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9 Theoretical Problems 3 Practical Problems tthhth 40 4400 9 theoretical problems 3 practical problems THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 THE FORTIETH INTERNATIONAL CHEMISTRY OLYMPIAD 12–21 JULY 2008, BUDAPEST, HUNGARY _________________________________________________________________________ THEORETICAL PROBLEMS PROBLEM 1 The label on a bottle containing a dilute aqueous solution of an acid became damaged. Only its concentration was readable. A pH meter was nearby, and a quick measurement showed that the hydrogen ion concentration is equal to the value on the label. 1.1 Give the formulae of four acids that could have been in the solution if the pH changed one unit after a tenfold dilution. 1.2 Could it be possible that the dilute solution contained sulfuric acid? Sulfuric acid: pKa2 = 1.99 Yes No If yes, calculate the pH (or at least try to estimate it) and show your work. 1.3 Could it be possible that the solution contained acetic acid? Acetic acid: pKa = 4.76 Yes No If yes, calculate the pH (or at least try to estimate it) and show your work. 1.4 Could it be possible that the solution contained EDTA (ethylene diamino tetraacetic acid)? You may use reasonable approximations. EDTA: p Ka1 = 1.70, p Ka2 = 2.60, p Ka3 = 6.30, p Ka4 = 10.60 Yes No If yes, calculate the concentration. _______________ SOLUTION 1.1 Any univalent, strong acid (HCl, HBr, HI, HNO 3, HClO 4) is acceptable. HF is not! THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1097 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 1.2 Yes No No, the first dissociation step can be regarded as complete in aqueous solutions, + thus [H ] > cacid. No text or calculations are needed. 1.3 Yes No Yes, but it can happen only in quite dilute solutions. c = [HA] + [A –] = [H +] [H +] = [A –] + [OH –] This means that [HA] = [OH –] [H+- ][A] [H ++ ]([H ]− [OH]) - [H +3 ] K = = =− [H]+ [HA] [OH]- K w The pH of the solution must be acidic, but close to 7. 6.5 is a good guess. [H]=+ 3 (K K ) A good approximation is: w [H]=+3 (K +[H]) + K The full equation can be solved through iteration: w Starting with a neutral solution two cycles of iteration give identical results: 5.64 ×10 –7 mol dm -3 as the required concentration. Exact pH is 6.25. 1.4 Yes No We can suppose that this solution would be quite acidic, so the 3 rd and 4 th dissociation steps can be disregarded. The following equations are thus true: – 2– + c = [H 4A] + [H 3A ] + [H 2A ] = [H ] + – 2– [H ] = [H 3A ] + 2 [H 2A ] 2– This means that [H 4A] = [H 2A ] [H+2 ] [HA 2- ] K K =2 =[H]+ 2 1 2 [H A] 4 (or pH = (p K1 + p K2 ) / 2 = 2.15) c = 0.0071 mol dm -3 THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1098 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 PROBLEM 2 Determine the structure of the compounds A - H (stereochemistry is not expected), based on the information given in the following reaction scheme: Pd radical ZnCl2 A B C D + 5 H − H O 2 oxidation (C H O) 2 ( C10H18 ) 10 18 1. O3 450°C 2. Zn/H+ Pd/C, 350°C 1. Pd/H ∆ 2 Na2CO3, H F − − G 2. NaBH E H2O, 8 H 4 − H2O Hints: • A is a well-known aromatic hydrocarbon. • A hexane solution of C reacts with sodium (gas evolution can be observed), but C does not react with chromic acid. 13 • C NMR spectroscopy shows that D and E contain only two kinds of CH 2 groups. • When a solution of E is heated with sodium carbonate an unstable intermediate forms at first, which gives F on dehydration. _______________ THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1099 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 SOLUTION A B C D OH H G F E HO O O O THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1100 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 PROBLEM 3 Vinpocetine (Cavinton®, Calan®) is one of the best selling original drugs developed in Hungary. Its preparation relies on a natural precursor, (+)-vincamine (C 21 H26 N 2O3), which is isolated from the vine plant, vinca minor . The transformation of (+)-vincamine to vinpocetine is achieved in two steps depicted below. 1. NaOH H N cat. conc. H 2SO 4 2. C2H5Br N A B (Vinpocetine) CH Cl C2H5OH HO 2 2 H CO C 3 2 C2H5 Vincamine 1 All compounds ( A to F) are enantiomerically pure compounds. • The elementary composition of A is: C 74.97%, H 7.19%, N 8.33%, O 9.55%. • B has 3 other stereoisomers. 3.1 Propose structures for the intermediate A and vinpocetine ( B). A study of the metabolism of any drug forms a substantial part of its documentation. There are four major metabolites each formed from vinpocetine ( B): C and D are formed in hydrolysis or hydration reactions, while E and F are oxidation products. Hints: • The acidity of the metabolites decreases in the order C >> E >> D. F does not contain an acidic hydrogen. • C and E each have 3 other stereoisomers, while D and F each have 7 other stereoisomers. • F is a pentacyclic zwitterion and it has the same elementary analysis as E: C 72.11 %, H 7.15 %, N 7.64 %, O 13.10 %. • The formation of E from B follows an electrophilic pattern. • The formation of D from B is both regio- and stereoselective. 3.2 Propose one possible structure for each of the metabolites C, D, E and F! 3.3 Draw a resonance structure for B that explains the regioselective formation of D and the absence of the alternate regioisomer in particular. _______________ THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1101 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 SOLUTION 3.1 H H N N N N MeO C EtO C 2 Et 2 Et A B 3.2 C D H H N N N N HO HO2C Et EtO2C Et apovincaminic acid ethyl vincaminate Both stereoisomers around the new chiral center are acceptable. E F HO – H H O N N N N + EtO C EtO2C 2 Et Et 10-hydroxyvinpocetine vinpocetine N-oxide THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1102 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 All aromatic positions for the OH are acceptable in E. 3.3 H + N N EtO2C – Et All aromatic positions for the OH are acceptable in E. THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1103 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 PROBLEM 4 A major transformation route for oxiranes (epoxides) is ring opening. This may be accomplished in various ways. On acid catalysis the reactions proceed through cation-like (carbenium ion-like) species. For substituted oxiranes the direction of ring opening (which C–O bond is cleaved) depends on the stability of the intermediate carbenium ion. The more stable the intermediate carbenium ion the more probable its formation. However, an open carbenium ion (with a planar structure) only forms if it is tertiary, benzylic or allylic. On base catalysis the sterically less hindered C–O bond is cleaved predominantly. Keep stereochemistry in mind throughout the whole problem. To depict stereochemistry use only the bond symbols and nothing else where necessary. 4.1 Draw the structure of the reactant and the predominant products when 2,2-dimethyl- oxirane (1,2-epoxy-2-methylpropane) reacts with methanol at low temperatures, catalysed by (i) sulphuric acid (ii) NaOCH 3. 2,2-dimethyloxirane NaOCH H+ 3 CH OH CH OH 3 3 4.2 Draw the structure of the predominant product when the epoxide ring of the following leukotriene derivative is opened with a thiolate (RS –). O COOCH 3 H C 3 CH3 1. RS - + H3C 2. H THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1104 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 Different porous acidic aluminosilicates can also be used to catalyse the transformation of alkyl oxiranes. In addition to ring opening, cyclic dimerisation is found to be the main reaction pathway producing mainly 1,4-dioxane derivatives (six-membered saturated rings with two oxygen atoms in positions 1,4). 4.3 Draw the structure(s) of the most probable 1,4-dioxane derivative(s) when the starting compound is ( S)-2-methyloxirane (( S)-1,2-epoxypropane). Give the structure of the reactant as well. 4.4 Draw the structure(s) of the substituted 1,4-dioxane(s) when the epoxide reacting is (R)-1,2-epoxy-2-methylbutane (( R)-2-ethyl-2-methyloxirane). Give the structure of the reactant as well. 4.5 Give the structure(s) of the substituted 1,4-dioxane(s) when this reaction is carried out with racemic 1,2-epoxy-2-methylbutane (2-ethyl-2-methyloxirane). _______________ SOLUTION 4.1 2,2-dimethyloxirane CH CH CH3 3 NaOCH 3 H+ 3 CH CH CH OCH 3 CH OH 3 CH OH OH 3 HO 3 3 O 3 H3CO 4.2 HO COOCH 3 CH SRH3 C 3 H3C THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1105 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 4.3 H C O O 3 H C 3 O or H3C O O CH3 H H3C ( S)-2-methyloxirane product 4.4 (R )-1,2-epoxy-2-methylbutane: O H3C C2H5 O O O H3C H5C2 H3C H5C2 H3C H5C2 CH3 C H C H O O 2 5 O 2 5 C2H5 CH 3 CH 3 H5C2 H3C H3C O CH3 O C H or or 2 5 or O C2H5 H5C2 O H3C O H5C2 O CH 3 C2H5 CH3 R,R S,S R,S 4.5 O O O H3C H5C2 H3C H C 5 2 H3C H5C2 CH3 C H C H O O 2 5 O 2 5 C2H5 CH 3 CH 3 H5C2 H C or or H3C or 3 O CH3 O C2H5 O C2H5 H5C2 O H C O H C O CH 3 5 2 3 C2H5 CH3 R,R S,S R,S THE COMPETITION PROBLEMS FROM THE INTERNATIONAL CHEMISTRY OLYMPIADS, Volume 2 1106 Edited by Anton Sirota, ICHO International Information Centre, Bratislava, Slovakia THE 40TH INTERNATIONAL CHEMISTRY OLYMPIAD, 2008 PROBLEM 5 A and B are white crystalline substances.
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