THEORETICAL EXAMINATION ANSWER SHEETS

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CEMISTRY: ART, SCIENCE, FUN TEORETICAL EXAMINATION ANSWER SEETS JULY 20, 2007 MOSCOW, RUSSIA

1 Quest. 1.1 1.2 2.1 3.1 3.2 3.3 3.4 Tot Points Student code: Marks 3 3 2 4.5 2 4 6 24.5 7 1.1.1 Structures: Propanedial 1 st isomer 2 nd isomer 1.1.2 Circle the acidic hydrogen atom O O C C C The acidity of propanedial is caused by a) the stability of a carbanion due to conjugation with two carbonyl groups b) weakness of C bond in a carbonyl group c) hydrogen bonds between two propanedial molecules The correct answer 1.2.1 The structures corresponding to minima on energy curve:

1 Quest. 1.1 1.2 2.1 3.1 3.2 3.3 3.4 Tot Points Student code: Marks 3 3 2 4.5 2 4 6 24.5 7 1.3.1 The probability density (a) t = 0 Ψ 2 ( x,0) = Ψ 2-0,06-0,04-0,02 0,00 0,02 0,04 0,06 Distance, nm (b) t = π/(2ω) x, π 2 ω 2 Ψ = Ψ 2-0,06-0,04-0,02 0,00 0,02 0,04 0,06 Distance, nm

1 Quest. 1.1 1.2 2.1 3.1 3.2 3.3 3.4 Tot Points Student code: Marks 3 3 2 4.5 2 4 6 24.5 7 (c) t = π/ω x π ω 2 Ψ, = Ψ 2-0,06-0,04-0,02 0,00 0,02 0,04 0,06 Distance, nm 1.3.2 The probability of finding the proton in the left well = 1.3.3 The time of proton transfer t = The proton mean speed v = 1.3.4 The uncertainty of proton position x =

1 Quest. 1.1 1.2 2.1 3.1 3.2 3.3 3.4 Tot Points Student code: Marks 3 3 2 4.5 2 4 6 24.5 7 The minimal uncertainty of proton velocity v = a) Proton is a rather heavy particle, and its tunneling in malonaldehyde can be described in classical terms of position and velocity b) Proton tunneling is a purely quantum effect; it cannot be described in classical terms c) Uncertainty of proton velocity is so large that tunneling cannot be observed experimentally d) Uncertainty of proton velocity is so small that tunneling cannot be observed experimentally The correct answer is

2 Quest. 1.1 1.2 2.1 2.2 3.1 3.2 3.3 3.4 3.5 Tot Points Student code: Marks 1 2 4 2 1 5 2 3 2 22 8 2.1.1 Thermodynamic data for the reaction (1): r G 0 (1) = K = 2.1.2 Equilibrium constant for the reaction (1) with cobalt nanoparticles: (a) K (r = 10 8 m) = (b) K (r = 10 9 m) =

2 Quest. 1.1 1.2 2.1 2.2 3.1 3.2 3.3 3.4 3.5 Tot Points Student code: Marks 1 2 4 2 1 5 2 3 2 22 8 2.2.1 Minimum water content in the mixture: (a) 2 O%(bulk Co) = (b) 2 O%(nanoparticles with r = 1 10 9 m) = 2.2.2 The correct answer is (mark the proper box): (a) (b) (c) 2.3.1 Standard molar Gibbs function of CoO (external layer) 0 G (CoO, r b) = 2.3.2 Standard molar Gibbs function of Co (internal layer): 0 G (Co, ra, r b) = 2.3.3 Standard Gibbs energy for the reaction (1) with the double-layered nanoparticles G (1, r, r ) = 0 r a b

2 Quest. 1.1 1.2 2.1 2.2 3.1 3.2 3.3 3.4 3.5 Tot Points Student code: Marks 1 2 4 2 1 5 2 3 2 22 8 2.3.4. Plot r G 0 (1,r 0 ) vs. r 0 (b) r G 0 (1,r 0 ) (a) r G 0 (1,r 0 ) r 0 r 0 r G(1,r 0 ) (c) r G(1,r 0 ) (d) r 0 r 0 The correct plot is (mark the proper box): (a) (b) (c) (d) 2.3.5 The correct answer is (mark the proper box): (a) (b) (c)

3 Quest. 1.1 1.2 2.1 2.2 3.1 4.1 Tot Points Student code: Marks 2 4.5 4 3 3 3 19.5 7 3.1.1 The overall reaction equation The kinetic equation for X d[x] dt = 3.1.2 The rate equation d[p] dt = Reaction orders: with respect to B (i): with respect to D (ii): overall (iii):

3 Quest. 1.1 1.2 2.1 2.2 3.1 4.1 Tot Points Student code: Marks 2 4.5 4 3 3 3 19.5 7 3.2.1 1) An open system, [X] 0 > k 2 /k 1 [X] [X] 0 t 2) An open system, [X] 0 < k 2 /k 1 [X] [X] 0 t 3.2.2 A closed system, [B] 0 = [D] 0, [X] 0 > k 2 /k 1 [X] [X] 0 t

3 Quest. 1.1 1.2 2.1 2.2 3.1 4.1 Tot Points Student code: Marks 2 4.5 4 3 3 3 19.5 7 3.3.1 X Y P C + X +... 2X 2 6 X + Y 2Y +... C + Y +... 2P 2 6 3.4.1 The highest possible temperature: T =

4 Quest. 1 2 3.1 3.2 3.3 4 5.1 5.2 5.3 Tot Points Student code: Marks 1 1 1.25 1.75 2.25 1 2 1 2.25 13.5 8 4.1. Equation: 4.2.1. Calculation of the T value: T = mg/ml 4.2.2. Calculation of the T value: T = mg/ml 4.2.3. Calculation of the T value: T = mg/ml

4 Quest. 1 2 3.1 3.2 3.3 4 5.1 5.2 5.3 Tot Points Student code: Marks 1 1 1.25 1.75 2.25 1 2 1 2.25 13.5 8 4.3. Equation(s): 4.4.1 Equation(s): 4.4.2. Equation: 4.4.3. The composition of the crystallohydrate is: Formula of the salt Fe 2 (SO 4 ) 3 x 2 O: x =

5 Quest. 1.1 1.2 1.3 2.1 2.2 3.1 3.2 Tot Points Student code: Marks 5 5 10 30 10 10 5 75 7.5 5.1.1 Structure of product D 5.1.2 Which class of organic compounds does D belong to? Check the appropriate box. Note! Only one checkmark is allowed. Several checkmarks will lead to 0 marks for this question. ketones ethers acetals esters alcohols aldehydes glycols 5.1.3 The expected yield of D The yield is equal to 85% ; lower than 85% ; greater than 85% yield = % 5.2.1 The structures of A, B, and C. A B C 5.2.2 Draw in the boxes intermediate compounds formed during the acidic hydrolysis of C, and basic hydrolysis of B. C + / 2 O t C 3 COO + C 2 5 O B O / 2 O O / 2 O C 3 COO + C 2 5 O

5 Quest. 1.1 1.2 1.3 2.1 2.2 3.1 3.2 Tot Points Student code: Marks 5 5 10 30 10 10 5 75 7.5 5.3.1 The structure of senecioic acid 5.3.2 The structure of E.

6 Quest. 1.1 1.2 2.1 2.2 2.3 2.4 3.1 3.2 Tot Points Student code: Marks 3 9 2 2 3 10 5 3 37 7 6.1.1 The net ionic equation accounting for the ability of LGL to set in air 6.1.2 Write down the net ionic equations matching the processes enumerated in the Table. For each process check the Yes box if it leads to changes of p. Otherwise check the No box. а) protonation of ortho-silicate ions leading to the formation of Si-O groups Reaction equation: b) formation of hydrated [SiO 4 ( 2 O) 2 ] 4- anions Reaction equation: Yes No Yes c) polycondensation of ortho-silicate ions leading to the formation of Si-O-Si bonds Reaction equation: Yes No No 6.2 For [Si 3 O 9 ] n- ion found in aqueous solution of silicates: 6.2.1 Determine the charge (n). Your justification 6.2.2 Determine the number of oxygen atoms bridging adjacent tetrahedra. Your justification n = 6.2.3 Depict the ion structure joining together several tetrahedra (1). Number of oxygen atoms =

6 Quest. 1.1 1.2 2.1 2.2 2.3 2.4 3.1 3.2 Tot Points Student code: Marks 3 9 2 2 3 10 5 3 37 7 6.2.4 The fragment of the layered structure joining 16 tetrahedra (1) Your justification Structure 6.3.1 p of 0.1 M aqueous solution of copper sulfate Your justification p = 6.3.2 Equation of a reaction between aqueous solutions of CuSO 4 and sodium metasilicate (LGL)

7 Quest. 1.1 1.2 2.1 2.2 2.3 2.4 2.5 Tot Points Student code: Marks 12 12 5 12 7 8.5 16 72.5 7.5 7.1.1 A number of reaction types is listed in the table below. All reactions involved in metabolism of MG-CoA to IPP are in the list. Choose those types of reactions which are catalyzed by Е1 and Е3 (put numbers in appropriate places). No Reaction type 1. Dehydration 2. Decarboxylation 3. Dephosphorylation 4. 4 electron reduction 5. Release of the reduced form of coenzyme A (CoA-S) 6. Monophosphorylation 7. Oxidation of hydroxyl group as the third stage of MG-CoA β-oxidation cycle Е1 Е3 7.1.2 Draw the structure of X with stereochemical details and indicate absolute configuration (R or S) of the stereocenter. 7.2.1 Write down the overall reaction equation for reductive ozonolysis of DAP with dimethyl sulfide used as the reducing agent. 7.2.2 Determine molecular formula of Y. Your justification Number of carbon atoms Number of hydrogen atoms Molecular formula:

7 Quest. 1.1 1.2 2.1 2.2 2.3 2.4 2.5 Tot Points Student code: Marks 12 12 5 12 7 8.5 16 72.5 7.5 7.2.3 Calculate the number of IPP and DAP molecules needed to give Y5. Your justification: Number of IPP molecules Number of DAP molecules 7.2.4 Draw the product of coupling reaction between one IPP molecule and one DAP molecule, subsequent reductive ozonolysis of which gives Y1, Y2 and one more product, the latter containing phosphorus. 7.2.5 Draw the structures of Y and Y4 with stereochemical details. Y4 Y

8 Quest. 1.1 1.2 2.1 2.2 2.3 3.1 3.2 3.3 3.4 Tot Points Student code: Marks 8 9 5 11 14 16.5 12 10 13.5 99 8 8.1.1 Expressions for the rates: v act = v p = v deact = v t = 8.1.2 Compare rates using operators <<,,,, >> v deact v act v deact v t v deact v p 8.2.1 Mass of the obtained polymer. Your justification: m = 8.2.2 Degree of polymerization of the obtained polymer. Your justification: DP =

8 Quest. 1.1 1.2 2.1 2.2 2.3 3.1 3.2 3.3 3.4 Tot Points Student code: Marks 8 9 5 11 14 16.5 12 10 13.5 99 8 8.2.3 Structure of the obtained polymer. 8.3.1 Fill in the right column with symbols (a-g) of 1 NMR signals corresponding to substructures in the left column. * C 2 * O C 2 * * * * * * Cl Cl 8.3.2 Composition and molecular weights of copolymers P1 and P2. Your justification: Your justification: n(c) = n(d) = M(P1) = M(P2) =

8 Quest. 1.1 1.2 2.1 2.2 2.3 3.1 3.2 3.3 3.4 Tot Points Student code: Marks 8 9 5 11 14 16.5 12 10 13.5 99 8 8.3.3. All possible reactions of activation P1: P2: 8.3.4 Structure of P1 and one of possible structures of P2 P1: P2: