CPT-26 ANSWERS 73. (1) 145. (4) 2. (1) 74. (3) 146. (4) 3. (3) 75. (2) 147. (3) 4. (2) 5. (4) 76. (4) 77. (2) 148. (2) 149. (3) 6. (3) 78.

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1 1 08/01/2018 COMMON PRACTICE TEST [PMT] : CPT-26 ANSWERS CODE GOL 1. (1) 37. (3) 73. (1) 109. (3) 145. (4) 2. (1) 38. (1) 74. (3) 110. (3) 146. (4) 3. (3) 39. (1) 75. (2) 111. (2) 147. (3) 4. (2) 5. (4) 40. (1) 41. (2) 76. (4) 77. (2) 112. (1) 113. (4) 148. (2) 149. (3) 6. (3) 42. (4) 78. (1) 114. (4) 150. (3) 7. (2) 8. (3) 43. (3) 44. (4) 79. (4) 80. (3) 115. (4) 116. (2) 151. (1) 152. (2) 9. (1) 10. (4) 45. (2) 46. (2) 81. (1) 82. (3) 117. (2) 118. (4) 153. (3) 154. (1) 11. (2) 47. (1) 83. (4) 119. (3) 155. (4) 12. (3) 13. (1) 48. (2) 49. (3) 84. (1) 85. (2) 120. (3) 121. (2) 156. (1) 157. (1) 14. (1) 50. (4) 86. (1) 122. (4) 158. (2) 15. (3) 51. (4) 87. (2) 123. (1) 159. (3) 16. (3) 52. (2) 88. (1) 124. (3) 160. (4) 17. (4) 53. (1) 89. (4) 125. (3) 161. (4) 18. (1) 19. (1) 54. (3) 55. (2) 90. (4) 91. (1) 126. (4) 127. (3) 162. (4) 163. (4) 20. (3) 56. (1) 92. (4) 128. (4) 164. (4) 21. (3) 57. (4) 93. (2) 129. (3) 165. (3) 22. (1) 23. (3) 24. (4) 25. (2) 26. (4) 27. (4) 28. (4) 29. (4) 30. (1) 31. (2) 32. (2) 33. (1) 34. (1) 35. (1) 36. (3) 58. (2) 59. (2) 60. (3) 61. (3) 62. (3) 63. (2) 64. (3) 65. (2) 66. (3) 67. (4) 68. (4) 69. (2) 70. (3) 71. (2) 72. (3) 94. (4) 95. (2) 96. (3) 97. (2) 98. (3) 99. (1) 100. (3) 101. (2) 102. (1) 103. (4) 104. (1) 105. (2) 106. (3) 107. (3) 108. (4) 130. (4) 131. (4) 132. (3) 133. (2) 134. (1) 135. (1) 136. (4) 137. (2) 138. (1) 139. (3) 140. (1) 141. (4) 142. (1) 143. (4) 144. (1) 166. (1) 167. (3) 168. (4) 169. (2) 170. (2) 171. (1) 172. (3) 173. (3) 174. (3) 175. (4) 176. (4) 177. (3) 178. (1) 179. (1) 180. (1)

2 2 PHYSICS 4. (2) f 3 5. (4) V rms 9. (1) C C 14. (1) p p v v T Solution C C R constant. 28. (4) Here, Diameter of oxygen molecule, d = 2 Å Radius of oxygen molecule, r = d 2 Å = 1 Å = m 2 2 Less stable, as (=) bond is not in conjugation with carbonyl group. 47. (1) Ethyl acetoacetate represents keto-enol isomerism. 48. (2) CH 3 CH = CHCH 3 and CH 3 CH 2 CH = CH 2 are position isomers. The two isomers differ in the position of double bond. 49. (3) Ethyl alcohol shows functional isomerism with dimethyl ether. C H OH CH O CH Alcohol Ether 50. (4) The acyclic isomers of compound having molecular formula C 4 Oare as Molecular volume = 4 3 r3 N A (where N A is Avogadro s number) = (10 10 ) l0 23 = m 3 Actual volume occupied by 1 mole of oxygen at STP = 22.4 litre = m (4) 7 3 Molecular volume m Actual volume m Cp Cv R R 1 C C C v v v 3 3 = (1) Volume increases (expansion) 41. (2) dq du PdV CHEMISTRY 46. (2) The enols of(3-dicarbonyl compounds are more stable because of conjugation and intramolecular hydrogen bonding. Thus, the order of stability is III > II > I. 51. (4) The following isomers are possible for C 4 H 7 OH. (stabilised by conjugation and hydrogen bonding)

3 3 52. (2) The possible isomerism of C 4 H 8 are as (iii) (iv) Hence, four primary amines are possible for the formula C 4 H 11 N. 58. (2) 53. (1) 54. (3) In vicinal-dihaldes, the two Cl atoms are present at adjacem positions. In case of C 3 H 6 C1 2 only one such structure is possible Hence, only one vicinaldihalide is possible for C 3 H 6 C (2) Compounds having bivalent functional group (like C == O, O, S etc) with at least 4 carbon atoms (in case of ether and thioether) or at least 5 carbon atoms (in case of ketones) exhibit metamerism. Hence, C 2 S C 2 will show metamerism. Since the compound has C n H 2n O type general formula, it must be a ketone (or aldehyde). It has the following metamers (the isomers having same functional group, but different alkyl group attached with the same functional group): Compound with molecular formula C 4 O has the following metamers 55. (2) 56. (1) CH 3 CH 2 OH Ethanol CH 3 O CH3 Dimethyl ether 60. (3) Hence, it has three metamers. 57. (4) C 6 COC 6 does not exhibit tautomerism as it contains no -H atom. For the formula C 4 H 11 N, the possible primary amines are (i) CH 3 CH 2 CH 2 CH 2 NH2 61. (3) The general formula for ethers and alcohols both is C n H 2n + 2 O but the functional group present in ethers is O and in alcohols is OH, so they are functional isomers. The possible chain isomers of hydrocarbon C 5 H 12 are (ii)

4 4 62. (3) The compound is C 3 H 6 Cl 2 and the number of possible isomeric compounds is 5. Hence, three isomeric ethers are possible. 63. (2) 65. (2) Empirical formula is C 2. Empirical formula mass = 29 Molar mass = 58 Molecular formula = (C 2 ) n = C 4 The molecular formula is C 4 and the isomers are two (chair isomers) in number. 66. (3) 64. (3) In acetyl acetone, the entol form is stabilised by H- bonding, hence it has more enol content then others. Isomers of C 4 O are as follows In methoxy methane and ethanol both molecular formula is same, but functional groups are different, so they are functional isomers. 67. (4) Tautomerism is observed in urea. 68. (4) (1) (2) (3)

5 5 (4) 69. (2) Two isomers are possible. 73. (1) When two compounds have similar molecular formula, but differ in the functional group then the isomerism is called functional group isomerism, i.e. 70. (3) Hence, three isomers are possible. The compounds which differ in the nature of carbon chain, are called chain isomers, e.g. 74. (3) Nitro alkanes exhibit tautomerism. In it, -H-atom is labile and form nitrolic acid. 75. (2) 2-pentanone and 3-methyl-2-butanone are chain isomers because they differ in carbon skeleton. 71. (2) Four primary alcohols of the formula C 5 H 12 O are possible. 76. (4) n-pentane and 2-methyl butane are consitutional isomers or chain isomers or skeletal isomers. 77. (2) The possible isomers of C4H9Cl are as follows (i) Structural isomers (1) (2) (3) 72. (3) (4)

6 6 78. (1) C 6 H 14 -hexane. Total of five isomeric hexanes are possible. 79. (4) Half-chair conformation is the most energetic conformation of cyclohexane. 80. (3) Anti-staggered conformation is most stable. Hence, option (3) is correct. 81. (1) Since, the energy difference between two conformations of ethane is not so large, so the eclipsed and staggered conformations interconvert rapidly. Thus, it is impossible to isolate these forms at room temperature. 82. (3) Hence, half-chair is less stable due to torsional and angle strain. 85. (2) The different arrangement of atoms in space that results from the carbon-carbon single bond free rotation by 360 ; are called conformations or conformational isomers and this phenomenon is called conformational isomerism. 86. (1) In the given alternates of conformation of butane staggered-anti conformation is the most stable 87. (2) The geometrical isomers must have (i) (ii) (1) (2) 88. (1) at least one carbon-carbon double bond. The two goups attached to same carbon atom should be different. Geometrical isomers exists in two structural forms cis and tram. In cis form the two group are present on the same side and in tram form the two same group are present on different sides. The isomers which form cis and trans structure, will shou geometrical isomerism. Draw all the choices to find which of them will show geometrical isomerism (3) Choice (a), (c) and (d) have two idnetical group (hydrogen) attached to same double bonded carbon atoms They don t exhibit geometrical isomerism Choice (b) will show geometrical isomerism. 89. (4) Since in staggered conformation, the groups are present as far apart as possible, it is the most stable conformation of n-butane. 83. (4) Planar hexagon conformer has considerable angle strain due to the fact that its bonds are not It also has torsional strain. Due to presence of these strains planar hexagon conformer of cyclohexane is least stable. 84. (1) The stability order of conformation of cyclohexane is chain > twist boat > boat > half chair. This can exhibit geometrical isomerism. 90. (4) As the number of alkyl group attached to double bonded carbon atoms increases, stability of alkene increases (due to hyperconjugation), trans form is more stable than cis form.