Chemistry-2.Open Test-Paper-1

CHEMISTRY Test-Paper-1 MULTIPLE CHOICE QUESTIONS (SINGLE OPTION CORRECT) 25. Reaction of para-chloro aniline with acetic anhydride in pyridine give a mixture of 94% of parachloroacetanilide , contaminated with 6% unreacted amine. Which of the following treatment would be best used to purify the amide? (A) React the unreacted amine with methyl iodide (B) Wash an ether solution of the crude product with concentrated brine (aq. NaCl) (C) Wash an ether solution of the crude product with 5% aqueous sulfuric acid (D) Wash an ether solution of the crude product with 5% aqueous sodium carbonate 25. C 26. A synthesis of 2,5-dimethyl-2-hexanol from 2-methyl propene requires the formation of two four carbon intermediates, X and Y. These intermediates combine to give the desired product after the usual hydrolysis work up. Select appropriated methods of preparing X and Y from 2-methyl propene. (A) X add HBr, then react with Mg in ether Y add 2-methyl propene (B) X add HBr (peroxides), then react with Mg in ether Y reacts with C6H5CO3H is CH2Cl2 (C) X add HOBr Y add B2H6 in ether, then NaOH (D) X add HOBr Y add HBr (peroxides), then react with Mg in ether 26. B 27. How many mL of 1.00 M NaOH must be added to 100 ml of 0.1 M H3PO4 solution tAITS-Opent Test-P1-CHo obtain a phosphate buffer solution with pH of about 7.2? (The pK values of H3PO4 are pK1 = 2.1, pK2 = 7.2, pK3 = 12 ) (A) 10.0 ML (B) 15.0 ML (C) 20.0 ML (D) 30 ML 27. B 27. Let millimole of NaOH further added are ‘x’ ⇒ x = 5 Therefore total millimoles of NaOH = 10 + 5 = 15 Let volume of NaOH is Vml 1  Vml = 15 Vml = 15 28. Two equivalents of Br2 add to one equivalent of 1,7-octadiene. How many stereo isomeric tetrabromides will be formed? (A) 1 (B) 2 (C) 3 (D) 4 28. C 28. The compound formed contains two chiral carbons and a plane of symmetry. Optical isomers = 22−1 + 21−1 = 3 29. The of H2CO3 are 6.4 and 10.3. The pKa of HOBr is 8.7. If equimolar amount of Na2CO3 and HOBr are dissolved in water what will be the predominant anionic species in the resulting solution. (A) and BrO− (B) Br− and OH− (C) and Br− (D) and 2BrO− 29. A 30. Heating benzene in a large excess of 80% D2SO4 in D2O results in what product? (A) C6H5SO3D (B) C6H5OD (C) C6H5SO3H (D) C6D6 30. D 30. 31. Fluorine can be stored in a metal container whereas other halogens can not be stored. The reason for this is (A) very low F−F bond dissociation enthalpy (B) fluorine does not react with metals (C) fluorine reacts with a metal to form a non-reactive metal fluoride film (D) None of these 31. C 32. Aluminium oxide exists in nature as gems with different colours. The reason for the difference in colour is that (A) the oxidation states of aluminium in these gems are different (B) the extent of crystallinity in these gem is different (C) the Al−O bonding is different in these gem structures (D) these are different transition metal ions present as impurities in these gems 32. D 33. Which of the following statement is correct? (A) Oxidation potential of lithium is highest among all alkali metals (B) Ionization energy of lithium is highest among all alkali metals (C) Li reacts gently with H2O while K reacts violently and catches fire (D) All the above are true 33. D 34. The kinetic molecular theory of gases predicts pressure to rise as the temperature of gas increases because (A) the average kinetic energy of the gas molecules decrease (B) gas molecules collide more frequently with the container walls (C) gas molecules collide less frequently with the container walls (D) gas molecules collide less energetically with the container walls 34. B 35. Which of the following postulates of Dalton’s atomic theory has been shown to be incorrect? (A) In a chemical reaction, matter is neither created nor destroyed (B) When the atoms of different elements combine to form compounds, they combine in simple whole number ratio (C) Matter consists of particles called atoms (D) All atoms of the same elements are identical in every way 35. D 36. How many diastereomers are possible for the complex ion [Cr(OH)3(NH3)2Cl]−? (A) 6 (B) 3 (C) 4 (D) 2 36. B 36. COMPREHENSION-I Read the paragraph and answer the questions (37 – 39) given below: For simplicity soluble salts in water equilibrium constant (solubility product) is given as Ksp = [A+][B−] Salt is assumed to be partially soluble and completely dissociated therefore the soluble amount of salt is present in the form of ions. The solubility of salt also depends on pH of solution and presence of other ions. Ksp of CH3COOAg is 10−8. 0.1 mole CH3COOAg is added in each of three different solution. (A) in 1 L of a buffer solution having pH = 3 (B) in 1 L of 0.1 M HCl solution (C) in 1 L of 0.1 M HNO3 solution Ka(CH3COOH) = 10−5, Ksp(AgCl) = 10−10 37. The concentrations of CH3COOH in solution A is (buffer solution does not have CH3COOH and CH3COO− ions) (A) (B) concentration of CH3COOH is approximately same of solubility of CH3COOAg in solution (C) Both A and B (D) cannot calculate by given data 37. C 38. What is the final pH of solution B? (A) 5 (B) 4 (C) 3 (D) 8 38. B 38. The net reaction is Let [H+] = [Cl−] = x x2 = 10−8 x = 10−4 Therefore final pH = − log10−4 = 4 39. The solubility of CH3COOAg in solution C is (A) 10−2 M (B) 10−3 M (C) 10−4 M (D) 10−5 M 39. A 39. x(x − y) = 10−8 Because the equilibrium constant of second reaction is 105 (very high), so x will be approximately equal to y. Let say x = y and x − y = t xt = 10−8 …(i) …(ii) We can neglect y as compared to 0.1. …(iii) Multiplying (i) and (iii) or y2 = 10−4 = x2 y = 10−2 = x COMPREHENSION-II Read the paragraph and answer the questions (40 – 42) given below: Compound (A) requires three equivalent of NaOH for neutralization. When (A) was subjected to acid hydrolysis, B (C30H16O4) and C (C6H10O7) were obtained in a 1 : 2 molar ratio. When A was methylated with methyl iodide at every possible site before hydrolysis, hydrolysis produced B’ (C31H48O4), D (C9H16O7) and E (C10H18O7). Methylation of D and E with MeI produced the same isomeric mixture of compound J. D was reduced with LiAlH4 to give K and L was produced by the reduction of K. Oxidative cleavage of vicinal diol of L with NaIO4 produced M and two equivalents of formaldehyde. Reduction of M produced N. 40. Which is the correct configuration of L? (A) (B) (C) (D) All are correct 41. What is the correct structure of D? (A) (B) (C) (D) A and B both are correct 42. What should be the correct structure of E? (A) (B) (C) (D) A and B both are correct 40. C 41. D 42. C COMPREHENSION-III Read the paragraph and answer the questions (43 – 45) given below: Compound (A) requires three equivalent of NaOH for neutralization. When (A) was subjected to acid hydrolysis, B (C30H16O4) and C (C6H10O7) were obtained in a 1 : 2 molar ratio. When A was methylated with methyl iodide at every possible site before hydrolysis, hydrolysis produced B’ (C31H48O4), D (C9H16O7) and E (C10H18O7). Methylation of D and E with MeI produced the same isomeric mixture of compound J. D was reduced with LiAlH4 to give K and L was produced by the reduction of K. Oxidative cleavage of vicinal diol of L with NaIO4 produced M and two equivalents of formaldehyde. Reduction of M produced N. When J was reduced to F and acid hydrolysis of F produced G. Reduction of G generated H and G was oxidized with NaIO4 to I with formation of one equivalent of formaldehyde. Z was obtained from I through reduction. Among all compound only Z was optically inactive. 43. Which statement is true about compound N? (A) Compound N is optically active and two optically active isomers are possible of this compound in nature (B) Compound N is optically active and four optically active isomers are possible of this compound in nature (C) Compound N is a racemic mixture (D) Compound N is a meso product 44. What should be the correct structure of Z? (A) (B) (C) (D) 45. What should be the structure of G? (A) (B) (C) (D) Answers 43. A 44. A 45. B COMPREHENSION-IV Read the paragraph and answer the questions (46 – 48) given below: Solutions containing H3PO4 and/or NaH2PO4 are titrated with a strong base standard solution. Associate the constant of these solutions with the titration curves (pH vs. volume of titrant) shown is the figure: For H3PO4: pK1 = 2.1, pK2 = 7.2, pK3 = 12.0. 46. The sample containing both in mole ratio H3PO4 : NaH2PO4, 1 : 2. (A) curve – i (B) curve – ii (C) curve – iii (D) curve – iv 46. C 47. The sample contains both in a mole ratio H3PO4 : NaH2PO4, 2 : 1. (A) curve – i (B) curve – ii (C) curve – iii (D) curve – iv 47. A 48. Sample contains NaH2PO4 only. (A) curve – i (B) curve – ii (C) curve – v (D) curve - vi 48. C