common practice test [pmt] : 2017-19€¦ · position isomers. the two isomers differ in the...

CPT-26-/XI-S/PMT/08-01-18

NARAYANA INSTITUTE : A-1/171 A, Janakpuri, New Delhi - 110058, Ph. 011-41576122 / 24 / 25

1

ANSWERS

COMMON PRACTICE TEST [PMT] : 2017-19

CPT-26

73. (1)

74. (3)

75. (2)

76. (4)

77. (2)

78. (1)

79. (4)

80. (3)

81. (1)

82. (3)

83. (4)

84. (1)

85. (2)

86. (1)

87. (2)

88. (1)

89. (4)

90. (4)

91. (1)

92. (4)

93. (2)

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)

37. (3)

38. (1)

39. (1)

40. (1)

41. (2)

42. (4)

43. (3)

44. (4)

45. (2)

46. (2)

47. (1)

48. (2)

49. (3)

50. (4)

51. (4)

52. (2)

53. (1)

54. (3)

55. (2)

56. (1)

57. (4)

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)

1. (1)

2. (1)

3. (3)

4. (2)

5. (4)

6. (3)

7. (2)

8. (3)

9. (1)

10. (4)

11. (2)

12. (3)

13. (1)

14. (1)

15. (3)

16. (3)

17. (4)

18. (1)

19. (1)

20. (3)

21. (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)

CODE

GOL

08/01/2018

109. (3)

110. (3)

111. (2)

112. (1)

113. (4)

114. (4)

115. (4)

116. (2)

117. (2)

118. (4)

119. (3)

120. (3)

121. (2)

122. (4)

123. (1)

124. (3)

125. (3)

126. (4)

127. (3)

128. (4)

129. (3)

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)

145. (4)

146. (4)

147. (3)

148. (2)

149. (3)

150. (3)

151. (1)

152. (2)

153. (3)

154. (1)

155. (4)

156. (1)

157. (1)

158. (2)

159. (3)

160. (4)

161. (4)

162. (4)

163. (4)

164. (4)

165. (3)

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)

CPT-26-/XI-S/PMT/08-01-18


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SolutionPHYSICS

4. (2)

f 35. (4)

rmsV T

9. (1)

p

v

C

C

14. (1)

p vC C R constant.

28. (4)Here,Diameter of oxygen molecule, d = 2 ÅRadius of oxygen molecule,

r = d 2

2 2 Å = 1 Å = 10–10 m

Molecular volume = 4

3r3NA

(where NA is Avogadro’s number)

= 4

3 × 3.14 × (10–10)3 × 6.023 × l023 = 25.2 × 10–7 m3

Actual volume occupied by 1 mole of oxygen at STP= 22.4 litre = 22.4 × 10–3 m3

7 3

3 3

Molecular volume 25.2 10 m

Actual volume 22.4 10 m

= 1.125 × 10–4

29. (4)

p v

v v v

C C R R1

C C C

39. (1)Volume increases (expansion)

41. (2)

dQ dU PdV

CHEMISTRY46. (2)

The enols of(3-dicarbonyl compounds are more stablebecause of conjugation and intramolecular hydrogenbonding. Thus, the order of stability is III > II > I.

(stabilised by conjugation and hydrogen bonding)

Less stable, as (=) bond is not in conjugation with

carbonyl group.47. (1)

Ethyl acetoacetate represents keto-enol isomerism.

48. (2)CH3CH = CHCH3 and CH3CH2CH = CH2 areposition isomers. The two isomers differ in the positionof double bond.

49. (3)Ethyl alcohol shows functional isomerism with dimethylether.

2 5 3 3Alcohol Ether

C H OH CH — O — CH

50. (4)The acyclic isomers of compound having molecularformula C4H10Oare as

51. (4)The following isomers are possible for C4H7OH.

CPT-26-/XI-S/PMT/08-01-18


3

52. (2)

The possible isomerism of C4H8 are as

53. (1)

In vicinal-dihaldes, the two —Cl atoms are presentat adjacem positions. In case of C3H6C12 only onesuch structure is possible Hence, only one vicinal-dihalide is possible for C3H6C12.

54. (3)

Compound with molecular formula C4H10O has thefollowing metamers

55. (2)

3 2Ethanol

CH — CH — OH 3 3Dimethyl ether

CH — O — CH

56. (1)

C6H5COC6H5 does not exhibit tautomerism as itcontains no -H atom.

57. (4)

For the formula C4H11N, the possible primary aminesare

(i) 3 2 2 2 2CH — CH — CH — CH — NH

(ii)

(iii)

(iv)

Hence, four primary amines are possible for theformula C4H11N.

58. (2)

Compounds having bivalent functional group (likeC == O, —O—, —S— etc) with at least 4 carbonatoms (in case of ether and thioether) or at least 5carbon atoms (in case of ketones) exhibitmetamerism. Hence, C2H5 — S — C2H5 will showmetamerism.

59. (2)

Since the compound has CnH2nO type generalformula, it must be a ketone (or aldehyde). It hasthe following metamers (the isomers having samefunctional group, but different alkyl group attachedwith the same functional group):

Hence, it has three metamers.

60. (3)

The general formula for ethers and alcohols both isCnH2n + 2O but the functional group present in ethersis —O— and in alcohols is —OH, so they arefunctional isomers.

61. (3)

The possible chain isomers of hydrocarbon C5H12are

CPT-26-/XI-S/PMT/08-01-18


4

62. (3)

The compound is C3H6Cl2 and the number ofpossible isomeric compounds is 5.

63. (2)

In acetyl acetone, the entol form is stabilised by H-bonding, hence it has more enol content then others.

64. (3)

Isomers of C4H10O are as follows

Hence, three isomeric ethers are possible.

65. (2)Empirical formula is C2H5.Empirical formula mass = 29Molar mass = 58

Molecular formula = (C2H5 )n = C4H10The molecular formula is C4H10 and the isomers aretwo (chair isomers) in number.

66. (3)

In methoxy methane and ethanol both molecularformula is same, but functional groups are different,so they are functional isomers.

67. (4)Tautomerism is observed in urea.

68. (4)

(1)

(2)

(3)

CPT-26-/XI-S/PMT/08-01-18


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(4)

69. (2)

Hence, three isomers are possible.

70. (3)

The compounds which differ in the nature of carbonchain, are called chain isomers, e.g.

71. (2)

Four primary alcohols of the formula C5H12O arepossible.

72. (3)

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.

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.

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)

(4)

CPT-26-/XI-S/PMT/08-01-18


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78. (1)C6H14-hexane.

Total of five isomeric hexanes are possible.79. (4)

Half-chair conformation is the most energeticconformation of cyclohexane.

80. (3)Anti-staggered conformation is most stable. Hence,option (3) is correct.

81. (1)Since, the energy difference between twoconformations of ethane is not so large, so the eclipsedand staggered conformations interconvert rapidly. Thus,it is impossible to isolate these forms at roomtemperature.

82. (3)

Since in staggered conformation, the groups are presentas far apart as possible, it is the most stableconformation of n-butane.

83. (4)Planar hexagon conformer has considerable anglestrain due to the fact that its bonds are not 109.5°. Italso has torsional strain. Due to presence of thesestrains planar hexagon conformer of cyclohexane isleast stable.

84. (1)The stability order of conformation of cyclohexane ischain > twist boat > boat > half chair.

Hence, half-chair is less stable due to torsional andangle strain.

85. (2)The different arrangement of atoms in space that resultsfrom the carbon-carbon single bond free rotation by360°; are called conformations or conformationalisomers and this phenomenon is called conformationalisomerism.

86. (1)In the given alternates of conformation of butanestaggered-anti conformation is the most stable

87. (2)The geometrical isomers must have(i) at least one carbon-carbon double bond.(ii) The two goups attached to same carbon atom

should be different. Geometrical isomers existsin two structural forms cis and tram. In cis formthe two group are present on the same side andin tram form the two same group are present ondifferent sides.The isomers which form cis and trans structure,will shou geometrical isomerism. Draw all thechoices to find which of them will showgeometrical isomerism

(1)

(2)

(3)

Choice (a), (c) and (d) have two idnetical group(hydrogen) attached to same double bonded carbonatoms They don’t exhibit geometrical isomerism

Choice (b) will show geometrical isomerism.88. (1)

89. (4)

This can exhibit geometrical isomerism.90. (4)

As the number of alkyl group attached to doublebonded carbon atoms increases, stability of alkeneincreases (due to hyperconjugation), trans form is morestable than cis form.

common practice test [pmt] : 2017-19€¦ · position isomers. the two isomers differ in the...

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