Write all the reaction equations. Problem 7. Simple equilibrium The gaseous substances A2 and B2 were mixed in a molar ratio 2:1 in a closed vessel at a temperature T1. When the equilibrium A2(g) + B2(g) = 2AB(g) was established the number of heteronuclear molecules in a gas phase became equal to the total number of homonuclear molecules. K1 1:1 T1? The equilibrium mixture obtained from the initial mixture A2 : B2 = 2 : 1 was heated so that equilibrium constant became K2 = K1 / 2. B2 A2 AB T1? Consider the reaction yield = neq(AB) / nmax(AB) as a function of the initial molar ratio A2 : B2 = x : 1 at any fixed temperature (nmax is the maximum amount calculated from the reaction equation). Answer the following questions qualitatively, without exact equilibrium calculations. ; 0? Now, consider the variable ratio A2 : B2 = x : 1 at a fixed total pressure. AB A
in 1 cm2 of the surface. Calculate TOF for the catalyst. This catalyst worked during 40 minutes and then became inactivated. Estimate TON. Figure 1b. The amount of the product NB as a function of time a) TOF is often used to describe the operation of deposited catalysts. To make a deposited catalyst one has to deposit atoms of metal on the inert surface. These atoms form catalytic sites. The dependence of the rate of the catalytic reaction upon the amount of metal atoms deposited on 1 cm2 of the surface (less than one monolayer) is shown in Fig. 2а. Calculate TOF. Figure 2a. The dependence of Nb on NCat b) Russian scientist professor Nikolay I. Kobozev has shown that the dependence of NВ on NCat can be much more complicated. The corresponding curve in Fig. 2b has maximum! According to the Kobozev’s theory (a simplified version) a structure consisted of n deposited atoms rather than a single atom form a catalytic site. Maximum rate of catalytic reaction was observed when From the data shown in Fig. 2b calculate n, the number of atoms forming a catalytic site. TOF for the point of maximum rate in Fig. 2b is given in SI units. Figure 2b. The dependence of Nb on NCat 4. Atoms of Au deposited on the Mo-TiOX support exhibit exceptional catalytic activity for the CO oxidation (M. S. Chen and D. W. Goodman, Science, v.306, p.254, 2004). The maximum rate of reaction r1 {mol/cm2/s} was observed for the bilayer atomic structure presented in Fig. 3a. Red and yellow spheres are atoms of Au. For the monolayer structure (Fig. 3b), the reaction was four times slower, r2 = ¼ r1. Calculate the ratio of TOF for the atoms of Au in the upper layer in Fig. 3а (all red spherical particles), to TOF for the monolayer in Fig. 3b (all yellow spherical particles). In the former case, every single Au atom is a catalytic site. The rate of the catalytic reaction on each yellow site in Fig.3a and Fig.3b is the same if the site is accessible to reactants and is equal to zero if the access is blocked.
a) b) Figure 3. Structure of the gold catalyst deposited on the Mo-TiO2 support. a) Bilayer structure; b) monolayer structure Problem 10. Kinetic puzzles Propose the mechanisms for the reactions given below. Prove that your mechanisms are consistent with the experimentally observed rate laws. Use proper approximations if necessary. 1. Oxidation of bromide ion by permanganate in acidic solution 2MnO4– + 10Br– + 16H+ = 2Mn2+ + 5Br2 + 8H2O a) at low concentrations of Br– and H+ r = kс(MnO4–)с2(Br–)с3(H+) b) at high concentrations of Br– and H+ r = kс(MnO4–)с(Br–)с(H+) where с are the total concentrations of reactants. In both cases с(MnO4–) <<с(Br–), с(H+). 2. Oxidation of benzamide by peroxydisulfate in the presence of Ag+ ions in water-acetic acid solution 2C6H5CONH2 + 2H2O + 3S2O82– = 2C6H5COOH + 6SO42– + N2 + 6H+ r = k[Ag+][S2O82–] 3. Oxidation of formate ion by peroxydisulfate in water solution HCOO– + S2O82– = CO2 + 2SO42– + H+ r = k[HCOO–]1/2[S2O82–] 4. Oxidation of azide ion by iodine in carbon disulfide solution I2 + 2N3– = 3N2 + 2I– r = k[N3–] 5. Condensation of aldehydes with acryl esters in the presence of the base – 1,4-diazabicyclo[2.2.2]octane (DABCO) in tetrahydrofurane solution
Problem 21. Cyclobutanes In 1894 Emil Fischer proposed the “lock and key” principle for interaction between a drug and its molecular target. The interaction is efficient only in case of substances having specific complementary geometry that fit exactly to the molecular target. According to this model, a potential drug should accept a definite conformation with the appropriately located functional groups. One of ways to achieve this goal is restriction of conformational mobility of molecules. Recently Ukrainian chemists reported synthesis of conformationally rigid diamines I and J according to the scheme below. The starting compound A was synthesized for the first time in 1958 by J.D. Roberts and F.F. Caserio (authors of the classical textbook on organic chemistry), according to the scheme: Another method for A synthesis is given below: Decipher the schemes. Write down the structural formulae of compounds A-P accounting for the following: C and D are isomers; J has two planes of symmetry; hydrocarbon K has a single type of hydrogen atoms; H = 10.0%; N and O are isomers; H = 3.8%; C = 22.9%. Starting from P, a very interesting compound W was synthesized: 2. Write down the structural formulae of Q-W. 3. Can W be resolved into enantiomers? Problem 22. Introduction to translation
6. Write down the nucleotide sequence of mRNA encoding the peptide Use A, U, G, and C for unambiguously decided positions, N1/N2 if any of two nucleotides is possible at a particular position, and N if any of four nucleotides is possible at a particular position (N1 and N2 can be any of A, U, G, and C). 7. Molecular weight of an E.coli protein is of about 51 kDa. Estimate the length of encoding mRNA (in nm, rounding to integer). Take the average molecular weight of an amino acid as 110 g/mol, and the average length of a ribonucleotide residue as 0,34 nm. How long will it take a cell to synthesize this protein if the ribosome reads 20 ribonucleotide residues per second? A group of researches accomplished protein synthesis in a cell-free system (in vitro). All required components (ribosomes, tRNAs, ATP, GTP, salts, amino acids, aaRS, translation factors, etc.) were added to the system. A synthetic polyribonucleotide consisting of only A and C in the ration of 1:5 was used as the messenger RNA (nucleotide residues are arranged randomly in the mRNA). 8. Determine the amino acid composition of the synthesized protein. What are the ratios between the amino acid residues in the protein? The 3D structure of a tRNA is depicted in Fig. 2. There are two key regions: the CCA3’ terminus which is linked to the amino acid, and the anticodon exactly matching to the mRNA codon. Fig. 2. The 3D structure of a tRNA 9. A mutant tRNATyr with anticodon specific to Ser codon (instead of Tyr codon) was introduced into the synthetic system described in i.8. What would be the resultant protein?
organisms are schematically given hereunder (left to right: Poxviridae host cell infected with fowlpox, Poxviridae host cell infected with canarypox virus, and human cell). Each square box in the pictures stands for a nucleotide residue with one of the canonical nitrogen bases: adenine (A), guanine (G), uracil (U) or cytosine (C). Hydrogen bonds are formed according to the complementary principle (Chargaff’s rule) between the bases with boxes opposite to each other. The only exceptions are: A-C G-U The mRNA triplet (codon) identical for all three sequences is circled. Fragments of mRNA sequences belonging to different organisms are given in the hereunder Table in an arbitrary order. These sequences contain Elements X depicted in the above images. № Nucleotide sequence (5´→3´) 1 …GCUGCUAAUGAAGAAAUGACUAUAAAUAGAUGGGUCAUGCCUGACACGCAAAG… 2 …AGGCACUCAUGACGGCCUGCCUGCAAACCUGCUGGUGGGGCAGACCCGAAAAUCCCAC… 3 …GACGAGAUAAUGAAGAAAUGGUCCUAAACAGAUGGGUCGUUCCUGACACCCCGG… X. SYMBOLS
Imines (nitrogen analogues of carbonyl compounds) are formed when any primary amine reacts with aldehyde or ketone under appropriate conditions. Mechanically, the amine first attacks the aldehyde with formation of an intermediate. Its subsequent dehydration gives the imine. Imine formation is like a biological reaction: it is fastest near neutrality. Many biological processes involve imine formation. Three outstanding examples are: synthesis of amino acids from oxoacids, transamination of α-amino acids and mechanism of vision. The former two processes include formation of an imino intermediate between an amino acid and vitamin B6 derivative (pyridoxal). The transformation of light energy into electric signal in our eyes includes the cis-trans-photoisomerization of a polyene retinal (an aldehyde), which is covalently linked to the protein (an amine) by the imine bond. Imines are also very important in organic synthesis as intermediates in the so-called “reductive amination” reaction allowing direct transformation of carbonyl compounds into amines. In this task you will prepare aniline derivative of benzaldehyde (I). Chemicals and Reagents Aniline Benzaldehyde 96% aqueous ethanol Table of Chemicals Compound State R-Ratings S-Provisions C6H7N, Aniline Liquid 23/24/25 40 41 43 48/23/24/25 50 68 26 27 36/37/39 45 46 61 C7H6O,Benzaldehyde Liquid 22 2 24 C2H6O, Ethanol, 96% aqueous solution Liquid 11 2 7 16 Equipment and Glassware Magnetic stirrer with heating Magnetic bar Glass beaker, 25 mL Round-bottom two necked flask, 50 mL Reflux condenser Laboratory stand with metal rings and clamps Adding funnel Separating funnel Filter flask Porous Shott’s glass filter Water- or vacuum pump
In this task you will prepare acetone derivatives of carbohydrate D-mannose (I) and α-amino acid L-cysteine (II). Chemicals and Reagents D-Mannose Iodine, crystalline Anhydrous acetone Na2S2O3 solution, dilute Chloroform Na2SO4, calcined L-Cisteine hydrochloride Ninhydrine reagent (0.3 % sol-n of ninhydrine in n-butanol cont. 3% of sodium acetate) Table of Chemicals Compound State R-Ratings S-Provisions C6H12O6, D-Mannose Solid - 28 I2 Solid 20/21 50 23 25 61 CH3C(O)CH3 Liquid 11 36 66 67 2 9 16 26 Na2S2O3 Aqueous solution - 24/25 CHCl3 Liquid 22 38 40 48/20/22 2 36/37 Na2SO4 Solid - - C3H8NO2SCl, L-Cisteine hydrochloride Solid 22 36/37/38 25 26 36/37/39 C9H6O4, Ninhydrine Solution 22 36/37/38 26 28A 37/39 C4H9OH, n-butanol Liquid 10 22 37/38 41 67 2 7/9 13 26 37/39 46 CH3COONa Solution - - Equipment and Glassware Magnetic stirrer with heating Magnetic bar Glass beaker, 50 or 100 mL (2 ea.) Round-bottom flask, 50 mL Reflux condenser Laboratory stand with metal rings and clamps Thermometer Adding funnel Separating funnel Filter flask Porous Shott’s glass filter (2 ea.) Rotary evaporator Water- or vacuum pump Analytical balance (± ) Pipette pump Capillary for melting point determination (2-3 ea.)
Increasing of the solvent-polymer affinity results in more expanded polymer coils, which, in turn, provides for higher resistance to the solution flow. Thus, the index of power (a) is growing with increasing of the solvent affinity towards the polymer. Usually a polymer sample is polymolecular (polydisperse), i.e. it contains macromolecules of different molecular weights. Accordingly, polymer samples are characterized by average molar masses (depends on the way of averaging). Thus, a viscosity-average molar mass Mv can be found from the Mark-Kuhn-Houwink equation using experimentally determined [] and reference data for K and a. Polydispersity (or heterogeneity) index of a polymer sample can be determined as the ratio of its viscosity-average molar masses found in solvents significantly differing in their affinity towards the polymer. In this task you will find the polydispersity index of a polystyrene sample by capillary viscometry using toluene (K=0.017 ml/g, a=0.69) and methyl ethyl ketone (K=0.039 ml/g, a=0.57). All constants are given for 25 °С. Chemicals and reagents: Polystyrene (number-average molar mass of about 100 000) solution in toluene, 10g/L, 25 mL Polystyrene (number-average molar mass of about 100 000) solution in methyl ethyl ketone, 10g/L, 25 mL Toluene, 50 mL Methyl ethyl ketone, 50 mL Table of Chemicals Compound State R-Rating S-Provision (C8H8)n, Polystyrene Solutions in toluene and methyl ethyl ketone - - C7H8, Toluene Liquid 11 38 48/20 63 65 67 2 29 36/37 46 62 C4H8O, Methyl ethyl ketone Liquid 11 36 66 67 2 9 16 Apparatus and glassware: Ubbelohde or other capillary viscometer Graduated cylinder, 10 mL 10 glass vials, 20 mL