Passage

SRI CHAITANYA IIT ACADEMY - VSPSEC : SUB : CHEMISTRY

Passage IHoffmann bromamide reaction involves conversion of a carboxylic acid amide into an amine with a loss of a carbon atom on treatment with aqueous sodium hypobromite. Thus Hoffmann results in shortening of a carbon chain

Mechanism of the reaction is :-

01.

(A) (B) (C) (D) 02. Which of the following will not give Hoffmann bromamide reaction.

(A) (B) (C) (D) 03.

(A) (B) (C) (D) NonePassage II

When a crystalline compound X is heated with and concentrated , a reddish brown gas A is evolved. On passing A into caustic soda, a yellow solution of B is formed. A yellow precipitate of C is

Sri Chaitanya Page 1 Vizag

obtained when a solution of B is neutralised with acetic acid and then treated with a lead acetate solution. When X is heated with NaOH, a colourless gas is evolved which, when passed into a solution of

, gives a reddish brown precipitate of D04. Compound (X) is :

(A) (B) (C) (D) 05. If the solution B is colourless, which of the following ions would not be present in the solid X?

(A) (B) (C) (D) 06. Which of the following is the composition of the brown precipitate?

(A) (B) (C) (D) Passage III

The apparatus consists of three temperature jacketed 1.000 L bulbs connected by stopcocks. Bulb A contains a mixture of H2O(g), CO2(g), and N2(g) at 25oC and a total pressure of 564 mm Hg. Bulb B is empty and is held at a temperature of -70oC. Bulb C is also empty and is held at a temperature of -190oC. The stopcocks are closed, and the volume of the lines connecting the bulbs is zero. CO2 sublimes at -78oC, and N2 boils at -196oC

07. The stopcock between A and B is opened, and the system is allowed to come to equilibrium. The pressure in A and B is now 219 mm Hg. What does bulb A contain?(A) CO2 (B) CO2 & N2 (C) N2 (D) CO2, N2 & Water

08. Both stopcocks are opened, and the system is again allowed to come to equilibrium. The pressure throughout the system is 33.5 mm Hg. What does bulb B contain?(A) CO2 (B) CO2 & N2 (C) N2 (D) CO2, N2 & Water

09. How many moles of CO2 are in the system?(A) 0.02 (B) 0.04 (C) 0.06 (D) 0.08

Passage IV


10. The compounds (P) and (R) are respectively

(A) (B) (C) (D) 11. The compound (T) is

(A) (B) (C) (D) Passage V

Consider the interconversion of nitrosotriacetoaminie into nitrogen, phorone and water.

The reaction is 1st order in each direction, with an equilibrium constant of 104. The activation energy for the backward reaction is 57.45 kJ/mol. Assume its Arrhenius preexponential factor as 1012 s-1

12. What is the expected forward constant at 300 K, if we initiate this reaction starting with only reactant?(A) 102 (B) 106 (C) 108 (D) 104

13. If the change in entropy of the reaction is 0.07 kJ K-1 mol-1 at 1 atm pressure. Calculate upto which temperature the forward reaction would not be spontaneous?(A) T < 285.7 K (B) T > 250 K (C) T < 340.2 K (D) T > 200 K

14. Calculate Kp of the reaction at 300 K(A) (B) (C) (D)

Passage VI

Compound (C) was prepared in a three step sequence from ethyl trifluoroacetate. The first step in a sequence involved treating ethyl trifluoroacetate with NH3 to give a compound (A), which on heating with (X) gives (B). (B) on treatment with an orangometallic (Y), followed by hydrolysis produces (C) . Based on above passage attempt the following questions:

15. Structure of (A), would be

(A) (B) (C) (D) 16. Structure of (B) , would be

(A) (B) (C) (D) 17. (X) , should be

(A) BaO2 (B) H2O2 (C) P4O10 (D) N2OPASSAGE VII

Diamond is a crystalline allotrope of carbon, which crystallizes in a 3-dimensional lattice. Each carbon forms four bonds with other carbon atoms. The unit cell of the diamond lattice can be thought of containing carbon atoms present at all CCP positions as well in alternate tetrahedral voids. Assuming contact between nearest atoms, answer the following

18. Number of carbon atoms per unit cell of diamond structure is(A) 4 (B) 8 (C) 12 (D) data insufficient

19. What is the fraction of length covered along the body diagonal of the cubic unit cell by carbon atoms?(A) 0.50 (B) 0.34 (C) 0.75 (D) 0.25


20. How much fraction of area of one face is covered by atoms?

(A) (B) (C) (D)

PASSAGE VIIIAn orange colored compound (X) on heating decomposes to form a solid and gases mixture which contains a gas (Y) and vapour. (Y) on treatment with gives (Z). (Z) on hydrolysis with hot water liberates a gas (P) which when added to results in the formation of black precipitate

21. The compound ‘X’ is

(A) (B) (C) (D)

22. The compound (Y) is(A) (B) (C) (D)

23. What would be the colour if solution of (P) is added in excess to solution(A) red (B) green (C) colour less (D) blue

PASSAGE-IX

24. major product

(A) (B)

(C) (D) 25. No. of possible structural isomers for F is

(A) 2 (B) 10 (C) 14 (D) 6

26. (Major product)

(A) (B)


(C) (D)

PASSAGE –XPinacol is a 1,2, diol which on treating with acid produces pinacalone(ketone). It is an intramolecular rearrangement. The reaction starts with the protonation of hydroxyl group followed by elimination of water and formation of carbocation. The carbocation is then stabilized by Whitmore1,2 shift

27.

(A) (B) (C) (D) 28. Which of the following a compounds, on pinacol-pinacalone rearrangement produces a compound

which gives a precrpitate with KOI ?

(A) (B) (C) (D) 29.


Which of the following is not correct about the compound X and Y ?(A) (X) is can be reduced to 1-phenylethane on treating with

(B) (X) on treating with followed by treating with PCl5 produces two amides

(C) (Y)- may be obtained by the crossed reductive coupling between acetone and benzophenone (D) (X) gives yellow precipitate with NaOI

PASSAGE –XI

30. V and P can be(A) (B) (C) (D)

31. Q and S can be

(A) (B) (C) (D)

32. R, u, T can be

(A) (B)

(C) (D) None

PASSAGE – XIITo find the reaction order with respect to each reactant, one should look at the exponents in the rate law, not the coefficients in the balanced chemical equation. And one method of determining the values of the exponents in a rate law is to carry out a series of experiments in which the initial rate of reaction is measured as function of different sets of initial concentrations. Consider for example the oxidation of iodide ion by hydrogen peroxide in an acidic solution

The rate of formation of the red-coloured trioxide ion can be determined by measuring the rate

of appearance of the colour. Following are initial rate data at250C.

Experiment Initial Initial Initial rate of

formation of1 0.100 0.100 2 0.100 0.2003 0.200 0.1004 0.200 0.200

33. The rate law for the formation of is

(A) Rate = (B) Rate =

(C) Rate = (D) Rate = Sri Chaitanya Page 6 Vizag

34. The value of the rate constant is

(A) (B) (C) (D)

35. What is the rate of formation of when the concentrations are , and ?

(A) (B) (C) (D) PASSAGE - XIII

36. The molecule in sequential reaction is

(A) (B)

(C) (D)

37.

(A) (B)

(C) (D)

38. In the sequential reaction, if instead of is used. Product will be

(A) (B) (C) (D) PASSAGE –XIV

When two forms of allotropes exist in dynamic equlibirium at a particular temperature, it is called enantiotropy. Monotropy is an allotropy in which one allotrope is stable under normal conditions, the other allotrope being less stable for ex:- O2 is more stable than O3. The allotropy of sulphur is much more complicated than for other elements except for carbon : e.g. liquid sulphur exists in two forms. The pale yellow mobile form and dark viscous form are in equilibrium with each other. The rapid and reversible gelation of liquid sulphur was observed in the temperature range . Variety of crystalline and amorphous forms of sulphur result by adopting suitable processes

39. Which of the following is cause of allotropy?i) It is due to difference in crystalline structureii) Different atomicity iii) It may be due to difference in nuclear spin(A) (B) (C) (D)

40. When sulphur is heated at453 K, if forms(A) Yellow mobile liquid (B) Amber coloured viscous liquid(C) Colourless mobile liquid (D) Reddish vapours

41. When molten sulphur is poured in cold water, the allotrope of sulphur formed is(A) Rhombic (B) Monoclinic (C) Plastic (D) colloidal

PASSAGE XV(E) ,a compound [(Z) – 6 - Heneicosen-11-one ] has been synthesized using following reaction sequence


The molecular formula of (C) is and is same for compound (E) too42. The structure of compound (B) is

(A) (B)

(C) (D) 43. The product (C) in the reaction sequence will be

(A)

(B)

(C)

(D) 44. The correct structure of compound (E) will be

(A)

(B)

(C)

(D) PASSAGE- XVI

For a solution of a salt at a specified concentration the product of the concentration of the ions, each

raised to the proper power is called the ionic product . Thus, for a saturated solution in equilibrium

with excess solid, the ionic product is equal to its solubility product . If the ionic product of the

solution is less than the corresponding solubility product, it means that the solution is unsaturated


45. The concentration of ions in a given saturated solution of ion per litre.

The solubility product of is(A) (B) (C) (D)

46. . What happens when solutions of 300 ml of and 500 ml of

are mixed ?(A) Precipitate will form (B) Precipitate will not form(C) Precipitation starts (D) Precipitation cannot be predicted

47. If is taken as . What will be the solubility in 0.1M ?

(A) (B) (C) (D) PASSAGE-XVII

48. M , S , T are

(A) (B) (C) (D)

49. N , P , R are(A) (B)

(C) (D) 50. O , U , V are

(A) (B) (C) (D) PASSAGE –XIX

Study the following road map and answer question

51. The compound (X), (the major product) is

(A) (B) (C) (D) 52. The reagent Y can be

(A) Baeyers Reagent (B) Lucas Reagent Sri Chaitanya Page 9 Vizag

(C) (D) 2,4-Dinitrophenyl hydrazine

53. The product Z is :

(A) (B) (C) (D) PASSAGE –XX

A chemist opened a cupboard to find four bottles containing water solutions, each of which had lost its label. Bottles 1, 2, 3 contained colourless solutions, while bottle 4 contained a blue solution. The labels from the bottles were lying scattered in the floor of the cupboard. They were :Copper (II) sulphate, Hydrochloric acid

Lead nitrate, Sodium carbonateBy mixing samples of the contents of the bottles, in pairs, the chemist made the following observations. Bottle 1 + Bottle 2 White precipitate Bottle 1 + Bottle 3 White precipitate Bottle 1 + Bottle 4 White precipitate Bottle 2 + Bottle 3 Colourless gas evolved Bottle 2 + Bottle 4 No visible reaction Bottle 3 + Bottle 4 Blue precipitate

54. Bottle 3 contains(A) copper (II) sulphate (B) hydrochloric acid (C) lead nitrate (D) sodium carbonate

55. When bottle 1 is mixed with bottle 4, white precipitate is observed, which is(A) PbSO4 (B) PbCO3 (C) PbCl2 (D) Pb(NO3)2

56. Which of the bottle will give distinctive deep blue colour with NH3?(A) Bottle 1 (B) Bottle 2 (C) Bottle 3 (D) Bottle 4

PASSAGE XXIConsider the following partially labelled figure for an ideal binary solution of benzene and toluene and answer the following questions.

57. What is (25oC) ?(A) 30 mm Hg (B) 100 mm Hg (C) 20 mm Hg (D) 40 mm Hg

58. The composition of liquid for which the vapour contains equal number of benzene and toluene molecules is nearly(A) (B) (C) (D)

PASSAGE XXIIWhen a liquid is completely miscible with another liquid, a homogeneous solution consisting of a single phase is formed. If such a solution is placed in a close evacuated vessel, the total pressure exerted by vapour, after the system attained equilibrium will be equal to the sum of partial pressures of the constituents. A solution is said to be ideal if its constituents follow Raoult’s law under all conditions of concentrations i..e, the partial pressures of each and every constituent is given by . Where pi is


the partial pressures of the constituent i, whose mole fraction in the solution is x i and is the corresponding vapour pressure of the pure constituent. The changes in the thermodynamic functions when an ideal solution is formed by mixing pure components is given by the following expression.

(1)

where ntotal is the total moles of all the constituents present in the solution.

(2)

(3)

(4)Since both the components of an ideal binary liquid system follow Raoult’s law of the entire range of the composition, the partial pressure exerted by the vapours of these constituents over the solution will be given by

(5) (6)

where xA and xB are the mole fractions of the two constituents in the liquid phase and and are the respective vapour pressures of the pure constituents. The total pressure (p) over the solution will be the sum of the partial pressures. The composition of the vapour phase (yA) can be determined with the help of Dalton’s law of partial pressures

59. For an ideal solution in which , the plot of total pressure (p) versus the mole fraction of A at constant temperature in the vapour phase is

(A) (B) (C) (D)

60. A plot of reciprocal of total pressure (1/p) (y-axis) versus yA (x-axis) gives

(A) a linear plot with slope = (B) a linear plot with slope =

(C) a linear plot with slope = (D) a linear plot with slope =

61. Two liquids A and B form an ideal solution at temperature T. When the total vapour pressure above the solution is 600 torr, the mole fraction of A in the vapour phase is 0.35 and in the liquid phase 0.70. The vapour pressures of pure B and A are(A) 800 torr; 1300 torr (B) 1300 torr, 300 torr (C) 300 torr, 1300 torr (D) 300 torr, 800 torr

PASSAGE XXIII

Assume that there were four possible values for the spin quantum number ms. Principal

quantum number n is defined as usual. However, quantum number l and m, are defined as follows:l : 1 to (n + 1) in integral steps ml : - l/2 to +l/2 (including zero, if any) in integral steps The orbitals corresponding to l = 1, 2, 3, … designated as A, B, C .. respectively

62. The number of elements that would be present in the second period of the periodic table is(A) 9 (B) 20 (C) 24 (D) 36

63. If Aufbau’s principle is not violated i.e. (n + l) rule must be followed, the outermost electronic configuration of an element with atomic number 100 would be


(A) 3B84A4 (B) 3C164A8 (C) 3C124B8 (D) 4B125A8

64. The number of sub-orbitals and the maximum number of electrons that can be filled in E-orbitals are respectively(A) 6, 24 (B) 5, 20 (C) 7, 28 (D) cannot be determined

PASSAGE XXIVA liquid (A) (mol.wt 120)has a normal boiling point of .When a 0.5g of solute (Mol.wt 130)is dissolved in 20g of A, it shows an elevation in boining point of 0.50 K. Answer the following

65. The vapour pressure of the solution at is nearly(A) 750 mm (B) 740 mm (C) 730 mm (D) 720 mm

66. The molal elevation constant of liquid (A) is nearly

(A) 1.6 (B) 2.6 (C) 3.6 (D) 4.667. If liquid (A) is to be distilled at without decomposition, what maximum pressure is to be maintained

in the flask nearly ?(A) 340 mm (B) 470 mm (C) 520 mm (D) 640 mm

PASSAGE XXVA compound (A) given below is kept in acidic medium for sometime. It was observed that (A) hydrolyses into (B) and (C). The properties of (A), (B) & (C) are tested with different reagents and tabulated

68. (B) formed is

(A) (B) (C) (D) Both A & B69. Compound (C ) is :

(A) (B) (C) (D)

70. (B) is formed due to(A) Hydrolysis followed by cleavage of C O bond and followed by esterification between and

group(B) Decarboxylation of group(C) Oxidation of into (D) Both A and B

PASSAGE XXVI


A pleasant smelling optically active ester (A) has molecular weight 186. It does not react with in . Hydrolysis of (A) gives two optically active compounds B and C. compound (C) gives positive iodoform

test and on warming with conc. gives D (saytzeff product) with no geometrical isomers. (C) on treatment with benzene sulphonyl chloride gives (E) which on treatment with NaBr gives optically active

F. When slat of (B) is treated with racemic mixture of F is formed71. Structure ofcompound B

(A) (B)

(C) (D) 72. Structure of compound C

(A) (B) (C) (D) 73. Structure of A is

(A) (B)

(C) (D) PASSAGE XXVII

Epoxides (or oxiranes) are three membered cyclic ethers and differ from other cyclic and acyclic ethers in that they are reactive to various reagents. The reason for this reactivity is the strained three membered ring. Reactions with nucleophiles can result in ring opening and relief of strain. Nucleophiles will attack either of the electrophilic carbons present in an epoxide by an SN2 reaction

74.

A and B are

(A) (B)

(C) (D)

75. The reaction of with RMgX followed by hydrolysis leads to the formation of


(A) (B) (C) (D) PASSAGE XXVIII

Hoffmann bromamide reaction involves conversion of a carboxylic acid amide into an amine with a loss of a carbon atom on treatment with aqueous sodium hypobromite. Thus Hoffmann results in shortening of a carbon chain

Mechanism of the reaction is :-

76.

(A) (B) (C) (D) 77. Which of the following will not give Hoffmann bromamide reaction

(A) (B) (C) (D) 78.

(A) (B) (C) (D) NonePASSAGE XXIX

Dicarboxylic acids contain two carboxylic acid groups. Acidity of carboxylic acid depends upon stability of intermediate formed. The hydrogen present in carboxylic acid is labile. Dicarboxylic acids on decarboxylation form monocarboxylic acids, alkanes and cyclic ketones depending upon conditions. Greater the symmetry, higher will be the melting point

79.


(A) o-methyl acetophenone (B)

(C) (D) 80. Which of the following statements are correct :

i) bromo acids on decarboxylation give unsaturated hydrocarbons.

ii) of fumaric acid is lower than, of maleic acid

iii) on reaction with gives corresponding anhydride

iv) on heating with gives a product which contain equal no. and bonds(A) i,ii, iii, iv (B) ii,iii, iv only (C) i , ii, only (D) i,iii, iv only

81. Pick up the correct statement from the followingi) On electrolysis of aqueous sodium formate liberates hydrogen gas both at cathode and anodeii) Aqueous sodium salt of butane dioic acid on electrolysis gives Ethylene iii) Aqueous sodium salt of Fumaric acid on electrolysis gives AcetyleneAqueous sodium salt of 1,6 hexane dioic acid on electrolysis give Cyclo butane as major product(A) i, ii, iii (B) ii, iii, iv (C) Only iii , iv (D) All are correct

Passage A frost diagram of an element X is a plot of for the couple X(N)/X(0) against the oxidation number, N of the element : The most stable oxidation state of an element in aqueous solution corresponds to the species that lies lowest in its frost diagram. Frost diagrams can be used to gauge the inherent stabilities of different oxidation states of an element and to decide whether particular species are oxidizing (or) reducing agents. The oxidizing agent in the couple with the more positive slope [the more positive ] is liable to undergo reduction. The reducing agent of the couple with the less positive slope [the most negative ] is liable to undergo oxidation. A species in a frost diagram is unstable with respect to disproportionation if its point lies above the line connecting the two adjacent species. The species will tend to comproportionate into an intermediate species that lies below the straight line joining the terminal species. Following is the frost diagram of different nitrogen compounds


82. Choose the correct statement(A) can be disproportionated into and NO

(B) can be disproportionated into NO and

(C) can not be disproportionated into

(D) can be disproportionated into ion and

83. Which of the following is the correct statement(A) Ammonium nitrate can undergo disproportionation reaction in acidic medium(B) Ammonium nitrate disporportionation has negative free energy change in acidic medium (C) Ammonium nitrate can undergo comproportionate reaction in acidic medium

(D) Ammonium nitrate can not undergo comproportionate reaction in acidic medium84. Choose the correct statement

(A) potassium nitrite can undergo disproportionation into NO and in basic medium

(B) potassium nitrite can undergo disproportionation into NO and in acidic medium completely

(C) potassium nitrite can undergo disporportionation into NO and in acidic medium incompletely (D) none of the above statements is correctPassage

Water can be thought of as hydrogen oxidized by oxygen. Thus hydrogen can be recovered by reduction of water, using an aqueous solution of sodium sulphate, at a platinum electrode connected to the negative terminal of a battery. The solution near the electrode becomes basic. Water can also be thought of as oxygen reduced by hydrogen. Thus, oxygen can be recovered by oxidation of water at the ‘Pt’ electrode connected to the positive terminal.When copper is used at both electrodes gas is generated only at one electrode during the initial stage of electrolysis

Another species in solution that can be reduced is sodium ion. The reduction of sodium ion to metallic sodium does not occur in aqueous solution, because water is reduced first.The electrode potential is affected by other reactions taking place around the electrode. The potential of the electrode in a 0.100 M solution changes as precipitates.

Precipitation of begins at [temperature = 25°C, ]

85. Based on the above observations, connect the following half reactions with the standard reduction potentials (in volts)


a) reduction of copper ion i) +0.34b) reduction of oxygen ii) -2.710c) reduction of water iii) -0.83d) reduction of sodium ion (Na+) iv) 0.000e) reduction of hydrogen ion v) +1.230(A) a-i, b-ii, c-iii, d-v, e-iv (B) a-i, b-v, c-ii, d-iii, e-iv(C) a-I, b-iii, c-ii, d-v, e-iv (D) a-I, b-v, c-iii, d-ii, e-iv

86. Standard Reduction potential for

(A) -0.431 V (B) -0.331V (C) -0.231 V (D) -0.131V87. Reduction potential of copper electrode at pH = 1

(A) +0.51 V (B) +0.41V (C) +0.21 V (D) +0.31VPassage

88. How many aromatic rings are there in F(A) 4 (B) 5 (C) 6 (D) 7

89. From the following correct statement/s arei) ‘A’ is aromaticii) Formation of B involved hydrolysis and addition – elimination reaction.iii) ‘B’ possess four intramolecular hydrogen bonds.iv) ‘B’ possess two intramolecular hydrogen bonds(A) i, ii, iv (B) i, ii, iii (C) ii, iv (D) i, iii

90. Choose correct statementsi) formation of ‘D’ involved attack ii) formation of ‘E’ involved claisen ester condensation and William son’s synthesis.iii) formation of ‘E’ involved claisen ester condensation but not William son’s synthesis.iv) ‘F’ can give test with Tollen’s reagent(A) i, iii (B) ii, iv (C) i, ii (D) i, ii, iv

Passage

The formation of the oxide per mole of consumed is written as

The free energy change of the reaction is given by .

If the enthalpy change and entropy change are assumed to be independent of temperature, the plot of versus T is linear such as shown in the following figure known as Ellingham diagram. Figure is widely used to discuss the principle involved in the extraction of metal from its oxide


The reduction of by carbon at may be represented as

The free energy of this reaction can be obtained from the following

two reactions.

91. Choose the correct statement(A) The slope of linear plot of is positive for with the exception when ‘M’ is carbon

(B) The slope of linear plot of is negative for (C) The slope of linear plot of is positive for CO

(D) The slope of plots of for the various oxides formed from are negative

92. Choose correct statements from the following statementsi) The slope of plot becomes more positive when the oxide is formed from M(g)ii) At , the stable oxide of ‘C’ is where as at , the stable oxide is CO.iii) Mercury (II) oxide decomposes spontaneously to its elements by heating alone.iv) Reduction of ZnO by CO starts at (A) i, ii, iii, iv (B) i, ii, iv (C) ii, iii, iv (D) i, ii, iii

93. Choose correct statements from the followingi) For the reduction of by ‘C’ [to be spontaneous] ii) decreasing ease of reduction of metal oxides by carbon is .iii) decreasing ease of reduction of metal oxides by carbon is .iv) Reduction of by Al is possible at all temperatures(A) i, iv (B) ii, iii (C) ii, iv (D) iii, iv

Passage :


94. Which of the following are correct statementsi) Second step in the formation of ‘A’, involved addition reaction only.ii) Ist step in the formation of ‘B’ is a second order reaction.iii) Formation of ‘C’ is a zero order reaction.iv) Formation of ‘C’ is involved a tetrahedral transition state and formation of ‘C’ is due to nucleophilic addition and elimination(A) i, ii (B) ii, iv (C) ii, iii (D) i, iv

95. Choose correct statements w.r.t formation of D from C and formation of H from G(A) formation of ‘D’ from C is involved nucleophilic addition and internal nucleophilic substitution and formation of H from G is Aldol condensation(B) formation of ‘D’ is involved nucleophilic addition only and formation of ‘H’ from ‘G’ is involved internal Claisen ester condensation

(C) formation of ‘D’ from ‘C’ is involved nucleophilic addition and internal Williamson’s synthesis and formation of H from G is internal Claisen ester condensation (D) formation of ‘D’ from ‘C’ is involved nucleophilic substitution only and formation of ‘H’ from ‘G’ is Dieckmann’s reaction

96. Choose incorrect statements(A) ‘M’ is thermodynamically more stable (B) ‘L’ is thermodynamically less stable(C) second step in the formation of ‘I’ from ‘H’ is involved a cyclic transition state

(D) ‘J’ is possessing two chiral centresPassage

Consider the follows reaction

The free energy of the reaction occurring at 298 k and 1atm has been plotted against the fraction of dissociated as shown adjacent

97. When two moles of NO2 change in to equilibrium mixture with the is(A) (B) (C) (D)

98. When one mole of change into equilibrium mixture with , The isSri Chaitanya Page 19 Vizag

(A) (B) (C) (D) 6.24 kJ

99. Mark out the correct statements(A) The conversion of into 2NO2 is spontaneous

(B) The conversion of into is non-spontaneous

(C) The attainment of equilibrium from and from both are equally spontaneous

(D) The attainment of equilibrium from2 moles of with is more spontaneous than the

conversion of into Passage

100. Organic compound ‘A’ is

101. The resolvable organic compound ‘C’ is

102. The resolvable organic compound, G is

Passage : Compound X is a trisaccharide which occurs principally in cottonseed meal. Compound X does not react with Bendict’s or Fehling’s solutions nor does it mutarotate. Acid catalysed hydrolysis gives three different D-hexoses, A, B and C. Compounds A and B as well as compound given below(figure 1), all give the same osazone upon reaction with excess acidic phenylhydrazine.


Compound C reacts with nitric acid to give optically inactive compound D. The Kiliani Fischer approach is used to establish the configurational relationship between D-glyceraldehyde and C. The intermediate aldotetrose which leads to C does not give a meso compound when oxidized by nitric acid. When A is treated with nitric acid, the dicarboxylic acid ( aldaric acid ) produced is optically active. Both A and B react with 5 moles of . One mole of A gives 5 moles methanoic acid and one mole of methanal while one mole of B gives 3 moles of methanoic acid., 2 moles of methanal and one mole of carbon dioxide. Both A and B are related to the same aldotetrose which is the diastereomer of the one to which C is related. Methylation of X followed by hydrolysis gives a 2, 3, 4-tri-O- methyl-D-hexose (E) ( derived from A),a 1,3,4,6,-tetra-O-methyl-D-hexose(F)(derived from B) , and a 2,3,4,6,-tetra-o-methyl-D-hexose (G) ( derived from C)

103. Observe the following molecules

Choose the correct match :(A) (B) (C) (D)

104. How many moles of acetic anhydride are consumed to make A completely acetylated?(A) 3 (B) 4 (C) 5 (D) 6

105. The trisaccharide ‘x’ is

(A) (B)


(C) (D) PASSAGE

An efflorescent salt is one that loses water to the atmosphere. This will occur if the water vapour pressure in equilibrium with the salt is greater than the water vapour pressure in the atmosphere. For example , could be efflorescent as it loses 2 molecules of and simultaneously forms 1 formula unit of .

If partial pressure of ( let it be x torr) is less than the vapour pressure of water at the same temperature it doesn’t mean will always effloresce. Only in that condition, when the partial pressure of moisture in the air is less than ‘x’. In other words, this will occur when the relative

humidity is less than ( The vapour pressure of water at 250C is 23.8 Torr)

Conversely, it can be said that CuSO4. 3H2O can reduce the moisture content of any confined volume of gas to ‘ ’ torr

106.

For the equilibrium, at is atm2. Then which of the following is the appropriate

relative humidity at which will effloresce essentially?(A) (B) (C) (D)

107. ;PH2O=

For the equilibrium, If the vapour pressure of water at 00C is 4.58 Torr, at which

relative humidities will be deliquescent (absorb moisture) when exposed to the air at 00C essentially?(A) (B) (C) (D)

108. Equilibrium constants are given for the following reactions at 00C ;

Which one is most effective drying agent at 00C ( The vapour pressure of at 00C is 4.58 Torr)?

(A) (B) (C) (D) All are equally effectivePASSAGE

A chemist opened a cubboard to find four bottles containing water solutions, each of which had lost its label. Bottle 1, 2 and 3 contained colourless solutions, while bottle 4 contained a blue solution. The labels from the bottles were lying scattered on the floor of the cupboard. They were:

Coper (II) sulphate Hydrochloric acidLead nitrate Sodium carbonate

By mixing samples of the contents of the bottles, in pairs, the chemist made the following observation:


109. Bottle 3 contains(A) copper (II) sulphate (B) hydrochloric acid(C) lead nitrate (D) sodium carbonate

110. When bottle 1 is mixed with bottle 4, white precipitate is observed, which is(A) (B) (C) (D)

111. Which of the following bottle will give distinctive colour with ?(A) Bottle 1 (B) Bottle 2 (C) Bottle 3 (D) Bottle 4

Passage Ideal gas equation is represented as PV=nRT. Gases present in universe were found to exhibit ideal or nearly ideal behavior in the Boyle temperature range only. They deviate from ideal gas behavior at high pressure and low temperature. The deviations are explained in terms of compressibility factor Z. For ideal behavior Z= PV/ n R T =1. The main cause to show deviations were due to wrong assumptions made about forces of attraction and volume occupied by the molecules. V in PV = n R T is supposed to be volume of gas or the volume of the container in which the gas is placed assuming that gaseous molecules do not have appreciable volume. Actually volume of the gas is that volume in which a gas molecule can move freely. If volume occupied gas molecules is not negligible, then the term V should be replaced by an ideal volume which is available for free motion of each moleculeof the i.e for 1 mole of gas Vactual = volume of the container – volume occupied by the molecules=V-bWhere b represents the effective volume or Co-volume or excluded volume occupied by the molecules present in one mole of gas. Vactual = V-nb ( for n-moles of gas). The excluded volume can be calculated by considering bimolecular collisions. The excluded volume is the volume occupied by the sphere of 2r for each pair of molecule=(4/c) π(2r)3 = 4x8 πr3 /3The excluded volume for a single molecules is (2/c)xx8 πr3.= 4x(4/c) πr3= 4V.Therefore the excluded volume for N molecules is =4NV=4b (NV=b) Tc, Pc and Vc are critical constants of a gas. Pc= a/27b2 a is van der waal’s constant, which is elated to attractive forces. Tc= 8a/27Rb Vc=3b

112. Which of the following statements are correct?1) Larger the value of Tc / Pc for a gas larger would be its excluded volume2) Larger is the excluded volume of gas , more will be its critical volume(Vc)3) The slope for an isochore obtained for a gas showing p(V-b) =RT is (R/V-b)4) The excluded volume for He is more than hydrogen(A) 1,2,3 (B) 1,2,4 (C) 2,3,4 (D) 3,4

113. As the pressure approaches zero i.e at very low pressures, the curves plotted between Z and P for n moles of gases have the following characteristics1) The intercept on y-axis lead to the value of unity2) The intercept on y-axis lead to a value of ‘n’3) The curves possess same slope for different gases at the same temperature4) The curves possess different slopes for different gases at the same temperature


5) The curves possess same slope for a gas at different temperature(A) 1, 4,5 (B) 2,3 (C) 2,3,4 (D) 2,3,5

114. The ratio of coefficient of thermal expansion α=(dV/dT)p/V and the isothermal compressibility K= (dV/dP)T /v for an ideal gas is(A) –P/T (B) P/T (C) T/P (D) –T/P

Passage Primary amines react with nitrous acid to form diazonium cation. It is an unstable specires and converts into carbo cation

115. Which of the following will be least stable ? (A) CH3-CH2 -+N N (B) C6H5-+N N(C) OCH30-C6H4--+N N (p) (D) O2N-C6H4--+N N (p)

116. In the given reaction : Cyclobutylaminomethane + NaNO2+ HCl product(A) cyclobutylmethanol (B) cyclopentanol(C) chlorocyclopentane (D) All these

117. Which daiazonium cation is most reactive for coupling reaction(A) C6H5-N2

+ (B) CH3 –C6H4-N2+ (p)

(C) O2N-C6H4-N2+ (p) (D) CH3-O-C6H4-N2

+(p)PASSAGE :

Dissociation of weak of electrolyte is expressed in terms of Ostwald dilution law. An acid is substances that furnishes a proton or an electron pair acceptor. Where as a proton acceptor or electron pair donoris a base. Strong acid has weak conjugate base. The dissociation constants of an acid (Ka) and it s conjugate base (Kb) are related by Kw=Kax Kb. pH +pOH =14.Buffer solution don’t show appreciable change in when a few drops of acid or base is added.Answer the following questions:

118. Which of the following statements are correct?1) pH of 10-10 molar NaOH is nearly 72) the degree of dissociation is given by 1/ 1+ 10pka-Ph

3) For weak electrolytes of polyprotic acid nature having no other electrolyte , the anion concentration produced in second step of dissociation is always equal to K2 at reasonable concentration of acid4) The concentration of amide ions produced during self ionization of ammonia is equal to concentration of ammonium ions5) Ostwald dilution law is valid for strong electrolytes(A) 1,2,3 and 5 are correct (B) 1,2,3,and 4 are correct(C) 1,3,4,and 5 are correct (D) 2,3,4,and 5 are correct

119. Which of the following statements are true ?1) Perchlorate ion is a weaker base than chlorate ion2) The degree of dissociation of water is 1.8 x10-9

3) the equilibrium constant for dissociation of water is 1.78 x10-16

4) phosphate ion is conjugate acid of monhydrogenphosphate ion(A) 1,2,and 3 are correct (B) 2, 3 and 4 are correct

(C) 1, 2 and 4 are correct (D) 1 and 2 are correct120. 0.16gm of hydrazine Kb=4x10-6 are dissolved in water and the total volume of solution is made

upto 500ml. The percentage of hydrazine that reacts with water is(A) 2% (B) 3% (C) 4% (D) 1%


PASSAGE :

E2 reaction is bimolecular reaction. E2 reaction is of two types . Hoffmann elimination end Saytzeff elimination

Answer the following questions based on the above mechanism121. In E2 reaction formation of Hoff amnn product takes place by formation of which intermediate

(TS) ? (A) I (B) II (C) III (D) I and III

122. Which one of these compounds will give Saytzeff product in E2 reaction?(A) CH3-CH2-CH(Br)-CH3 (B) CH3-CH2-CH(F)-CH2- CH3

(C) CH3-CH2-CH2 –N+(CHc)3OH- (D) CH3-CH2-CH(OCOCHc)-CH3

123. In E2 reaction Ts has character(A) alkene (B) Carbonium ion (C) carbanion (D) any othese

PASSAGE :An aqueous solution of salt (A) gives white crystalline precipitate (B) with NaCl solution. The filtrate gives a black ppt. (C) when H2S is passed in it. Compound (B) is dissolved in hot

water and the solution gives a yellow ppt. (D) on treating with NaI and cooling. The compound (A) does not give any gas with dil. HCl but liberated reddish brown gas on heating124. The compound (A) in the above passage is

(A) Pb(NO3)2 (B) PbS (C) Cu(NO3)2 (D) NaCl125. The compound (C) in the above passage is

(A) NaI (B) PbS (C) PbO (D) PbCl2126. The compound (D) is

(A) PbI2 (B) PbO (C) PbS (D) NO2

PASSAGE All the four colligative properties of solutions depend solely on the total number of solute particles present in solution. Various electrolytes ionizes and yield more than one particle

per unit in solution. The properties of these electrolytes can be correlated by the use of a factor called van’t Hoff factor (i)

127. The van’t Hoff factor for NaCl is 1.9. The degree of dissociation is(A) 45% (B) 100% (C) 90% (D) 60%

128. A 0.01 M solution of K3Fe (CN)6 is 50% dissociated at 270C, then the osmotic pressure of solution will be(A) 0.02 atm (B) 0.61 atm (C) 0.78 atm (D) 1.29 atm

129. The ratio of the elevation of boiling point for NaCl solution to that for sugar of same concentration is(A) 1 (B) 2 (C) 3 (D) 0.5

PASSAGE A solution which remains in equilibrium with undissolved solute, in contact, is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a


given temperature, is called solubility product (KSP). For the electrolyte AxBy with solubility

S, AxBy (s) xAy+ + yBx-

While calculate the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to

that of strong electrolyte. If in a solution, the ionic product of an electrolyte exceeds its KSP value at a particular temperature, then precipitation occurs130. The solubility of PbSO4 in water is 0.0303 g/l at 250C, its solubility product at that

temperature is(A) 10-4 M2 (B) 9.18 x 10-4 M2 (C) 10-8 M2 (D) 9.18 x 10-8 M2

131. The solubility of BaSO4 in 0.1 M BaCl2 solution is (KSP of BaSO4 = 1.5 x 10-9)(A) 1.5 x 10-9 M (B) 1.5 x 10-8 M (C) 2.25 x 10-16 M (D) 2.5 x 10-18 M

132. If S0, S1, S2 and S3 are the solubilities of AgCl in water, 0.01M CaCl2, 0.01 M NaCl and 0.0.5 M AgNO3 solutions respectively, then(A) S0 > S1 > S2 > S3 (B) S0 > S2 > S1 > S3 (C) S0 > S1 = S2 > S3 (D) S0

> S2 > S3 > S1

PASSAGE A metal complex having composition Cr(NH3)4Br2I was isolated in two forms (X) and (Y). Form (X) reacts with AgNO3 to give a pale yellow precipitate which is partially soluble in

excess of NH4OH whereas (Y) gives a greenish yellow precipitate which is insoluble in NH4OH133. Select the correct statement

(A) The formula of (X) and (Y) are and respectively(B) The formula of (X) and (Y) are [Cr(NH3)4IBr] Br and [Cr(NH3)4Br2] I respectively(C) The formula of (X) and (Y) are both [Cr(NH3)4I]Br I

(D) The formula of (X) and (Y) are [Cr(NH3)3IBr2] (NH3)2

134. Both the (X) form and (Y) form show(A) linkage isomerism (B) coordination isomerism (C) Ionization isomerism (D) none of these

135. Select the correct statement(A) (X) – cis form optically inactive, (Y) – cis form optically active(B) (X) – cis form optically inactive, (Y) – trans form optically active (C) The cis and trans forms of both X and Y are optically active

(D) The cis and trans form of both X and Y are optically inactivePASSAGE :

An optically active compound (A) gives an optically inactive compound (B) on

hydrogenation. (A) gives no precipitate with and gives optically inactive (C) with in presence of Nickel Boride

136. Compound (B) is

(A) (B)

(C) (D)

137. Compound (A) is

(A) (B)

(C) Sri Chaitanya Page 26 Vizag

(D)

138. Compound (C) is

(A) (B)

(C)

(D)

PASSAGE : A yellow coloured powder (A) reacts with acetic acid to form a white crystalline solid (B) which is used for curing skin diseases. Comp (B) on heating gives solid (A), a gas (C) and a compound (D). comp.(D) on reacting with and forms a yellow precipitate of compound (E). comp. (B) when treated with a

solution of sodium carbonate gives a white ppt. of (F) and when is passed into solution of (B), it forms a black ppt, of (G)

139. Identify (D)(A) (B) (C) (D)

140. Compound (A) is(A) (B) (C) (D)

141. The white ppt. (F) is(A) (B) (C) (D)

PASSAGEA neutral compound (A), gives and on refluxing with dil. Alkali

followed by acidification. (B) liberates from saturated sodium bicarbonate solution. Compound (C) on dehyration yields 2-Butene as the major product. Compound (B) on treatement with osmium tetraoxide followed by reductive hydrolysis gives (D),

142. Compound (D) when treated with lead tetra acetate furnishes acetone and (E) . (E) is acidic and reduces tollen’s reagent compound. (D) is

(A) (B)

(C) (D)

143. Compound (B) is

(A) (B)

(C) (D)


(A)


(B)

(C)

(D) None of thesePASSAGE

A waxy crystalline solid (A) with garlic odour is obtained on burning a white solid (E) in stream of air. (A) reacts vigorously with hot water giving a gas (B) and an acid (C). Gas (B) has unpleasant odour of rotten fish and is neutral towards litmus. When gas (B) is passed through a blue solution of compound(F) it produces a black precipitate of compound (D). Compound (F) gives chacolate color ppt. Of (G) with

145. Compound (G) is(A) (B) (C) (D)

146. The gas (B) is(A) (B) (C) (D)

147. The acid (C) is(A) (B) (C) (D)

Passage : Alkyl derivatives of aceto acetic ester can undergo two types of hydrolysis, ketonic and acid hydrolysis. The scheme of these hydrolysis reactions are as follows.Ketonic hydrolysis

Acid hydrolysis

The above names are in agreement to the types of products obtained148. What is the final product S in the given reaction

(A) (B)

(C) (D)

149. Which reaction sequence can prepare succinic acid as final product

(A)

(B)

(C)

(D)

150.


The final product is

(A) (B)

(C) (D)

Passage : A system of greater disorder of molecules is more probable. The disorder of molecules is reflected by the entropy of the system. A liquid vaporizes to form a more disordered gas. When a solute is present, there is additional contribution to the entropy of the liquid due to increase randomness. As the entropy of solution is higher than that of pure liquid, there is weaker tendency to form the gas. Thus, a solute (non volatile) lowers the vapour pressure of a liquid, and hence a higher boiling point of the solution.Similarly, the greater randomness of the solution opposes the tendency to freeze. In consequence, a lower the temperature must be reached for achieving the equilibrium between the solid (frozen solvent)

and the solution. Elevation of B.Pt and depression of F.Pt of a solution are the colligative

properties which depend only on the concentration of particles of the solute, not their identify. For dilute solutions, and are proportional to the molality of the solute in the solution.

= Ebullioscopic constant =

And = = Cryoscopic constant = (M = molecular mass of the solvent)

The values of and do depend on the properties of the solvent. For liquids, is almost

constant .[Troutan’s rule, this constant for most of the unassociated liquids (not having any strong bonding like Hydrogen bonding in the liquid state) is equal to 90 J/mol.]For solutes undergoing change of molecular state is solution (ionization or association), the observed T values differ from the calculated ones using the above relations. In such situations, the relationships are modifies as ;Where i = Van’t-Hoff factor, greater than unity for ionization and smaller than unity for association of the solute molecules

151. Depression of freezing point of which of the following solutions does represent the cryoscopic constant of water?(A) 6% by mass of urea is aqueous solution (B) 100g of aqueous solution containing 18g of glucose (C) 59g of aqueous solution containing 9g of glucose (D) 1 M glucose solution in water

152. Dissolution of a non-volatile solute into a liquid leads to the(A) Decrease of entropy (B) Increase in tendency of the liquid to freeze (C) Increase in tendency to pass into the vapour phase (D) Decreases in tendency of the liquid to freeze

153. To aqueous solution of NaI, increasing amounts of solid is added. The vapour pressure of the solution(A) decreases to a constant value (B) increases to a constant value (C) increases first and then decreases (D) remains constant because is sparingly soluble in water.

Passage


Werner performed two experiments as given below Experiment 1: He prepared a compound (x) by reacting KCl with . The compound (x) did not give any precipitate with but gave electrical conductance corresponding to 3 ions.Experiment 2: He took 0.319 g of and passed through a cation exchange resin and the acid coming out required 20.5 ml of 0.125 M NaOH

154. The correct formula of the compound (x) is(A) (B (C) (D)

155. The number of moles of AgCl which will be precipitated with which will be precipitated with

in . is(A) 2 (B) 3 (C) 1 (D) 0

156. The complex . can be rightly represented as

(A) (B)

(C) (D)

PASSAGE An octahedron can be drawn within a cube by choosing the centre of its six faces. The centre of cube constitutes an octahedral void. In close packing two anions touch each other along the line obtained by joining the centres of two adjoining face and one cation (occupying an octahedral void) and two anions (occupying at the centre of two opposite faces) touch each other along the line joining the centres of two opposite faces

157. The ratio of for the above passage is(A) 0.225 (B) 0.732 (C) 0.414 (D) 1

158. The coordination number of atom present in the void formed in the above passage is(A) 4 (B) 6 (C) 3 (D) 12

159. The number of octahedral voids present in 234g of NaCl is(A) NA (B) 4 NA (C) 16NA (D) NA/4

PASSAGEA compound (P) which consists of a benzene ring gave the following tests.(a) gave characteristic colour with neutral (b) gave Tollen’s test but not Fehling’s test(c) showed the presence fo one methoxy group. Compound (P) on oxidation followed by

decarboxylation followed by reaction with Br2 gave Compound (P) gave only three mono substituted product when electrophilic substitution reaction was done. Also when (P) is heated with Zn dust it gives a compound (A) which gives 3 mononitro derivatives. Compound (P) on treatment with HBr gave compound (Q) which on reaction with sodium acetate in the presence of acetic anhydride to give compound (R). compound (P) on reaction with CH3CHO gave a compound (S) which on heating gave compound (T)

160. The compound (A) must be

(A) (B) (C) (D) 161. The compound (P) must beSri Chaitanya Page 30 Vizag

(A) (B)

(C) (D) 162. The compounds (R) and (S) must be

(A) and (B)

(C) (D) AllPassage

The proton , neutron and electron are the three sub-atomic particles important in an atom .The particles occupy two regions . Protons and neutrons occupy the central place of the nucleus and electron the vast space out side the nucleus. Neutron transfer does not takes place in ordinary chemical reactions.Proton transfer constitutes acid-base reactions .Electron transfer constitutes redox reactions .Redox reactions are essential for life. Photo synthesis and Respiration are two prime examples .Redox reactions also allow key thermo dynamic quantities to be measured as demonstrated in this problem .Given the following information


163. What is x in the above Latimer diagram ?

(A) 1.634V (B) 1.098V (C) 3.075V (D) 1.414 V164. A galvanic cell using standard hydrogen electrode as an anode is constructed in which the over all reaction

is . Silver ions are added until AgBr precipitates at the

cathode and reaches 0.06M.The cell voltage is measured to be 1.721 V . for the galvanic cell is

approximately (A) 1.734 V (B) 2.19 V (C) 1.065V (D) 0.32V

165. Using the data given in the passage and /or in the above two questions ,which of the following can be calculated?( i ) Solubility of AgBr in a 0.1 M aqueous solution of ammonia at

( ii )

( iii ) Solubility of bromine in the form of in water at

( iv) of at (A) iii and iv (B) i,ii and iii (C) i , iii and iv (D) i ,ii ,iii and iv

PassageA salt (A) when heated with and conc. liberates a gas which is absorbed in solution.The solution turns yellow.When this solution is acidified with acetic acid and lead acetate solution is added , a yellow precipitate (B) is formed.

When (A) is mixed with and heated with conc. ,a gas (C) is evolved which turns starch-iodide paper blue

166. The acidic radical present in the salt (A) is(A) (B) (C) (D)

167. The compound formed which turns solution yellow is(A) (B) (C) (D)

168. What is the colour of gas which is evolved when salt(A) is heated with and (A) Violet (B) Brown (C) Greenish yellow (D) Blue

Passage Hoffmann bromamide reaction involves conversion of a carboxylic acid amide into an amine with a loss of carbon atom on treatment with aqueous sodium hypo bromite

The mechanism is,

169. If the amide is optically active and the stereogenic centre is directly linked to carbonyl group then step (2) occurs


(A) With retention of configuration (B) With complete inversion of configuration (C) With partial inversion of configuration (D) By a reversible reaction

170. Which of the following will not give Hoffmann bromamide reaction ?

(A) (B) (C) (D) 171.

X and Y in the above sequence of reactions are respectively

(A) (B)

(C) (D) No reaction in both the two pathsPASSAGE

For the reaction , products,rate= where ‘K’ is rate constant.If the reaction involves

different reactants with the same concentration at all times ,then the general form is

,which on integration gives .If the concentrations of one or more than one reactants are

in large excess than the other or one of the reactant acts as a catalyst,the concentrations of these species remain same and can be taken as constant and hence reduction in order of reaction .In heterogeneous catalysis, it is assumed that the reaction occurs on the surface and that the adsorbed species obey the Langmuir adsorption isotherm.The surface reaction rate is proportional to the fraction adsorbed . For

,

172. A gaseous substance decomposes according to the over all equation , .The variation of

the partial pressure of with time (starting with pure ) is given below at

The decomposition is irreversible .The rate constant is nearly(A) (B) 66 mm Hg (C) 0.1386 (D)

173. At and at a constant pH of 5, the inversion of sucrose proceeds with a constant half life of 500 min.At the same temperature ,but a pH of 4,the half life would be(A) 500 min (B) 50 min (C) 5000 min (D) 400.min


174. The platinum catalysed decomposition of HI obeys the rate law, at high pressure with

at . At low pressures, the rate law becomes .pHI with at .Calculate the HI pressure at which the value of rate should be at

(A) 10 mm Hg (B) 2 mm Hg (C) 5 mm Hg (D) 50 mm HgPASSAGE

Organic compounds undergo different types of substitution reactions .Aromatic systems generally undergo electrophilic substitution reactions. Nucleophlic substitution reactions may also occur in them if only electron withdrawing groups are present as substituents .Compounds like alkyl halides usually undergo nucleophilic substitution reactions .These reactions are influenced by electronic factors mainly.There is the posibility of attack by an electricphile occuring on the ring carbon to which the substituent is already attached

175. Which of the following is correct with respect to higher yield of the product ?

(A) (B)

(C) (D) 176. Which of the following is the main product of the given reaction?

(A) (B) (C) (D) 177. Which of the following compounds favour reaction more readily?

(A) (B) (C) (D) PASSAGE

A factory, producing methanol, is based on the reaction: CO + 2H2 CH3OH Hydrogen & carbon monoxide are obtained by the reaction CH4 + H2O CO + 3H2 Three units of factory namely, the “reformer” for the H2 and CO production, the

“methanol reactor” for production of methanol and a “separator” to separate CH3OH from CO and H2 are schematically shown in figure

CH4 + H2O

CO + H2

CO + H2 + CH3OH

CH3OH


Reformer MethanolReactor

Separator

Four positions are indicated as , , and . The flow of methanol at position is 103 mol/sec. The factory is so designed that of the CO is converted to CH3OH. Excess

of CO and H2 at position are used to heat the first reaction. Assume that the reformer reaction goes to completion. At the position () mole ratio of CO to H2 is

CO + 2H2 CH3OH ; Hr = 500 R178. What is the flow of CO and H2 at position ()?

(A) CO : 1500 mol/sec ; H2 : 3000 mol/sec(B) CO : 1500 mol/sec ; H2 : 4500 mol/sec (C) CO : 1000 mol/sec ; H2 : 2000 mol/sec (D) CO : 1500 mol/sec ; H2 : 2000 mol/sec

179. What is the flow of CO and H2 at position ()?(A) CO : 500 mol/sec ; H2 : 1000 mol/sec(B) CO : 500 mol/sec ; H2 : 1500 mol/sec (C) CO : 500 mol/sec ; H2 : 2000 mol/sec (D) CO : 500 mol/sec ; H2 : 2500 mol/sec

180. Amount of energy released in methanol reactor in 1 minute ?(A) 1200 kcal (B) 6000 kcal (C) 12000 kcal (D) 60000 kcal

PASSAGE Covalent compounds undergo hydrolysis via SN1 (unimolecular nucleophilic substitution) or SN2 (Bimolecular nucleophilic substitution) mechanism, for SN2 mechanism within the molecule atom should have at least one vacant orbital, if it is not there then hydrolysis takes place via SN1 mechanism (dissociative step) in

drastic condition181. What are the hydrolysis products of BeCl2?

(I) [Be(OH)4]2 (II) Be(OH)2 (III) HCl (IV) BeH2

(A) I, III (B) II, III, IV (C) I, II, III (D) II, III182. Least probable product formed on hydrolysis of BCl3 is

(A) HCl (B) [B(OH)4] (C) B(OH)3 (D) None183. CCl4 is inert towards hydrolysis under ordinary conditions because (I) No vacant orbital on attacking site of C-atom (II) CCl4 is non polar and does not react with polar H2O molecule (III) Bond dissociation energy of C Cl bond is very high (IV)H2O molecule cannot approach the anti-bonding M.O. of C Cl bond due to steric

crowding. Select correct code:

(A) I, II and IV (B) I and IV (C) I and II (D) I, II, III and IVPASSAGE :

Keto-enol interconversion is also called ketoenol tautomerization or enolization. The interconversion of the tautomers can be catalysed by either acids or bases. In a

basic solution, hydroxide ion removes proton from the carbon of the keto isomer, forming an enolate ion. Protonation on oxygen forms the enol tautomer, whereas protonation on the carbon reforms the tautomer.

Basecatalyzed ketoenol interconversion.


In an acidic solution, the carbonyl oxygen of the keto tautomer is protonated and water removes a proton from the carbon, forming the enol.

Acidcatalyzed ketoenol interconversion.

184.

Final product P is

(A) (B) (C) (D) 185.

Above interconversion takes place in(A) Acidic medium (B) Basic medium (C) Both (D) Neutral medium (pH=7 at 298K)

186. Decreasing order of enol content of the following compounds in liquid phase

(A) 4>3>2>1 (B) 2>1>3>4 (C) 1>2>3>4 (D) 3>1>2>4

PASSAGE

When the reactants are mixed in a chemical system, the system will proceed spontaneously to a position of lower free energy, and the system will eventually achieve equilibrium. At any point along the way from the pure reactants to


equilibrium, the reactants are not at standard states. The change in Gibbs free energy for a reversible reaction under these non standard conditions, rG is related to rG0 by the equation

rG = rG0 + RT ln Qwhere Q is the reaction quotient. Further more, as long as G is negative – that is the reaction is “descending” from the free energy of the reactants to the equilibrium position – the reaction is spontaneous. Eventually the system reaches equilibrium because no further change in concentration of reactants, products is seen at this point. G must be zero; that is there is no further change in free energy in system. Substituting G = 0 and Q = K in the equation,0 = G0 + RT ln K or G0 = - RT ln K

187. The standard free energy of formation of ) is . Calculate for the reaction

(A) (B) (C) (D)

188. At 300K, the standard free energy of formation of is . Calculate for the following reaction

(Given : ln 2 = 0.7, R = 8 J/K mol)(A) (B) (C) (D)

189. What is partial pressure of HCl (g) above an aqueous solution that is 2M in H+ and

1.5M in . Reaction is

Given : R = 8.0J/ mol. K : fGo (HCl, g) = -95 kJ / mol, fGo (Cl-, aq) = -131.848kJ / mol. All data at K.

(A) (B) (C) (D)

Passage

Preparation of Grignard reagent

R – X + Mg (Organo magnesium halide)

Grignard reagent may be made from primary, secondary and tertiary alkyl halide as well as from vinyl halide and aryl halides. Grignard reagent is strong nucleophile and strong base

190.

Major product (P) is

(A) (B) (C) (D) 191. In which of the following reaction tertiary alcohol will not be obtained as a product


(A) (B)

(C) (D) 192.

End product (Q) is

(A) (B) (C) (D)

Passage :A,B and C are three complexes of chromium (III) with the empirical formula . All the three

complexes have water and chloride ions as ligands. Complex A does not react with concentrated , where as complexes B and C loses 6.75% and 27.02% of their original weight respectively on treat with concentrated H2SO4

193. The complex B is

(A) (B)

(C) (D)

194. Select the correct statement(A) conductance in these complexes is in order C<B<A(B) EAN of chromium is not identical (C) Nuclear spin of chromium is identical (D) Non- reactivity of A with conc.H2SO4 is due to absence of outside co-ordination sphere

195. The Complexes can show(A) Coordination Isomerism (B) Optical isomerism (C) Linkage isomerism (D) Hydrate isomerism

Passage :58. W.H Nernst derived a mathematical relationship between the emf of a cell (Ecell) and the concentration of reactants and products in a redox reaction under non-standard conditions. The nernst equation for a redox reaction of the type

is

At 298 K .

Where Q is the reaction quotient. At equilibrium , there is a no net transfer of electron, so E = 0 and Q= K, where K is the equilibrium constant. We can also apply nernst equation for a half cell to calculate oxidation or reduction potential

196. What is the EMF of represented cell at 298 k.

Given :

(A) 0.77V (B) -0.829V (C) -0.77V (D) None of these197. Find the solubility product of a saturated aqueous solution of Ag3PO4 at 298 k if the emf of the cell


(A) (B) (C) (D)

198. What would be the reduction potential of an electrode at 298 K, which originally contained 1 M K2Cr2O7 reduce all to Cr3+. Assume pH=1.0 and which was treated with 50% of the Sn necessary to reduce

all to Cr3+. Assume pH of solution remains constant.

Given :

(A) 1.285 V (B) 1.193 V (C) 1.187 V (D) None of thesePassage :

Aromatic compound are these that meet the following criteria(i) The structure must be cyclic, containing electrons(ii) Each atom in the ring must have unhybridized p- orbitals(iii) Unhybridized p-orbitals must overlap orbitals(iv) Structure must be planarIn anti-aromatic compound electron are delocalization, but delocalization of pi electrons over the ring increasing the electronic energy

199. Which of following compound is aromatic

(A) (B) (C) (D) 200. Which of following is anti aromatic

(A) (B) (C) (D) 201. In which of following reaction product formed is aromatic

(A) (B) (C) (D) AllPASSAGE

A white crystalline compound soluble in water, has no reaction with dil. H2SO4. The salt on strong heating gave an acid gas that is paramagnetic which on cooling transformed into diamagnetic molecule, leaving behind on yellow residue, changed to white on cooling. The salt solution gave a white precipitate with ammonia, dissolved in excess ammonia forming a complex. The salt when heated with cobalt nitrate gave a green coloured mass. The salt solution gave a bluish white precipitate with potassium ferrocyanide solution

202. The acid gas liberated is(A) (B) (C) (D)

203. From the data most probable salt(A) (B) (C) (D)

204. The colourless complex formed is

(A) (B) (C) (D)

Passage :


Arndt – eistert reaction consists in increasing the length of the carbon chain by one methylene group in carboxylic acid. . However carboxylic acid should be converted first to an acid chloride. The sequence of steps may be shown as

205. The product obtained on treatment of ‘C’ with Ag2O and will be

(A) (B) (C) (D)

206. If is used propanoic acid gives

(A) (B)

(C) (D)

207. The application of this reaction of Malonic acid gives(A) oxalic acid (B) succinic acid (C) glutaric acid (D) adipic acid

Passage :The electromotive force (emf) is the maximum voltage of a voltaic cell. It can be directly related to maximum work that can be done by the cell. The standard free change, standard emf and equilibrium constant are all related. An electrode potential depends on concentration of the electrode substance but not on the stoichiometry of the electrode reaction. According to Nernst equation

at 250C emf of the cell is zero at equilibrium. Free energy is related to the electrode

potential as

208. Given E0 = 0.45V

E0 = -0.04V

What is the standard electrode potential of

(A) 0.41V (B) 1.27V (C) 0.49V (D) -1.27V

209. E0 = 0.52V

E0 = +0.16V

in KJ/mol for the disproportion of Cu+ is(A) -3.474 (B) -69.48 (C) -34.74 (D) +3.474

210. For the reactionsE0 = 1.51V

E0 = 1.23VThen for the reaction

The equilibrium constant is [ Anti log 0.271=1.866](A) 1.866 x 1086 (B) 1 (C) 1.866 x 1085 (D) 1.866 x 10-86

Passage :On dissolving 0.0264 g of a compound (A) in 3.0 g of camphor, the melting point of camphor is lowered by 40C. and undergoes acetylation on treatment with acetyl chloride. On reacting (A) with HCl and ZnCl2 a dense oily layer is quickly formed. Also on passing over alumina at 3500C, (A) yields a compound (B) which on ozonolysis gives (C) and (D) give a yellow precipitate with 2,4 – dinitrophenylhydrazine but only (C) reduces Fehling’s solution. (C) also forms a resinous substance with concentrated NaOH(Kf for camphor = 40 K kg mol-1)


211. The molecular formula of ‘A’ is

(A) (B) (C) (D) 212. C, D respectively are

(A) (B)

(C) (D) 213. the product (s) may be

(A) - hydroxy aldehyde (only) (B) - hydroxy ketone (only) (C) - hydroxy aldehyde + - hydroxy ketone

(D) formate ion + 20 alcoholPassage :

In an elementary act of a chemical reaction does not occur each time reacting molecules collide only those molecules react that have sufficient energy to break the bonds in the initial particles and thus create the possibility of formation of products. Consequently every reaction is characterized by a definite energy barrier. This is known as threshold energy. To surmount it, certain excess energy (in comparison with the average energy of the molecules at a given temperature) is needed known as activation energy to result in formation of products. The dependence of rate constant ‘K’ of a reaction on activation energy ‘Ea’ is expressed by Arrehenius equation

where A is frequency factor, T is absolute temperature and R is gas constant. Substance A is transformed into substance C according to the reaction

, The energy diagram of the process is given as

214. Which of the following is correct(A) (B)

(C) (D) 215. In the above diagram, activation energy of is 30 kJ/mole and the average energy difference

Between A and B is 20 kJ/mole. The equilibrium constant at is (assume Frequency factor is Constant)(A) (B) 2.82 x 10-5 (C) (D)

216. In the same transformation if K1>K2 and K2>K3 which of the following statement is correct(A) The threshold energy of and are same and 2nd step is endothermic(B) The threshold energy of and are same and 2nd step is exothermic (C) The threshold energy of and are different and 2nd step is exothermic (D) The threshold energy of and are different and 2nd step is endothermic

Passage

x and y are two isomers having molecular formula (C6H12O)


217. If ‘Y’ is tertiary alcohol with five carbon ring. Then it on treating with above series of the reagents then the product obtained is

(A) (B) (C) (D) 218. What is the structure of (x)

(A) (B) (C) (D) 219.

W is

(A) (B) (C) (D) Passage

Real gases deviate from ideal behaviour and the extent of deviation is expressed in terms of compressibility factor (Z) which is the ratio of the observed molar volume to the ideal molar volume. Z is a function of pressure & temperature for real gases. If Z=1, then it is an ideal gas and if Z>1, then the gases are less compressible and for easily liquefiable gases, Z<1

220. Which of the following statements is correct(A) the compressibility factor for ideal gases depends on temperature and pressure

(B) for real gases is independent of pressure

(C) for all real gases have same value (D) for different real gases have different values


221. Which of the following statement is correct for gas A having molar mass 16g and density 0.75g/L at 2 atm

and 270C temperature(A) Force of attraction are dominating than force of repulsion among the gas molecules(B) Force of repulsion are dominating than force of attraction among the gas molecules (C) Force of attraction equals to force of repulsion (D) None

222. The graphs between compressibility factor and pressure can be explained by Van der Waal’s equation

. The value of Z< 1 by

(A) ignoring the constant a (B) ignoring the constant b (C) ignoring both a & b (D) retaining both a & b

Passage

223. Type of hybridization in compound (K)(A) (B) (C) (D)

224. Which of the following set of reagent is suitable to get (F) from (G)(A) H2O2 in neutral (B) H2O2 in basic medium (C) H2O2 in acidic medium (D) none

225. Which of the following gas is turning the filter paper dipped into solution brownish black(A) gas D (B) gas J (C) gas C (D) none

PassageDuring the distillation of binary liquid solution the following points are helpful1. A liquid boils at a temperature at which its vapour pressure becomes equal to surrounding atmospheric pressure.2. A liquid boils at a constant temperature when the composition of the liquid and of vapour are same and vice versa3. On progressively raising the temperature of liquids, liquid with high vapour pressure reach its boiling point earlier4. The residue left behind after a part of liquid is removed as vapour has a boiling point which is either equal to or more than boiling point of initial liquid5. The component whose further addition increases the vapour pressure is known as vapour pressure increasing component (VPI)Fractional distillation of ideal and non-ideal binary liquid solution of liquids A and B are given below


226. According to the figure I, when first condensate is removed, the composition of A is(A) Rich in condensate than original liquid (B) Poor in condensate than original liquid (C) Does not change (D) None

227. From the figure II, which is correct statement(A) In the region of curves from B to C, A is the VPI Component(B) In the region of curves from B to C, B is the VPI component (C) In the region of curves from A to C, B is VPI component (D) In the region of curves from A to C, both A&B are VPI components

228. From figure I the boiling point of first condensate C1, First liquid residue(L1) and second condensate (C2) are in the order(A) (B) (C) (D)

Passage Different special arrangements of atoms that result from restricted rotation about a single bond are called conformation. Stability of conformer depends on torsional strain, steric strain, dipole-dipole attraction and intramolecular hydrogen bonding

229. Which one of the following statements is not correct?(A) Gauche form of n-butane is less stable than anti form due to gauche interaction(B) n-butane has four different conformers (C) Gauche form of n-butane is less stable than the eclipsed form (D) Energy barrier between anti and fully eclipsed form is 4-5 kcal/mole

230. Which one of the following conformers of 3-amino-2-butanol is most stable?

(A) (B)


(C) (D) 231. Arrange stability of different conformers of cyclohexane in decreasing order

1) Chair 2) Half chair 3) Boat 4) Twist boat(A) 1, 4, 3, 2 (B) 1, 3, 4, 2 (C) 1, 2, 3, 4 (D) 1, 4, 2, 3

Passage A certain metal (A) is boiled in dilute nitric acid to give a salt (B) and an oxide of nydrogen (C)An aqueous solution of (B) with brine solution gives a precipitate (D) which is soluble in ammonium hydroxide . On adding aqueous solution of (B) to hypo solution, a white precipitate (E) is obtained . (E) on standing turns to a black compound (F). Metal A is also present in pyrargyrite

232. When precipitate (D)is dissolved in excess of , a soluble complex is obtained . The hybridization of metal in that complex is

(A) (B) (C) (D)

233. The formula of black precipitate (F) is(A) (B) (C) (D)

234. When ppt D is reacted with . It gives a coloured ppt which is

(A) , orange red (B) ; brick red

(C) ; brick red (D) ;dark redPassage

VBT describes the bonding in terms of hybridized orbitals of the central atom or ion. The theory mainly deals with the geometry (i.e., shape) and magnetic properties of the complex.The salient features of the theory are(1) The central metal loses requisite number of electrons to form the ion. The number of electrons lost is the valency of the resulting cation. In sum cases, the metal atom does not lose electrons.(2) The central metal ion or atom makes available a number of empty s, p and d atomic orbitals equal to its coordination number. These vacant orbitals hybridized together to form hybrid orbitals which are same in the number as the atomic orbitals which are same in the number as the atomic orbitals hybridizing together. They are equivalent in energy and have definite geometry

235. has ___________ hybridization

(A) (B) (C) (D) None of these

236. Which complex has zero magnetic moment (spin only)

(A) (B) (C) (D) All of these

237. Which of the following cyano complex would exhibit the lowest value of paramagnetic behavior ?

(A) (B) (C) (D)

PassagePhenol undergoes reactions, which are normally not shown by alcohols.e.g., when is treated with

, it is converted into benzene. Formylation of phenols with chloroform in alkaline solution is known as Reimer Tiemann reaction. In this reaction, it is believed that and react to from carbene which reacts with phenol in its ketonic form to give product


238.

A is

(A) (B) (C) Both (A) and (B) (D) 239.

In the above reaction , X is

(A) (B) (C) (B) is major , (A) is minor (D) None of these240. In the above reaction, Y is

(A) (B)

(C) (D) Passage :

A complex compound of chromium contains five molecules, one nitro group and two chloride ions

for one cation. One molecule of this compound produces three ions in aq.solution, on reacting with

excess of solution, two moles of get precipitated.241. The formula of the complex compound is

(A) (B)

(C) (D) 242. The types of isomerism shown by the complex compound is

(A) Geometrical, ionization (B) Ionization , linkage (C) Linkage , optical (D) Geometrical , optical

PASSAGE : An optically active compound (A) gives an optically inactive compound (B) on

hydrogenation. (A) gives no precipitate with and gives optically inactive (C) with in presence of Nickel Boride

243. Compound (B) isSri Chaitanya Page 46 Vizag

(A) (B)

(C) (D)


(A) (B)

(C)

(D)

245. Compound (C) is

(A) (B)

(C)

(D)

PASSAGEA yellow coloured powder (A) reacts with acetic acid to form a white crystalline solid (B) which is used for curing skin diseases. Comp (B) on heating gives solid (A), a gas (C) and a compound (D). comp.(D) on reacting with and forms a yellow precipitate of compound (E). comp. (B) when treated with a

solution of sodium carbonate gives a white ppt. of (F) and when is passed into solution of (B), it forms a black ppt, of (G)

246. Identify (D)(A) (B) (C) (D)

247. Compound (A) is(A) (B) (C) (D)

248. The white ppt. (F) is(A) (B) (C) (D)

249.(A) (B) (C) (D)

250.(A) (B) (C) (D)

251.(A) (B) (C) (D)

252.(A) (B) (C) (D)

253.(A) (B) (C) (D)

254.(A) (B) (C) (D)

255.(A) (B) (C) (D)


256.

(A) (B) (C) (D)257.

(A) (B) (C) (D)258.

(A) (B) (C) (D)259.

(A) (B) (C) (D)260.

(A) (B) (C) (D)261.

(A) (B) (C) (D)262.

(A) (B) (C) (D)263.

(A) (B) (C) (D)264.

(A) (B) (C) (D)265.

(A) (B) (C) (D)266.

(A) (B) (C) (D)267.

(A) (B) (C) (D)268.

(A) (B) (C) (D)269.

(A) (B) (C) (D)270.

(A) (B) (C) (D)271.

(A) (B) (C) (D)



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