India 2023 — INChO2023-Question.pdf

Fonte: gare di altri paesi/India/chimica/INChO2023-Question.pdf · Apri PDF · apri PDF p.2 Cluster: Termodinamica Soluzioni (stessa cartella): · · · · · · · · · · · · · ·

Problema 1

Problem 1

19 marks Water quality in aquaculture Aquaculture/fish farming is the controlled process of cultivating aquatic organisms for human consumption. India is the second largest fish producing country in the world. The success of fish/prawn farming in ponds depends on water quality. Unutilized feed, fertilizers, and chemicals added to pond and excretions from aquatic organisms lead to degradation of water quality. Pond water is typically characterized by levels of Dissolved Oxygen (DO), alkalinity, salinity, pH, etc. Aquatic organisms require minimum 5 ppm (parts per million; mg/L) dissolved oxygen for healthy growth. Below 2 ppm, they are not able to survive. This requires regular monitoring of DO level in aquaculture ponds and aerators are operated when DO level goes down. 1.1 In a pond, DO concentration (mark the correct option(s) with X) i) increases with increase in temperature ii) decreases with water depth iii) increases during night

For DO measurement in a pond, a sample bottle is completely filled with the water sample (without leaving any space for air). A solution of MnSO4 and alkaline iodide azide (KOH + KI + NaN3) (AIA) are added using pipette to bottom part of the bottle. During this addition, about 4 mL of water sample from the top overflows out. Then the bottle is closed and sealed. Under alkaline conditions, the dissolved oxygen in the water oxidizes Mn2+ and quantitatively converts it into manganese oxy di-hydroxide (stable precipitate). Next, the whole content of sample bottle is mixed with conc. H2SO4, the precipitate dissolves and I2 is formed. The I2 generated is titrated against standard sodium thiosulphate (Na2S2O3, HYPO) solution using starch indicator. 1.2 Write the balanced equations for the chemical reactions involved in the above procedure of DO measurement.

1.3 How many moles of HYPO are equivalent to one mole of DO in this procedure?

1.4 A sample bottle filled initially with 250 mL of pond water sample consumed 6 mL of $0.025{\text{mol L}}^{-1}$ HYPO. Calculate the DO concentration in ppm in pond water.

The amount of MnSO4 and AIA (KOH + KI + NaN3) to be added to a sample is based on maximum possible DO concentrations in these ponds. For coastal regions having moderate climate, this value can be taken as ~12.5 ppm. [The value corresponds to saturated DO of distilled water assuming regions having minimum (winter) temperature of $6{}^{\circ }\text{C}$. Also, waters with dissolved salts have lower DO than distilled water.] 1.5 Assuming that 10% extra reagents are added, calculate the moles of Mn2+ which must be added in the sample bottle.

Wastes from various organisms in water generate nitrites in water and NO2– can react with chemical reagents used for DO estimation. Sodium azide (NaN3) under acidic conditions reacts with nitrite producing N2O as one of the products. 1.6 Write the equation for the reaction between NaN3 and NO2– under acidic conditions. The optimum pH for healthy growth of most aquatic animals is in the range 6.5 - 8.5. Waste products from organisms including gases such as CO2 tend to make water acidic. Ability of water to neutralize acidic substances and thus, resist changes in pH is known as alkalinity (measured in terms of ppm of CaCO3). Generally, it can be due to OH–, CO32– and HCO3 – ions present in water. A certain minimum level of alkalinity is desired for proper growth of aquatic animals. Aerator Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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For carbonic acid (H2CO3): pKa1 = 6.4 and pKa2 = 10.3 1.7 Write the reactions of acid (represent by H+) with these three ions responsible for alkalinity. To determine alkalinity, water sample is titrated with standard acid first using phenolphthalein indicator. The titration is continued in the same solution by adding methyl orange indicator. The acid equivalents used till phenolphthalein and methyl orange end points, respectively are known as phenolphthalein (P) alkalinity and methyl orange (M or total) alkalinity (both expressed as ${\text{mmol L}}^{-1}$ equivalent to standard acid consumed). Indicator pH of the solution Colour Indicator pH of the solution Colour Phenolphthalein 8.3 – 10.0 Pink Methyl orange 2.9 - 4.6 Red < 8.3 Colourless

4.6 Yellow

1.8 Depending on the ions present in water sample, the P and M alkalinity values can be related to each other. For each of the following relationship between P and M, identify the ion/s responsible.

1.9 Calculate the pH of water containing $0.44{\text{mg L}}^{-1}$ of dissolved (hydrated) CO2 when alkalinity is 5 ppm. (assume that all of the alkalinity exists as bicarbonate; alkalinity of 50 ppm $\equiv$ $50{\text{mg L}}^{-1}$ CaCO3 $\equiv 10^{-3}$ mol HCO3 – /L)

Dissolved calcium in pond water is required for successful embryonic development of aquatic animals. Minimum calcium hardness of 50 mg/L as CaCO3 is desirable for freshwater aquaculture ponds. There are situations where hardness is to be increased without significantly affecting pH and alkalinity. 1.10 Of the following, the substance(s) which can be used for this purpose is/are (Mark X against the correct option(s)) i) Ca(OH)2

ii) CaCO3

iii) CaSO4 iv) CaO Bleaching powder is widely used in aquaculture for disinfection (chlorination) to kill pathogens and thereby prevent diseases. Bleaching powder having the composition Ca(OCl)2.Ca(OH)2.CaCl2.2H2O generates Cl2 on reaction with water.

Ca(OCl)2.Ca(OH)2.CaCl2.2H2O + 2H2O $\to$ 3Ca(OH)2 + 2Cl2 The amount of bleaching powder/chlorine required to disinfect a water sample is known as its chlorine demand (CD) expressed in units of ppm (milligram of Cl2 per L of water). CD is determined by adding a certain dosage to the water sample and determining the residual chlorine remaining after disinfection. Residual chlorine = Chlorine Dosage – Chlorine Demand 1.11 Calculate the amount of bleaching powder required in kg to treat water in a pond of area 1 hectare and depth 1 m ($10^7$ L of water), if chlorine demand is 6 ppm and residual chlorine is 2 ppm.

Topic: Thermodynamics, Modern-Quantum Physics, Astrophysics Metodi: Physical Modeling, Dimensional Analysis, Ideal Gas Law Competenze: Mathematical Modeling, Physical Reasoning, Unit Conversion Fonte: Testo (PDF) — p.2

Problema 2

Problem 2

23 marks Maillard reaction in cooking

Fried and baked foods such as chips, biscuits, breads, namkeens etc, and many other cooked foods have a distinctive flavour which makes them taste and smell good. This flavour is produced due to a series of non- Relationship P = M 2P = M M > 0; P = 0 P > M/2 P < M/2 Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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enzymatic reactions between amino acids and reducing sugars, which increase with temperature and proceed rapidly around 140 to $165{}^{\circ }\text{C}$. These reactions use water in the food product as a reaction medium (solvent). This set of reactions are the major cause of browning of food during cooking and is known as the Maillard reaction, reported first by Louis-Camille Maillard in 1912. Maillard reaction between a pair of reducing sugar and amino acid molecule may proceed via multiple pathways producing a range of compounds. Two of the categories of compounds formed in this reaction are enaminols (compounds containing alkene, amine and alcoholic functionality) and $\alpha$-dicarbonyls (containing two carbonyl groups on adjacent carbons in the molecule). 2.1 Draw the first product, A, of the Maillard reaction as glucose reacts with an amino acid from proteins in food (represent as Protein-NH2). Draw four tautomeric structures of A (stereochemistry not required).

2.2. Enaminols can produce $\alpha$-dicarbonyl compounds via multistep reaction. A 1,2 enaminol B (obtained from glucose) transforms via intermediate C to produce 3-deoxyosone as shown below. Draw the structures of B and C.

2.3 Reactivity of amino acids with sugars in Maillard reactions is pH dependant. Amino acid asparagine has the following structure. Draw the structure of prominent species of aspargine at pH = 7.

pKa values: –COOH: 2.14 $\alpha$–NH2: 8.72

Maillard reaction products generally enhance taste, color, and flavor of food. However, longer cooking of food (to the stage of charring) can lead to production of toxic compounds having acrid flavours. One such example is reaction of $\alpha$-dicarbonyl compound with asparagine (available in most foods) to form acrylamide. 2.4. Draw the intermediates (D, E, F, G) formed during reaction of asparagine and $\alpha$-dicarbonyl product (shown below) under cooking conditions to form acrylamide.

Protein Glucose Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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Acrylamide is a known carcinogen. As it is absorbed by our body, it is converted to glycidamide by the enzyme cytochrome P450 (CYP450).

Glycidamide can react with DNA bases by covalently modifying them, thus showing carcinogenic properties. 2.5 Draw the modified guanine site S within a DNA chain due to reaction with glycidamide. In a DNA chain, guanine is H-bonded with cytosine.

2.6 A tripeptide glutathione (common in many foods, and also synthesized by our body) acts as a natural inhibitor to glycidamide and acrylamide. Both acrylamide and glycidamide get attached covalently to glutathione, and get excreted in the urine. Draw the major product T as glycidamide reacts with glutathione. Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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2.7 Some commercial food products are modified with additives such as flavonoids, vitamins, etc which trap the Malliard intermediates or trap acrylamide during overcooking. i) Trapping acrylamide: Niacin (vitamin B3) can react with acrylamide making it unavailable for further absorption in the body. This reaction happens via an addition mechanism, in which a nucleophile approaches an α,β-unsaturated carbonyl compound at the position. Given below is a generalized form of this mechanism.

Draw the structure of U in the reaction of acrylamide with niacin.

ii) Trapping intermediate species such as alpha-dicarbonyls: The following pharmaceuticals and food ingredients trap $\alpha$-dicarbonyls (during cooking conditions). Draw the major product of their reactions with methyl glyoxal (MGO). Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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Some general observations about reactions in food are: Fact I: The first step of Maillard reaction generates water as a bi-product, and excess water may favour backward reaction. Fact II: NaCl enhances the hydrolysis rate of sucrose (a non-reducing sugar) to glucose and fructose in aqueous solution. Fact III: The first step of Maillard reaction may proceed slowly, and the forward reaction may be facilitated by either H+ or high temperatures. Fact IV: The amines are less likely to be found as –NH3+ with increase in pH. Fact V: Metal ions in food such as Na+, Ca2+, etc, can increase the rate of Maillard reaction (by stabilizing negative charges in intermediates).

2.8 Given below are a few statements about the Maillard reaction in different foods. Indicate if they are true (T) or false (F). For each statement, identify which of the above fact(s) explain or correlate with your answer. Write “Fact I”, “Fact II”, “Fact III” “Fact IV”, “Fact V” or “None” in the box provided.

a) The rate of the initial step of the Maillard reaction is decreased at pH values lower than the pKa value of the amino group. b) Very low pH decreases the reactivity of carbonyl group of the carbohydrate. c) In recipes of stir frying, onions brown faster with lemon/tamarind juice than when stir fried alone. d) Higher salt concentration promotes the Maillard reaction in food dishes containing sucrose. e) During cooking, potatoes kept earlier in salt water tend to brown faster than potatoes kept earlier in water. f) In soups, vegetables such as carrots, cauliflowers, etc, get less browned during boiling in a pressure cooker than during open vessel boiling.

Topic: Thermodynamics, Nuclear & Particle Physics Metodi: First Law of Thermodynamics, Thermodynamic Cycle Analysis, Ideal Gas Law, Radioactive Decay Law Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.3

Problema 3

Problem 3

20 marks Historical alum production and dyeing Ancient civilizations including Egyptian, Chinese and Indians, had found aluminium sulphate useful in dyeing of clothes and tanning of leather. Pure crystalline aluminium sulphate was not abundant in nature. Aluminium is abundant in earth’s crust as alumina or exists as sulphates with other metal sulphates, which all were called alums. Only two of these: potassium aluminium sulphate [KAl(SO4)2.12H2O] and ammonium aluminium sulphate [NH4Al(SO4)2.12H2O] could be easily obtained as colourless crystals of high purity. These two alums became of great economic importance, due to their better dyeing properties. 3.1 A 1 M solution of ammonical alum would have pH in the range.

i) 0 – 1.0 ii) 1.0 – 6.0 iii) 6.0 – 8.0 iv) 8.0 – 13.0 v) 13.0 – 14.0

Molar mass (${\text{g mole}}^{-1}$) Solubility at $20{}^{\circ }\text{C}$ (g per 100 g water ) Solubility at $60{}^{\circ }\text{C}$ (g per 100 g water) FeSO4.7H2O 278.02 25.6 101 Al2(SO4)3.18H2O 666.43 36.4 59.2 KAl(SO4)2.12H2O 474.40 6.0 33.3 NH4Al(SO4)2.12H2O 453.33 6.6 21.1 Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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The first large scale industry to produce synthetic alum was set up in the 17th century in Whitby, England, in the regions having shale rocks with some fossils in it. The rocks had aluminosilicates (Al2O3.xSiO2), pyrites (FeS2) and carbon, which were essential ingredients to prepare alum. Alum obtained had to be free from iron salts for using as mordant. The steps involved in this process of producing alum were: I. Mineral rocks were piled on wooden logs which were burned for months. Carbon (fossils) in rocks also supported the fire. II. The burned rocks were soaked in rainwater to dissolve the sulphates generated by oxidation. The soluble salts were dissolved from the charred rock. The silicates were left behind as slag. FeSO4 + 2Al2O3.xSiO2 + 6H2SO4 (aq) $\to$ FeSO4 (aq) + 2Al2(SO4)3 (aq) + 2x SiO2 $\downarrow$ + 6H2O III. The supernatant liquor was separately collected, and concentrated by evaporation at low heat. IV. The concentrated fluid was then treated with stocked human urine (containing urea which decomposes to give ammonia) or roasted sea weed (containing potassium oxide) as an alkali source to produce ammonium alum or potash alum, respectively.

3.2 Two of the reactions in Step-I are: FeS2(s) + O2(g) $\to$ FeSO4(s) + A

Eqn 1 FeS2(s) + O2(g) + H2O (g) $\to$ FeSO4(s) + B Eqn. 2 i) Identify A and B, and balance the equations.

ii) During rainy season, rotten egg smell is observed in shale rocks (both burnt and unburnt). Write a balanced equation for the reaction giving rise to rotten egg smell. iii) In step III, a fresh egg was added to the heated solution at frequent intervals. If the egg sank, the heating was continued. If the egg floated, heating was stopped, and the next step (Step IV) was carried out. This technique of using egg helped to: (Mark X against the correct option(s))

a) stop the reaction that was producing rotten egg smell. b) ensure homogeneization of solution during evaporation. c) prevent precipitation of iron sulphate during cooling. d) nucleate/seed more number of alum crystals under hot conditions. e) prevent loss of sulphate as SO2 gas. f) consistently obtain saturation point of Al2(SO4)3 in solution. iv) For step IV, write the balanced chemical reactions which led to production of potash alum and ammonical alum, respectively.

Potash alum and ammonia alum crystals are isomorphic and look alike. So by physical appearance, are hard to distinguish. Quicklime (CaO) is used to distinguish the two. 3.3 Write balanced chemical equation(s) for reaction(s) of quicklime which can differentiate potash and ammonia alum.

3.4 In the alum crystals obtained, a very small amount of Fe2+ may remain which may not be detectable with eyes. One way to determine its presence in alum is by the addition of a reagent which will develop a dark colour with it. Among the following substances available in past, identify the reagent useful to detect Fe2+ and write the corresponding balanced equation for the reaction.

i) CaCO3

ii) K3[Fe(CN)6] iii) Ca(OH)2

iv) CuSO4.5H2O Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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The crude alum crystals obtained in step IV were recrystallized using water as solvent to separate any residual Fe2+ from the alum. The steps are shown below.

3.5 The statement(s) correct for the above process of recrystallization is/are (Mark X against the correct option(s)) i) Hot solvent melted the solid. ii) More Fe2+ was present in hot solvent than in cold solvent. iii) The impurities settled faster than the pure crystals on cooling. iv) Stage 3 contained saturated solution of alum In fabric dyeing, mordants impact dye brightness, wash-fastness, and colour by aiding dye-fiber bonding. For example, animal fibres (silk, wool) have amino and carboxyl groups. Mordants bind hydrated metal ions to fibre amino and carboxyl groups as shown below:

[Note that the protein chains are non-planar.] Alizarin dye, whose molecular structure is illustrated below, is a bidentate ligand that binds with the mordanted fibre. Usually one alizarin molecule binds to one metal ion.

3.6 i) Show the full structure of the species formed around Al3+ ion as mordant- dye-fiber bonding and show the overall charge of the complex species. ii) Write the number of optically active isomers for the species.

3.7 The metal ions left in the effluents from dyeing industry are of environmental concern. Aluminum sulphate has replaced potash alum as the standard mordant since it produces identical results in terms of fabric colour at similar mass usage. The suggested quantity of both mordants required is 15% of the mass of the dry cotton cloth. A study shows that after use as mordant, 92% of aluminium is present in effluent water. Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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i) What will be the mass of hydrated aluminium sulfate or potash alum required if a dyer mordantes 2 kg fabric? ii) Compare the mass of aluminium left in effluent water, using hydrated aluminium sulfate vs potash alum.

Topic: Thermodynamics, Geometric Optics, Fluid Mechanics Metodi: First Law of Thermodynamics, Thermodynamic Cycle Analysis, Ideal Gas Law, Ray Tracing Competenze: Mathematical Modeling, Diagrammatic Reasoning Fonte: Testo (PDF) — p.7

Problema 4

Problem 4

23 marks The Odyssey of Match sticks

While use of fire has been synonymous with human civilization, easy and safe generation of fire had always been a challenge. An ideal fire generation device should produce a flame instantaneously from a very fast reaction/process, which requires: i) a (easily combustible) fuel

ii) an oxidizer iii) a mechanism to initiate burning of fuel In this process, additionally fumes, hot gases, sparkles, and ashes are likely to be produced depending on the substances used. Today, we have safety matchboxes, where risks from fumes, sparkles, falling ash and spontaneous combustion of match sticks have been minimized. A binder is used to keep the various chemical components of a match stick together as a solid mass on a wooden stick or stiff paper. The current state is achieved through an interesting historical journey as described below.

A. The start of a long journey The discovery of phosphorous in 1669 created new possibilities to generate fire. In 1781, a phosphoric taper was introduced in France as a fire-making device based on a 4-inch glass vial (Figure 1). A piece of white phosphorous was placed at the bottom of the vial along with a thin wax candle (taper/wick) under air-tight conditions. When this vial was dipped in warm water ($>40{}^{\circ }\text{C}$) followed by breaking of the seal, the taper lighted up in flame.

Figure 1. Phosphoric Taper 4.1 Identify the initiator fuel and oxidizer in this set up. 4.2 Write a balanced chemical equation for the reaction initiating fire. Draw the chemical structure of the final phosphorous compound formed in this reaction. 4.3 Dipping of the vial in warm water before breaking the seal resulted in a better and reliable burning of the taper, as this process (mark X against the correct option(s)) i) initiates phosphorous burning with bright white light ii) triggers white phosphorous to red phosphorous isomerism iii) melts the phosphorous and adheres it to the taper iv) reduces the inactive phosphorus oxide surface layer to phosphorus v) minimizes sparkling during phosphorous burning Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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B. Instantaneous light box In 1805, instantaneous match boxes were invented in France that soon became popular in Europe. These match boxes had two parts: (i) a stoppered bottle of concentrated sulfuric acid, and (ii) wooden sticks with a mixture of potassium chlorate, sugar, and binder (gum) on the tips. The stick tip when dipped into the acid solution and withdrawn, caused it to catch fire. 4.4 Identify the initiator fuel and oxidizer in this set up.

4.5 A lot of char formed on the stick at the end of the reaction. Name the possible source(s) of the char. 4.6 After multiple usage of the match box and/or further storage for a long period of time, the match sticks did not burn, nor did it form any char in acid. The most likely reason for this inactivity is (mark X against the correct option)

i) KClO3 reacts with sugar over time ii) H2SO4 absorbs atmospheric water and gets diluted iii) KClO3 absorbs atmospheric water and converts to HClO3 iv) The sugar absorbs atmospheric water and hydrolyzes

C. Friction light In 1820s, friction stick light, also known as Lucifer, was developed in England. Here, the matchstick head contained KClO3 and antimony (III) sulfide. The matchstick head was kept between folds of sandpaper and pulled to create fire. KClO3 present in the matchstick decomposed easily due to heat produced by friction. Here, the ignition was typically followed by splattering of sparks and burning particles in all directions. 4.7 Identify the initiator fuel and oxidizer in this set up.

4.8 i) Write the balanced equation for the decomposition reaction of potassium chlorate. Assuming that the friction caused the temperature to rise to 450 K, determine ii) the enthalpy change per mole and iii) the free energy change per mole for the decomposition reaction at this temperature. The standard enthalpy of formation ($\Delta H_f^{\circ }$) and standard entropy ($S^{\circ }$) of the substances (both quantities can be assumed to be independent of temperature) are given below:

4.9 The decomposition reaction of KClO3 happens easily on slight heat (generated by friction), but not at room temperature. This is because the reaction has (mark the correct option with X)

i) positive $\Delta S$

ii) low $\Delta H$

iii) negative $\Delta H$

iv) high activation energy

Substance $\Delta H_f^{\circ }$ (${\text{kJ mol}}^{-1}$) $S^{\circ }$ (${\text{J K}}^{-1}{\text{ mol}}^{-1}$) O2 (g) 0 205.2 KCl (s) – 436.5 82.6 KClO3 (s) – 397.7 143.1 Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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D. The current model: Safety Matches Current safety match boxes have two parts: i. a matchstick, and ii. a striking surface on the side of the container box. The match stick head contains a mixture of potassium chlorate, antimony sulfide, paraffin wax, ammonium phosphate, and glue. The striking surface contains a mixture of red phosphorous and an abrasive such as powdered glass. When the match stick head is struck on the abrasive surface, the red phosphorus is converted to white phosphorous and transferred to the matchstick head. The white phosphorus reacts in air and initiates the conversions of KClO3 and antimony (III) sulfide, which finally produce heat for the ignition of the match stick. 4.10 Burning of the match stick produces antimony (III) oxide and a species that smells of gunpowder. Write the balanced chemical equation for this reaction.

4.11 The presence of wax in match stick head (mark the correct option with X)

i) ensures steady burning of the match stick over a few seconds ii) prevents fume formation via condensation of the hot gases produced iii) absorbs toxic gases and prevents health hazards iv) prevents spontaneous decomposition of KClO3

Consider the matchstick head as a sphere with a diameter of 0.4 cm (Figure 2) in which 50% volume is occupied by the chemical mixture containing potassium chlorate and antimony sulfide with mass percentages 40% and 18% respectively. Given estimated values- Match stick head mixture, Density = $4.0{\text{g cm}}^{-3}$; Specific heat = $2.1{\text{J g}}^{-1}{}^{\circ }{\text{C}}^{-1}$ Specific heat of the wood = $1.76{\text{J g}}^{-1}{}^{\circ }{\text{C}}^{-1}$ $\Delta H^{\circ }$ of white phosphorous oxidation = $-2984{\text{kJ mol}}^{-1}$ $\Delta H^{\circ }$ of potassium chlorate decomposition in air = $-78{\text{kJ mol}}^{-1}$ $\Delta H^{\circ }$ of antimony sulfide decomposition in air = $-700{\text{kJ mol}}^{-1}$

Figure 2 4.12 If $5.0\mu \text{g}$ of white phosphorous is transferred to the matchstick head, how much energy it can generate from its reaction in air? If all this energy heats up $1\text{mm}\times 0.5\text{mm}\times 0.1\text{mm}$ region of match stick head, estimate the spot temperature at this specific region. Assume room temperature to be 300 K.

4.13 Determine the temperature rise of wood in contact with matchstick head after completion of reactions in the chemical mixture on stick head. Assume that the heat released from combustion is available to 1/10th of the length of the stick and 50% heat escapes as hot gases and flame. Mass of wooden stick = 0.1 g.

Topic: Thermodynamics Metodi: First Law of Thermodynamics, Thermodynamic Cycle Analysis, Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning, Estimation & Approximation Fonte: Testo (PDF) — p.10

Problema 5

Problem 5

21 marks Chemistry and Brain

Our brain communicates with and exerts control over other organs of the body through the central nervous system (CNS), consisting of neurons. Neurons send a message through neurotransmitters, chemicals which cross the synapses at boundary of two cells. Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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A cell membrane is made of bilayers of lipids and has an oil like environment within the membrane bilayer, and the cytoplasm is mainly aqueous. Small molecules are able to enter into these bilayers depending on their structural features. 5.1 For each of the following pairs of molecules of biological origin, choose molecule (A or B) that has more affinity to cell membranes than cytoplasm.

5.2 Give the IUPAC name of compound B in 5.1 i).

Catecholamines are a group of neurotransmitters that are either secreted by neurons or the adrenal gland. One such example is dopamine, which controls pleasure or euphoria in our brain. The amount of dopamine in the cells are regulated by way of several biosynthetic mechanisms and finally converted to homovanillic acid and excreted through urine. Other examples are epinephrine (G) and norepinephrine that are derived from the aminoacid phenylalanine. 5.3 The following is a chemical synthesis route to epinephrine (G) starting from catechol. Identify the intermediates D, E, F and propose a structure of G. G was stable to acid/alkaline hydrolysis conditions.

5.4 Draw the structures of the stereoisomers of G in Fischer projection. Another group of neurotransmitters are derived from lipids. Anandamide is a lipid, first isolated and identified by William A Devane in 1992. It is present in the brain cells, and binds with the receptor proteins of brain cells. It is responsible for calmness, dream states and the mood “Ananda” as in Sanskrit. Hence, the name anandamide.

Anandamide has short duration neurological effect and has attracted scientists to use its structural variants as therapeutic agents. A series of anandamide analogues were synthesized by solid phase synthesis technique, in which the starting material is attached to a solid support. After a sequence of reactions building larger molecules, the product can be easily separated from the reaction mixture by simple filtration. Then the support is removed by an appropriate reagent. Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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In a scheme of synthesis shown below, several analogues P1 - P4 of anandamide (represented as Pj) were produced. These were obtained by varying number of methylene units (n) and functional groups R1 in reagent in the conversions L Mi and by varying Z in the reagent in the conversions Ni Oj in the scheme given below. Solid support is represented as

5.5 Draw the structure of I, K, L, Mi, Ni, and Oi formed using reagents having n = 5, R1 = H and Z = OH during the conversions L Mi and Ni Oj .

5.6 Draw the structures with correct stereochemistry of Pj obtained from sets of reagents having the given n, R1 and Z.

P1 n = 2, R1 = OH and Z = CH3 P3 n = 5, R1 = H and Z = CH3 P2 n = 2, R1 = OH and Z = OH P4 n = 5, R1 = H and Z = OH

5.7 Draw the structure of Pj obtained from set of reagents having n = 5, R1 = H and Z = OH, when Pt in the presence of excess H2 was used as the reducing agent for Oj in place of Lindlar’s catalyst. Reagent Function Reagent Function DIC (N,N’- Diisopropylcarbodimide) a dehydrating agent TFA (Trifluroaceticacid) a strong acid DMF (N,N’- Dimethylformamide) a solvent DCM (Dichloromethane) a solvent DMAP (4- dimethylaminopyridine) a base CuI + NaI a catalyst Indian National Chemistry Olympiad 2023 ©HBCSE January 28, 2023

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One of the receptor proteins of anandamide in neurons has a pocket like binding site (dashed line shows its cross section) formed by seven protein helices (indicated with roman numerals, helices I and IV are not shown in the view). The binding site has a polar region and a hydrophobic non-polar region. Unique structure of anandamide provides the correct geometry for it to bind in this pocket.

5.8 Draw the structure of anandamide showing the correct stereochemistry and appropriate geometry suitable for binding with the above protein receptor. Dashed line shows the boundary of the binding pocket.

5.9 Among the following, molecule(s) more lipophilic than anandamide is/are (mark X against the appropriate option(s))

The protein receptors in brain cells are also the main targets of aromatic polycyclic phenols called cannabinols — narcotic molecules present in Hashish and Marijuana. These molecules do not break down during metabolic functions of the brain cells and stay longer in the binding pocket. This causes longer mood effects, addiction to the narcotic and permanently damages the functions of brain cells. Anandamide gets degraded fast leaving a very short duration mood effect. One of the degradation pathway of anandamide is enzymatic hydrolysis to a fatty acid Q. 5.10 Draw the structure of Q.

Fatty acid Q is a precursor in the biosynthesis of molecules critical to metabolism. Also, Q is a biosynthetic product of linoleic acid. In mammals, Q becomes an essential fatty acid if there are problems in its biosynthesis. 5.11. Among the following, the false statement/s is/are (mark X against the appropriate option(s)) i) Saturated fatty acids are essential to maintain a good level of Q. ii) Linoleic acid is an essential fatty acid. iii) Dietary supplementation of Q is good for health and for good mood. iv) Cannabinols are good replacements of fatty acids for good health. Polar region 1 H hydrogen 1.008 [1.0078, 1.0082] 1 18 3 Li lithium 6.94 [6.938, 6.997] 4 Be beryllium 9.0122 11 Na sodium 22.990 12 Mg magnesium 24.305 [24.304, 24.307] 19 K potassium 39.098 20 Ca calcium 40.078(4) 37 Rb rubidium 85.468 38 Sr strontium 87.62 38 Sr strontium 87.62 55 Cs caesium 132.91 55 Cs caesium 132.91 56 Ba barium 137.33 87 Fr francium 88 Ra radium 5 B boron 10.81 [10.806, 10.821] 13 Al aluminium 26.982 31 Ga gallium 69.723 49 In indium 114.82 81 Tl thallium 204.38 [204.38, 204.39] 6 C carbon 12.011 [12.009, 12.012] 14 Si silicon 28.085 [28.084, 28.086] 32 Ge germanium 72.630(8) 50 Sn tin 118.71 82 Pb lead 207.2 7 N nitrogen 14.007 [14.006, 14.008] 15 P phosphorus

30.974 33 As arsenic 74.922 51 Sb antimony 121.76 83 Bi bismuth

208.98 8 O oxygen 15.999 [15.999, 16.000] 16 S sulfur 32.06 [32.059, 32.076] 34 Se selenium 78.971(8) 52 Te tellurium 127.60(3) 84 Po polonium 9 F fluorine 18.998 17 Cl chlorine 35.45 [35.446, 35.457] 35 Br bromine 79.904 [79.901, 79.907] 53 I iodine 126.90 85 At astatine 10 Ne neon 20.180 2 He helium 4.0026 18 Ar argon 39.95 [39.792, 39.963] 36 Kr krypton 83.798(2) 54 Xe xenon 131.29 86 Rn radon 22 Ti titanium 47.867 22 Ti titanium 47.867 40 Zr zirconium 91.224(2) 72 Hf hafnium 178.49(2) 104 Rf rutherfordium

23 V vanadium 50.942 41 Nb niobium 92.906 73 Ta tantalum 180.95 105 Db dubnium 24 Cr chromium 51.996 24 Cr chromium 51.996 42 Mo molybdenum 95.95 74 W tungsten 183.84 106 Sg seaborgium 25 Mn manganese 54.938 43 Tc technetium 75 Re rhenium 186.21 107 Bh bohrium 26 Fe iron 55.845(2) 44 Ru ruthenium 101.07(2) 76 Os osmium 190.23(3) 108 Hs hassium 27 Co cobalt 58.933 45 Rh rhodium 102.91 77 Ir iridium 192.22 109 Mt meitnerium 28 Ni nickel 58.693 46 Pd palladium 106.42 78 Pt platinum 195.08 110 Ds darmstadtium 29 Cu copper 63.546(3) 47 Ag silver 107.87 79 Au gold 196.97 30 Zn zinc 65.38(2) 48 Cd cadmium 112.41 80 Hg mercury 200.59 111 Rg roentgenium 112 Cn copernicium 114 Fl flerovium 113 Nh nihonium 115 Mc moscovium 117 Ts tennessine 118 Og oganesson 116 Lv livermorium 57 La lanthanum 138.91 58 Ce cerium 140.12 59 Pr praseodymium 140.91 60 Nd neodymium 144.2

Topic: Modern-Quantum Physics, Astrophysics Metodi: Photon Energy Relation, Bohr Model & Quantization, de Broglie Relation, Newton’s Law of Gravitation Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.12