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58 Chemistry Olympiad th F i n a l c o m p e t i t i o n (31 Mar 2012) Theoretical tasks and solutions TASK 1 Complexes with cyclam 1,4,8,11-tetraazacyclotetradecane (cyclam) is a macrocyclic ligand with four coordination sites at nitrogen atoms. This ligand can form complexes with heavy metal cations (with 1:1 stoichiometry). These complexes are applicable as catalysts, they can be also applied in medicine, as they affect important processes of biological significance. NH HN NH HN cyclam These complexes are stable in pH range limited by disadvantageous effect of nitrogen atoms protonation (only neutral molecule exhibits complexing properties) and possibility of heavy metal hydroxides deposition. This task concerns general conditions of stability of complexes with cyclam, in solutions of constant pH = 7. Basing on obtained data it would be possible to predict stability of complexes of specific metal cations. By solving this task, please assume predomination of some protonated forms of the ligand. Problems: a. Calculate concentration of the neutral ligand in solution with total concentration of cyclam (protonated and non-protonated forms) equal to 0.01 M (pH = 7). b. Determine the ratio of stability constant, , and so-called „conditional” stability constant, ’, where: ’ = [ML]/([M]cL), [ML] is concentration of the complex, [M] is concentration of free metal ions (in symbols M, ML the charge is omitted), cL is total concentration of cyclam (protonated and non-protonated forms) not bound with metal ions. c. Derive equation and calculate the minimal value of stability constant, , for which in solution of pH = 7 at least 99.9% of metal ions is complexed, for cL = 0.01 M. d. Heavy metal ions of charge 2+ can form hydroxide deposits. What is the minimal value of stability constant, , to avoid hydroxide deposition in solution containing cyclam and metal ions (at pH = 7). Express the stability constant, , as a function of solubility product of the hydroxide, Ks0, and other selected equilibrium constants, given in the task, and concentrations ML and cL. e. Are the conditions from sections (c) and (d) fulfilled for cations Cu2+ i Ni2+ for [ML] = 0.01 M and cL = 0.01 M. For the case of the complex with Ni(II) ions calculate the concentration of noncomplexed nickel ions. How does this concentration change after 2-fold dilution of the solution ? Dissociation constants of acids: Ka1 (for H4L4+) = 4.103; Ka2 (for H3L3+) = 3.102; Ka3 (for H2L2+) = 3.1011; Ka4 (dla HL+) = 3.1012 (L is the form presented in the figure). Stability constants of complexes with cyclam, ; for Cu2+: 2.1027, for Ni2+: 2.1022. Solubility products, Ks0; for Cu(OH)2: 3.1019, for Ni(OH)2: 2.1015. 1 TASK 2 Phosphorus oxide – acidic or alkaline? Compound A is formed in the reaction of white phosphorus with oxygen carried out at lowered pressure and at a temperature below 50 °C. Reactive compound A can be isolated via vacuum distillation from the products mixture in the form of white crystalline substance. The compound melts at ca. 24 °C and can be easily dissolved in a series of organic solvents e.g. in benzene. Only one signal was found in the 31P NMR spectrum of its benzene solution. It was also observed that the solution obtained by dissolution of 0.185 g of compound A in 10.00 g of benzene freezes at a temperature that is by 0.43° lower than melting point of pure benzene. The cryoscopic constant Et for benzene equals 5.12 K · kg/mol. Compound A reacts very easily with water to form a solution that contains acid B which forms a precipitate of an anhydrous barium salt upon addition of the excess of Ba(OH)2 solution. A sample of compound A with a mass m1 = 1.26 g was reacted with a substantial excess of nickel carbonyl Ni(CO)4. The reaction was proceeding at room temperature and a colourless gas was being evolved. It was found that, after compound A had been consumed, the mass of the reaction mixture decreased by 0.642 g. 4.5 g of pure compound C was obtained after removal of the unreacted carbonyl. The 31P NMR spectrum of this compound comprised only one signal as well. Compound A in the solid state (3.08 g) was also reacted with gaseous diborane (B2H6) at room temperature. After 24 hours products of the reaction were dissolved in n-pentane and left for crystallisation. 2.67 g of pure crystalline compound D were collected. It was found that this compound reacts violently with water and in the course of reaction colourless gas is liberated and the resulting solution contains acids B and E. In a carefully controlled reaction 0.147 g of compound D were used and ca. 80 mL (equivalent for 0 °C and atmospheric pressure) of gas were given off. Compound D reacts easily with nickel carbonyl too and using an excess of carbonyl leads to the previously described compound C. Problems: a. Determine compound A formula and confirm it with appropriate calculations. Write the equation of white phosphorus reaction with oxygen. b. Draw Lewis electron structure of compound A molecule taking into account all of the lone valence electron pairs. Justify your answer. c. Write the chemical equation in molecular form (reagent form) of compound A reaction with water. d. Write the chemical equation in molecular form of acid B reaction with barium hydroxide. Draw the molecular structure of anion present in the structure of barium salt. Justify your answer. e. Determine the formula of compound C and write the equation of its formation reaction. Justify your answer and confirm it with calculations f. Draw and describe the structure of nickel coordination sphere in compound C. Justify your answer. g. Determine the formulae of compounds D and E. Write the equation of compound D reaction with water. Justify your answer and confirm it with calculations h. Basing on the appropriate acids and bases definition determine the chemical character of reagents in the formation reaction of compound D and calculate the reaction yield. i. Draw and describe the boron coordination sphere in compound D. Justify your answer. Use the following values of molar masses in your calculations (g/mol): B – 10.81 C – 12.01 H – 1.008 Ni – 58.69 O – 16.00 P – 30.97 and the molar volume of gases at 0 °C and under atmospheric pressure Vm = 22.41·10−3 m3/mol 2 TASK 3 Determination of stoichiometry and stability of complexes with an NMR method. Application of approximations in limiting conditions One of the problems in the 2nd stage of the 58 Chemistry Olympiad was related to the crown ethers ability to complex metal ions. Now we present the results of 1 H NMR study on formation reaction of a 12-crown-4 ether (labelled as 12C4) with Na+ cation introduced into the system in a form of sodium thiocyanate, NaSCN. Three independent experiments have been carried out: the first and the second at temperature 23°C, and the third one at temperature –50°C. In Fig. 1. Molecular the two first experiments the spectra were recorded at varying NaSCN formula of a 12C4 concentration and a single averaged signal coming from both the crown ether. complexed molecules and the free ether was monitored. The signal chemical shift, δobs, was dependent on the ratio of sodium ions [Na+]0 and crown ether [E]0 concentrations. The appearance of the averaged signal proves that the complexation reaction is a so called fast exchange reaction. Let us remind that for such reactions the observed chemical shift, δobs, is an average of characteristic chemical shifts of the ligand, δE, and the complexes, δEnM, multiplied by corresponding concentration ratios of ligands unbound and bound in complexes to the total concentration of ligands in the system. By writing the general complex formula as EnM (where E means ether, and M means metal), the relationship can be presented as follows: [ E M] [E] [EM ] [ E M] obs E EM 2 2 E2M 3 3 E3M (1) [E] 0 [E]0 [E]0 [E]0 The total concentration of ligands in the system: [E]0 = [E] + [EM] + 2·[E2M] + 3·[E3M]+ . . . . Experiment 1 The characteristic chemical shift, δE = 2.761 ppm, has been determined from the NMR spectrum of the pure crown ether. Then, small amounts of solid sodium thiocyanate were added to a 12C4 solution (in deuterated methanol, CD3OD) with concentration [E]0 = 0.219 mol/dm3, and the chemical shift, δobs, was recorded as a function of total concentrations [NaSCN]/[12C4]. The measurement results are shown in Fig. 2, and additionally, the data for points within the linearity range, i.e., at [NaSCN]/[12C4] ≤ 0,3, are given in Table 1. The experimental points form a curve showing two characteristic bends, which prove that two complexes, k1 and k2, with different stoichiometry and different overall complex formation constants for metal ion and for ether, βn, are present in the system Based on the plot given in Fig. 2 one can determine the stoichiometry of the two complexes being formed and remaining in equilibrium. complex 2 Fig. 2 3 complex 1 Table 1. Results of experiment 1: [E]0 = 0.219 mol/dm3 [NaSCN]/[12C4] δobs / ppm 0 0.045 0.096 0.150 0.240 0.300 2.761 2.846 2.937 3.025 3.190 3.282 Experiment 2 In all measurements the initial concentration of the crown ether was kept constant at [E]0 = 0.050 mol/dm3, while the thiocyanate concentration varied but was much higher than that of the emerging complex. The sodium ion concentrations, c, and the corresponding resultant chemical shifts, δobs, are given in Table 2. Table 2. Results of experiment 2: [E]0 = 0.050 mol/dm3 Total Na+ ion concentration c / (mol/dm3) Observed chemical shift δobs / ppm 4.437 4.558 4.660 4.757 4.821 4.856 0.204 0.271 0.359 0.495 0.646 0.768 Experiment 3 The spectrum of the pure 12C4 ether was recorded again at temperature –50°C, then a small amount of NaSCN was added, and the spectrum was recorded again. The spectrum showed distinctly resolved ether and complex peaks (the fast exchange reaction does not occur here). It turned out that the chemical shift of the pure ether virtually did not change with decreasing temperature. One can therefore assume that the chemical shift of the k2 complex that is stable in the presence of a large excess of ether, is the same at both 23°C and at –50°C, and is δk2 = 4.575 ppm. Using the results of the three experiments described above, determine the stoichiometry of complexes that are formed in the system, calculate their stability constants and characteristic chemical shift of the k1 complex, by following the directions given below. In the solution one should assume the following symbols for the major quantities occurring in the problem: [E]0 + – total concentration of the crown ether, constant in each experiment, [Na ] = c – total concentration of sodium ions, varying during titration, δE – chemical shift of the pure 12C4 crown ether, δk1, δk2 – chemical shifts of two complexes, ck1, ck2 – concentrations of EnM complexes, respectively, in equilibrium. To simplify the notation of equations, the following symbols can be also introduced: obs obs E , k1 k1 E and k2 k2 E . 4 Problems: a. Based on the plot in Fig. 2 determine the stoichiometric coefficients of the resulting complexes. Justify the answer with a short comment. b. Give chemical equations for reactions taking place in the system and the formulae for stability constants of the complexes, by writing them down in a form containing the initial ether concentration, [E]0, and the concentration of sodium ions, c. c. Using appropriate equations determine complex concentrations ck1 and ck2 by introducing the simplifications proposed in Note 1. d. Give an equation describing the dependence of the averaged chemical shift, δobs, on corresponding chemical shifts of the ether, δE (data), and the two complexes: δk1 (this should be treated as an unknown) and δk2 (data). Present the equation as a dependence of corresponding differences of chemical shifts, that is obs , k1 and k2 . Write the two equations in two forms corresponding to the limiting cases, by introducing corresponding complex concentrations determined in Direction c. e. Calculate β2 from the simplified equation from Direction d. for the limiting case [E]0 >> c by determining the slope of the straight line (δobs–δE) = f(c) from two selected points from Table 1 and using known value of δk2 (see Note 2). f. Solve the equation from Direction d. in the high Na+ concentration limit (c), by converting the equation so as to get a linear dependence of 1/(δobs – δE) on 1/c. Calculate β1 and δk1 constants (see Note 3). Notes: 1. The equations for equilibrium constants can be simplified if the experiment conditions can be considered as the limiting ones. In experiment 1 [E]0 >> c, so the equilibrium is strongly shifted towards the k2 complex and we can assume that [E]0 >> ck2 >> ck1 ≈ 0. In experiment 2 [E]0 << c, so the equilibrium is strongly shifted towards the k1 complex and we assume that c >> ck1 >> ck2 ≈ 0. Corresponding approximations are best introduced by determining ck1 and ck2 as functions of complex stability constants β1 and β2. 2. Fig. 3 shows linear dependence of (δobs – δE) on c in the limit [E]0 >> c. The plot should facilitate the selection of points of which the coordinates allow to calculate the slope of the straight line. Experiment 1 Fig. 3 5 3. To solve the problem in the high Na+ concentration limit ([E]0 << c), it is necessary to “linearise” the dependence of the corresponding difference of chemical shifts (δobs–δE) on varying [Na+] = c concentration. The dependence of 1/(δobs–δE) on 1/c should be linear (1/(δobs–δE) = a·(1/c) + b) in the limit of large values of the ratio c/[E]0, i.e., in the conditions of experiment 2. Fig. 4 shows the relationship plotted using the data from Table 2. Experiment 2 Fig. 4. TASK 4 Biologically active indole derivatives Tryptamines are derivatives of indole (a heterocyclic compound consisting of two condensed rings, a benzene ring and a pyrrole ring), abundantly present in living organisms. To tryptamines belong, inter alia, neurotransmiters as important as melatonin or serotonin. Due to that also other tryptamine derivatives are not neutral to human organism, and may have medicinal use or exert hallucinogenic effects. The phosphorylated tryptamine derivative F, initially isolated from plant sources, has been obtained through the reaction sequence depicted in Scheme 1. The starting material for the synthesis was 4hydroxyindole. It is known that the most reactive position of indole is the C-3 within the pyrrole ring, which easily undergoes aromatic electrophilic substitution, but may also be involved in reactions typical for enamines. First, the phenol group of 4-hydroxyindole was protected by reacting it with benzyl chloride in the presence of sodium methanolate. This led to compound A, which was then subjected to FriedelCrafts acylation with oxalyl chloride leading to compound B. Subsequently, B without isolation, was reacted with dimethylamine, which resulted in compound C. In the next step, C was reduced to D (molecular formula C19H22N2O) by means of LiAlH4 and then catalytically hydrogenated to yield compound E and toluene as a by-product. In the last step, E was converted into F by applying following reagents: 1. n-buthyllithium 2. tetrabenzyl pyrophosphate (TBPP) 3. catalytic hydrogenation (Pd/C, H2). A homologue of compound E, known as 4-hydroxygramine and having molecular formula of C11H14N2O, may be obtained in a multi-component reaction (Mannich reaction), by reacting compound A with formaldehyde and dimethylamine, followed by a catalytic hydrogenation. Exploiting a similar multi-component reaction, but from different starting materials, one can obtain compound X, as a mixture of several stereoisomers (Scheme 2 depicts only one of them). 6 O Cl X Cl H3C OH O Cl NH [B] A N H C MeOH/MeOLiAlH4 TBPP = BnO O O Bn P P BnO O Bn O O F 1. n-BuLi 2.TBPP 3. H2, Pd-C E Scheme 1 Chiral O H2, Pd-C D C19H22N2O 1 O H3C 2 3 NH 4 CH3 N O O Scheme 2 Problems: a. Draw structural formulas for compounds A-F and 4-hydroxygramine. b. Draw structural formulas of the starting materials for the synthesis of compound X. c. Compound X could be obtained from the same starting materials, but in a stereochemically pure form (as depicted in Scheme 2). Choose, which conditions should be applied to achieve this: I. the reaction temperature should be increased; II. the reagents should be mixed in a certain order; III. the reaction should be carried out in the presence of L-proline; IV. the reaction mixture should be stirred always in the same direction. d. Determine the absolute configuration of all stereogenic centers in compound X. e. Explain the reactivity of the C-3 position of indole by means of appropriate mesomeric structures. TASK 5 The Diels-Alder Reaction Cycloaddition reactions, including Diels-Alder reaction, in most cases are stereospecific, what means that geometry of substrates (e.g. dienophile or diene) determine the structure of the formed product. + X X Y Y General Scheme of the Diels-Alder Reaction I. The carboxylic acid with molecular formula C4H4O4 as well as its dimethyl esters (A , B and C) exist as two geometric isomers and one regioisomer, and all of them undergo the Diels-Alder reaction. The reaction of geometric isomers A and B mixture with diene D, composed entirely of carbon and hydrogen atoms, provides diastereomeric products E1, E2 and F. Ozonolysis of E and F mixture with zinc dust form product G (diastereomeric mixture) and subsequent reduction with LiAlH4 excess leads to the mixture of diastereomers of polihydroxylic alcohol H. 7 D AiB E1 + E2 i F (diastereomers) 1) O3 2) Zn 1) O3 2) Zn HO OH LiAlH4 G HO mixture of three diastereomers OH H mixture of three diastereomers D I J Compounds A and B after ozonolysis with zinc dust form the same product I, which also undergo cycloaddition reaction with diene D. The reaction leads to the cyclic ether J. The molecular mass of J is equal to the sum of mass of reactants D and I. II. Stereochemistry in Diels-Alder reaction: Isomer A in the reaction with diene D forms the mixture of two achiral diastereomers E1 and E2. The reduction of diastereomeric mixture of E1 and E2 with hydrogen on palladium catalyst leads to only one product K. The Diels-Alder reaction of isomer B with diene D provides exclusively racemic mixture of cycloadduct F. The product F after ozonolysis with zinc dust and subsequent reduction with LiAlH4 excess leads to one diastereomer of polihydroxylic alcohol H (as a racemic mixture). Problems: I a. Draw the structure of geometric isomer A or B and structure of regioisomer C b. Draw the structure of diene D c. Determine general structure of Diels-Alder product E/F (no stereochemistry required) d. Draw the general structure of ozonolysis product G (no stereochemistry required) e. Draw the structure of compound I and general structure of J (no stereochemistry required) II a. Determine unambiguously geometry of isomers A and B b. Draw stereochemical structures of compounds E (E1 and E2) and their reduction product K c. Draw the stereochemical structure of compound F and stereochemical structure of corresponding diastereomer of polihydroxylic alcohol H d. Draw all possible stereoisomers of the product J. Determine enantiomers and diastereomers. SOLUTIONS SOLUTION OF TASK 1 a. The total concentration of cyclam, cL = [L] + [HL+] + [H2L2+] + [H3L3+] + [H4L4+]. Using acidic dissociation constants, after rearrangement one can write: cL = [L]{1 + [H+]/Ka4 + [H+]2/(Ka4 Ka3) + [H+]3/(Ka4 Ka3 Ka2) + [H+]4/(Ka4 Ka3 Ka2 Ka1)}. Taking into account the values of dissociation constants, it can be assumed that at pH = 7 the form H2L2+ predominates in the solution, i.e. the above equation can be simplified to the form: cL = [L][H+]2/(Ka3 Ka4), thus [L] = cL Ka3 Ka4 /[H+]2. After introducing numerical results, for cL = 0.01 M and [H+] = 107 M, the obtained [L] = 9.1011 M. 8 b. Because stability constant = [ML]/([M][L]), the ratio /’ = cL/[L]. Basing on previous data, cL/[L] = 0.01 / 9.1011 = 1.1.108. P O c. Because [ML]/[M] = 99.9 / 0.1, ’ = [ML]/([M]cL) = 99.9 / (0.1 0.01) = 1 10 . . . Thus, = ’ 1.1 10 = 1.1 10 . . . 8 . 13 O O 5 P O P P O O d. = [ML]/([M][L]) = [ML][H ] /([M]cL Ka3 Ka4). Assuming that a deposit of hydroxide does not precipitate: Ks0 = [M][OH]2, the above equation can be written in the form: = [ML][H+]2[OH]2/(cL Ks0 Ka3 Ka4) = KW2 ([ML]/cL) /( Ks0 Ka3 Ka4). + 2 e. For [ML] = cL = 0.01 M, using appropriate constants, the minimal values of fulfilling the conditions from (d) are: 4.1012 for the Cu2+ complex and 6.108 for the Ni2+ complex. Comparison of these results with experimental data shows that Cu2+ and Ni2+ complexes obey the conditions from (c) and (d). Because for Ni 2+ - cyclam complex ’ = [ML]/([M]cL), for [ML] = cL, [M] = 1/’ is obtained. For the Ni 2+ complex ’ = 2 . 1022 / 1.1 . 108 = 2 . 10 14 . Thus, [M] = [Ni 2+ ] = 1/ (2 . 10 14 ), i.e. [Ni2+] = 5.1015 M. 2-fold dilution results in the same value of the [ML]/cL ratio, therefore Ni2+ ions concentration does not change as well. SOLUTION OF TASK 2 a. White phosphorus is built of P4 molecules. Depending on the reaction conditions, it may form P4On (n = 6, 7, 8, 9 or 10) oxides in the reaction with oxygen whose structures stem from P 4 molecules structure. Freezing point depression of benzene allows one to calculate the molality of oxide A solution (cmA) and consequently its molar mass: cmA mA Tt 0.43 0.185 0.084 mol/kg, czyli M A 220 g/mol. Et 5.12 mb c mA 0.010 0.084 This value corresponds to P4O6 molar mass (219.88 g/mol). Thus, compound A is phosphorus(III) oxide, P4O6. Phosphorus reaction with oxygen proceeds according to the following equation: P4 + 3O2 P4O6 b. The P4O6 molecule exhibits a cage structure stemming form the structure of P4 molecule. There is one lone electron pair on every phosphorus atom and oxygen atoms have two lone electron pairs. This structure agrees with 31P NMR spectrum indicating all phosphorus nuclei are chemically equivalent c. P4O6 is phosphoric(III) acid anhydride and reacts with water according to the following equation: P4O6 + 6H2O 4H3PO3 d. Phosphoric(III) acid contains P–H bond and is a diprotic acid of an average strength. It forms, therefore, two series of salts and sparsely soluble barium salt containing HPO32− anion will be formed with the excess of barium hydroxide: H3PO3 + Ba(OH)2 2− 3 3 + 2H2O HPO3 anion exhibits tetrahedral structure (sp hybridisation of phosphorus). There are three oxygen ligands and hydrogen atom in the vertices of a deformed tetrahedron. 9 2 H O O 2 H P O or O O P O e. Due to the presence of lone electron pairs on phosphorus atom P4O6 oxide can act as Lewis base in chemical reactions. Therefore, Lewis adduct is formed in the reaction with nickel carbonyl in which phosphorus forms a chemical bond with nickel replacing one of the CO ligands. Assuming that the only gaseous product of the reaction is CO, the molar ratio of the evolved CO to the used P4O6 equals 0.642 g 1.26 g 0.02292 : 0.00573 4 :1 which indicates that four nickel atoms 28.01 g/mol 219.88 g/mol were bound by a P4O6 molecule. The reaction proceeds according to the following equation: 4Ni(CO)4 + P4O6 P4O6·4Ni(CO)3 + 4CO The formula of compound C is P4O6·4Ni(CO)3 (or P4O6[Ni(CO)3]4). The molar mass of compound C equals 790.76 g/mol. Such a reaction course is confirmed by the approximate molar mass of compound C, calculated from the reaction product mass: MC mC 4.5 785 g/mol nP4O6 0.00573 f. It follows from the fact that all of phosphorus nuclei are chemically equivalent in the molecule of compound C that every phosphorus atom is bound to one nickel atom and three oxygen atoms. The coordination centres in the form of nickel atoms satisfy the 18 electron rule (10 valence electrons in 3d orbitals + 6 electrons from CO ligands + 2 electrons coming from phosphorus). Therefore, the structure of nickel coordination sphere is tetrahedral and ligands bound to nickel atom are localised in vertices of a slightly deformed tetrahedron. O O O C P O Ni C C O O Nickel coordination sphere g. Similarly as in the reaction with nickel carbonyl, P4O6 phosphorus oxide serves as a Lewis base in the reaction with B2H6 forming an adduct with acid (BH3). As a result of the reaction with water, the adduct is decomposed and its hydrolysis leads to phosphoric(III) acid (acid B) and H3BO3 boric acid (acid E). The liberated gas is molecular hydrogen and its quantity depends on the amount of reacted diborane. nB2H6 VH2 Vm 6 80 0.595 103 mole which correspond to B2H6 mass equal to 3 22.4110 6 mB2H6 nB2H6 M B2H6 0.595 103 27.67 0.0165 g. The molar ratio of P4O6 to B2H6 in the Lewis adduct equals: (0.147 0.0165) g : 0.595 103 0.594 : 0.595 1:1 219.88 g/mol Only two phosphorus atoms are involved in bond formation with BH3 and the formula of compound D is P4O6 · 2BH3 (or P4O6 · B2H6). The hydrolysis reaction proceeds according to the following equation: P4O6 · 2BH3 + 12H2O 4H3PO3 + 2H3BO3 + 6H2 10 h. The reaction between P4O6 and B2H6 leading to the formation of compound D: (P4O6 + B2H6 P4O6·2BH3) is an acid-base reaction according to the Lewis definition. Phosphorus oxide, being electron pair donor, is a Lewis base and B2H6 (acceptor) is a Lewis acid. The reaction product, compound D, is a Lewis adduct. The yield of compound D formation reaction equals: w% nD mD / M D 2.67 / 247.55 100% 100% 100% 77.0% nP4O6 mP4O6 / M P4O6 3.08 / 219.88 i. There is a donor-acceptor bond formed between phosphorus and boron atoms in adduct D. Ligands bound to boron atom (sp3 hybridisation) are localised in vertices of a slightly deformed tetrahedron: H H O B H H H O B P O O or H P O O SOLUTION OF TASK 3 a. The experimental curve δobs as a function of the concentration ratio [NaSCN]/[12C4] shows clear changes of the slope at [NaSCN]/[12C4] = c/[E]0 ≈ 0.5 and at c/[E]0 ≈ 1.0. This means that complexes with E/M = 2 and E/M = 1, i.e., E2M and EM, are formed in the system. In the range c/[E]0 >> 1.0 (thiocyanate excess), the equilibrium is shifted towards the EM complex, and in the range c/[E]0 << 0.5 towards the E2M complex. b. The complex formation reactions occurring at the equilibrium are as follows: (k1): E + M EM and (k2): 2E + M E2M. The complex formation equilibrium constants, k1 and k2, also referred to as the complex stability constants, can be written as: [ET ] ck1 , (1) β1 [E] [T] ([E]0 ck1 2ck2 ) (c ck1 ck2 ) [E 2 T] ck2 . (2) β2 2 [E] [T] ([E] 0 ck1 2ck2 ) 2 (c ck1 ck2 ) c. From the equations for the equilibrium constants β1 and β2, ck1 and ck2, respectively, are to be determined by introducing the simplifications suggested in Note 1. In equation (1), we assume that c >> ck1 >> ck2, and consequently c−ck1−ck2 ≈ c. We can further neglect subtraction of ck2 in the first term of the denominator and get: c k1 . (3) β1 ([E]0 ck1 ) c In equation (2) we assume that [E]0 >> ck2 >> ck1 and we can assume that [E]0 − ck2 − 2ck1 ≈ [E]0, and in the second term of the denominator we can neglect subtraction of ck1 and we get: ck2 . (4) β2 2 [E]0 (c ck2 ) Then, from equations (3) and (4) we determine: [E]0 c β1 (5) ck1 1 c β1 11 and d. ck2 [E]02 c β 2 . 1 [E]02 β 2 (6) The resultant chemical shift resulting from chemical shifts of the crown ether, δE, and the complexes, δk1 and δk2, can be written as follows: [E]0 ck1 2ck2 c c (7) δobs δE k1 δk1 2 k1 δk2 , [E]0 [E]0 [E]0 remembering that [E]0 = [E] + ck1 + 2 · ck2 . By performing corresponding multiplications, reducing terms in equation (7) and introducing substitutions to simplify the notation: obs obs E , k1 k1 E and k2 k2 E , we obtain: obs ck1 c k1 2 k2 k2 . [E]0 [E]0 (8) Equation (8) should be written as two equations for limiting conditions, that is by substituting equations (5) and (6) to corresponding terms of equation (8) we obtain: c β1 obs_k1 k1 (9) 1 c β1 and e. obs_k2 2 [E]0 c β 2 k2 . 1 [E]02 β 2 (10) In the case when [E]0 >> c and the equilibrium is strongly shifted towards the k2 complex (E2M) (experiment 1) we solve equation (10), written as a function of c: [E] β obs_k2 2 0 2 2 k2 c . (11) 1 [E]0 β 2 One can easily notice that the factor in front of c is equal to the slope Δδobs_k2 = m·c. The factor m can be determined from the point that is farthest away from the {0,0} point but still belonging to the line, that is point 5 in Table 1. One only has to recalculate the coordinates of the point to {c, Δδobs_k2}. For point {0.24, 3.190} we obtain the values c = 0.0526 mol/dm3 and Δδobs = 0.429 ppm as well as m = 8.155 ppm/mol/dm3. From equation (11) we determine β2 and having substituted the data we get β2 = 20.1 (mol/dm3)-2. f. Solution in the case c >> [E]0 (experiment 2) boils down to the case solved in Folder B of the 58 Chemistry Olympiad, i.e., the solution of equation (9). To convert equation (9) into a linear function one must write the equation as an equality of reciprocals, as suggested by Note 4 (by rising both sides of the equation to the power −1): 1 obs_k1 E 1 1 c β1 1 1 1 , c β1 k1 E c β1 k1 E And convert to the form: 11 1 1 1 . obs_k1 E β1 k1 E c k1 E (12) (13) 1 = a·(1/c)+b, so comparing obs_k1 E corresponding parameters of the straight line with equation (13) we get the following relationships: The plot in Fig. 4 shows ideal linearity of data 12 a 1 β1 ( k1 E ) and b 1 . k1 E For two extreme points from Table 2 we calculate corresponding reciprocals, as given in the Table below. c / (mol/dm3) 0.204 0.768 (1/c) / (mol/dm3)-1 4.902 1.302 δobs / ppm 4.437 4.856 (1/(δobs – δE) / ppm-1 0.597 0.477 Then, we calculate parameters of the straight line: a = 0.0332 ppm-1·(mol/dm3) and b = 0.434 ppm-1, and substitute them to formulae: δk1 =1/b+ δE and β1= b/a, obtaining: δk1 = 5.065 ppm and β1= 13.1 (mol/dm3)-1. SOLUTION OF TASK 4 a. A C B E D F 4-hydroksygramina b. Reactants for the synthesis of X: c. Conditions for stereoselective reaction: III d. Absolute configuration of X: 3S, 4S e. Reactivity of position C-3 of indole comes from high electron density in this position In contrary to position 2 (structure III) localization of negative charge in position 3 (structure II) is not connected with dearomatization of the benzene ring, which would be energetically unfavorable. struktura I struktura II 13 struktura III SOLUTION OF TASK 5 Part I CO2Me CO2Me Ia MeO2C or CO2Me CO2Me MeO2C C (A or B) Ia Ic D Ib Ie Id O CO2Me CO2Me or CO2Me O O OMe CO2Me I CO2Me CO2Me O E or F G Ie O O or CO2Me CO2Me J Part II II a II b CO2Me CO2Me CO2Me CO2Me MeO2C CO2Me A CO2Me CO2Me or (E)-isomer (Z)-isomer + B H II b CO2Me CO2Me H H CO2Me CO2Me + H or CO2Me CO2Me K CO2Me CO2Me CO2Me CO2Me + CO2Me CO2Me IIc two diastereomers (achiral) CO2Me or CO2Me CO2Me II c HO CO2Me H or cycloadducts cis E (E1 i E2) (from ester A) MeO2C OH MeO2C H H CO2Me H HO CO2Me CO2Me MeO2C CO2Me MeO2C OH or HO HO enantiomers OH OH diastereomer of compound H obtained from cykloadduct F cycloadduct trans F (obtained from ester B) 14 58 Chemistry Olympiad F i n a l c o m p e t i t i o n (30th March 2012) Practical tasks and solutions TASK 1 Aluminium polychloride analysis Suspensions are removed during the purification of water and sewage using the so-called coagulants. They form sols having high surface which detain suspension particles and after addition of flocculants they form precipitates. Aluminium polychloride (PAC) is one of the coagulants and it is obtained via a reaction between aluminium hydroxide and hydrochloric acid, but only some of its hydroxyl groups are substituted with chloride ions and, as a result, its composition is variable. A PAC sample, whose mass (m1) is given on a 100 mL volumetric flask labelled A, was dissolved in water with a small amount of nitric acid and a clear solution was obtained. 50,00 mL of EDTA solution, whose concentration is given on a bottle, was added. The resulting solution’s pH was adjusted to ca. 4.5 using methyl orange as an indicator. Afterwards, the solution was heated and boiled for 10 minutes, while the pH was corrected using diluted ammonia solution. After cooling, the solution was transferred to flask A and water was added to the graduation mark. A second PAC sample, whose mass (m2) is given on a 100 mL volumetric flask labelled B, was dissolved in 10 mL of 2 mol/L nitric acid solution. The solution was transferred to flask B and water was added to the graduation mark. Glassware and reagents at your disposal: burette two Erlenmeyer flasks with ST beaker 25 mL volumetric pipette graduated cylinder 10 mL volumetric pipette small funnel wash bottle with distilled water ca. 0.050 mol/L KSCN solution ca. 0.025 mol/L MgCl2 solution Reagents at disposal of all participants: EDTA solution (concentration given on a bottle) ca. 0.05 mol/L AgNO3 solution eriochrome black T mixed with NaCl and a spatula ammonium buffer with pH = 10 10 % NH4Fe(SO4)2 solution chloroform with a pipette Attention: The ratio of AgNO3 solution volume to KSCN solution volume equals S and is given in your answer sweet. Additional information: The EDTA complex of aluminium exhibits lower conditional stability constant that the MgY2− complex, but the aluminium complex is inert. The eriochrome black T magnesium complex is much less stable than the MgY2− complex. The AgNO3 solution is acidified with nitric acid (its concentration is 0.5 mol/L). The AgCl solubility product is higher than that of AgSCN which is higher than AgBr Kso. Problems: a. Propose an analysis plan to determine the percentage of aluminium in the PAC sample using the procedures given below and the information contained in the problem. b. Propose an analysis plan to determine the percentage of chloride ions in the PAC sample using the information contained in the problem, the procedures given below and the available reagents. c. Write the equations (at least 6) of chemical reactions taking place during sample analysis as well as of the reactions carried out before and described in the problem. d. Derive the formulae to calculate the concentrations of titrants necessary for the completion of tasks a. and b. e. Give the determined concentrations of solutions form task d. f. Determine the percentage of aluminium in PAC. g. Determine the percentage of chloride ions in PAC. h. Determine the stoichiometric formula of PAC, Al(OH)xCly. i. Why does the solution of the PAC sample with EDTA turn red upon heating (as in the problem description)? j. Why is chloroform introduced into the flask during chloride ions determination? Is it necessary to do it when determining bromide ions? Justify your answer. Procedures Complexometric determination of aluminium Transfer an accurately measured portion of aluminium-ions-containing solution into an Erlenmeyer flask. Introduce a precisely known amount of EDTA solution to the flask, excessive with respect to the expected amount of aluminium. Add one drop of methyl orange and diluted ammonia solution until the colour changes to yellow-orange. Heat the resulting solution and boil it for 10 minutes, correcting its pH with diluted ammonia solution. After cooling, add 5 mL of ammonium buffer solution with pH = 10, a pinch of eriochrome black T and titrate the navy solution quickly with MgCl2 solution until the colour changes to violet-blue. Repeat the titration. Attention! The titration has to be carried out quickly and as soon as a noticeable change of colour is observed the titrant’s volume has to be read. The titration must not be continued even though the analyte solution may recover the blue colour. Argentometric determination of chloride ions Transfer an accurately measured portion of chloride-ions-containing solution into an Erlenmeyer flask with ST. Add a precisely known volume of AgNO3 solution, excessive with respect to the expected amount of chloride ions. If necessary add nitric acid so that its concentration lies in the range 0.2 – 0.5 mol/L. Add ca. 2 mL of chloroform, ca. 1 mL of NH4Fe(SO4)2 solution, close the flask with a stopper and shake it. Open the flask, wash the stopper with distilled water and titrate the obtained solution with KSCN solution until slightly orange colours is obtained (when the precipitate settles down). Repeat the titration. 2 TASK 2 Distinguishing surface active agents (surfactants) Owing to their characteristic structures and specific behaviour in aqueous solutions, surface active agents (surfactants) are used diversely in analytics. The interactions between cationic surfactants and dyes chelating metal ions bring about significant colour changes (batochromic shift) and if the chelates are not very stable (e.g. complexes with metal titration indicators), they may be decomposed. Metal chelates which are insoluble in water may be dissolved in aqueous solutions of non-ionic surfactants due to the fact that they form micelles at appropriate concentrations. Cationic and anionic surfactants may form precipitates that are soluble in the excess of surfactant with suitable dyes (acidic or alkaline). The mentioned phenomena make it possible to differentiate between cationic, anionic and non-ionic surfactants. Test tubes A-J The solutions of substances given in the table below are placed in test tubes labelled A-J. The solutions concentrations are also given. Substance Concentration Iron(III) chloride 210-5 mol/L Mercury(II) nitrate 210-5 mol/L Dithizone, HDz 210-4 mol/L Eriochrome cyanine R, ECR 210-4 mol/L Safranin T, SFT 210-4 mol/L Rose Bengal, RB 210-4 mol/L Potassium palmitate, PK 110-2 mol/L Sodium dodecyl sulphate, SDS 110-2 mol/L Triton X-100, TX 110-2 mol/L Cetyltrimethylammonium chloride, CTA 110-2 mol/L Dithizone is present in a slightly alkaline aqueous solution and is used for extractionspectrophotometric determination of mercury or silver (it forms orange and yellow chelates, respectively, in strongly acidic solutions). In slightly acidic solutions., eriochrome cyanine R forms violet complexes with aluminium and iron(III) ions, which turn blue under the influence of cationic surfactants. Safranin T is an alkaline, whereas rose Bengal acidic non-chelating dye. Triton X-100, poly(ethelene oxide) substituted octylphenol, is a non-ionic surfactant. Glassware and reagents at your disposal: 8 empty test tubes sulphuric acid, 1mol/L 6 polyethylene Pasteur pipettes indicator paper wash bottle with distilled water You may use the solutions for problem 1. 3 Problems: a. Give the probable arrangement of substances in test tubes A-J taking into account solutions’ colour and their pH and carrying out simple tests for the presence of surfactants. b. Derive a plan that will allow you to identify substances in the solutions. c. Identify the substances present in the solutions in test tubes A-J, using the available reagents and the given procedure. d. Give justification for your identification, confirming it with two observations. e. Can magnesium be determined by titration with EDTA and eriochrome black T as an indicator, if the analysed solution contains cationic surfactants? Justify your answer with appropriate observations. Investigating the influence of surfactants on coloured systems: Transfer ca. 1 mL of the analysed dye solution into a test tube and introduce the same amount of metal solution. Add surfactant solution drop by drop, shake the test tube and carefully watch what is taking place in the solution. Carry out a blind test for comparison. USE YOUR SOLUTIONS ECONOMICALLY. DO NOT USE PORTIONS LARGER THAN 1 ML FOR THE TESTS. Eriochrome cyanine R, ECR Rose Bengal, RB Safranin T, SFT Dithizone, H2Dz 4 SOLUTION OF PROBLEM 1 a. Analysis plan for the determination of aluminium percentage in PAC sample Upon heating of the m1 sample with EDTA solution, according to the procedure, aluminium ions reacted with some of the EDTA forming AlY . The remaining EDTA has to be titrated with MgCl2 solution. In order to determine the number of aluminium moles in the sample, a portion of 25.00 mL from flask A has to be titrated with MgCl2 solution using V1 mL of the titrant. Taking into account the fact that metal ions react with EDTA in a 1:1 molar ratio and the relative volumes of the flask and pipette, one may write: n Al nEDTA 4 nMgCl2 50 c EDTA 4 V1 cMgCl2 b. Analysis plan for the determination of chloride ions percentage in PAC sample Upon dissolution of m2 sample in nitric acid chloride ions are transferred to the solution and the resulting mixture in flask B is acidic. In order to determine the amount of chloride ions, one has to take 25.00 mL of the solution from flask B, add 25.00 mL of acidified AgNO3 solution with a known concentration (one gets the appropriate concentration of acid according to the given procedure) and determine chloride ions according to the procedure using V2 mL of KSCN solution. Taking into account the relative volumes of the flask and pipette, one gets the following relationship: n Cl 8 (n AgNO3 n KSCN ) 8 (25 c AgNO3 V2 c KSCN ) c. Equations of the reactions taking place during the chemical analysis 3 PAC reaction with nitric acid: Al(OH)x Cl y xH Al yCl xH 2 O Aluminium ions reaction with EDTA: Al3 H 2 Y 2 AlY 2H Titration of EDTA excess with magnesium chloride solution: H 2 Y 2 Mg2 MgY 2 2H H OH H 2 O Mg2 HIn MgIn H H NH 4 OH NH 4 H 2 O Reactions taking place during chloride ions determination: Cl Ag AgCl SCN Ag AgSCN m1 = 0.2718 g S = 0.800 m2 = 0.5895 g EDTA concentration 0.04990 mol/L SCN Fe3 FeSCN 2 Titration Conc. of MgCl2 V0 Determination of Al V1 10.00; 9.90 average 9.95 Conc. of AgNO3 and KSCN V3 18.70; 18.80 average 18.75 Determination of Cl V2 16.05; 16.15 average 16.10 5 Titrant volumes, mL 20.00; 19.90 average 19.95 d. Derivation of the formulae for solutions concentrations Titration of 10.00 mL of EDTA solution, ammonium buffer solution, eriochrome black T, using V0 mL of n MgCl2 10 c EDTA MgCl2 solution: c MgCl2 V0 V0 Titration of 10.00 mL of MgCl2 solution + 25.00 mL of AgNO3 solution + chloroform + iron(III) solution, using V3 mL of KSCN solution: n AgNO3 n Cl n KSCN 25 c AgNO3 2 n MgCl2 n KSCN 2 10 c MgCl2 V3 c KSCN From the information in the problem: 25 c 25 c KSCN c AgNO3 KSCN S S 2 10 c MgCl2 2 10 S c MgCl2 c KSCN 25 25 - S V3 - V3 S 25 c AgNO3 c KSCN 2 10 c MgCl2 V3 c KSCN S e. Concentrations of MgCl2, KSCN and AgNO3 solutions c MgCl2 0.02501 mol/L cKSCN = 0.04002 mol/L c AgNO3 0.05003 mol/L f. Determination of aluminium percentage in PAC PAC mass in flask A equals m1 = 0.2718 g mAl = (500.04990 - 49.950.02501)26.98 = 40.47 mg Al percentage = 14.89% g. Determination of chloride ions percentage in PAC PAC mass in flask B equals m2 = 0.5895 g mCl = 8(250.05001 – 16.100.04003)35.45 = 172.0 mg Cl percentage = 29.18% h. Stoichiometric formula of PAC PAC formula – Al(OH)xCly x+y=3 molar ratio f Al: 14.89/26.98 = 0.5517 molar ratio of Cl: 29.18/35.45 = 0.8232 y = 0.8232/0.5517 =1.492 x = 3 – 1.49 = 1.508 Al (OH)1.51Cl1.49 i. Explanation of the colour change during heating of PAC with EDTA In the reaction of disodium salt of ethylenediaminetetraacetic acid with aluminium ions oxonium cations are formed, acidifying the solution: H 2 Y 2 Al3 AlY 2H H+ ions have to be neutralised by adding ammonia solution, so that the reaction equilibrium is shifted towards AlY complex formation. j. Justification of chloroform usage Chloroform is added to separate AgCl precipitate from the solution. Owing to the fact that AgCl is more soluble than AgSCN, the unwanted reaction could take place: AgCl SCN AgSCN Cl . When determining bromide ions, it is not necessary to introduce chloroform as AgBr is less soluble than AgSCN. 6 SOLUTION OF PROBLEM 2 An exemplary arrangement of solutions: Substance Substance A Potassium palmitate, PK F Iron(III) chloride B Sodium dodecyl sulphate, SDS G Mercury(II) nitrate C Triton X-100, TX H Eriochrome cyanine R, ECR D Cetyltrimethylammonium chloride, CTA I Safranin T, SFT E Dithizone, HDz J Rose Bengal, RB a. Probable arrangement of substance in test tubes A-J. Colourless solutions, having nearly neutral pH may contain metal ions or surfactant solutions. They are in test tubes A, B, C, D, F and G. Soap solution i.e. potassium palmitate solution may be slightly opalescent and alkaline– test tube A. Dithizone, eriochrome cyanine R, Rose Bengal and safranin T are orange-red (test tubes E, H, I and J) and the HDz solution is alkaline – test tube E. Surfactant solutions froth upon shaking which is visible for test tubes B, C and D. b. Identification plan Reaction with sulphuric acid: with frothing solutions–PK identification (palmitic acid insoluble in water – the only one from surfactants); with coloured solution dithizonate anion reacts with acid to form dark precipitate of H2Dz, Rose Bengal changes its colour to light yellow. Mercury(II) ions may be identified with dithizone (one of the colourless, not frothing solutions) – formation of the red-orange precipitate soluble in any surfactant. The other colourless, not frothing solution, containing Fe(III) ions (may be identified with potassium thiocyanate solution from problem 1) allows one to identify ECR – formation of violet solution. This allows one to identify CTA – colour change to violet. CTA gives precipitate with Rose Bengal as a result of ion-pair adduct formation. The remaining coloured solution (SFT) may be used for SDS identification basing on precipitate formation with a small amount of surfactant. Remaining TX does not form precipitates with any of the dyes. 7 c. and d. identification of substances in test tubes A-J and justification. Identification A B C D E PK Justification Opalescent and slightly alkaline solution, Froths when shook with distilled water + K → precipitate, the only one of surfactants + MgCl2 or running water → white precipitate SDS Colourless and neutral solution, Upon shaking with distilled or running water froths abundantly + K → no changes, SFT (tt I) + SDS → precipitate, + SDS → precipitate dissolution RB (tt J) + SDS → no changes; Fe-ECR + SDS → no changes TX Colourless and neutral solution, Upon shaking with distilled or running water froths abundantly + K → no changes, Dissolves dithizone, mercury and silver dithizonates precipitates SFT (tt I) or RB (tt J) + TX → no changes CTA Colourless and neutral solution, Upon shaking with distilled or running water froths abundantly + K → no changes, Fe-ECR (violet) + CTA → blue,, the only one of surfactants RB (tt J) + CTA→ precipitate, + CTA → precipitate dissolution HDz Orange and slightly alkaline solution, + K → brown precipitate, soluble in TX, (SDS, CTA) Hg(II) + HDz +K → orange↓, soluble in TX (SDS, CTA) Ag(I) + HDz +K → orange↓, soluble in TX (SDS, CTA) Colourless and slightly acidic solution, does not froth + KSCN (problem 1) → orange solution F FeCl3 + ECR (tt H) → violet solution + CTA → blue solution + HDz +K → brown precipitate, soluble in TX, (SDS, CTA) Colourless and slightly acidic solution, does not froth + KSCN (problem 1) → no changes G Hg(NO3)2 + ECR (tt H) → no changes + HDz +K → orange precipitate, soluble in TX, (SDS, CTA) 8 Identification H I J Justification ECR Orange and neutral solution + K → no changes + Fe(III) → vilet solution, + CTA → blue solution + Al(III) (diluted flask B) → violet solution, + CTA → blue solution SFT Red and neutral solution + K → no changes + SDS (1-2 drops)→ precipitate, + SDS → precipitate dissolution + CTA → no changes + TX → no changes RB Red and neutral solution + K → turns colourless + SDS → no changes + CTA (1-2 drops)→ precipitate, + CTA→ precipitate dissolution + TX → no changes e. Complexometric determination of Mg(II) in the presence of CTA The presence of cationic surfactant renders magnesium determination via EDTA titration with eriochrome black T as an indicator impossible. In the ammonium buffer solution the violet magnesium complex with eriochrome black T decomposes upon addition of CTA and the solution colour changes to blue, just as at the end of magnesium titration with EDTA solution. Used abbreviations: + K – addition of sulphuric acid, + CTA – addition of cetyltrimethyl ammonium solution, + SDS – addition of sodium dodecyl sulphate solution , + Fe(III) – addition of iron(III) chloride solution, + Hg(II) – addition of mercury(II) nitrate solution, + TX – addition of Triton X-100 solution, tt – test tube 9 Comments to the solution of task 2 b. Addition of sulphuric acid allows one to identify potassium palmitate (palmitic acid is sparsely soluble in water), this is the only surfactant solution that becomes cloudy. Dithizonate anion, one of the orange solutions, forms dark precipitate of H2Dz upon reaction with acid. The second of the orange dyes, Rose Bengal, turns practically colourless upon reaction with acid. The remaining dyes do not change their colour upon acidification. One may identify mercury(II) ions (one of the not frothing colourless solutions) using dithizone which form complexes with them in acidic solutions, as opposed to iron(III) ions. Dithizone and mercury dithizonate precipitates are dissolved upon addition of surfactants. The remaining clear, not frothing solution contains iron(III) ions which can be identified in a reaction with potassium thiocyanate solution from problem 1 (reddish colour). Using iron solution and ECR one may identify CTA (violet-red solution of Fe(III) with ECR turns violet-blue upon addition of CTA). One may find the acidic dye using CTA, thanks to the forming precipitate – Rose Bengal. The other red-orange, alkaline dye, safranin T, can be identified as it forms precipitate with SDS and remains in the solution upon addition of non-ionic surfactant – Triton X-100. c., d. The presence of potassium palmitate in test tube A is confirmed by the precipitation of magnesium palmitate (with MgCl2 solution from problem 1) or the precipitation of magnesium and calcium palmitate precipitates from tap water. Other surfactants do not give such precipitates. The presence of dithizone in test tube E can be confirmed by the reaction with diluted AgNO 3 solution – formation of yellow chelate precipitate, soluble in surfactants (or chloroform). Silver nitrate solution allows one to identify CTA and FeCl3 by the precipitation of faint AgCl precipitate. Triton X-100 is the only of the clear surfactant solutions that does not form precipitate with any of the dyes. SDS does not form precipitate with Rose Bengal, neither does it affect iron(III) or aluminium complexes (from flask B after substantial dilution with water). Mercury(II) dithizonate is not decomposed upon addition of sulphuric acid and chloride anions (MgCl2) as opposed to silver dithizonate. 10