Fonte: gare di altri paesi/India/chimica/INChO-2008.pdf · Apri PDF · apri PDF p.3
Cluster: Termodinamica
Soluzioni (stessa cartella): · · · · · · · · · · · · · ·
Problema 1
Problem 1
13 Marks Potential energy curves and molecular orbitals J. G. Dojahn, E. M. C. Chen and W.E. Wenthworth [J. Phys. Chem. 100, 9649 (1996)] examined the potential energy diagrams of X2 and X2 where X is a halogen. The diagram for F2 and F2 is reproduced on the next page. 1.1 From this diagram, the dissociation energies (in eV) of F2 and F2 are (Mark X in the correct box)
(1 mark) 1.2 Using this diagram, the electron affinities of the fluorine atom and F2 molecule can be calculated. Mark X in the correct box for the values (in eV) given below.
(2 marks) 1.3 What are the appropriate equilibrium internuclear distances for F2 and F2 ?
(choose from the following values : 1.2, 1.4 , 1.6, 1.7, 1.9 and 2.1 Å)
(1 mark) 1.4 The vibrational frequencies for F2 and F2 reported by Dojahn et al. on fitting the potential energy curves were 917 and 450 cm 1. What is the ratio of the corresponding force constants?
(2 marks) F2
1.1 1.3 1.5 1.6 1.7
F2 1.2 1.3 1.6 1.8 1.9
F atom 3.2 3.4 3.6 3.7 3.8
F2 molecule 2.8 3.0 3.2 3.3 3.4
F2:
F2 : Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
4 F + F 1 g + F + F U (eV) 3
2
1
0
-1
-2
-3
-4
-5 -6
1 2 u + 2 3 4 5 6 7 Internuclear distances, R (Å) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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The above observation regarding the force constants can be qualitatively explained on the basis of MO theory. 1.5 Fill in the appropriate number of electrons (as and ) in the figure below and label the MO‟s in the boxes provided.
Choose the labels from the list: 2p, 2p, 2p, 2p, 2s and 2s
(2 marks) 1.6 The bond orders in F2 and F2 are respectively
F2:
F2 :
(1 mark) 1.7 On the basis of qualitative MO theory, predict the paramagnetic species from the following list (Mark X in the correct box/es)
N2
F2
F2
O2 2
(2 marks)
F2 2s 2s 2p 2p F2 Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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1.8 The valence photoelectron spectrum of F2 was measured by A.W. Potts and W.C. Price with the X-ray photon of energy 21.2 eV. In this spectrum two broad peaks corresponding to the electronic kinetic energies of approximately 2.3 and 5.6 eV were seen. What are the corresponding ionization energies I1 and I2?
(1 mark) 1.9 These ionization potentials can be identified as the negative of the respective MO energies, viz I = . Write down the labels of MO‟s that correspond to I1 and I2 . (refer to 1.5)
I1 : MO label
I2 : MO label
(1 mark)
Topic: Modern-Quantum Physics Metodi: Photon Energy Relation, Physical Modeling Competenze: Diagrammatic Reasoning, Physical Reasoning Fonte: Testo (PDF) — p.3
Problema 2
Problem 2
19 Marks Unsaturated compounds Alkenes and alkynes are collectively referred to as unsaturated compounds, as they contain less hydrogen atoms compared to the corresponding alkanes. Alkenes are also called olefins, a term derived from olefiant gas, meaning oil forming gas. This term originated due to the oily appearance of alkene derivatives. Compared to alkenes, alkynes are not so common in nature, but some plants use alkynes to protect themselves against diseases or predators. 2.1 Give the IUPAC names with stereochemical descriptors (E/Z) for each of the
following compounds.
a)
b)
(2 marks) 2.2 The following compound A can be prepared by the addition of HBr to either of two alkenes B and C. Give the structures of B and C.
(1 mark)
C=C H CH2CH(CH3)2 OHC H C=C CH CH3 CH2CH2CH2C H Br A CH3 Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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and Br Br CH3 CH3 and OH CHO OH OH OH OH CHO OH HO HO C C C C C C Me H Me Me Me H H H and i) ii) iii) d) CH3-C=CHCH3 b) CH3-CH=C=CHCH3 c) CH2=C-CH2CH3 a) CH2=CHCH2CH=CH2 e) CH2=CH-CH=CHCH3 CH3 CH3 i) - 226 ii) -119 iii) -295 iv) -252 v) -113
2.3 Draw the energy profile diagram of the reaction of HBr with either B or C in question 2.2. Label the intermediate/s and transition state/s.
(2 marks) 2.4 Indicate the relationship between the compounds in each of the following pairs, as homomers (H), enantiomers (E) or diastereomers (D). Write the corresponding alphabet in the box.
(1.5 marks) 2.5 Heats of hydrogenation indicates the relative stability of alkenes. Match the list of compounds given in column A with the corresponding heats of hydrogenation listed in column B
Column A
Column B (kcal mol 1)
(2.5 marks) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
8 O O O O K +
2.6 Draw the structure of the product/s formed when the following compound is subjected to ozonolysis.
(1 mark) 2.7 An optically active alkyne D has 89.52% carbon. Compound D can be catalytically hydrogenated to n-butylcyclohexane. Treatment of D with C2H5MgBr liberates no gas. Catalytic hydrogenation of D over Pd/C in the presence of quinoline (a catalyst poison) and treatment of the product with ozone and then H2O2 gives an optically active tricarboxylic acid E (C8H12O6). Compound E on heating undergoes dehydration by loss of a molecule of water to give F. Draw the possible structures of D, E and F.
(3 marks) 2.8 When equimolar amounts of 1,3-butadiene and bromine are reacted at high temperature, two compounds G (major) and H (minor) with the formula C4H6Br2 are formed. Compound G reacts with more Br2 to form compound I (C4H6Br4) which proved to be a meso compound. Compound H reacts with more Br2 to form both I and a diastereomer J. Draw the structures of G and H. Draw the Fischer projection for I and J indicating the stereochemistry (with R/S designation) of the chiral carbons.
(4 marks) 2.9 Diels-Alder reaction of 2,5-dimethylfuran and maleic anhydride gives compound K which can exist in two stereoisomeric forms. Draw the structures of the two isomers.
(1mark) 2.10 Compound K undergoes acid catalysed dehydration to give L (C10H8O3). Draw the structure of L.
(1 mark) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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Topic: Conservation of Energy, Thermodynamics, Geometric Optics Metodi: Physical Modeling Competenze: Diagrammatic Reasoning, Physical Reasoning Fonte: Testo (PDF) — p.6
Problema 3
Problem 3
15 Marks Aromatic compounds In 1825, Michael Faraday isolated benzene for the first time from the oily mixture that condensed from illuminating gas (the fuel burnt in gas lights). Subsequently many compounds related to benzene were discovered. These compounds had typical odours (aroma) and hence this group of compounds was called aromatic. 3.1 As per the Hückel‟s rule of aromaticity, a cyclic compound is aromatic if it is conjugated, planar and has (4n+2) electrons, where, n is a positive integer, including zero. Similar compounds possessing (4n) electrons are highly unstable and are called antiaromatic compounds. It is interesting that compounds adjust their 3D and electronic structures to lower their energies. Answer the following by marking X in the correct box.
Yes
No
Aromatic
Antiaromatic Nonaromatic Aromatic
Antiaromatic Nonaromatic NH N .. .. Compound is ii) has dipole moment iv) Compound is i) The preferred structure of cyclooctatetrane is iii) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
10 O O (a) (b) CH OH H3C CH2
Acidic Basic Neutral
(vi) The compound having higher dipole moment is
(3 marks) 3.2 Alcohols dehydrate to alkenes, which is an acid catalyzed reaction. Arrange the
following alcohols according to increasing rate of dehydration.
( I )
( II )
( III )
(1 mark) 3.3 The local anesthetic proparacaine is synthesized by the following sequence of
reactions. Deduce the structure of the product formed at each step.
(5 marks) O H COOH SOCl2 HNO3 K base CH3CH2CH2Cl L M a) is v) O (C2H5)2NH N O + H2/Pd/C P M b)
CH O2N OH CH2 C OH MeO CH3 CH2 Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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A
D
3.4 Draw the structure of the major product of bromination of the following compounds using Br2/ FeBr3
(1.5 marks) 3.5 Normally benzene undergoes bromination only in the presence of a Lewis acid. However, bromination of benzene can also be brought about by addition of a small amount of pyridine (Q). Q catalyses the reaction by involving the lone pair of electrons on nitrogen. This is an example of „nucleophilic catalysis‟. Identify the missing intermediate/products A, B, C and D.
A
B
(4 marks) 3.6 In the above reaction (3.5) pyridine acts as a catalyst because it is (mark X in the correct box) i) more nucleophilic than benzene and is a poor leaving group. ii) less nucleophilic than benzene and is a poor leaving group. iii) less nucleophilic than benzene and is a good leaving group. iv) more nucleophilic than benzene and is a good leaving group.
(0.5 mark) O O NO2 OMe Br2 / FeBr3 Br2 / FeBr3
N Q Br2
B
C H+ Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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Topic: Thermodynamics, Electromagnetic Induction, Nuclear & Particle Physics Metodi: Physical Modeling Competenze: Diagrammatic Reasoning, Physical Reasoning Fonte: Testo (PDF) — p.9
Problema 4
Problem 4
15 Marks Reaction stoichiometry, kinetics and thermodynamics Nitrosyl chloride (NOCl), is a yellow gas that is most commonly encountered as a decomposition product of aqua regia. It is toxic and irritating to the lungs. On heating NOCl decomposes as 2NOCl 2NO + Cl2. The enthalpy change ) for the formation of 1 mole of Cl2 by the decomposition of NOCl is 75.3 kJ between 100 K to 600 K. The standard entropies (S0 298K) of different species are as given below: Substance NOCl NO Cl2 S0 298 J/K 264 211 223
4.1 Calculate Kp of the above decomposition reaction at 298 K.
(2 marks) 4.2 Calculate the temperature at which Kp will be double the value at 298 K.
(1 mark) 4.3 Calculate the temperature above which the reaction will become spontaneous. (1 mark) 4.4 A gaseous mixture of NO, Cl2 and NOCl with partial pressures (in bar) 1.5, 0.88 and 0.065 respectively was taken at 475 K. Deduce whether the net reaction will lead to increase in NOCl concentration.
(2.5 marks) 4.5 The initial rates of the reaction 2NO +Cl2 2NOCl at different initial concentrations of the reactants are given below. [NO(g)] (mol dm-3) [Cl2(g)] (mol dm-3) Initial Rate (mol dm-3 s-1) 0.250 0.250 1.43 x 10-6 0.250 0.500 2.86 x 10-6 0.500 0.500 11.4 x 10-6 Write the kinetic rate law.
(1 mark) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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4.6 The rate constant for the formation of NOCl at 400 K is 2.0 104 times the value at 300 K. Calculate the activation energy for the reaction.
(1.5 marks) 4.7 The following mechanisms (I and II) were proposed for the formation of NOCl.
Mechanism I NO + NO N2O2 (fast equilibrium)
N2O2 + Cl2 2NOCl (slow) Mechanism II NO + Cl2 NOCl2 (fast equilibrium)
NO + NOCl2 2NOCl (slow)
Deduce which mechanism/s is/are consistent with the rate law arrived at in 4.5. (3 marks) 4.8 The extent of a reaction is defined as (ni – ni 0)/ i where ni 0 and ni are the number of moles of reactant or product (species i) present respectively at the commencement of the reaction (t = 0) and at any time, t. i is the stoichiometric coefficient of the substance “i” in the balanced equation. For the calculation of the extent of reaction, the stoichiometric coefficient is considered to be positive for products and negative for reactants. The extent of the reaction is the same for all reactants and products. For the reaction, 2NO + Cl2 2NOCl, when 0.39 mol, 0.28 mol and 0.13 mol of NO, Cl2 and NOCl respectively were taken initially, after certain time 0.18 mol of Cl2 was found to remain in the reaction mixture. Calculate the extent of reaction with respect to NO and NOCl.
(1.5 marks) 4.9 When reactants are not taken in stoichiometric proportions, a reaction will go to completion when one of the reactants is completely consumed. This reagent is known as the limiting reagent which will have the lowest value for ni 0 / | i|. What is the limiting reagent in (4.8)? Calculate the extent of the reaction in (4.8), with respect to NOCl, when the reaction goes to completion. (1.5 marks) k1
k 1 k2
k1
k 1 k2 Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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Topic: Thermodynamics Metodi: First Law of Thermodynamics, Thermodynamic Cycle Analysis, Ideal Gas Law Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.12
Problema 5
Problem 5
14 Marks Phase equilibrium Phase diagram of a one-component system (S), is shown below. Answer the following questions with the help of this diagram.
5.1 What kind of phase change will take place if solid S is kept in the open under normal condition?
(Mark X in the correct box)
a) Sublimation
b) Melting
c) Evaporation
d) No change
(1 mark) 5.2 Under which condition, are all the three phases of the system S in equilibrium? (Mark X in the correct box). a) T > 304.2 K and P > 72.9 atmosphere
b) T = 216.8 K and P = 5.2 atmosphere
c) T > 304.2 K
d) P > 72.9 atmosphere
(1mark) 5.3 Temperature of the system at X1 is increased at constant pressure to reach X4. What are the phase/s of the system at the four different states X1 to X4? (2 marks)
194.7 216.8 298.15 304.2 T/K
A O C B
X1 X2
X3 X4 72.9
67.0
5.2 1.0
P /atm X5 Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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5.4 Show graphically the heating curve (Temperature vs. Time) for the process in the above problem (5.3) with appropriate labeling of physical states. (2 marks) 5.5 If pressure is increased, the melting temperature of solid S will
(Mark X in the correct box) a) Not change
b) Increase
c) Decrease
(1mark) 5.6 With the help of Clapeyron equation for phase change ( V d d
/ / ), find what happens to the volume of the system at X2 on heating? (1 mark) 5.7 What is the state of the system S, if it is heated in a sealed container above the critical temperature (304.2 K)?
(1mark) 5.8 What happens to the system S at X5, if pressure is changed slowly keeping temperature constant?
(Mark X in the correct box) a) At a higher pressure, the system will become liquid. b) At a lower pressure, the system will become liquid. c) The system will be in the same phase. (1 mark) Solutions 5.9 Raoult‟s law states that the partial pressure ) ( 1 p of a solvent over a solution is given by the vapor pressure ) ( 0 1 p of the pure solvent times the mole fraction ( 1 ) of the solvent. Derive an expression relating lowering of vapor pressure with solute mole fraction.
(1 mark) 5.10 The vapor pressure of water at 200C is 17.54 mm of Hg and that of 10% (w/w) solution of an organic solute at the same temperature is 16.93 mm of Hg. Calculate the molecular weight of the solute using the expression derived in 5.9. (1mark) 5.11 A solute is dissolved in two immiscible liquids in contact. Nernst‟s distribution law states that at equilibrium, the ratio of the concentration of the same molecular species Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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in the two phases is constant (KD) at constant temperature. Solute A is present as monomeric species A in solvent S1 and associates to form An in solvent S2. Species A and An are in equilibrium with an equilibrium constant K. a) What is the concentration of A in solvent S2, if concentration of An is C2?
(1 mark)
b) If the concentration of A in solvent S1 is C1, calculate the distribution constant KD of the solute. (1mark)
Topic: Thermodynamics Metodi: First Law of Thermodynamics, Ideal Gas Law Competenze: Mathematical Modeling, Diagrammatic Reasoning Fonte: Testo (PDF) — p.14
Problema 6
Problem 6
13 Marks Chemistry of Phosphorus and its compounds Elemental phosphorus is recovered from the minerals flurapatite and hydroxyapatite by carbon arc reduction. The resulting white phosphorus is a solid consisting of P4 molecules. 6.1 Draw the geometrical and Lewis structures of P4.
(1 mark) 6.2 White phosphorus reacts vigorously with aqueous sodium hydroxide resulting in the evolution of a gas and formation of sodium hypophosphite. Write the balanced equation for this reaction.
(1 mark) 6.3 Sodium hypophosphite and sodium phosphite contain oxoanions of phosphorus. a) Draw structures of these oxoanions. b) State whether these oxoanions will act as an oxidizing agent or a reducing agent. (Mark X in the correct box)
Oxidising agent Reducing agent hypophosphite
phosphite
c) Give appropriate reasons for your answers.
(3 marks) 6.4 H3PO4 can be synthesized by the reaction of hydroxyapatite (Ca5(PO4)3F) with H2SO4. Write balanced equation for this reaction.
(0.5 mark) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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Complete combustion of phosphorus yields phosphorus(V) oxide P4O10, that has a cage structure. 6.5 Draw the structure of P4O10.
(2 marks) 6.6 Calcium oxide reacts with P4O10 to form calcium phosphate. Calculate the amount of CaO required (in grams) to react with 426 g of P4O10.
(1 mark)
Halides of phosphorus play an important role in the synthesis of a variety of phosphorus compounds. 6.7 Write balanced equation/s for the formation of triethyl phosphate from phosphorus trichloride.
(1 mark) 6.8 Show the electronic configurations of P atom in i) ground and ii) excited states. State the hybridization of orbitals of P atom in PCl3 and PCl5 molecules. Predict the geometry of PCl5.
(2.5 marks) 6.9 PCl5 can react with NH4Cl to form cyclophosphazines (NPCl2)n where n 3. Write the balanced equation for the formation of cyclophosphazine with n = 3 and draw the structure of this product.
(1 mark)
Topic: Thermodynamics, Gravitation Metodi: Physical Modeling Competenze: Diagrammatic Reasoning, Physical Reasoning Fonte: Testo (PDF) — p.16
Problema 7
Problem 7
11 Marks Cobalt complexes A pink solid compound (A) has the formula CoCl3.5NH3.H2O. An aqueous solution of this salt, which is also pink, on titration with AgNO3 gives three moles of AgCl per mole of A. Solid A on heating above 1200C gives a purple solid (B) with the same ratio of NH3:Cl. Compound B on titration with AgNO3 gives two moles of AgCl per mole of B. 7.1 Write the electronic configuration of cobalt in compound A. (0.5 mark) 7.2 Write the molecular formulae of A and B with their IUPAC names.
(1.5 marks) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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7.3 Valence bond theory is useful in predicting shapes of complexes. Show the arrangement of the electrons for cobalt in the low spin purple complex B. Predict the hybridization and the shape of this complex.
(1.5 marks)
Although Valence bond theory is successful in predicting the shapes of complexes, it is unable to explain magnetic properties of coordination complexes. Crystal field theory (CFT) not only explains the magnetic properties but also accounts for color and spectra of these complexes. CFT is based on splitting of the d orbitals of the central metal in presence of ligands. 7.4 a) Using CFT draw the d orbital energy level diagram for the purple complex B. Label the energy levels and show the electron distribution.
(1 mark)
b) Complex B is
(Mark X in the correct box) paramagnetic
diamagnetic
(0.5 mark)
7.5 Using CFT, show the arrangement of the electrons on the central metal atom in the complex ion [Co(NH3)6]2+. State with justification, whether [Co(NH3)6]2+ is readily oxidized or not.
(1.5 marks) 7.6 For the complex [Co(NH3)3Cl3], draw the structures of possible stereoisomers and label them with stereochemical descriptors.
(2 marks) 7.7 A set of equivalent protons in a molecule gives one signal in the 1H NMR spectrum. The number of signal/s for the isomers drawn in 7.6 will be (1 mark) 7.8 Draw the structures of the following complexes and predict which one will be chiral. a) cis [CoCl2(ox)2]3
b) trans [CoCl2(ox)2]3 where ox = oxalate ligand (represent the same as o o in the answer).
(1.5 marks) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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Topic: Conservation of Energy, Thermodynamics, Magnetism Metodi: Physical Modeling Competenze: Diagrammatic Reasoning, Physical Reasoning Fonte: Testo (PDF) — p.17
Problema 8
Problem 8
10 Marks Electrochemical cells A secondary cell is one that can be recharged. Lead battery or the common car battery is an example of a secondary cell. Similarly, a battery using an alkali is known as the Edison cell. The Edison cell is represented as follows.
Fe(s) | FeO(s) , 20% KOH, Ni2O3 (s), NiO(s) | Ni (s) 8.1 Write the half cell reactions and overall cell reaction for the above cell.
(1 mark) 8.2 If a student adds distilled water to the cell so that the concentration of KOH reduces to half of its original value, E cell will (Mark X in the correct box)
(i) get doubled
(ii) be halved
(iii) remain unchanged
(1 mark) 8.3 A student was asked to set up the following cell. Fe(s) Fe3+ (aq) Ag+(aq) Ag (s)
c = 0.05M c = 0.1M
Calculate E0 for this cell using the following data; E0 Fe 3+ (aq) , Fe 2+ (aq) = 0.771 V ; E0 Fe 2+ (aq) / Fe(s) = 0.440 V; E0 Ag + (aq) / Ag(s)= 0.799 V Hence calculate Ecell at 298 K.
(4 marks) 8.4 In a hurry, the student interchanged the electrodes, i.e. he placed the silver electrode in the solution of ferric ions (left hand half cell) and iron electrode in the solution of silver ions (right hand half cell). Write the reactions that may occur in each of the half cells due to this interchange.
(1 mark) 8.5 Assuming that the reaction in the left hand half cell in 8.4 reaches equilibrium, calculate (i) equilibrium constant for the reaction. (ii) concentration of all the ions in the left hand half cell.
(3 marks) Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
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Topic: Oscillations & Waves Metodi: Physical Modeling Competenze: Mathematical Modeling, Physical Reasoning Fonte: Testo (PDF) — p.19
Problema 9
Problem 9
10 Marks Polyamides Several synthetic and naturally occurring polymers have amide linkage. Proteins which serve many functions in the biological systems are naturally occurring polyamides, made up of several amino acid residues. 9.1 Identify the product/s of complete hydrolysis of the compound shown below.
(0.5 mark)
9.2 ‘Quina‟ which has the following structure is a synthetic polymer that feels very much like silk.
i) Is Quina a nylon or a polyester? (0.5 mark) ii) Identify the monomers used to synthesize Quina.
(1 mark) 9.3 Classify the following peptides as acidic, basic or neutral and state whether the charge on the peptide will be positive, negative or zero at pH = 6.0 (Refer to the table provided on page no.39).
(3 marks)
Peptide Nature Charge Acidic Basic Neutral Positive Negative Zero Gly-Leu-Val
Leu-Trp-Lys-Gly-Lys
Arg-Ser-Val
N H NH O O HN CH2 NH C O (CH2)6 C O NH CH2 NH Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
21 H3N-CH-CONH-CH-CONH-CH-COO CH2 S S CH2C6H5 CH3 CH2C6H5 CH2 H3N-CH-CONH-CH-CONH-CH-COO CH3 + +
9.4 a) For the following peptide give the products of complete hydrolysis
(1.5 marks)
b) One of the above hydrolysis products is reduced in liver. Write the reaction showing the reduction.
(1 mark) 9.5 The artificial sweetener aspartame is a methyl ester of synthetic dipeptide Asp-Phe i) How many stereoisomers of aspartame are possible?
(0.5 mark)
ii) Draw the structure of aspartame. (1 mark) Electrophoresis is a method for separation of a mixture of amino acids. In this method, a sample of an amino acid mixture is placed at the centre of a piece of filter paper or a gel. The paper or the gel is then placed in a buffered solution between two electrodes and an electric field is applied. Depending on the net charge, an amino acid will move either to the anode or to the cathode. 9.6 A mixture of arginine (pI=10.76), alanine (pI=6.02) and aspartic acid (pI=2.98) is separated by electrophoresis at pH = 5 (pI is the isoelectric point). Identify the amino acids A, B and C in the chromatogram given below.
(+)
( )
(1 mark)
A
B
C Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
22 CH2CH2CH2NHCNH2 NH H CO2H H2N CH2SH H CO2H H2N CH2 H CO2H H2N N H N CH3 H CO2H H2N CH2COOH H CO2H H2N H H CO2H H2N CH2CH2CH2CH2NH2 H CO2H H2N H2N-CH-CH2-CH-CH3 COOH CH3 CH2 H CO2H H2N Table1. Structure and pKa values of a few -amino acids
-Amino acid Structure pKa value -amino group -carboxylic group Side chain Alanine (Ala)
9.7 2.3
Arginine (Arg)
9.0 2.2 12.5 Aspartic acid (Asp)
9.8 2.1 3.9 Cystenine (Cys)
10.8 1.8 8.3 Glutamic acid (Glu)
9.7 2.2 4.2 Glycine (Gly)
9.6 2.3
Histidine (His)
9.2 1.8 6.0 Leucine (Leu)
9.6 2.4
Lysine (Lys)
9.0 2.2 10.0 Phenylalanine (Phe)
9.1 1.8
H2N-C-CH2-CH2-COH COOH H O Indian National Chemistry Olympiad Theory 2008 HBCSE, 2nd February 2008
23 CH2OH H CO2H H2N CH H CO2H H2N CH3 CH3 CH2 H CO2H H2N NH Serine (Ser)
9.2 2.2
Tryptophan (Trp)
9.4 2.4
Valine (Val)
9.6 2.3
Metodi: Physical Modeling Competenze: Physical Reasoning Fonte: Testo (PDF) — p.20