National German Competition

47. International

Chemistry Olympiad

Azerbaijan 2015

Volume 21

www.ShimiPedia.ir Chemistry Olympiad 2015

National German Competition 2015, Volume 21 Translated and published by Wolfgang Hampe

Contact addresses: IPN University of Kiel, z.H. PD Dr. Sabine Nick tel: +49-431-880-3116 Olshausenstraße 62 fax: +49-431-880-5468 24098 Kiel email: [email protected]

IPN University of Kiel, z.H. Monika Barfknecht tel: +49-431-880-3168 Olshausenstraße 62 fax: +49-431-880-5468 24098 Kiel email: [email protected]

Wolfgang Hampe tel: +49-431-79433 Habichtweg 11 24222 Schwentinental email: [email protected]

Association to promote the IChO (Association of former participants and friends of the IChO) Internet address: www.fcho.de

This booklet including the problems of the 47th IchO and the latest statistics is available as of September 2015 from http://www.icho.de ("Aufgaben")

4 www.ShimiPedia.ir Chemistry Olympiad 2015

Contents

Part 1: The problems of the four rounds

First round (problems solved at home) ...... 6 Second round (problems solved at home) ...... 10 Third round, test 1 (time 5 hours) ...... 18 Third round, test 2 (time 5 hours) ...... 24 Fourth round, theoretical test (time 5 hours) ...... 31 Fourth round, practical test (time 5 hours) ...... 42

Part 2: The solutions to the problems of the four rounds

First round ...... 47 Second round ...... 53 Third round, test 1 ...... 60 Third round, test 2 ...... 66 Fourth round, theoretical test ...... 72

Part 3: The problems of the IChO

Theoretical problems ...... 81 Practical problems ...... 95 Solutions ...... 110

Part 4: Appendix

Tables on the history of the IChO ...... 120

www.ShimiPedia.ir 3 Chemistry Olympiad 2015

4 www.ShimiPedia.ir Problems

Part 1

The problem set of the four rounds

www.ShimiPedia.ir 5 Problems Round 1

First Round (homework)

Problem 1- 1 Quite Old! Natural carbon consists of three isotopes. a) Which are these isotopes? How do they differ in their atomic composition?

One of these isotopes is radioactive with a half-life of t½ = 5730 years. b) How is this isotope generated in nature? Give a formation equation! The radioactive carbon isotope is an  emitter. c) Write down the decomposition equation. d) What is the law for the radioactive decay? e) What is the meaning of "half-life" of an isotope? Derive a general formula for the half-life, starting with the law for the radioactive decay. f) After which time does radioactive material stop to disintegrate? g) What is the influence of temperature, pressure and similar conditions on the rate of radioactive decay? Radiocarbon dating is an important method to determine the age of carbon containing material. h) Explain this method! Give the reason why examined materials have to be former living objects or their derivatives. The paper of a treasury map found in 2013 has a decay rate of 14.48 decays per g of carbon per minute. In natural carbon this decay rate is 15 decays per g of carbon per minute. i) Calculate the age of the map? Should the finder go on a treasury hunt? j) Account for the fact that the age of bones of dinosaurs cannot be determined by the method of carbon dating.

Problem 1-2 Fats and Oils Fats and oils are esters of carboxylic acids with glycerol. They can be solid, semisolid and liquid. Fats which are liquid at room temperature are called oils. Depending on their origin fats have a characteristic composition of fatty acids. a) Explain what "characteristic composition of fatty acids" means and show such a characteristic for an example chosen by you. There is a simple code to characterize fatty acids by giving the proportion of the total number of carbon atoms (m) and the number of double bonds (n) (as it is done in sport results). The code for oleic acid is 18:1. H H H H H H H H H H H H H H H H H O H C C C C C C C C C C C C C C C C C C OH H H H H H H H H H H H H H H H Fig. 1: Structure of oleic acid

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b) Write down the empirical formulae and plot the structures of the fatty acids below. Give their code. i) Capric acid ii) Linoleic acid iii) Linolenic acid iv) Erucic acid

Natural fats which contain (poly) unsaturated fatty acids are especially valuable for dietary intake. c) Explain by considering intermolecular interaction why saturated fats are solid while unsaturated fats are rather liquid. More than 90 % of fats produced worldwide serve as foodstuff for men or animal feed, a small part is raw material for producing other chemicals. There are special regions in a fat molecule where a chemical attack is possible. OH O

O CH OH O 2

O CH OH O

O CH 2 Fig. 2: Structural formula of ricinus oil d) Mark the regions where a chemical attack is possible and give a name to it. A possible reaction of a fat is the saponification with a solution of sodium hydroxide to give glycerol and soap. e) Give the reaction equation of the saponification of a fat (using structural formulae). Show the detailed mechanism of the saponification of one of the three fatty acid residues. Explain why this equilibrium reaction drifts to the products. The saponification of copra oil needs several hours while the saponification of 3-nitrobenzoic-acid methyl ester is finished after only ten minutes. f) Account for this fact. Soaps belong to the tensides. g) What characterizes a tenside? Explain the cleaning effect of tensides. h) To which kind of tensides do soaps belong? Which other kinds of tensides do exist? Give an example to each kind and draw the structural formula.

Problem 1-3 Oxygen ... Oxygen is the most frequent element on earth having a mass ratio of 49.4 %. a) Give five natural deposits of oxygen. The production of oxygen from air is technically performed by a special procedure in Germany known as Linde operation. b) How can oxygen be produced in a laboratory? Write down the names of three methods and the equations of the reactions.

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Besides oxygen nitrogen (consisting of N2 molecules) is the major component of air c) Are there differences between O2 und N2 molecules concerning their magnetic properties? Ac- count for your answer using MO diagrams.

Oxygen molecules (O2) are denoted as triplet oxygen. Moreover there is a more reactive species, the so called singlet oxygen. These names arise from the total spin multiplicity M which can be calculated with the formula M = 2·S + 1 (S = total spin). d) Account for the names of these two "kinds" of oxygen using this formula! O O In text books you often find the Lewis structure for dioxygen (O2) shown on the right hand side. Lewis structure of O2 e) Does this Lewis structure show the correct distribution of electrons in the O2 molecule? Account for your answer.

Problem 1-4 ... and Oxides Oxygen forms binary compounds with nearly all other elements. a) From which elements no isolatable oxygen compound is known until now? b) Write down the equations of the reactions of oxygen with i) White phosphor ii) Yellow sulfur iii) Lithium iv) Calcium Which acid/base properties do the aqueous solutions of the products show? In its covalent compounds oxygen has in most cases the coordination number 1, 2 or 3. c) Give one example for each of these three coordination numbers plotting its Lewis structure. Besides some exceptions the oxidation number –II is assigned to oxygen. d) Give four compounds in which oxygen has another oxidation number than –II. Do not name more than one compound for each different oxidation number. In quantitative analysis oxides can be used to determine the amount of other elements for example iron, cobalt, nickel, tin and aluminum. To determine the iron(III) content in a solution it is precipitated with ammonia, filtered through ash- free filters, washed with water and at the end with ammonium nitrate solution. The filter with the precipitate is given into a porcelain crucible and heated with a Bunsen burner, at first slowly and then up to maximal 700 °C. You have to pay attention that there are no reduction processes and that besides Fe2O3 no Fe3O4 is formed which would falsify the result of the analysis.

A sample of iron(III) chloride is weighed into a 250.0 mL measuring flask which then is filled with water up to the calibration mark. Three samples of 50.0 mL each are taken and treated as described above. Weight of the ash: 0.2483 g / 0.2493 g / 0.2488 g. e) i) Write down the equation for the formation of iron(III) oxide from iron(III) hydroxide.

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ii) What is the reason for washing with ammonium nitrate? Why should ammonium chloride not be used in the last step? Account for your answer. iii) Why is ash-free filter paper used?

iv) With which simple physical measurement could the unwanted formation of Fe3O4 be detected? v) Does the formation of Fe3O4 lead to a higher or to a lower calculated content of iron? Ac- count for your answer. vi) Calculate the mass of the iron(III) chloride sample which was given into the measuring flask.

Problem 1-5 ... another Oxide - An Elaborate Determination

290.0 mg of a metal oxide (MeO2), concentrated hydrochloric acid as well as a solution of potassium iodide are placed in an apparatus as shown below.

Fig. 3: Apparatus for a quantitative determination of a metal oxide The hydrochloric acid in the dropping funnel is pressed into the reaction vessel A using carbon dioxide. This mixture is heated to maintain light boiling for 30 minutes. During this time the apparatus is continuously purged with small amounts of carbon dioxide to remove the volatile content of vessel A. When the reaction in vessel A has finished the content of the washing flask C is quantitatively transferred into the Erlenmeyer flask B. The content of the Erlenmeyer flask B is titrated with a standard solution of sodium thiosulfate (c = 0.1 mol /L) until the mixture is nearly discoloured. Then some drops of solution of starch are added and the titration is continued until the solution is totally discoloured. 24.25 mL of the thiosulfate solution are consumed for the titration. a) Write down reaction equations for all reactions in the apparatus as well as for the reactions during the following titration. b) Why are drops of starch solution added? c) Account for the reason of such an elaborate procedure. What is the influence of air and water (moisture) on the accuracy of the determination? d) Determine the metal in the oxide.

www.ShimiPedia.ir 9 Problems Round 2

Second Round (homework)

Problem 2-1 Afraid of Water! The nitrate of M, dissolved in a small amount of water at 50 °C, is added to a 7.5-fold excess of a warm (also at 50 °C) aqueous solution of X. A solid Y forms which converts to compound Z · n H2O when cooled down. The mass ratio of the metal M in Z amounts to 33.04 %. The following information is given:  The metal M dissolves is nitric acid of the concentration of c = 8 mol/L but not in sulfuric acid or hydrochloric acid of the same concentration.  The metal M dissolves in a hot solution of sodium hydroxide.  Finely dispersed M ignites in air.  If a solution of sodium hydroxide is added to an aqueous solution of M a precipitate forms which dissolves in an excess of sodium hydroxide solution.  M in an aqueous solution does not form a complex compound with ammonia.  When dissolving X in water the solution cools down.  Y dissolves slightly in water.  As a solid Y has a layer structure.  The conductivity of solid Y rises when heated.  Y shows thermochromism.

 Z · n H2O can be dehydrated in a drying oven at 150 °C.

 Z and Z · n H2O decompose in water.

 When Z · n H2O is heated a gaseous compound G is formed besides water vapor and a metallic mirror of M is formed  X and G react in an aqueous solution with each other to yield compound H. a) Determine M, X, Y and Z. Account for your decision. b) Write down the reaction equation for the two step formation of Z using the compounds mentioned above. c) Determine H. Write down the equation of its formation from X and G! Draw the Lewis structure of H. Which structure do you expect according to the VSEPR model. Account for your answer.

The water containing compound Z · n H2O can be dehydrated in a drying oven. The image shows the

thermography (TG) plot.

Mass percentage Mass

Temperature/°C 10 www.ShimiPedia.ir Problems Round 2

d) Determine the number of water molecules n (n integer) in one formula unit of Z · n H2O using the observed decrease of mass.

Beside Z another compound was found, which consists of the same elements with the same oxidation numbers as in Z. The mass ratio of the metal M, however, has only the value of 26.13 %. e) Give the empirical formula of this compound. f) Give at least three compounds which lead to a decrease of temperature when dissolved in water. Give an explanation of this phenomenon! g) Explain the formation of the metallic mirror when Z · n H2O is heated. Write down the reaction equation and apply oxidation numbers. h) Give the equation for the reaction of an aqueous solution of the nitrate of M with an excess of an ammonia solution.

Problem 2-2 Organic Chemistry – Short and Crisp 5-Methylfuran-3-on (2) is needed as a reagent for the synthesis of a natural compound. It can be produced in four steps starting with acetylacetone (1): O O O EtE3t3, Nhexane,, Hexan , i) LDA, THF, NBS,NBS, DMCDCM, Et2O, TMSCl, RT, -78 °C, Rürefluxckfluss , K2CO3, A B 2 h C 48 h 30 min RT, 12 h O ii) TMSCl, C11H24O2Si2 0 °C, 1 h 1 2

Hints: - In the reaction of 1 to A two isomers form. 1 - In the H-NMR spectrum (A in CDCl3 at 298 K) of the mixture of isomers of A six singlets are found. The intensities are written below the signals, the chemical shiftabove the signals.

www.ShimiPedia.ir 11 Problems Round 2

- In the mass spectrum of C the following values for the molecular peak m/c are found (intensities in brackets): 177.96 (100.0%), 179.96 (97.3%), 178.97 (5.6%), 180.96 (5.3%) a) Draw the structural formulae of A, B and C and show how you derived the structures from the hints and the reaction equation. b) Which kind of isomers are formed in the reaction 1  A? Draw the structural formulae! c) Analyze the 1H-NMR spectrum of A: Assign all hydrogen atoms of b) to the corresponding signals. Determine the ratio of the two isomers of A. Calculate the intensities of the different signals using the ratio of isomers. d) Why do you find two peaks with approximately equal intensities at 177.96 and 179.96 in the mass spectrum of C? Explain! e) In the first step of the synthesis of 1  A triethylamine, Et3N, is used. What is it needed for? Explain! f) In the second step of the synthesis of A  B lithium diisopropylamide, LDA, is used. Is it possible

to use Et3N here analogue to the reaction of 1  A? Account for your answer. g) Would a reaction of acetylacetone with N-bromosuccinimide, NBS, in the presence of a base lead to C, too? Why is the way via A and B used in the synthesis? Give an explanation!

In an aldol reaction followed by an elimination of H2O product 2 is linked to another structural element. The unit 3 is formed which you find in the natural compound, too.

In the natural compound solved in a deuterized solvent an H/D exchange is observed which can be explained by keto-enol tautomerism. h) Show a mechanism of the H/D exchange at the methyl group of the structural unit 3. How many of H/D exchange reactions are possible?

Problem 2-3 Hückel Theory The Hückel theory can be used to describe π electron systems qualitatively. The p orbitals which are involved in the π bonds (per definition pz orbitals) have to be taken into consideration. The π molecular orbitals (MO) are linear combinations of them. By doing so bonding and antibonding MOs are formed. If they are accommodated similarly a non-bonding interaction occurs. A simple example is ethene (Fig. 2). The energy levels of ethene result from bonding and antibonding interaction and contain the empirical parameters  (energy of the electron in the isolated atom) and β (coupling of the atom orbitals in the molecule, β > 0). The total π energy of ethene results from the occupation of the energy levels with two electrons as follows

E = Σi ni εi = 2 (α + β) + 0 (α – β) = 2 α + 2 β,

with ni = number of electrons in MO i and εi = associated energy of the MO.

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antibonding

bonding

Fig 2: π Molecular orbitals of ethene following the Hückel theory

A Hückel MO scheme can also be established for cyclic and planar π systems. There is a simple “trick” for working out the orbital energies: Frost-Musulin diagrams. Starting from the energy level of the atom orbitals () a circle with the radius 2β is drawn. The molecular framework of an n-cyclic system is then drawn as a polygon into the circle with one atom put at the bottom. The atomic positions then map on to the energy level diagram of the π system. Fig. 3 shows the Frost-Musulin diagram of benzene.

Fig. 3: Frost-Musulin diagram of benzene If you compare the π bonding energy of benzene with that of hexatriene you find stabilization by delocalization of 1.0β. This is often interpreted as the so called aromatic stabilization energy: Ben- zene gains additional bond energy and is called an "aromatic" compound. Anitaromatic compounds do not show such an effect, they are unstable. a) Design the Frost-Musulin diagram of a planar 4- and 7-cyclus. Give the charge of the molecule for different electron configurations (4-membered ring: 2, 4 and 6 π electrons; 7-membered ring: 6 and 8 π electrons) and calculate the stabilization β compared to the open chain versions of the molecules (ions). Are these compounds (ions) aromatic? (The π energy levels of the aliphatic versions are: Butadiene:  ± 1,618 · β;  ± 0,618 · β; Heptatrienyl cation: ;  ± 1,848 · β;  ± 1,414 · β;  ± 0,765 · β). b) Which are the conditions a compound has to fulfill in order to be called aromatic?

www.ShimiPedia.ir 13 Problems Round 2

c) Indicate which of the following molecules and ions are aromatic and which are not. Account for your decision by using the conditions of b)- i) Pyrrole ii) Allyl anion iii) Azulene iv) 1H-Pyrrolium cation v) Pyridinium cation vi) Caffeine.

The shape of the wave function of a special energy level can be found by linear combination of the pz orbitals. The result for the cyclopentadienyl anion (Cp–) is shown in fig. 4:

Fig. 4: Frost-Musulin diagram of cyclopentadienyl anion with the MOs These MO diagrams are of great importance in coordination chemistry. The coordination mainly takes place by HOMO (highest occupied molecular orbital) - LUMO (lowest unoccupied molecular orbital) interaction. d) Write down the HOMO and LUMO of the cyclopentadienyl anion! An iron(III) cation is coordinated by a cyclopentadienyl anion (Fig. 5).

Fig 5: Orientation of the η5-cyclopentadienyl iron(III) cation e) Determine the interactions between the valence orbitals of the iron cation (3d, 4s und 4p) and the frontier orbitals of the cyclodienyl anion. Use the orientation in the coordination system as shown in fig. 5. Record your results in a table as shown below (x = interaction expected, – = no interaction expected). Account for your assignment by one example for HOMO and LUMO each with the help of orbital plots which illustrate the interaction.

- - Orbital Fe HOMO (Cp ) LUMO (Cp ) 3dx2-y2 3dz2 ... In ferrocene two Cp- ligands are coordinated to one iron(II) cation. In this complex iron has 18 valence electrons and thus a stable noble gas configuration. This "magic" number is found in lots of coordination compounds.

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5 5 f) Which compound is more stable: [Rh(η -Cp)2] or [Ru(η -Cp)2]? Which redox property should the less stable compound have according to the configuration of the valence electrons at the metal center? Account for your answer! 5 g) The complex [(η -Cp)2Ru2(CO)4] may have three diasteromeric structures. Draw 3-D structures and give the number of valence electrons at each metal center.

Problem 2-4 Disproportionation of Copper 0.168 g of copper(II) nitrate is dissolved in water to yield 100 mL solution. The pH of this solution is 4.4. a) Why does an aqueous solution of copper(II) nitrate react acidic? Write down the reaction equa-

tion! Calculate the pKa value of the first step of protolysis. b) Determine the pH from which copper hydroxide precipitates in a solution with c(Cu2+) = 1.03·10–2 –19 mol/L. (Ksp, 25°C = 1.6 · 10 ) There are two redox equilibria for Cu+ ions: + – o Cu + e Cu E 1 = 0.52 V 2+ – + o Cu + e Cu E 2 = 0.16 V c) Write down the reaction equation for the disproportionation of Cu+ ions and calculate the equilibrium constant for this reaction at 22 °C. d) Which oxidation state of copper ions should be the most stable according to its electron configuration? Which one is the most stable in an aqueous solution? Account for your answer. e) 10 mmol of copper(I) nitrate are dissolved in 1 L of water at 22 °C. Calculate the composition of the copper containing species in mol/L. (Use K =1.72 · 106 instead of the result in c). ) Copper(I) oxide is suspended in a copper(II) solution of c(Cu2+) = 0.01 mol/L at 22 °C. –15 (Ksp (CuOH) = 1.0 ·10 ) f) Calculate the pH from which copper(I) oxide is stable in an aqueous solution. Which influence does the temperature have in the range from 0 °C to 100 °C? Draw a graph!

www.ShimiPedia.ir 15 Round 3 Test 1

Problems Round 3

The top 60 of the participants of the 2nd round are invited to the 3rd round, a one-week chemistry camp.

Test 1 Göttingen 2015: Problems 3-01 to 3-09 Test 2 Göttingen 2015: Problems 3-11 to 3-20

time 5 hours. your name write it on every answer sheet. relevant calculations write them down into the appropriate boxes. otherwise you will get no points atomic masses use only the periodic table given. constants use only the values given in the table. answers only in the appropriate boxes of the answer sheets, nothing else will be marked. draft paper use the back of the pages of the problem booklet, but everything written there will not be marked. problem booklet you may keep it.

Good Luck

16 www.ShimiPedia.ir Problems Round 3 Test 1 + 2

Useful formulas and data

G0 = H0 - T·S0 G0 = - E·z·F G0 = - R·T·ln K

−H0 G = G0 + R · T· ln Q ln (Kp /Kp ) = ·(T -1 - T -1) 1 2 R 1 2 p·V = n·R·T for ideal gases and osmotic pressure R ·T Nernst equation : E = E0 + ·ln (c /c ) z ·F Ox Red R ·T for metals E = E0 + ·ln (c(Mez+/c0) z ·F R ·T for non-metals E = E0 + ·ln (c0/c(NiMez-) z ·F + 0 0 R ·T c(H )/c for hydrogen E = E + ·ln 0 1/2 z ·F (p(H2)/p ) with c0 = 1 mol/L, p0 = 1.000∙105 Pa

Rate laws 0. order c = co - k·t

k1 t 1. order c = co· e -1 -1 2. order c = k2·t + co Arrhenius equation: k = A ∙ e-Ea/(R∙T) A pre-exponential factor

Ea activation energy Law of Lambert and Beer: A = ·c·d A absorbance  molar absorption coefficient d length of the cuvette c concentration I I Transmission T = Absorbance A = lg 0 with I = intensity I0 I n Freezing point depression T = K · m(Solvent) n amount of particles dissolved K cryoscopic constant Speed of light c = 3.000∙108 ms-1 Gas constant R = 8.314 JK-1mol-1 Faraday constant F = 96485 Cmol-1 23 -1 Avogadro constant NA = 6.022·10 mol po = 1.000·105 Pa 1 atm = 1.013·105 Pa 1 bar = 1·105 Pa 1 Å = 10-10 m

A periodic table was provided

www.ShimiPedia.ir 17 Round 3 Test 1

Third Round Test 1

Problem 3-01 Basic Knowledge A Fill in the missing numbers: a)  Na2S2O3 +  I2  NaI +  Na2S4O6 2+ - - - - b)  Ba +  MnO4 +  CN +  OH  BaMnO4 +  CNO +  H2O - + - - c)  ClO3 +  H3O +  Br  Br2 +  Cl +  H2O

B Match the possible colors with the given solid compounds and solutions of ions. Some colors may be matched with more than one substance; some may not come into consideration. Ions dissolved in water (c = 1 mol/L) Sample of solid of Possible colors Fe2+ in acidic solution Iron sulfide (FeS) Black, white, colorless, 3+ Al in acidic solution Copper sulfate (CuSO4) yellow, yellowish brown, Cu2+ in acidic solution Silver iodide (AgI) light green, 2+ Cu in ammoniac solution Potassium sulfate (K2SO4) blue, deep blue, - Cl in basic solution Potassium permanganate (KMnO4) red, violet, + Na Potassium chromate (K2CrO4)

C Which of the following compounds are sparely soluble? Silver bromide, potassium nitrite, lead sulfate, calcium chloride, sodium fluoride, iron(II) sulfide.

D Write down the formulae of the following compounds: Barium nitrate, potassium oxalate, aluminum oxide, potassium manganate(VI), potassium aluminum sulfate, sodium carbonate decahydrate.

E Given are sample of the elements below. Which sample of these are totally soluble in an excess of dil. hydrochloric acid (c = 2 mol/L)? Potassium, lead, aluminum, copper, zinc, silicium.

F Given are sample of the elements below. Which samples of these are totally soluble in an excess of dil. nitric acid (c = 2 mol/L)? Potassium, lead, aluminum, copper, zinc, silicium.

Problem 3-02 Given the following galvanic cell (T = 298 K): Cu(s) | Cu2+(aq) c = 1.00 mol/L || Ag+(aq) c = x mol/L) | Ag(s). a) Write down the equation for the cell reaction. The voltage U for different values of x has been measured: x 0.1000 0.0500 0.0100 0.0050 0.0010 U in V 0.403 0.385 0.344 0.326 0.285

18 www.ShimiPedia.ir Round 3 Test 1

b) Plot U as a function of lg x! c) Write down the cell reaction for x = 0.0200 mol/L and determine the voltage. d) Calculate the equilibrium constant K for the cell reaction. 3.00 g of potassium iodide are dissolved in water, the solution is filled up to 50.0 cm3. 50.0 cm3 of silver nitrate solution (c = 0.200 mol/L) is added. If this solution replaces the silver nitrate solution in the galvanic cell the copper electrode becomes the cathode and a voltage of 0.420 V is measured. e) Calculate the solubility product of silver iodide.

Cu2+ + 2 e– Cu E° = 0.34 V Ag+ + e– Ag E° = 0.80 V

Problem 3-03 Interhalogen Compounds Compounds between different halogens are called interhalogen compounds. Besides the diatomic compounds XY there are compounds with more than two atoms. Their formulae are generally XYn, where n = 1, 3, 5 or 7, and X is the less electronegative of the two halogens. The tendency to form interhalogen compounds with more than two atoms rises with increasing atom mass of X and decreasing mass of Y. a) Give an example for an interhalogen compound with more than two atoms which should be existent and an example which should rather not be existent.

Diatomic interhalogens can be formed from the elements. All combinations are known. There are the following natural isotopes of the halogens in existence: 19F, 35Cl, 37Cl, 79Br, 81Br, 127 I. b) Write down the empirical formulae of the interhalogens XY and give the number of molecular peaks in the mass spectrum of each compound XY. c) Which shape should the molecules XY3, XY5 and XY7 have following the VSEPR model? Sketch 3-D figures.

If exposed to water interhalogens disproportionate as halogens do, too. But also without water a disproportionation of many of these compounds may take place. d) Write down the equation of the reaction of XY with water (X is less electronegative than Y). e) Write down the equation of a possible disproportion reaction of XY in the absence of water (X is less electronegative than Y).

Problem 3-04 Hess's Law and Reaction enthalpies A thermally very well insulated calorimeter was filled with water of 22.55 °C. When adding 1.565 g of zinc sulfate the temperature went up to 23.52 °C after zinc sulfate had totally dissolved. In a second experiment the same calorimeter was filled with water of 22.15 °C. After adding and dissolving 13.16 g of zinc sulfate heptahydrate the temperature went down to 21.84 °C. In both cases the heat capacity of the system was 0.900 kJ/K.

www.ShimiPedia.ir 19 Round 3 Test 1

a) Calculate the enthalpy HR of the reaction ZnSO4 + 7 H2O  ZnSO4 · 7 H2O b) Calculate the enthalpy of formation of nitrous acid (HNO2) in aqueous solution at constant pressure and temperature from the given reaction enthalpies.

(1) NH4NO2(s)  N2(g) + 2 H2O(l) H1 = -307.4 kJ/mol

(2) 2 H2(g) + O2(g)  2 H2O(l) H2 = -571.7kJ/mol

(3) N2(g) + 3 H2(g) + aq  2 NH3(aq) H3 = -161.7 kJ/mol

(4) NH3(aq) + HNO2(aq)  NH4NO2(aq) H4 = -38.1 kJ/mol

(5) NH4NO2(s) + aq  NH4NO2(aq) H5 = +25.1 kJ/mol

Problem 3-05 Lithium Lithium is the lightest metal and the least dense solid element. The small density is traced back to the fact that it has as well as the other alkali metals a body centered cubic structure. face centered cubic (fcc) body centered cubic (bcc)

a) Calculate the packing fraction (in percent) of an fcc and of a bcc cell. b) What is the percentage difference of the density between these two structures if the atoms are assumed to be of the same kind?

Spodumene, LiAlSi2O6, is an important source of lithium. It is chemically opened up with CaCO3 and converted into the oxides of aluminum and lithium. When treated with water lithium hydroxide crys- tallizes as a monohydrate. c) Write down the equation of the reaction of spodumene with CaCO3.

Using hydrochloric acid lithium hydroxide can be converted into lithium chloride which is taken to obtain pure lithium by fused-salt electrolysis. In another process lithium can be obtained by electrolysis of a solution of lithium chloride in pyridine or acetone. This solubility can be used to separate it from sodium chloride and potassium chloride. d) Give the reason why lithium chloride is soluble in solvents as pyridine, acetone and alcohols in contrast to the other alkali chlorides.

In the qualitative analysis lithium is very difficult to detect because it forms nearly no poorly soluble compound. Quantitatively it can be determined gravimetrically as sulfate or aluminate (formal: x

Li2O · y Al2O3). In literature (H. Grothe, W. Savelsberg (1937). Über die analytische Bestimmung des Lithiums. Z. analyt. Chem. 110, 81 – 94) the following instruction is given (translated from German):

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"III. Specification of a procedure to determine lithium. A. Precipitating agent: 50 g of potassium aluminum sulfate are dissolved in 900 mL of warm water. The solution is cooled down and a conc. solution of 20 g of sodium hydroxide is added while stirring and cooling, until the formed precipitate is totally dissolved. After a long time (over- night) the solution is filtered and the pH is brought to 12.6. Then the solution is filled up to 1 L. B. Procedure of detection: The solution of lithium is brought to pH = 3. Then for each 10 mg of lithium 40 mL of the precipitating agent is added. The solution is brought to pH = 12.6 again by using some drops of sodium hydroxide solution (c = 1 mol/L). After a short time the solution can be filtered. The precipitate is decanted with a lot of cold water, brought on a filter paper and washed with cold water until the washing water does not redden phenolphthalein. The filter

paper is burned off and the residue (x Li2O · y Al2O3) weighed." e) What could cause the turbidity when potassium aluminum sulfate ((KAl(SO4)2 · 12 H2O) is dissolved? Why can it be dissolved again by using sodium hydroxide solution? Give the equations for the relevant reactions. f) Why does the pH in this determination have be kept at exactly 12.6?

Exactly 0.1980 g of the residue of the ashing is brought into solution. The content of aluminum is detected by complexation. 50.00 mL of a solution of Na2EDTA (c = 0.10 mol/L) is added to the solution and the Na2EDTA which has not reacted is titrated with zinc sulfate solution (c = 0.10 mol/L) with xylenolorange (Ind) as indicator. Consumption: 15.25 mL. 2– g) Write down the equation for the complexation reaction of aluminum and Na2EDTA . Use H2Y for the EDTA component. h) Why are complexometric determinations often executed in buffer solutions (NH3/NH4Cl or AcOOH/NaOOAc)? i) Arrange the complex compounds of this determination (EDTA – Zn2+, EDTA – Al3+, Ind – Zn2,+ Ind – Al3+) in the direction of decreasing stability. j) In the original instruction the stoichiometric factors are replaced by x and y. Determine the empirical formula of lithium aluminate in the residue of the ashing.

Problem 3-06 The dependence of the equilibrium constant on temperature of the reaction

PCl5 PCl3 + Cl2 can be expressed by the equation log Kp = -4374/(T/K) + 1.75·log(T/K) + 3.78. a) Calculate Kp at 200 °C. The reaction proceeds under isothermal-isobaric conditions at a temperature of 200 °C and a pressure of 150 kPa in a vessel with variable volume until equilibrium is reached. b) Calculate p(PCl5) and p(PCl3 in equilibrium. Use in this case Kp= 0.200. c) Calculate the rate of conversion of PCl5 (in %).

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Phosphorus pentachloride is a molecular compound only if heated above 160 °C. Below this temperature it forms an ionic solid. d) Which are the ions in the solid? Give the Lewis structure of these ions and draw their 3-D image.

Compounds as PCl5 and PF5 have a trigonal bipyramidal shape following the VSEPR model. In a trigonal bipyramid you can distinguish axial and equatorial positions. Nevertheless only one single fluo- 19 rine signal in the F NMR spectrum of PF5 is detected. e) Account for this fact.

Problem 3-07 Isomerism

Draw the structures of all isomers with the empirical formula C3H6O and write down their names. (Ignore stereo isomerism)

Problem 3-08 Aromatic Compounds Given the following reaction scheme: KMnO B 4 C

i / ii

iv / v i / iii A

Isomeri- H / Pd/C 2 zation

vii F

iv / v CH2O

1. NaNO2 /

vi - H2O 2. KI G E

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a) Complete the structural formulae A – G as well as the empirical formulae of the reagents i - vii. The isomerization is known by a special name, which one? b) Which range of temperature has to be taken to produce G? Which product is formed if the solution (without KI) is heated to boiling? c) What is the resonance effect of the substituent F in case of a second substitution? Show with the help of resonance structures in which position the substituent directs. How does your result match with the product which forms in the reaction of F with iv/v? Explain why the product shown is formed.

Problem 3-09 Michael Reactions α,β-unsaturated carbonyl compounds are typical Michael systems showing an interesting reactivity. On one hand they decolor an aqueous solution of bromine as alkenes do and, like ketones, reactions with nucleophiles occur. On the other hand the conjugated system of a keto group and an alkene double bond shows some unique reactions. a) Show the mechanism of the reaction of but-3-en-2-one with bromine. Give the (general) name of the intermediate cation! b) Write down the equation for the reaction of hydrogen cyanide (HCN) with the keto group of the butenone.

The Grignard reaction of cyclohexenone and butyl magnesium bromide gives 3-butylcyclohexane-1- one in addition to the classical product 1-butyl cyclohex-2-en-1-ol:

c) Using resonance structures account for the fact that there can be an addition not only at the carbonyl group but also at the double bond. Is it an electrophilic or a nucleophilic addition?

Besides Grignard compounds many other organometallic reagents are used to form a C-C linkage e.g.

BuCeCl2 and Bu2CuLi both of which are produced in situ. The concept of hard and soft acids and bases (HSAB concept of Pearson) allows a good valuation whether the addition proceeds directly (at the carbonyl carbon atom) or rather conjugated (at the β-carbon atom). The gist of this theory is that soft acids react faster and form stronger bonds with soft bases, whereas hard acids react faster and form stronger bonds with hard bases, all other factors being equal. Cer compounds are known to be very hard Lewis acids while copper(I) compounds are rather weak. d) Explain which C atom in cyclohex-2-en-1-one is harder, which one is weaker. At which C atom

would you expect a reaction with BuCeCl2 and Bu2CuLi, respectively?

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Third Round Test 2 Problem 3-11 Multiple Choice With one or more correct answers even if the question is written in singular. a) A member of which group of compounds is generated by the oxidation of a secondary alcohol?

A Aldehyde B Carboxylic acid C tert. Alcohol D Peroxide E Ketone b) Which compound contains an element with the same oxidation number as chromium in

K2Cr2O7? 3- + A Cl2O2 B Fe(CN)6 C VO2 D K2MnO4 E H2S2O8 c) Which of the following mixtures is a buffer solution?

A CH3COOH (50 mL; 0.1 mol/L) + NaOH (50 mL; 0.1 mol/L) B CH3COOH (50 mL; 0.1 mol/L) + NaOH (50 mL; 0.05 mol/L) C CH3COOH (50 mL; 0.05 mol/L) + NaOH (50 mL; 0.1 mol/L) D CH3COOH (50 mL; 0.05 mol/L) + NaOH (50 mL; 0.05 mol/L) d) Which ion has unpaired electrons at its disposal? A Cu+ B As4+ C Zn2+ D Ag+ E Mn5+ e) Which compound does puff up cakes?

A CaCO3 B (NH4)2CO3 C Ca(COO)2 D NaHCO3 E Yeast f) The image1 shows the phase diagram of water. Which of the following statements is correct?

With rising pressure

A the boiling temperature decreases and the fluid

Pressure in kPa in Pressure melting temperature increases slightly. solid B the boiling temperature increases slightly and the melting temperature decreases. C the melting and boiling temperature increase. gaseous D the melting and boiling temperature decrease. E the melting and boiling temperature do not change. Temperature in °C g) Which of the following formulae represent more than one compound?

A CH4O B C2H2Cl2 C Pt(NH3)2Cl2 D CuSO4·5H2O E C2H6O h) Which of the following molecules and ions are planar? 2- A C2H4 B PH3 C COCl2 D PtCl4 E CH4

1 Image from "chemikerboard.de" 24 www.ShimiPedia.ir Problems Round 3 Test 2

Problem 3-12 Buffer Action and Acidity

A buffer solution with pH = 5.8 has to be made from a diluted acid (pKS = 6.5) and its sodium salt. a) Calculate the ratio of amounts of acid and conjugated base. b) Calculate the concentration of formiate ions at pH = 4.2.

There is 1 L of a buffer solution which contains 0.1 mol of NH3 and 0.1 mol of NH4Cl. The pH value of this solution is 9.25. c) Which volume of hydrochloric acid (c = 1 mol/L) and sodium hydroxide solution (c = 1 mol/L), respectively, can be added with the result that the pH does not change more than 1?

A solution of 2.895 g of an unsubstituted weak carboxylic acid X in 500 g of water shows a freezing- point depression of 0.147 K. If 0.957 g of sucrose (C12H22O11) is dissolved in 100 g of water a freezing- point depression of 0.052 is detected. d) Which saturated carboxylic acid was used? e) Determine the acidity constant Ka of this acid and , the degree of protolysis.

Problem 3-13 Reactions

There are six aqueous solutions of the following compounds: NH4Cl, BaCl2, Na2S, Pb(NO3)2,

Na2SO4 and AgNO3. a) Record in the table on the answer sheet whether a reaction takes place (e.g. "yellow prec." if a yellow precipitate forms or "gas formation" ect.) or "n.r." if nor reaction takes place. b) Write down the equations for all reactions (indicate the aggregate state and the hydration by using (s), (l), (g), (aq))! Is it possible to identify the original solutions only by the results of the reactions without using more utilities? Account for your decision!

Problem 3-14 Double-Contact Process The double contact process is used to synthesize sulfuric acid on an industrial scale. During the decisive step in this process sulfur dioxide is oxidized to sulfur trioxide:

2 SO2 + O2  2 SO3 (1). Given are the following thermodynamic values* at 25 oC and po = 1.000·105 Pa (Standard pressure). 0 -1 o -1 -1 -1 -1 Hf in kJ·mol S in J·mol ·K Cp in J·mol ·K

SO2 (g) -297.00 249 39,9

O2 (g) 0 205 29,4

SO3 (g) -396.00 257 50,7 * values taken from Atkins, Physical Chemistry 3rd Edition

Cp is the molar heat capacity at constant pressure. You can use it to determine the enthalpy of formation and the entropy of formation of a compound at a temperature differing from the standard 0 0 0 temperature:  Hf (T) =  Hf + Cp·(T – 298 K), S(T) = S + Cp·ln(T/298 K).

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O a) Determine Kp at 600 C by using the Cp values.

To produce SO2 at first sulfur is burnt at 1400 °C to1500 °C under oxygen deficiency conditions. Af- terwards it is oxidized completely to SO2 at 700 °C in an excess of air. b) Why do they work at first at 1500 °C under oxygen deficiency conditions and why is the total oxidation completed at 700 °C?

In doing so you get a mixture of 10% (V/V) of SO2, 11% (V/V) of O2 and 79% (V/V) of N2 which is practically free of SO3. This mixture is led through the contact reactor. At 600 °C and standard pressure the equilibrium is established. c) Calculate the volume percentage of the components of the gas mixture at equilibrium (in %).

Determine the degree of conversion of SO2 (in %). (Use Kp = 65.00 here. At the end of the calculation there will arise an equation of third order with only one real solution. The result should have two decimals.

Problem 3-15 A Well Near a Volcano

The water taken from a volcano well is tested for hydrogen sulfide. A sample of 10.0 mL is taken and all hydrogen sulfide is removed with a stream of carbon dioxide. It is absorbed in bromine water and then the excess of bromine is removed. The acidity of the sample is titrated with a solution of NaOH (c = 0.100 mol/L) using methyl red as indicator. 19.95 mL are being consumed. a) Write down balanced equations for all reactions of this method. Calculate the hydrogen sulfide content of the water in g/L.

Another method to measure the hydrogen sulfide content works without a stream of carbon dioxide using a different redox process. If a known amount of iodine is produced in a solution it will oxidize hydrogen sulfide. The excess of iodine can be titrated with thiosulfate.

6.50 g of KIO3 are dissolved in water to give a solution of 1L. 10.0 mL of this solution are taken and 0.5 g of KI and 5 drops of starch solution are added. Then 10.00 mL of the volcano water are added, too, and the mixture acidified with 10 mL of 10 % hydrochloric acid. After 5 minutes the mixture is titrated with a solution of Na2S2O3 (c = 0.100 mol/L) until the blue colour disappears. (Hint: Hydrogen sulfide is oxidized by bromine to form a sulfate and by iodine to form sulfur.) b) Write down balanced equations for all reactions of this second method. Which volume of the

Na2S2O3 solution do you expect to be consumed in the titration? c) The silver bracelet of the technician who worked with the samples has blackened. Explain by a reaction equation.

26 www.ShimiPedia.ir Problems Round 3 Test 2

Problem 3-16 Gaseous Compounds A An airbag is a safety device in vehicles. It is an occupant restraint system consisting of a flexible fabric envelope or cushion designed to inflate rapidly during an automobile collision. Older airbag formulations contained sodium azide and other agents including potassium nitrate and silicium dioxide. An electronic controller detonates this mixture during an automobile crash forming sodium and nitrogen. After the nitrogen inflated the cushion the temperature of the gas is appr. as low as 150 °C due to the expansion. a) Write down the reaction equation of the decomposition of sodium azide. b) Draw the Lewis formula of the azide ion.

Since sodium metal is highly reactive the KNO3 and SiO2 react and remove it, in turn producing more

N2 gas following the (not arranged) equations (2) and (3):

Na + KNO3  K2O + Na2O + N2(g) (2)

K2O + Na2O + SiO2  K2SiO3 + Na2SiO3 (silicate glass) (3) c) Arrange the equations (2) and (3). d) Calculate the mass of sodium azide necessary to fill a cushion of 50.0 L (at 150 °C, 1300 hPa).

– Until 1999 N2 and N3 were the only stable nitrogen containing particles which could be produced in + a larger scale. In 1999 the discovery of another such a non-cyclic particle was published: N5 . The following structure was shown: + (K.O.Christie u.a. : N5 ; ein neuartiges 1,33 A homoleptisches Polystickstoff-Ion mit 108° 166° 1,12A hoher Energiedichte, Angew. Chem 111 (1999) 2112 – 2117) e) Draw three mesomeric resonance structures of this particle which are consistent with this structure.

B 20 cm3 of a gas X are filled into a measuring tube. 80 cm3 of oxygen are added and the mixture is ignited. When the pressure and the temperature of the beginning are restored you observe a de- cline of volume of 10 cm3. There is some oxygen left in the mixture after the reaction. f) Which of the following gases could be X? Hydrogen, ammonia, carbon monoxide, ethene, methane. Distinguish ϑ > 100 °C and ϑ < 100 °C.

Problem 3-17 Decay of Arsine

At 500 °C arsine, AsH3, decomposes quickly and totally in a reaction of first order to give arsenic and hydrogen. a) Write down the reaction equation and the respective rate equation.

The kinetics of the decomposition was studied at a lower constant temperature in a closed tube.

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In the beginning of the experiment there was pure gaseous arsine in the tube with a pressure of p0 = 86.1 kPa.

After 120 minutes the pressure has increased to p120 = 112.6 kPa. b) Determine the pressure at the end of the decomposition. c) Determine the rate constant and the half-life of the reaction. d) What amount of time is necessary to decompose 99 % of arsine? (Use in this case k = 1.3·10-4 s-1.)

Problem 3-18 Epoxides Epoxides are very reactive heterocyclic compounds and therefore of great importance in the organic synthesis. a) Give a reason for the great reactivity of the epoxides.

They are divided into symmetric and asymmetric epoxides:

O O

A B

ssymmetricymmetrisc h a s yasymmetricmmetrisch

Compound B can be synthesized in the following sequence of reactions starting from dimethyl sulfide: O

Base H CH3I O S C D -HI -S(CH3)2 B b) Complete the structural formulae of C and D (D is a sulfur ylide, a neutral dipolar molecule containing a formally negatively charged atom attached to a heteroatom with a formal positive charge). c) Show the mechanism of the reaction of D to the epoxide B using structural formulae.

Epoxides can be opened by a lot of nucleophiles to form bifunctional alkanes. The reaction of oxirane A with water, for example, catalyzed by an acid gives ethylene glycol. This reaction proceeds with another mechanism than the reaction without an acid. d) Show a possible mechanism of the ring-opening catalyzed by acid using structural formulae. e) Complete the structural formulae of the products E – J of the following reactions:

28 www.ShimiPedia.ir Problems Round 3 Test 2

O NH O H S 3 E 2 H

O CH OH / H+ O HBr 3 F I

1. RMgX + O LiAlH4 O 2. H O/H G 2 J

If an epoxide is treated with Lewis acids (LSkat), such as BF3 or MgI2 isomerization to a carbonyl compound takes place. Thereby a symmetric epoxide forms one isomer while asymmetric epoxides form more of them. f) Draw the structural formulae of K, L and M.

LS O kat. K

LS O kat. L M

Problem 3-19 Amino Acids L-α-amino acids play an important role in all creatures. Histidine (on the right) has a pKa2 value of 6.00. It is the only amino acid which contributes to the capacity of the blood buffer (pH = 7.40). Glutamic acid (2-aminopentane diacid) and aspartic acid (2-aminobutan diacid) are important neurotransmitter. Cysteine (2- amino-3-mercapto propanoic acid, mercapto = thiol) plays an important role in the structure of proteins because it can form the stabilizing disulfide bridges. Two cysteine molecules can be linked together with a disulfide bridge to form dicysteine (cystine). a) Which species of histidine are present in the equilibrium at the pH values of 1.82 and 9.17, re-

spectively, which are the other pKa values of histidine? Only one of the two nitrogen atoms in the ring can be protonated, which one? Rationalize your answer. b) Draw L-glutamic acid in the Fischer projection. c) Draw 3-D structural formulae of S-aspartic acid as well as of R-cysteine. ((Hint: in front of the paper plane, behind the paper plane, in the paper plane)

In proteins and peptides the amino acids are linked with peptide bonds – the resulting compound is in chemistry called a carboxylic acid amide. In some text books the reaction is shown in the following way:

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This reaction proceeds actually in this way only under extreme conditions because the carboxylic acid is not sufficiently reactive towards a nucleophilic attack (low carbonyl activity). d) Which electronic effect is responsible for the low carbonyl activity of the carboxylic acids?

In order to raise the reactivity of the carboxylic acids they can be converted into halides or anhy- drides. e) Show the total mechanism of the formation of an amide between a reactive species R-COX and

an amine R‘NH2. Which geometrical structure shows the reaction center directly after the attack of the amino group? f) Write down the equation for the formation of dicysteine from cysteine (with structural formulae without stereochemical concerns). Which type of reaction is it?

Problem 3-20 Organic Puzzle a) Draw the structural formulae of the following compounds. A Acetaldehyde B Acetone C Acetophenone D Methanoic acid E Aniline F Benzaldehyde G Nitrile of benzoic acid H 3-Nitrobenzoic acid I Phenol b) Assign each compound (using the letters of the table above) to a cell of a 3x3 table given on the answer sheet. Consider thereby the criteria below.

1. Line: The aqueous solution shows an acidic reaction. 2. Line: The compound can react with itself in an aldol reaction. 3. Line: It in a monosubstituted aromatic compound. 1. Column: The compound contains an aldehyde group as a functional group. 2. Column: It is an aromatic compound, the substituent(s) of which directs newly incoming substituents into meta position. 3. Column: Compounds without such criteria.

30 www.ShimiPedia.ir Problems Round 4 (theoretical)

Fourth Round (theoretical problems) (A periodic table and the same list of formulae and data as in the third round were provided) The top 15 of the 3rd round are the participants of the 4th round, a one-week practical training.

Problem 4-01 Traces of Water

The compound KPbI3 can be used to detect minimal amounts of water qualitatively. To detect water quantitatively other methods are used.

Method A: Heating / Annealing The substance is heated until its mass is constant. a) Record preconditions of the substance when using this method.

Method B: Gaseous (reaction) water The enclosed water is bound chemically or physically. The increase of mass of the absorbing agent such as sulfuric acid and anhydrous calcium chloride is determined. This method is also used to dry gasses. b) Give the reason why hydrogen sulfide, hydrogen iodide and ammonia should not be dried by sulfuric acid. Write down respective reaction equations. c) Give the reason why ammonia may not be dried by calcium chloride.

Method C: Calcium carbide (CaC2) The water containing sample is brought to reaction with calcium carbide and the reaction product is led into an ammoniac copper(I) chloride solution. The red precipitate is filtered off, dried to mass constancy and weighed. d) Draw the Lewis structure of the carbide anion. Give the formulae of three isoelectronic species. e) How does copper(I) chloride exist in an ammoniac solution? f) Write down the equation of all reactions of this method.

Method D: Iodometric (Karl Fischer method) In 1935 Karl Fischer published a method to determine water based on the reaction between iodine, sulfur dioxide and water previously published by Bunsen. g) Give the equation of the "Bunsen" reaction!

In the Fischer method the water containing sample is brought to reaction with methanol, pyridine, sulfur dioxide and iodine following formally the equation

H2O + SO2 + 3 C5H5N + I2 + CH3OH C5H5NHCH3OSO3 + 2 C5H5NHI . The endpoint of the titration is reached when a permanent brown colour occurs. h) What is responsible for the brown colour? Why can the iodine-starch reaction not to be used? i) Which function has the pyridine?

www.ShimiPedia.ir 31 Problems Round 4 (theoretical)

To determine the water content the sample is added to a solution of iodine and sulfur dioxide in water free methanol which contains pyridine and then titrated with a solution of iodine in alcohol. As the change of colour at the endpoint (bright yellow  brown) is difficult to detect visually the change is nowadays detected coulometrically. During the titration iodine is produced in an electro- chemical cell until no more iodide is formed. The water content can be calculated by the used amount of charge.

A B The titer of the Karl-Fischer solution is given in water equiva- Sample 1 1.65 mL 1.45 mL lents in mg/mL and amounts to t = 4.8 mg/mL. Samples of 10 g of two different food oils are investigated. The results are given Sample 2 1.62 mL 1.43 mL in the table on the left. Sample 3 1.60 mL 1.45 mL j) Calculate the mass percentage of water in the oils.

In another oil C a water content of 0.09 % is found in a coulometric titration. k) Which amount of charge (in coulomb) is consumed in an investigation of 10 g of oil C?

Problem 4-02 Electrochemistry A Latimer diagrams are plots of reduction potentials of half reactions including each of the different oxidation states of one element. Normally the species with the highest oxidation state is placed on the left, going to the right the oxidation state decreases. The different states are connected with arrows on which the reduction potential of the half reaction is written. These may refer to standard conditions (25 °C, pH = 0, c = 1 mol·L−1) or to any other condition (e.g. pH = 14).

– 0.926 V – 1.154 V Example: [AuCl4]  [AuCl2]  Au (Standard conditions)

– a) Calculate the standard potential x = E°([AuCl4] / Au).

Gold does not react with nitric acid but does with aqua regia, a 3:1 mixture of conc. hydrochloric acid and conc. nitric acid, which was developed by alchemists to "dissolve" gold. – In this reaction with aqua regia the complex [AuCl4] is formed. c([AuCl ]−)/c0 b) Calculate the complex formation constant of [AuCl ]–, K = 4 using the 4 co (c(Au3+)/c0) ·(c(Cl−)/c0)4 result of a) and E°(Au3+/Au) = 1.50 V.

In an acidic solution (pH = 0) the following standard potentials are existent: – – – – – – ClO4 / ClO3 ClO3 / ClO2 ClO2 / HClO HClO/ Cl2 Cl2/ Cl E° in V 1.20 1.18 1.65 1.63 1.36 c) Draw the Latimer diagram. – – – Check whether ClO3 disproportionates under these conditions to form ClO4 and Cl . Write down the equation for a disproportionation reaction if the case may be.

32 www.ShimiPedia.ir Problems Round 4 (theoretical)

B d) In which span of pH-values can hydrogen peroxide oxidize Cer3+ions? 4+ 3+ + E°(Ce /Ce ) = 1.61 V E°(H2O2, H /H2O) = 1.76 V

On the other hand hydrogen peroxide can be oxidized by strong oxidation reagents such as potassium permanganate in acidic solution. e) Write down the equation of this reaction.

Problem 4-03 Metals and Electrons When atomic orbitals are filled up with electrons the lowest available orbital is fed first and then the following in the direction of rising energies. Thereby Hund's rule as well as the Pauli principle has to be considered. There are exceptions from these rules. a) Give two examples for elements with electron configurations (in the ground state) other than the expected regular one. Write down the expected and the observed configuration. (Use the abbreviation for full electron shells such as [He] for 1s2 or [Ne] for s22s22p6 etc.)

The reason for this deviation is the stability of certain electron configurations. b) Which electron configurations are energetically especially favoured?

These favoured electron configurations play a decisive role in the electron configuration of metal cations. c) Write down the electron configurations of the following metal cations. (Use the abbreviation for full electron shells as mentioned above.) i) Fe3+ ii) Mn3+ iii) Pd4+ iv) Cr3+ v) Fe2+ vi) Pb2+ vii) Au3+ viii) Co2+ ix) Cu+ x) Ti2+

Discrete metal ions as given in part c) exist only formally. In reality e.g. in solids or complex compounds they always have a coordination sphere and are surrounded by other particles which form a regular coordination polyhedron. You often find the coordination numbers four and six. d) Draw 3-D figures of the polyhedrons for the coordination numbers four and six. (Don't consider a hexagon and a three-sided prism) e) Draw the theoretically possible stereoisomers of the complex compounds MX2Y2, MX4Y2 and

MX3Y3 and give their names (M: central particle; X and Y: monodentade ligands).

All degenerate systems try to reduce orbital degeneracy. The degenerated d-levels of an isolated metal cation split as soon as they come under the influence of a ligand field into levels of higher and lower energy. f) Describe the splitting off of the d-electron energy levels in an octahedral ligand field. What are the effects on the different d-orbitals? Draw an energy scheme which shows this issue as accurately as possible. g) With which d-electron configurations high- and low-spin configurations occur at all? h) Give the number of unpaired electrons which exist in a high-spin and in a low-spin state for the cations of part c) in an octahedral ligand field.

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Problem 4-04 Calculations around Kinetic and Energetic In a gas flow substance A exists with a partial pressure of p(A) = 8.9·10-4 bar. A is in equivalence with 3 A2 (2 A A2) which is gaseous, too, with Kp = 2.1·10 . a) Determine the ratio p(A)/p(A2).

keff Given the reaction A + B  C + D with the rate constant keff. The following reaction sequence is assumed to be the reaction mechanism k A + B 1 AB k2 C + D . k-1 -1 The reaction rates v1 and v-1 are approximately of the same size. Additional is k1/k-1 = 15 (mol/L) -1 and k2 = 25 s . b) Calculate keff using the assumption of steady state equilibrium. c) Which is the necessary precondition to use the steady state approximation?

The hydrolysis of urea follows the equation

(NH2)2CO(aq) + H2O(l)  2 NH3(aq) + CO2(aq) . d) Calculate the equilibrium constant K for this reaction at 298 K.

(NH2)2CO(aq) H2O(l) NH3(aq) CO2(aq) H° in kJ/mol -317.7 -286.0 -80.9 -413.1 S° in Jmol-1K-1 176 68 110 121 e) Calculate G if the following concentrations exist at 298 K

c((NH2)2CO(aq)) = 0,85 mol/L c(CO2(aq)) = 0,097 mol/L c(NH3(aq))= 0,02 mol/L

k3 In the equilibrium D + E F the reaction as well as the back reaction are elementary reac- k- tions. c(D), c(E) and c(F) are time3 dependent concentrations in the process of the spontaneous reaction of D and E. f) Determine the dependency of G on v3 and v-3 which are the rates of the reaction and the back reaction.

k3 At a certain time in the course of the reaction D + E F from part e) the observed reaction k-3 rate is defined as vob = v3- v-3 . At a certain time let be

vob/ v3 = 0.5 c(D) = 0.4 mol/L c(E) = 0.9 mol/L c(F) = 1.8 mol/L. g) Calculate the equilibrium constant K for this reaction at 298 K.

Problem 4-05 Three Compounds

Given are the compounds AB3, CA and CB2. The mass percentage of A in AB3 is 23.81 %, that of B in

CB2 73.14 %. a) Calculate the mass percentage of C in CA.

34 www.ShimiPedia.ir Problems Round 4 (theoretical)

b) Determine the elements A, B and C. Write down the formulae and names of the given compounds.

Problem 4-06 Complexes and More A Insoluble Prussian blue is a coordination compound. The crystal structure and the empirical formula can be derived from one octant (on the left) of the cubic unit cell (drawn with iron ions only).

(on the edges) (on the edges) (H2O in the middle of the octant)

Indicate in all of the questions a) to d) the oxidation states of the iron ions as FeII, FeIII.

a) Determine the empirical formula of Prussian blue using the octant above. Show shortly how you found your result.

Insoluble Prussian blue forms only at high concentrations of iron ions. At lower concentrations so- III II called soluble Prussian blue K[Fe Fe (CN)6] is formed. It originates from the combination of an iron(II) solution with a solution of potassium hexacyanoferrate(III) as well as from an iron(III) solution with a solution of potassium hexacyanoferrate(II). b) Write down the equation for these two reactions. c) Taking these reactants find an equilibrium reaction which explains the formation of the identical product.

Comparable with the reaction of part b) there is a reaction between a solution of iron (II) with potassium hexacyanoferrate(II) which leads to a white precipitate. d) Write down an equation of the reaction which leads to the white precipitate. Explain why this compound is not colored in contrary to soluble Prussian blue.

The orientation of the cyano groups in the compound of part a) (see fig. above) can be derived with the help of the HSAB concept. It says that the combinations of weak with weak and of hard with hard acids and bases lead to more stable adducts than mixed combinations. Iron(III) is classified as hard, iron(II) as weak. The hardness of bases decreases within the row F > O > >N, Cl > Br, H > S, C > I, Se > P, Te > As > Sb.

www.ShimiPedia.ir 35 Problems Round 4 (theoretical)

e) Fill the cyano groups in their expected orientation following the HSAB concept into the empty circles on the answer sheet (C for carbon, N for nitrogen). Determine the coordination spheres of iron(II) und iron(III) in the compound.

B A very diluted solution of copper(II) chloride in water has a light blue color. By adding hydrochloric acid the solution turns green and with rising concentration of hydrochloric acid intensively green- brown. f) Which species of copper(II) chloride exist predominantly in the diluted aqueous solution and in the solutions in half-concentrated and concentrated hydrochloric acid?

In analytical chemistry copper is detected in the form of a deep blue colored ammine complex. But it can be recognized with a borax bead as turquoise metaborate (Cu(BO2)2), too. In literature you find different formulae for borax with the empirical formula B4H20Na2O17:

As decahydrate Na2B4O7 · 10 H2O, octahydrate Na2[B4O5(OH)4] · 8 H2O and as mixed oxide decahydrate Na2O · 2 B2O3 · 10 H2O. g) Show with the help of a Lewis structure why the octahydrate is the most reasonable. Attach charges to the atoms if necessary.

In the separation scheme of cations copper(II) ions are precipitated with hydrogen sulfide as black copper sulfide. However, this compound is not copper(II) sulfide but a mixed compound of copper(I) and copper(II) ions which contains another sulfur containing anion besides sulfide anions (S2–). h) Which other anion does the compound contain besides the sulfide anion? Give a more exact formula than CuS.

In the Deacon process to produce chlorine by oxidizing hydrogen chloride using oxygen from the air copper(II) chloride is used as catalyzer. i) Write down the equation for the total reaction of the Deacon process. j) Give reaction equations which show the catalytic effect as well as the regeneration of copper(II) chloride.

Problem 4-07 Many Questions Concerning Thermodynamics 2 mol of oxygen at 273 K have a volume of 11.2 dm3. The gas can be regarded as perfect with the -1 -1 heat capacity of CV = 21.1 Jmol K which is supposed to be independent of temperature. a) Calculate the pressure of the gas. b) Give the different meanings of Cp and CV. Explain why these values have to be different.

Calculate Cp.

The sample of gas mentioned above is heated reversibly to 373 K at constant volume. c) How much work is done to the system?

36 www.ShimiPedia.ir Problems Round 4 (theoretical)

d) Calculate the rise of the internal energy. e) Calculate the heat which was added to the system. f) What is the final pressure? g) What is the increase in enthalpy H?

The gas sample (2 mol at 373 K in 11.2 dm3) is allowed to expand maintaining the temperature against a piston that supports a pressure of 2 atm. h) Calculate the work done by the expansion. i) What is the change in internal energy and in enthalpy of the gas? j) Calculate the heat absorbed by the gas.

The dependence of the boiling point of methane on pressure is well described by the empirical equa- 443 tion: log(p/bar) = 3.99 – Ts/K - 0,49 k) Determine the boiling point of methane at a pressure of 3 bar.

The difference in internal energy of liquid and gaseous methane at the boiling point of 112 K at atmospheric pressure is 7.25 kJ/mol.

An object is cooled by the evaporation of CH4(l). l) What volume of CH4(g) at 1.000 atm must be formed by the liquid to remove 32.5 kJ of heat from the object?

Problem 4-08 Synthesis of (–)-Muscone (–)-Muscone is the primary contributor to the odor of musk. It is an oily liquid with a characteristic smell. Natural muscone was originally obtained from musk, a glandular secretion of musk deer, which has been used in perfumery and medicine for thousands of years. Since obtaining natural musk requires killing the endangered animal, nearly all muscone used in perfumery today is synthetic. One synthesis of (−)-muscone begins with commercially available (+)-citronellal, a monoterpene aldehyde, which is mainly found in citrus fruits. The scheme on the next page shows the way of synthesis. Hints: - In this scheme compound A reacts exclusively to compound B and compound C exclusively to D aas well as compound G exclusively to (–)-muscone. - Compound V is a side product. - Compound G exists as a mixture of two isomers.

- (–)-Muscone has the empirical formula C16H30O.

www.ShimiPedia.ir 37 Problems Round 4 (theoretical)

10-bromo-dec-1-ene

then r.t. then r.t.

Grupps cat. I

(-)-Muscone r.t. = room temperature

- TBDMSCl Protective reagent for hydroxy groups

Si Cl - TBAF splits up silyl ether

+ N F-

- THF O Solvent

- Grupps cat. I catalyzes the metathesis of olefins, e.g.: P(Cy)3 Cl Ru R1 R1 Cl Ph P(Cy)3 R R R1 R2 2 2 Kat.Cat. Cy = Cyclohexyl- Rest 2 + R R 3 3 R3 R4

R R 4 4

38 www.ShimiPedia.ir Problems Round 4 (theoretical)

a) Determine and draw the structural formulae of the compounds A to G as well as the structural formula of (-)-muscone.

During the reaction of (+)-citronellal to compound A in a so-called carbonyl-ene-cyclisation compound V (isopulegol) forms as a side product. Under the terms given in the scheme above the Lewis acid X is formed which catalyzes the carbonyl-ene-cyclisation (the stereoselectivity of this reaction stays unaccounted): O

H OH X

(+)-Citronellal V, Isopulegol b) Give the empirical formula of X and write down the reaction equation of the formation of X. c) Map the mechanism of the carbonyl-ene-cyclisation taking the catalyzing effect of X into consideration.

The formation of F succeeds with a (deep) red chromium(VI) species Y which is formed in situ. It oxidizes E to compound F and is itself reduced to a green compound. Y is formally the anhydrate of the chromium acid. d) Write down the equations for the reduction and the oxidation as well as the complete equation for the redox reaction between the compounds E and Y. Use the abbreviations R and R' for the substituents outside the reactive center.

Step F to G is a so-called cyclisation metathesis. A gaseous compound Z forms besides compound G. e) Give the name of compound Z?

Problem 4-09 Small but Powerful Carbocyclic compounds are often found as basise of biologically active reagents and functional materials. Many ways are known to synthesize the especially stable five- and six-membered rings but the preparation of larger and smaller rings is a synthetic challenge. The reason for the difficulties with smaller rings is their low stability and thus their high reactivity. The bonding situation in a cyclopropane ring is described by bent bonds. The overlapping orbitals between two carbon atoms can't point directly to each other (as they do in alkanes); rather, they overlap in an angle. The result is a banana-shaped bond shown on the next page. a) Compare the bond angel and the bond length of cyclopropane and the more stable cyclohexane. Draw a cyclopropane and a cyclohexane ring (in chair conformation with hydrogen atoms) in a 3-D plot. Account for the different stability of the C-C bonds in cyclopropane and cyclohexane.

www.ShimiPedia.ir 39 Problems Round 4 (theoretical)

Banana-shaped bond in cyclopropane

Because of the higher π-character of the C-C bonds in cyclopropane rings addition reactions may proceed analogue to those of alkanes. The electrophilic hydrobromation is an example. b) Write down the equation of the reaction of methyl cyclopropane with hydrogen bromide. Give the mechanism of the hydrobromation and use it to explain which product is formed pref- erentially. What is the name of the rule of this regioselective formation? c) Draw an energy profile of this reaction (ΔRH° < 0, energy depending on the reaction coordinate) and attach reactants, products, transition states and intermediates (if existent).

Generally cyclopropane rings are generated via cycloaddition of carbenes and carbenoids to olefins. Carbenoids are substances similar to carbenes and show a comparable reactivity. d) Give a common form of the Lewis structure of a carbene (with substitutes = R). What is the hybridization state of the carbon atom in the center of the carbene?

The best method for preparing cyclopropanes is by a process called the Simmons-Smith synthesis:

An addition of CH2-Zn-I to a C=C double bond. In this reaction the new C-C bonds form in cis position of the original double bond. e) Draw the structural formula and give the name of the product of the Simmons-Smith reaction of (Z)-1,2-diphenylethene.

Another compound to synthesize cyclopropane rings starting from olefins is diazomethane (CH2N2), which is also used for the preparation of the hydrocarbon 4 (C13H20). 4 has a tetrahedral shape. The 1H- and 13C-NMR spectra show two resp. three signals. The scheme of this reaction is shown below: . Ph3P Br2/DCM CO2Et LiAlH4/Et2O -15° 20°C, 6h 1 2 99% 95%

CH2N2 (10 eq), Pd(OAc)2 tBuOK/DMSO -20° 25°C 20°C, 6h srepeatedechsfach sixwi etimesderho lt 3 4 60% 92%

DCM = dichloromethane, DMSO = dimethyl sulfoxide

40 www.ShimiPedia.ir Problems Round 4 (theoretical)

f) Give the structural formulae of the compounds 1 to 4. g) Mark in a figure of 4 those C and H atoms which are responsible for the NMR signals.

Problem 4-10 Carbonyl Compounds as Reactants and Products Carbonyl compounds are valuable resources for organic synthesis because they react with many different compounds, often selectively. In the following scheme some reactions starting with a butyl alcohol (1) are depicted:

iPr: Isopropyl cHex: Cyclohexyl DMP: m-CPBA: a) Draw the structural formulae A to G . b) Give a possible method for i) to form B directly from butyl alcohol.

Compound F is an epoxide. c) Which reagents ii) are needed to synthesize the diol 2 and then to convert the diol 2 with iii) to A? Give the name of the reaction of diol 2 to A. (Help: Compound iii) contains a halogen atom in the highest oxidation state.) d) Which product do you expect if compound C is brought to reaction with MeLi? Draw the structural formula. e) Why is it not possible to get compound D by bringing B directly into reaction with MeLi?

Finally compound D is converted to 5. f) Propose the reaction conditions iv) for the reaction of D to 5. g) Show the mechanism for this reaction and give the names of the class of substances of the intermediate and the product.

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Fourth Round (practical problems)

Problem 4-11 Gravimetric Determination of Zinc as Zn(NH4)PO4 Equipment: 2 x 400 mL beaker, 25 mL pipette with pipette control, 50 mL graduated cylinder, 100 mL narrow- necked bottle, Bunsen burner with tripod and tile, glass rod, 2 glass filter crucibles, suction flask with rubber ring, pressure tubing, vacuum attachment, desiccator with drying agent, precision balance, pen Chemicals:

Test solution (100 mL volumetric flask)

Solution of ammonia, c(NH3) = 2 mol/L

Ammonium chloride, NH4Cl(s),

diluted hydrochloric acid, c(HCl) = 2 mol/L

Solution of diammonium hydrogen phosphate, w((NH4)2HPO4) = 10 %

Indicator solution of methyl red in Ethanol, w(C15H15N3O2) = 0,05 %

Demineralized water

Procedure: The test solution has to be filled up with demineralized water to the calibration mark and mixed well. 25 mL of this solution are pipetted into a 400 mL beaker. Approximately 150 mL of demineralized water are added. Then 25 mL of diluted hydrochloric acid, 2-3 full spatulas of ammonium chloride, 25 mL of a solution of diammonium hydrogen phosphate and some drops of methyl-red indicator solution are added. Heat to boiling and add dropwise diluted ammonia until a colour change to yellow-orange can be observed. Then stir with the glass rod until the precipitate has formed crystals or is well sedimented. While the solution cools down to room temperature the glass filter crucibles are marked with the pen, weighed and the values are listed. The precipitate is sucked through a glass filter crucible, washed with a small portion of cold demineralized water and then dried at 130°C for one hour in a drying oven. The crucible is cooled down to room temperature for about 20 to 30 minutes in an exsiccator and then weighed again. a) Write down the number of your sample on the answer sheet. b) Write down the results of your weighing.

42 www.ShimiPedia.ir Problems Round 4 (practical)

c) Calculate the mass concentration of zinc in mg/L in your test solution. d) What kind of compound is zinc ammonium phosphate, a mixed crystal, a mixture of crystals, a double salt or an alloy? Account for your decision. e) What happens if you anneal the precipitate? Write down the reaction equation.

Problem 4-12 Standardization of a Na2EDTA Solution This solution will be used in problem 4-13. Equipment:

250mL volumetric flask with a solution of Na2EDTA (concentration unknown), stopper, 20 mL pipette with pipette control, 2 Erlenmeyer flasks (wide mouth), spatula, 25 mL burette with funnel and clamp, stand. Chemicals:

Solution of Na EDTA, (c(Na EDTA) ≈ 0,1 mol/L when filled up) 2 2

Solution of ammonia, w(NH3) = 25 % (in the hood)

Indicator buffer pills

Standardized solution of zinc sulfate, c(ZnSO4) = 0.097 mol/L

Demineralized water

Procedure:

The flask with the solution of Na2EDTA of unknown concentration has to be filled up with demineralized water up to 250 mL. Mix thoroughly. Transfer exactly 20 mL of this solution into an Erlenmeyer flask and fill up to 100 mL. Add an indicator buffer pill and - after it is dissolved - 2 mL of the ammonia solution (w(NH3) = 25 %). Titrate speedily with the standardized solution of zinc sulfate (c(ZnSO4) = 0,097 mol/L). The end-point is given by the color change from green to red. Problems: a) Write down the label code of your volumetric flask on the answer sheet. b) Note the consumption of the standardized solution of zinc sulfate. c) Calculate the concentration of your Na2EDTA solution.

www.ShimiPedia.ir 43 Problems Round 4 (practical)

Problem 4-13 Complexometric Determination of Nickel Equipment and Glassware: Volumetric flask (100 mL) with stopper with test solution, volumetric pipette (20 mL), pipettes control, 1 burette (25 mL) with funnel, 1 measuring cylinder (50 mL), stand and clamp, spatula, 2 conical (Erlenmeyer) beakers (300 mL, wide mouth) Chemicals:

Test solution containing nickel a volumetric flask

Dil. solution of ammonia, c(NH3) = 2 mol/L

Trituration of murexide indicator

Solution of Na2EDTA·2 H2O, c(Na2EDTA)  0.1 mol/L (from problem 4-12) Demineralized water Procedure: The flask with the test solution has to be filled up to 100 mL. The solution has to be mixed well. 20 mL of this solution are transferred with a pipette to a conical beaker (300 mL, wide mouth) and 15 mL of the solution of ammonia are added. Drops of the indicator solution are added until intense yellow color occurs (approximately 8 drops). If the solution in orange the pH value is not high enough and additional solution of ammonia has to be added. The mixture is filled up with demineralized water to about 100 mL and then titrated with the solution of Na2EDTA. There will be a sharp change of colors from yellow to violet. This color has to persist. Problems: a) Write down the label code of your volumetric flask on the answer sheet. b) Record the consumption of the standard solution of Na2EDTA. c) Calculate the mass concentration  of nickel in your tested solution (in mg/L).

Problem 4-14 Qualitative Analysis You find the following mixtures of salts in seven beakers:

BaCl2/NaCl - AgNO3/Cu(NO3)2 – FeCl3/CuCl2 – KSCN/KI – KIO3/K2SO4 – Na2CO3/NaOAc –

(NH4)2SO4/FeSO4. Equipment: 25 test tubes, test tube holder, 3 Pasteur pipettes, pH paper, spatula Chemicals:

Seven beakers with mixtures of salts (labelled from A to G)

44 www.ShimiPedia.ir Problems Round 4 (practical)

Dil. solution of ammonia, c(NH3) = 2 mol/L

Dil. nitric acid, c(HNO3) = 2 mol/L

Demineralized water

Procedure: Dissolve each mixture with about 30 mL of demineralized water. Combine at any time two of the solutions in a test tube and note the observation. Repeat this procedure with all possible combinations. You may use the dil. solutions of ammonia and nitric acid as aids as well as the pH paper. Problems: a) Write down the label code of your test mixture. b) Report your observations on the answer sheet. c) Assign the salt mixtures to the beakers. The following reduction potentials are given (pH = 0): E° in V E° in V Cu2+ / Cu+ 0.16 Ag+ / Ag 0.80 2+ Cu / Cu 0.34 HIO3 / HIO 1.13 – I2 / I 0.54 Cl2 / Cl 1.34 Fe3+ / Fe2+ 0.77 Take possible acid/base reactions into consideration. Additional safety precautions: The mixtures contain toxic heavy metals, such as barium. Do not pour the solutions into the sink.

www.ShimiPedia.ir 45 Answers

Part 2

The answers to the problems of the four rounds

The solutions are more detailed than expected from the pupils. That may facilitate their comprehension in case of mistakes. Furthermore future participants should use this booklet to become acquainted with the problems of the competition and their solutions.

46 www.ShimiPedia.ir Answers Round 1

Answers Round 1 Solution to problem 1-01 12 13 14 a) Natural isotopes: 6퐶 , 6퐶 und 6퐶. They differ in the number of neutrons in the nucleus. 14 14 1 14 1 b) C is produced from nitrogen by cosmic radiation in the atmosphere: 7N + 0n → 6C + 1H 14 14 – c) 6C → 7N + e –푡 d) 푁 = 푁0 · 푒

N = number of nuclei at time t, N0 = number of nuclei at the begin (t = 0), λ = decay constant e) The half-life is the time it takes for the number of nuclei to fall to half its initial value. It is constant for a nuclide and characteristic for it. N ln 2 0,693 The formula for half-life can be derived by setting N = N /2: 0 = N · e–t1/2  t = = 0 2 0 ½   f) Never, N = 0 will never be reached. g) Carbon has two stable, nonradioactive isotopes: 12C and 13C, and one radioactive isotope, 14C. Radiocar- bon dating is a radiometric dating technique that uses the decay 14C to estimate the age of organic materials. The concentration of 14C in the atmosphere might be expected to reduce over thousands of years. However, 14C is constantly being produced in the lower stratosphere and upper troposphere by cosmic rays. Thus the proportion of radioactive to non-radioactive carbon in the atmosphere is constant and in living organic material too. Once an organism dies the natural carbon exchange is not continued and the ratio of 14C in the material decreases. The comparison of the actual content of 14C in some organic material with its content in the atmospheric carbon leads to the age of the organic material. In doing this you may compare the number of decays of 1 g of carbon per minute. h) The decay rate is dependent on the number of atoms. N 1 N = N · e–t  t = ln 0 · 0 N  ln 2 5730 15 and with  = : t = years · ln = 292 years 5730 ln2 14,48 2013 – 292 = 1721 The treasury map originates from the year 1721, it could be authentic. i) After 65 m years the ration of 14C is so small that hardly a decay can be detected. Furthermore it cannot be assumed that the ratio of carbon isotopes has been stable in such a long period. Remark: Nevertheless the age of the bones of dinosaurs can be estimated by the decay of radioactive isotopes of other elements with a longer half-life such as 238U, 87Rb und 40K. Solution to problem 1-2 a) The "characteristic composition of fatty acids of a fat" shows the mean distribution of fatty acids in a fat. For example, a fat could consist of the following three esters O O O

O CH O CH O CH O 2 O 2 O 2

O CH O CH O CH O O O

O CH O CH O CH 2 2 2 3 x Oleic acid 2 x Oleic acid 1 x Stearic acid 1 x Oleic acid, 2 x Stearic acid Then the characteristic composition of fatty acids is 66.7 % of oleic acid and 33.3 % of stearic acid. b) Capric acid COOH 10 : 0

C10H20O2 Linoleic acid COOH 18 : 2

C18H32O2

47 www.ShimiPedia.ir Answers Round 1

Linolenic acid COOH 18 : 3

C18H30O2 Erucic acid COOH 22 : 1

C22H42O2 c) The residues of the fat molecules experience van der Waals forces. Because van der Waals forces operate only at short distances they are strongest in molecules which chains can pack together closely. Saturated residues can arrange linearly which leads to strong intermolecular forces. These fats are solid. Residues with double bonds are mostly cis configurated. Thus a linear arrangement is not possible and the interactions are lower. These fats are rather liquid. d)

e) O O O

R1 O CH R1 O CH R1 O CH O 2 O 2 O 2 + NaOH R2 O CH R2 O CH R2 O CH O O – O – Na+ 3 3 3 – CH2 R O CH2 R O CH2 R OH O OH – OH – OH

R1 O CH O O 2 O 1 + NaOH 2 + H O R O CH2 R O CH 2 3 O R – + + O O Na – 2 – H2O – CH2 – OH R O CH 3 O R O– Na+ CH HO 2 O H H

1 2 3 R , R , R = any alkyl- or alkenyl residue of a fatty acid The deprotonation of a carboxylic acid by hydroxide ions is almost irreversible, thus the reaction runs towards the carboxylates. f) The attack of the hydroxide is hindered by the long chain ester. g) A tenside has two different ends, a hydophobic and a hydrophilic end. So they have the property to ac- cumulate at the boundary surface of liquids and thus decrease the surface tension. When tensides are dispersed in water the long hydrocarbon tails cluster together on the inside of a tan- gled, hydrophobic ball, while the ionic heads on the surface of the cluster stick out into the water layer. These mostly spherical clusters are called micelles. If there is a higher concentration cylindrical micelles and block micelles may be formed.

Tenside molecule

48 www.ShimiPedia.ir Answers Round 1

Spherical micelle Cylindrical micelle Block micelle Grease and oil droplets are solubilized in water when they are coated by the nonpolar tails of the tenside in the center of micelles. Once solubilized, the grease and oil can be rinsed away. h) Soaps are anionic tensides. There are four kinds of tensides. Anionic tensides: a long hydrophobic hydrocarbon tail is connected – with a hydrophilic negatively charged group. Cationic tensides: a long hydrophobic hydrocarbon tail is connected + with a hydrophilic positively charged group. Ampholytic tensides: a long hydrophobic hydrocarbon tail is connected +/– with an ampholytic group (positively and negatively charged). Nonionic tensides: a hydrophobic residue is connected with an un- charged group of polyglycol ethers. Anionic tensides R CH COONa Soaps 2 Alkylbenzene sulfonates (ABS, LAS) R1

CH SO3Na 2 R Alkane sulfonates (AS) R1

CH SO3Na 2 R  R CH CH CH2 SO3Na -Olefin sulfonates (AOS) Ester sulfonates (ES) R CH COOMe SO Na 3 Fatty alcohol sulfate (FAS) R CH2 O SO3Na

Fatty alcohol ether sulfate (FAES) R O CH2 CH2 O SO3Na n Cationic Tensides

H37C18 CH3 Distearyl dimethylammonium chloride (DSDMAC) N + Cl– H C CH 37 18 3 H3C CH3 Dodecyl dimethylbenzyl ammonium chloride N + Cl–

H25C12 CH2

Esterquats (EQ) CH2 CH2 O C R + H3C N CH2 CH2 OH CH CH O C R Cl– 2 2 O Amphoteric Tensides CH3 + – Betaines R N CH2 COO CH 3 R = C12 – C18

49 www.ShimiPedia.ir Answers Round 1

Sulfobetaines CH3 + – R N CH2 CH2 SO3 CH 3 R = C12 – C18 Nonionic Tensides

Fatty alcohol polyglycol ether (FAE) R O (CH2 CH2O)n H

Alkylphenol polyglycol ether (APE) R O (CH2 CH2O)n H

Fatty alcohol polyethylenglycol polypropylenglycol ether H

R (CH2 CH2O)n (CH2 CO)m H CH 3 Fatty acid ethanolamide O CH2 CH2 OH R C N CH CH OH 2 2

Solution to problem 1-3 a) Air, oxides, carbonates, silicates, sea water, water, biosphere or other compounds (or individual compounds) b)

Electrolysis of water: 2 H2O 2 H2 + O2 700 °C Thermic decomposition of peroxides: 2 BaO 2 BaO + O 2 500 °C 2

Catalytic decomposition of hydrogen peroxide (Pt, 2 H2O2 2 H2O + O2

MnO2):

Catalytic decomposition of oxygen containing com- 2 Ag2O 4 Ag + O2  > 160°C

pounds: 2 Au2O3 4 Au + 3 O2  > 160°C

4 KClO3 3 KClO4 + KCl  > 400°C

KClO4 KCl + 2 O2  > 300°C

the decomposition of KClO3 proceeds already at

150 °C with MnO2 as catalyst:

KClO3 KCl + 1,5 O2

KClO3 + 3 MnO2 KCl + 3 MnO3

3 MnO3 3 MnO2 + 1,5 O2

c) Nitrogen is diamagnetic, oxygen paramagnetic. The reason is the existence of unpaired electrons in the oxygen molecule.

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Energy Energy

Oxygen Nitrogen d) Triplet oxygen Singlet oxygen 2 unpaired electrons with equal spin 2 unpaired electrons with different spin Total spin S ½ + ½ = 1 –½ + ½ = 0 Multiplicity M 2 · 1 + 1 = 3 2 · 0 + 1 = 1 The names follow the multiplicity: M = 1 Singlet, M = 2 dublet, M = 3 triplet, M = 4 quartet etc. Remark: The spin of the inner electrons does not have to be considered. They cancel each other because of their opposite algebraic sign in double occupied orbitals. e) Part c) and d) demonstrate that dioxygen is a diradical. The Lewis structure does not show any unpaired electrons. Thus it does not show the correct distribution of electrons. Solution to problem 1-4 a) Helium, neon, argon b) P4 + 5 O2 P4O10 acidic reaction 4 Li + O2 2 Li2O basic reaction 1 /8 S8 + O2 SO2 acidic reaction 2 Ca + O2 2 CaO basic reaction c) Examples (R = Alkyl group) KZ 1: KZ 2: KZ 3: + O C O O O H H O C O R R , H H H   O O N O N O Cl Cl , R H d)

Oxidation number Examples Oxidation number Examples -I H2O2, Na2O2 +½ O2PtF6

-½ KO2 +I O2F2

-1/ NaO +II OF 3 3 2 0 HOF

T e) i) 2 Fe(OH)3 Fe2O3 + 3 H2O

51 www.ShimiPedia.ir Answers Round 1

ii) Ammonium nitrate shows an acidic reaction. It should remove absorbed hydroxide ions and wash the precipitate to be neutral. In the presence of chloride ions iron(III) chloride may evaporate and thus the result is falsified.

iii) The filter shall be burned without any residue which could be weighed with Fe2O3 to give a wrong result.

iv) Fe3O4 is magnetic and could be identified with a magnet.

v) M(Fe2O3) = 159.69 g/mol M(Fe3O4) = 231.54 g/mol

1 mol of iron(III) ions form 159.69g/2 = 79.855 g Fe2O3

or 231.54/3 g = 77.18 g Fe3O4, respectively

 If Fe3O4 is formed the mass after annealing is too low. Thus the formation of Fe3O4 leads to a content of iron which is too low.

vi) Mean result of Fe2O3 after annealing: (0.2483 g + 0.2493 g + 0.2488 g) : 3 = 0.2488 g . Amount of iron in 50 mL solution:

0.2488 g ·2 0.2488 g ·2 -3 n(Fe)50 = = = 3.116 · 10 mol M(Fe2O3) 159,69 g/mol  Mass of iron(III) chloride in 250.0 mL: -3 –1 m(FeCl3) = 5 · 3.116 · 10 mol · M(FeCl3) = 0.01558 mol · 162.21 g · mol

m(FeCl3) of the sample = 2.527 g. Solution to problem 1-5 a) MeO2 + 4 HCl MeCl2 + Cl2 + 2 H2O – – Cl2 + 2 I I2 + 2 Cl – + 2– I2 + 2 Na2S2O3 2 I + 4 Na + S4O6 b) Starch solution is added to realize the change more easily. Iodine forms a clathrate with starch. c) The oxygen of the air can oxidize iodide to iodine. Thus the consumption of thiosulfate would be too high because more iodine has to be titrated. In contact with steam chlorine might disproportionate into chloride and hypochlorite. Even if you assume that the equilibrium of the formation of hypochlorite in the presence of hydrochloric acid lies on the side of chlorine it cannot be excluded that some hypochlorite forms. As this reaction takes place in the vessel with the metal oxide chlorine set free by the metal oxide would be lost to oxidize iodide to iodine. This would lead to a smaller consumption of sodium thiosulfate. d) c(thiosulfate) = 0.1 mol /L = 0.1 mmol/mL

24.25 mL of thiosulfate solution ≙ 0.1 mol/mL · 24.25 mL/2  1.21 mmol I2

 290 mg of MeO2 produce 1.21 mmol of I2

1 mmol of MeO2 ≙ 1 mmol of I2  1.21 mmol of MeO2 ≜ 290 mg

1 mol of MeO2 ≜ 290 g/1.21 = 239.7 g M(Me) = (239.7 - 2 · 16.00) g/mol = 207.7 g/mol.  Me = Pb

52 www.ShimiPedia.ir Answers Round 2

Answers Round 2 Solution to problem 2-1 a) M: Pb, X: KI, Y: PbI2, Z: KPbI3 There are several ways to solve this problem. Only one of these will be described here. Specification of M: The information about the solubility in acids indicates a passivation in the presence of sulfate and chloride ions. These facts point to lead which forms a sparely soluble sulfate and a sparely soluble chloride. The information about the solubility in sodium hydroxide solution strongly limits the number of metals as only a few metals can be dissolved in this solution (examples: beryllium, aluminum, tin, iron, lead). The information of the pyrophoric property limits the selection, too. Metals which can ignite when finely dispersed are magnesium, titanium, nickel, cobalt, iron and lead as well as rare earth metals of the inner transition series. When a sodium hydroxide solution is added, the aqueous solution of the cations forms an insoluble hydroxide which dissolves in an excess of it. Examples are aluminum, lead, zinc, beryllium, chromium, galli- um, indium, copper and gold. With ammonia there is no formation of an ammine complex: iron(III), aluminum, beryllium, titanium, zinc, tin, lead. Finding of X and M: The endothermic solvation hints to compounds which are asked for in part f). Com- pound X must be able to react with the gaseous compound G. Possibilities are the formation of car-

bonates (from CO2), oxides (from O2), triiodates (from I2). An aspect in favor for the latter is the formation of a metallic mirror when G is formed (reducing properties of iodine). X or parts of X and the metal M must form a sparely soluble compound Y which has to have broader properties: layer structure, properties of a semiconductor, thermochromism. Possibilities are sparely soluble sulfates, chlorides, iodides. Taking all statements into account only lead is remaining. Y must be lead(II) iodide. X may be sodium or potassium iodide (see f)).

Part e) hints to a double salt. With the formation of the simplest double salt XPbI2 can the alkali metal be determined by means of a calculation of the given mass ratio. b) 2 KI + Pb(NO3)2 PbI2 + 2 KNO3 – PbI2 + KI KPbI3 I c) H = KI KI + I KI + 3 2 3 K I

– H is an ionic compound so the cation and the anion are shown separately. In the I3 anion the central iodine atom is surrounded by 5 electron pairs resulting in a trigonal bipyramidal arrangement. There are two bonded and three lone pairs. The lone pairs are equatorial arranged (according to VSEPR it is an

AX2E3 system). All these facts result in a linear structure of H. d) The TG plot shows a difference in mass of 5 – 6 %. The molar mass of the water free compound is

M(KPbI3) = 627 g/mol. Then the mass loss is 627 g/mol · 5.5/94.5 = 36.5 g/mol  n = 2. e) As the oxidation numbers and the elements are the same as in Z the wanted compound can only be

K2PbI4 with a mass ratio of lead of 207.2/793 · 100 % = 26.13 % f) NH4Cl, KI, CaCl2 · 6 H2O, Na2SO4 · 10 H2O, NaI, KNO3, NaNO2.

53 www.ShimiPedia.ir Answers Round 2

The lattice energy is needed in order to dissolve a salt, the solvation energy is set free. As soon as the needed energy (-lattice energy) is higher than the solvation energy the solution provides the difference and cools down. Condition for cooling-down: Lattice energy < Solvation energy or |Lattice energy| > |Solvation energy| g) +I,+II,–I +I,–II +I,–II +I,–I 0 0 T KPbI3 · n H2O n H2O + KI+ Pb + I2 h) Pb(NO3)2 + 2 NH3 + 2 H2O Pb(OH)2 + 2 NH4NO3 or as ionic equation: 2+ – Pb (aq) + 2 OH Pb(OH)2(s) Remark: Lead hydroxide is formed. There is no formation of an ammine complex in an aqueous solution. Solution to problem 2-2 a) In the reaction to A two isomers are formed. Acetylacetone is deprotonated by triethylamine to yield a carbanion which is stabilized by mesomerism to an enolate. In doing so the two isomers are formed, which are trapped by TMSCl.

free rota- tability

Afterward the two isomers are deprotonated by LDA and again a carbanion or rather an enolate is formed which is trapped again by TMSCl:

Both isomers of B react with NBS to yield C. The reaction takes place at the outer double bond.

54 www.ShimiPedia.ir Answers Round 2

b) E/Z-Isomers are formed:

c) Assignment and ratio of isomers: δ = 5.24 ppm: 1 proton at position 2 (Isomer 1) δ = 4.95 ppm: 1 proton at position 2' (Isomer 2) Remark: The assignment to isomer 1 or isomer 2 is arbitrary and could be done the other way round but it should be the same as in the following signal assignments. The signal at 5.24 refers to the Z-isomer, that at 4.95 to the E-isomer but the students are not expected to know that. From the ratio of isomer 1 to isomer 2 you find 2.11:1. The signals of the hydrogen atoms of the methyl groups at position 3 and 3' are almost identical as well as those of the methyl groups at position 1 und 1': δ = 1.93 ppm: 6 protons at 3 and 3' δ = 1.79 ppm: 6 protons at 1 and 1' . The remaining signals belong to the hydrogen atoms of the trimethylsilyl groups.

δ = 0.00 ppm: 9 protons of the Si(CH3)'3 group

δ = -0.04 ppm: 9 protons of the Si(CH3)'3 group Calculation of the intensities: The protons at positions 3 and 3‘ as well as those at 1 and 1‘ are so similar that their signals are almost identical and thus the signals appear as a singlet.  The intensity for both singnals at δ = 1.93 ppm and  = 1.79 ppm is 3 · 2.11 + 3 · 1 = 9.33 For isomer 1 the intensity of 9 · 2.11 = 18.99 is expected for isomer 2 the intensity of 9 · 1 = 9 is expected.  δ = 0.00 ppm: 9 · 1 = 9 (Isomer 2) δ = 0.04 ppm: 9 · 2.11 = 18.99 (Isomer 1) d) Compound C contains a bromine atom. Natural bromine consists of two isotopes, 79Br und 81Br (=2), with nearly the same abundance. Thus in the mass spectrum two nearly identical peaks with =2 are found at 177.96 and 179.96. e) Et3N is used as a base. It deprotonates acetylacetone at the CH2 group between the two carbonyl groups, so an enolate can be formed. This enolate reacts with TMS-Cl to yield TMS-enolether. f) The base strength of Et3N does not suffice for a second deprotonation. Thus the stronger Base LDA is used. g) The reaction of acetylacetone with NBS in the presence of a base is not regiospecific and would not lead

to the desired product because the deprotonation of acetylacetone would take place at the CH2 group between the two carbonyl groups. The synthesis via A and B guarantees that only the desired regiospecific compound forms.

55 www.ShimiPedia.ir Answers Round 2

The H/D exchange may happen up to three times so that a completely deuterized CD3 group may be obtained.

Solution to problem 2-3 a) Cyclobutadiene 2 π electrons (charge: +2): 4 π electrons (charge: 0):

1P 1P

Remark: The degeneracy of 2 and 3 leads to a diradical electron structure. 6 π electrons (charge: –2):

Tropylium ions 6 π electrons (charge: +1): 8 π electrons (charge: –1):

Remark: The degeneracy of 2 and 3 leads to a diradical electron structure.

56 www.ShimiPedia.ir Answers Round 2

The calculation of the total energy is done as follows:  The energy of the π system is compared with the aliphatic version. Therefore the respective energies of the occupied orbitals have to be calculated with the given formulae and added.  The energy levels of the Frost circle result from the following considerations:  The height of the respective energy level with respect to the AO level (E = ).  The difference from the origin to the energy level is 2β.  The angle  is 360°/n; the angle between the energy level and the height amounts to m · , where m = number of the energy level – 1 and n = number of carbon atoms in the n-cyclic system.  To find the height the cosine function is used: cos(m · ) = H/2β  H = cos(m · ) · 2β   E =  + H

The calculation for cyclobutadiene gives = 90°. Number of electrons Bond energy Aliphatic Difference 2 E = 2 · (+2β) = 2.0+4.0β E = 2.0+3.2β ΔE = 0.8β 4 E = 2 · (+2β) + 2 ·  = 4.0+4.0β E = 4.0+4.5β ΔE = –0.5β 6 E = 2 · (+2β) + 4 ∙  = 6.0+4.0β E = 6.0+3.2β ΔE = 0.8β  The cyclobutadienyl dication and the cyclobutadienyl-dianion are aromatic because of E > 0. The calculation for tropylium gives  = 51.429°. Number of electrons Bond energy Aliphatic Difference 6 E = 2 · (+2β) + 4 · (+cos () ∙ 2β) = 6.0+9.0β E = 6.0+8.1β ΔE = 0.9β 8 E = 2 · (+2β) + 4 · (+cos() · 2β) + 2 · (+cos(2) · 2β) = 8.0+8.1β E = 8.0+8.1β ΔE = 0  The tropylium cation is aromatic because of E > 0. b) The compound has to be 1. cyclic, 2. planar, 3. totally conjugated 4. fulfill the Hückel rule: it must have 4n + 2 π- electrons. c) The following compounds (ions) fulfill all conditions and are aromatic: (i) Pyrrole (6 π electrons), (iii) azulene (10 π electrons), (v) pyridinium cation (6 π electrons), (vi) caffeine (10 π electrons and 6 π electrons in the five membered ring). Non-aromatic ore antiaromatic are: (ii) Allyl anion (4 π electrons; violates rules 4 and 1), (iv) 1H-pyrrolium cation (N protonated pyrrole) (violates rule 3) d) HOMO: 2 and 3; LUMO: 4 and 5 Remark: The orbitals are degenerate (having the same energy). 57 www.ShimiPedia.ir Answers Round 2

e) The decision is based on the number of nodal planes in the bond axis of the respective orbital. Orbitals of same symmetry show interaction. – – HOMO (Cp ), (2,3 ) LUMO (Cp ), (4,5)

3dx2-y2 – x 3d – – z2 x = interaction expected 3dxy – x

3dxz x – - = no interaction expected

3dyz x – 4s – –

4px x –

4py x –

4pz – –

Orbital interaction between Fe and Cp– is possible in each column.

5 5 f) [Ru(η -Cp)2] is more stable as it fulfills the 18 electron rule. [Rh(η -Cp)2] has 19 electrons and should act as a reducing reagent. g) Possible structures:

18 Valence electrons each 18 Valence electrons each 18 Valence electrons each Remark: The numbers of valence electrons at the metal centers are found as the sum of all electrons assigned to the metal. Each bond of the terminal CO groups counts for 2 electrons, the bridging CO groups provide 1 electron for each metal center and the cyclopentadienyl ring 6 electrons. The 7 d-electrons of the metal centers are added  Σ = 18. Solution to problem 2-4 2+ + + a) [Cu(H2O)6] + H2O [Cu(H2O)5(OH)] + H3O (The tetraaqua complex is also accepted) c1 c2 c3 –3 M(Cu(NO3)2) = 187.56 g/mol  c(Cu(NO3)2) = 8.96 · 10 mol/L.

c2/(1 mol/L) · c3/(1 mol/L) Ka = c1/(1 mol/L) –4.40 –3 c2 = c3 =10 mol/L c1 + c2 = 8.96 · 10 mol/L –3 –4.40 –3 c1 = 8.96 · 10 mol/L – 10 mol/L = 8.92 · 10 mol/L –7 Ka = 1.78· 10  pKa = 6.75

K · (1 mol/L)3 b) K = c(Cu2+)·c2(OH–)/(1 mol/L)3  c(OH–) = √ sp .25°C sp. 25°C c(Cu2+)

58 www.ShimiPedia.ir Answers Round 2

1.6 ·10−19 c(OH–) = √ mol/L c(OH–) = 3.94 · 10–9 mol/L  pOH = 8.40 pH = 5.6 1.03 ·10−2 At this pH the fraction of protolysed Cu2+ ions can be neglected. c) 2 Cu+ Cu2+ + Cu RT RT E = Eo + · ln c(Cu+) E = Eo + · ln (c(Cu2+) / c(Cu+)) 1 1 F 2 2 F F In equilibrium: E = E .  ln K = (Eo – Eo ) · T = 295 K 1 2 1 2 RT 96485 ln K = (0.52 – 0.16) · = 14.16 K = 1.41 · 106 8.314·295 d) According to the electron configuration copper(I) should be most stable because it has a closed electron shell (d10 configuration). In contrast copper(II) has a d9 system. The stability of copper(II) in aqueous solutions is caused by the high hydration enthalpy which is set free when the small, highly charged copper(II) ions are dissolved. This effect overbalances the unfavorable electron configuration. c(Cu2+)/(1 mol/L)2 e) K = c2(Cu+)/(1mol/L) c(Cu+) = x mol/L c(Cu2+) = y mol/L K = y/x2 x + 2 y = 0.01  y = ½ (0.01 – x) K = (0.01 – x) / 2 x2 x2 + (x / 2 · K) – (0.01 / 2 · K) = 0 K = 1.72 · 106 x = 5.38 · 10–5  c(Cu+) = 5.38 · 10–5 mol/L, y = 4.97 · 10–3  c(Cu2+) = 4.97 · 10–3 mol/L f) Copper(I) oxide can only dissolve when disproportionation takes place. This will not be if E2 > E1 (see c)). + – + –15 – + + c(Cu ) = KL· (mol/L)²/ c(OH )  c(Cu ) = 10 · (mol/L)²/c(OH )  c(Cu ) = 0.1 · c(H3O ) c(Cu2+) = 0.01 mol/L RT RT E = 0.52 + · ln [0.1 · c(H O+)/( mol/L)] E = 0.46 + · ln [c(H O+)/(mol/L)] 1 F 3 1 F 3 RT RT E = 0.16 + · ln[0.01 / (0.1 ·c(H O+)/(mol/L))] E = 0.10– · ln [c(H O+)/(mol/L)] 2 F 3 2 F 3 −0.36 ·F E > E  ln [c(H O+)/(mol/L)] < ln [c(H O+)/(mol/L)] <–7.13 2 1 3 2 ·R ·T 3 + –4 c(H3O ) < 8.01 · 10 mol /L  pH = 3.1 The higher the temperature the lower the pH has to be. −0.36 ·F With ln [c(H O+)/(mol/L)] < you get the following table: 3 2 ·R ·T + + Temp. c(H O) c(H3O) ln 3 pH in °C 1 mol/L 1 mol/L 0 -7.65 4.76·10-4 3.32 3.3 -4 10 -7.38 6.24·10 3.20 3.1

-4 20 -7.13 8.01·10 3.10 2.9 30 -6.89 1.02·10-3 2.99 2.7 40 -6.67 1.27·10-3 2.90 pHvalue 2.5 50 -6.47 1.55·10-3 2.81 -3 2.3 60 -6.27 1.89·10 2.72 70 -6.09 2.27·10-3 2.64 80 -5.92 2.69·10-3 2.57 -3 Temperature in °C 90 -5.75 3.18·10 2.50 100 -5.60 3.70·10-3 2.43

59 www.ShimiPedia.ir Answers Round 3 Test 1

Answers Round 3 Test 1 Solution to problem 3-01

A a) 2 Na2S2O3 +1 I2 2 NaI + 1 Na2S4O6 2+ – – - – b) 2 Ba + 2 MnO4 +1 CN + 2 OH 2 BaMnO4 + 1 CNO + 1 H2O – + – – c) 1 ClO3 + 6 H3O + 6 Br 3 Br2 + 1 Cl + 9 H2O B Fe2+ in acidic solution light green Iron sulfide (FeS) black 3+ Al in acidic solution colorless Copper sulfate (CuSO4) white Cu2+ in acidic solution blue Silver iodide (AgI) yellow 2+ Cu in ammoniac solution deep blue Potassium sulfate (K2SO4) white - Cl in basic solution colorless Potassium permanganate (KMnO4) violet/black + Na colorless Potassium chromate (K2CrOI4) yellow C Silver bromide, lead sulfate, iron(II) sulfide.

D Ba(NO3)2, K2(COO)2, Al2O3, K2MnO4, KAl(SO4)2, Na2CO3·10H2O E Soluble are potassium, aluminum, zinc. F Soluble are potassium, lead, copper, zinc.

Solution problem 3-02 a) Cu + 2 A g+  Cu2+ + 2 Ag (Only if c(Ag+) ≤ 1.7∙10-8 mol/L the reaction runs in the reverse direction.) b)

U/V

0.45

0.40

0.35

0.30

-3.0 -2.5 -2.0 -1.5 -1.0 -0.5 0.0 log x c) (Remark: The electrode at which a positive charge carrier goes from the metal into the solution or a negative charge carrier comes out of the solution on to the metal (oxidation) is called anode, in the opposite case it is the cathode.) Cathode: 2 Ag+ + 2 e-  2 Ag Anode: Cu  Cu2+ + 2 e- Cell reaction: 2 Ag+ + Cu  2 Ag + Cu2+ U = E E = E(cathode) – E(anode) 8.314 JK−1mol−1 ·298 K E = 0.80 V + · ln 0.02 – 0.34 V E = 0.36 V 96485 Cmol−1 (or taken from the plot at lg x = -1.70)

60 www.ShimiPedia.ir Answers Round 3 Test 1

d) At equilibrium: E(cathode) = E(anode) R ·T R ·T 0.80 V + · ln(c(Ag+)/c0) = 0.34 V + · ln(c(Cu2+)/c0) with c0 = 1 mol/L F 2 · F c(Cu2+) · c0 (0.80−0.34)푉 ·2 · 퐹  ln K = ln = c(Ag+)2 R ·T ln K = 35.8 K = 3.6 · 1015 e) Determination of the concentration of the silver ions: R ·T E = E(cathode) – E(anode)  0.420 V = 0.34 V – [0.80 V + · ln(c(Ag+)/c0)] F  c(Ag+) = 1.31·10-15 mol/L Determination of the concentration of the iodide ions: – m(KI) = 3.00 g M(KI) = 166 g/mol  n0(I ) = 0.0181 mol after addition of the silver nitrate solution: - - + + n(I ) = n0(I ) - n0(Ag ) n0(Ag ) = 0.05 L · 0.200 mol/L = 0.010 mol n(I-) = 0.0181 mol - 0.010 mol n(I-) = 8.1 · 10-3 mol c(I-) = 8.1 · 10-3 mol/0.1 L c(I-) = 8.1 · 10-2 mol/L + 0 - 0 -15 -2  Ksp = c(Ag )/c · c(I )/c Ksp = 1.31·10 · 8.1 · 10 -16 Ksp = 1.1 · 10

Solution to problem 3-03 a) 1 example of the known compounds ClF3, BrF3, IF3, ClF5, BrF5, IF5, IF7, (ICl3)2 or of the also possible ClF7, BrCl3,5,7, ICl5,7, IBr3,5,7 not possible following the statements: FCl3,5,7, FBr3,5,7, FI3,5,7, ClBr3,5,7, ClI3,5,7, BrI3,5,7 b) ClF: 2, BrF: 2, IF: 1, BrCl: 3, ICl: 2, IBr: 2 c) XY3: T-shaped XY5: square pyramidal XY7: pentagonal bipyramidal

= X, = Y, ---- = free electron pair d) XY + H2O HY + HXO e) 3 XY XY3 + X2 or 5 XY XY5 + 2 X2 or 7 XY XY7 + 3 X2

Solution to problem 3-04 a) ZnSO4(s)  ZnSO4(aq): Enthalpy of solution Hsol (ZnSO4) = -0.900 kJ/K · (22.52 – 22.55) K / 1.565 g = -0.873 kJ / 1.565 g -1 With M(ZnSO4) = 161.48 g/mol: Hsol (ZnSO4) = 161.48 g mol · (-0.873 kJ / 1.565 g)

Hsol (ZnSO4) = -90.08 kJ/mol

ZnSO4 · 7 H2O(s)  ZnSO4(aq): Hsol(ZnSO4 · 7 H2O) = -0.900 kJ/K · (21.84 – 22.15) K / 13.16 g = 0.279 kJ / 13.16 g -1 With M(ZnSO4 · 7 H2O) = 287.59 g/mol: Hsol(ZnSO4 · 7 H2O) = 287.59 g mol · (0.279 kJ / 13.16 g)

Hsol(ZnSO4 · 7 H2O) = 6.10 kJ/mol

 HR = Hsol (ZnSO4) - Hsol(ZnSO4 · 7 H2O) = -90.08 kJ/mol - 6.10 kJ/mol

HR = -96.18 kJ/mol b) HR of reaction ½ H2(g) + ½ N2(g) + O2(g)  HNO2(aq) has to be calculated:

NH4NO2(aq)  NH3(aq) + HNO2(aq) - H4 61 www.ShimiPedia.ir Answers Round 3 Test 1

+ NH4NO2(s) + aq  NH4NO2(aq) + H5 + N2(g) + 2 H2O(l)  NH4NO2(s) - H1 + NH3(aq)  ½ N2(g) + 1.5 H2(g) + aq - ½ H3 + 2 H2(g) + O2(g)  2 H2O(l) + H2

½ H2(g) + ½ N2(g) + O2(g)  HNO2(aq) - H4 + H5 - H1 – ½ H3 + H2

HR = (38.1 + 25.1 + 307.4 + 161.7/2 – 571.7) kJ/mol = -120.3 kJ/mol

Solution to problem 3-05 a) 4  r 3 Volume of the cube: V = a3 Volume of the sphere V = = 4,2 r3 3 fcc bcc

Face diagonal of the cube: dface = a 2 Space diagonal of the cube: dspace = a 3 Length of the face diagonal = 4 r Length of the space diagonal = 4 r 4 r = a  a = 2 r 4 r = a  a = 4 r / 3 3 Volume of the cube = (2 r ) = 22.6 r ≙ 100 % Volume of the cube = (4 r/ )3 = 12,3 r3 ≙ 100 % 3 Z = 4  total volume of the atoms = 16.8 r Z = 2  total volume of the atoms = 8.4 r3 Filled space = (16.8 / 22.6) · 100 % = 74 % Filled space = (8,4 / 12.3) · 100 % = 68 %

74−68 b) The difference in density for the same kind of atoms is about · 100 %  8 %. 74 c) 2 LiAlSi2O6 + 4 CaCO3 Li2O + 4 CaSiO3 + Al2O3 + 4 CO2 d) Due to its small size and its charge the lithium cation solvated stronger than the other alkali metals. Thus it dissolves in polar, non-aqueous solvents, too. e) When potassium aluminum sulfate is dissolved hydrated aluminum cations form. These tend to hydrolysis and aluminum hydroxide is formed which can be dissolved again by a solution of sodium hydroxide: 3+ 2+ + + + + [Al(H2O)6] [Al(H2O)5(OH)] + H [Al(H2O)4(OH)2] + 2 H [Al(H2O)3(OH)3] + 3 H – – [Al(H2O)3(OH)3] + OH [Al(OH)4] + 3 H2O f) Otherwise there is the danger that aluminum hydroxide precipitates and falsifies the result.

Remark: At pH = 12.6 the concentration of dissolved aluminum has the highest value.

Dependency of the aluminum(III) concentration of a saturated aqueous solution of pH 25 °C 2– 3+ – + g) H2Y + Al AlY + 2 H h) During the complexation with EDTA protons are set free. An excess of protons in the progression of the titration would shift the equilibrium in the direction of the non-complex metal (or the metal-EDTA complex is less stable). i) EDTA – Al3+ > EDTA – Zn2+ > Ind – Zn2+ > Ind – Al3+

62 www.ShimiPedia.ir Answers Round 3 Test 1

j) Consumption of Na2EDTA = (50.00 – 15.25) mL = 34.75 mL  n(Al) = 3.475 mmol n(Al2O3) = ½ n(Al)  m(Al2O3) = ½ n(Al) · M(Al2O3) m(Al2O3) = ½ · 3.475 mmol · (101.96 mg/mmol) = 177.16 mg m(Li2O) = (198.0 – 177.16) mg = 20.84 mg n(Li2O) = m(Li2O) / M(Li2O)  n(Li2O) = 20.84 mg / (29.88 mg/mmol) = 0.70 mmol

½ · 3.475 mmol Al2O3 : 0.70 mmol Li2O  2.48 : 1  5 : 2  x = 2 und y = 5

Empirical formula: 2 Li2O · 5 Al2O3

Solution to problem 3-06 a) log Kp = -4374/473+ 1.75·log 473 + 3.78 log Kp· = -0.786 Kp = 0.164 0 0 p(PCl3)/p ·p((Cl2)/p 0 b) Kp = 0 with p = 1.000 bar p(PCl3) = p(Cl2) und p(PCl5) + 2· p(PCl3) = 1.50 bar p(PCl5)/p x2 Let x = p(PCl )/1 bar: 0.200 =  x2 + 0.400 ·x 0150 – 0.300 = 0 3 1.5 −2 ·x

x1 = 0.383 (x2 = -0.783)

p(PCl3) = p(Cl2) = 0.383 bar = 38.3 kPa

p(PCl5) = 0.734 bar = 73.4 kPa p(PCl ) n(PCl ) 383 c) 3 = 3 = (1) p(PCl5) n(PCl5) 734

In equilibrium: n(PCl5) = n0(PCl5) – n(PCl3). Assume that n0(PCl5) = 1 mol

 n(PCl3) = 1 mol - n(PCl5) and inserted in (1): n(PCl3) 1 mol− n(PCl5) 383 383 = =  (1 + ) · n(PCl5) = 1 mol n(PCl5) = 0.657 n(PCl5) n(PCl5) 734 734

i.e. (1 - 0.657) mol of PCl5= 0.343 mol of PCl5 ≙ 34.3 % of the original PCl5 have reacted. + – d) PCl5 as a solid consists of PCl4 and PCl6 ions:

Cl + Cl – Cl Cl Cl P Cl P Cl Cl Cl Cl e) There must be a rapid exchange between the axial and the equatorial position which is faster than the NMR experiment. Thus only one signal is detected. Remark: The exchange of the positions takes place in the so called Berry pseudoration. Both, the axial and the equatorial constituents move at the same rate of increasing the angle between the other axial or equatorial constituent. This forms a square based pyramid. The frequency of the rotation for PF5 is about 100000 s–1 (298 K) and cannot be detected by the slower NMR experiment.

X = F, Cl

63 www.ShimiPedia.ir Answers Round 3 Test 1

Solution to problem 3-07 O Systematic name: Propan-2-one Other names: Acetone, Propanone, 2-Propanone, dimethyl ketone, β-Ketopropane OH Systematic name: prop-1-en-2-ol Other names: 1-Propen-2-ol, propen-2-ol,2-hydroxypropene OH 1-Propen-1-ol

OH 2-Propene-1-ol, allyl alcohol, 2-propenol

O Systematic name: Epoxypropane Other names: 1,2-Epoxypropan, 2-Methyloxiran, propylene oxide, propene oxide, CH3 1,2-propylene oxide, methyl oxirane, methyl ethylene oxide, methyleth-

ylene oxide OH Systematic name: Cyclopropanol Other names: Cyclopropyl alcohol, hydroxy cyclopropane

O Systematic name: Oxetane Other names: oxacyclobutane, trimethylene oxide, 1,3-propylene oxide, 1,3-epoxypro-pane Systematic name: Methyl vinyl ether O Other names: methoxyethene, ethyl methyl ether

O Systematic name: Propanal

Other names: Propionaldehyde

Solution to problem 3-08 a)

A: *B: *C: D:

E: F: G:

*i: AlCl3 *ii: CH3Cl iii: CH3COCl iv: HNO3/ v: H2SO4/ vi: HCl vii: Br2

H2SO4 HNO3 * there are more possible solusions Isomerization: Keto-enol tautomerism. b) During the formation of G the temperature must be kept below 5 °C. When heated to boiling 3- nitrophenol is formed:

64 www.ShimiPedia.ir Answers Round 3 Test 1

Remark: A diazonium coupling does not occur because diazonium salts can only attack strongly activated aromatic compounds. The aromatic compound at hand however is deac-

tivated because of the strong –M effect of the NO2 group.

c) The amino group shows a +M effect and is ortho/para directing. For the directing effect of the substituent see organic text books The meta-substituted product does not match to the +M effect of the amino group. Under these strongly acidic reaction conditions the amino group is protonated and therefore without a free electron pair. Thus the +M effect cease to exist and the –I effect works only which leads to a preference of the meta substitution.

Solution to problem 3-09: a) Electrophilic addition (AE) of Br2 at the double bond of the butenone:

Bromonium cation b) Nucleophilic addition (AN) of HCN at the keto group and formation of a cyanohydrin:

c) There is a resonance structure in which the β-C atom is positively charged. There the Grignard reagent may perform a nucleophilic attack:

d) The C atom directly bonded to oxygen is harder because the oxygen attracts the electrons of the C=O bond and thus the carbonyl carbon is less polarizable. The β-carbon atom, however, is less concerned by the electron attraction of the oxygen atom and thus more polarizable, weaker. The hard Cer reagent prefers to attack the harder carbonyl group, the copper reagent the weaker β- carbon.

65 www.ShimiPedia.ir Answers Round 3 Test 2

Answers of Round 3 Test 2 Solution to problem 3-11 a) E b) (Chromium has the oxidation number +6) D, E (Peroxodisulfuric acid) c) B d) B, E e) B, D, E f) B g) B, C (cis-trans), E h) A, C, D Solution to problem 3-12 c(HA) c(HA) a) pK = pH + lg  = 10pKs−pH = 106.5 – 5.8 = 5 : 1 S c(A−) c(A−) b) HCOOH + NaOH  HCOONa + H2O c(HCOOH) = 0.1 mol/L – c(NaOH) and c(HCOO–) = c(NaOH) 0.1 mol/L− c(HCOO−) pK = pH + lg  c(HCOO–) = (0.1 mol/L) / (10pKs−pH + 1) S c(HCOO−) c(HCOO–) = 0.074 mol/L 0.1 mol/L c) It is pK (NH +) = pH + lg  pK (NH +) = 9.25 S 4 0.1 mol/L S 4 It has to be 10-10.25 < c(H+) < 10-8.25 0.1 mol−n(NaOH) 0.1 mol−n(NaOH) - lg K = - lg c(H+) + lg  c(H+) = K · S 0.1 mol+n(NaOH) S 0.1 mol+n(NaOH) 0.1 mol+n(NaOH) Addition of NaOH solution: 10-10.25 < 10-9.25 · 0.1 mol−n(NaOH) 0.1 · (0.1 mol + n(NaOH) < 0.1 mol - n(NaOH) n(NaOH) < 0.0818 mol V(NaOH) = n/c < 81.8 mL The following calculation can be left out and the result immediately given because it is a symmetrical problem. 0.1 mol−n(HCl) Addition of HCl solution: 10-8.25 < 10-9.25 · 0.1 mol+n(HCl) n(HCl) < 0.0818 mol V(HCl) = n/c < 81.8 mL n d) Equation for the freezing-point depression: T = K · m(solvent) n = amount of all dissolved particles Determination of the cryoscopic constant of water: m(solvent) 0.1 kg K = T· K = 0.052 K · m(sucrose)/M(sucrose) 0.957 g/(342.3 g/mol ) K = 1.86 K·kg·mol-1 Determination of the carboxylic acid X: HA H+ + A– + – n = n(HA) + n(H ) + n(A ) n = n0·(1 - ) + n0· + n0· = n0·(1 + ) 2.895 g ·(1+ ) 0.147 K = 1.86 K·kg·mol-1 · M(X) · 0.5 kg It is a weak acid  (1 + )  1  M(X) = 73.3 g/mol X = R–COOH M(X) = 73.3 g/mol = (28.3 +45) g/mol

 R = CH3–CH2 X = propionic acid with M = 74.08 g/mol n ·(1+) T ·m(solvent) e) T = K · 0   = - 1 m(solvent ) K · n0 -1 n0 = 2.895 g /(74.08 g/mol) K = 1.86 K·kg·mol T = 0.147 K m(solvent) = 0.5 kg 0.147 K · 0.5 kg · 74.08 g·mol−1  = -1  = 0.0112 1.86 K·kg·mol−1 ·2.895 g

66 www.ShimiPedia.ir Answers Round 3 Test 2

2 c0 ·  · c0 ·  0  0 0 KS = /c KS = c0 · /c c = 1 mol/L 500 g of water ≙ 0.5 L of water c0 (1− ) 1−  2 −1 2.895 g ·0.0112 ·1 L ·mol -6 K = K = 9.92 · 10 S 74.08 g ·mol−1 ·0.5 L · (1−0.0112) S

Solution to problem 3-13 a) Solution of  NH4Cl BaCl2 Na2S Pb(NO3)2 Na2SO4 AgNO3  weak gas NH Cl n. r. white prec. n.r. white prec. 4 formation

BaCl2 white prec. white prec. white prec. white prec.

Na2S black prec. n.r. black prec.

Pb(NO3)2 white prec. n. r.

Na2SO4 n. r.

AgNO3 b) NH4Cl (aq) + Na2S (aq) NH3 (g) + NaCl (aq) + NaHS (aq)

2 NH4Cl (aq) + Pb(NO3)2 (aq) 2 NH4NO3 (aq) + PbCl2 (s)

NH4Cl (aq) + AgNO3 (aq) NH4NO3 (aq) + AgCl (s)

BaCl2 (aq) + 2 AgNO3 (aq) Ba(NO3)2 (aq) + 2 AgCl (s)

BaCl2 (aq) + 2 Na2S (aq) Ba(OH)2 (s) + 2 NaCl (aq) + 2 NaHS (aq)

BaCl2 (aq) + Pb(NO3)2 (aq) Ba(NO3)2 (aq) + PbCl2 (s)

BaCl2 (aq) + Na2SO4 (aq) BaSO4 (s) + 2 NaCl (aq)

Na2S (aq) + Pb(NO3)2 (aq) PbS (s) + 2 NaNO3 (aq)

Na2S (aq) + 2 AgNO3 (aq) Ag2S (s) + 2 NaNO3 (aq)

Pb(NO3)2 (aq) + Na2SO4 (aq) PbSO4 (s) + 2 NaNO3 (aq) Die Lösungen lassen sich eindeutig identifizieren, da sich das Reaktionsverhalten bei allen unterscheidet, wie aus der folgenden Zusammenstellung sichtbar wird.

Solution of NH4Cl BaCl2 Na2S Pb(NO3)2 Na2SO4 AgNO3 1 x gas Results of the 1 x gas 3 x white prec. 2 x white prec. 4 x white prec.. 1 x white prec.. 2 x white prec.. experiments 2 x white prec. 1 x black prec. 1 x black prec. 2 x black prec.

Solution to problem 3-14 a) G = H – T·S Cp = -7.8

H(T) = H(298 K) + Cp·(T – 298 K) S(T) = S(298 K) + Cp·ln(T/298K) H(873 K) = -198 kJ/mol – 7.8 JK-1mol-1·575 K S(873) = -187 JK-1mol-1 – 7.8·ln(873 K/298 K) H(873) = -202.49kJ mol-1 S(873) = -195.38 JK-1 mol-1 –(G/RT)  G(873) = -31.92 kJ/mol Kp = e K(873) = 81.27 b) By doing so the formation of nitrogen oxides is avoided. c) Assumption: 100 mol mixture at the beginning x: reacted amount of O2 in mol o 5 Kp = 65.00 p = 1.000·10 Pa 67 www.ShimiPedia.ir Answers Round 3 Test 2

SO2 O2 SO3 N2  Amount at the begin 10 11 0 79 100 Amount at equilibrium 10 – 2x 11 – x 2x 79 100 – x 10− 2푥 11−푥 2푥 79 Molar fraction at equilibrium 1 100−푥 100−푥 100−푥 100−푥 10−2푥 11−푥 2푥 79 Partial pressure at equilibrium · p0 · p0 · p0 · p0 p 100−푥 100−푥 100−푥 100−푥 o

4 푥2 ·(100−푥)  65.00 = (10−2 푥)2 ·(11−푥)  256 x3 – 5060 x2 + 35100 x – 71500 = 0 x3 - 19.77 x2 + 137.11 x – 279.30 = 0 x  3.5 (solved by iteration) Amount at equilibrium: (100 – x) mol = 96.5 mol

Ration of volume SO2: (10 – 2·3.5)/96.5·100% = 3.1 % SO3: 7.3 % O2: 7.8 % N2: 81.9 % Conversion of SO2: 2·3.5/10·100% = 70.0 % Solution to problem 3-15 + + + a) H represents H3O bzw. H (aq). – 2- + H2S + 4 Br2 + 4 H2O  8 Br + SO4 + 10 H + – H + OH  H2O n(H+) = n (OH–) = 19.95·10-3 L · 0.100 mol/L = 19.95·10-4 mol -4 + -5 19.95·10 mol H ≙ 19.95·10 mol H2S -5 -5 -3 19.95·10 mol H2S in 10.0 mL volcano water ≙19.95·10 mol/0.0100 L = 19.95·10 mol/L Mass contend = 19.95·10-3 mol/L · 34.086 g/mol = 0.680 g/L – – + b) IO3 + 5 I + 6 H  3 I2 + 3 H2O – – [ggf. I2 + I  I3 ] – + I2 + H2S  2 I + S + 2 H 2- – 2- I2 + 2 S2O3  2 I + S4O6 6.5 g Amount of prepared iodine (I ): n(KIO ) = = 3.04 · 10-4 mol 2 3 100 ·214.0 g/mol -4 n(I2) = 3 · n(KIO3) = 9.12 · 10 mol 0.0100 L ·0.800 g/L Amount of H S: = 2.35 · 10-4 mol 2 34.086 g/mol -4 -4 2.35 · 10 mol of H2S are oxidized by 2.35 · 10 mol of I2 -4 -4 -4 (9.12 ·10 – 2.35 · 10 ) mol = 6.77· 10 mol of I2 are not consumed -4 -4 2- -3 2- 6.77· 10 mol I2 consume 2 · 6.77· 10 mol S2O3 = 1.354 · 10 mol S2O3 in the titration V = n/c  V = 1.354 · 10-3 mol / (0.100 mol/L) = 1.354 · 10-2 L  13.5 mL c) H2S + Ag  Ag2S + H2

Solution to problem 3-16 a) 2 NaN3  2Na + 3 N2 (1) b) – N N N N N N N N N c) 10 Na + 2 KNO3  K2O + 5 Na2O + N2(g) (2)

K2O + Na2O + 2 ϑSiO2  Na2O3Si + Na2O3Si (3) d) Amount of nitrogen in 50 L at T = 423 K and p = 1.3·105 Pa: p · V 1.3 ·105 Pa · 50 ·10−3 m3 n(N ) = n(N ) = n(N ) = 1.85 mol 2 R · T 2 8.314 JK−1mol−1 · 423 K 2

68 www.ShimiPedia.ir Answers Round 3 Test 2

n(N2) = 1.5 · n(NaN3) + 0.1 · n(NaN3) (1) und (2)

1.85 mol = 1.6 · n(NaN3)  m(NaN3) = 1.85 mol/1.6 · 65.0 g/mol m(NaN3) = 75.2 g e)

There are more resonance structures but they have a higher separation of charge. f) V for ϑ > 100 °C and ϑ < 100 °C have to be considered: Reaction, volume in cm3 (ϑ > 100 °C/ ϑ < 100 °C) V for ϑ > 100 °C V for ϑ < 100 °C Hydrogen 2 H + O  2 H O 2 2 2 V = -10 cm3 V =  -30 cm3 20 10 20/ 0 Ammonia 1 4 NH + 3 O  2 N + 6 H O 3 2 2 2 V = 5 cm3 V =  -25 cm3 20 15 10/10 30/ 0 Ammonia 2 4NH + 7 O  4 NO + 6 H O 3 2 2 2 V = -5 cm3 V =  -35 cm3 20 35 20/20 30/ 0 Carbon 2 CO + O  2 CO 2 2 V = -10 cm3 V = -10 cm3 monoxide 20 10 20/20 Ethene C H + 3 O  2 CO + 2 H O 2 4 2 2 2 V = 0 cm3 V =  -40 cm3 20 60 40/40 40/ 0 Methane CH + 2 O  CO + 2 H O 4 2 2 2 V = 0 cm3 V =  -40 cm3 20 40 20/20 40/ 0 At ϑ > 100 °C it could be hydrogen, at both conditions carbon monoxide.

Solution to problem 3-17 a) AsH3  As + 3/2 H2 v = k · c(AsH3) b) pend = 1.5 · 86.1 kPa = 129.15 kPa. -kt c) For a reaction of first order it is c = c0 · e . -kt Using c = n/V and p · V = n · R · T you get with constant V and T: pt(AsH3) = p0(AsH3) · e

 k = ln[pt(AsH3) / p0(AsH3)] / (-t) k = ln[p120(AsH3) / 86.1 kPa] / (-120 min) (1)

Determination of p120(AsH3): p120(AsH3) = p0(AsH3) - x

For each amount of AsH3 which decomposes a 1.5 fold amount of H2 forms:

p0  p120 = p0(AsH3) – x + 1.5 · x = p0(AsH3) + 0.5 · x  86.1 kPa + 0.5 · x = 112.6 kPa x = 53.0 kPa

 p120(AsH3) = 86. 1 kPa – 53.0 kPa = 33.1 kPa (1) k = ln[33.1 kPa / 86.1 kPa] / (-120 min) k = 0.00797 min-1

−푘· 푡1/2 -1 Half-life: ½ p0 = p0 · 푒  t1/2 = ln 2 / k t1/2 = ln 2 / 0.008 min t1/2 = 87.0 min −4 −1 −1.3 ·10 푠 · 푡99% -4 -1 d) 0.01 = 푒  t99% = - ln 0.01 / (1.3·10 s ) t99% = 35424 s t99% = 590 min

Solution to problem 3-18 a) The structural element of epoxides is a three membered ring which has a high ring strain. Thus ring- opening reactions are favored which explains the reactivity of epoxides.

69 www.ShimiPedia.ir Answers Round 3 Test 2

b) O

H3C CH3 H3C CH2 CH I S I Base S H O S 3 -HI CH3 CH3 -S(CH3)2

C D B c)

1,3 Elimination

e) E: OH H: OH H2N HS F: OH I: OH O Br

G: OH J: OH R f) O O LSkat. H K LS O kat. H O

L/M L/M L/M M/L

Solution to problem 3-19 a) Only the nitrogen atom with a double bond adjacent to a carbon atom is protonated. The free electron pair of the other N atom is part of the aromatic system and therefore only weakly basic.

70 www.ShimiPedia.ir Answers Round 3 Test 2

b) c) O

SH OH

OH OH

H2N H2N

O O S-Aspartic acid R-Cysteine (S)-Asparaginsäure (R)-Cystein d) The OH group of the carboxyl function performs a +M effect on the carbonyl atom and reduces its electrophilic power. e)

Tetrahedral structure f) It is a redox reaction (exactly: oxidative coupling):

O NH 2 O NH2

HS OH S OH + + H HO SH HO S 2

NH O 2 NH2 O

Solution to problem 3-20 a) A Acetaldehyde B Acetone C Acetophenone

D Methanoic acid E Aniline F Benzaldehyde

G Nitrile of benzoic acid H 3-Nitrobenzoic acid I Phenol

b) Aldehyde as a Aromatic compound with meta- – functional group directing substituent(s) Acidic reaction in water D H I Reacts with itself in an aldol reaction A C B is a monosubstituted aromatic compound F G E

71 www.ShimiPedia.ir Answers Round 4 (theoretical)

Answers Round 4 (theoretical)

Solution to problem 4-01 a) - the substance is not decompose producing gaseous products - the substance is to produce any other volatile components besides water - the substance is not react with components of the air - the water has to be released totally - the content of water is not change under standard conditions b) The following reactions may occur because sulfuric acid is strongly acidic and may react oxidizing: 1 3 H2S + H2SO4 SO2 + /8 S8 + 2 H2O (potentially 2 H2S + SO2 /8 S8 + 2 H2O)

2HI + H2SO4 I2 + SO2 + 2 H2O

2 NH3 + H2SO4 (NH4)2SO4 (resp. NH3 + H2SO4 NH4HSO4) c) Calcium chloride forms ammine complexes (ammoniates) of different composition [Ca(NH3)n]Cl2 (n = 1, 2, 4, 8). d)

2– – + C C Isoelectronic species: CO, CN , NO , NN

e) [Cu(NH3)2]Cl f) CaC2 + H2O CaO + H2C2

H2C2 + 2 [Cu(NH3)2]Cl Cu2C2 + 2 NH4Cl + 2 NH3 g) I2 + SO2 + 2 H2O 2 HI + H2SO4 – h) I3 ions generate the brown colour. The use of starch solution is not possible because it has to be worked aprotic (free of water) in this reaction. i) Pyridine acts as a base and thus shifts the equilibrium towards the products. j) Average consumption: Oil A = 1.62 mL, oil B = 1.44 mL

m Vconsumption· t Water content in % = water ∙ 100 % = ∙ 100 % mweighed portion mweighed portion 1.62 mL · 4.8 mg/mL Water content of oil A = ∙ 100 % = 0.078 % 10000 mg 1.44 mL ·4.8 mg/mL Water content of oil B = P 100 % = 0.069 % 10000 mg

– k) 1 mol H2O ≙ 1 mol I2 ≙2 mol e 0.09 9 ·10−3 g m(H O) = · 10 g = 9 · 10-3 g n(H O) = = 5 · 10-4 g ≙ 1 · 10-3 mol e– 2 100 2 18 g/mol Q = n(e–) · F Q = 1 · 10-3 mol · 96485 C/mol  96.5 C

Solution to problem 4-02 a) x = (2·0.926 V + 1.154 V):3 = 1.002 V 3+ – b) (1) Au + 3 e  Au E1° = + 1.50 V ΔG1°= -3∙F∙1.50 V – – – (2) [AuCl4] + 3 e  Au + 4 Cl E2° = + 1.00 V ΔG2°= -3∙F∙1.00 V

We have to find the equilibrium (= complex formation) constant Kco for the reaction

72 www.ShimiPedia.ir Answers Round 4 (theoretical)

3+ –  – (3) Au + 4 Cl  [AuCl4]

(3) = (1) – (2) ΔG3 = -3∙F∙(1.50 V – 1.00 V) 25 ΔG3 = - R∙T∙ln Kco ln Kco = 58.42 Kco =2.34∙10

– 1,20 V – 1,18 V – 1,65 V – 1,63 V 1,36 V – c) ClO4  ClO3  ClO2  HClO  Cl2  Cl – y – – Disproportionation: 4 ClO3  3 ClO4 + Cl – – – – E°(ClO4 / ClO3 ) = 1.20 V E°(ClO3 /Cl ) = (2·1.18 + 2·1.65 + 1.63 + 1.36) V / 6 = 1.44 V y = (1.44 – 1.20) V = 0.24 V > 0  G° = -n·F·y < 0  The reaction disproportinates spontaneously. R·T d) It has to be E(H O , H+/H O) = 1.76 V + · ln[(c(H+)/co)2] > 1.61 V (with co = 1 mol/L) 2 2 2 2·F 8,314·298  V· ln[c(H+)/c°]> – 0.15 V ln[c(H+)/c°] > -5.84 c(H+) > 2.91·10–3 pH < 2.54 96485

+ 2+ + e) 5 H2O2 + 2 KMnO4 + 6 H  5 O2 + 2 Mn + 8 H2O + 2 K

Solution to problem 4-03 a) Two examples from the following table. Expected Observed Expected Observed Cr [Ar]3d44s2 [Ar]3d54s1 Gd [Xe]4f86s2 [Xe]4f75d16s2 Cu [Ar]3d94s2 [Ar]3d104s1 Pt [Xe]4f145d86s2 [Xe]4f145d96s1 Nb [Kr]4d35s2 [Kr]4d45s1 Au [Xe]4f145d96s2 [Xe]4f145d106s1 Mo [Kr]4d45s2 [Kr]4d55s1 Ac [Rn]5f17s2 [Rn]6d17s2 Ru [Kr]4d65s2 [Kr]4d75s1 Th [Rn]5f27s2 [Rn]6d27s2 Rh [Kr]4d75s2 [Kr]4d85s1 Pa [Rn]5f37s2 [Rn]5f26d17s2 Pd [Kr]4d85s2 [Kr]4d10 U [Rn]5f47s2 [Rn]5f36d17s2 Ag [Kr]4d95s2 [Kr]4d105s1 Np [Rn]5f57s2 [Rn]5f46d17s2 La [Xe]4f16s2 [Xe]5d16s2 Cm [Rn]5f87s2 [Rn]5f76d17s2 Ce [Xe]4f26s2 [Xe]4f15d16s2 b) Especially favoured are totally occupied (noble gas configuration, d10, f14, but also s2, p6) as well as half occupied (d5, f7, but also p3) electron shells. c) i) Fe3+: [Ar]3d5 ii) Mn3+: [Ar]3d4 iii) Pd4+: [Ar]4d6 iv) Cr3+: [Ar]3d3 v) Fe2+: [Ar]3d6 vi) Pb2+: [Xe]4f145d106s2 vii) Au3+: [Xe]4f145d8 viii) Co2+: [Ar]3d7 ix) Cu+: [Ar]3d10 x) Ti2+: [Ar]3d2 d)

73 www.ShimiPedia.ir Answers Round 4 (theoretical)

f) The degenerate orbitals in an octahedral ligand field split off into three levels of lower and two levels of

higher energy. The energy of the d-orbitals which lie between the axes of the coordinate system (dxy, dyz,

dyz) is lowered, that of the d-orbitals on these axes is lifted because of the stronger interaction with elec-

trons of the ligands.

Energy

g) d4 to d7 h) Fe3+ Mn3+ Pd4+ Cr3+ Fe2+ Pb2+ Au3+ Co2+ Cu+ Ti2+ high-spin 5 4 4 3 4 0 2 3 0 2 low-spin 1 2 0 3 0 0 2 1 0 2

Solution to problem 4-04

p(A )/p° a) 2 = 2.1·103 p(A ) = 2.1·103 · (8.9·10-4)2 bar p(A ) = 1.66·10-3 bar (p(A)/p°)2 2 2 -4 -3 p(A)/p(A2) = 8.9·10 bar/1.66·10 bar p(A)/p(A2) =0.54 k1 b) Steady state of the equilibrium: k1 · c(A)· c(B) = k–1 · c(AB)  c(AB) = · c(A)· c(B) k−1

v = keff · c(A)· c(B) = k2 · c(AB) k1 k1 -1 -1 -1 -1 keff · c(A)· c(B) = k2 · · c(A)· c(B) keff = k2 · keff = 25 s · 15 (mol/L) = 375 s (mol/L) k−1 k−1 c) The formation of the intermediate and its reaction back to the reactant have to be much faster than the reaction of the intermediate to the product. d) G° = H° - T · S° G° = 28.8 kJ/mol – 298 K · 97 Jmol-1K-1 = -106 Jmol-1 G° = - R·T·ln K  ln K = 106 Jmol-1 /(R·T) ln K = 0.0428 K = 1.04

c(CO ) c2(NH ) 2 · 3 푐0 (푐0)2 0 e) G = G° + R·T·ln Q = G° + R·T·ln c(NH ) CO c(H O) (c = 1 mol/L) 2 2 · 2 푐0 푐0 0.097 ·0.022 G = -106 J/mol + R·T·ln G = -34.8 kJ/mol 0.85 ·1000/18 c(F)/c° f) Es ist G = G° + R·T·ln (1) c(D)/c° · c(E)/c° As reaction and back reaction are elementary in the equilibrium (eq) it follows that k3 ceq(F) k3 · ceq(D) · ceq (E) = k–3 · ceq (F)  = k−3 ceq(D) ·ceq(E) ceq(F)/c° k K =  K = 3 · c° ceq(D)/c° · ceq(E)/c° k−3 G° = - R·T·ln K inserted in (1): c(F)/c° k c(F)/c° G = - R·T·ln K + R·T · ln G = - R·T · ln 3 · c° + R·T·ln c(D)/c° · c(E)/c° k−3 c(D)/c° · c(E)/c° k c(F) v G = R·T·(ln –3 · ) G = R·T·ln 3 k3 c(D)·c(E) v−3

74 www.ShimiPedia.ir Answers Round 4 (theoretical)

v v − v v v g) ob = 3 −3 = 1 – −3 = 0.5  −3 = 0.5 v3 v3 v3 v3

v−3 k−3·c(F) k−3 1.8 mol/L k−3 k3 o = = · = 0.5  o = 0.1 · c = K = 10 v3 k3· c(D)·c(E ) k3 0.4 mol/L· 0.9 mol/L k3·c k−3

Solution to problem 4-05 M(A) a) w(A in AB ) = = 0.2381  M(A) = 0.9375 · M(B) 3 M(A) + 3 ·M(B) 2 · M(B) w(B in CB ) = = 0.7314  M(C) = 0.7345 · M(B) 2 M(C) + 2 ·M(B)

M(C) 0.7345 · M(B) w(C in CA) = = x  x = = 0.4393 M(C) + M(A) 0.7345 · M(B) + 0.9375 · M(B) The mass percentage of C in CA is 43.93 %. b) On account of the empirical formulae only monovalent and divalent nonmetals have to be considered for B: Hydrogen, halogens and chalcogens. The solution can be found in a process of elimination by inserting the respective molar masses. You find for B = bromine: M(B) = M(Br) = 79.90 M(C) = 0.7345 · M(B) = 58.68  Element C is nickel (M(Ni) = 58.69) M(A) = 0.9375 · M(B) = 74.91  Element A is arsenic (M(As) = 74.92)

AB3 = AsBr3 (arsenic tribromide) CA = NiAs (nickel arsenide, a semiconductor) CB2 = NiBr2 (nickel bromide) Remark: Neither hydrogen nor fluorine, chlorine and an element of the chalcogen group give reasonable solutions. Only iodine provides elements and compounds if you allow small deviations: You find for B = iodine: M(B) = M(I) = 126.91 M(C) = 0.7345 · M(B) = 93.22  Element C could be niobium (M(Nb) = 92.91) M(A) = 0.9375 · M(B) = 118.98  Element A could be tin (M(Sn) = 118.71)

 CA = NbSn, AB3 = SnI3 and CB2 NbI2. None of these compounds is known.

Solution to problem 4-06 a) Octant Unit cell (8 x octant) 1 III III 4 x /8 Fe 4 Fe  Empirical formula: 1 II II 3 x /8 Fe 3 Fe III II Fe 4Fe 3C18N18H28O14 or 3 x ¼ O + 1 x O 14 O (H2O) III II Fe 4Fe 3(CN)18 · 14 H2O 9 x ¼ CN 18 CN

2+ III III II + b) Fe + K3[Fe (CN)6] K[Fe Fe (CN)6] + 2 K 3+ II III II + Fe + K4[Fe (CN)6] K[Fe Fe (CN)6] + 3 K 2+ III 3– 3+ II 4– c) Fe + [Fe (CN)6] Fe + [Fe (CN)6] 2+ II II II + d) Fe + K4[Fe (CN)6] K2[Fe Fe (CN)6] + 2 K

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The reaction product contains only iron(II) and thus an electron transfer between iron ions of different oxidation states via the cyanide ions, which causes the color, is not possible. e) Coordination spheres:

4– Iro(II) is surrounded by six cyano groups: formal [Fe(CN)6] 3– one iron(III) is surrounded by six cyano groups: formal [Fe(NC)6] three iron(III) are each surrounded of four cyano – groups and two water molecules : formal [Fe(H2O)2(NC)4]

2+ f) Diluted solution: [Cu(H2O)6] – half-concentrated hydrochloric acid: [Cu(H2O)4Cl2] (or [Cu(H2O)3Cl3] ) 2– 2– concentrated hydrochloric acid: [CuCl4] (or [CuCl4(H2O)2] ) g)

O H O B” O H O B O B O H O B” O

O H 2– h) The other anions are disulfide anions (S2 ). Cu2S · CuS2 (= 3 CuS). i) 4 HCl + O2 2 Cl2 + 2 H2O j) CuCl2 + ½ O2 CuO + Cl2

CuO + 2 HCl CuCl2 + H2O

Solution to problem 4-07 a) p = n · R · T/V p = 2 mol · 8.314 Jmol-1K-1 · 273 K /11.2·10-3 m3 p = 405.3 · 103 Pa b) Cp: Heat capacity at constant pressure

CV: Heat capacity at constant volume

Cp > CV because when heating at constant pressure work is done by the system. Since work is done in the

case of Cp it follows that for a given amount of heating the temperature rise is less and therefore that Cp -1 -1 is greater. Cp - CV = R  Cp = 29.4 Jmol K c) Zero -1 -1 d) U = n · CV · T U = 2 mol · 21.1 Jmol K · 100 K U = 4.22 kJ e) 4.22 kJ f) p = n · R · T/V p = 2 mol · 8.314 Jmol-1K-1 · 373 K /11.2·10-3 m3 p p = 553.8 · 103 Pa g) U = H – p · V U = H - (p · V) H = 4.22 kJ + 11.2 · 10-3 m3 (553.8 · 103 Pa - 405.3 · 103 Pa) H = 4.22 · 103 J + 11.2 m3 · 148.5 Pa H = 5.899 kJ -1 -1 (or H = n · Cp · T H = 2 mol · 29.4 Jmol K · 100 K H = 5.88 kJ) h) Work done = p,V-work W = - p · V 2 mol ·R ·373 K W = - 2 atm · 101.3 kPa/atm ( – 11.2 · 10-3 m3) W = -3.93 kJ 2 ·101,3 kPa i) U = H = 0 j) Q = 3.93 kJ 443 443 k) lg 3 = 3.99 -  TS/K = 0,49 + TS = 126.6 K = -146.6 °C 푇푆/K − 0,49 3,99 −lg 3

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l) U = H - (p · V)  H = U + (p · V)

V(CH4(l)) ≪ V(CH4(g))  V(CH4(l))  0 L

 with p · V = n · R · T for 1 mol CH4: (p · V) = R · 112 K – 0 J/mol = 931 J/mol H = U + (p · V) H = 7.25 kJ/mol + 0.931 kJ/mol  8.18 kJ/mol

32.5 kJ are removed  32.5/(8.18 kJ/mol) = 3.97 mol CH4 must evaporate 3,97 mol ·R ·112 K that are V = = 36.5·10-3 m3 = 36.5 L CH 101300 Pa 4

Solution to problem 4-08 a)

A B C D

E F G (–)-Muscone

b) X = MgBr2

Et2O Mg 2 MgBr MgBr2

d) Y = CrO3

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e) Z: Ethene

Solution to problem 4-09 a) In cyclohexane the C-C bond angles are very near to 109.5°, the ideal tetrahedral value. In Cyclopropane these angles are with 60° clearly smaller. The C-C bond length in cyclopropane is a bit smaller than in cyclohexane. The better the overlapping of the orbitals the stronger the bond. The eclipsed position of the hydrogen atoms in cyclopropane weakens the bond compared with the not eclipsed position in cyclohexane.  The C-C bonds in cyclopropane are weaker than in cyclohexane. H H H b) H H H

Mechanism of the reaction:

secondary carbenium ion (more stable) Mainly the more stable secondary carbenium ion is formed and consequently 2- bromobutane is formed. The observed re- gioselectivity follows the rule of Markovni- kov. primary carbenium c) ion (less stable)

Energy Transition state I Intermediate: carbenium ion Transition state II d)

Reactants Hybridization state of the carbon 2 Products atom in the center: sp

Reaction coordinate

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e) Simmons-Smith reaction of (Z)-1,2-Diphenylethene (Remark: The reaction equation is not asked for):

(1R,2S)-1,2-Diphenyl cyclopropane resp. (1R,2S)-1,2-Diphenyl cyclopropane f) OH Br

1 2 3 4 g)

Remark: The methylene hydrogen atoms in the cyclopropane rings are chemically equivalent.

Solution to problem 4-10 a) A B C D

E F G

b) i): Oxidation with K2Cr2O7, dil. H2SO4 (Jones Oxidation)

or with CrO3, dil. H2SO4, acetone or with KMnO4 + - c) ii): Acidic (H3O ) or basic (OH ) opening of the epoxide

iii): Periodate cleavage with NaIO4 d) e) MeLi would only react violently accompanied by deprotonation of the acid. f) Ethane diol, H+ g)

Hemiacetal

Acetal

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20. – 29. July 2015

Examinations

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IChO Baku: Theoretical Test

Theoretical Test

Physical Constants, Units, Formulas and Equations

Universal gas constant R = 8.3145 J∙K–1∙mol–1 Standard pressure p = 1 bar = 105 Pa = 750 mmHg Atmospheric pressure 1 atm = 1.013105 Pa = 760 mmHg Zero of the Celsius scale 273.15 K

Reversible adiabatic process for an ideal gas pV 1/RCV = const

Work made by an ideal gas in an adiabatic process W = nCV (T1 – T2)

Dependence of internal energy on temperature U(T2) = U(T1) + CV (T2 – T1) Gibbs energy G = H – TS  Relation between equilibrium constant and standard G K = exp Gibbs energy RT 

aprod G =  G  RT ln , Dependence of Gibbs energy of reaction on concen- areag tration or pressure a = c / (1 mol/L) for the substances in solution, a = p / (1 bar) for gases

Change of Gibbs energy in time for the system with G Syst  G r   G r two chemical reactions t 1 1 2 2

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Problem 1. New and well-forgotten old refrigerants The problem of choosing a refrigerant for refrigeration and air conditioning systems attracted the attention of scientists and technologists throughout the last century. It has been suggested that during this time refrigerants progressed through four generations. Ammonia, which was ascribed to the first generation, had been used in most of the oldest refrigeration units. It was later replaced by chlorofluorocarbons (CFCs) – derivatives of methane and ethane with the hydrogen atoms replaced by fluorine and chlorine. In Baku, at "Bakkonditsioner" factory, production of the first So- viet serial household air conditioners BK-1500 had been launched. A second-generation refrigerant chlorodifluoromethane CHF2Cl was used in them. In this problem, we compare various refrigerants in terms of thermodynamics.

First air conditioner of Baku factory in a souvenir shop in the Old City (“Icheri Sheher”)

Thermodynamic properties of various refrigerants

–1 –1 –1 Refrigerant “Generation” ΔHvap / kJ·mol (at 280 K) Cv (gas) /J·K ·mol

NH3 1 21.3 26.7

CHF2Cl 2 20.0 48.8

CF3CH2F 3 22.1 79

CF3CF=CH2 4 19.1 120

Consider a model refrigeration cycle consisting of 4 steps schematically shown below in the pressure (p) – internal energy (U) coordinates. p

3 2 (T2) p2 liquid liquid + gas gas

p1 0 (T1) 1 (T1) U

During the first step of the cycle (line 0-1 in diagram), a liquid refrigerant is boiling at constant pressure p1 and temperature T1 (boiling temperature) until it completely evaporates. At this step, the refrigeration unit absorbs heat from surrounding objects. At the second step, the refrigerant under- goes reversible adiabatic compression and heats up to temperature T2 (line 1-2). After that the compressed refrigerant is cooled in a condenser at constant pressure p2 (line 2-3) and then returns to the initial state (line 3-0).

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Let the cycle involve 1 mole of refrigerant, which is initially (point 0) completely liquid, T1 = 280 К, T2 = 380 К, assume that the vapor of any refrigerant behaves like an ideal gas. The thermodynamic characteristics of refrigerants are listed in the table above. 1.1. For each of refrigerants, ammonia and chlorodifluoromethane, calculate the amount of heat Q absorbed by refrigeration unit during heat exchange (line 0-1) and the work W required to com- press its vapor (line 1-2). 1.2. Which quantity(ies) (U, H, S, G, V) remains constant during the compression step? Indicate it by the circle.

To compare the energy efficiency of refrigeration cycles with different parameters and refrigerants, the coefficient of performance (COP) is used, which is defined as a ratio of heat removed from a cooled system to the work of compressor: COP = Q/W. 1.3. Calculate the values of COP in a considered cycle for ammonia and chlorodifluoromethane. 2.1. Why was ammonia replaced by CFCs in household refrigeration units? (Choose only one option) a) to increase the energy efficiency of refrigeration cycles b) to reduce the cost of refrigerant c) for environmental care d) for user security reasons

A search for replacement of CFCs as refrigerants started when it was shown that their use can cause irreparable damage to the protective ozone layer of the atmosphere. The third, ozone-friendly generation of refrigerants came on the scene. Its typical representatives are fluoroalkanes. 2.2. What is the cause of the damage made by CFCs to the ozone layer? (Choose only one option) a) ozone molecule easily adds to C–F bond b) C–F bond is easily broken by radiation, which leads to the formation of free radicals c) ozone molecule easily adds to C–Cl bond d) C–Cl bond is easily broken by radiation, which leads to the formation of free radicals

However, under the 1997 Kyoto Protocol, fluoroalkanes also have to be replaced because they ac- cumulate in the atmosphere and rapidly absorb infrared radiation, causing a rise in temperature of the atmosphere (the greenhouse effect). The refrigerants of the fourth generation such as 2,3,3,3- tetrafluoropropene CF3CF=CH2 have been suggested and are coming into use. 2.3. Why does this compound enhance the greenhouse effect less than fluoroalkanes? (Choose only one option) a) it absorbs less infrared radiation b) it is more reactive and easier to decompose c) it easily reacts with ozone d) it is better soluble in water

3. Calculate the values of the COP in the refrigeration cycle considered above for two refrigerants

of the third and fourth generations – CF3CH2F and CF3CF=CH2. Did the energy efficiency improve

in comparison with CF2Cl2? Choose “Yes” or “No”.

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Unlike household appliances, industrial refrigeration systems are often still using ammonia. It does not contribute to the greenhouse effect nor does it destroy the ozone layer. Industrial units can have a huge size and a large cost. Prior to their construction, they should be carefully modeled taking into account many different factors. In real systems, some part of the refrigerant at the start of the heat exchange with the environment is in the vapor phase (point 0 in the diagram below), and at the end (point 1) it is always overheated above the boiling point. p

3 (T3) 2 (T2) p2 liquid liquid + gas gas

p1 0 (T0) 1 (T1) U

Consider a cycle with 1 mole of ammonia. Its thermodynamic properties are the following: enthalpy –1 of vaporization ΔHvap = 23.35 kJ·mol at Tvap = 239.8 К (boiling temperature at 1 bar pressure). Heat –1 –1 –1 –1 capacity of the liquid phase Cv(liq) = 77 J·K ·mol , of the gas phase Cv(gas) = 26.7 J·K ·mol . Assume that the heat capacities are temperature-independent and the vapor behaves like an ideal gas. The temperature dependence of the saturated vapor pressure of ammonia can be described by the empirical equation: log (p/bar) = 4.87 – 1114 / (T/K – 10.4). During the first step of the cycle (line 0-1 in diagram), the equilibrium mixture of liquid refrigerant and its vapor receives heat from the environment at constant pressure p1 = 3.0 bar. The refrigerant completely evaporates and overheats up to the temperature T1 = 275 K. In the beginning of the process (point 0), the molar fraction of gaseous ammonia is x = 0.13.

4.1. Calculate the initial temperature of refrigerant T0, its volume change ΔV and the amount of heat Q absorbed by refrigeration unit during this step.

Then the refrigerant is reversibly and adiabatically compressed. It heats up to the temperature T2 = 393 К (line 1-2). 4.2. Find the work W required for compression and the COP of the system.

At the next step corresponding to the line 2-3 in diagram, the compressed refrigerant is cooled in a condenser at constant pressure. Then it returns to the initial state through adiabatic expansion with zero work (line 3-0).

4.3. Determine the temperature T3 at point 3 to which the refrigerant is cooled in a condenser.

In the production of refrigeration units it is necessary to consider climatic factors. If a condenser is cooled by atmospheric air, the temperature T3 is higher as higher is the air temperature.

4.4. How will the COP change if T3 increases while T0, T1, T2 remain the same? a) Increase b) Remain the same c) Decrease

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Problem 2. Coupling of chemical reactions

I.Prigogine (left) N. Shilov W. Ostwald When in the system one reaction allows another one to proceed they say that these two reactions are coupled. Ilya Prigogine, Nobel prize winner in chemistry (1977) in his books widely used the concept of “coupled reactions”. Coupling of reactions is an essential feature of living systems, including human body. How one reaction makes another one to occur? In this problem we are going to discuss several possible mechanisms of coupling. (I) “Chemical coupling” “On Chemical coupling” was the title of the dissertation defended by Russian chemist N.Shilov in 1905. N. Shilov was the graduate student of famous professor W. Ostwald from Germany. Dr. Shilov described the following set of reactions. The substance А does not react with Ac. In the presence of the third reagent (called inductor), In, however, the reaction of А with Ac takes place: AAс In the absence of In no reaction! (1)

In the presence of In A Ас  Р1 (2) In is not a catalyst! Its concentration decreases in the course of the reactions. According to the mechanism proposed by Shilov, Ас reacts not with A itself, but with the intermediate product R of the reaction of А with In: ()a A   In  R  P 2 (3) ()b R Ас  P1  and  are stoichiometric coefficients. Other stoichiometric coefficients and all partial reaction or- ders in both reactions are equal to unity. n In the Shilov’s experiments the ratio of the consumed amounts of Аc and In, I  Ас increased up nIn to the constant value with the increasing initial concentration [Ac]0 at [In]0 = const.

1.1. What was this limiting constant value of I at [Ac]0  , [In]0 = const?

1.2. Plot on the graph the dependence of I vs [In]0 at [Ac]0 = const. Assume that In was completely consumed and Аc was in excess. Use the steady-state approximation if necessary.

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What if Shilov’s mechanism is not valid and In is a conventional catalyst of the reaction (2)? Simulta- neously In reacts with А and its concentration decreases. The reaction mechanism in this case is

()a A   In  P2 In, catalysis (4) ()b A Ас  P1

1.3. What is the limiting value of I for the reaction mechanism (4) at [Ac]0  , [In]0 = const?

(II) «Kinetic coupling» The Gibbs energy of the gas-phase reaction k 5 Br H2  HBr H (5) k5 is positive, G(5) = 66 kJmol–1 at Т = 600 К and standard pressures of all reactants and products. r 2.1. What is the ratio of the rates of forward and reverse reactions, 5 , under such conditions? r5

Reaction (5) proceeds in the forward direction due to the reaction (6) which simultaneously occurs in the system: k 5 Br H2  HBr H k5

k6 H Br2  HBr  Br (6) k5, k–5, k6 are rate constants of forward and reverse reaction (5) and forward reaction (6), respectively. This is the kinetic coupling of two reactions.

Let pressures of neutral molecules keep standard values p(H2) = p(Br2) = p(HBr) = 1 bar, and pressures of radicals p(H), p(Br) reach quasi-stationary values. Rate constant k6 is 10 times larger than k–5.

2.2. Calculate G(5) and under such conditions.

2.3. Compare the results obtained in 2.1 and 2.2. How does kinetic coupling affect the reaction (5)? (a) It makes forward reaction to proceed faster,

(b) It changes the sign of the observed rate robs = r5 – r–5 , (c) It makes reverse reaction to proceed slower. Choose only one

(III) ”Second law of thermodynamics restricts coupling” According to the Second Law of thermodynamics occurring of two simultaneous chemical reactions

GSyst in the system should decrease the system’s Gibbs energy Gsyst,  0 . t One of these reactions may have positive Gibbs energy and still proceed in the forward direction due to the coupling with the second reaction. This second reaction must have negative Gibbs energy and the requirements of the Second law must be fulfilled! Consider the example. The synthesis of urea

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–1 2NH3 + CO2 = (NH2)2CO + H2O (7) G(7) = 46.0 kJmol is supposed to be coupled with the complete oxidation of glucose –1 –8 –1 – 1/6 C6H12O6 + O2 = CO2 + H2O (8) G(8) = –481.2 kJmol , r(8) = 6.010 M min 1. Both reactions are presented schematically. No other reactions are considered. 3.1. What is the maximum rate of the reaction (7) permitted by the Second Law?

Assume that partial pressure of CO2 in the system was twice increased. 3.2. By how many times will the maximum permitted rate of the reaction (7) be changed under such conditions? Temperature is 310 К.

Problem 3. Two binding centers – competition or cooperation? Many chemical reactions in living organisms include the formation of “host-guest” complexes where the host molecule reversibly binds one or several guest molecules. Consider a host molecule H with two binding centers – say, a and b which have different affinity to guest molecules:

 []HGa H + G  HGa K = a [HG ][ ] []HG H + G HGb K = b Kb  Ka. b [HG ][ ] where Ka,b are the binding constants for the centers a and b, brackets denote molar concentrations. Attachment of one G molecule to H can change the binding ability of the second centre. This change is described by the “interaction factor”  which reflects the influence of one binding center on another and is defined as follows: []HG HGa + G HG2 2 = Kb [HGa ][ G ] where HG2 is the completely bound complex. 1.1. Determine the range of values (or one value, if necessary) of  which correspond to three possible ways of interaction between binding centers: a) cooperation (binding by one center facili- tates subsequent binding); b) competition (first binding complicates the second); c) independence (no interaction).

1.2. Find the equilibrium constant for the process: HGb + G HG2.

The solution was prepared with the initial concentrations [H]0 = 1 M and [G]0 = 2 M. After the reactions were completed, the concentration of H decreased by 10 times and that of G – by 4 times. For these host and guest, Kb = 2Ka.

2.1. Determine the concentrations of all other species in the solution and find the binding constant Ka and the factor .

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2.2. Determine the relative Gibbs energies of H and all host-guest complexes2 at [G] = 1 M and  = 3. In the scheme on the answer sheet, write the corresponding formula near every line.

Some amount of G was added to 1 mole of H and the mixture was dissolved in water to obtain 1 liter of the solution. The number of the totally bound molecules HG2 in the solution is equal to the total number of single-bound molecules HG.

2.3. Find the initial amount of G (in mol). The constants Ka and Kb and the factor  are the same as in question 2.1. 2.4. What would be the equilibrium composition of the solution if: a)  = 0; b)  is very large (  ).

The constants Ka and Kb as well as the initial concentrations of H and G are the same as in question 2.1.

Problem 4. From one yellow powder to another: A simple inorganic riddle

The yellow binary compound X1 was completely dissolved in concentrated nitric acid by heating, the gas evolved is 1.586 times denser than air. Upon adding an excess of barium chloride to the solution formed a white solid X2 precipitates. The filtrate reacts with an excess of silver sulfate solution forming a precipitate of two solids X2 and X3, also separated by filtration. To the new filtrate the solution of sodium hydroxide was being added drop-wise until the solution became nearly neutral. At this time a yellow powder X4 crystallized from the solution. The mass of X4 is nearly 2.4 times larger than that of X2.

1. Determine the chemical formulae of X1 – X4. 2. Write down all reaction equations in molecular form.

3. In the structural unit of X1 all atoms of one element are in equivalent positions. Draw the struc-

ture of X1.

4. Predict the products of X1 interaction with: a) excess oxygen; b) excess concentrated sulfuric acid;

c) solid KClO3 at friction. Write down the reaction equations.

Problem 5. Indispensable glucose Carbohydrates are the most important providers of energy for living cell. Monosaccharide glucose is a source of energy for the living cell, but for persons who suffer from diabetes glucose may be dan- gerous. High level of glucose may lead to hyperglycemia and even death. That is why people avoid consuming too much carbohydrates and glucose particularly.

1. Determination of reducing sugars in fruit juice One of the technique for determination of reducing sugars in different samples includes the use of Fehling's reagent. This reagent was prepared by mixing 50.00 mL of 0.04000 M copper sulfate (solution A) and potassium-sodium tartrate and sodium hydroxide (solution B). A 10.00-mL aliquot of fruit

2 Assume for simplicity that G(G) = 0

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juice was transferred into a titration flask and Fehling's reagent was added. Solution C thus obtained, was then heated and red precipitate was formed. 1.1. Write the balanced ionic equation of chemical reaction occurring upon heating of the solution C. Use Cu2+ for copper.

After that 10 mL of 10% solution of potassium iodide and 1 M sulfuric acid were added to the flask. The mixture was covered with watch glass and was then placed in a dark place. An excess of iodine was then titrated with 0.05078 М sodium thiosulphate solution. 11.87 mL of the titrant was required to reach the endpoint. 1.2. What happened within the flask? Write the balanced equation(s) in molecular or ionic form. 1.3. Assuming the initial sample contained only glucose and fructose, and all fructose was transformed into glucose under the experimental conditions, calculate the total mass content of sugars (in g/L) in a fruit juice.

A new 10.00-mL aliquot of the same juice was treated with a 10.00-mL portion of acidified potassium iodate solution (0.01502 М) and 10 mL of 10 % solution of potassium iodide. After the mixture has darkened, an excess of sodium hydroxide solution was added. The flask was then covered with a watch glass and put into a dark place. The obtained solution was acidified and titrated with 0.1089 M solution of sodium thiosulphate. The average titrant volume used for titration was 23.43 mL. 1.4. Write the balanced equation(s) for the described reactions in molecular or ionic form. 1.5. Why is it better to synthesize iodine in situ (in the same titration flask) than to add its standard solution? Choose one answer. a) The aqueous iodine solution is unstable b) Iodine evaporates c) Iodine precipitates d) Standard solution of iodine cannot be obtained 1.6. Calculate the mass content of each sugar (in g/L) in the juice. 1.7. One bread exchange unit (1 BEU) corresponds to the content of 12 g of digestible carbohydrates in product. How many BEU are in one glass (200 g) of juice?

2. Diagnosis of diseases The derivative of glucose, 2-deoxy-2-(18F)fluoro-D-glucose (FDG), is the most common radiopharma- ceuticals for diagnosis of cancer using positron emission tomography. The first step of FDG preparation is to produce a radionuclide fluoro-18 by nuclear reaction in a cyclotron. The next step is the radiochemical synthesis. Fluorine-18 is introduced into D-glucose molecule by nucleophilic substitution. 18F-2-deoxy-D-glucose once injected into the patient actively accumulates in cells of malignant tumors; this process is accompanied by decomposition of fluorine-18. This radionuclide is a β+ emitter – nucleus emits a positron (anti-electron). Positron interacts with an electron and after that an- nihilation occurs, which can be detected. This allows determining precisely the tumor location.

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2.1. Complete the nuclear reactions leading to various fluorine isotopes. a) 18O + p  n + … 2 18 b) … + 1 D  F +  c) 19F +  20F + … b) 16O + …  18F + p + n

The decay mode of unstable light nuclei depends on the ratio between neutrons and positrons in them. If this ratio is greater than that for a stable isotope then the nucleus decays in a β–-mode, if it is smaller – in a β+-mode. 2.2. Determine the type of decay for the following nuclei: 11С, 20F, 17F and 14C.

When nuclear reaction (a) is used for fluorine-18 preparation, the target material is presented as 18 16 water enriched with H2 O. The presence of usual water H2 O leads to a side nuclear reaction with 16O, leading to the formation of isotope 17F. 2.3. It is known that within five minutes after completion of irradiation the target the ratio of radioactivities of 18F and 17F is 105. Assuming that the radioactivity of each isotope is proportional to the nuclear reaction yield and the mole fraction of a component in the irradiated target, calcu- 18 18 17 late the mass fraction of H2 O in the target. T1/2( F) = 109.7 minutes, T1/2( F) – 65 seconds. The ratio between nuclear reactions yields is η(18O→18F) / η(16O→17F) = 144.7. 2.4. Calculate the yield of labeling D-glucose with fluorine-18, if initial radioactivity of a fluorine-18 sample was 600.0 MBq and radioactivity of the obtained 18F-2-deoxy-D-glucose is 528.2 МBq. Synthesis time is 3.5 minutes. 2.5. Biological half-life of 18F-2-deoxy-D-glucose is 120.0 minutes. How much radioactivity (in MBq) will remain in the patient one hour after injection of FDG with the initial radioactivity of 450 MBq.

Problem 6. Bread is the staff of life When you pass by the bakery, you are stopped by the smell of freshly baked bread. The hero of one of the novels said on a similar occasion: "If you tell me that this is not perfect, you are my enemy forever." The principal bread flavor component was identified in 1969 as compound X which occurs in equilibrium with its tautomer Y in a 2:1 ratio. Unfortunately, both forms are labile, and after some hours bread has no the same nice smell. This tautomeric mixture of X and Y was synthesized in 1993 from piperidine by the reaction sequence given in Scheme 1. Scheme 1.

It is noteworthy that the initial ratio of X and Y was 1:4; on standing this ratio gradually changed to the equilibrium one.

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Compound A which is characterized by 3-fold axis of symmetry (i.e., rotation by 120 results in a molecule indistinguishable from the original) occurs in equilibrium with its diastereomer B. The in- terconversion of these two forms proceeds via very reactive intermediate E which is also intermediate in A and B formation as well as their transformation to C. Compounds A, B, and E have the same elemental composition: C = 72.24%, H = 10.91%, N = 16.85%. 1. Write down the structural formulae of compounds A-E, X, Y.

Treatment of compound D with MeLiLiBr complex in Et2O at 0 C failed to produce the target products X and Y. Instead, a yellow precipitate F was initially formed. Aqueous workup of this precipitate led to the mixture of compound D and its tautomer G. 2. Write down the structural formulae of compounds F and G.

Another approach to compound X is based on the use of pipecolinic acid derivartive H. It was shown that X can be synthesized by reaction sequence presented in Scheme 2. Scheme 2.

3. Write down the structural formulae of compounds I and J

Problem 7. Not by bread alone Pomegranate is called in Azerbaijan, which is famous for its vegetables, as the “king of all fruits”. Pomegranate is honored in various religions as a “fruit of Paradise”, symbol of righteousness, wealth, hope for eternal life. In 1878 alkaloid pelletierine was isolated from the bark of pomegranate tree (Punica granatum L., Lythraceae). This alkaloid is traditionally used as an anti-helminthic drug. Initially a wrong structure of (3-(piperidin-2- yl)propionaldehyde) XW was proposed for pelletierine. But now it is accepted that natural pelletierine is (S)-1-(piperidin-2-yl)propan-2-one (XS).

1. Write down the structural formulae of XW and XS (the latter – with the stereochemical information).

The synthesis of natural pelletierine (XS) based on the transformation of nortropanol A was recently described.

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2. Write down the structural formulae of compounds B-G with the stereochemical information. 3. Nortropanol A was used in this reaction as a single stereoisomer. How many stereoisomers can exist for compound A?

Enantiomer of XS was synthesized using chiral tert-butanesulfinamide (H):

4. Write down the structural formulae of compounds I-L with the stereochemical information.

Problem 8. Oil for Life and Life after Oil Azerbaijan is known for its vast oil and gas fields. The first drilling for oil was done in Bibi-Heybat in 1846, 13 years before establishment of the first commercial oil well in Pennsylvania (USA). This remarkable date in the history of Azerbaijan is regarded as a starting point of contemporary oil industry, the leading sector of today’s world economy. Currently, on- land and shelf sea oil production is being developed in Azerbaijan. Though serious precautions are taken, there is always a risk of hydrocarbon pollution of the environment during production, transportation, and processing of oil. In this task we will consider diverse technologies of oil spills clean up and specific features of metabolic pathways involved. Application of complex solvents (dispersants) leading to capture of marine oil spills is among most promising clean up approaches. Organic substance X (11.94% of H by mass) is a typical component of such dispersants. Safety of X to human is fiercely debated. X1 (54.53% of carbon by mass) composed

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of three elements and excreted with urine is the major metabolite of X in humans. The numbers of atoms of different elements in X1 are three consecutive terms of a geometric progression, whereas the sum of these numbers does not exceed 25.

1a. Decide on the relationship (tick the correct variant) between the numbers of carbon and oxygen atoms in X1 n(C) > n(O)  n(C) < n(O)  n(C) = n(O)  Data insufficient 

1b. Derive the empirical formula of X1 (hereafter always show your work where required).

The biotransformation of X into X1 occurs in two enzymatically catalyzed steps according to the hereun- der reaction equations (NAD+ and NADH are the oxidized and reduced forms of nicotinamide adenine dinucleotide, respectively): Х + NAD+ → X0 + NADH + H+ (1) + + X0 + NAD + H2O → X1 + NADH + H (2) 1c. Derive the molecular formula of X.

A minor metabolic transformation of X is catalyzed by cytochrome P450-dependent monooxygenase. This reaction leads to two compounds X2 (51.56% of oxygen and 9.74% of hydrogen by mass) and X3. 1d. Derive the molecular formula of X2 and draw its structure.

X contains only primary and secondary carbon atoms. X0 and X3 contain common functional group. 1e. Draw the structural formulae of X, X1, and X3.

In a medical study, personnel working with X-based solvents without proper protection was found to have a stationary concentration of X in blood.

2. Choose the graph of X1 daily mass content in the body of a volunteer participated in this experiment. Write down the number of the correct graph.

1 2 3

mass

mass mass

0 6 12 18 24 0 6 12 18 24 0 6 12 18 24 time, h time, h time, h

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4 5 6

mass

mass mass

0 6 12 18 24 0 6 12 18 24 0 6 12 18 24 time, h time, h time, h

The use of different bacteria is also considered as a promising way for the removal of hydrocarbon (even aromatic) contaminants from sea water and soil. Under aerobic conditions, benzene under- goes biodegradation as follows (first three steps are balanced):

Under the same conditions, a monocyclic aromatic hydrocarbon P (91.25% of carbon by mass) un- dergoes the following transformation (first three steps are balanced):

P3 gives a positive iodoform test. A 100 mg sample of P3 requires 6.41 mL of 0.100 M KOH solution for complete neutralization. 3. Derive the structures of P–P3.

Microorganisms Alicycliphilus are capable of biodegradation of aromatic hydrocarbons even under anaerobic conditions. The process requires a suitable electron acceptor such as inorganic anion Y1 (in soil).

The intermediate anion Y2 is enzymatically decomposed according to the reaction equation: Y2(aq) = Y3(aq) + Y4(g), wherein each of Y3 and Y4 is composed of atoms of only one element. T2 does not contain identical functional groups, and by contrast to Y3 gives a precipitate when treated with the ammonia solution of Ag2O. 4. Deduce and draw the structures of Y1-Y4, T1-T2.

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PRACTICAL EXAMINATION JULY 23, 2015

List of Chemicals Concen- Name State Quantity Labeled Hazard Codes tration Task 1 3-Methyl- Solution in 3-methylthio- 4g/8 mL 4 g H225, H302, H332 thiophene CCl4 phene in CCl4 1-Bromo-2,5-pyrro- NBS Solid - 7.3g H302, H314 lidinedione (NBS) 7,3 g Carbon tetrachlo- H301, H331, H311, H317, Liquid - 24 mL CCl ride 4 H351, H372, H402, H412

Unknown catalyst in CCl4 Catalyst Potassium car- Solid - 0.02 g K CO H315, H319 bonate 2 3 Task 2 Test solution con- Aqueous To be de- 100 mL Test solution H302, H312, H314, H332 taining VO2+ and Cr3+ solution ter-mined Aqueous H290, H302, H314, H332, Sulfuric acid 1 M ~ 500 mL 1M H SO solution 2 4 H351 Potassium perman- Aqueous 0.03 M 15 mL 0.03 M KMnO H272, H302, H400, H410 ganate solution 4 Aqueous Oxalic acid 0.03 M 30 mL 0.03 M Н С О H314, H318 solution 2 2 4 Phenylanthranilic Aqueous 0.1 % 5 mL Indicator acid solution Ammonium iron(II) Aqueous Read from 100 mL Mohr's salt H302, H315, H319, H335 sulfate solution the label Aqueous Silver nitrate 0.3 % 5 mL 0.3 % AgNO H272, H302, H314, H410 solution 3 Ammonium persul- Aqueous H272, H302, H315, H317, 10 % 70 mL 10 % (NH ) S O fate solution 4 2 2 8 H319, H334, H335 Task 3 Diclofenac containing Aqueous To be de- 5 mL Control H301 medicine solution ter-mined Potassium perman- Aqueous KMnO 610-3 M ~ 30 mL 4 H272, H302, H400, H410 ganate solution 6×10-3 M Sulfuric acid (in the Aqueous H290, H302, H314, H332, same bottle as for 1 M ~ 500 mL 1M H SO solution 2 4 H351 Task 2) Diclofenac sodium Aqueous ~ 600 DCF ~ 20 mL H301 salt solution mg/L 600 mg/L

The hazard statements description was given.

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List of labware and equipment Item Quantity Located On the tables for common use Refractometer Refracto 30GS 1-2 / 1 lab Under the hood Napkins for refractometer cleaning Under the hood Wash bottle “Cleaning solvent” for the refrac- Under the hood tometer Aluminum foil for wrapping 1-2 rolls / 1 lab On lab assistants’ table Balances 1-3/ 1 lab On separate tables Gloves (S, M, L) On lab assistants’ table

Large bottle labeled “H2O dist.” Near the sink Napkins for general purposes 1 Pack / 1 row Near the sink

Item Quantity As labeled on Fig. 1, 2, 5 On each working place, to be used in more than one task Hot-plate magnetic stirrer 1 Waste bottle labeled “Waste” 1 Cotton gloves 1 pair

Wash bottle, 500 mL, labeled “H2O distilled” 1 Pipette pump, 10 mL, green 1 Pipette pump, 2 mL, blue 1

Graduated cylinder, 25.0 mL for H2SO4 only 1 Safety goggles 1 Napkins for general purposes 1 pack Task 1 Laboratory stand 2 1 Round-bottom three-necked flask, 100 mL 1 2 Reflux condenser, connected to water supply 1 3 Glass ground joint stopper 6 (with your student code) 4 Dropping funnel, 50 mL 1 5 Oval magnetic stir-bar (big) 1 6 Pear-shaped round-bottom flask for distillation, 1 7 50 mL Claisen distillation adapter 1 8 Thermometer with fixed ground joint tube 1 9 Buchner type fritted glass filter 1 10 Rubber spacer for vacuum filtration 1 11 Liebig (downward) condenser 1 12 Distilling receiver cow 1 13 Receiver flask, 10 mL 4 (with your student code) 14 Receiver flask, 50 mL 1 15 Adjustable lab jack lift support 1 16 Oval magnetic stir-bar (small) 1 17 Plastic beaker, 50 mL, labeled “For the receiver 1 with the product”

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Teflon sleeves for ground tapered joints 12 Large funnel, 65 mm, with short stem 1 Joint clips 5 18 Grey clamp 1 19 Red clamp 1 20 Permanent marker 1 Glass beaker, 25 mL 1 Plastic container labeled “Used glassware” 1 Plastic container labeled “Ice bath” 1 Digital manometer 1 Cotton wool 3 Spatula 1 Glass rod 1 Ruler 1 Pencil 1 Task 2 Laboratory stand 1 Clamp for burette 1 Plastic beaker, 100 mL, labeled “Waste” 1 Glass beaker, 150 mL 1 Volumetric flask with a stopper, 100 mL 1 Small funnel, 45 mm 1 Medium-size funnel, 55 mm 1 Watch glass 1 Burette, 25.00 mL, clamped in the stand 1 Volumetric pipette, 10.00 mL 1 Graduated pipette, 5.00 mL 1 Erlenmeyer flask, 150 mL 2 Graduated cylinder, 100.0 mL 1  Pasteur pipette 2  White paper sheet 1

Task 3 Photometer, 525 nm 1 1 Thermostat with adaptor 1 2 Spectrophotometer cell with 3.5 cm optical 2 3 path length Magnetic stirrer 1 4 Magnetic stir-bar (medium-size) 1 Netbook with adaptor and mouse 1 Volumetric flask with a stopper, 100 mL 1 Graduated pipette, 2 mL 2 Memory stick 8 Gb 1 Black magnet 1

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TASK 1. Tuning bromination selectivity by catalysis Selectivity of chemical reactions is one of the most challenging problems of contemporary research. In many cases, reaction conditions and the catalysts applied are keys to achieving high selectivity of organic reactions. In this task, you will study one of such cases. 3-Methylthiophene can theoretically be transformed into four monobrominated derivatives T1-T4, which have been actually synthesized and characterized in detail. Structures of T1-T4 and the values of refractive indexes are given in Ta- ble 1. Table 1. Structures and refractive indexes of monobrominated thiophenes. Designation A B T3 T4 Br

Structure Br Br S S S Br S 20 nD 1.5961 1.5706 1.5786 1.5795

The selective synthesis of each of T1-T4 can be performed using 3-methylthiophene as the starting material. T1 and T2 can be obtained by direct bromination using different catalysts, whereas T3 and T4 are the products of “one pot” multistep synthesis (see Scheme 1).

NBS, CCl4 NBS, CCl4 T2 T1 AIBN cat. HClO4 cat.

1. 3.5 eq. Br , 2 S NaOAc, 1. BuLi, TMEDA,

H2O, 100 °C Et2O, rt T4 T3 2. Zn dust 2. CBr4, -70 °C

NBS = N Br AIBN = NC N N CN

N TMEDA = N

Scheme 1. Selective synthesis of monobrominated thiophenes.

Q1. Assign the structures given in Scheme 1 with T1, T2 to the structures given in the Table 1. In this task, you will: - Synthesize a monobrominated thiophene derivative using one of the catalysts from the list below;

- Measure the product refractive index (nD) - Compare the results obtained with literature data and decide on the product structure and the catalysts given.

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List of possible catalysts: HClO4 in CCl4, AIBN in CCl4

PROCEDURE - Apparatuses used in this task are shown in Fig. 1 and 2. - Always equip every joint with the Teflon sleeve. Immediately place every piece of the used glassware in the corresponding container. Always keep the container tightly closed. - You can use cotton gloves when handling hot things!

Step 1. Clamp the three-necked flask on the laboratory stand on top of the hot-plate magnetic stirrer. (Fig.1). Apply the dropping funnel and the reflux condenser to the corresponding necks and put the big magnetic stir-bar into the flask through the open neck. Ask your lab assistant to switch on water supply in the reflux condenser (Do not do it yourself!). Transfer NBS quantitatively into the flask using spatula and big plastic funnel. Transfer ~15 mL of CCl4 into the 25 mL glass beaker. Pour ~2/3 of the solvent volume from the beaker into the flask. Shake the Catalyst and quantitatively add it through the same plastic funnel into the flask. Add the rest of the solvent from the beaker to the flask. Close the open neck with the stopper. Put the flask into the ice bath filled with water and ice to ~ 2/3 of its volume. Start stirring the mixture.

Fig. 1. Set up needed to implement Steps 1-4 of the synthesis. Refer to page 4-5 for the numbers

Step 2. Using the big plastic funnel quantitatively transfer the solution of 3-methylthiophene to the dropping funnel with tap closed. Apply a piece of the cotton wool to the open end of the dropping funnel and reflux condenser. Under vigorous stirring, add the solution of 3-methylthiophene dropwise during ~ 3 min. Replace the dropping funnel by a glass stopper. Use the Teflon sleeve. Re- move the ice bath. Dry the plate and bottom of the flask with napkin. Step 3. Wrap up the flask with aluminum foil. Switch on the heater (position 3). Bring up the mixture to boiling and boil it for 10 min. Prepare the ice bath (~2/3 of the volume) while the mixture boils. Step 4. Switch off the heater and carefully (hot!) remove the hot-plate magnetic stirrer aside. Dip the flask equipped with the condenser and stoppers into the ice bath for 3-5 min. Keep gently swirling the flask from time to time to provide the faster cooling. Then remove the reflux condenser and load 0.02 g of K2CO3 using the big funnel applied to the open neck. Close the neck with a glass stopper and shake the flask several times. Turn off the water supply and unscrew the adaptors of the tubings from the reflux condenser. Let the residual water leak out of the condenser and immediately

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place it into the container for the used glassware. Remove the clamp keeping the flask in the ice bath. Step 5. Weigh the 10 mL receiver flask for product with the glass stopper, both marked with your student code. Write down the value in the answer sheet. Put the small magnetic stir-bar in the 50 mL pear-shaped distillation flask. Screw on the adaptors with tubings to the Liebig condenser and fix it on the stand with the red clamp. Turn on the water supply yourself and make sure there is no water leakage. Step 6. Assemble the distillation apparatus as shown on Fig. 2 supplying all the joints with the teflon sleeves and clips. First, attach two 10 mL and one 50 mL receiver flasks to the distilling receiver cow. Then connect the vacuum hose to the cow and complete assembling. Fix the apparatus over the magnetic stirrer to adjust it on height. Use the adjustable lab jack lift support.

Fig. 2. Set up needed to implement Steps 5-10 of the synthesis. Refer to page 4-5 for the numbers

Step 7. Remove the hot-plate magnetic stirrer aside. Insert the fritted glass filter into the Claisen distillation adapter using the rubber spacer. Turn on the water-jet pump and switch on the digital manometer. Remove the three-necked flask from the ice bath and dry it with a napkin. Carefully transfer the reaction mixture from the three-necked flask to the filter (Attention! If you do it too fast, the mixture can be partially sucked into the curved part of the adaptor). When finished, turn off the pump and replace the filter with a glass stopper, use the teflon sleeve. Step 8. Tightly wrap up the flask and distillation adaptor with aluminum foil up to the thermometer joint. Bring back the magnetic stirrer and turn on stirring and heating (position 6). Do not turn on the water-jet pump! Collect the distilled solvent into the 50 mL receiver. When the solvent distillation is over, turn off heating and stirring and carefully (hot!) remove the hot-plate magnetic stirrer aside. Replace the receiver flask containing the distilled solvent by a new 10 mL receiver. Close the 50 mL flask with a glass stopper and deliver it to your lab assistant. Step 9. Remove the foil and put the pear-shaped bottom flask into the ice bath for 2-3 min to bring the temperature below ambient. Remove the ice bath; dry the flask with a napkin. Bring back the magnetic stirrer under the flask (Attention! The hot-plate may be still hot!). Turn on stirring. Wrap up the flask tightly with aluminum foil. Switch on the water-jet pump. When vacuum is stabilized (follow the reading of the digital manometer), turn on the heater (position 6). Observe the ini-

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tial stage of the targeted product distillation and collect the first 3-5 drops into an attached receiver flask not labeled with your student code. Then rotate the cow and collect the targeted product into the receiver flask with your student code. Write down the product boiling point and pressure reading from the digital manometer into the answer sheet. Step 10. When the targeted product is collected, turn off the heater, remove the foil and carefully (hot!) remove the hot-plate magnetic stirrer aside. Cool down the apparatus to ambient temperature using the ice bath. Ask your lab assistant to disconnect the vacuum line. Disconnect the receiver flask with the targeted product and immediately close it with the glass stopper labeled with your student code. Do not attempt to drag the teflon sleeve out if it remains in the receiver. Place the flask into the 50 mL plastic beaker labeled “For the receiver with the product”. Immediately attach a new receiver instead of the removed one and apply the joint clip. Leave the apparatus as it is. Step 11. Measure the refraction index (before weighing) following the instruction below. Record the temperature from the refractometer. Weigh the receiver with the targeted product closed with the labeled stopper. Calculate the mass and yield of the product (take the mass of the teflon sleeve equal to 149 mg). The molar masses of 3-metylthiophene and the product equal 98 and 177 g mol-1, respectively. Q2. Write down all the result in the followingtable . # Parameter /Characteristics Value Units 1 Mass of the receiver flask with the glass stopper with student code g 2 Mass of the product g 3 Yield of the product % 4 Refraction index for the product - 5 Temperature recorded from the refractometer °C 6 Boiling point of the product °C 7 Pressure at the Boiling point mmHg

REFRACTO 30GS – OPERATING INSTRUCTIONS

Fig. 3. Using the Refracto 30GS

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1. To switch Refracto 30GS on, press and hold “ESC” button (1) until the display appears. The instrument is ready for operation. It switches off automatically if not operated for 10 min. 2. Clean the cell and glass rod with a napkin wetted with the solvent from the washing bottle labeled “cleaning solvent”. Dry both with another napkin. 3. Make sure the sample to be measured has reached ambient temperature and is homoge- neous. 4. Apply 2-3 drops of the sample onto the measuring cell (2) using the glass rod. 5. To start the measurement press and hold the ok button (3) until the beep. 6. Take the value of the refraction index and the temperature from digital display (4) and write down the result in the answer sheet. 7. Clean up the cell and the glass rod.

Q3. By comparing the obtained and literature data, draw the structure of the product and catalyst given.

Q4. Draw the structure of the 3-methylthiophene-based reactive intermediates behind the selectivity in the case of T1 and T2.

Q5. Write down the product (T1 or T2) formed as a result of direct bromination of 3- methylthiophene with NBS under the given conditions / catalyst used.

ZnBr2 Dibenzoyl peroxide LiBr in AcOH Visible light or UV light

Q6. In the synthetic pathways to T3 and T4, draw the structures of the compounds formed in the first steps of each pathways shown on Scheme 1.

TASK 2. Analysis of the solution of a chromium – vanadium alloy Antiferromagnetic materials show a good prospect in the development of memory devices for ultra- high-density data storage, the world’s smallest magnetic memory bit using only 12 atoms being one of prime examples. Vanadium – chromium alloys exhibit antiferromagnetic properties at subzero temperatures. It is obvious that composition of alloys used in various hi-tech applications should be accurately controlled. In this task, you will analyze an aqueous solution simulating the product of digestion of vanadium – chromium alloy sample. The task consists of two parts: 2+ - I. Oxidation of vanadyl (VO ) to vanadate (VO3 ) in the test solution using potassium permanganate, followed by determination of vanadium (note that chromium (III) is not oxidized under these conditions).

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II. Oxidation of the test solution with ammonium persulfate, followed by titrimetric determination of the total content of vanadium and chromium with Mohr’s salt (Ammonium iron(II) sulfate).

Procedure - The amount of vanadium and chromium should be calculated and reported in mg per 100 mL of the test solution. - Start doing this task with Part A, since you will need time to oxidize the test solution to be analyzed in Part C. - The 10.00-mL volumetric pipette has two graduation lines. You should pipette a volume between the two lines.

Part A. Preparation of the solution for determination of vanadium and chromium total content 1. Transfer a 10.00-mL aliquot of your test solution into the 150-mL beaker and add 20 mL of 1M sulfuric acid using the 25-mL graduated cylinder. 2. Add 6–8 drops of the 0.3% solution of silver nitrate (the catalyst) and heat the mixture on the hotplate to 70–80°С (position 3), until condensate on the beaker wall appears.

3. Add 20 mL of the 10% ammonium persulfate solution to the heated mixture using the 100- mL graduated cylinder. 4. Continue heating and observe the appearance of yellow color, indicating the formation of dichromate. Note! You can perform the determination of vanadium (Part B, 1 – 6), while the mixture is being heated.

5. Keep heating the mixture for 10-15 min (position 3) after appearance of the yellow color to decompose the excess of ammonium persulfate (the decomposition is over when you see no small bubbles in the solution). 6. Cool the solution to ambient temperature. 7. Transfer quantitatively the solution from the 150-mL beaker into the 100-mL volumetric flask, dilute to the mark with distilled water, stopper the flask and mix thoroughly.

Part B. Titrimetric determination of Vanadium 1. Transfer a 5.00-mL aliquot of the test solution into an Erlenmeyer flask using the graduated pipette. Note! The 5.00-mL graduated pipette is self-draining.

2. Carefully add 0.03 M potassium permanganate solution dropwise, shaking the flask after adding each drop until light pink color appears. Make sure that the light pink color is stable. Remove the excess of potassium permanganate by adding 0.03 M oxalic acid solution dropwise. Shake the flask after each drop until the light pink color changes to pale blue. Let the solution stand for about 1 min to make sure that the pink color has disappeared completely.

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3. Transfer 10 mL of the 1M H2SO4 solution into the Erlenmeyer flask using the 25-mL graduated cylinder. 4. Add 2–3 (not more!) drops of the indicator into the Erlenmeyer flask and shake it vigorously. Let the flask stand for 2–3 min and observe the purple color appearance. 5. Fill the burette with the Mohr’s salt solution. Use the 100-mL plastic beaker labeled “Waste” to drain the excess of Mohr’s salt solution from the burette, record the initial reading. 6. Titrate the solution in the Erlenmeyer flask with the Mohr’s salt solution until the color changes to pure light green through brownish-grey one. 7. Take the final reading of the burette. Repeat as necessary.

Q1. Show your results on the answer sheet.

Part C. Titrimetric determination of vanadium and chromium total content in the test solution 1. Wash the 10.00-mL volumetric pipette with distilled water, rinse with the solution prepared in 100-mL volumetric flask (obtained in part A).

2. Pipette a 10.00-mL aliquot into an Erlenmeyer flask, add 10 mL of 1M H2SO4 solution using the 25-mL graduated cylinder. 3. Add 3–4 drops of the indicator. Vigorously shake the flask and let it stand for 3–4 min. Ob- serve appearance of red color. 4. Fill the burette with the Mohr’s salt solution. Use the 100-mL plastic beaker labeled “Waste” to drain the excess of Mohr’s salt solution from the burette, record the initial reading. 5. Titrate the solution in the flask with the Mohr’s salt solution until the color changes to light yellow-green. 6. Take the final reading of the burette. Repeat as necessary.

Q2. Show your results on the answer sheet.

Part D. Questions and Data Analysis Q3. Write down the balanced chemical equations for the reactions that take place upon: a) oxidation of the test solution with potassium permanganate b) titration of vanadate with Mohr’s salt

Q4. Write down the balanced chemical equations for the reactions that take place upon: a) oxidation of the test solution with ammonium persulfate b) titration of the oxidized test solution with Mohr’s salt

Q5. Calculate the a) V(IV) and b) Cr(III) concentrations in the test solution. Calculate the amount of the metals in mg per 100 mL of test solution.

Q6. This protocol can not be applied to the determination of vanadium and chromium in steels, if the steel was digested by conc. HCl. Give equations of two reactions to explain the reasons behind.

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TASK 3. Kinetic determination of Diclofenac (DCF) Kinetic methods with spectrophotometric detection for assaying drugs have been intensively developed during the last decade due to a number of obvious advantages, including inherent simplicity, cost-effectiveness, availability in most quality control laboratories, and improved selectivity. In this task you will:  Perform kinetic determination of Diclofenac (DCF) in a medicine by following the progress of the drug oxidation reaction.  Determine the reaction order with respect to DCF

Q1. Spectral changes in the course of DCF oxidation with KMnO4 are given in Fig. 4, (1 to 10 reflects the reaction progress). Complete the table below suggesting which wavelengths can be applied for photometric kinetic determination of DCF. In each case, indicate the direction of the absorbance changes (denote increasing with  and decreasing with ).

Fig. 4. DCF oxidation with KMnO4 # Wavelength, nm Yes or No and direction 1 420 2 480 3 520 4 580 5 610

Procedure Part A. Assembling of laboratory equipment Assemble the laboratory equipment as shown in Fig. 5. Connect the photometer (1), 525 nm (fixed wavelength) and thermostat (2) to the Netbook via USB slots. Connect the thermostat to the cable labeled “Thermo” to the power supply at your work place via the power adapter. Put the optical cuvette (3) on top of the magnetic stirrer (4), pass the cuvette through the photometer from aside (not possible from top down) and place the thermostat over the cuvette from top down (Fig. 5b).

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Fig. 5. Laboratory equipment

Hints! - Plug in your Netbook to the mains before switching on. - Plug in all the equipment (the photometer and thermostat) before switching on the Net- book. Switch on the mouse. - If only one window (hereafter referred to as Pattern) instead of two appears after launch- ing the software, quit and re-launch the program. - Do not unplug ANY device from the USB slot while carrying out the measurements. If it still happens, you will see a warning on the screen. Quit and re-launch the program. - If your Netbook falls asleep, click the «Setup» button in the Measurements window on the absorbance plot pattern when reverting to the measurements. - In case you see chaotic temperature changes on the screen, stop and re-start the measurement.

Part B. Plotting of the calibration curve

All measurements needed to plot the calibration curve are carried out at 30 °C with constant KMnO4 and H2SO4 initial concentrations. The DCF concentration is varied by using 4 different aliquots (of 0.2, 0.4, 0.6, and 0.8 mL) of the DCF stock solution.

1) Transfer 5 mL of 1M H2SO4 solution using the graduated cylinder and 0.2 mL of DCF stock solution using the 2 mL pipette into the 100 mL volumetric flask, dilute to the mark with distilled water, stopper the flask and mix thoroughly. 2) Carry over the flask contents into the cuvette, put the medium-size stir-bar and switch on the magnetic stirrer. Adjust the stirring speed regulator to the mark shown on Fig. 5a to provide for intensive mixing. 3) Launch the «Chemistry-Practicum» software on the Netbook. The software will detect the external devices (sensors) automatically. You will see two plot patterns (that of absorbance/extinction/optical density, D vs. t, s; and that of temperature, T °C vs. t, s) on the display. 4) Set the following parameters in the Menu bars of the corresponding plot patterns (Fig. 6):

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- Click the icon next to the button («Fixes X-axis maximum on screen») on the absorbance plot pattern. The entire plot will always fit to the screen;

- Click the button («Sets the Y range») on the absorbance plot pattern and set the absorbance range (the ordinate axis) from -0.1 to 1.1. - Type “2” (instead of “1”) in the box of the measurements interval on the absorbance plot pattern. - Choose «Precisely» in the «Precisely/Roughly» window on the temperature plot pattern, then click on the «T = X» button and set the required temperature of 30 C in the pop-up window. - Calibrate the photometer by clicking the «Setup» button in the Measurements window on the absorbance plot pattern.

Fig. 6. “Chemistry-Practicum” software interface

Note! Setting the parameters (step 4) is needed only prior to the first measurement.

5) Click the button («Start measure for chosen sensors») to switch on the thermostat and observe the lamp heating up the solution in the cuvette. Follow the current temperature reported in the line above the plot. Wait until the thermostat lamp switches off, reflecting the set up temperature is attained. Stop the measurements by clicking button (is activated and turns to red-orange when the measurement is on).

6) Click any part of the absorbance plot pattern to activate it. Take 2 mL of the KMnO4 solution using the 2 mL pipette. Click the button («Start measure for chosen sensors») in the Menu bar of the Measurements window and quickly blow out (press the pipette piston) the permanganate solution from the pipette into the cuvette.

Note! Make sure the temperature in the cuvette equals 30 C before adding the KMnO4 solution!

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7) Observe the progress of the kinetic curve on the screen. Continue measurement for 50 s

after adding the KMnO4 solution, then terminate the measurement by clicking the «Stop measurements» button. 8) Save the data by clicking the button («Export all the data collected in an external file») in the Menu bar of the absorbance plot pattern, choose the Desktop and type the file name “DCF2” (change the name to “DCF4”, or “DCF6”, or “DCF8” in the subsequent experiments).

Note! - Use only the names of the given format! - Always save the data on your Desktop before starting the next experiment, otherwise the current data set will be lost after the next click on the button. - Make sure absorbance plot pattern is active when exporting the data. Otherwise, you will export invalid results. In case no pattern is chosen, you will get a warning.

9) Empty the cuvette into the Waste bottle, wash thoroughly the cuvette with distilled water. Use black magnet from the outer side of the cuvette to avoid your stir-bar being dropped into the Waste bottle while washing. Wipe carefully the external surfaces of the cuvette with the napkin. Also, use the napkin to dab the thermostat lamp. 10) Repeat the steps 1), 2) 5)-9) with the other volumes of the DCF stock solution.

Part C. 1. Studying of the DCF containing medicine (“Control”) 1) Wash the volumetric flask and prepare the mixture as described above using a 0.4 mL aliquot of the medicine (“Control”) instead of the DCF stock solution. 2) Repeat the steps 1), 2), 5)-9) described in Part B. When saving the data, name the file “DCFmed”. 3) Repeat the measurement of the “Control” as necessary.

2. Experimental data analysis 1) Open the Excel file on your memory stick in Excel. One by one open your saved data files in Notepad by double clicking on them on Desktop. Choose Edit/Select All in the Menu bar, then right click and copy the selected data into the Excel sheet with the corresponding name (the volume of DCF added or “DCFmed”) and choose Edit/Paste in the Menu bar. You will see the experimental data on the Excel sheet (time, s, in column A, and absorbance in column B). 2) Ignore the values before the maximum. Select columns A and B, and plot the data. Use the “Insert Scatter” icon shown on Fig. 7.

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Figure 7. Position of the “Insert Scatter” icon

3) Choose the initial linear section of the remaining curve (15 to 20 data points), apply linear approximation by adding the linear trend line and bring the parameters to the chart area. Make sure that the R2 value exceeds 0.98. If needed, decrease the number of the experimental data points plotted removing later data points. Still always search for the most wide range of the experimental data providing for the target R2 value. Determine the value of the

initial rate of absorbance change, v0. Note! You will get zero point for this part of the task if less than 12 values are included in the plotted data range. 4) Analyze similarly the experimental data obtained with the other DCF concentrations and with the medicine solution “Control” (“DCFmed” file). 5) Calculate the DCF concentrations in the reaction mixtures (in mg/L). Write down the DCF concentrations and initial rates in appropriate cells of the “Results” Excel sheet. 6) Plot the calibration graph on the “Results” sheet and use it to determine the DCF concentration in the analyzed mixture prepared from the medicine (“Control”). Fill in the appropriate cells of the “Results” Excel sheet with the coefficients of linear approximation of the calibration graph. Calculate the DCF concentration in the medicine. 7) Write down the accepted value in the cell F10 of the “Results” sheet. 8) On the “Results” Excel sheet, graphically determine the reaction order with respect to DCF and write down the exact obtained value in the cell I3. 9) Once finished, save your file and invite your Lab assistant to demonstrate that you have got experimental data in the Excel file. Sign and get the Lab assistant’s signature. Note! Only the data saved on the memory stick will be considered as the result of the Task.

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Solution to the theoretical problems Solution to problem 1

1.1. Ammonia Q = νΔHvap = 21.3 kJ; W = νCv(gas)(T2 – T1) = 2.67 kJ Chlorodifluoromethane Q = νΔHvap = 20.0 kJ; W = νCv (gas)·(T2 – T1) = 4.88 kJ. 1.2. U H S G V 1.3. Ammonia COP = Q/W = 7.98 Chlorodifluoromethane COP = Q/W = 4.10 2.1. d 2.2. d 2.3. b

3. CF3CH2F COP = ΔHvap / (Cv(gas)(T2 – T1)) = 2,80

CF3CF=CH2 COP = ΔHvap / (Cv(gas)(T2 – T1)) = 1,59 Yes No

4.1. T0 = 10,4 + 1114 / (4,87 – lg p1) = 264 K

ΔV = (νRT1 / p1) – (xνRT0 / p1) = 6,7 L

Q = ν (1–x)(ΔHvap – RTvap + (Cv(gas)–Cv(liq))(T0–Tvap)) + νCv(gas) (T1–T0) + p1ΔV = 19,8 kJ or

Q = ν (1–x)(ΔHvap + (Cv(gas)+R–Cv(liq))(T0–Tvap)) + ν(Cv(gas)+R)(T1–T0) = 19,8 kJ

4.2. W = νCv (gas) (T2– T1) = 3,15 kJ COP = Q/W = 6,3 4.3. The internal energies of the refrigerant are equal in points 0 and 3. Thus,

x·(ΔHvap – RTvap + (Cv (gas) – Cv (liq))(T0– Tvap)) + Cv (liq) (T0– T3) = 0,  T3 = 298 К. 4.4. c (because the length of 0-1 line decreases) or (because x (see 4.3) increases and less liquid is in the

equilibrium mixture at T0, so less heat Q is necessary to evaporate it!) Solution to problem 2

1.1. The value of I should increase with the increase of [Ac]0 at [In]0 = const, because the larger fraction of the intermediate product R will enter the reaction (3b). The maximum value of I will be achieved if all R reacts

in (3b),  I = 1/β. 1.2. Shilov’s mechanism includes the initial reaction A + In  R (3a')

and two competitive reactions R + Ac  P1 (3b)

R  P2 (3a'')

The rates of conversion of In and Ас are determined by the rates of the reactions (3а') and (3b), respectively: 푘(3푎′)[퐴][퐼푛] 푘(3푏)[퐴푐]· 푟(3푏) 푘(3푏)[푅][퐴푐] 푘(3푏)[퐴푐]+푘(3푎′′) 푘(3푏)[퐴푐] = = = 푟(3푎′) 푟(3푎′)[퐴][퐼푛] 푘(3푎′)[퐴][퐼푛] 푘(3푏)[퐴푐]+푘(3푎′′) in steady-state approximation for [R]. We see that the ratio of two rates does not depend on the initial

concentration [In]0 and I will also not depend on it. This gives the straight line parallel to the [In]0 axis on the graph.

0 [Ac] = const

[Ac]0 = const

= =

I I

0 [In[In]] 0

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1.3. In this case I will permanently increase with the increase of [Ac]0   at [In]0 = const. The rate of the reaction (4b) may be so high that conversion of In in reaction (4а) will be negligible.

Hence I  ∞ if [Ac]0   at [In]0 = const.

2.1. The standard Gibbs energy of reaction (5) at 600К is 66 kJ/mol. The equilibrium constant is -66000/8,314/600 -6 K = e = 1.8·10 = k5/k-5 . Reaction is considered at standard pressures of all the reactants and products. The ratio of the rates of forward and reverse reactions is r k [Br][H ] k 5 = 5 2 = 5 = 1.8·10-6 r−5 k−5[HBr][H] k−5

2.2. The steady-state condition is the same for both radicals, e.g. for radical Н d[H] = k ·[Br][H ] - k ·[HBr][H] - k ·[H][Br ] = 0 dt 5 2 -5 6 2 The concentrations of all the neutral molecules are the same (they correspond to the pressure of 1 bar), [H] k [H ] k /푘 1.8·10−6  = 5 2 = 5 −5 = = 1.6·10-7 [Br] k−5 + k6 1 + k6/푘−5 1 + 10 The Gibbs energy of reaction (5) under such conditions is: [H][Hbr] -3 -7 G(5) = G°(5) + RT·ln = [66 + 8.314·10 ·600·ln(1.6·10 ] kJ/mol= -12 kJ/mol [Br][H2] The ratio of rates is: r k [Br][H ] k [퐵푟] k 1 + k /푘 k 5 = 5 2 = 5 · = 5 · 6 −5 = 1 + 6 = 11,3 r−5 k−5[HBr][H] k−5 [퐻] k−5 [k5/푘−5] k−5 2.3. b (Arguments – not required from the student. The statement (β) is true beyond any doubt r−5 G/(RT) rbeob = r5 – r–5 = r5 · (1 - ) = r5 · (1 - e ) r5 ΔG is positive in (2.1) and negative in (2.2), the expression in brackets consecutively is negative in

(2.1) and positive in (2.2). The observed rate robs changes its sign from negative to positive due to

chemical coupling as r5 is always positive.

The statement (a) may not be true as r5 = k5[H2][Br] and change of [Br] is not specified in the prob-

lem. k5 and [H2] are the same in (2.1) and (2.2). The forward reaction in (2.2) may be faster or slower. The statement (c) may not be true for the similar reason.) 3.1. According to the Second law the following condition has to be met: -7 -1 -1 GSyst/T = G(7) · r7 + G(8) · r8 ≤ 0  r7 ≤ r8 · (-G(8))/ G(7) = (481.2/46) · 6.3·10 mol L min This is the maximum possible rate of the coupled reaction. 3.2. In this case Gibbs energies for both reactions will change: G'(7) = G(7) – RT·ln2 = (46.0 – 1.8) kJmol-1 = 44.2 kJmol-1 G'(8) = G(8) + RT·ln2 = (-481.2 + 1.8) kJmol-1 = -479.4 kJmol-1 The observed rate of the reaction (8) will be decreased due to the increase of the rate of reverse reaction (8). However with such a huge negative Gibbs energy of the reaction (8) the rate of reversed reaction is negligible. The maximum rate of the reaction (7) is -8 -1 -1 -7 -1 -1 r7 ≤ r8 · (-G(8'))/ G(7') = (479.4/44.2) · 6.0·10 mol L min = 6.5·10 mol L min The relative increase of maximum rate is practically negligible: -7 -7 r7'/r7 = (6.5·10 )/ (6.3·10 ) = 1.03  r7(max. 3.2) / r7(max. 3.1) = 1.03

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Solution to problem 3 1.1. Cooperation: β > 1 Competition: 0 < β < 1 Independence: β = 1

[H퐺2] [H퐺2] [H퐺푎] 퐾푎 1.2. K = = · = β·Kb · = β·Ka [HGb][G] [HGa][G] [HGb] Kb

2.1. Kb = 2 Ka  [HGb] = 2 [HGa]

Material balance with respect to H: [H] + [HGa] + [HGb] +[HG2] = [H]0 = 1 mol/L

or 0.1 + 3[HGa] + [HG2] = 1 mol/L

Material balance with respect to G: [G] + [HGa] + [HGb] +2[HG2] = [G]0 = 2 mol/L

or 0.5 + 3[HGa] +2 [HG2] = 2 mol/L

Solving the system of two equations, we find: [HGa] = 0.1 mol/L, [HG2] = 0.6 mol/L,

hence [HGb] = 0.2 mol/L)

[H퐺푎] 0.1 [H퐺2] 0.6 Ka = = = 2 β = = = 3 [H][G] 0.1 ·0.5 [HGa][G]퐾푏 0.1 ·0.5·4 2.2.

HGa

HGb

HG2

2.3. 1) [HG2] = [HGa] + [HGb] = 3[HGa] [HG2] 3 = β∙Kb = 12 = 12 [G] = 0.25 mol/L [HGa][G] [G]

2) Material balance with respect to H: [H] + 3[HGa] + [HG2] = 1 mol/L

[H] + 6[HGa] = 1 mol/L [H] + 12[H][G] = 1 mol/L [H] = 0.25 mol/L.

3) [HGa] = Ka[H] [G] = 0.125 mol/L.

[HG2] = 3[HGa] = 0.375 mol/L.

4) Material balance with respect to G: [G]0 = [G] + 3[HGa] + 2[HG2] = 1.375 mol/L

n0(G) = 1.375 mol/L

2.4. a) β = 0: In this case, no HG2 is formed.

Material balance with respect to H: [H] + [HGa] + [HGb] = 1 mol/L [H] + 3[HGa] = 1 mol/L

Material balance with respect to G: [G]0 = [G] + [HGa] + [HGb] = 2 mol/L [G] + 3[HGa] = 2 mol/L [HG ] Equilibrium constant: K = a = 2 a [H][G] Solving the system of three equations, we get:

[H] = 0.129 mol/L [G] = 1.129 mol/L [HGa] = 0.290 mol/L [HGb] = 0.580 mol/L [HG2] = 0

b)    In this case, formation of HG2 is practically irreversible, so only HG2 is present in the solution.

[H] = 0 [G] = 0 [HGa] = 0 [HGb] = 0 [HG2] = 1 mol/L

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Solution to problem 4

1. The precipitate X2 formed by addition of barium chloride in acidic medium is barium sulfate BaSO4.

The precipitate X3 formed by addition of silver sulfate is silver chloride AgCl.

The yellow precipitate X4 formed by addition of alkali can be mercury oxide HgO or silver phosphate

Ag3PO4. The mole ratio n(X4) : n(X2) is 0.931 for n(HgO) : n(BaSO4) which is not valid and 1.798 for

n(Ag3PO4) : n(BaSO4) which gives 2.4 being multiplied by 4/3. So, the molar ratio is nAg3PO4) : n(BaSO4) =4

:3 which corresponds to n(P) : n(S) = 4:3, i.e. to formula of X1 P4S3.

X1 = P4S3 X2 = BaSO4 X3 = AgCl X4 = Ag3PO4

2. The gas evolved has a molar mass 1,586 · 29 g/mol = 46 g/mol that is NO2

P4S3 + 38 HNO3  4 H3PO4 + 3 H2SO4 + 38 NO2+ 10 H2O

H2SO4 + BaCl2  BaSO4+ 2 HCl

Ag2SO4 + 2 HCl  2 AgCl + H2SO4

BaCl2 + Ag2SO4  BaSO4 + 2 AgCl

H2SO4 + 2 NaOH  Na2SO4 + 2 H2O

2 H3PO4 + 6 NaOH + 3 Ag2SO4  2 Ag3PO4 + 3 Na2SO4 + 3 H2O

3. Phosphorus sulfide P4S3 is a molecular cage

4. a) P4S3 + 8 O2  2 P2O5 + 3 SO2

b) P4S3 + 16 H2SO4  4 H3PO4 + 19 SO2 + 10 H2O

c) 3 P4S3+ 16 KClO3 16 KCl + 6 P2O5 + 9 SO2 Solution to problem 5 2+ – – 1.1. C6H12O6 + 2 Cu + 5 OH  C6H11O7 + Cu2O+ 3 H2O 2+ – 1.2. 2 CuSO4 + 4 KI  2 CuI + I2 + 2 K2SO4 or 2 Cu + 4 I  2 CuI + I2 – – KI +I2  KI3 or I + I2  I3 2– – 2– 2 Na2S2O3 + I2  2 NaI + Na2S4O6 or 2 S2O3 + I2  2 I + S4O6 1.3. Total amount of copper(II): 50.00 mL · 0.0400 mol/L = 2.0000 mmol Obviously, there is an excess of iodine and the remaining iodine was titrated with sodium thiosulphate: 11.87 mL · 0.05078 mol/L = 0.6028 mmol 2.0000 mol - 0.6028 mol = 1.3972 mmol of copper(II) was required to oxidize the sugars. n(cugars) = ½ · n(Cu2+) = 0.6986 mmol c(sugars) = 0.6986 mmol/10.00 L = 0.06986 mol/L mass content = 180.16 g/mol · 0.06986 mol/L = 12.6 g/L

1.4 KIO3 + 5KI + 3H2SO4 = 3I2 + 3K2SO4 + 3H2O – – + IO3 + 5I + 6H = 3I2 +3H2O – – – Only glucose was oxidized with iodine: C6H12O6 + I2 + 3OH  C6H11O7 + 2I + 2H2O or

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3NaOH 2NaI I2 2H2O

2 Na2S2O3 + I2  2 NaI + Na2S4O6 1.5. b – 1.6. n(I2) = 3n(IO3 ) = 3·0.01502 M · 10 mL = 0.4506 mmol

n(I2) = n(glucose) = 0.4506 mmol c(glucose) = 0.4506 mmol/10.00 mL = 0.04506 mol/L Mass content of glucose = 180.16 g/mol · 0.04506 mol/L = 8.12 g/L Mass content of fructose = (12.6 – 8.12) g/L = 4.48 g/L 1.7. 0.2 L·8.12 g/L = 1.6 g of digestible carbohydrates, that is 0.14 BEU. 2.1. a) 18O + p  n + … 18F 2 18 20 b) … + 1퐷  F +  Ne 19 2 20 c) F + 1퐷  F + … p d) 16O + …  18F + p + n  2.2. Nucleus 11С 20F 17F 14C Decay mode β+ β- β+ β– 2.3. The initial ratio of radioactivities: 18 18 18 18 18 A0( F) η( O  F ) χ(H2 O) χ(H2 O) 17 = 16 17 · 16 = 144.7 · 16 A0( F) η(( O  F ) χ(H2 O) χ(H2 O) After 5 minutes the ratio changed due to radioactive decay of fluorine: 푙푛2 18 A (18F)·exp (−5· ) 18 18 A300( F) 0 109.7 A0( F) χ(H2 O) 5 17 = 푙푛2 = 23.75 · 17 = 3437 · 16 = 10 A300( F) A (17F)·exp (−300· ) A0( F) χ(H2 O) 0 65 18 χ(H2 O) 16 = 29 χ(H2 O) 29 ·20 Mass fraction of H 18O ist ω(H 18O) = = 0.97 ≙ 97 % 2 2 29 ·20+18 2.4. During the synthesis. the radioactivity will decrease:

A3.5 = A0 · exp(- 3.5) = 581.3 MBq η = 528.2/581.3 = 0.909 ≙ 90.9 % 2.5. Radioactivity is excreted by radioactive decay and through the excretory organs (e.g. kidneys). The excre- tion process may be considered as two competitive first-order reactions. Activity after one hour is: ln2 ln2 A = A · exp(-( +  )·t) = 450 · (-(exp( + ) · 60) = 218 Bq 60 0 1 2 109.7 120

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Solution to problem 6 1. A B C D

E X Y

2. F G

3. I J

Solution

1-2. From the given contents of C, H, and N we can calculate the molecular formulas of A, B and E as (C5H9N)n. This formula corresponds to isomeric piperideines (piperidines containing one double bond). However, piperideines themselves have no 3-fold axis of symmetry. Thus, we conclude that A is a symmetric trimer of piperideine. B is another trimer of piperideine (B is diastereomer of A). Analysis of both Scheme 1 and Scheme 2 enables us to conclude that X and Y are piperidine derivatives. Therefore, compound E should be also piperidine derivative; most probably, this is a monomeric piperideine. It is consistent with the fact that E is an intermediate in both the formation of A, B and their transformation into compound C. The easy trimerization of E and its reactivity against HCN demonstrates the presence of a highly polarized bond. From thee isomeric piperideines only 1-piperideine (3,4,5,6-tetrahydropyridine) has a highly polarized C=N bond; both the spontaneous trimerization and reaction with HCN seem to be impossible for other isomers. Indeed, the chlorination of piperidine produces N-chloropiperidine which eliminates HCl under the treatment with base affording 3,4,5,6- tetrahydropyridine. The compound A is characterized by 3-fold axis of symmetry. It is possible only if A is an all-cis trimer of E. Compound B is diastereomer of A, i.e. B is a cis,trans-trimer of E.

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The compound C is the product of addition of HCN to a C=N bond of 1-piperideine E. Accounting for polariza- tion of this bond, it is 2-cyanopiperidine. It was treated with t-BuOCl and base analogously to the treatment of piperidine at the first step of the discussed synthesis. It is reasonable to conclude that the same reagents should lead to the same type of transformation, i.e. D is 2-cyano-3,4,5,6-tetrahydropyridine. This conclusion is consistent with N for compound D. If D has two nitrogen atoms, the molecular weight of D is 108. It corresponds to the molecular formula C6H8N2. On the contrary to E, this molecule is stable. This stabilization is provided by the conjugation between the C=N and CN bonds; this fact enables to discriminate 2-cyano-3,4,5,6- tetrahydropyridine from unstable 2-cyano-1,2,3,4-tetrahydropyridine where such stabilization is absent. Methylmagnesium iodide can react with carbon atoms of both C=N and CN bonds. The triple CN bond is more polar; a positive charge on the carbon atom of the nitrile group is higher. Attack on this carbon atom has also lower steric demands. Therefore, MeMgI should react with CN bond faster than with C=N bond. Moreo- ver, this conclusion can be made without this analysis as attack onto C=N bond should produce 2-methyl-2- cyanopiperidine. The tautomeric equilibrium for this compounds is impossible. The hydrolysis of the formed imine anion affords a ketone moiety. According to the problem, this molecule occurs in equilibrium with its tautomer; Y being the kinetic product and X being the thermodynamic product. The keto-enol tautomerism is inappropriate here as enols are usually much less stable than ketones. However, the imine-enamine tautomerism is also possible in this molecule. Both tautomers have the conjugation between two double bonds; the enamine tautomer seems to be more stable due to the hydrogen bond between N–H group and the proximal ketone function. Moreover, compound D is a 1-piperideine derivative. The kinetic product of its transformation should be 1-piperideine derivative too. It is compound Y. Its enamine tautomer is compound X. The analysis of Scheme 2 supports this conclusion.

Similarly, tautomer of the compound D is 2-cyano-1,4,5,6-tetrahydropyridine (G). To form the mixture of compound D and its tautomer G, intermediate F should have the possibility to react with water by two ways (by carbon atom and by nitrogen atom). It is possible if F contains an anion obtained by deprotonation of D. The coloring of the reaction mixture is an additional argument in favor of the conjugate anion formation.

3. It is not necessary for the students to know the first reaction. The comparison of molecular formulae of compound I and the starting compound H demonstrates clearly that this transformation leads to the substitution of an oxygen atom by a CH2 group. There are two possibilities for this substitution: 1) the transformation of the ester group into enol ether and 2) the transformation of the less reactive carbamate function into the

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corresponding 1-alkoxy-1-aminoalkene. In the first case, the hydrolysis of I should produce 2-acetylpiperidine (or its N-Boc derivative). In the second case, the product of hydrolysis should be 1-acetylpipecolinic acid (or its ethyl ester). From the Scheme 1 it is clear that X and Y have no acyl substituent at the nitrogen atom. Moreo- ver, the method of synthesis of X and Y described in Scheme 1 does not correspond to the presence of an ester function. Oxidation cannot remove Boc function. So, compound J is 2-acetylpiperidine. Its oxidation produces 2-acetylpiperideines X and Y.

Solution to problem 7 1.

XW XS

2. B C D

E F G

3. The number of possible stereoisomers of A: 4 4. I J K L

Solution 2. The first step is acylation of nortropanol. The comparison of molecular formula of C with the formula of A shows that transformation of B into C results in a loss of two hydrogen atoms. It is possible to conclude that this step is alcohol oxidation to ketone (this oxidation was discussed in Preparatory problem). Therefore, the first step is acylation of nitrogen atom. During the transformation of E into F N,O-dimethylhydroxylamine substitutes –OH group activated by transformation into chloride after treatment with SOCl2. The comparison of molecular formulae of F and pelletierine shows that the formation of pelletierine ((S)-1-(piperidin-2-yl)propan- 2-one) includes the loss of Cbz protecting group (its substitution by hydrogen atom) and substitution

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of -N(CH3)OCH3 fragment by methyl group. Analysis of all these steps enables to conclude that G is Cbz- protected pelletierine, E is 2-(N-Cbz-piperidin-2-yl)acetic acid and F is its N-methoxy-N-methylamide. Acid E is formed from the compound D by addition of two hydrogen atoms (compare molecular formulae of D and E). In turn, compound D is formed by introduction of one oxygen atom into ketone C. Even without knowledge of these reactions it is possible to conclude that oxidation of C produces lactone (intramolecular ester) corresponding to acid E (the Baeyer-Villiger oxidation). Treatment of this lactone with Et3SiH and BF3OEt2 leads to the cleavage of the ester and reduction of semiaminal moiety –NCH(R)OH to the corresponding saturated amine. Another way for solution is a comparison of structure of C with structure of pelletierine. This comparison shows that C(O)–CH bond should be broken. Therefore, during C-to-D transformation oxygen should be introduced into this bond. Product contains the C(O)–CH3 and the CH–H fragments. Therefore, from two possible ways of adding two hydrogen atoms we should select the way that leads to the CH–H fragment and –COOH moiety. Other arguments are similar to the above ones.

3. Nortropanol A has 3 chiral centers. However, due to bicyclic structure of A only 4 stereoisomers can exist.

4. The knowledge of molecular formula of pelletierine allows to conclude that the last step is the removal of Boc protecting group. In other words, compound L is Boc-protected pelletierine. The step of K formation is, evidently, acylation of nitrogen atom in 2-allylpiperidine. So, K-to-L transformation is the oxidation of terminal alkenyl group into the corresponding ketone (the Wacker process). The formula of the intermediate, the structure of which is given in the scheme, enables to conclude that the base-induced I-to-J transformation is the elimination of HBr and the next step is the removal of tert-butylsulfinyl group. Then, the J formation includes a cyclization producing piperidine ring. Therefore, we can write the following structures of compounds I–L.

118 www.ShimiPedia.ir IChO Baku: Solutions Theoretical Test

Solution to problem 8 1a. n(C) > n(O)  n(C) < n(O)  n(C) = n(O)  Data insufficient  1b. With regard to 1a, three variants (n(H)>n(C)>n(O), n(C)>n(H)>n(O), and n(C)>n(O)>n(H)) are possible for X1. For each inequality, one can write down the corresponding formula using elements of a geometric progression (q is the progression common ratio), equations for calculation of mass fractions of carbon and its roots

Inequality Formula Equation 1. root (q1) 2. root (q2) 12.01·푞·푛 n(H)>n(C)>n(O) С H O = 0.5453 2.00 7.93 qn q2n n 12.01·푞·푛 +1.008·푞2·푛 + 16.00·푞·푛

12.01·푞2·푛 n(C)>n(H)>n(O) Сq2nHqnOn = 0.5453 –1.21 1.32 12.01·푞2·푛 +1.008·푞·푛 + 16.00·푞·푛

12.01·푞2·푛 n(C)>n(O)>n(H) Сq2nHnOqn = 0.5453 –0.06 1.66 12.01·푞2·푛 +1.008·푛 + 16.00·푞·푛

There is only one positive integer root, thus the empirical formula is X1 = C2H4O

1c. Since (1) and (2) are the reaction equations, one can write down the formula of X as C2nH4nOn + 2H  1O =

C2nH4n+2On-1. With an account for the known mass fraction of hydrogen: 1,008·(4푛+2) = 0.1194  n = 3 and X = C H O 12,01·2푛 +1,008·(4푛+2)푛 + 16,00·(푛−1) 6 14 2 1d. X2 is formed from X composed of three elements (C, H, and O) via a monooxygenase catalyzed reaction: 100−51.56−9.74 9.74 51.56 n(C) : n(H) : n(O) = : : = 1 : 3: 1 12 1.008 16.00

Since the number of hydrogen atom is necessarily even, the molecular formula of X2 is C2H6O2. Other variants with a higher even number of hydrogen are not valid. Ethylene glycol (structure: HO-CH2-CH2-OH) is the only stable substance with the molecular formula deciphered above. 1e.

2. Number of graph: 1 (The number of kinks of the curve depends on that of urinations per day (4 to 6 times in norm); the am- plitude of 2-butoxyacetic acid content in organism can vary; the metabolite is always (even after urine passage) found in organism in nonzero concentration.) 3. Dioxygenase incorporates two oxygen atoms in vicinal positions of the substrate, which can be followed

by chemical bonds reorganization. The empirical formula of the hydrocarbon Р is C7H8

n(C) : n(H) = (91.25/12.01) : ((100.91.25)/1.008) = 7 : 8  P = C7H8 . Toluen The molar mass of P3 equivalent containing acidic group(s) is 100 : (6.41 · 0.100) g/mol = 156 g/mol.

Two dioxygenase steps suggest the composition of C7H8O4.

P3 must be a monocarboxylic acid if it still contains seven carbon atoms. Fragments containing a СH3CO–

group (or a СH3CH(OH)– group further transforming into СH3CO– one) are involved into the iodoform reaction. This suggests splitting of the benzene moiety during the second oxygenase step at the carbon connected to the methyl group.

119 www.ShimiPedia.ir IChO Baku: Solutions Theoretical Test

P P1 P2 P3

4. Thorough consideration of the transformations brings to the conclusion that Y1, Y2, and Y3 are anions of the same charge. Dioxygenase reaction suggests oxygen as Y4. The identity of the first two steps shows that Y1 contains by one oxygen atom more than Y2 (Т1 is phenol). So, Y2 must have a composition of n- Y3O2 . Formally Y3 is a halogenide or chalcogenide. Still silver chalcogenides, bromides and iodides do not dissolve even in concentrated ammonia solutions. Fluorine does not afford stable oxygen-containing anions. Thus all Y particles contain chlorine. There are four ways of pyrocatechol ring splitting as a result of the dioxygenase reaction, still only one of these leads to the product without identical functional groups:

Y1 Y2 Y3 Cl–

Y4 T1 T2

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About the History of the IChO

About the history of the International Chemistry-Olympiads

The idea of chemistry olympiads was born 1968 during an Czechoslovakian national olympiad that was attended by observers from Poland and Hungary. These three countries participated in the first IChO 1968 in Prague. The number of teams attending the IChO in the following years is shown in the plot below.

Number of teams attending the IChO

40 Number of Teams Teams of Number

30 Numberof Teams

0 1970 1980 1990 2000 2010 YearYear of 0f Olympiad Olympiad

The participating countries are shown in the following table.

121 www.ShimiPedia.ir About the History of the IChO

Participating Delegations •= host. + = participant. o = observer

Year  6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 Country  8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Argentina + + + + + Armenia Australia o + + + + + + + + + + • + Austria + + + + + + + + + + + + + + + + + + + + + + + + Azerbaijan o o Belarus + + + + Belgium + + + + + + + + + + + + + + + + + + + + Brazil o o + Bulgaria + + + + + + + + + + • + + + + + + + + + + + + + + + + + + Canada o o + + + + + + + + + + + • + + China + + + + + + + + • + + + + Chinese Taipei + + + + + + + +

Costa Rica Croatia o o Cuba + o + + + + + + + + + + + Cyprus o + + + + + + + + + + Czech Rep. + + + + + + + Czechoslovakia • + + + + + + + • + + + + + + + • + + + + + + + Denmark + + + + + + + + + + + + + + + + + + DDR o + + + + + • + + + + + + + + + + + • + Egypt El Salvador Estonia + + + + + + Finland o + + + + + + + + + + • + + + + + + + + + + + France o + + + + + + + + + • + + + + + + + + + fYROM (Macedonia) Georgia Germany o + + + + + + + + + • + + + + + + + + + + + + + + + Greece + + + + + + + + + + + + + + + Hungary + + • + + + • + + + + + + + + + + + • + + + + + + + + + + + + Iceland India o o + Indonesia o + + + Iran + + + + + + + Ireland o o + + Israel  Country 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 Year  8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9

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Participating Delegations •= host. + = participant. o = observer

Year  0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 Country  0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 Argentina + + + + + + + + + + + + + + + + Armenia o o + + + + + + + + + Australia + + + + + + + + + + + + + + + + Austria + + + + + + + + + + + + + + + + Azerbaijan + + + + + + + + + + + + + + + • Belarus + + + + + + + + + + + + + + + + Belgium + + + + + + + + + + + + + + + + Brazil + + + + + + + + + + + + + + + + Bulgaria + + + + + + + + + + + + + + + + Canada + + + + + + + + + + + + + + + + China + + + + + + + + + + + + + + + Chinese Taipei + + + + + • + + + + + + + + + +

Costa Rica o o + + + + + + Croatia + + + + + + + + + + + + + + + + Cuba + + + + + + + + + + + + + Cyprus + + + + + + + + + + + + + + + + Czech Rep. + + + + + + + + + + + + + + + + Czechoslovakia Denmark • + + + + + + + + + + + + + + + DDR Egypt o o + + + + + + + o El Salvador o o + + + Estonia + + + + + + + + + + + + + + + + Finland + + + + + + + + + + + + + + + + France + + + + + + + + + + + + + + + + fYROM (Macedonia) o o + + + Georgia o o + + Germany + + + + • + + + + + + + + + + + Greece + + + • + + + + + + + + + + + + Hungary + + + + + + + + • + + + + + + + Iceland o o + + + + + + + + + + + + + + India + • + + + + + + + + + + + + + + Indonesia + + + + + + + + + + + + + + + + Iran + + + + + + + + + + + + + + + + Ireland + + + + + + + + + + + + + + + + Israel o o + + + + + + + + + +  Country 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 Year  0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5

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Participating Delegations •= host. + = participant. o = observer

Year  6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 Country  8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Italy + + + + + o o + + + + + + + + + + + + + Japan Yugoslavia + + + + + + + + + + Kazakhstan o o + + Kenya o Korea + + + + + + + + Kuwait o o + + + + + + + + + + + Kyrgyzstan o o + Liechtenstein Latvia + + + + + + + + + Lithuania + + + + + + + + + Malaysia Mexico + + + + + + + + Moldova Mongolia Montenegro Netherlands + + + + + + • + + + + + + + + + + + + + New Zealand + + + + + + + + Nigeria Norway o + + + + + + + + + + + + • + + + + + Pakistan Oman Peru Philippines o Poland + • + + + + + + + • + + + + + + + + + + + + • + + + + + + + + Portugal Romania + + + • + + + + + + + + • + + + + + + + + + + + + + + + + GUS/Russ.Fed + + + + • + + + Saudi Arabia Serbia Singapore o + + + + + + + + + + + Slovakia + + + + + + + Slovenia + + + + + + + + + South Africa Spain o + + + +

 Country 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 Year  8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9

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Participating Delegations •= host. + = participant. o = observer

Year  0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 Country  0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 Italy + + + + + + + + + + + + + + + + Japan o + + + + + + + • + + + + + Yugoslavia o Kazakhstan + + + + + + + + + + + + + + + + Kenya o Korea + + + + + + • + + + + + + + + + Kuwait + + + + + + + + + + + + + + Kyrgyzstan + + + + + + + + + + + + + + + + Latvia + + + + + + + + + + + + + + + + Liechtenstein o o o o o + o Lithuania + + + + + + + + + + + + + + + + Malaysia o + + + + + + + + + + Mexico + + + + + + + + + + + + + + + +

Moldova o o + + + + + + + + + Mongolia o o o + + + + + + + + + + Montenegro o o + + Netherlands + + • + + + + + + + + + + + + + New Zealand + + + + + + + + + + + + + + + + Nigeria o o o + + + + Norway + + + + + + + + + + + + + + + + 0man o o + Pakistan o o + + + + + + + + + + Peru o o + + + + + + + + + + + Philippines o Poland + + + + + + + + + + + + + + + + Portugal o o + + + + + + + + + + + + + Romania + + + + + + + + + + + + + + + + GUS/Russ.Fed + + + + + + + • + + + + + • + + Saudi Arabia o o + + o o + + + + + Serbia o o + + + + Singapore + + + + + + + + + + + + + + + + Slovakia + + + + + + + + + + + + + + + + Slovenia + + + + + + + + + + + + + + + + South Africa o o Spain + + + + + + + + + + + + + + + +

 Country 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 Year  0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5

125 www.ShimiPedia.ir About the History of the IChO

Participating Delegations •= host. + = participant. o = observer

Year  6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 Country  8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 Sweden + + + + + + + + • + + + + + + + + + + + + + + + + + Switzerland o + + + + + + + + + + + + + Syria Tajikistan Thailand o + + + + + + + + + + Turkey o + o + + + + + + Turkmenistan o UdSSR + • + + + + + + • + + + + + + + + + + Ukraine + + + + + + United Kingdom o o + + + + + + + + + + + + + + + + + United States o o + + + + + + + + • + + + + + + + Uruguay o o + Uzbekistan Venezuela o o + + + + + + + Vietnam + + + +  Country 6 6 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 Year  8 9 0 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9

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About the History of the IChO

Participating Delegations •= host. + = participant. o = observer

Year  0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 Country  0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 Sweden + + + + + + + + + + + + + + + + Switzerland + + + + + + + + + + + + + + + + Syria o o + + + + + + Tajikistan o o + + + + + + + + + + + Thailand + + + + + + + + + + + + + + + + Turkey + + + + + + + + + + + • + + + + Turkmenistan o o + + + + + + + + + + + + + UdSSR Ukraine + + + + + + + + + + + + + + + + United Kingdom + + + + + + + + + • + + + + + + United States + + + + + + + + + + + + • + + + Uruguay + + + + + + + + + + + + + + + + Uzbekistan o o + + + Venezuela + + + + + + + + + + + + + + + Vietnam + + + + + + + + + + + + + + • +  Country 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 Year  0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5

127 www.ShimiPedia.ir About the History of the IChO

Inofficial ranking since 1974

(set up by adding the points of the teams. up to position 50)

1974 1975 1976 1977 1978 1979 1980 1981 1982 1983 1984 1985 1986 1987 1988 IChO held in RO H DDR CS PL SU A BG S RO D CS NL H FIN 1 SU SU DDR CS SU PL PL H CS RO D SU NL SU RC . RO H SU SU PL SU D CS D SU CS CS PL RC D . CS PL H H D RO DDR PL PL D SU D D RO USA . H BG PL PL DDR CS H BG NL CS H A SU CS PL 5 PL RO A S CS A A A A H A NL A D GB . DDR DDR RO A H S RO D SU A GB H USA F DDR . BG S BG D A H BG DDR H F PL DDR H GB N . YU CS CS DDR RO D CS RO BG DDR USA PL BG PL RO . S A S RO S BG S SU DDR PL RO USA F H H 10 D* D D BG BG FIN FIN NL S NL DK F RO DDR SU . YU YU YU TR DDR NL FIN F BG S GB CS NL I . B B B FIN I S FIN GB NL RO GB USA NL . B F N N FIN BG S BG BG . I RO DK F N DDR A CS 15 * hors concourse DK FIN BG S CDN S AUS . YU S N FIN N FIN SGP . I I I YU DK N F . YU GR B B DK A . YU GR FIN I FIN 20 B DK GR GR CDN . C KWT C DK . YU B C . YU S . CDN B 25 CH CH . KWT KWT

(List of abbreviations see page 132)

128 www.ShimiPedia.ir About the history of the IChO

1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 IChO held in DDR F PL USA I N RC RUS CDN AUS T DK 1 DDR RC RC RC RC RC RC IR H SGP USA RC . D PL RO H TPE GB IR RC D USA ROK RUS . RC D H PL USA USA RO RUS TR ROK RC USA . BG USA PL USA I A A A TPE RC IR H 5 SU CS NL A GUS SGP D D IR H RO TPE . H RO USA GUS H ROK GB USA RUS RA H A . PL F I D D TPE SK UA ROK RUS TPE SK . RO A D RO CDN CZ TPE CZ RC AUS UA BY . CS DDR N F SGP GUS I H SGP D PL VN 10 I H GB I CZ IR CZ RO PL GB AUS TR . NL GB CS SGP A D RUS GB USA PL VN SGP . GB I SU CS RO H H TPE UA A D D . A AUS A AUS P RO AUS BY AUS RO RA ROK . USA SGP AUS NL NZ DK SGP SGP CDN TPE BY IR 15 S NL DK DK ROK I F RA RO SK T CZ . F N SGP ROK LV T TR TR A NL F FIN . N DK CDN GB IR NZ PL F T IR TR T . AUS T BG CH DK UA USA I EST UA SGP MEX . CDN FIN F T AUS AUS DK AUS CZ VN IND GB 20 DK CDN S LV NL F RA ROK VN LT GB AUS . FIN BG T NZ LT PL ROK EST F TR RUS IND . B C CH S SK NL UA CDN S BY MEX CDN . C S LV LT F SK LT T BY F A RA . GR CH LT N C CDN T VN NZ I IRL UA 25 CH B FIN CDN GB LT NL SK LV T NZ PL . KWT GR C SLO T S CH CH RA FIN I NZ . KWT GR BG BG N BG NL SLO CZ CDN BG . CY B TPE B BG S NZ GB CDN LT F . CY B S FIN NZ DK SK S NL DK 30 SLO FIN FIN EST EST PL LT BG SK NL . GR SLO LV CDN SLO I N BG B . CY GR CH MEX MEX DK MEX KZ RO . MEX MEX MEX N LV NL CH DK KZ . N SLO SLO N IRL SLO CH LT 35 CH B LV CY N EST CZ CH . YV CY CY BG MEX CY FIN SLO . CY GR B S CH LV B EST . KWT TR GR LT CY DK S S . YV FIN E E NZ CY YV 40 C YV B FIN GR EST CY . KWT KWT GR BG KZ LV HR . C FIN YV E SLO I . YV GR IRL YV RI . C B B BR N 45 KWT RI KS E AZ . KWT YV N IRL . C RI RI E . GR LV . ROU GR 50 C BR

(List of abbreviations see page 132)

www.ShimiPedia.ir 129 About the History of the IChO

2001 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 IChO held in IND NL GR D TPE ROK RUS H GB J TR USA 1 RC RC RC RC ROK RC RC RC TPE RC RC TPE . ROK T IR ROK VN TPE RUS RUS RC T ROK ROK . USA TPE ROK RUS IR ROK TPE UA ROK ROK RUS RUS . RUS ROK T UA RUS RUS PL ROK RUS J RI IND 5 IR A BY D AZ VN ROK T SGP TPE USA RC . TR UA RUS PL TPE T D BY J H T SGP . IND USA IND TPE T J T VN USA CZ SGP J . AUS PL SGP H RA Pl IND TPE H SGP CDN D . TPE IND D TR D IND H H IR USA H H 10 T D TPE VN IND D SK SGP GB IR IR UA . SGP IR UA IND A SK LT KZ RO RUS TR RI . PL H PL IR CZ DK USA A T TR IND USA . RO RUS CDN RO UA SGP VN PL D LT CZ BY . F CDN CZ LT PL BR GB IR IND D F VN 15 SK TR RO CZ AUS CDN BY IND PL PL J RO . H AUS KZ USA TR AZ EST RO AUS GB TPE LIT . VN GB VN SGP H UA UA AUS A IND D CZ . CZ SGP EST CDN SK USA RI D BY RI SK KZ . RA E GB AZ USA H IR SK VN RO KZ RA 20 BY SK AUS AUS GB CZ RO TR F A AUS PL . C BY H KZ RO AUS AUS LT RI VN VN SK . D VN SK GB BY IRL A EST TR SK RO IR . GB FIN USA J SGP F KZ I LT CDN GB A . UA F YV A J IR SGP GB UA EST BY GB 25 A LT IND BY RI A NZ CDN EST AUS PL AUS . MEX CZ F SK LV TR CZ NZ CZ UA A IL . DK KZ A T BG RI F BR SK F LT HR . CDN LV I RA HR GB TR USA CDN RA EST BR . EST NL TR EST MEX RO J LV I NZ RA CDN 30 RI RO AZ F KZ NL ARM RI RA BY UA NZ . HR RA MEX NZ LT HR SLO F NZ KZ FIN TR . I EST LT SLO F LT RA CZ TM BR SLO EST . N HR NL HR EST KZ BR J MEX IL I LV . BG BG FIN LV CDN SLO CDN DK KZ HR BR F 35 CY NZ HR NL I EST I RA IL SLO HR ARM . KZ I J I DK RA MAL MEX BR FIN NZ I . B DK DK CH SLO BR IL SLO HR DK TM NL . LT SLO RA FIN FIN TJ IRL IL AZ NL LV TM . NZ N GR RI NL LV NL AZ DK E S DK 40 CH YV LT S IRL MAL CH HR S I NL TJ . E MEX E BG GR S S TM LV LV PE YVA . FIN BR TM KS NZ IRL LV BG IRL BG PK BG . SLO S BR E KS IL DK MGL FIN CR TJ SLO . NL RI BG GR S FIN MD IRL N CH E CH 45 LV TM CH BR B IS E MAL E IRL MEX FIN . BR B NZ TM BR I BG E NL MEX CH MEX . S IRL IS CY CH CY TM S MGL MGL MGL MGL . YV CH IRL YVA P N HR NL PE MAL IL T . IRL C CY IRL IS TM PK CH PK N CY PK 50 GR CY KS IS N CH N ROU SLO S BG AZ

(List of abbreviations see page 132)

130 www.ShimiPedia.ir About the history of the IChO

2013 2014 2015 2016 2017 2018 2019 2020 2021 2022 2023 2024 IChO held in RUS VN AZ 1 RC SGP RC . ROK UA ROK . TPE RUS TPE . USA VN SGP 5 H TPE RO . SGP RC RUS . RUS USA J . PL TR IND . UA RO USA 10 IND T PL . VN IR TR . T PL UA . BY ROK T . J RI KZ 15 KZ J IR . IR BY CS . SK GB VN . CZ D BY . RI LT SK 20 D IND GB . RO SK SRB . A CZ A . LIT H LT . AUS AUS H 25 GB UZ EST . TR CDN CDN . NZ SRB RI . HR RA D . F MEX LV 30 DK A I . MD NZ RA . CDN EST AUS . LV KZ BR . SLO MAL BG 35 RA KSA MRL . SRB HR PE . BR DK DK . EST BR KSA . UZ NL CH 40 AZ PK MD . I F F . E I NZ . IL BG IL . CY E UZ 45 N SLO SLO . ARM TM PK . PK LV FIN . CH CH AZ . BG PE KS 50 TJ N NL

(List of abbreviations see page 132)

www.ShimiPedia.ir 131 About the History of the IChO

List of abbreviations

A Austria LV Latvia ARM Armenia LT Lithuania AUS Australia MAL Malaysia AZ Azerbaijan MD Moldova B Belgium MEX Mexico BG Bulgaria MGL Mongolia BR Brazil N Norway BY Belarus NL Netherlands C Cuba NZ New Zealand CDN Canada P Portugal CH Switzerland PE Peru CS Czechoslovakia PK Pakistan CY Cyprus Republic PL Poland CZ Czech Republic RA Argentina D Germany RI Indonesia DDR German Democratic Republic RC China DK Denmark RO Romania E Spain ROK South Korea EAK Kenya ROU Uruguay EST Estonia RUS Russian Federation ET Egypt S Sweden F France SGP Singapore FIN Finland SK Slovakia GB United Kingdom SLO Slovenia GR Greece SRB Serbia GUS Commonwealth of Independent States SU Soviet Union H Hungary T Thailand HR Croatia TJ Tadzhikistan I Italy TM Turkmenistan IL Israel TPE Chinese Taipei IND India TR Turkey IR Iran UA Ukraine IRL Ireland USA United States of America IS Iceland UZ Uzbekistan J Japan VN Vietnam KS Kyrgyzstan WAN Nigeria KSA Saudi Arabia YU Yugoslavia KWT Kuwait YV Venezuela KZ Kazakhstan

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