India 2018 — INChO2018-Question.pdf

Fonte: gare di altri paesi/India/chimica/INChO2018-Question.pdf · Apri PDF · apri PDF p.1 Cluster: Termodinamica Soluzioni (stessa cartella): · · · · · · · · · · · · · ·

Problema 1

Question No 1 2 3 4 5 Total Marks 20 25 21 19 22 107 Marks Obtained

Signature of Examiner 1 IA 11A 18 VIIIA 8A 2 IIA 2A 13 IIIA 3A 14 IVA 4A 15 VA 5A 16 VIA 6A 17 VIIA 7A 3 IIIB 3B 4 IVB 4B 5 VB 5B 6 VIB 6B 7 VIIB 7B 8 9 VIII 8 10 11 IB 1B 12 IIB 2B Periodic Table of the Elements Lanthanide Series Actinide Series © 2013 Todd Helmenstine chemistry.about.com sciencenotes.org 1 H Hydrogen 1.008 3 Li Lithium 6.941 4 Be Beryllium 9.012 11 Na Sodium 22.990 12 Mg Magnesium 24.305 19 K Potassium 39.098 20 Ca Calcium 40.078 21 Sc Scandium 44.956 22 Ti Titanium 47.88 23 V Vanadium 50.942 24 Cr Chromium 51.996 25 Mn Manganese 54.938 26 Fe Iron 55.933 27 Co Cobalt 58.933 28 Ni Nickel 58.693 29 Cu Copper 63.546 30 Zn Zinc 65.39 31 Ga Gallium 69.732 13 Al Aluminum 26.982 5 B Boron 10.811 32 Ge Germanium 72.61 14 Si Silicon 28.086 6 C Carbon 12.011 33 As Arsenic 74.922 15 P Phosphorus 30.974 7 N Nitrogen 14.007 34 Se Selenium 78.09 16 S Sulfur 32.066 8 O Oxygen 15.999 35 Br Bromine 79.904 17 Cl Chlorine 35.453 9 F Fluorine 18.998 36 Kr Krypton 84.80 18 Ar Argon 39.948 10 Ne Neon 20.180 2 He Helium 4.003 37 Rb Rubidium 84.468 38 Sr Strontium 87.62 39 Y Yttrium 88.906 40 Zr Zirconium 91.224 41 Nb Niobium 92.906 42 Mo Molybdenum 95.94 43 Tc Technetium 98.907 44 Ru Ruthenium 101.07 45 Rh Rhodium 102.906 46 Pd Palladium 106.42 47 Ag Silver 107.868 48 Cd Cadmium 112.411 49 In Indium 114.818 50 Sn Tin 118.71 51 Sb Antimony 121.760 52 Te Tellurium 127.6 53 I Iodine 126.904 54 Xe Xenon 131.29 55 Cs Cesium 132.905 56 Ba Barium 137.327 57-71 72 Hf Hafnium 178.49 73 Ta Tantalum 180.948 74 W Tungsten 183.85 75 Re Rhenium 186.207 76 Os Osmium 190.23 77 Ir Iridium 192.22 78 Pt Platinum 195.08 79 Au Gold 196.967 80 Hg Mercury 200.59 81 Tl Thallium 204.383 82 Pb Lead 207.2 83 Bi Bismuth 208.980 84 Po Polonium [208.982] 85 At Astatine 209.987 86 Rn Radon 222.018 87 Fr Francium 223.020 88 Ra Radium 226.025 89-103 57 La Lanthanum 138.906 58 Ce Cerium 140.115 59 Pr Praseodymium 140.908 60 Nd Neodymium 144.24 61 Pm Promethium 144.913 62 Sm Samarium 150.36 63 Eu Europium 151.966 64 Gd Gadolinium 157.25 65 Tb Terbium 158.925 66 Dy Dysprosium 162.50 67 Ho Holmium 164.930 68 Er Erbium 167.26 69 Tm Thulium 168.934 70 Yb Ytterbium 173.04 71 Lu Lutetium 174.967 89 Ac Actinium 227.028 90 Th Thorium 232.038 91 Pa Protactinium 231.036 92 U Uranium 238.029 93 Np Neptunium 237.048 94 Pu Plutonium 244.064 95 Am Americium 243.061 96 Cm Curium 247.070 97 Bk Berkelium 247.070 98 Cf Californium 251.080 99 Es Einsteinium [254] 100 Fm Fermium 257.095 101 Md Mendelevium 258.1 102 No Nobelium 259.101 103 Lr Lawrencium [262] 104 Rf Rutherfordium [261] 105 Db Dubnium [262] 106 Sg Seaborgium [266] 107 Bh Bohrium [264] 108 Hs Hassium [269] 109 Mt Meitnerium [268] 110 Ds Darmstadtium [269] 111 Rg Roentgenium [272] 112 Cn Copernicium [277] 113 Uut Ununtrium unknown 114 Fl Flerovium [289] 115 Uup Ununpentium unknown 116 Lv Livermorium [298] 117 Uus Ununseptium unknown 118 Uuo Ununoctium unknown Indian National Chemistry Olympiad 2018

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1 Reaction at Cathode:

Reaction at Anode:

Overall discharge reaction

Problem 1

20 Marks Lead Acid Batteries Part A: Electrochemical processes in a lead acid cell Conventional lead acid batteries used in cars and invertors are the most common rechargeable batteries in the market. A rechargeable lead acid cell consists of a positive electrode of Pb(s)/PbO2(s), and a negative electrode of Pb(s), both immersed in aqueous H2SO4 electrolyte.

Figure 1: Schematic of a conventional lead acid cell Standard electrode potentials for some half-cell reactions at 298 K are given below.

1.1 Write the half-cell reactions that take place at cathode and at anode in above cell during discharge at 298 K. Write the overall discharge reaction and calculate the $E^{\circ }$ of this cell. Molar Enthalpies of formation at 298 K of some species are given below.

Half-Cell Reaction Eo (V) PbO2(s) + SO4 2 aq) + 4H+ (aq) + 2e PbSO4(s) + 2 H2O(l) +1.69 PbO2(s) + 4 H+ (aq) + 2e Pb2+ (aq) + 2 H2O(l) +1.46 Pb2+ (aq) + 2e Pb(s) 0.13 PbSO4(s) + 2e Pb(s) + SO4 2 (aq) 0.36 Species (kJ mol 1) PbO2(s) 277.4 H2O(l) 285.8 Pb2+ (aq) 1.6 SO4 2 (aq) 909.3 PbSO4(s) 920.0 ) H ( f Indian National Chemistry Olympiad 2018

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1.2 Calculate the standard enthalpy change and standard Gibbs energy change

for the overall discharge reaction of the cell in 1.1.

In a galvanic cell, the chemical energy released during the discharge reaction drives the electrical work. The Gibbs energy change of the discharge reaction represents the maximum electrical work wmax(el) that can be extracted from the cell during discharge. $w_{\text{max(el)}}=\Delta G_{\text{rxn}}^{\circ }$ Maximum work is obtained when the cell is operated reversibly, i.e., at a very low current. Assume that the cell is designed to exchange heat with the surrounding to prevent overheating or overcooling. 1.3 For reversible operation of the cell at 298 K and all species in the cell in their standard states, a) determine the amount of heat absorbed by the cell from the surrounding per mole of reaction as per the cell equation in 1.1. b) what fraction of wmax(el) is obtained from the heat exchanged reversibly with the surroundings? The concentration c of a species in a solution is related to its activity a as $a=\gamma \times c$, where $\gamma$ is the activity coefficient. For highly concentrated solutions, the reaction quotient Q must be expressed in terms of activities of the species (i.e., $\gamma \ne 1$). In commercial lead-acid batteries, highly concentrated H2SO4 (aq.) is used. Conventionally, the liquid and solid components in a chemical reaction are assumed to be in their ‘standard states’

) (

rxn H

) (

rxn G Indian National Chemistry Olympiad 2018

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with activity as 1. However, due to the high concentration of sulfuric acid in the electrolyte, the activity of water also cannot be taken as 1. 1.4 For the discharge reaction of cell in 1.1, write the expression for the cell EMF in terms of activities of the species involved in the reaction. Consider H2SO4 (aq.) to be completely dissociated and assume that only solids are in their standard states. The lead-acid cell is considered to be discharged when the current that can be drawn from it decreases below a certain value. Discharge is also accompanied by a decrease in H2SO4 concentration. Activities of water and products of activities of relevant sulphuric acid species at 298 K in a lead acid cell are listed below at two concentrations of sulphuric acid.

1.5 Assuming the discharged and charged states correspond to 1.00 molal and 5.00 molal H2SO4, respectively, find the drop in the cell EMF values when it is discharged at 298 K.

Following are certain facts about the cell in Fig. 1. (i) PbSO4(s) predominantly deposits on the electrodes and not in other parts of the cell. (ii) PbSO4(s) deposits on both the electrodes. (iii) Resistivity of PbSO4(s) deposits ($3\times 10^9\Omega$ cm) is much higher than that of PbO2(s) ($2\times 10^{-4}\Omega$ cm). (iv) Resistivity of PbO2(s) ($2\times 10^{-4}\Omega$ cm) is close to that of Pb(s) ($0.2\times 10^{-4}\Omega$ cm). No. H2SO4 Molality (m)

1 1.00 0.9618 0.0018 2 5.00 0.7032 0.8847 Indian National Chemistry Olympiad 2018

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4 1.6 Which of the above facts makes the cell rechargeable? (Write Statement number(s) i - iv, as applicable)

Electrolysis of water may become a competing process during charging of the cell in Fig. 1. The relevant half-cell reactions are given below.

1.7 For the operation of the cell in Fig.1 at 298 K, indicate whether the given statement is correct or incorrect. (Mark X in the appropriate box)

a. During charging of this cell,

Topic: Thermodynamics, Circuits Metodi: First Law of Thermodynamics, Thermodynamic Cycle Analysis, Electric Potential Method Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.1

Problema 2

Problem 2

25 marks

When Rain meets the Soil Part I: The fragrance of soil When rain falls on soil after a dry period, a sweet earthy smell is produced known as the fragrance of first rain, which can be detected by camels even at a distance of 50 km. This scent has been captured in an increasingly famous scent “mitti ka itir” produced in Kannauj, India. This smell is due to the compounds geosmin (A) and methyl-isoborneol (B) produced by the actinobacteria present in soils. On combustion, 0.455 g of ‘A’ yields 1.318 g of CO2. 2.1 Calculate the percentage of carbon in ‘A’. Show the calculations involved.

‘A’ undergoes dehydration easily under acidic condition to form an odourless hydrocarbon Argosmin ‘C’. 0.455g of A reacts with sodium to liberate hydrogen gas which occupies 28.01 cm${}^3$ at STP. 2.2 Calculate the empirical formula of A. Show the calculations involved.

C undergoes ozonolysis giving a single compound D which is further converted to E as follows. i. O , ( eq.) ii. H O, Zn i. NaOH - I ii. H O+ Indian National Chemistry Olympiad 2018

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/reflux (ii) KCN (iii) (iii) H O+ (i) H /Pd ( eq.) (i) CH MgI ( eq.) (i) PCl (ii) H O+ E (i) (ii) Mg, I (trace) (ii) H O+ E can be synthesized by another route using cyclohexane-1,2-dione as shown below. 2.3 Draw the structures of the intermediate compounds and hence of E in the boxes provided.

2.4 Draw the structures of D and C.

2.5 A has three chiral centres. Draw all possible structural isomers of A consistent with the above information. (no stereo isomers are required)

D

C Indian National Chemistry Olympiad 2018

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G G forms an anhydride easily on heating The Cactaceae (cactus) family of plants grows in extremely hot and dry conditions and contain the compound dehydrogeosmin „F‟, which has ten times more odour than geosmin. It is believed to play a significant role in pollination in these plants. F is converted to A by addition of one mole of H2. The ozonolysis product of F gives a red precipitate with Fehling‟s‟ solution but does not give yellow precipitate with NaOH – I2. The dehydration product of F, when subjected to ozonolysis gives the following products: F

• X i. Dehydration ii. Ozonolysis

X also does not give yellow precipitate with NaOH – I2. 2.6 Draw the structure of F and hence A.

The compound „B‟ can be synthesized by the following route. 2.7 Draw structures for G and H in the boxes provided.

i. MeMgI ii. H O+ B i. Ca(OH) , excess ii.

F

A

H Indian National Chemistry Olympiad 2018

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Part II: If the rain is $acidic,\dots$ As the water in clouds and rain drops equilibrates with atmosphere, it dissolves gases from the atmosphere. Usually, rain is slightly acidic because of dissolved CO2 leading to a pH sometime as low as 5.6. The acidity of rain water in industrial areas is higher with pH as low as 3.5 to even 2 in some regions. Natural water bodies often have carbonates of Na+, K+, Ca2+, and Mg2+, dissolved from carbonate rocks, whose buffering action neutralizes acid in the rain. This buffering mechanism is important for sustaining aquatic life. The carbonate buffer system is composed of various species and the equilibria between these species are governed by the following reactions. CO2 (g) CO2 (aq) CO2 (aq) + H2O H2CO3 (aq) H2CO3 (aq) + H2O H3O+ (aq) + HCO3 (aq) $K_{a1}=4.3\times 10^{-7}$ HCO3 (aq) + H2O H3O+ (aq) + CO3 2 (aq)

$K_{a2}=4.8\times 10^{-11}$

The ability of a buffer to resist change in pH depends on the total concentration of the buffering species as well as their concentration ratio. Buffer capacity of a given solution is defined as the number of moles of a strong acid or a strong base that changes the pH of 1 L of the buffer by 1 unit. A laboratory experiment was designed to simulate the buffering action of carbonate solutions, in which 300 mL of 0.004 M Na2CO3 solution was titrated against sulphuric acid and the following graph was obtained.

2.8 Based on the graph, answer the following questions: a. For what pH range(s), we cannot obtain buffers based on carbonate species?

b. What volume of 0.03 M H2SO4 would change the pH of 1 L 0.004 M Na2CO3 by 1 unit.

0 2 4 6 8 10 12 0 10 20 30 40 50 60 70 pH Volume of 0.03 M H2SO4 Indian National Chemistry Olympiad 2018

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c. Buffer capacity of a Na2CO3 solution would i) Increase with the concentration of Na2CO3(aq)

ii) Remain unaffected by the concentration of Na2CO3(aq) iii) Decrease with the concentration of Na2CO3(aq)

Next, let us do some simple calculations to understand the buffering action of a CaCO3 solution. A saturated solution of CaCO3 (solubility = $1.1\times 10^{-4}$ M) was prepared in distilled water. Assume that the distilled water was free of any dissolved gases and any other cations.

2.9 Calculate the pH of the saturated CaCO3 solution. (Hint: amount of H2CO3 (aq) formed under these conditions would be negligible.)

2.10 Determine what volume of 0.03 M H2SO4 would be required to lower the pH of 300 mL of saturated CaCO3 solution by 1unit. Assume that the volume change of the CaCO3 solution due to acid addition is small. Indian National Chemistry Olympiad 2018

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2.11 If the saturated solution of CaCO3 is in contact with excess solid CaCO3, the buffer capacity of this system would be (Mark X in the appropriate box)

a. same as that of the saturated solution.

b. more than that of the saturated solution. c. lower than that of the saturated solution.

2.12 Comparing the saturated solutions of MgCO3 ($K_{sp}=6.8\times 10^{-6}$) and CaCO3 ($K_{sp}=2.8\times 10^{-9}$), the correct statements is (Mark X in the appropriate box) a. Buffer capacities of both the solutions are equal

b. Buffer capacity of MgCO3 solution will be more than that of CaCO3 c. Buffer capacity of CaCO3 solution will be more than that of MgCO3

2.13 Among Na2CO3, MgCO3, and CaCO3, identify the salts that fit the following descriptions and write the formula in the boxes provided. (i) For equal volumes of saturated solutions of the three salts taken separately, solution of this salt will provide the best buffering action against acid.

(ii) After heavy rains, this salt is found only in dissolved forms in water bodies, no deposits are found in soils.

(iii)In regions receiving very high rainfall annually, deposits of this salt remain for centuries, and provide a major source of carbonate buffering against acidic rain for the longest period of time.

Acidic rain in regions lacking sufficient buffering substances lead to acidification of the water bodies, which is often detected by the loss of the sweet smell of geosmin from the region. Indian National Chemistry Olympiad 2018

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Topic: Thermodynamics Metodi: First Law of Thermodynamics, Ideal Gas Law Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.9

Problema 3

Problem 3

21 Marks Hydrogen Bonding and Water of Crystallization

Part 1 Water is the most abundant molecule on the earth’s surface. Water makes a major fraction of the mass of the biological as well as many non-biological entities. Water is liquid at room temperature because of extensive intermolecular hydrogen-bonding. A molecule containing H covalently bonded to an electronegative atom is known as a hydrogen bond donor, whereas a molecule with electron-rich atom which makes H-bond with an H atom is known as a hydrogen bond acceptor. 3.1 Identify the molecules in the figure below which would act as H-bond donors, H-bond acceptors or both (write the corresponding number/s in the box/es)

(a) Only hydrogen bond acceptors (b) Only hydrogen bond donors (c) Both, as hydrogen bond donors and acceptors

Hydrogen bond is represented by a dotted line (----) while a covalent bond is represented by a solid line. For some compounds, intermolecular H-bonding leads to formation of porous frameworks with specific pore sizes. Such frameworks can be used to selectively trap small molecules like NH3, CH4, SO2, etc. in an atmosphere. Xenon is a modern anaesthetic that has been successfully used in surgeries, which shows its ability to interact with other molecules. One challenge with this use is high cost of production, and ease of storage and transport. Interestingly, Xe can be reversibly trapped in porous threedimensional frameworks of molecules such as water, quinol, etc, held together by H-bonds. Slow evaporation of an aqueous solution of p-quinol saturated with Xe yields crystal of p-quinol framework. X-ray crystallography of these crystals indicates six p-quinols organized around one Xe atom. The phenyl rings in p-quinols are in cyclic arrangement positioned alternately above and below a plane. This arrangement generates hexagonal cavities in the structure. The unit cell for Xe-encapsulated p-quinol crystal is shown below. Two molecules of p-quinol are centered at two opposite face centers, 8 molecules are centered at the edge centers, while the rest are inside the cell body. Indian National Chemistry Olympiad 2018

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A view of the above unit cell as seen along the c-axis is given in the box below, with C, O, and Xe atoms labelled (H atoms are not labelled in the structure). 3.2 Show the hydrogen bonds that define the hexagonal cavities in the structure given below.

Centers of molecules are at the edge centers Centers of molecules are at the face centers Indian National Chemistry Olympiad 2018

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15 3.3 By referring to the unit cell diagrams given above, calculate the number of p-quinol molecules and the number of Xe atoms present per unit cell. Show the steps involved in arriving at the answer.

The cell parameters for the above unit cell are: a = 16.58 Å, b = 16.58 Å, c = 5.42 Å, $\alpha$ = $\beta$ = $90^{\circ }$, $\gamma$ = $120^{\circ }$. 3.4 Determine the density of this crystal (in kg m${}^{-3}$), assuming all the hexagonal cavities in the crystal are occupied by Xe atoms.

3.5 If all the hexagonal cavities in the crystal are occupied by Xe atoms, calculate the volume that Xe trapped in 1 cm${}^3$ of p-quinol framework would occupy when released at 298 K and 1 atm.

Part B: Water of crystallization Water of crystallization is the amount of water necessary for certain salts to crystallize from their solutions. Studies have shown that water of crystallization forms bonds with the ions of the salt and stabilizes the crystal structure. This water also affects the shapes and colour of their crystals. Indian National Chemistry Olympiad 2018

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16 Blue-green crystals of CuCl2.2H2O on heating lose water giving brown coloured anhydrous salt. The unit cell of anhydrous CuCl2 is shown below where each copper is in square planar coordination and all Cu-Cl distances are equal.

3.6 This ionic arrangement is stabilized because (Mark X in the correct box(es)):

(i) Every Cu2+ is stabilized by two Cl ̄ present perpendicular to the CuCl4 plane. (ii) Every Cl ̄ is stabilized by two Cu2+ present perpendicular to the CuCl4 plane. (iii) This structure has FCC arrangement of the Cu2+ ions. (iv) Layers of Cl ̄ ions are next to each other and are sandwiched between the layers of Cu2+ ions. When CuCl2 is exposed to moisture, water enters the space between the ions forming CuCl2.2H2O. The water molecules induce many changes such as shift of Cu-Cl units with respect to each other and changes in Cu-Cl distances. CuCl2.2H2O crystallizes in an end-centered orthorhombic lattice stabilized by hydrogen bonds. The Cu2+ ions occupy all the vertices and centres of two opposite faces. The following table shows four possible framework positions of Cl ions with respect to Cu2+ ions. Out of the four frameworks given, three are not observed because of one or more of the following reasons: I. The structure would have high inter-ionic repulsions, which cannot be effectively minimized by the two molecules of water per Cu2+ ion. II. The cell shown is not a unit cell because it cannot be repeated to generate a lattice. III. The ratio of Cu2+:Cl ions is not 1:2 per unit cell. 3.7 Identify the impossible frameworks by writing the correct reasoning statement (give number I, II or III) against the structure. In the correct framework, show water molecules with appropriate bonds with the ions present. Indian National Chemistry Olympiad 2018

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17 Framework Reason/s for impossible framework wherever applicable Indian National Chemistry Olympiad 2018

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Materials which can store and release thermal energy by absorbing water molecules are promising heat storage systems. Salts like CuCl2.2H2O can be dehydrated by keeping in the sunlight, which is an endothermic process. The anhydrous salts then can be used to heat a room by exposure to water vapours. These salt-based devices can be dehydrated and hydrated again and again. In search of a thermal energy storage material, a student checked the amount of energy released by anhydrous CuCl2 on exposure to water. The student added 2.5 g anhydrous CuCl2 to 50 mL water (initially at $25{}^{\circ }\text{C}$) in a covered polystyrene cup and noted $3.2{}^{\circ }\text{C}$ rise in temperature. 3.8 Assuming that the polystyrene cup is a closed and thermally insulated system and the specific heat of water (with dissolved salts) is $4.18\text{J}{}^{\circ }{\text{C}}^{-1}{\text{g}}^{-1}$, calculate the amount of heat released (enthalpy of hydration) by anhydrous CuCl2 in kJ mol${}^{-1}$.

3.9 To heat a given room in winter for one night, 6,000 kJ of energy is used. Determine the amount of anhydrous CuCl2 (in kg) required to be stored in the rechargeable device to achieve this amount of heating. Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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Topic: Thermodynamics, Kinetic Theory Metodi: First Law of Thermodynamics, Ideal Gas Law, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.15

Problema 4

Problem 4

19 Marks Lignin Lignin is a biopolymer that constitutes 15 - 40% of the dry weight of plants. Its major function is to provide structural support to the plant. Lignin has the potential of yielding several industrially important chemicals, a concept known as Bio refinery (similar to petroleum refinery). A major challenge in this conversion is the depolymerization of lignin. A representative structure of lignin is:

4.1 In the above structure of lignin, identify the functional groups present. (Mark X against the correct option/s) a) Diaryl ether

b) Aryl alkyl ether

c) Ester

d) 1o Amino

e) Lactone

f) 2o Alcohol

Direct chemical depolymerization of lignin is a difficult process. Oxidized forms of lignin can be depolymerized, but it is expensive and involves hazardous reagents. In 2014, a group led by an American scientist Shannon S. Stahl showed that formic acid in the presence of sodium formate can be used efficiently to depolymerize the oxidized forms of lignin. To understand the mechanism of this process, model compounds D and E that represent structural units of lignin and oxidized lignin, respectively, were prepared. Following scheme illustrates the synthesis of D. . Lithium diisopropylamide (LDA, a Base) . THF, $-{}^{\circ }\text{C}$ THF/Water ( : ) Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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4.2 Identify structures of B, C and D.

D on oxidation using CrO3/H2SO4 gave an unexpected product E and two minor products. E gives no reaction with Tollen’s reagent or with saturated NaHCO3.

4.3 Identify structure of E.

Both model compounds D and E were treated with excess formic acid in the presence of aqueous sodium formate at room temperature to give products F and G, respectively. 4.4 Identify structures of F and G.

excess HCOOH/HCOONa ${}^{\circ }\text{C}$ Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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excess HCOOH/HCOONa ${}^{\circ }\text{C}$

On heating with excess formic acid in aqueous sodium formate at $110{}^{\circ }\text{C}$, D gave only F, whereas, E gave a mixture of G, H, I, and J. For this reaction of E at $110{}^{\circ }\text{C}$, variation in the amounts of E, G, and J over time is presented in Plot 1. When G was heated with formic acid in aqueous sodium formate to $110{}^{\circ }\text{C}$, H, I and J were obtained. For this reaction, variation in the amounts of G, H and J with time are given in Plot 2. The rates of formation of I and J were same. 4.5 Identify the curves corresponding to E, G and J in Plots 1, and for J in Plot 2. (Write the appropriate letter in the corresponding boxes in the plots)

Plot 1

Plot 2

4.6 From the curve for H in plot 2, we can infer that (Mark X in the correct box/es)

i) H is formed at a very low rate.

ii) H is formed by the reaction of G and J.

iii) H is in equilibrium with G.

iv) H is formed as well as consumed during the reaction. Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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C-H bond and its Deuterium analogue, C-D bond, differ in their bond energies. This difference in bond energies also affects the rate of reactions where these bonds are involved. A new sample of compound E was prepared by substituting a Deuterium in place of Hydrogen at the chiral carbon in E. This substitution decreased the rate of reaction of E with formic acid/sodium formate at $110{}^{\circ }\text{C}$. 4.7 Choose the correct inference of the reaction of E with formic acid/sodium formate at $110{}^{\circ }\text{C}$. (Mark X against the correct option/s) a) The rate decreased because the bond energy of C-H bond is more than of C-D bond b) A change in rate due to Deuterium substitution suggests breaking of a C-H bond during the reaction and it is the rate determining step c) The change in rate suggests that the rate determining step involves a base. d) The change in rate suggests a protonation/deuteration step by formic acid and it is the rate determining step.

Among H, I & J, compound H decolorized acidic KMnO4 solution and compound J gave a positive test with aqueous ferric chloride. I gives the iodoform test. 4.8 Identify compounds H, I and J.

4.9 The major conclusion/s of the above studies with model compounds D and E is/are: (Mark X in the correct box/es) a) Depolymerisation of oxidized lignin is via cleavage of C-C bonds in

units. b) Oxidized lignin undergoes depolymerisation by formic acid/sodium formate by cleavage of ether linkages. c) Selective oxidation of 2o alcohol groups on lignin is essential for an efficient depolymerisation reaction by formic acid/sodium formate. d) Every C-O bond cleavage consumes one molecule of formic acid

A naturally occurring popular flavor compound K used initially by Aztecs in Central America was introduced to the Europeans in 1520. In later periods, 90% of the supply of this compound was sourced from a plant cultivated in Madagascar. It was also produced from small molecules obtained from waste lignin from the paper industry. Currently, >99% of K is produced using petrochemicals. Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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Protocatechualdehyde with dimethyl sulphate in the presence of NaOH gives K along with L and M. In this reaction, the product ratios can be controlled by the reaction conditions. With excess of NaOH and excess of dimethyl sulfate, product L was formed exclusively.

4.10 Identify L.

When one equivalent of NaOH and one eqivalent of dimethyl sulphate were used, K was formed in 87% yield. The same reaction when carried out with excess of NaOH and only one equivalent of dimethyl sulfate, M was formed in 95% yield. 4.11 Identify K and M.

Plants use phenylalanine to make several small aromatic organic molecules which are later used in the formation of the lignin framework. Some of these small molecules get attached to the carbohydrates in the cell wall. Compounds R and S shown below are formed as a result of sunlight induced photochemical reaction of P.

Protocatechualdehyde Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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R and S were chemically synthesized by the following scheme. Air oxidation of N in presence of an acid gave O. , H Pyridine/piperidine UV Light S + R H , Friedel Crafts reaction

4.12 Identify N, O, P and Q Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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Topic: Thermodynamics Metodi: Physical Modeling, First Law of Thermodynamics Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.21

Problema 5

Problem 5

22 marks Keto-Enol Tautomerism: Kinetics and Thermodynamics

Two compounds having the same molecular formula but different structural arrangement of atoms in molecules are known as structural isomers, e.g., C2H6O represents both ethyl alcohol and dimethyl ether. One form of structural isomerism is tautomerism, in which the same molecule dynamically interconverts between two or more structural forms. Carbonyl compounds exhibit keto-enol tautomerism, represented as follows:

5.1 For the following compound, write the possible tautomers.

Many carbonyl compounds with acidic alpha hydrogen are important for industrial synthesis of many chemicals. Ethyl acetoacetate (EAA), i.e., ethyl-3-oxo butanoate is one such widely used compound. In solution, EAA is an equilibrium mixture of the keto and an enol form. 5.2 Draw the structures of possible enol forms for EAA and identify the most stable enol form. Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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The most stable enol form is observed experimentally. To study the kinetics of the following reaction, a detailed experiment was conducted. EAA(Keto) EAA(Enol)

(1) Four flasks (A - D) were taken, each containing 25 mL solution S of EAA (15 g L${}^{-1}$) in anhydrous methanol. The flasks were maintained at $20{}^{\circ }\text{C}$ throughout this study. Excess of freshly prepared bromine solution in methanol (25 mL) at $20{}^{\circ }\text{C}$ was added rapidly to each flask at t = 0 s. The enol form reacts extremely rapidly with bromine while the keto form does not. This reaction results in an acidic solution. Bromination of enol is a very fast reaction compared to the enolization of EAA. The mono-brominated ester formed is stable and doesn’t react further under these reaction conditions. 5.3 Write the balanced chemical equation for the reaction that would take place on addition of bromine to solution S.

Excess of 2,4,4-trimethyl-1-pentene was added rapidly to flasks A, B, C and D at t = 10, 40, 75, and 100 s, respectively. 5.4 The 2,4,4-trimethyl-1-pentene added will (Mark X in the appropriate box(es)) (i) increase the rate of keto to enol conversion. (ii) consume unreacted bromine. (iii) react with monobrominated ester and hence ensure complete bromination of EAA. (iv) stop the enolization reaction.

After addition of 2,4,4-trimethyl-1-pentene, about 10 mL of 10% aqueous KI solution was added to each of the flasks A - D and allowed to stand for 30 minutes. The reaction mixture was still acidic. During this time, brominated ester is converted back to EAA. Then, the liberated iodine is titrated with 0.1 M sodium thiosulphate solution using starch indicator. Assume that there is no reaction between iodine and any form of EAA. 5.5 Write the balanced equation of the reaction for the liberation of iodine.

$k_F$ $k_B$ Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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5.6 Write the balanced equation for reaction of iodine with sodium thiosulphate.

The experiment can be summarized as follows:

Following volumes of sodium thiosulphate solution were consumed in titration with the solutions in the flasks A, B and C at 20 oC. Flask Volume of Na2S2O3 solution used (mL) A 3.78 B 4.02 C 4.30

5.7 From this data, find $n_{\text{keto}}$, the number of moles of keto form of EAA at the time of addition of 2,4,4-trimethyl-1-pentene, for the flasks A, B and C. Show detailed calculations for flask A. You may assume volumes to be additive during mixing of solutions in this experiment.

t = 0 A B C D Solution S (25 mL) + Bromine solution (25 mL) t = 10 s t = 40 s t = 75 s t = 100 s

• 2,4,4-trimethyl-1-pentene
• 10% KI (aq.) titration with Na2S2O3 (0.1 M) 30 min Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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The following graph indicates the plot of $-\ln (n_{\text{keto}})$ versus time (t). Point corresponding to flask D is indicated on the graph. 5.910 5.912 5.914 5.916 5.918 5.920 5.922 5.924 5.926 5.928 5.930 5.932 5.934 5.936 0 25 50 75 100 $-\ln (n_{\text{keto}})$ t (s)

5.8 Plot the values corresponding to flasks A, B and C in the graph above. Using the graph, find the following for conversion of EAA (keto) to EAA (enol) in reaction (1). a. The order of the reaction. b. Rate constant $k_F$ for the reaction. Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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c. Equilibrium constant $K_{eq}$ for the reaction. Report the answer upto 5 significant figures.

When the above experiment was carried out at $40^{\circ }\text{C}$, the values of $K_F$ and $K_{eq}$ were found to be $5.25\times 10^{-4}{\text{s}}^{-1}$ and $0.06272$, respectively. 5.9 Using the above information, find the values of $\Delta H^{\circ }$ (in kJ mol${}^{-1}$) and $\Delta S^{\circ }$ (in kJ mol${}^{-1}$ K${}^{-1}$) for the enolization reaction (1). Assume $\Delta H^{\circ }$ and $\Delta S^{\circ }$ to be independent of temperature. Show necessary calculations.

Reaction kinetics is very important for the industrial use of a reaction, as the reaction rates affect the time in which a certain amount of product can be synthesised. This time of production in turn affects the cost of the final product. Indian National Chemistry Olympiad 2018 Roll No HBCSE, 27th January 2018

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An industrialist has 8.5 kg of EAA and hopes to produce compound Y (Molar mass = 174 g), which is obtained in equimolar amount from the Enol form of EAA. EAA (Keto) EAA(Enol) Y 5.10 For what value of $k^{\prime }$, the industrialist would obtain a certain amount of Y in minimum amount of time? (Mark X in the appropriate box(es))

(i) $k^{\prime }\to 0$

(ii) $k^{\prime }\to \infty$

5.11 The industrialist did not know the value of $k^{\prime }$, but wanted to know the minimum time that would be required to produce 10 kg of Y using the available EAA in methanol as solvent at $40^{\circ }\text{C}$. Estimate this time assuming that (i) the kinetics of the enolization reaction of EAA in this large scale reaction is same as determined in the laboratory experiment described above and (ii) other effects such as changes in temperature and solution volume during the progress of reaction are negligible.

$k^{\prime }$

$k_F$ $k_B$ HBCSE, 27th January 2018

Rough Page HBCSE, 27th January 2018

Rough Page

Topic: Thermodynamics, Kinetic Theory Metodi: First Law of Thermodynamics, Thermodynamic Cycle Analysis, Graph Linearization Competenze: Mathematical Modeling, Experimental Data Analysis, Graph Linearization Fonte: Testo (PDF) — p.27