7 From Table 4c, estimate \(\lambda_{\max}\) (nm) of A.

    4.8 Detail analysis of B shows that it contains \(\ce{Cr^{3+}}\) ion. Calculate the ‘spin-only’ magnetic moment of this compound.

    4.9 Compound B is a 1 : 1 type electrolyte. Determine the empirical formula of B and the values of x, y, z in \(\ce{CrL_xCl_y(ClO4)_z * H2O}\).

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  5. Problem 5: Glycoside A from Rosaceae seeds

    Glycoside A (\(\ce{C20H27NO11}\)), found in seeds of Rosaceae gives a negative test with Benedicts’ or Fehling’s solutions. Enzymatic hydrolysis of A yields (-) B, \(\ce{C8H7NO}\) and C, \(\ce{C12H22O11}\), but complete acid hydrolysis gives as organic products, (+) D, \(\ce{C6H12O6}\) and (-) E, \(\ce{C8H8O3}\).

    C has a \(\beta\)-glycosidic linkage and gives positive test with Benedicts’ or Fehling’s solution. Methylation of C with \(\ce{MeI/Ag2O}\) gives \(\ce{C20H38O11}\), which upon acidic hydrolysis gives 2,3,4-tri-O-methyl-D-glucopyranose and 2,3,4,6-tetra-O-methyl-D-glucopyranose.

    (±)B can be prepared from benzaldehyde and \(\ce{NaHSO3}\) followed by \(\ce{NaCN}\). Acidic hydrolysis of (±)B gives (±)E, \(\ce{C8H8O3}\).

    5.1 Write structures of A – D with appropriate stereochemistry in Haworth projection, except for B.

    Glycoside A is found to be toxic and believed to be due to extremely toxic compound F, liberated under the hydrolytic conditions. Detoxification of compound F in plant may be accompanied by the reactions (stereochemistry not shown).

    Figure 1

    A small amount of compound F in human being is believed to be detoxified by a direct reaction with cystine giving L-cysteine and compound I, \(\ce{C4H6N2O2S}\) which is excreted in urine (stereochemistry not shown).

    Figure 2

    Compound I shows no absorption at 2150-2250 \(\mathrm{cm^{-1}}\) in its IR spectrum but a band at 1640 \(\mathrm{cm^{-1}}\) and the bands of carboxyl group are observed.

    5.2 Write molecular formula for compounds F and G, and structural formula for compounds H and I and indicate stereochemistry of H. (Table 5.1 may be useful for structure identification.)

    (-)1-Phenylethane-1-d, \(\ce{C6H5CHDCH3}\) can be prepared in optically active form and the magnitude of its rotation has the relatively high value, \([\alpha]_{\mathrm D}\) is equal to -0.6.

    Figure 3

    The absolute configuration of (-)1-phenylethane-1-d is related to (-) E according to the following reactions.

    Figure 4

    Compound (-) M can also be obtained from compound N as follows.

    Figure 5

    5.3 Deduce the absolute configuration of (-) E and the structure with configuration of each intermediate (J – O) in the sequence with the proper \(R,S\)-assignment as indicated in the answer sheet.

    5.4 Choose the mechanism involved in the conversion of compound O to 1-phenylethane-1-d.

    Table 5.1 Characteristic Infrared Absorption

    Stretching Vibration Region (\(\mathrm{cm^{-1}}\)) Stretching Vibration Region (\(\mathrm{cm^{-1}}\))
    \(\ce{C-H}\) (alkane) 2850-2960 \(\ce{O-H}\) (free alcohol) 3400-3600
    \(\ce{C-H}\) (alkene) 3020-3100 \(\ce{O-H}\) (H-bonded alcohol) 3300-3500
    \(\ce{C=C}\) 1650-1670 \(\ce{O-H}\) (acid) 2500-3100
    \(\ce{C-H}\) (alkyne) 3300 \(\ce{C-O}\) 1030-1150
    \(\ce{C#C}\) 2100-2260 \(\ce{NH}\), \(\ce{NH2}\) 3310-3550
    \(\ce{C-H}\) (aromatics) 3030 \(\ce{C-N}\) 1030, 1230
    \(\ce{C=C}\) (aromatics) 1500-1600 \(\ce{C=N}\) 1600-1700
    \(\ce{C-H}\) (aldehyde) 2700-2775, 2820-2900 \(\ce{C#N}\) 2210-2260
    \(\ce{C=O}\) 1670-1780

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  6. Problem 6: Sequence of peptide A

    Peptide A has a molecular weight of 1007. Complete acid hydrolysis gives the following amino acids in equimolar amounts: Asp, Cystine, Glu, Gly, Ile, Leu, Pro, and Tyr (see Table 1). Oxidation of A with \(\ce{HCO2OH}\) gives only B which carries two residues of cysteic acid (Cya which is a cysteine derivative with its thiol group oxidized to sulfonic acid).

    6.1 How many sulphonic acid groups are formed from oxidation of a disulfide bond ?

    Partial hydrolysis of B gives a number of di and tri-peptides (B1-B6). The sequence of each hydrolysis product is determined in the following ways.

    The N-terminal amino acid is identified by treating the peptide with 2,4-dinitrofluorobenzene (DNFB) to give DNP-peptide. After complete acid hydrolysis of the DNP-peptide, a DNP-amino acid is obtained which can be identified readily by comparison with standard DNP-amino acids.

    6.2 B1, on treatment with DNFB followed by acid hydrolysis gives a product, DNP-Asp. This suggests that B1 has aspartic acid at the N-terminus. Write down the complete structure of DNP-Asp at its isoelectric point (no stereochemistry required).

    Next, the C-terminal amino acid is identified by heating the peptide at \(100\,^\circ\mathrm{C}\) with hydrazine, which cleave all the peptide bonds and convert all except C-terminal amino acids into amino acid hydrazides, leaving the C-terminal carboxyl group intact.

    In this way N- and C-terminal amino acids are identified and the complete sequences of B1-B6 are as shown :

    B1 Asp-Cya B4 Ile-Glu
    B2 Cya-Tyr B5 Cya-Pro-Leu
    B3 Leu-Gly B6 Tyr-Ile-Glu

    Hydrolysis of B with an enzyme from Bacillus subtilis gives B7-B9 with the following compositions:

    B7 \(\ce{Gly-NH2}\) (Glycinamide)
    B8 Cya, Glu, Ile, Tyr
    B9 Asp, Cya, Leu, Pro

    6.3 Write down the sequence of B8, if DNP-Cya is obtained on treatment of B8 with DNFB followed by complete acid hydrolysis.

    6.4 If the N- and C-terminal amino acids of B9 are identified as Asp and Leu respectively, write down the sequence of B9.

    6.5 Write down the complete structure of A using abbreviation in Table 1, indicating the position of the disulfide bond.

    However, the calculated molecular weight of A based on the above sequence is 2 mass units higher than the experimental value. On careful observation of the mixture from complete acid hydrolysis of A, 3 molar equivalents of ammonia are also produced in addition to the amino acids detected initially.

    6.6 Suggest the revised structure of A and circle the site(s) of the structure to indicate all the possible source of ammonia.

    6.7 Using the information in Table 2, calculate the isoelectric point of A.

    Table 1: Formulae and symbols of common amino acids at isoelectric point

    Name Formula Three-letter symbol
    Alanine \(\ce{CH3CH(NH3+)CO2-}\) Ala
    Arginine \(\ce{H2NC(=NH)NH(CH2)3CH(NH3+)CO2-}\) Arg
    Asparagine \(\ce{H2NCOCH2CH(NH3+)CO2-}\) Asn
    Aspartic Acid \(\ce{HO2CCH2CH(NH3+)CO2-}\) Asp
    Cysteine \(\ce{HSCH2CH(NH3+)CO2-}\) Cys
    Cystine \(\ce{[SCH2CH(NH3+)CO2-]2}\) –
    Glutamic Acid \(\ce{HO2CCH2CH2CH(NH3+)CO2-}\) Glu
    Glutamine \(\ce{H2NCOCH2CH2CH(NH3+)CO2-}\) Gln
    Glycine \(\ce{^{+}H3NCH2CO2-}\) Gly
    Histidine (figure below) His
    Isoleucine \(\ce{CH3CH2CH(CH3)CH(NH3+)CO2-}\) Ile
    Leucine \(\ce{(CH3)2CHCH2CH(NH3+)CO2-}\) Leu
    Lysine \(\ce{H2N(CH2)4CH(NH3+)CO2-}\) Lys
    Figure 1

    Table 1 (continued)

    Name Formula Three-letter symbol
    Methionine \(\ce{CH3SCH2CH2CH(NH3+)CO2-}\) Met
    Phenylalanine \(\ce{PhCH2CH(NH3+)CO2-}\) Phe
    Proline (first figure below) Pro
    Serine \(\ce{HOCH2CH(NH3+)CO2-}\) Ser
    Threonine \(\ce{CH3CH(OH)CH(NH3+)CO2-}\) Thr
    Tryptophan (second figure below) Trp
    Tyrosine (third figure below) Tyr
    Valine \(\ce{(CH3)2CHCH(NH3+)CO2-}\) Val
    Figure 2
    Figure 3
    Figure 4

    Table 2: \(\mathrm{p}K_\mathrm{a}\) of some important groups in amino acids

    Groups Equilibrium \(\mathrm{p}K_\mathrm{a}\)
    Terminal carboxyl (first figure below) 3.1
    Asp /or Glu side-chain carboxyl (second figure below) 4.4
    His side-chain (third figure below) 6.5
    Terminal amino (fourth figure below) 8.0
    Cys side-chain (fifth figure below) 8.5
    Figure 5
    Figure 6
    Figure 7
    Figure 8
    Figure 9

    Table 2 (continued)

    Groups Equilibrium \(\mathrm{p}K_\mathrm{a}\)
    Tyr side-chain (first figure below) 10.0
    Lys side-chain amino (second figure below) 10.0
    Arg side-chain (third figure below) 12.0
    Figure 10
    Figure 11
    Figure 12

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