CUET Chemistry

ANSWERS 213 7. P-BLOCK ELEMENTS MULTIPLE CHOICE QUESTIONS 1. (a) 2. (b) 3. (c) 4. (a) 5. (b) 6. (a) 7. (d) 8. (c) 12. (a) 13. (b) 14. (c) 15. (c) 16. (b) 9. (d) 10. (d) 11. (c) 20. (a) 21. (c) 22. (a) 23. (c) 24. (a) 28. (c) 29. (c) 30. (b) 31. (c) 32. (a) 17. (c) 18. (c) 19. (a) 36. (c) 37. (c) 38. (b) 39. (c) 40. (c) 44. (d) 45. (c) 46. (b) 47. (d) 48. (c) 25. (a) 26. (c) 27. (b) 52. (d) 53. (b) 54. (c) 55. (d) 56. (a) 60. (c) 61. (a) 62. (a) 63. (a) 64. (b) 33. (b) 34. (a) 35. (b) 68. (b) 69. (b) 70. (d) 71. (c) 72. (b) 76. (b) 77. (a) 78. (a) 79. (a) 80. (b) 41. (a) 42. (c) 43. (c) 84. (c) 85. (a) 86. (c) 87. (a) 88. (c) 92. (b) 93. (c) 94. (b) 95. (c) 96. (b) 49. (c) 50. (b) 51. (c) 100. (d) 101. (d) 102. (b) 103. (a) 104. (b) 108. (b) 109. (a) 110. (a) 111. (c) 112. (d) 57. (a) 58. (d) 59. (c) 116. (d) 117. (d) 118. (b) 119. (b) 120. (c) 124. (c) 125. (c) 126. (c) 127. (c) 128. (c) 65. (c) 66. (b) 67. (b) 132. (b) 133. (a) 134. (d) 135. (d) 136. (c) 140. (b) 141. (a) 142. (b) 143. (a) 144. (d) 73. (a) 74. (c) 75. (b) 148. (a) 149. (b) 150. (d) 151. (a) 152. (c) 156. (c) 157. (a) 158. (d) 159. (b) 160. (a) 81. (b) 82. (c) 83. (d) 164. (d) 165. (b) 166. (a) 167. (c) 168. (b) 172. (d) 173. (c) 174. (c) 89. (c) 90. (c) 91. (b) 5. (b) 6. (d) 97. (d) 98. (d) 99. (b) 5. (b) 6. (d) 105. (b) 106. (c) 107. (a) 113. (d) 114. (d) 115. (a) 121. (a) 122. (a) 123. (c) 129. (c) 130. (b) 131. (c) 137. (d) 138. (a) 139. (c) 145. (c) 146. (c) 147. (b) 153. (b) 154. (d) 155. (a) 161. (b) 162. (a) 163. (a) 169. (d) 170. (b) 171. (a) CASE BASED QUESTIONS Case Study 1 1. (b) 2. (c) 3. (a) 4. (c) 4. (d) Case Study 2 2. (c) 3. (c) 4. (c) 1. (d) 2. (b) 3. (c) Case Study 3 1. (a) HINTS/SOLUTIONS 1. Among the given species SiO2 is solid at room 6. SnCl2 has relatively larger ionic character and temperature. solid at room temperature. SnCl4 has larger covalent character. Hence it has low m.p. and is 2. Carbide of tungsten (WC) is an intestitial compound. liquid at room temperature. 3. PbCO3.Pb(OH)2 is called white lead and is used 7. SiC is a network covalent solid. in white paints. 8. Metallic character increases down the group. 4. SiO2 is weakly acidic as well as solid at room Hence Bi is most metallic. temperature. 9. It has small size and hence weak van der waal’s 5. PbO2 is oxidant as well as basic. forces.

214 CHEMISTRY 10. Phosphorus does not show the oxidation state 35. The hybridisation of N in NO2+ is sp, that in of – 2. NO3– is sp2 and in NH4+ is sp3. 11. Among N, P, As, the ionic radius of As3– ion is 36. The boiling point of water (H2O) is highest due maximum. Bi forms trivalent cation Bi3+ but not to H-bonding. Bi3–. 37. The statement c does not apply to P4 molecule. 12. N has high electronegativity and hence easily forms N3–. 38. HPO3 is metaphosphoric acid which exists as cyclic trimer. 13. Inertness of nitrogen is attributed to the high bond dissociation energy of N ≡ N bond. 39. It exists in cyclic trimeric form. 14. Yellow phosphorus dissolve in CS2 while red 40. In trisilyl amine, there is a formation of dπ – pπ phosphorus does not. bond between Si and N atom. This bond refers to as back π donation. 15. BiF3 is most ionic. 16. The most volatile hydride is PH3. In fact boiling 41. NH4NO2 ⎯→ N2 + 2H2O. 42. Yellow phosphorus is more reactive than red points vary as NH3 > BiH3 > SbH3 > AsH3 > PH3. phosphorus. 17. The bond angle is highest around N in ammonia (107°). 43. 2NH3 + NaOCl ⎯→ NH2⎯NH2 + NaCl + H2O. 18. PF5 molecule has two different bond lengths, the 44. In POCl3, pπ – dπ bond is formed between O axial P⎯F bonds are relatively longer than the and P. equatorial P⎯F bonds. 45. N2O5 and P2O5 have entirely different 19. NH3 is the strongest base among the given structures. species. 46. The decreasing order of acid strength is : 20. N2H4 is strongest reducing agent. 21. HN3 is hydrazoic acid and forms corresponding HNO3 > H3PO4 > H3AsO4 > H3SbO4. azides. 47. NH3 + 3Cl2 ⎯→ NCl3 + 3HCl 22. NO2 is dark brown gas. excess 23. BiCl5 does not exist. Bi does not have O.N. of 2NH3 + 3I2 ⎯→ NI3 . NH3 + HI. + 5 due to inert pair effect. 48. Anhydride of H3PO3 is P4O6 24. PF3 cannot be hydrolysed because P⎯F bond is stronger than P⎯O bonds. P4O6 + 6H2O ⎯⎯Hea⎯t → 3H3PO4 + PH3. 49. In H3PO4, O.N. of P is + 5 i.e., highest oxidation 25. Down the group as the size increases the electronegativity decreases and hence the acidic state which cannot further increase. character also increases. 50. P4 ⎯⎯Lim⎯ite⎯d → P4O6 ⎯⎯4Cl⎯2 → 2POCl3 26. 2NF3 + 3H2O(g) ⎯→ 6HF + N2O3. 3O2 Heat + 2PO2Cl Metaphosphoryl chloride 27. KNO2 + NH4Cl ⎯⎯⎯→ KCl + N2 + 2H2O. 28. Its colour depends on heating due to its 51. 2H3PO4 ←⎯H2⎯O⎯ H4P2O7 ⎯⎯875⎯K⎯→ 2HPO3 decomposition into NO2. + H2O 29. The basicity of orthophosphoric acid (H3PO4) is 52. Solid PCl5 exist in ionic form as (PCl4)+ (PCl6)– 53. The structure of H4P2O7 is: 3 because it contains 3 O⎯H bonds. OO 30. H2N2O2 is called hyponitrous acid. 31. The old name of HCl is muriatic acid. ⏐⏐ ⏐⏐ HO⎯ P ⎯O⎯ P ⎯OH 32. BiOCl is called pearl white ⏐⏐ ⏐⏐ 33. [H2SO4 ⎯→ SO2 + H2O + (O)] × 10 OH OH P4 + 10(O) + 6H2O ⎯→ 4H3PO4 P4 + 10H2SO4 ⎯→ 10SO2 + 4H3PO4 + 4H2O 54. When mixture is thrown in water, CaC2 produces C2H2 which burns with luminous flame 34. 2NH3 + 3CuO ⎯→ N2 + 3H2O + 3Cu. and Ca3P2 produces PH3 which causes ignition on coming in contact with air. Heat 55. H3PO2 is hypophosphoric acid and its basicity is 1.

ANSWERS 215 56. 6FeSO4 + 2KNO3 + 4H2SO4 ⎯→ Fe2(SO4)3 + 74. The correct structure of sulphuryl chloride is : K2SO4 + 4H2O + 2NO. O 57. It is due to small size of oxygen. S 58. In H2O2, the O.N. of O is – 1. Cl Cl 59. In FeS2, O.N. of S is – 1 and also in Na2O2, O.N. O of oxygen is – 1. 60. Oxygen does not have vacant d-orbitals and it 75. S2O62– ion is OO and is called || || exhibits a maximum valency of 2. O− ⎯ S ⎯ S ⎯O− || || 61. Plastic sulphur contains irregular chains of OO sulphur atoms. dithionate ion. 62. H2O is least acidic and at the same time possess highest thermal stability. 76. H2SO4 is similar to selenic acid H2SeO4. 77. The oxidation states of S in S8, S2F2 and H2S 63. All the hydrides of group 16 elements possess V shape or bent shape. are respectively 0, + 2 and – 2. 64. Na2S2O3 is antichlor i.e., it is used to remove 78. See the structures of corresponding acids in unused chlorine during bleaching process. Important terms, facts and formulae. 65. Among the given species SO2 is gas while SeO2 79. In contact chamber catalytic oxidation of SO2 and TeO2 are solids. takes place. 66. Both in H2S and S8, the sulphur atoms assume 80. S does not form pπ – pπ bond due to its larger sp3 hybrid states. size. 67. The structure of H2S2O7 is: 81. H2S2O7 is oleum and is obtained by dissolving SO3 in H2SO4. OO || || 82. Data reveals that the element is S and A and B H⎯O⎯S ⎯O⎯S⎯O⎯H are SO2 and H2S || || SO2 + 2H2S ⎯→ 2H2O + 3S. OO 83. Cl2 does not react with sulphuric acid. Thus, it contains one S⎯O⎯S link and two 84. Both I and IV statements are correct. O⎯H bonds. 85. The reaction involved is 2SO2 + O2 ⎯→ 2SO3. 68. 2FeCl3 + H2S ⎯→ 2FeCl2 + 2HCl + S Thus, according to starting moles and also the (Colloidal form) number of moles consumed and formed, A is SO3; 69. S⎯O⎯O⎯S link is present in Marshall’s acid B is SO2 and C is O2. (H2S2O8), which is peroxydisulphuric acid. 86. 109% oleum by mass means that 100 g of sample 70. H2SO4 forms two series of salts i.e., sulphates and bisulphates. Hence, it can exhibit basicity reacts with 9.0 g of H2O to form 109 g of pure H2SO4. Now water absorbs only free SO3 present of 1 and 2. in oleum 71. S8 + 24F2 ⎯→ 8SF6 (Octahedral). 72. S2Cl2 is generally used in vulcanisation of SO3 + H2O ⎯→ H2SO4 rubber. 80 g 18 g 9 g of water can react with SO3 = 80 × 9 = 40 g 18 O Thus, SO3 present in 100 g of sample is 40 g or || it is 40% by mass. 73. H2SO4 is H⎯O⎯ S ⎯O⎯H i.e., it contains 87. The reaction involved is: || O 1 SO2 (g) + 2 O2 ( g) ⎯→ SO3 (g) ; ΔH = –ve 2-OH bonds. Thus, low temperature and high pressure favour the forward reaction.

216 CHEMISTRY 88. SO2 + H2O ⎯→ H2SO3 . The structure of H2SO3 109. HF dissolves glass producing ultimately H2SiF6, .. i.e., hydro-fluorosilicic acid. S 110. Fluorine, due to absence of d-orbitals in the valence shell does not form polyhalides. is HO OH. Thus, it has two –OH groups. 111. Chloro and fluoro sulphonic acids are called O superacids. 89. Trend of electron affinity is Cl < F < Br < I. 112. Carnallite is KCl.MgCl2.6H2O which contains traces of bromine as KBr and MgBr2. 90. Fluoride ion is not oxidised by MnO2. 91. Bond energy of F–F is less than Cl⎯Cl bond due 113. 3OCl– ⎯→ 2Cl– + ClO3–. 114. Fluorine does not form hypohalite ion. to small size of flourine atom. 115. In ClO2–, the chlorine atom assumes sp3 92. ΔHf° of various halide ions in gaseous state in hybridisation. kJ mol– 1 are as under 116. Iodide (I–) ion is strongest reducing agent. F– = 1 ΔHdiss. + EA1 159 − 333 = – 235.5 117. Chlorine being less electropositive cannot 2 =2 replace iodine from KIO3. Cl– = 1 ΔHdiss. + EA1 = 243 − 348 = – 226.5 118. 3H2SO4 + 3KClO3 2 2 ⎯→ 3KHSO4 + 2ClO2 + HClO4 + H2O. Br– = 1 ΔH vap. + 1 ΔHdiss. + EA1 119. CH3COO– ion is not a pseudo halide ion. 2 2 120. Caliche contains iodine as NaIO3. 121. I– ion is a ligand while I2 molecule is acceptor of = 31 193 − 324 = – 212 2 +2 lone pair from I– ion to form I3– ion. 122. H2PtCl6 + NH3 ⎯→ (NH4 )2PtCl6 + H2O. I– = 1 ΔHfus + 1 ΔHvap. + 1 ΔHdiss. + EA1 2 2 2 Ammonium chloroplatinate (Yellow ppt.) = 15 + 44 + 151 − 295 = – 190. 2 2 2 123. CaOCl2 converts ethanol and acetone into chloroform which is anaesthetic agent. 93. Fluorine because of its highly reactive nature. 124. Step 2 : 3Cl2 + 6KOH 94. HCl has lowest boiling point. ⎯⎯Hot⎯→ KClO3 + 5KCl + 3H2O 95. H⎯I bond length is maximum and H⎯F has Step 3 : KClO3 + I2 ⎯→ KlO3 + Cl2. highest bond energy. 125. Iodine react with Na2S2O3 to form colourless NaI and Na2S4O6. 96. Ionic character depends on the difference of 2Na2S2O3 + I2 ⎯→ 2NaI + Na2S4O6. electronegativity. 126. NaClO + 2HCl ⎯→ NaCl + H2O + Cl2. 127. HF combines with F– to form (HF2)–. 97. NaCl has no reaction with iodine. 128. Ease of liquification increases with size of the noble gas due to increase in strength of 98. KI + I2 KI3 . I2 and I– form complex ion I3– interparticle forces (van der Waal’s forces). 99. IF5 is square pyramidal. 129. Interparticle forces in noble gases are van der Waal’s forces. 100. 2Cl2 + 2HgO ⎯⎯⎯→ HgCl2.HgO + Cl2O. 130. Argon. 101. 2NaClO3 + SO3 + H2SO4 ⎯→ 2ClO2 + 2NaHSO4. 131. Helium. 102. 2HIO3 ⎯→ I2O5 + H2O. 132. Xe+ [PtF6]–. 103. HClO4 is strongest acid and thus gives most 133. XeF2 is fluorine donor whereas SbF5 is fluoride stable corresponding anion, (ClO4–). acceptor. 104. 2ClO3– + I2 ⎯→ 2IO3– + Cl2. 134. No compound of He, Ne and Ar is known. 105. ICl2– is not a pseudohalide ion. 106. More the number of oxygen atom more is its oxidising strength. 107. 2HCN + NO2 ⎯→ (C(yCanNog)2en) + NO + H2O Cyanogen is a pseudohalogen. 108. All oxoacids of halogens are monobasic.

ANSWERS 217 135. In XeF4, Xe is linked to four atoms and there 154. I2 is coloured but on reaction with Na2S2O3 it are two lone pairs around it. produces colourless products NaI and Na2S4O6. 136. XeO3 is an explosive compound. 155. In gaseous state PCl5 is covalent but in solid 137. The statement (d) is the correct explanation. state it exists as [PCl4]+ [PCl6]– i.e., ionic 138. PtF6 is a good oxidising agent. It even oxidises 156. Among the given elements sulphur forms dioxygen. dπ – pπ bonding in its oxides such as SO2 and 139. He is highly inert in character and lightest SO3. 140. He has the smallest surface area. 157. Tin dissolves in conc. HNO3 to form metastannic acid and also in NaOH to form sodium stannite 141. Helium. or sodium stannate. 142. Xenon atom is largest in size and hence is most easily polarisable. 158. SF4 has see-saw shape, CF4 has tetrahedral shape and XeF4 has square planar shape. The 143. 2XeF6 + SiO2 ⎯→ 2XeOF4 + SiF4. which number of lone pairs in SF4 is one, in CF4 is zero and in XeF4 is two. 144. Xe+ [PtF6]–. 159. Tri-n-butyl phosphate is not a wax. Hence, this 145. XeF6 on hydrolysis produces XeO3 explodes. matching is wrong. 146. Atomic radii increase from top to bottom in a 160. H2S + Pb(CH3COO)2 ⎯→ PbS + 2CH3COOH. group. Black 147. XeF4 has two lone pairs and 4-bond pairs. 161. The answer is based on the fact that smaller 148. Extent of adsorption on coconut charcoal the size of cation larger will be the covalent character. increases with increase in atomic number of the noble gas. 162. Acid strength decreases as Cl2O7 > SO2 > P4O10. 163. Nitric acid oxidises antimony (Sb) to antimonic 149. Helium. acid (H3SbO4). 150. XeF6 + CsF ⎯→ Cs+ [XeF7]–. 164. 3KClO3 + 3H2SO4 Here XeF6 acts as fluoride acceptor. ⎯→ 3KHSO3 + 2ClO2 + HClO4 + H2O. 151. Boiling points of noble gases increases as the size increases. 165. SiH4 + 2O2 ⎯→ SiO2 + 2H2O Silane 152. For monoatomic gases cp/cv ratio is 1.66. 166. A is P4O6. It dissolves in water to give H3PO3(B) which gives white ppt. of Ag3(PO4) with AgNO3. 153. SO2 bleaches by reducing the colouring matter whereas Cl2 bleaches by oxidising the colouring matter. 8. d- AND f-BLOCK ELEMENTS MULTIPLE CHOICE QUESTIONS 1. (c) 2. (c) 3. (c) 4. (b) 5. (a) 6. (d) 7. (c) 8. (b) 12. (d) 13. (a) 14. (c) 15. (b) 16. (d) 9. (c) 10. (b) 11. (a) 20. (c) 21. (c) 22. (a) 23. (b) 24. (c) 28. (a) 29. (d) 30. (b) 31. (d) 32. (c) 17. (b) 18. (d) 19. (d) 36. (c) 37. (a) 38. (a) 39. (b) 40. (c) 44. (b) 45. (d) 46. (a) 47. (c) 48. (c) 25. (c) 26. (d) 27. (d) 52. (d) 53. (b) 54. (b) 55. (b) 56. (b) 60. (b) 61. (d) 62. (b) 63. (c) 64. (d) 33. (c) 34. (d) 35. (a) 68. (a) 69. (b) 70. (b) 71. (d) 72. (b) 76. (d) 77. (b) 78. (d) 79. (b) 80. (d) 41. (d) 42. (d) 43. (a) 49. (a) 50. (d) 51. (c) 57. (b) 58. (d) 59. (c) 65. (b) 66. (d) 67. (a) 73. (a) 74. (b) 75. (a)

218 CHEMISTRY 81. (c) 82. (a) 83. (b) 84. (c) 85. (b) 86. (b) 87. (a) 88. (a) 89. (a) 90. (d) 91. (b) 92. (a) 93. (b) 94. (a) 95. (c) 96. (a) 97. (d) 98. (d) 99. (d) 100. (b) 101. (a) 102. (a) 103. (c) 104. (d) 105. (a) 106. (a) 107. (d) 108. (c) 109. (b) 110. (c) 111. (a) 112. (b) 113. (d) 114. (a) 115. (a) 116. (c) 117. (c) 118. (b) 119. (c) 120. (c) 121. (d) 122. (b) 123. (a) 124. (d) 125. (d) 126. (b) 127. (b) 128. (d) 129. (b) 130. (d) 131. (b) 132. (c) 133. (b) 134. (c) 135. (a) 136. (b) 137. (b) 138. (c) 139. (b) 140. (a) CASE BASED QUESTIONS Case Study 1 1. (c) 2. (c) 3. (b) Case Study 2 1. (b) 2. (b) 3. (a) Case Study 3 1. (c) 2. (c) 3. (c) HINTS/SOLUTIONS 1. (n – 1) d1 – 10 ns1 – 2. 19. Osmium (Os). 2. Zn exhibits only +2 state in its compounds. 3. Melting point along a series of transition 20. 2Cu + Cl2 ⎯→ Cu2Cl2. Copper has oxidation state of + 1. elements increases up to chromium sub group. 4. + 3 is the most common oxidation state of 21. Pyrolusite is MnO2. 22. Magnetic moment is proportional to the number lanthanides. 5. Iron. of unpaired electrons. The number of unpaired 6. Mercury belongs to group-12. electrons is maximum in 3d5. 7. Oxides of transition metals in highest oxidation 23. Technetium (Z = 43) is the first man-made are acidic in nature. The oxidation state of element. It belongs to second (4d) transition Cr = + 6. series. 8. Sc3+ (3d0) has no electrons in d-subshell and hence d-d transitions are not possible. 24. Fe3+ : [Ar] 3d5, has five unpaired electrons. 9. Fe3+ (3d5) has five unpaired electrons. 25. During conversion of Cu to Cu2+, two electrons 10. In the first transition series Mn shows the are lost. highest oxidation state (+ 7). 26. Oxidation state of manganese in KMnO4 is + 7 11. Technitium. whereas in MnSO4, MnO2 and K2MnO4 is + 2, + 4 and + 6 respectively. 12. Oxidation number of Cr is + 6 in K2CrO4. 13. Nickel sulphide (NiS) is black in colour. 27. The statement (d) is the correct explanation. 14. Cadmium exhibits oxidation state of + 2 only. 28. Mn2O7 is highly explosive compound. 15. Tungsten. 29. Mn2+ formed during the reaction catalyses the 16. In Fe2+ (3d6) , d-d transitions are possible reaction between KMnO4 and oxalic acid whereas in others d-subshell is fully filled and solutions. hence d-d transitions are not possible. 30. MnO–4 + 8H+ + 5e– ⎯⎯→ Mn2+ + 4H2O ] × 2 17. Species with unpaired electrons are S2– ⎯⎯→ S + 2e– ] × 5 paramagnetic whereas species with all their electrons paired are diamagnetic. 2MnO–4 + 5S2– + 16H+ ⎯⎯→ 2Mn2+ + 5S + 8H2O 18. Iron is used as catalyst in Haber’s process 5 moles of sulphide react with 2 moles of KMnO4. whereas molybdenum acts as promoter in the ∴ 1 mole of sulphide would react with 2/5 process. moles of KMnO4.

ANSWERS 219 31. The statement (d) is not correct. Interstital 56. The oxides with intermediate oxidation states compounds are rather relatively less reactive. are amphoteric in character. 32. CdS is yellow in colour. ZnS is white whereas 57. (NH4)2 Cr2O7 ⎯→ Cr2O3 + N2 + 4H2O. NiS and HgS are black in colour. 58. Osmium has electronic configuration 5d6 6s2. It 33. Cobalt is present in vitamin B-12. exhibits highest oxidation state. 34. Titanium is known as wonder metal. 59. Zr and Hf have almost equal ionic radii because of lanthanoid contraction. 35. μ = n(n + 2) B.M. 60. Species with unpaired electrons are 36. Zn2+ (3d10) has fully filled d-subshell and hence paramagnetic. d-d transitions are not possible. 61. MnO4– + 8H+ + 5e– ⎯→ Mn2+ + 4H2O. Mn2+. 37. Chromite is FeCr2O4. and in 38. [Ar] 3d5 4s2. In acidic medium KofMMnnOi4niMs nreOd4u–ciesd+t7o Oxidation number 39. [Xe] 4f7 5d1 6s2. Mn2+ is + 2. 40. The expected increase in size is Hf is cancelled 62. Ag (Z = 47) and Cd (Z = 48). due to decrease in size because of lanthanoide contraction. 63. Cd2+ (5d10) has no unpaired electrons and hence diamagnetic. 41. MnO4– has a tetrahedral structure ∴ Mn is sp3 hybridised. 64. Mn2+ (3d5) has five unpaired electrons whereas Zn2+ (3d10), Cu+ (3d10) and Sc3+ (3d0) have all 42. (NH4)2 Cr2O7 ⎯→ N2 + Cr2O3 + 4H2O. their electrons paired. 43. HMnO4. 44. 65. Oxalates are oxidised to CO2 by acidified KMnO4 solution. E= Formula mass . 66. Chromic acid is H2CrO4. 67. Mn2+ (3d5), Fe3+ (3d5). No. of electrons lost or gained per molecule 68. Cd2+ (4d10) and Ag+ (4d10) have fully filled d- subshell and hence d-d transitions are not 45. x −2 −1 . possible. Cr O2 Cl2 x + 2(– 2) + 2(– 1) = 0 or x = + 6. 69. AgI has all paired electrons and hence [Ag(NH3)2]+ would be diamagnetic. 46. It does not contain d-electrons in +3 oxidation 70. CrO5 has two O–O bonds. state. 47. Tendency of an element to form complexes does O not depend on whether the element shows variable oxidation states or not. OO 48. 2MnO2 + 4KOH + O2 ⎯⎯Hea⎯t → Cr 2K2MnO4 + 2H2O. OO 49. Mn2+ has 5 unpaired electrons in 3d-subshell. 71. Oxides of transition metals in highest oxidation 50. Zn has greater ionisation enthalpy than alkali state are acidic in nature. and alkaline earth metals and hence its Oxidation state of Mn in Mn2O7 is + 7. hydroxide would be weak base. 72. I– + 2MnO4– + H2O ⎯→ IO3– + 2MnO2 + 2OH–. 51. AgNO3 is sensitive to light. 73. V is in its high oxidation state in V2O5 and hence 52. d-d transitions become possible in incomplete acidic in character. d-subshell. 74. MnO2 + 2KOH + O ⎯→ K2MnO4 + H2O 53. Oxidation state of Mn is + 7stianteKMshnoOw4n. Up to Mn the highest oxidation by a Potassium transition element is equal to the number of manganate electrons in ns and (n – 1) d orbitals. 75. CuCl2 + Cu ⎯→ 2CuCl. 54. Paramagnetic nature is due to unpaired 76. Cr2O72– + 6I– + 14H+ ⎯→ 2Cr3+ + 3I2 + 7H2O. electrons. 77. CrO2Cl2 + 4NaOH 55. Lutetium is the last element of lanthanoids. Due ⎯→ Na2CrO4 + 2NaCl + 2H2O. to lanthanoid contraction it has the smallest 78. CrO3 + 2HCl ⎯→ CrO2Cl2 + H2O. size.

220 CHEMISTRY 79. In Cu(II) (3d9) there is one unpaired electron 101. 2/5 moles. and hence it is paramagnetic. It is coloured due 2MnO4– + 5SO32– + 6H+ ⎯→ to of d-d transitions. 2Mn2+ + 5SO42– + 3H2 O. 80. In alkaline solution KMnO4 is reduced to MnO2. 102. The order of densities is There is a decrease in oxidation number by 3 alkali metals < alkaline earth metals < (from + 7 to + 4) per molecule of KMnO4. transition metals. Scandium is the lightest transition element. Molecular mass E= . 103. Dichromate ion. Change in oxidation number per molecule M O O O– =3 O– O 81. 4FeCr2O4 + 16NaOH + 7O2 ⎯→ Cr Cr 8Na2CrO4 + 2Fe2O3 + 8H2O. OO 82. Ce4+ has tendency to go over to Ce3+ and hence acts as oxidising agent. There are 8 Cr—O bonds. 104. Unlike p-block elements, the various oxidation 83. Lu3+ being smallest (due to lanthanoid contraction) has maximum tendency to form states of a transition element differ by one unit. complexes. 105. Cr2O72– + 4H2O2 + 2H+ ⎯→ 2CrO5 + H2O 84. Oxidation state of Fe in [Fe(CN)6]3– is + 3. (Blue colour in ether) Fe3+ : [Ar] 3d5. It has 5 unpaired electrons. 106. Cu(II) has 3d9 configuration. It has one unpaired 85. Chromyl chloride is CrO2Cl2. electron. 86. Cu2+ + 4NH3 ⎯→ [Cu(NH3)4]2+. 87. 2MnO4– + 5C2O42– + 16H+ ⎯→ Magnetic moment = n(n + 2) B.M., 2Mn2+ + 10CO2 + 8H2O. i.e., 3 B.M. = 1.73 B.M. 88. CuF2 is ionic in nature. 107. M(s) ⎯→ Mn+ (aq) + ne–. 89. Cr2O72– + 3Sn2+ + 14H+ ⎯→ The above change involves sublimation, 2Cr3+ + 3Sn4+ + 7H2O. ionisation and hydration. 90. Transition elements exhibit variable valencies. 108. Cr2O72– + 14H+ + 6e– ⎯→ 2Cr3+ + 7H2O. 91. K2Cr2O7 is reduced to Cr3+. E = M. mass = M . 92. 5C2O42– + 2MnO4– + 16H+ ⎯→ Change in O.N. 6 2Mn2+ + 10CO2 + 8H2O. 109. Acidic solution favours dichromate whereas 93. In CuF2 copper is in + 2 state. alkaline solution favours chromate. 94. CrO3 + 2OH– ⎯→ CrO42– + H2O. 95. In Zn complexes d-d transitions are not possible. 110. KMnO4 + 3H2SO4 ⎯→ 96. KNKMMO2nn–OO, S44.2–toanMd Cnl2–+.uCndOe3r2g–ohoaxsidnaotiorneaacntidonredwuitche K+ + MnO3+ + H3O+ + 3HSO4– 97. 4d and 5d series of transition elements have (green) similarities due to lanthanoid contraction. 111. 2KMnO4 + 3H2SO4 ⎯→ 98. Ni2+ (3d8) has only 2 unpaired electrons. Mn2O7 + 2K+ + H3O+ + 3HSO4–. 99. bFeec3a+ uasnedFSeOa4n2d– ions cannot be oxidized further S are in their highest oxidation (explosive) state. 112. Among Cr3+, Fe2+, Cu2+ and Zn2+, Fe2+ has 100. In neutral and alkaline medium: maximum number of unpaired electrons. MnO4– + 2H2O + 3e– ⎯→ MnO2 + 4OH– 113. Cu2+ + 2KCN ⎯→ Cu(CN)2 + 2K+ In acidic medium : 2Cu(CN)2 ⎯→ 2CuCN + (CN)2 CuCN + 3KCN ⎯→ K3 [Cu (CN)4]. 114. Ti4+ (3d0) and Cu+ (3d10) cannot undergo d-d transitions. MnO4– + 8H+ + 5e– ⎯→ Mn2+ + 4H2O.

ANSWERS 221 115. [FeC2O4 ⎯→ Fe3+ + 2CO2 + 3e–] × 5 CASE BASED QUESTIONS [MnO4– + 8H+ + 5e– ⎯→ Mn2+ + 4H2O] × 3 Case Study 1 5FeC2O4 + 3MnO4– + 24H+ 1. 4KI + 2CuSO4 ⎯→ 5Fe3+ + 10CO2 + 12H2O + 3Mn2+ ⎯⎯→ 2K2SO4 + Cu2I2 ↓ + I2 116. Oxidation state of V in VOSO4 is + 4, V(IV) has one electron in d-subshell. white ppt. 117. Fe3+ salts give blood red colour with KSCN. 2. 2Na2S2O3 + I2 ⎯⎯→ Na2S4O6 + 2NaI 118. In dichromate ion the six terminal Cr–O bonds Oxidation number of S in Na2S2O3 is + 2 and in Na2S4O6 is + 2.5 are equivalent due to resonance. 3. Cu2+ is reduced to Cu+. Hence, oxidation number of copper decreases from + 2 to + 1. 9. CO-ORDINATION COMPOUNDS MULTIPLE CHOICE QUESTIONS 1. (c) 2. (b) 3. (b) 4. (b) 5. (d) 6. (b) 7. (c) 8. (c) 12. (a) 13. (b) 14. (c) 15. (c) 16. (b) 9. (b) 10. (a) 11. (c) 20. (c) 21. (c) 22. (d) 23. (d) 24. (a) 28. (a) 29. (c) 30. (a) 31. (b) 32. (d) 17. (c) 18. (d) 19. (b) 36. (b) 37. (b) 38. (c) 39. (a) 40. (d) 44. (d) 45. (c) 46. (d) 47. (b) 48. (d) 25. (a) 26. (c) 27. (c) 52. (b) 53. (c) 54. (c) 55. (c) 56. (c) 60. (b) 61. (d) 62. (d) 63. (b) 64. (a) 33. (c) 34. (c) 35. (a) 68. (a) 69. (c) 70. (b) 71. (b) 72. (d) 76. (c) 77. (d) 78. (c) 79. (d) 80. (b) 41. (c) 42. (b) 43. (c) 84. (d) 85. (a) 86. (a) 87. (a) 88. (b) 92. (a) 93. (b) 94. (b) 95. (a) 96. (b) 49. (c) 50. (b) 51. (c) 100. (a) 101. (b) 102. (d) 103. (c) 104. (b) 108. (b) 109. (c) 110. (a) 111. (a) 112. (d) 57. (b) 58. (a) 59. (c) 116. (a) 117. (d) 118. (d) 65. (a) 66. (d) 67. (c) 5. (c) 6. (a) 73. (d) 74. (a) 75. (b) 5. (d) 81. (b) 82. (a) 83. (c) 89. (c) 90. (c) 91. (a) 97. (b) 98. (a) 99. (b) 105. (c) 106. (d) 107. (b) 113. (d) 114. (c) 115. (b) CASE BASED QUESTIONS Case Study 1 1. (c) 2. (b) 3. (b) 4. (d) Case Study 2 1. (d) 2. (a) 3. (d) 4. (c) Case Study 3 1. (a) 2. (c) 3. (a) Case Study 4 1. (b) 2. (d)

222 CHEMISTRY HINTS/SOLUTIONS 1. 2, 2-Dipyridyl is a bidentate ligand and also 22. [Pt (en)2Cl2] will show two geometrical forms and causes chelation. cis form will show optical isomerism. 2. K2SO4. Al2 (SO4 )3. 24H2O 23. Cr does not form polynuclear carbonyl. (Potash alum) 24. Cisplatin whose structure is given in ‘a’ is effective against cancers. ⎯⎯Wa⎯ter⎯→ 2K+ + 2Al3+ + 4SO42– + 24H2O Thus 8 mol of ions are produced per mole of 25. The trichlorohydroxogallate (III) ion is potash alum. [Ga(OH)Cl3]–. The total charge on the ion is 3 + (– 1) + (– 3) = – 1. 3. Glycenato is an unsymmetrical bidentate ligand. 4. O.N. of Fe is + 2.x + 5(0) + 0 = + 2 or x = + 2. 26. In the given complex oxidation state of Pt is 3(– 1) + x = – 1 or x = + 2. 5. Since O.N. of cobalt is + 3 and that of Cd is + 2. Thus charge on [Cd Clx]3– should be – 3. Hence 27. Secondary valencies do not represent O.N. x + 2 = – 3 or x = 5. 28. The correct relation is Δt = 4 Δ o . 6. It is phenanthroline with two coordinating sites. 9 7. Choice (c) represents the correct name. 29. It forms complex with lead. 8. These are coordination isomers. III III 9. The given species are linkage isomers of each 30. Fe[ Fe (CN)6 ] is called Berlin green. other. 31. IUPAC name is given correctly in ‘d’. 10. The two ionisation isomers are 32. The complex pertains to the general formula (Mabcd) and consequently it shows three [Co(NH3)5 SO4]Br and [Co(NH3)5Br]SO4. geometrical isomers. The later will give white precipitate of BaSO4 with BaCl2 solution but former will not give 33. Small magnitude of charge on the central metal precipitate. atom does not help in formation of stable 11. K4 [PtCl6] ionises as 4K+ + [PtCl6]4– ions but does complexes. not contain free Cl– ions. 34. The structure ‘c’ given in the question is chiral 12. CuCl + 3KCl ⎯⎯Wa⎯ter⎯→ (SKolu3b[lCe cuoCmlp4le]x) and thus, it exhibits enantiomerism. 13. [Fe(CN)6]4– is inner orbital complex. 35. B, C, D do not have metal carbon bond. 14. The ion is [Cu(Cl)4]3–. It is a trivalent anion. 15. It is not a π-bonded complex. 36. It does not have partly filled d-orbitals. 16. It is a π acid ligand. 37. The formula ‘b’ is correct represention of the given name. 17. The given complex is square planar and Cu assumes dsp2 hybridisation in it. 38. [Ni(CN)4]2+ has square planar shape. 39. ‘a’ choice represents the correct IUPAC name. 18. AgCl + 2NH4OH ⎯→ [Ag(NH3)2] Cl + H2O. 19. Y is K[PtCl3(C2H4)] a Zeise’s salt. 40. The possible isomers are 20. [(NH3)5 Cr–O–O–Cr(NH3)5] is a binuclear [Cu(NH3)4] [PtCl4] ; [Cu(NH3)3Cl][PtCl3(NH3)] complex entity. [Pt(NH3)3Cl] [CuCl3(NH3)] and [Pt(NH3)4][CuCl4]. 21. cis-[Co(NH3)2(en)2]3+ can exhibit two 41. CuSO4 + 4NH3 ⎯→ [Cu(NH3)4]SO4. enantiomeric forms as shown below. 42. [Cr(en)3]3+, among the given species exhibits optical isomerism. en en 43. By definition a mixed complex contains more en Co Co en than one type of ligands. NH3 NH3 H3N NH3 44. It is evident from the structure of Ni-dmg complex that co-ordination number of Ni is 4. (Enantiomeric forms) 45. II ]2+ II Cl4 ]2− [Pt(py)4 [Pt

ANSWERS 223 46. C.N. is equal to the number of co-ordinate bonds 66. CN– ligands form relatively more stable between central metal atom and surrounding complexes as compared to halo groups. ligands. 67. Both [Co(en)3]3+ and [Co(Ox)3]3– exhibit optical 47. O.N. of Ni in Ni(CO)4 is 0 ; in K2 [NiF6] it is + 4. isomerism. In [Ni(NH3)6] (BF4)2 it is + 2 and in K4 [Ni(CN)6] it is + 2. 68. [Co(Ox)3]3– cannot show geometrical isomerism. 69. Zn2+ + K4 [Fe(CN)6] → Zn2 [Fe(CN)6] + 4K+. 48. There are 12 CO groups and 3 Mn atoms. 49. Choice itself is the answer. Zinc ferrocyanide is white precipitate with bluish 50. The compounds represent resonating forms. tinge. 51. The arrangement III and IV are same. 70. [Pt(NH3)5Cl]Cl3 would ionise as 52. They represent co-ordination position [Pt(NH3)5Cl]Cl3 Water [Pt(NH3)5Cl]3+ + 3Cl–. isomerism. 71. In ligand isomerism, the ligand itself exhibit 53. [Cr(H2O)4 Cl2] Cl.2H2O ⎯→ [Cr(H2O)4Cl2]+ + Cl– + 2H2O. isomerism and thus cause difference in the 54. Cu assumes dsp2 hybrid state and there is 1 structures. unpaired electron. 72. 2KI + HgI2 ⎯→ K2HgI4 (Nessler’s reagent). 55. Ni(CO)4 does not have any unpaired electron. 73. The nickel-dmg complex is as given below 56. See bonding. Both [Fe(CN)6]4– and [Ni(CN)4]2– H have five unpaired electrons. OO 57. In [Fe (CN)6]3– iron assumes d2sp3 hybrid state. CH3—C N N C—CH3 58. The magnetic moment of [FeF6]3– is highest Ni because of highest number of unpaired electrons. It has 5 unpaired electrons. CH3—C N N C—CH3 59. Acetylacetonato ion is OO O O– H It is monovalent anion and bidentate ligand. Thus, it has four chelates around Ni atom. III II 74. In Mn2(CO)10, there is one Mn-Mn bond Electrons in 2 Mn = 50 60. Prussian blue is K Fe[ Fe (CN)6 ]. Electrons donated by 10 CO = 20 61. [Cu(NH3)4]2+ is square planar structure. Electrons in one Mn-Mn bond = 2 62. All d-orbitals are completely filled. Total = 72 63. AgBr unaffected by light is removed as soluble ∴ EAN per Mn = 72 = 36. complex entity. 2 AgBr + 2 NaH2ySp2oO3 ⎯→ Na3[Ag(S2O3)4 ] + NaBr. 75. This is a conceptual question. Soluble complex 76. The complex entity [Zn(NH3)4]2+ is tetrahedral as zinc assumes sp3 hybrid state. 64. Tenhehagsivtewnoiodnoniosr[Csiot(eesn. )T2Chlu(sOtHh)e]+c1o. -Soirndcineaetaiochn number of Co is 6. 77. Ziese’s salt does not furnish Cl– ions in solution. 65. d7 (high spin) ­ ­ eg three unpaired e 78. It has unpaired electrons hence, it is paramagnetic. ­¯ ­¯ ­ t2g 79. As the complex has no unpaired electron. This d9 (high spin) ­¯ ­ eg one unpaired e means the hybrid state is dsp3. d6 (low spin) ­¯ ­¯ ­¯ t2g 0 unpaired e 80. AgCl + 2Na2S2O3 ⎯→ Na3[Ag(S2O3)2] + NaCl. d4 (low spin) eg two unpaired e 81. Ni(CO)4 is tetrahedral and [Ni(P Ph)3Cl2] is ­¯ ­¯ ­¯ t2g ­¯ eg square planar. ­ ­ t2g

224 CHEMISTRY 82. MnO4– does not have any d-electron in central [Pt(NH3)5Cl] Cl3 [Pt (NH3)5Cl]+ + 3Cl– metal atom. 3Cl– + 3AgNO3 ⎯→ 3AgCl + 3NO3–. 83. It pertains to formula [Co(NH3)Cl3]. Hence it 96. Charge on the complex ion is 2(+ 3) + 2(– 1) = + 4. does not ionise in aqueous solutions. Since charge on each sulphate ion is – 2. Hence, 84. It can also be named as there should be two sulphate ions. The value of Tri-μ-carbonylbis(tricarbonyl) iron (III) x = 2. 85. The complex is [(NH3)5Cr – OH – Cr(NH3)5]Br5. 97. The chemical reaction involved is 86. Tollen’s reagent is [Ag(NH3)2]OH. Thus, O.N. NiSO4 + 4Py + 2NaNO2 ⎯→ Na2SO4 of Ag = + 1 ; CN = 2 and EAN = 47 – 1 + 4 = 50. + [Ni(Py)4(NO2)2]. 87. Tetrachloropalladium (II) ion is [PdCl4]2– 98. The cation is [Co(en)2CO3]+. It will show geometrical isomerism, and its cis-isomer will ∴ EAN = 48 – 2 + 8 = 54. 88. II and III are optical isomers and not geometrical also show optical isomerism. isomers. 99. As per data the formula of complex is [MCl.(H2O)5]Cl2. 89. Co(NO2)3 + 3KNO2 ⎯→ K3 [Co(NO2)6 ] Now 200 ml of 0.01 M solutions contains Cl– ions Potassium cobaltinitrite 90. Their molar conductivities are different because = 2 × moles of complex they contain different number of ions = 2(200 × 10–3 × 0.01) [Cr(H2O)6]Cl3 ⎯→ [Cr(H2O)6]3+ + 3Cl– but [Cr(H2O)5Cl]Cl2.H2O = 4 × 10–3 Moles of AgNO3 required = moles of Cl– ⎯→ [Cr(H2O)5Cl]2+ + 2Cl– + H2O. 91. [ZnCl4]2– and [MnCl4]2– are tetrahedral = 4 × 10–3 Volume of 0.1 AgNO3 required = moles/molarity complexes. = 4 × 10−3 L 92. [FeF6]3– and [CoF6]3– are outer orbital 0.01 complexes. = 0.04 × 1000 = 40 ml. 3d 4s 4p 100. Hybrid state assumed by Pt is dsp2 hence the 93. Fe2+ : arrangement is square planar. Since there are no unpaired electrons the complex is Fe(C5H5)2 : diamagnetic. . . . . . . . . . . . . d2sp3 101. Na2 [Fe(CN)5NO] C.N. of Fe = 6 six e pairs from two C5H5− ions O.N. of Fe = + 3 94. All the statements except ‘b’ are incorrect. The two geometrical forms are. EAN of Fe = 26 – 3 + 12 = 35. Cl NH3 102. They are ionisation isomers but their molar conductivity is different due to different Cl NH3 Cl NH3 magnitude of charge on the cation and anion. Co Co 103. For d8 ion to adopt square planar ligand field, Cl NH3 Cl Cl the number of unpaired electrons are nil. Hence the magnetic monoment is zero. NH3 NH3 104. The formation constant β3 is given as fac isomer mer isomer β3 = k1 × k2 × k3 or 95. The formula which corresponds to above data is log β3 = log k1 + log k2 + log k3 = 9.70 [Pt(NH3)5Cl]Cl3. Its ionisation in aqueous β3 = Antilog(9.70). solution is shown as

ANSWERS 225 10. HALOALKANES AND HALOARENES MULTIPLE CHOICE QUESTIONS 1. (a) 2. (b) 3. (b) 4. (a) 5. (c) 6. (d) 7. (d) 8. (d) 12. (c) 13. (c) 14. (c) 15. (a) 16. (d) 9. (b) 10. (d) 11. (c) 20. (d) 21. (c) 22. (c) 23. (c) 24. (a) 28. (a) 29. (b) 30. (c) 31. (c) 32. (b) 17. (c) 18. (c) 19. (c) 36. (c) 37. (b) 38. (d) 39. (c) 40. (d) 44. (d) 45. (b) 46. (a) 47. (d) 48. (c) 25. (b) 26. (c) 27. (c) 52. (d) 53. (c) 54. (b) 55. (b) 56. (d) 60. (b) 61. (b) 62. (d) 63. (d) 64. (d) 33. (d) 34. (b) 35. (d) 68. (c) 69. (b) 70. (b) 71. (c) 72. (d) 76. (c) 77. (d) 78. (d) 79. (a) 80. (c) 41. (d) 42. (c) 43. (b) 84. (b) 85. (b) 86. (d) 87. (d) 88. (a) 92. (c) 93. (b) 94. (d) 95. (d) 96. (c) 49. (b) 50. (a) 51. (d) 100. (b) 101. (a) 102. (b) 103. (b) 104. (b) 108. (d) 109. (b) 110. (a) 111. (a) 112. (a) 57. (d) 58. (d) 59. (b) 116. (b) 117. (b) 118. (c) 119. (b) 120. (a) 124. (d) 125. (d) 126. (b) 127. (a) 128. (c) 65. (d) 66. (c) 67. (c) 132. (d) 133. (a) 134. (c) 135. (d) 136. (c) 140. (a) 141. (b) 142. (a) 143. (a) 144. (b) 73. (d) 74. (b) 75. (a) 148. (d) 149. (a) 150. (a) 151. (c) 152. (d) 156. (d) 157. (a) 158. (a) 159. (a) 160. (b) 81. (a) 82. (b) 83. (c) 164. (a) 165. (b) 166. (b) 89. (d) 90. (a) 91. (b) 5. (b) 97. (b) 98. (a) 99. (b) 105. (c) 106. (d) 107. (a) 113. (c) 114. (b) 115. (b) 121. (b) 122. (a) 123. (d) 129. (c) 130. (a) 131. (b) 137. (d) 138. (b) 139. (b) 145. (b) 146. (d) 147. (a) 153. (c) 154. (b) 155. (d) 161. (a) 162. (d) 163. (b) CASE BASED QUESTIONS Case Study 1 1. (b) 2. (c) 3. (d) 4. (d) Case Study 2 1. (b) 2. (c) 3. (b) 4. (d) Case Study 3 1. (c) 2. (b) 3. (c) 4. (c) HINTS/SOLUTIONS 1. It has the longest chain of atoms and hence has 5. Elimination (dehydrohalogenation) converts an the highest boiling point. alkyl halide into alkene. 2. The reaction involves attack of Cl+ (an 6. Ease of hydrolysis is electrophile) on benzene ring and replacement R—I > R—Br > R—Cl > R—F. of one hydrogen atom by Cl atom. Thus, it is an electrophilic substitution reaction. 7. Allylic halides undergo substitution very easily. 8. Density of alkyl halides is in the order 3. The reaction is initiated by addition of H+ (an electrophilic) and is followed by subsequent R—I > R—Br > R—Cl > R—F. addition of Br–. 9. Sandmeyer reaction. 10. —Cl is ortho- and para- directing group. The 4. Under the given reaction conditions, a benzylic hydrogen is replaced by Cl atom and the product reaction is Friedel Craft alkylation. is benzyl chloride.

226 CHEMISTRY 11. CH3 – CCl2 – CH3 29. Electrophilic addition, because the reaction is initiated by addition of H+, an electrophilic. OH ⏐ 30. Aniline. ⎯⎯aq.⎯KO⎯H → CH3 ⎯ C ⎯CH3 31. Cyclopropane is formed through Freund ⏐ OH reaction. O OH ⏐⏐ ⎯⎯− H⎯2O⎯→ CH3 ⎯ C ⎯CH3 32. CCl3 . CHO + H2O ⎯⎯→ CCl3 –CH 12. Poisonous gas phosgene is formed OH 2CHCl3 + O2 ⎯⎯Sun⎯lig⎯ht → 2COCl2 + 2HCl 13. 3° halides undergo substitution more readily. 33. C5H5Cl + NaOC2H5 14. CH3CH2CHCl2 ⎯⎯aq.⎯KO⎯H → CH3CH2CHO Williamson synthesis CH3CCl2CH3 ⎯⎯aq.⎯KO⎯H → CH3 ⎯ C ⎯CH3 ⏐⏐ ⎯→ C2H5—O—C2H5 + NaCl O 34. CHCl3 + HNO3 ⎯→ Cl3C . NO2 + H2O 15. C2H5Br + AgOH ⎯⎯→ C2H5OH + AgBr. Chloropicrin O 35. Freon, CF2Cl2 is used as refrigerant. ⏐⏐ 36. Iodoform (CHI3) is formed. 16. HCCl3 + CH3 ⎯ C ⎯CH3 37. Ethyl bromide on reaction with alcoholic solution OH of AgNO2 would yield nitroethane as well as ⏐ ethyl nitrite. However nitroethane is the major ⎯→ CH3 ⎯ C ⎯CH3 product. ⏐ 38. CCl4. CCl3 39. * Chloretone CH3 ⎯CH⎯CH2 ⎯CH3 17. CCl3 . CHO + NaOH ⎯→ CHCl3 + HCOONa ⏐ Pure chloroform Cl 18. Wurtz-Fittig reaction. sec-Butyl chloride contains an asymmetric 19. Aluminium carbide on reaction with water carbon atom and hence is chiral. liberates methane. 40. SOCl2 is the best reagent for converting alcohols 20. Diphenylmethane is formed as product. into chlorides. 21. 1, 2, 3, 4, 5, 6-Hexachlorocyclohexane. 22. Cl3C – CHO + NaOH ⎯⎯→ CHCl3 + HCOONa 41. iso-Propyl chloride is a secondary halide. 23. CCl4 24. Mustard gas is dichlorodiethyl sulphide 42. Ulmann reaction. CH2CH2Cl 43. FeCl3 + Cl2 → FeCl4− + Cl+. 44. Chloral. S 45. Finkelstein reaction. CH2CH2Cl 46. C2H5Br ⎯⎯EMth⎯ger⎯→ C2H5MgBr Mustard gas ⎯⎯CH⎯3OH⎯→ C2H6 + MgBr(OCH3). 25. CS2 + 3Cl2 ⎯→ CCl4 + S2Cl2. 47. Among halides density decreases with increase 26. Teflon is a polymer of tetrafluoroethylene. 27. The order of bond strength is C–F > C–Cl > in size of alkyl group. 48. Carbylamine reaction. C–Br > C–I. 49. —Cl group is ortho- and para- directing group. 28. Tetraethyl lead is the product. 50. CH3CH2CH2Br ⎯⎯+ 2H⎯→ CH3CH2CH3 + HBr. 51. 2C2H5I + Ag2O ⎯→ C2H5 – O – C2H5 + 2AgI. 52. Vinyl chloride has least percentage of chlorine among the given compounds. 53. Cl– < Br– < I– I– has maximum nucleophilicity due to poor solvation.

ANSWERS 227 54. Weaker bases are better leaving groups. The CH3 ⏐ basic strength of the given groups is in the order: 68. CH3 ⎯ C ⎯CH2 ⎯CH3 ⎯⎯alc.⎯KO⎯H⎯→ ⏐ – OMe (II) > – OAc (I) > – OSO2 Me(III) > Cl – OSO2CF3 (IV) 55. CH3—C+ H—CH3 Addition of HCl to alkenes, even in the presence CH3 ⏐ of peroxides, takes place in accordance with CH3 ⎯ C == CH⎯CH3 Markownikov’s rule. More substituted alkene is the major product (Saytzeff rule). 56. CH2 ⎯CH2 ⎯⎯aq.⎯KO⎯H → CH2 ⎯CH2 69. For tertiary halides elimination product is the major product on reaction with base. ⏐ ⏐ ⏐⏐ 70. Propanal does not contain CH3CO— Cl or CH3 – CH(OH)– group. Cl OH OH 71. In the presence of FeCl3, substitution takes place at the ring. Ethylene chloride —CH3 group is ortho-para directing group. Cl 72. CH3CH2CH2Cl ⎯⎯KC⎯N → CH3CH2CH2CN ⏐ CH3 ⎯CH ⎯Cl ⎯⎯aq.⎯KO⎯H → CH3 ⎯CHO Butane nitrile Ethylidene chloride 73. 57. Hunsdiecker reaction, the alkyl halide formed Mg —MgBr ¾C¾2H¾5O¾H ® has one carbon less than the starting compound. —Cl ¾¾® 58. HC ≡ CH ⎯⎯+ 2H⎯Br⎯→ H3C ⎯CH ⎯ Br 74. Hunsdiecker reaction. ⏐ 75. CH3Cl ⎯⎯KC⎯N → CH3CN Br ⎯⎯H2O⎯/H⎯+ → CH3COOH. ⎯⎯ZnH⎯edauts⎯t → CH3 – CH = CH – CH3. 76. Butane, pentane and hexane. 77. Inversion of configuration indicates SN2 59. Propane is not formed. 60. (CH3)3 C—CH2—CH2—C (CH3)3 mechanism. 78. CHCl3 + 4KOH ⎯→ HCOOK + 3KCl + 2H2O. A 79. Boiling points of alkyl halides are in the order : ⏐ Wurtz reaction RI > RBr > RCl > RF. 80. In Williamson synthesis new C—O bond is (CH3)3CCH2Cl + ClCH2 C (CH3)3 61. HI > HBr > HCl formed. 62. 4C2H5Cl + NH3 ⎯→ [N(C2H5)4]+Cl– + 3HCl 81. —CH2 – CH – CH3 63. C2H5Br + 2H ⎯⎯CZ2Hn⎯-5COu⎯H → C2H6 + HBr ⏐ 64. X = CH3CH2CN, Y = CH3CH2CONH2 Cl Z = CH3CH2COOH. ⎯⎯⎯→ —CH = CH—CH3 65. Tertiary halides on treatment with base, such 1-Phenylpropene as sodium methoxide, readily undergo elimination resulting in the formation of In 1-phenylpropene double bond is in alkenes. conjugation with benzene ring and hence is more stable. 66. trans-1, 2-Dichlorocyclopropane does not have any plane of symmetry and hence is non- superimposable on its mirror image. 67. H—C ≡ C—H + 2 HCl ⎯⎯→ CH3 – CHCl2 ⎯⎯dil.⎯Na⎯OH⎯→ CH3CHO.

228 CHEMISTRY 82. Trihalogenated methane (such as chloroform) 101. and primary amines. —OH PBr3 —Br NH3 83. Aryl and vinyl halides do not undergo nucleophilic substitution easily. ⎯⎯→ ⎯⎯→ 84. Silver salts of carboxylic acids. Base (NH3) brings about dehydrohalogenation of bromocyclohexane. 85. CH3CH2CH2Cl ⎯⎯alc.⎯KO⎯H⎯→ CH3CH = CH2 102. The one enantiomeric pair is d and l-2-bromo- 3-methylbutane. ⎯⎯HB⎯r → CH3 ⎯CH ⎯CH3 ⎯⎯Na/⎯eth⎯er → 103. Polychlorides are heavier than water whereas monochlorides are lighter than water. ⏐ 104. Monochlorination of neo-pentane yields only one product (CH3)3C CH2Cl. So it can be obtained in Br good yield. CH3 CH3 ⏐⏐ 105. CH3 – CH = CH2 ⎯⎯Cl2⎯(hν⎯) → ClCH2CH = CH2 CH3—CH ⎯ CH ⎯ CH3 ⎯⎯acNe⎯taoIn⎯e → ICH2CH = CH2 86. —CH3 ¾C¾l2 ® —CH2Cl ¾Na¾OH® The first step is allylic substitution and the second step is Finkelstein reaction. hn 106. The ease of elimination is in the order —CH2OH 3° > 2° > 1°. 87. Freund reaction. 107. CH3CH = CHCl is a vinylic halide and hence 88. n-Butane gives 1-chlorobutane and would not undergo substitution easily. 2-chlorobutane as monochloro products. 108. —CH3 ¾C¾l2 ® —CCl3 ¾Na¾OH® 89. Reduction of the main compound does not take place during iodoform reaction. —COOH 90. In vinyl chloride C—Cl bond has partial double 109. SO2Cl2 or Cl2 may be used for benzylic bond character and hence is stronger. chlorination. 91. The reaction of diazonium salts with CuCl or 110. 3° > 2° > 1°. Because the order of stability of CuBr is known as Sandmeyer reaction. carbocations is 3° > 2° > 1°. 92. C—Cl bond in chlorobenzene has partial double 111. R—I > R—Br > R—Cl > R—F. bond character due to resonance. This can be explained in terms of C—X bond 93. Iodobenzene is formed as the major product. strength. 94. Tetraethylammonium bromide. 112. Reaction through SN1 mechanism leads to racemization whereas reaction through SN2 95. Strong bases react through SN2 mechanism. : mechanism leads to inversion of configuration. OR is the strongest base among the given bases. 113. Stronger nucleophile reacts faster through SN2. 114. Elimination takes place in anti fashion. 96. 3° > 2° > 1°. 115. Bromides and iodides are heavier than water. 116. The three isomers are 1, 1-dibromoethene. 97. Presence of electron withdrawing groups on ring makes the substitution easier by stabilizing the cis-1, 2-dibromoethene and trans-1, intermediate carbanion. 2-dibromoethene. 98. C—Cl bond in chloroethene is least polar because in vinyl halides electron withdrawing inductive effect of halogen atom is opposed by electron releasing mesomeric effect. 99. On reaction with alc. KOH both ethylene dichloride and ethylidene dichloride yield ethyne. 100. 2-Chloropropane on further substitution with chlorine atom would yield 1, 2-dichloropropane and 2, 2-dichloropropane.

ANSWERS 229 117. * 125. The eight isomers are : 1-Chloropentane, CH3 ⎯CH⎯CH2 ⎯CH3 2-chloropentane, ⏐ 3-chloropentane, 1-chloro-3-methylbutane, Cl 2-chloro-3-methylbutane, 2-Chloropentane contains an asymmetric carbon 2-chloro-2-methylbutane, 1-chloro-2-methylbutane, atom and hence is chiral. 1-chloro-2, 2-dimethylpropane. 118. Triphenylmethane. 119. cis -1, 2-dichloroethylene has net dipole moment because individual bond dipoles do not cancel each other. 120. SN1 reactions occur through a two step 126. CH3CH2CH2Cl + AlCl3 ⎯→ CH3CH2 + H2 mechanism. C 121. Polychloroalkanes are heavier than water. + AlCl4– 122. Chlorobenzene is less reactive than benzene due to electron withdrawing inductive effect of —Cl. 1, 2-Hydride shift 123. —NO2 group withdraws electrons at ortho- and + para- positions and stabilizes the intermediate carbanion. CH3 C H⎯CH3 124. 2-Chloro-2-phenylethane ionizes slowly to form iso-propyl carbocation carbocation. The carbocation is then attacked equally from both sides forming both the 127. Strong nucleophile favours SN2 and not SN1. enantiomers resulting in racemization. CH3 CH3 +⏐ ⏐ 128. CH3 ⎯CH⎯ C ⎯CH3 ⎯⎯− B⎯r− → CH3 ⎯C H⎯ C ⎯CH3 ⏐ ⏐⏐ CH3 Br CH3 ⎯⎯⎯→1, 2-Methyl shift OH CH3 ⎯ CH ⎯ ⏐ ⎯CH3 ←⎯OH⎯⎯ CH3 ⎯ CH ⎯ + ⎯ CH3 C C ⏐⏐ ⏐⏐ CH3 CH3 CH3 CH3 2,3-Dimethyl-2-butanol 129. 3-Chloro-1-butene is an allylic halide. CH3 CH3 CH3 ⏐ ⎯CH2Cl ⎯⎯− C⎯l− →CH3 ⎯ ⏐ + 1, 2 − Methyl shift ⏐ C C 130. CH3 ⎯ ⎯C H2 ⎯⎯⎯⎯⎯⎯⎯⎯→ CH3 — C — CH2 —CH3 ⏐⏐ CH3 CH3 + –H+ CH3 ⏐ CH3—C = CH ⎯CH3 2-Methylbut-2-ene

230 CHEMISTRY 131. IV < I < II < III. 138. X brings about elimination to form alkene. 132. The statement (d) is not correct. Vinyl chloride Addition of γ (HBr) to alkene gives 2-bromopropane. does not undergo nucleophilic substitution easily due to stronger C – Cl bond. 139. R—I > R—Br > R—Cl > R—F. This order can be explained on the basis of 133. CH3CH2CHCl2 ⎯⎯aq.⎯KO⎯H → [CH3CH2 CH(OH)2] strength of the C—X bond. ⎯⎯− H⎯2O⎯→ CH3CH2CHO 140. CH2 = CH—CH2—Cl ⎯⎯− H⎯Cl⎯→ CH2 == C==CH2 ⎯⎯alc.⎯KO⎯H⎯→ CH3C ≡ CH Propadiene ⎯⎯Am⎯mCuo⎯nCila⎯cal⎯→ CH3C ≡≡ CCu 141. Wurtz reaction 142. Dipole moments are in the order red ppt. CH3Cl > CH2Cl2 > CHCl3 > CCl4 134.CH3– is a very strong nucleophile. 143. In vinyl chloride (CH2 = CH—Cl), the C—Cl 135.The alkyl halide having weaker C—X bond has bond has partial double bond character due to greater reactivity. The C—X bond strength is in resonance. the order : 144. CH3—CO—CH2CH2Br on dehydrohalogenation gives an alkene in which C—C double bond is in C—I < C—Br < C—Cl < C—F conjugation with carbonyl group. Hence, the reactivity is in the order R—I > R—Br > R—Cl > R—F. CH3 C CH2CH2Br ⎯⎯alc.⎯KO⎯H⎯→ CH3 C CH== CH2 136. Self-explanatory answer. ⏐⏐ ⏐⏐ 137. A single stereoisomer having configuration OO opposite to that of the reactant is formed through SN2 reaction. 11. ALCOHOLS, PHENOLS AND ETHERS MULTIPLE CHOICE QUESTIONS 1. (c) 2. (d) 3. (d) 4. (d) 5. (d) 6. (a) 7. (c) 8. (c) 12. (b) 13. (a) 14. (c) 15. (b) 16. (d) 9. (b) 10. (b) 11. (d) 20. (d) 21. (c) 22. (c) 23. (c) 24. (c) 28. (c) 29. (d) 30. (c) 31. (d) 32. (a) 17. (d) 18. (c) 19. (a) 36. (b) 37. (c) 38. (a) 39. (a) 40. (d) 44. (a) 45. (c) 46. (d) 47. (d) 48. (c) 25. (b) 26. (d) 27. (b) 52. (d) 53. (c) 54. (c) 55. (a) 56. (b) 60. (b) 61. (c) 62. (c) 63. (c) 64. (c) 33. (b) 34. (a) 35. (d) 68. (c) 69. (b) 70. (b) 71. (b) 72. (b) 76. (c) 77. (a) 78. (b) 79. (b) 80. (c) 41. (a) 42. (c) 43. (b) 84. (c) 85. (b) 86. (d) 87. (b) 88. (d) 92. (b) 93. (a) 94. (d) 95. (b) 96. (d) 49. (d) 50. (b) 51. (a) 100. (a) 101. (c) 102. (a) 103. (d) 104. (c) 108. (a) 109. (b) 110. (b) 111. (b) 112. (c) 57. (b) 58. (c) 59. (c) 116. (d) 117. (c) 118. (a) 119. (a) 120. (b) 124. (d) 125. (d) 126. (b) 127. (c) 128. (b) 65. (d) 66. (c) 67. (b) 132. (d) 133. (b) 134. (c) 135. (a) 136. (b) 140. (c) 141. (a) 142. (a) 143. (b) 144. (d) 73. (a) 74. (b) 75. (c) 148. (c) 149. (c) 150. (d) 151. (d) 152. (d) 156. (d) 157. (a) 158. (d) 159. (c) 81. (d) 82. (c) 83. (b) 89. (b) 90. (a) 91. (a) 97. (b) 98. (b) 99. (a) 105. (d) 106. (a) 107. (a) 113. (a) 114. (c) 115. (b) 121. (c) 122. (c) 123. (d) 129. (c) 130. (d) 131. (c) 137. (b) 138. (c) 139. (a) 145. (a) 146. (d) 147. (a) 153. (c) 154. (c) 155. (d)

ANSWERS 231 CASE BASED QUESTIONS Case Study 1 1. (d) 2. (b) 3. (d) 4. (b) 4. (b) Case Study 2 1. (c) 2. (d) 3. (a) HINTS/SOLUTIONS 1. Here, —OH is attached to the carbon next to 15. CH3CH2CH2OH, CH3CHCH3 the ring. ⏐ OH 2. The boiling point increases with increase in molecular mass of the alcohol. Pentan-1-ol has Position of —OH is different. the highest molecular mass. 16. Chlorobenzene cannot form hydrogen bonds with 3. m-Chlorophenol is the most acidic due to —I water. effect of —Cl. 17. The alcohols are weaker acids than water. 18. Saponification of oil produces soap and glycerol. 4. C is salicylic acid and D is aspirin. 19. In this case the side products are gaseous. 20. Salol is phenyl salicylate. 5. 5-Chlorohexan-2-ol. —OH group gets preference over —Cl while numbering. OH O | 6. —NO2 group is electron withdrawing group whereas –OCH3 group is electron releasing CO group. | Salol 7. —NO2 and —Cl are electron withdrawing groups | and hence they destabilize the intermediate 21. Zymase converts glucose to ethanol. carbocation. —NO2 has stronger electron 22. 95.87% C2H5OH and 4.13% H2O forms the| withdrawing effect and hence (ii) is the least reactive and (i) is the most reactive. azeatrope that is obtained during distillation of| aqueous solution of ethanol.| 8. 3° > 2° > 1°. This order of reactivity depends 23. Phenolphthalein is the product. upon the relative stability of intermediate carbocations formed from these alcohols. 24. Starch Diastase Maltose. 9. A is ethanal and B is ethanol. ⎯⎯⎯⎯→ 10. CH3Cl has greater dipole moment than CH3F 25. Azeotropic distillation of rectified spirit in the due to longer C—Cl bond length. presence of benzene yields absolute alcohol. 11. Solubility of alcohols in water decreases with increase in molecular mass. 26. Aspirin is obtained by acetylation of salicylic acid 12. Methanol does not give iodoform on reaction and hence is also known as acetyl salicylic acid. with NaOH and I2 whereas ethanol gives. 27. Cyclohexanol 13. C6H5OH is the strongest acid among the given compounds. OH + 3H2 ¾¾® OH. 14. CH3CHO + CH2OH dry HCl Cyclohexanol ⏐ CH2OH ⎯⎯⎯⎯→ 28. Reimer Tiemann reaction. − H2O 29. 3° alcohols are the most reactive towards reaction with HCl. O —CH2 CH3CH 30. 2-Pentanol, 3-methyl-2-butanol and 2-methyl-1- butanol are the three alcohols, with molecular O —CH2 formula C5H12O, which are chiral. Acetal

232 CHEMISTRY 31. —CH3 ¾C¾l2 ® —CH2Cl 47. Diethyl ether does not react with Grignard reagent. It is rather used as solvent for preparing Light Grignard reagents. ¾N¾a¾OH¾(a¾q) ® —OH 48. Two —OH groups on the same carbon are unstable. Benzyl alcohol 49. 2-Hydroxypropane is a secondary alcohol. It 32. The common name of phenol is carbolic acid. would be oxidized to corresponding ketone. 33. CH3MgI. ⏐ 34. —OH group is highly activating and ortho-para 50. CH3OH does not contain CH3—CH —OH group. 51. 1–Ethoxy-2-methylbutane. directing group. 52. Pyridinium chlorochromate (PCC) is a specific 35. A mixture of o- and p-nitrophenols can be reagent for oxidation of primary alcohols to separated by steam distillation. o-Nitrophenol aldehydes. is steam volatile. 53. The reaction proceeds via more stable benzylic carbocation. 36. Alkyl group in tert-butyl alcohol is maximum 54. —NO2 group being electron-withdrawing group branched. stabilizes the phenoxide ion, thus making the corresponding phenol stronger acid. 37. Lucas reagent is equimolar mixture of conc. HCl 55. C6H5—OCH3 + HI ⎯⎯→ C6H5OH + CH3I and ZnCl2. C6H5—O bond has partial double bond character due to resonance and hence, is difficult to cleave. 38. Esterification. 56. (CH3)2 SO4 + 2C2H5ONa ⎯→ 39. Ketones on reaction with Grignard reagent yield Na2SO4 + CH3OC2H5 tertiary alcohols. 57. Bond angle is slightly greater than tetrahedral 40. Methyl salicylate is known as oil of winter green. angle due to repulsion between bulky alkyl groups. 41. CH3CH2CHCH3 [O] 58. Ethers and alcohols are functional isomers. 59. Williamson synthesis involves reaction of alkyl ⎯⎯→ CH3CH2 ⎯ C⎯ CH3 halides with sodium alkoxides or phenoxides. 60. n-Butyl alcohol and iso-butyl alcohol. ⏐ ⏐⏐ 61. :CH3 is the conjugate base of the weakest acid OH O (CH4). 2-Butanol Ethyl methyl ketone 62. Lower members of alcohols are soluble in water and the solubility decreases with increase in CH2OH + 2HIO4 ⎯⎯→ 2HCHO + 2HIO3 molecular mass. 42. ⏐ 63. Tertiary alcohols undergo dehydration when CH2OH heated with reduced copper. + 2H2O. 64. Ease of dehydration is : 3° > 2° > 1°. 43. Ethers are polar in nature and hence 65. 2, 4, 6-Trinitrophenol (Picric acid) is formed interparticle forces in them are dipole-dipole. when phenol is reacted with conc. HNO3 . 66. p-Bromophenol is the major product. 44. CO + 2H2 ZnO ⎯Cr2O3 CH3OH. ⎯⎯⎯⎯⎯→ 475 K 45. Primary alcohols on reaction with acidified K2Cr2O7 are oxidised to carboxylic acids with same number of carbon atoms. 46. C—O—C bond in ethers can be cleaved by acids. CH3 CH3 67. CH3—CH ⎯CH3 Conc. H2SO4 CH3CH = CH2 HBr CH3— CH —CH3 Na ⏐⏐ ⏐ CH3—CH ⎯ CH ⎯CH3 . ⎯⎯⎯⎯⎯⎯→ ⎯⎯→ ⎯⎯⎯→ Heat ⏐ ether OH Br

ANSWERS 233 68. Rate of reaction with Lucas reagent is : 87. During hydroboration—oxidation of alkenes 3° > 2° > 1°. addition of water to the double bond takes place in anti-Markownikov fashion. 69. C6H5MgBr + CH3CHO ⎯⎯⎯→ 88. Diethyl ether is formed. C2H5CH ⎯OMgBr H2O C6H5CHOH 89. Periodic acid test is given by vicinal diols. ⎯⎯⎯→ 90. neo-pentyl alcohol is a primary alcohol. ⏐⏐ 91. R ⎯CH ⎯ R ⎯⎯⎯→ R—C —R + H2 CH3 CH3 ⏐ || OH O 70. p-Hydroxybenzene sulphonic acid is the major 92. The choices (a), (c) and (d) are 3° alcohols which product at 373K. are not oxidized by the reagent. 71. —NO2 group is a very strong electron 93. Picric acid has three electron withdrawing — withdrawing group. It stabilizes the NO2 groups. corresponding phenoxide ion by dispersal of negative charge and increases the acid strength. 94. The reaction is hydroboration-oxidation. 95. The product is formed through rearrangement 72. On vigorous oxidation C—C bond undergoes cleavage. of the intermediate carbocation. 96. Carboxylic acids are reduced to 1° alcohols with 73. LiAlH4. Br | 97. HI > HBr > HCl OH + 3Br2 ¾¾® HO Br + 3HBr. | | | | | Br COOH COOH 74. 1° > 2° > 3°. | OH | 75. Carbocations are formed as intermediates. 98. + Zn ¾¾® 76. Periodic acid test. + ZnO. 77. RMgX + CH2—CH2 ¾H¾2O¾/H®+ RCH2CH2OH 99. Brown or red ⎯⎯→ green ⎯⎯→ deep blue. 100. This is mercuration-demercuration reaction. O This leads to hydration. The product formed is 78. Hydroboration-oxidation yields anti-product. in accordance with Markownikov’s rule. 79. Presence of two oxygen atoms indicates that 101. Chloro group is electron withdrawing group. C3H6 O2 is an acid. Primary alcohols on oxidation 102. yield carboxylic acids. 80. CH3MgI + CH2 —CH2 ⎯⎯→ CH3CH2CH2OMgI O CH2 ¾¾® HI |O— | | | | — H2O CH3CH2CH2OH. ⎯⎯⎯→ 81. Ethanal is the ozonolysis product of 2-butene. OH + CH2I 2-Butanol on dehydration yields 2-butene. 82. Reimer Tiemann reaction. 103. The order of reactivity of alcohols towards HBr is 3° > 2° > 1°. 83. Both phenol as well as benzyl alcohol donot react 2-Methylpropan-2-ol is a 3° alcohol. with NaHCO3. 84. Dehydration of glycerol yields acrolein or prop- 104. CH3—CH ⎯CH2OH ⏐ 2-enal. CH3 ⎯⎯→ CH3—CH ⎯CHO + H2 85. CH3—O—CH3 does not contain any acidic ⏐ hydrogen. CH3 86. 2-Butanol (CH3CHCH 2CH3) contains 105. For C4H10O, four isomeric alcohols and three ⏐ isomeric ethers are possible. OH CH3—CH ⎯ group. ⏐ OH

234 CHEMISTRY 106. The order of reactivity of alcohols towards 112. Anisole is methoxybenzene esterification is : CH3OH > 1° > 2° > 3°. OH + CH2N2 ¾H¾F® OCH3 + N2. 107. Acid strength among alcohols varies as : | | CH3OH > 1° > 2° > 3°. 108. CH3CH2CHO + CH3MgI ⎯→ + | Propanal 113. o-Nitrophenol is steam volatile due to intramolecular H-bonding. CH3CH2—CH ⎯OMgI H2O/ H + CH3CH2 CHCH3 ⎯⎯⎯⎯→ 114. Cl—group is electron withdrawing group. It increases the acid strength of phenol. ⏐⏐ 115. CH3 OH C2H5—O— HI¾¾® C2H5I + Butan-2-ol OH 109. o- and p-bromophenol is formed with para Phenyl—O bond is stronger and hence is not cleaved. isomer being the major product. 116. Reduction of ketones yields secondary alcohols 110. Esters (except that of formic acid) react with as product. excess Grignard reagent to form 3° alcohols. 111. Phenols are more acidic than alcohols. 117. (i) Hydroboration CH3 ⎯C = CH2 ⎯⎯⎯⎯⎯⎯→ CH3 ⎯CH ⎯CH2OH (ii) Oxidation ⏐⏐ CH3 CH3 Anti-Markownikov product is obtained. CH3 CH3 CH3 118. ⏐ + H+ ⏐+ 1, 2-Methyl shift. ⏐ CH3 ⎯C⎯⎯ CH ⎯CH3 CH3 ⎯C⎯⎯ CH ⎯CH3 CH3—C⎯ CH ⎯CH3 ⎯⎯⎯→ ⎯⎯⎯⎯⎯⎯⎯→ +⏐ ⏐⏐ ⏐ − H2O CH3 CH3 OH CH3 CH3 + Cl− ⏐ ⎯⎯⎯⎯⎯→ CH3—C⎯ CH ⎯CH3 ⏐⏐ Cl CH3 119. tert-butoxide is the conjugate base of tert-butyl alcohol which is the weakest acid among ethanol, 2-propanol, methanol and tert-butyl alcohol. CH3 CH3 CH3 120. ⏐ + H+ ⏐+ 1, 2-Methyl ⏐ CH3 ⎯C⎯⎯ CH ⎯CH3 CH3 ⎯C⎯⎯ CH ⎯CH3 CH3—C⎯ CH ⎯CH3 ⎯⎯⎯→ ⎯⎯⎯⎯⎯⎯⎯→ ⏐⏐ ⏐ +⏐ − H2O shift CH3 OH CH3 CH3 CH3 − H+ ⏐ ⎯⎯→ CH3—C = C⎯CH3 ⏐ CH3 121. Reduction of aldehydes and ketones with Zn(Hg)/HCl yields alkanes. 122. o, m and p-cresols, anisole and benzyl alcohol. 123. Kolbe’s reaction. 124. —NO2 group is electron withdrawing group whereas —CH3 is electron releasing group.

ANSWERS 235 + Br2 BrCH2—CH2Br 125. C2H5OH Conc. H2SO4 CH2 = CH2 CH3 = CH2 ⎯⎯⎯⎯⎯→ + H2 CH3—CH3 126. Acid strength of alcohols is in the order : 1° > 2° > 3°. 127. 2, 3-Dichloropentane has two non-equivalent chiral carbon atoms. Number of enantiomers = (2)2 = 4. 128. —CH3 group is electron releasing while —NO2 group is electron withdrawing group. Electron with- drawing effect of —NO2 group is more at para position. 129. It is the weakest acid among the given choices. 130. CH2 = CH2 O2, Ag 473K ⎯⎯⎯→ ⎯⎯⎯→ CH2 —CH2 CH2 ⎯ CH2 steam |—— ⏐⏐ |O OH OH | 131. CH3 ⎯CH ⎯CH3 . | ⏐ —— OH | 132. It is a primary alcohol.|| 133. Stabilization of phenoxide ion through delocalization of negative charge shifts the equilibrium in favour of ionized form C6H5OH + H2O C6H5 – O– + H3O+ 134. In the given structure (c), there are 10 electrons around nitrogen. 135. CH3CH2CH2OH Conc. H2SO4 CH3CH = CH2 HBr CH3CH2CH2Br Na CH3 (CH2)4 CH3. ⎯⎯⎯⎯⎯⎯⎯→ ⎯⎯⎯⎯→ ⎯⎯→ Peroxide ether 136. CH3C ≡ CH HgSO4 CH3— C ⎯ CH3 LiAlH4 CH3— CH ⎯ CH3 ⎯⎯⎯⎯→ ⏐⏐ ⎯⎯⎯→ dil.H2SO4 ⏐ O OH 137. (b) MgBr + CH2 —CH2 ¾® CH2CH2OMgBr ¾H¾2O¾/H®+ CH2CH2OH 138. The three secondary alcOohols are : 2-pentanol, 3-pentanol and 3-methyl-2-butanol. 139. CH3—CH ⎯CH = CH2 (i) Oxymercuration CH3—CH ⎯ CH —CH3 ⏐ ⏐⏐ CH3 ⎯⎯⎯⎯⎯⎯⎯⎯⎯→ (ii) Demercuration CH3 OH Alcohol formed is in accordance with Markownikov rule. 140. X= C|H CH3, Y= |C| CH3 OH O O H || CCl2- 141.

236 CHEMISTRY 12. ALDEHYDES, KETONES AND CARBOXYLIC ACIDS MULTIPLE CHOICE QUESTIONS 1. (a) 2. (c) 3. (d) 4. (b) 5. (b) 6. (b) 7. (c) 8. (c) 12. (b) 13. (c) 14. (a) 15. (a) 16. (c) 9. (d) 10. (b) 11. (d) 20. (c) 21. (b) 22. (d) 23. (b) 24. (d) 28. (b) 29. (c) 30. (b) 31. (a) 32. (c) 17. (a) 18. (c) 19. (b) 36. (c) 37. (a) 38. (b) 39. (b) 40. (b) 44. (c) 45. (c) 46. (d) 47. (d) 48. (d) 25. (c) 26. (b) 27. (a) 52. (d) 53. (c) 54. (b) 55. (b) 56. (b) 60. (d) 61. (b) 62. (c) 63. (d) 64. (c) 33. (a) 34. (d) 35. (d) 68. (b) 69. (b) 70. (c) 71. (d) 72. (d) 76. (b) 77. (c) 78. (b) 79. (d) 80. (c) 41. (b) 42. (b) 43. (a) 84. (b) 85. (a) 86. (c) 87. (c) 88. (b) 92. (a) 93. (d) 94. (d) 95. (a) 96. (d) 49. (b) 50. (d) 51. (b) 100. (d) 101. (b) 102. (a) 103. (a) 104. (a) 108. (d) 109. (b) 110. (d) 111. (c) 112. (b) 57. (d) 58. (a) 59. (c) 116. (d) 117. (b) 118. (b) 119. (c) 120. (c) 124. (b) 125. (d) 126. (a) 127. (c) 128. (b) 65. (a) 66. (c) 67. (c) 132. (c) 133. (c) 134. (a) 135. (c) 136. (a) 140. (a) 141. (a) 142. (b) 143. (a) 144. (a) 73. (b) 74. (a) 75. (b) 148. (c) 149. (d) 150. (b) 151. (a) 152. (b) 156. (d) 157. (a) 158. (c) 159. (c) 160. (a) 81. (a) 82. (a) 83. (a) 164. (b) 165. (b) 166. (a) 167. (d) 168. (b) 172. (a) 173. (b) 174. (d) 175. (a) 176. (d) 89. (a) 90. (d) 91. (c) 180. (c) 181. (a) 182. (a) 183. (c) 184. (d) 188. (d) 189. (a) 190. (c) 191. (c) 192. (b) 97. (b) 98. (a) 99. (d) 196. (b) 197. (c) 198. (a) 199. (b) 200. (d) 204. (c) 205. (c) 206. (c) 207. (c) 208. (b) 105. (b) 106. (b) 107. (b) 212. (a) 213. (a) 214. (a) 215. (d) 216. (b) 220. (b) 221. (c) 222. (a) 223. (d) 224. (a) 113. (b) 114. (d) 115. (b) 228. (b) 229. (d) 230. (a) 231. (a) 232. (a) 236. (b) 237. (b) 238. (c) 239. (d) 240. (b) 121. (c) 122. (b) 123. (b) 244. (b) 245. (a) 246. (d) 247. (b) 248. (a) 252. (c) 253. (c) 254. (b) 255. (b) 256. (d) 129. (b) 130. (d) 131. (d) 5. (c) 137. (d) 138. (a) 139. (c) 5. (a) 145. (c) 146. (b) 147. (d) 153. (a) 154. (a) 155. (a) 161. (a) 162. (c) 163. (b) 169. (d) 170. (b) 171. (c) 177. (b) 178. (a) 179. (b) 185. (b) 186. (c) 187. (a) 193. (a) 194. (a) 195. (c) 201. (a) 202. (a) 203. (c) 209. (d) 210. (b) 211. (c) 217. (b) 218. (a) 219. (b) 225. (d) 226. (c) 227. (d) 233. (c) 234. (a) 235. (b) 241. (a) 242. (c) 243. (a) 249. (d) 250. (c) 251. (d) 257. (b) 258. (b) 259. (d) CASE BASED QUESTIONS Case Study 1 1. (b) 2. (b) 3. (a) 4. (b) Case Study 2 1. (c) 2. (b) 3. (d) 4. (c) Case Study 3 1. (b) 2. (a) 3. (d)

ANSWERS 237 HINTS/SOLUTIONS 1. alkyl benzene is oxidised by using CrO2Cl2. 17. When different substance condense it is known as crossed aldol condensations. KOH (aq) 18. 2, 2-dimethylpropanal does not have α-H atom 2. CH3CHCl2 ⎯⎯⎯⎯→ CH3CH(OH)2 and hence undergo cannizzaro’s reaction. (unstable) 19. It is M.P.V. reduction. − H2O 20. Phenylethanal (C6H5CH2CHO) does not have –CHO group directly attached to phenyl group, ⎯⎯⎯→ CH3CHO i.e., it is not a aromatic aldehyde. Ethanal 21. It gives pinacol as the product. 3. 2CH3COCl + (C2H5)2Cd ⎯→ 2CH3COC2H5 22. Benzaldehyde undergoes benzoin condensation. + CdCl2. 23. 4. Both are methods of preparation of benzaldehyde. C = O + H2NR ¾® C = N—R H 5. Equimolar mixture of CO4 and HCl. H 6. (CH3COO)2Ca + Ca(COOCH)2 Schiff ’s base ⎯⎯→ CH3COCH3 + CH3CHO + HCHO 7. Reformatsky reaction is used to prepare β- Hydroxyester by using α-bromoester and aldehydes. 8. Stephen’s reduction converts alkyl cyanides to CH3 O CH3 aldehydes. Zn-Hg/HCl ¾¾¾¾¾® + H2O SnCl2 /HCl H3O+ 24. 4[H] R—CN ⎯⎯⎯→ R—CH == NH2 ⎯⎯→ RCHO Methyl cyclopentone 9. Acetophenone is least soluble because of its H—C = O extremely poor tendency to form H-bonds with Zn-Hg/HCl 4[H] water molecules. 10. Benedict’s solution reacts with aldehydes but not with ketones. 25. Friedel Craft’s reaction is quite helpful in 11. synthesis of acetophenone by using benzene and COOH CH3COCl in the presence of anhydrous AlCl3. C6H5—CHO + H2C COOH HO Malonic acid ⏐ conc. HNO3 ⏐⏐ 26. C6H5 ⎯ C ⎯C6H5 ¾–C¾O®2 C6H5CH == CHCOOH. ⏐ ⎯⎯⎯⎯→ C6H5 ⎯ C ⎯C6H5 –H2O Cinnamic acid 2(O) H Diphenyl ketone (i) NH3 Diphenyl methane 12. CH3CHO ⎯⎯⎯⎯→ CH3CH == NH2 + H2O O C6H5 ⏐⏐ ⏐ (ii) Heat (Acetaldimine) 27. C6H5 ⎯ C⎯CH3 + O ==C⎯CH3 13. Benzophenone (C6H5COC6H5) does not contain Acetophenone ⏐ CH3— CO grouping. Hence, it does not give iodoform test positive. – H2O Aluminium (i) EtMgI ter. butoxide O CH3 14. HCHO ⎯⎯⎯⎯⎯⎯→ Et.CH2OH + Mg(OH)I C6H5—C—CH == C—C6H5 (ii) HOH 1−Propanol Dypnone 15. Mineral acid like HCl has no reaction with 28. CH3CO—C6H5 (acetophenone) will show aldehydes and ketones. iodoform test positive but C6H5—CO—C6H5 will not give iodoform test positive. 16. Ethanal and C2H5OH react in the presence of conc. H2SO4 to give acetal.

238 CHEMISTRY Zn /alcohol 46. 2 —CHO ¾co¾nc¾. N¾aO¾H® ⎯⎯⎯→ NO2 29. C6H5CO—C6H5 2[H] C6H5—CH(OH) ⎯C6H5 Benzhydrol —CH2OH + —COONa 30. C6H5—C == O H2NOH NO2 NO2 H+ C6H5 .. C6H5 OH 47. The two functional isomers are propanone and H propanal. C=N + H C = N.. 48. CH3CH2CH2COCH3 OH and (CH3)2—CH—CO—CH3 are metamers. Anti and Syn forms 49. Aldehyde formed is CH3CH2CHO. Therefore, ketone must have four carbon atoms. The ketone with four carbon must be ethyl methyl ketone. 31. Zn/ether is used. O C2H5 32. —COCH3 group is electron withdrawing. Hence ⏐⏐ (i) EtMgBr ⏐ it is m-directing. 50. C2H5—C —C2H5 C2H5—C⎯ OH 33. It is known as Tischenko reaction. ⎯⎯⎯⎯⎯→ ⏐ 35. HCHO do not contain + I group and CHO group (ii) H2O is not Sterically hindered. C2H5 36. 6HCHO + 4 NH3 ⎯⎯→ (CH2 )6N4 + 6H2O + Mg(OH)Br Urotropine H OO OO CC CC OH 51. CH3 CH2 CH3 CH3 CH CH3 OH− ⏐ Enol is stabilised 37. CH3CHO + HCHO ⎯⎯→ H2C⎯CH2CHO by H-bonding β-Hydroxypropionaldehyde 52. Schiff’s reagent is rosaniline hydrochloride while 39. Among the given compounds, CH3CHO responds Schiff’s base is N-alkylimine a product of to iodo-form test but does not give cannizzaro’s reaction because of the presence of α-H atom in reaction between aldehydes or ketones with 1° its molecule. amines. 40. CH2 = CHOH and CH3CHO are tautomers. 41. One molecule gets oxidised and the other gets OO ⏐⏐ ⏐⏐ Ba(OH)2 reduced. 53. CH3—C —CH3 + CH3—C —CH3 ⎯⎯⎯⎯⎯→ 42. Benzaldehyde does have α-carbon atom but it OH O does not contain and α-H atom. ⏐ ⏐⏐ CH3 ⎯ C⎯CH2 ⎯ C⎯CH3 43. C6H5CHO + NH3 ⎯⎯→ hydrobenzamide. ⏐ 44. Paraldehyde is a hypnotic i.e., produce deep CH3 sleep induces. Diacetone alcohol 45. It is cumene peroxide method for preparation of acetone. 54. CCl3 . CHO + H2O ⎯⎯→ Cl3C − CH(OH)2 Chloral hydrate 55. 3 moles of acetone cyclize in the presence of conc H2SO4 to form mesitylene. See reactions of acetone.

ANSWERS 239 O OH ⏐⏐ α | 56. CH3CH2C —H + H2C —CHO ⎯⎯NaO⎯H⎯(dil⎯.) → CH3—CH2C ⎯ CH ⎯ CHO ⏐ || H CH3 CH3 O ⏐⏐ NH2 ⎯ NH2 57. C6H5—C —C6H5 ⎯⎯⎯⎯⎯→ C6H5—CH2—C6H5 KOH 58. As intermediate is a carbocation therefore 70. It is used in Wittig reaction. racemic mixture is formed. 71. All the given reactions will produce 59. It is cannizzaro’s reaction. acetophenone. 60. Reaction with ammonia produce different 72. [CH2== CH ⎯ CH = O ↔ + H2—CH==CH—O– products. C Resonating forms CrO3 NaOH d+ d– 61. C6H5CH3 ⎯⎯→ C6H5COOH ⎯⎯→ CH2 ......... CH ......... C ......... O is hybrid form. AC2O Tollen’s reagent C6H5COONa 73. CH3CHO ⎯⎯⎯⎯⎯⎯→ CH3COOH 62. Jones reagent is chromic acid used to convert 1° AgNO3 / NH4OH alcohols to acids and 2° alcohols to ketones. COCH2Cl Conc. 74. 63. CH3 ⎯CH2 —COCH3 ⎯⎯→ 2CH3COOH (Pentan-2-one) HNO3 (Acetic acid) 75. Q must be ketone. 64. It is claisen condensation. 76. Y must be aldehyde with three carbon atoms. Hence it is CH3CH2CHO. 65. To bring about nucleophilic addition reaction nucleophile is generated by abstraction of CH3CH2 C==O ¾NH¾2O¾H® α-hydrogen H H+ CH3 O ⏐ ⏐⏐ CH3CH2 OH CH3CH2 66. Mesityl oxide is CH3— C == CH ⎯ C —CH3. .. .. βa C == N + C==N Evidently, it is and α, β unsaturated ketone. H H OH 68. Geometrical isomers O 77. It is Reformatsky reaction (see summary). CH3CH2C—O CH3 CH3CH2C—O O Dry distill 78. 3 C ºº CH ¾con¾. H¾2S¾O®4 Ca ¾¾¾¾® CH3CH2 CH3 H3C CH3 C == O + CaCO3 Mesitylene CH3CH2 79. The compound corresponding to above data is 3-Pentanone C3H7 Cl3C – CHO, i.e., chloral. OH 69. CH2 == O ¾(i) ¾C3¾H7¾Mg¾Br¾® H2C 80. self explanatory (ii) H2O n-butyl alcohol

240 CHEMISTRY 81. Enolic form of acetone has Lewis structure. 92. NaBH4 does not reduce C = C bond. .. HgSO4 ⎯⎯⎯⎯⎯⎯→ :OH 93. CH ≡≡ CH CH3CHO H2SO4 ⏐ Han2dC = C⎯CH3 . Thus, it has 9 σ bonds, 1π bond two lone pairs. LiAlH4 SeO2 CH3C —C —CH3 82. CH3CCH2CH3 ⎯⎯⎯⎯⎯⎯→ ⏐⏐ ⏐⏐ P / Br2 ⏐⏐ (CH3COOH) CH3CH2Br ←⎯⎯ CH3CH2OH O OO (Ethyl bromide) O HOCH2 OH O || ⏐ 94. CH3CHCH3 oxid. ⏐⏐ 83. C6H5CH2— C —CH3 + HOCH2 ⎯⎯→ ⎯⎯⎯⎯⎯→ CH3 ⎯C⎯CH3 (Acetone) (X) C6H5CH2 O—CH2 (i) MeMgBr OH | C ⎯⎯⎯⎯⎯→ H3C O—CH2 (ii) H2O CH3 ⎯C⎯CH3 2-Benzyl-2-methyl dimethyl-1, 3-dioxolane | 84. CH3 (Y) —C ºº CH + H2O ¾H¾g2+¾® —C = CH2 50% NaOH OH 95. 2C6H5CHO ⎯⎯⎯⎯⎯⎯→ C6H5COONa + C6H5CH2OH —C —CH3 96. According to the data given, A should be a ketone O with CO group carrying no methyl group. Thus, Acetophenone it should be CH3CH2COCH2CH3, i.e., 3- CHO pentanone. OH OH CH3 HCN ⏐ H3O+ ⏐ Me2 C ⎯ CN Me2 C ⎯ COOH 97. Me2C = O ⎯→ ⎯→ 85. ¾C¾rO¾2C¾l2 ® (X) (Y) CH3 CHO CH3 CCl4 CH2== ⏐ H2SO4 C ⎯ COOH ←⎯⎯⎯⎯⎯ 86. C6H5COCH3 + Br2 ⎯⎯→ C6H5COCH2Br + HBr 445 K Phenacyl bromide (β-Elimination) 87. 3 moles of acetone condense in presence of HCl 98. It is p-rosaniline hydrochloride which is made to give phorone (see reactions of acetone) colourless by using H2SO3 88. slowest step of reaction is the transfer of hydride 99. ion to carbonyl group. 89. It forms para form and not paraldehyde. CHO COOK CH2OH ¾KO¾H¾(co¾nc®.) + 90. It is friedal craft acylation. 2 (i) O3 HC == O | || Cl Cl Cl 91. CH ≡≡ CH ⎯⎯⎯⎯⎯→ ⏐ HC == O (ii) H2O/Zn Glyoxal m-Chlorobenzaldehyde Potassium m-Chlorobenzyl internal m-Chlorobenzoate alcohol Cannizzaro CH2OH 50% KOH reaction ⏐ COOK Potassium glycolate

ANSWERS 241 100. This is an example of aldol type condensation. 109. This is an example of cyclic ketal formed by diethyl ketone and ethylene glycol. O O O H H 110. H + ¾¾¾® Cl OH C6H5CH ¾K¾OH® C6H5CH OH ¾– H¾2O® Cl OH O Heat – H2O C6H5CHO CH3COONa ⎯⎯⎯⎯⎯⎯→ C6H5CH = CHCOOH (B) (CH3CO)2 O Cinnamic acid (C) 101. Both propyne and acetone will give trimerised 111. C3H8O is alcohol while C3H6O is ketone which product mesitylene. shows iodoform test positive. The reaction H2O OH O ⏐ (O) ⏐⏐ 102. CaC2 ⎯⎯⎯⎯⎯→ Ca(OH)2 + CH ≡≡ CH sequence is CH3CH CH3 ⎯→ CH3C CH3 (Ethyne) CH ≡≡ CH H2O / H2SO4 CH3CHO I2 / NaOH ⎯⎯⎯⎯⎯→ (Ethanal) ⎯⎯⎯→ CHI3 + CH3COONa. HgSO4 112. It reaction of alkenes with water gas (CO + H2) OO in presence of metalcarbonyls to form higher CC aldehydes. 103. CH3 CH2 – CH2 CH3 Dil. NaOH O OH OH ¾¾CH¾Cl¾3 ® CHO O NaOH 113. ¾¾¾® Salicylaldehyde Heat H3C OH CH3 KOH(50%) O 3-Methylcyclopent-2-en-1-one OH OH CH2OH COOK COCH3 + 104. ¾C¾lC¾CH¾3® + HCl COCHO SeO2 114. This is an example of crossed Cannizzaro’s reaction. Phenyl glyoxal CH3 105. Oxygen being electronegative cannot carry ⏐ positive charge. 115. CH3—CH== C⎯CH == CH2 106. This is an acylation reaction and CH3OCH3 is (i) O3 CH3CHO + CH3 + HCHO not an acylating agent. | ⎯⎯⎯⎯⎯→ Ethanol O == C⎯CHO Methanal 107. A is acetone (ii) Zn/H2O Methyl NH2 −NH2 /KOH glyoxal CH3COCH3 ⎯⎯⎯⎯⎯⎯⎯→ CH3CH2CH3 + H2O 116. The given reaction is Benzillic acid Propane rearrangement which involves the treatment of benzil with strong base (NaOH) to form sodium 950−1030 K salt of benzilic acid. 108. CH3COCH3 ⎯⎯⎯⎯→ CH4 + CH2 == C == O Methane Ketene

242 CHEMISTRY OO acid is stronger acid than phenol and phenol is more acidic than ethanol. C6H5—C—C—C6H5 ¾N¾aO¾H® OH O OO Benzil ⏐⏐ ⏐⏐ 125. AcOAc is R—C —O—C —R. Thus, it represents OO C6H5—C—C—OH ¾¾¾® C6H5—C——C—ONa acid anhydrides. C6H5 C6H5 126. KMnO4 —COOH Sodium salt of benzillic acid = C2H5 ⎯⎯⎯⎯→ 117. The given reaction is oppenaur oxidation which [O] Benzoic acid causes oxidation of C—OH group to C = O group 127. COOH COCl ⏐ without oxidising C = C bond. Py COCl ⏐ + SOCl2 ⎯⎯⎯→ + SO2 COOH 118. CaC2 + H2O ⎯⎯→ Ca(OH)2 + CH ≡≡ CH (A) Oxalyl chloride Hg 2+ 128. ClCH2COOH + PCl5 ⎯⎯→ ClCH2COCl. 129. (b) ClCH2COOH CH ≡≡ CH ⎯⎯→ CH3 CHO H2SO4 (B) SOCl2 (X) Ag2O ⎯⎯⎯⎯→ CH3CHCl2 ⎯⎯⎯⎯→ HOCH2COOH Py H2O α-Hydroxyethanoic acid OH MnO Reduction ⏐ 130. CH3COOH + HCOOH ⎯⎯⎯→ CH3CHO + CO2 + H2O. 119. CH3COC2H5 ⎯⎯⎯⎯⎯⎯→ CH3 CH CH2CH3 131. Tollen’s reagent is reduced by formic acid. (A) LiAlH4 (B) Formic acid is the only carboxylic acid that can reduce Tollen’s reagent other acids can’t do it. H2SO4 CH3 CH == CHCH3 ⎯⎯⎯⎯⎯→ (C) 443 K (i) O3 132. (CH3CH2COO)2Ca + SO2Cl2 ⎯→ CaSO4 CH3CHO + CH3CHO ←⎯⎯⎯⎯⎯⎯ + (CH3CO)2O. (ii) Zn /H2O O ⏐⏐ NaOCl 120. O == + (C6H5)3P == CH2 133. CH3C CH3 ⎯⎯⎯→ CHCl3 + CH3COONa. 134. CH3CH2CONH2 + HNO2 Ylide ⎯→ C2H5COOH + N2 ↑. ¾¾¾® == CH2 + (C6H5) P = O 135. It refers to Hoffmann’s bromamide reaction. Methylenecyclohexene The reaction is called WITTIG REACTION. O OO ⏐⏐ 136. RCOCl + NaN3 ⎯⎯⎯→ R—C—N3 + NaCl 121. — — ¾V¾igo¾ro¾us® CH3— —CH3 Acyl azide Oxid. 137. It refer to mechanism of esterification. HOOC COOH (i) NaOCl 138. Urea is a feeble monoacidic base (Kb = 1.5 × 10–14) which forms crystalline white precipitate (ii) H3O+ of ureanitrate with conc. HNO3 122. Reactivity is governed by (i) + I effect and (ii) (H2N)2CO + HNO3 ⎯⎯→ O = C(NH2) . HNO3. steric factors. OH Ureanitrate ⏐ 123. CH2—C⎯CH3 139. Acid chlorides are also known as acyl chlorides 124. Benzoic acid is stronger acid as compared to and hence acylation. acetic acid due to –I effect of C6H5-group. Acetic 140. It is a monoacid base + NH2CONH2 NH2CON H + H+

ANSWERS 243 −H C4H10 O3 = 190 – 106 = 84. ∴ The number of –OH groups acylated 141. —OH ⎯⎯⎯⎯→ – OCOCH3 = 84/42 = 2. + CH3CO − 142. p-Chlorobenzoic acid is strongest among the The change in molecular mass for acylation of given acids. one —OH group = 42 a.m.u. The change in molecular mass for acylation of OH NaOH (aq) ⏐ − 2H2O HCl 143. CH3CCl3 ⎯⎯⎯→ CH3C — OH ⎯⎯→ CH3COONa ⎯⎯⎯→ CH3COOH. ⏐ (A) NaOH − NaCl OH 144. NBS causes decarboxylation of α-amino acids in aqueous medium. NBS CH3CH(NH2)COOH ⎯⎯→ CH3CHO + NH3 + CO2. 145. Picric acid is trinitro phenol and this gives purple colour with FeCl3. O Dry distill ⏐⏐ 146. 2[(CH3)2CHCOO]2Ca ⎯⎯⎯⎯⎯→ (CH3)2 CH C CH(CH3)2 + CaCO3. O ⏐⏐ 147. C6H5OH + ClCOCH3 ⎯⎯→ C6H5O—C—CH3 . Phenyl acetate 148. It is amphoteric because it can neutralise strong acid as well as strong base. OO ⏐⏐ − HCl ⏐⏐ 149. Cl— C —Cl + 2NH3 ⎯⎯⎯→ H2N—C—NH2 . Carbamide or urea 150. C6H5COOC2H5 does not have α-H atom. Hence O it cannot undergo self Claisen condensation. ⏐⏐ 155. R—C —NHBr is ore of the intermediate formed. 151. H3C— —COCl + NaOOC— —CH3 H3C— H2O H3C— –NaCl 156. HCOOC2H5 + 2MeMgI ⎯⎯⎯→ —CO O Me —CO ⏐ 152. In transesterification the alcohol in esters is H—C—OH + Mg(OH)I + Mg(OC2H5)I. replaced by alcoholic attacking compound. ⏐ Me 153. Acids with presence of α-H atoms will give HVZ reaction. Isopropyl alcohol 154. Both ethyl formate and methyl methanoate on 157. treatment with excess of ethyl magnesium iodide will produce 3-pentanol. —COOH + HNO3 ¾H¾2S¾O4® —COOH + H2O NO2

244 CHEMISTRY OO O ⏐⏐ ⏐⏐ ⏐⏐ 158. CH3CH2C —OH + HO C —CH2CH3 ⎯⎯⎯→ CH3CH2CCH2CH3 + CO2 + H2O. Pentanone−3 159. CH3COOH + CH2N2 O ⏐⏐ ⎯⎯⎯→ CH3C — OCH3 + N2 Methyl ethanoate O C2H5 ⏐⏐ ⏐ ⎯⎯⎯⎯→ 160. CH3 C —OC2H5 + 2C2H5MgBr CH3 —C—OH + Mg(OH)Br + Mg(OC2H5) Br. H2O ⏐ C2H5 3−Methylpentan−3ol 161. The functional isomers are CH3COOH and HCOOCH3. Out of these CH3COOH will neutralise NaHCO3 giving effervesence due to evolution of CO2. HVZ KOH (aq.) ⎯⎯⎯⎯→ 162. CH3CH2COOH CH3 CHCOOH (X) ⎯⎯⎯⎯→ CH3CHCOOH . Cl2 /P ⏐ ⏐ Cl (X) OH Lactic acid 163. It is a reaction of ester in the presence of sod ethoxide. 164. See methods of preparation of acid anhydrides. 165. CH3O— —COOH is weakest among the given acids. Hence, it has lowest value of Ka. OO ⏐⏐ ⏐⏐ 166. R—C —Cl + R′MgX ⎯⎯⎯→ MgXCl + RC R′ O OH ⏐⏐ ⏐⏐ R—C —R′ + R′MgX ⎯⎯⎯→ R—C—R′ (3° alcohol). ⏐ R′ 167. Among the carboxylic acids, HCOOH is strongest reducing agent which can reduce Tollen’s reagent as well. 168. It is ~− 10%. Heat 169. CH3COONH4 ⎯⎯→ CH3CONH2 + H2O. Acetamide 170. Soaps are sod salts of long chain fatty acids. O ⏐⏐ 171. CH3—C —NH2 + HNO2 ⎯→ CH3COOH + N2 + H2O. 172. HgO reacts with acetamide to form mercury salt.

ANSWERS 245 O 176. CH3CH2CCl2COOH is strongest acid among the given compounds due to the presence of 2- ⏐⏐ 173. CH3COCl + HONH2 ⎯⎯→ CH3 —C—NH—OH electron withdrawing Cl atoms at the α-position. hydroxamic acid 177. C = O does not undergo nucleophilic attack easily CH2COOH due to adjacent NH2 group 174. ⏐ is not hydroxy acid. L OR — C = 0 ←⎯⎯→ R — C — O− M P⏐ ⏐⏐ CH2COOH NMM PPQNH2 COOH Heat + NH2 175. ⏐ ⎯⎯→ HCOOH + CO2. 178. It is called Blanc’s reaction. COOH Oxalic acid 179. HI converts R—COOH to R—CH3. COOH HNO3 COOH 180. C12H22O11 ⎯⎯⎯→ + CO2 + H2O. H2C ¾H¾ea¾t ® CH3COOH + CO2 ⏐ (O) COOH COOH 181. It is Myricyl palmitate. Malonic acid CH2 —COOH Heat CH2 – CO O. 182. Methyl salicylate is one of component of oil of ⏐ ⎯⎯→ winter green and it is called oil of winter green. CH2 —COOH CH2 – CO 183. Isoamylacetate is also called banana oil. Succinic acid Succinic anhydride O BF3, 150° ⏐⏐ 184. CH3CH2OCH2CH3 + CO ⎯⎯⎯⎯→ CH3CH2— C —OC2H5 500 atm COOH 185. CH2 ¾H¾ea¾t ® CH3—COOH + CO2. – CO2 COOH Acetic acid 186. CH3—CH ⎯COOAg + Br2 ⎯⎯⎯→ CH3CH—Br + CO2 + AgBr. CCl4 ⏐ ⏐ CH3 CH3 2− bromopropane 187. β-keto acids can readily undergo decarboxylation. 188. CH3CONH2 is least reactive towards acyl substitution. O PCl5 CH3CH2 COCl ⏐⏐ ⎯⎯CA6l⎯CHl⎯63 → CH3CH2— C —C6H5 189. CH3CH2COOH ⎯⎯⎯→ Phenyl ethylketone 190. KMnO4 converts – CH3 in toluene to —COOH while HI reduces COOH group to – CH3. 191. CH3Cl Mg CH3MgBr CH3COCl (CH3)3COH ⎯⎯⎯⎯→ ⎯⎯⎯⎯→ Tert. butyl alcohol Ether (i) CO2 ⎯⎯⎯⎯→ 192. C2H5MgBr C2H5COOH. Thus, value of n = 2. (ii) H2O OH OH HCN ⏐ H3O+ ⏐ ⎯⎯H3O⎯+⎯→ 193. CH3CHO ⎯⎯→ CH3CH CN ⎯⎯→ CH3CH COOH CH2 = CHCOOH Acrylic acid

246 CHEMISTRY NaOI 204. Addition of bromine across the C = C occurs against Markownikoff rule in the presence of 194. CH3COC2H5 ⎯⎯⎯→ CHI3 + C2H5COONa peroxide. (Yellow) NH3 P2O5 2CHI3 + Ag ⎯⎯⎯→ CH ≡ CH + 6AgI. (Y) 195. The correct IUPAC name is 4-methoxycarbonyl 205. C6H5COCl ⎯⎯→ C6H5CONH2 ⎯⎯→ C6H5CN. benzoic acid. − HCl Heat 196. CH ≡ CH ⎯⎯⎯→ CH3CH(OCOCH3) Hence, Y is P2O5 or it can be PCl5 also. Hg2+ Ethylidene diacetate (A) 206. CH3COOH HI / P CH3CH3 . CH3CHO + (CH3CO)2O ←⎯⎯⎯ ⎯⎯⎯→ Ethane Distill 207. C2H5OH < C6H5OH < CH3COOH (B) (C) 197. 2-Methylbutanoic acid has chiral structure < ClCH2COOH. hence, it shows optical activity. 208. 423 K CH2COOH conc. CH2COOH ⏐ H2SO4 ⏐ 198. CH2(COOH)2 ⎯⎯⎯→ CH3COOH + CO2 HO—C—COOH ⎯⎯⎯→ C = O (Y) Fuming P2O5 + CO + H2O. CH2(COOH)2 ⎯⎯⎯→ O = C = C = C = O + ⏐⏐ 2H2O. CH2COOH CH2COOH (X) 209. It is a HVZ reaction. 199. (NH2)2 CO + 2HNO2 ——→ 2H2O + 2H2 + H2CO3. Rh 200. (CH3)2CHCOOH is a weakest acid. Hence, its 210. CH3OH + CO ⎯⎯⎯→ CH3COOH. conjugate pair (CH3)2CHCOO– is a strongest base among the given species. O Heat ⏐⏐ 211. 2H2NCONH2 ⎯⎯→ H2N—CO— NH—C— NH2 . 303 K Biuret OO OH NO2 ⏐⏐ NaOH ⏐⏐ 201. Cl3C—C —CH3 ⎯⎯⎯⎯→ CHCl3 + CH3C – ONa O2N 212. Styphenic acid is (X) (Y) H3O+ OH NO2 CH3COONa ⎯⎯⎯→ CH3COOH + NaCl. (Y) 213. CH3COOH + HN3 202. COONa COONa H3O+ ¾B¾r2/¾KO¾H® NH2 COOH ⎯⎯→ CH3NH2 + N2 + CO2. Methanamine CONH2 PCl5 H3O+ 214. CH3CH —COOH ⎯⎯⎯⎯→ CH3—CH—COCl ⏐⏐ OH OH NH2 Cs2 anthranilic acid 215. CH3COOAg + I2 ⎯⎯⎯→ CH3I + AgI 203. Refluxing with acid causes cleavage of CH3I + AgOCOCH3 ⎯⎯⎯→ CH3OCOCH3 . O Methylacetate ⏐⏐ —N—C — bond (dotted line). 216. See the reactivity of acid derivatives. 217. Acid catalysed esterification involves the nucleophilic attack of alcohol molecule on the protonated acid molecule

ANSWERS 247 LM OP+ + H2 /Pt M PR — C = OH ↔ R — C — OH . The crowding of M P⏐ ⏐ 226. CH3CH2CONH2 ⎯⎯⎯→ CH3CH2CH2NH2 NMM PQPOH OH Propanamine (X) electron releasing alkyl groups at the α-C atom makes the acid less susceptible to nucleophilic Br2 /KOH attack. Hence correct order of esterification is X > Y > Z. ⎯⎯⎯⎯⎯→ CH3CH2NH2 218. Electron withdrawing effect decreases as Ethanamine (Y) – NO2 > —CN > —F > —Br. Hence, the acid strength is 3 < 2 < 4 < 1. Thus, X and Y are 1° amines and they are homologues. KOH Zn CH3Cl 219. C6H5COO CH3 ⎯⎯⎯→ C6H5 COOK + CH3OH 227. C6H5OH ⎯⎯⎯→ C6H6 ⎯⎯⎯→ C6H5CH3 (X) AlCl3 (Y) HCl KMnO4 /OH− C6H5 COOK ⎯⎯⎯→ C6H5COOH + KCl ⎯⎯⎯⎯⎯→ C6H5COOH. (ppt.) 228. Benzoic acid > Phenol > Cyclohexanol. Other Esters will produce acids which are not solid at room temperature. Δ P2O5 220. Cinnamic acid (C6H5 – CH = CH – COOH) does not contain —OH group. 229. CH3COONH4 ⎯⎯→ CH3CONH2 ⎯⎯⎯→ 221. Losser is + I effect more is rate of esterification. (X) Heat 222. Moles of KOH required = 20 × 10–3 × 1 = 10–3 CH3CN H3O+ CH3COOH . 20 ⎯⎯⎯→ (Y) (Z) 230. C6H5COOC2H5 (i) CH3Mg Br excess (A) ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯ CH3 (ii) H2O →⎯ CH3 ⏐ H2SO4 ⏐ C6H5 —C = CH2 ←⎯⎯⎯⎯ C6H5 —C—OH ⏐ (B) − H2O CH3 10–3 mol KOH neutralise acid = 0.122 moles (i) O3 GMM (B) ⎯⎯⎯⎯⎯→ C6H5C = O + HCHO. (ii) H2O/Zn ⏐ 0.122 CH3 1 mol KOH neutralise acid = GMM × 10−3 moles Alc. NaCN Since 1 mol KOH should neutralise 1 mol of acid 231. BrCH2CH2CH2Br ⎯⎯⎯⎯→ NC – CH2CH2CH2 CN ∴ 0.122 × 103 = 1 or GMM GMM H3O+ = 0.122 × 103 = 122 g. ⎯⎯⎯→ HOOC – CH2CH2CH2 – COOH Thus, the acid should be C6H5COOH with GMM Gluteric acid = 122 g. 223. Ka of ethanol = 10–16 – 10–18 whereas that of Dil. NaOH CH3COON = 10–5. 224. p-methoxy benzoic acid is called p-anisic acid. 232. C6H5CONH2 ⎯⎯⎯⎯→ C6H5COONa NaOH C6H6 AlCl3 C6H5 − CH3 + HCl. ⎯⎯⎯→ ⎯⎯⎯→ Toluene CaO CH3Cl 233. RCOCH3 X2 /OH− CHX3 + RCOO– ⎯⎯⎯→ H+ ⎯⎯→ RCOOH. (i) NaOH HCl Heat 225. CH3CCl3 ⎯⎯⎯→ CH3COONa ⎯→ CH3COOH 234. CH2 —CH2 —COOH ⎯⎯→ ⏐ 575 K (X) (Y) (Z) CH2 —CH2 —COOH ⎯→ Adipic acid CH3CN ⎯⎯⎯⎯⎯⎯⎯ = O + CO2 + H2O. H3O+

248 CHEMISTRY PCl5 KCN H2 /Pd − BaSO4 235. CH3OH ⎯⎯→ CH3Cl ⎯⎯→ CH3CN 237. CH3COCl ⎯⎯⎯⎯⎯⎯⎯→ CH3CHO 2C2H5OH ⎯⎯⎯⎯⎯→ CH3CH(OC2H5)2 . + H2SO4 Acetal H3O CH3COOH 238. A and B are non-super impossible mirror images. OO 239. NaBH4 reduces > C = O group to > CHOH but ⏐⏐ ⏐⏐ does not reduce – COOC2H5 group. 236. H—C —OH also has free H—C —group and hence reduces Tollen’s reagent. 13. ORGANIC COMPOUNDS CONTAINING NITROGEN MULTIPLE CHOICE QUESTIONS 1. (d) 2. (a) 3. (c) 4. (a) 5. (c) 6. (b) 7. (b) 8. (b) 12. (c) 13. (a) 14. (b) 15. (c) 16. (c) 9. (c) 10. (b) 11. (d) 20. (d) 21. (d) 22. (b) 23. (a) 24. (a) 28. (a) 29. (b) 30. (d) 31. (a) 32. (d) 17. (b) 18. (b) 19. (c) 36. (d) 37. (a) 38. (b) 39. (a) 40. (a) 44. (c) 45. (b) 46. (b) 47. (d) 48. (d) 25. (b) 26. (a) 27. (a) 52. (d) 53. (b) 54. (b) 55. (a) 56. (a) 60. (a) 61. (c) 62. (b) 63. (b) 64. (b) 33. (c) 34. (b) 35. (a) 68. (b) 69. (c) 70. (b) 71. (c) 72. (d) 76. (c) 77. (a) 78. (d) 79. (a) 80. (b) 41. (c) 42. (d) 43. (c) 84. (a) 85. (c) 86. (a) 87. (a) 88. (b) 92. (c) 93. (c) 94. (b) 95. (c) 96. (b) 49. (d) 50. (b) 51. (b) 100. (c) 101. (a) 102. (b) 103. (d) 104. (c) 108. (b) 109. (d) 110. (c) 111. (d) 112. (b) 57. (b) 58. (c) 59. (c) 116. (c) 117. (a) 118. (b) 119. (a) 120. (a) 124. (d) 125. (b) 126. (d) 127. (c) 128. (b) 65. (a) 66. (d) 67. (c) 132. (a) 133. (d) 134. (c) 135. (b) 136. (c) 140. (b) 141. (a) 142. (c) 143. (b) 144. (c) 73. (a) 74. (c) 75. (c) 148. (a) 149. (d) 150. (c) 151. (b) 152. (c) 156. (b) 157. (b) 158. (a) 159. (d) 160. (c) 81. (c) 82. (d) 83. (a) 164. (b) 5. (b) 89. (c) 90. (a) 91. (a) 5. (d) 97. (b) 98. (a) 99. (d) 105. (a) 106. (a) 107. (b) 113. (b) 114. (b) 115. (c) 121. (b) 122. (b) 123. (b) 129. (b) 130. (b) 131. (b) 137. (c) 138. (c) 139. (a) 145. (b) 146. (c) 147. (d) 153. (d) 154. (c) 155. (c) 161. (b) 162. (d) 163. (a) CASE BASED QUESTIONS Case Study 1 1. (a) 2. (c) 3. (b) 4. (c) Case Study 2 1. (b) 2. (c) 3. (d) 4. (b) Case Study 3 1. (a) 2. (c) 3. (a)

ANSWERS 249 HINTS/SOLUTIONS 1. Chemical reaction I and III are correct whereas NaOH reaction II and IV are incorrect. In reaction II substitution will occur resulting in the formation 19. H2N—CH2—COONa ⎯⎯→ H2N ⎯ CH3 of propane-2-ol. In reaction IV product should be alcohol instead of alkyl diazonium salt. CaO Methyl amine 2. Only 1° amines give carbylamine reaction. + Na2CO3. 20. [NH3OH]+Cl− cannot act as nucleophile because K2Cr2O7 /H2SO4 lone pair of electron of N atom has been used in salt formation. 3. —NH2 ⎯⎯⎯⎯⎯⎯⎯→ O= =O H2SO4 / H2O [O] 21. C6H5 N2Cl ⎯⎯⎯⎯⎯→ C6H5OH + N2 + H2O. p-benzoquinone. Warm 4. Hinsberg’s reagent is phenyl sulphonyl chloride 22. In nitrobenzene, nitro group deactivates the ring C6H5SO2Cl. and thus, makes it less succeptible to electrophilic attack. 5. CH3CH2CH2NH2 and (CH3)2CHNH2 have same functional group. Sn /HCl 7. —NO2 group is ambidentate group. 23. C2H5—O—N = O ⎯⎯→ C2H5OH 8. NaNO2/HCl gives HNO2 which gives different [H] products with primary and secondary amines. + NH2OH.HCl. 9. Aniline couples with diazonium salt in very mild acidic conditions. Highly acidic conditions 24. Among the given species ‘a’ represents convert aniline to anilinium ion which does not couple similarly basic conditions convert picramide. diazonium ion into compound which also does not couple. +– − H+ + H+ + 10. The given data refers to Hinsberg test for 1° 25. C6H5 N H ←⎯⎯ C6H5NH2 ⎯⎯→ C6H5 N H3 . amino (–NH2) groups. Conjugate Aniline Conjugate acid 11. CH3CONH2 is weakest because CO group withdraws electron density from the N atom base of aniline of aniline O: :O:– Br + 26. —NH2 ¾Ba¾qr2. ® Br— —NH2. CH3—C—NH2 ¬¾® CH3—C = NH2. : Br H3PO2: 27. NO2+ is not a nucleophile. 28. Only statement I about aniline is correct. Others 12. C6H5N2Cl ⎯⎯H⎯3O⎯+ → C6H6 + N2 + Cl− .: 13. Aromatic amino group gives coupling reaction are wrong. with Benzenediazonium chloride but aliphatic 29. Heating CH3COONH4 in the presence of P2O5. amino group does not. 30. Electrophilic substitution is represented by (d). 14. Nitrobenzene is also known on oil of mirbane. 31. R — N →= C reduction R ⎯ N ⎯CH3 . ⎯⎯⎯⎯⎯⎯⎯→ Isocyanides 2° Amine 32. 3° nitro compounds donot have α-Hatom. 33. R—N →= C 2HOH HCOOH + RNH2. ⎯⎯⎯⎯⎯⎯→ 1° Amine 15. Only primary amines (—NH2) give N2 with 34. Electron withdrawing effect of carbonyl group HNO2. is more pronounced in amides than that of phenyl group in aniline 16. It is due to absence of labile H-atom in tertiary nitro compounds. R—C O O– N+ H2 17. HNO2 (produced from NaNO2/HCl) reacts ¬¾¾® R—C differently with 1° amine and 2° amines. N.. H2 18. A base having lower pKb but higher Kb is relatively stronger.

250 CHEMISTRY 35. C6H5NH2 + S==C==S 51. Alkyl cyanides have higher dipole moment than those of corresponding isomeric isocyanides. HgCl2 C6H5 ⎯ N == C == S . 52. Phenol does not produce salt with mineral acids. ⎯⎯⎯→ Phenyl isothiocyanate 36. Aldehydes and ketones react only with primary 53. Due to delocalisation of electrons on the N atom amines and form Schiff’s bases. with the benzene ring. 37. N is surrounded by 3 bond-pairs and one lone 54. The process is referred to as Hoffmann’s pair. Hence, its, hybrid state should be sp3. ammonolysis. 38. Coloured dyes are produced by the reaction of 55. 2-Methyl-2-nitropropane does not have phenols and benzene diazonium chloride hydrogen at the α-position. through coupling reaction. Benzyl alcohol does not contain phenolic group. Hence, it does not 56. Allylcyanide is CH2 = CH—CH2—C ≡ N. It form coloured compound with diazonium salt. contains 9σ and 3π-bonds. 39. 57. In aqueous medium dimethyl amine is most basic because its protonated cation is highly stabilised due to hydration. H2N— + – 58. C6H5 ⎯ NH2 CHCl3 /KOH C6H5 ⎯ N =→C —SO3H ¾¾® H3N— —SO3. ⎯⎯⎯⎯⎯⎯⎯→ Benzenamine Benzene isonitrile p-Aminobenzene Zwitter ion or The reaction refers to carbylamine reaction. sulphonic acid Internal salt structure 41. Tertiary amines do not react with acetyl chloride NH2OH or benzoyl chloride. 59. (CH3)2C = O ⎯⎯⎯⎯⎯⎯⎯→ (CH3)2C = NOH 42. Amides are amphoteric in nature. Acetoxime 43. Aniline gives crystalline precipitate of aniline hydrochloride but is not soluble in water. LiAlH 4 (CH3)2CH — NH2 + H2O 44. Sec. amines do not give N2 gas with HNO2. ⎯⎯⎯→ 45. In anilinium ion (C6H5NH3)+ the lone pair is not 2-Amino propane present on the N atom. It has NH2 group, hence a primary amine. 60. Lowest pKb refers to largest Kb and hence greatest basic strength. Among the given species CH3CH2NH2 is most basic and thus has lowest pKb. + 61. The weakest Bronsted base is aniline due to delocalisation of p-electrons of nitrogen atom on 46. In the formation of anilinium ion C6H5NH3 the the ring. lone pair of N atom is used to form bond with H+ ion. 62. H CH3CHCONH2 ⎯B⎯r2⎯/Na⎯O⎯H → CH3 — CH — NH2 ⏐ Ω Ω 47. CH3CH2—N — CH2CH3 has following metamers C6H5 C6H5 H CH3 H 2-Phenylpropanamide 1-Phenylethanamine ⏐ ⏐⏐ CH3CH2CH2—N —CH3 and CH3—CH ⎯ N —CH3. 63. CH3NH2 + HNO2 ⎯⎯→ CH3OH + H2O + N2 ↑ 48. Due to bulky CH3 groups around the N atom 64. The most basic amine among the given the bond angle slightly becomes greater than compounds is (CH3)2NH. Hence, it will be most 109°-28′. reactive towards dilute HCl. 65. 49. R—N→= C + 2HgO ⎯⎯→ R⎯N = C = O+ Hg2O. CH3CO Alkyl isocyanate R—NH2 + O 575 K CH3CO HO 50. 2CH3NO2 ⎯⎯⎯→ 2CO2 + 3H2 + N2. ¾¾® R—N—C—CH3 + CH3COOH N-Alkylethanamide

ANSWERS 251 66. The most basic amine among the given R 77. C = NOH is nitrolic acid and it gives red O2N compounds is N—H because here, the lone colour with NaOH. pair of N atom is most easily available for 78. (CH3)3C—NH2 contains—NH2 group. Hence it protonation. is primary amine. CO 79. CH3CH2CH2CH2NH2 ; (CH3)2CHCH2NH2 ; 67. I. R—X + NK ¾¾® (CH3)3C—NH2 ; CH3CHC2H5 . ⏐ –KX NH2 CO CO + N—R ¾¾¾® 80. [C2H5 N H3] OH− + FeCl3 ⎯⎯→ CO NaOH 3[C2H5 + H3] Cl− + Fe(OH)3. N Brown ppt. COONa 81. R—NH2 + NOCl ⎯⎯→ R—Cl + N2 + H2O. + RNH2 Reduction COONa 82. C2H5N →= C ⎯⎯⎯⎯→ C2H5NH—CH3 CH3 II. R—CN ⎯⎯LiA⎯lH⎯4 → R—CH2—NH2 III. R—CONH2 ⎯⎯KO⎯H/⎯B⎯r2 → R—NH2 HNO ⏐ IV. R—CONH2 ⎯⎯4Al⎯H⎯4 → R—CH2 NH2 2 C2H5 ⎯ N ⎯ NO2. 68. The main product of oxidation of aniline with N-Methyl ⎯⎯⎯→ N-nitrosoethanamine MnO2/H2SO4 is benzoquinone. 83. NO2— —NH2 < Br— —NH2 < 69. C2H5—N ® C + 1 C2H5 ⎯ N = C = S . H3C— —NH2 8 S8 ⎯⎯→ Ethyl Ethyl isothiocyanate carbylamine NO2— is electron withdrawing group. While CH3— group is electron repelling group. Br 70. Among the given species —NH2 group activates group has –1 effect but + R effect. the ring and hence it is most reactive towards electrophilic attack. 84. As the number of alkyl group on the N atom increases the basic strength in general also 71. The main product of this reaction is increases among the alkyl substituted anilines. p-bromoacetanilide. 72. C6H5COCl + HNHC6H5 ⎯⎯→ CH3 H3C C6H5CO—NHC6H5. 85. —NH2 —NH2 73. Among the given species pyridine has planar arrangement, 5H atoms and sp2-hybrid N atom. –10 –10 Kb = 2.6 × 10 Kb = 5 × 10 74. Tertiary amines do not react with H3C— —NH2. diethyloxalate. 75. Aqueous solutions of C2H5NH2 contains Kb = 12 × 10–12 +− 86. CH3— —NH2 H3CO— —NH2 C2H5N H3 and OH ions, which will react with (III) (II) zinc ions (Zn2+) to form white precipitate of Kb = 12 × 10–10 Kb = 12 × 10–10 Zn(OH)2. 76. CH3NH2 + COCl2 ⎯⎯→ CH3 ⎯ N = C = O H2N— —NH2. Methyl isocyanate + 2HCl. (I) Kb = 140 × 10–10

252 CHEMISTRY 99. Alkyl cyanides on hydrolysis give carboxylic acids. 87. —NH2 —NH2 + + CuCN 100. C6H5N2 Cl ⎯⎯→ C6H5CN NO2 NO2 LiAlH 4 HNO2 C6H5CH2OH . Kb = 0.00006 × 10 –10 Kb = 0.029 × 10 –10 ⎯⎯⎯→ C6H5CH2NH2 ⎯⎯⎯→ (Benzyl alcohol) 101. O2N— —NH2. CONH2 NH2 Kb = 0.001 × 10–10 ¾¾¾B¾r2¾/KO¾H¾¾® 88. CH3CH2CH2—N H2 is most basic among these (Bromamide reaction) Br and thus, has lowest pKb value. Br (m-Bromobenzamide) (m-Bromoaniline) 102. + K2CO3 + 2KBr + H2O. 89. C6H5 ⎯NH2 HNO2 ⎯⎯⎯→ C6H5 ⎯ OH + N2 + H2O. 1° Amine H3C Alcohol H3C 90. p-anisidine is H2N— —OCH3 –+ –+ N— —H + ClN2— —SO3Na ¾¾® N, N-Dimethylaniline Diazonium salt The statements b, c, d are wrong about this of sodium p-amino structure. sulphonate 91. (CH3)2C = O + H2NOH ⎯⎯→ (CH3)2C = NOH H3C N— —N = N— –+ H3C —SO3Na (X) Methyl orange . LiAlH 4 (CH 3 )2 CH — NH2 . ⎯⎯⎯→ 92. The electron pair on N atom gets delocalised to 103. Various isomeric structures are form cyclic π cloud. CH3 CH3 CH3 +– +– NH2 H2N NH3Cl NH3NO3 93. ¾¾Ag¾NO¾3 ® + AgCl o-Toludine p-Toludine NH2 white ppt. m-Toludine p-chloroaniline does not give white ppt with CH2NH2 H | AgNO3. N—CH3 LiAlH4 . 94. (CH3)2CH = NOH ⎯⎯⎯→ (CH3)2CH ⎯ NH2 . 1° Amine Acetoxime Benzylamine N-Methylaniline 95. Cu2+ + 4C2H5NH2 ⎯⎯→ [Cu(C2H5NH2)4 ]2− . 104. —CF3 group causes withdrawal of electrons creating deficiency of electrons on the N atom, Bluish Tetrakis [ethylamine] thus, making it least basic among the given species. copper (II) ion 105. In general, alicyclic and aliphatic amines are (Deep blue complex) more basic than ammonia but aromatic amines are less basic than ammonia. 96. The major product is m-nitroaniline. Initially anilinium ion is formed which directs the NO2 group to m-position. 97. The correct basic strength of methyl amines is as given in ‘b’. 98. —NO2 group is meta directing. m- Dinitrobenzene is produced. The presence of two deactivating NO2 group in m-dinitrobenzene checks further nitration.

ANSWERS 253 Cl2 ( g) 106. CH2 = CH—CH3 ⎯⎯⎯⎯→ CH2 = CH—CH2Cl CO CH2CONH2 785 K | 117. NH NaOH ¾¾¾® NH3 CH2 = CH — CH2NH2. CO CH2COOH ⎯⎯⎯→ (A) Excess Allylamine 107. Gabriel synthesis is employed to prepare only ¾B¾r2/K¾O¾H® CH2NH2 | primary amine in which —NH2 group is linked CH2COOH. to primary C atom. b-Alanine (B) 108. Only 2-butanamine has chiral structure and 118. CH3CHO + H2NC2H5 ⎯⎯→ thus, it shows optical activity. 109. Isobutyronitrile will produce 2-Methyl-1- H propanamine on reduction with H2/Pt. CH3 ⎯CH = N ⎯C2H5 H2 / Pt | CH3CH2 — N — C2H5. ⎯⎯→ NO2 NH2 N-Ethyl aldimine Diethyl amine 110. ¾¾¾(N¾H4¾)2S¾x ¾® . CH3 CH3 NO2 Partial reduction ⏐⏐ NO2 119. CH3 ⎯ C = O + H2NOH ⎯⎯→ CH3 ⎯ C = NOH Oxime 111. In fluoroderivatives, the electron withdrawing CH3 effect holds the electron pair of N atom more tightly and cause base weakening effect. Reduction ⏐ + H2O. CH3 ⎯ CHNH2 112. 1, 4-Dinitrobenzene has μ = 0, and is not soluble ⎯⎯⎯⎯→ in acid or base. Isopropylamine 113. 120. Acrylo nitrile is CH2 = CH—C ≡ N . There are 5σ bond and 3π bonds. Hence, number of σ electron = 12 and π-electrons = 6, and lone pair = 1. P2O5 Reduction —NH2 + COCl2 ¾–¾HC¾l ® C2H5 – N = C = O. 121. C6H5COONH4 ⎯⎯→ C6H5—C≡N ⎯⎯⎯⎯→ Phenyl isocyanate Heat LiAlH4 HNO2 C6H5 ⎯CH2NH2 ⎯⎯→ C6H5CH2OH + N2 + H2O. 114. (b) Benzyl amine Benzylalcohol 2 —Cl + 2NH3 ¾C¾u2¾O ® 2 —NH2 122. During acylation of —NH2 group, there is + 2CuCl + H2O. increase in the molecular mass by 42 units for each acyl CH3CO—) group added Cl – H R—NH C=O R—NH2 + C = O ¾¾¾¾® + (CH3CO–) MMLNMMM OQPPPPP115. CH3 CH3 + CH3 Net increase in mol. mass after acylation ⏐ (i) Ag2O (ii) Heat CH3 ⎯ N ⎯CH2 CH3 I− = 348 – 180 = 168. æææHæoffæmaænnææÆ ⏐ ∴ Number of (CH3CO—) groups added Elimination CH3 168 = = 4. (CH3)3N + CH2 = CH2. 42 116. AgCl + 2C2H5NH2 Hence, the compound has four—NH2 groups. ⎯⎯→ [Ag(C2H5NH2)2]Cl . Reduction H3O+ Bis-(ethylamine)silver (I) chloride (soluble complex) 123. R2CHNH2 KMnO4 (Z) [R2CH = NH] ⎯⎯→ (Y) R2C = O + NH3. Ketone (X)

254 CHEMISTRY 124. In I and II lone pair of N is in sp3 hybrid orbital 132. but in III lone pair is less easily available due to CN CONH2 ¾¾H2¾O2¾/O¾H–® electronegative O atom present in the ring skeleton. Hence III is less basic than I. II and Benzonitrile Benzamide IV have lone pair of N in sp2 hybrid orbital. Thus, they are less basic than I and III. In IV, the lone CN pair is delocalised over the entire cyclic structure ¾¾S¾OC¾l2¾® (Huckel rule). Hence it is least basic. Thus, the –H2O . order of basic strength is I > III > II > IV. 125. For molar-mass 72. The value of n in Benzonitrile CnH2n+1 NH2 is 4. NH2 N2Cl NH2 NH2 ¾N¾aN¾O2¾/H¾Cl® KI Br Br 133. 280 K ¾¾® NO2 126. ¾B¾r2® NO2 I (X) Br (Y) Br Cu Powder Br . ¾N¾aN¾O2® +– ¾¾¾¾¾¾¾® HCl N2 Cl (Ulmann reaction) Br Br ¾E¾tO¾H® NO2 Boiling NO2 NO2 Br Br —. (Z) 3, 3′-Dinitrobiphenyl NH3 134. 127. C6H5COCl ⎯⎯→ C6H5CONH2 P2O5 C6H5CN H2 /Ni C6H5CH2NH2 . NO2 NH2 ⎯⎯⎯→ ⎯⎯⎯→ Benzyl amine ¾E¾lec¾tr¾oly¾sis® . OH − H2O Strong acidic (Y) medium 128. C2H5NH2 + HONO ⎯→ C2H5OH + N2 + H2O ¾H¾NO¾3/¾H2¾SO¾4 ® (1 mol) (1 mol) (X) 333 K Thus, 1 mol of C2H5NH2 gives out N2 = 1 mol = 22.4 L at 0°C and 1 atm pressure. 129. C6H5NH2 NaNO2 /HCl C6H5OH NaOH 135. C6H5COOH SOCl2 C6H5COCl NH3 ⎯⎯⎯→ ⎯⎯→ ⎯⎯⎯⎯⎯→ ⎯⎯→ (A) Py 50°C (X) C2H5I P2O5 H2 / Pd C6H5CONH2 ⎯⎯→ C6H5CN ⎯⎯→ C6H5CH2NH2. C6H5ONa ⎯⎯⎯→ C6H5OC2H5 . (B) (C) Benzylamine (Y) Phenetole (D) (Z) 131. Among the given species, it is alkyl isocyanate 136. (R—N = C = O) which is one of the intermediate in bromamide reaction. CH3— —NH2 ¾N¾aN¾O¾2/¾H¾Cl® CH3— —N2Cl ¾H¾2¾O ® CH3— —OH. Heat p-cresol

ANSWERS 255 NaHSO3 /HCl C6H5NH ⎯ NH2HCl . HNO2 137. C6H5N2Cl ⎯⎯⎯⎯⎯⎯→ (Phenyl hydrazine 4. (CH3)2CHNH2 ⎯⎯⎯→ (CH3)2CHOH or SnCl2 /HCl + N2 + H2O. hydrochloride) 5. The reaction leading to formation of B is called 138. The statement ‘c’ is correct about these Hoffman’s bromamide reaction. structures, N atom cannot be pentavalent. Case Study 2 139. H2N— ¾¾CH¾Cl¾3 ® CN— 1. In aqueous medium KOH Et2NH > Et3N > EtNH2 > n-buNH2 ¾¾H¾Cl¾® H2N— + HCOOH. This order is because of hydration effect on Hydrolysis protonated amine. 2. C6H5CONH2 is very feeble base and has (X) highest value of pKb. 3. C6H5NH2 does not dissolve in water but forms 140. For a compound to show optical activity, it salt with HCl. should possess chiral structure. Among the given formula only C6H5CH(NH2)CH3 possesses chiral .. structure. 4. HO. . —group at p-position increases basic 141. HNO3 is weaker than H2SO4 and hence acts as a base in its presence to liberate NO+ 2 ion. strength and thus, increases Kb, 142. Hexadeutrobenzene C6D6 contains heavier 5. m-phenylene diamine is —NH2 . Its isotopes of hydrogen and thus have lower rate of nitration. H2N basic strength is marginally more than aniline CASE BASED QUESTIONS consequently, the value of pKb becomes smaller. Case Study 1 Case Study 3 1. The amine B is, NH2 . 1. The reagent that we use for conversion of I to 2. The molar mass of ‘A’ is 87 g mol–1. II is Br2/NaOH. 3. C is a reduction product of A. The reducing 2. The rate determining step is formation of IV. agent can be LiAlH4, Zn-Hg/alc, etc. This step involves rearrangement process. 14. BIOMOLECULES MULTIPLE CHOICE QUESTIONS 1. (d) 2. (b) 3. (d) 4. (a) 5. (a) 6. (d) 7. (b) 8. (d) 12. (a) 13. (b) 14. (d) 15. (a) 16. (c) 9. (c) 10. (d) 11. (d) 20. (c) 21. (d) 22. (a) 23. (c) 24. (d) 28. (d) 29. (b) 30. (a) 31. (d) 32. (c) 17. (c) 18. (b) 19. (b) 36. (b) 37. (c) 38. (c) 39. (b) 40. (b) 44. (d) 45. (c) 46. (d) 47. (c) 48. (d) 25. (b) 26. (d) 27. (c) 52. (b) 53. (a) 54. (c) 55. (a) 56. (b) 60. (b) 61. (c) 62. (b) 63. (a) 64. (a) 33. (b) 34. (c) 35. (b) 68. (c) 69. (a) 70. (c) 71. (c) 72. (a) 76. (b) 77. (c) 78. (c) 79. (c) 80. (a) 41. (c) 42. (d) 43. (a) 84. (d) 85. (b) 86. (b) 87. (b) 88. (b) 92. (b) 93. (c) 94. (d) 95. (d) 96. (b) 49. (b) 50. (d) 51. (c) 57. (a) 58. (a) 59. (a) 65. (d) 66. (d) 67. (c) 73. (b) 74. (c) 75. (b) 81. (d) 82. (b) 83. (c) 89. (c) 90. (d) 91. (b)

256 CHEMISTRY 97. (d) 98. (d) 99. (b) 100. (b) 101. (d) 102. (b) 103. (a) 104. (b) 105. (d) 106. (b) 107. (d) 108. (a) 109. (a) 110. (a) 111. (b) 112. (d) 113. (a) 114. (d) 115. (c) 116. (c) 117. (d) 118. (b) 119. (d) 120. (c) 121. (b) 122. (c) 123. (b) 124. (b) 125. (a) 126. (a) 127. (c) 128. (b) 129. (b) 130. (b) 131. (b) 132. (c) 133. (b) 134. (c) 135. (d) 136. (d) 137. (b) 138. (b) 139. (c) 140. (a) 141. (d) 142. (d) 143. (b) 144. (c) 145. (d) 146. (d) 147. (c) 148. (c) 149. (a) 150. (a) 151. (d) 152. (a) 153. (b) 154. (a) 155. (a) 156. (b) 157. (a) 158. (b) 159. (b) 160. (b) 161. (c) 162. (c) 163. (a) 164. (d) 165. (c) 166. (b) 167. (c) 168. (b) 169. (d) 170. (b) 171. (c) 172. (c) 173. (b) 174. (d) 175. (d) 176. (d) 177. (b) 178. (d) 4. (d) 5. (b) CASE BASED QUESTIONS 4. (c) 5. (c) Case Study 1 4. (d) 1. (d) 2. (d) 3. (b) Case Study 2 1. (d) 2. (b) 3. (c) Case Study 3 1. (d) 2. (b) 3. (c) Case Study 4 1. (a) 2. (b) 3. (d) HINTS/SOLUTIONS Case Study 2 2. Glucose catabolism yields a total of 38 ATP. 38 ATP × 7.3 kcal/mol ATP = 262 kcal. Glucose has 686 kcal. Thus the efficiency of glucose metabolism is 262/686 × 100 = 38%. 15. POLYMERS MULTIPLE CHOICE QUESTIONS 1. (b) 2. (a) 3. (d) 4. (b) 5. (a) 6. (a) 7. (c) 8. (b) 12. (c) 13. (b) 14. (d) 15. (c) 16. (b) 9. (b) 10. (d) 11. (a) 20. (d) 21. (a) 22. (c) 23. (d) 24. (a) 28. (a) 29. (d) 30. (d) 31. (d) 32. (c) 17. (a) 18. (b) 19. (a) 36. (c) 37. (a) 38. (c) 39. (c) 40. (b) 44. (c) 45. (b) 46. (b) 47. (a) 48. (b) 25. (b) 26. (a) 27. (a) 52. (d) 53. (d) 54. (c) 55. (a) 56. (b) 60. (b) 61. (d) 62. (d) 63. (c) 64. (d) 33. (b) 34. (c) 35. (d) 68. (b) 69. (d) 70. (c) 71. (b) 72. (c) 76. (d) 77. (d) 78. (c) 79. (d) 80. (b) 41. (b) 42. (c) 43. (a) 84. (a) 85. (d) 86. (b) 87. (a) 88. (a) 92. (c) 93. (b) 94. (b) 95. (a) 96. (d) 49. (a) 50. (d) 51. (a) 4. (d) 5. (c) 57. (d) 58. (a) 59. (d) 65. (b) 66. (c) 67. (a) 73. (c) 74. (c) 75. (d) 81. (b) 82. (d) 83. (a) 89. (c) 90. (c) 91. (d) CASE BASED QUESTIONS Case Study 1 1. (a) 2. (c) 3. (b)

ANSWERS 257 16. CHEMISTRY IN EVERYDAY LIFE 8. (c) 16. (a) MULTIPLE CHOICE QUESTIONS 24. (a) 32. (a) 1. (a) 2. (a) 3. (d) 4. (a) 5. (a) 6. (a) 7. (c) 40. (b) 12. (a) 13. (b) 14. (a) 15. (a) 48. (c) 9. (c) 10. (a) 11. (a) 20. (a) 21. (b) 22. (b) 23. (a) 56. (b) 28. (c) 29. (c) 30. (c) 31. (a) 64. (d) 17. (b) 18. (a) 19. (c) 36. (c) 37. (b) 38. (a) 39. (b) 72. (c) 44. (b) 45. (b) 46. (a) 47. (a) 25. (c) 26. (d) 27. (d) 52. (a). 53. (d) 54. (c) 55. (c) 60. (b) 61. (b) 62. (d) 63. (b) 33. (b) 34. (b) 35. (c) 68. (d) 69. (d) 70. (a) 71. (c) 41. (b) 42. (d) 43. (c) 4. (d) 49. (c) 50. (a) 51. (c) 57. (c) 58. (b) 59. (a) 65. (b) 66. (c) 67. (c) CASE BASED QUESTIONS Case Study 1 1. (c) 2. (b) 3. (d)



CUET MOCK TEST PAPER 1. In amorphous solid, amorphous meaning in 7. Factors affecting the conductivity of an Greek is electrolytic solution are (a) Break (b) no form A. distance between the electrodes (c) no range (d) short form B. nature of the electrolyte 2. Which one of the following solids shows both C. concentration of electrolyte Frenkel and Schottky defects? D. power of AC source (a) NaCl (b) KCl Choose the correct option given below (c) AgBr (d) AgI (a) A and B only (b) B and C only 3. The values of Van’t Hoff factors for KCl, MgSO4, (c) C and D only (d) A and D only and K2SO4, respectively, are 8. Which of the following statement is incorrect about the order of a reaction? (a) 1, 1 and 2 (b) 1, 2 and 3 (c) 2, 2 and 3 (d) 2, 3 and 3 4. Match List I and List II. (a) It is an experimentally determined quantity. (b) It can be a zero or fractional number. List I List II (c) It is applicable to elementary and complex A. Binary solution reactions. I. A solution whose osmotic B. Isotonic solution pressure is less than that of (d) It is always equal to the sum of the another. stoichiometric coefficients of reactants in the balanced chemical equation for a reaction. II. Solution having the same osmotic pressure at a given 9. Given below are two statements temperature as that of a given solution. Statement I: The possibility of molecularity being three is very rare. C. Hypotonic III. A solution that contains the solution maximum amount of solute Statement II: The probability of more than that can be dissolved in a given three molecules colliding simultaneously is very D. Saturated amount of solvent at a given small. solution temperature. In the light of the above statement, choose the IV. Solution with two components most appropriate answer from the options given below (a) A-IV, B-II, C-I, D-III (a) Both Statement I and Statement II are correct. (b) A-II, B-IV, C-I, D-III (c) A-IV, B-II, C-III, D-I (b) Both Statement I and Statement II are incorrect. (d) A-I, B-II, C-IV, D-III 5. The boiling point of 4 1iquids A, B, C and D are (c) Statement I is correct and Statement II is incorrect. 170°C, 100°C, 200°C and 30°C, respectively. The correct increasing order of their vapour (d) Statement I is incorrect and Statement II is correct. pressure is (a) D < B < A < C (b) C < A < B < D 10. An emulsion cannot be broken by (c) A < C < B < D (d) C < A < D < B (a) freezing 6. The quantity of charge required to obtain one (b) heating mole of copper from CuO is (c) centrifuging (a) 1 F (b) 2 F (d) adding emulsifying agent (c) 3 F (d) 4 F M-1

M-2 CHEMISTRY 11. The extent of H2 gas is adsorbed on activated 18. Which of the following elements does not belong charcoal is very low because of to the actinoid series? (a) very weak van der Waal’s interaction (a) Th (b) Pu (b) very strong van der Waal’s interaction (c) Tm (d) No (c) very high critical temperature 19. The corret IUPAC name of [Pt(NH3)2Cl(NO2)] is (d) No interaction 12. The main reaction occurring in the blast furnace (a) Diamminechloridonitrito-N-platinum (II) during the extraction of iron from haematite is (b) Diamminechloridonitrito-N-platinum (IV) (a) FeO + SiO2 ⎯→ FeSiO3 (c) Diamminechloridonitrito-N-platinum (0) (b) 3Fe2O3 + CO ⎯→ 2Fe3O4 + CO2 (c) Fe2O3 + 3C ⎯→ 2Fe + 3CO (d) Diamminechloridonitrito-O-platinum (0) (d) 2FeS + 3O2 ⎯→ 2FeO + 2SO2 20. The CFSE for the tetrahedral complex is 8000 13. The number of sigma and pi bonds in a cm–1. The CFSE for octahedral complex for the same metal and ligand will be cyclotrimetaphosphoric acid molecule (a) 8000 cm–1 (b) 18000 cm–1 (c) 16000 cm–1 (d) 20000 cm–1 respectively are (a) 12 and 1 (b) 15 and 3 21. [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are the example of (c) 12 and 3 (d) 15 and 1 14. Match List I and List II. (a) geometrical isomers List I List II (b) ionization isomers (c) coordination isomers A. XeO3 I. Linear (d) linkage isomers B. XeF4 II. Square planar C. XeOF4 III. Pyramidal 22. The correct order of increasing crystal field D. XeF2 IV. Square pyramidal splitting energy is (a) A-II, B-IV, C-III, D-I (a) [Cr(NH3)6]3+ < [Cr(Cl)6)3– < [Cr(CN)6]3– (b) [Cr(Cl)6]3– < [Cr(CN)6)3– < [Cr(NH3)6]3+ (b) A-II, B-III, C-IV, D-I (c) [Cr(Cl)6]3– < [Cr(NH3)6]3+ < [Cr(CN)6]3– (d) [Cr(NH3)6]3+ < [Cr(CN)6]3– < [Cr(Cl)6]3– (c) A-IV, B-I, C-III, D-I 23. The reagent used in the following reaction is (d) A-III, B-II, C-IV, D-I CH3CH2CH2CH3 ⎯→ CH3CH2CH2CH2Cl 15. Phosphinic acid is + CH3CH2CHClCH3 (a) Cl2/UV light (a) H3PO2 (b) H3PO3 (b) Cl2 in dark (c) Cl2 in the presence of copper in dark (c) H4P2O5 (d) H4P2O6 (d) KCl + H2SO4 24. In the following reaction, the product formed 16. Which of the following statement is correct? (a) All P—Cl bond lengths in the PCl5 molecule will be in the gaseous state are equal. (b) All the three N—O bond lengths in HNO3 CH3 are equal. + X2 (c) In peroxosulphuric acid (H2SO5) sulphur is ¾F¾e ® in a +6 oxidation state. dark (d) Change in enthalpy is positive for the (a) CH3 preparation of SO3 by catalytic oxidation of only SO2. (b) X X 17. In the given reaction, the oxidation state of Mn CH3 are only 3MnO42– + 4H+ ⎯→ 2MnO4– + MnO2 + 2H2O (a) +3, +4, +7 only (b) +2, +4, +7 only (c) +4, +6, +7 only (d) +4, +5, +7 only

CUET MOCK TEST PAPER M-3 CH3 CH3 (c) Statement I is correct and Statement II is (c) + incorrect. XX (d) Statement I is incorrect and Statement II is correct. X (d) only 31. Which one of the followings is used as a nitrogen source in Gabriel’s synthesis of amines? 25. Which of the following compound is chiral in (a) Potassium phthalimide nature? (a) 1-Bromobutane (b) 2-Bromobutane (b) Sodium azide (c) 2-Bromopropane (d) 2-Bromopropan-2-ol (c) Sodium nitrite 26. Ethanal can be formed by ethanol through (a) treatment with LiAlH4 (d) Potassium cyanide (b) treatment with pyridinium chlorochromate (c) treatment with KMnO4 32. Methylamine reacts with nitrous acid to form (d) catalytic hydrogenation (a) CH3—O—N=O (b) CH3—O—CH3 27. The common name of the given compound is (c) CH3OH (d) CH3CHO Br O 33. Which one of the following reagents is not used for the reduction of nitro compounds into amines? (a) H2/Pd (b) Sn + HCl (c) Fe + HCl (d) Fe + NaOH H3C—CH—CH2—C—H 34. Azo coupling reaction with benzene diazonium (a) β-Bromobutyradehyde chloride will not be shown by (b) Cinnamaldehyde (c) Vanillin (a) Anisole (b) Aniline (d) Isobutyraldehyde 28. The given reaction is (c) Phenol (d) Nitrobenzene 35. In the Gabriel synthesis, the following amines can be prepared A. Isobutyl amine CHO B. 2-Phenylethylamine ¾¾C¾O, H¾C¾l ¾® C. N-methylbenzylamine Anhyd. AlCl3/CuCl D. Aniline (a) Etard reaction Choose the correct option given below: (b) Gatterman reaction (a) A and B only (b) B and C only (c) Gatterman-Koch reaction (d) Friedel-Crafts acylation reaction (c) C and D only (d) B and D only 29. The reactant and reagent used to prepare the 36. Which one of the following statements is following aldehyde CH3—CH=CH—CHO is (a) CH3CH=CHCN, H2/Pd incorrect about glucose? (b) CH3CH=CHCH3, LiAlH4 (c) CH3CH=CHCH3, SnCl2/HCl (a) It is aldohexose. (d) CH3CH=CH—CN, DIBAL-H (b) On heating with HI, it forms n-hexane. 30. Given below are two statements (c) It is present in pyranose form. Statement I: Ethanal is a volatile liquid. (d) It gives a 2,4-DNP test. Statement II: Ethanal and propanone are miscible with water in all proportions. 37. Match List I and List II. In the light of the above statement, choose the List I List II most appropriate answer from the options given below: A. Zymase I. Hydrolysis of cane sugar (a) Both Statement I and Statement II are B. Urease II. Conversion of glucose into ethyl correct. alcohol (b) Both Statement I and Statement II are C. Maltase III. Hydrolysis of maltose into incorrect. glucose D. Invertase IV. Decomposition of urea into NH3 and CO2

M-4 CHEMISTRY Choose the correct option 42. The structure of the compound I is (a) A-II, B-IV, C-III, D-I Ph CH3 H3C Ph (a) Ph (b) Ph (b) A-I, B-III, C-IV, D-II H H (c) A-IV, B-III, C-II, D-I (d) A-II, B-IV, C-I, D-III 38. Which of the following statement is correct Ph CH3 H3C CH3 about tranquilizers? (c) CH2Ph (d) H (a) Do not affect the message transfer from a H Ph nerve to a receptor. 43. When (J) treated with acetic anhydride, in the (b) Inhibit the enzymes which catalyze the presence of the corresponding salt of an acid, degradation of noradrenaline. the product obtained is (c) These are narcotic drugs. (a) cinnamic acid (d) Can relieve pain and fever. (b) crotonic acid 39. Artific ial sweetener which is stable at cooking (c) maleic acid temperature and does not provide calories. (d) benzylic acid The correct option is 44. The structure of compounds J, K, and L, (a) Sucrose (b) Sucrolose respectively, are (c) Glucose (d) Aspartame (a) PhCOCH3, PhCH2COCH3, 40. Which of the following is not used as food and PhCH2COO–K+ preservatives? (b) PhCHO, PhCH2CHO, and PhCOO–K+ (a) Table salt (b) Sucrose (c) PhCOCH3, PhCH2CHO, and CH3COO–K+ (d) PhCHO, PhCOCH3, and PhCOO–K+ (c) Benzoic acid (d) Salts of sorbic acid 45. Which of the following statements is correct for Read the passage given below and answer the questions (41 to 45): compound (K)? A tertiary alcohol H upon acid-catalyzed dehydration (a) It reacts with alkaline KMnO4 followed by gives a product I. Ozonolysis of I leads to compounds acidic hydrolysis and forms benzoic acid. J and K. Compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on (b) It reacts with iodine and NaOH to form reaction with KOH gives only M. triiodomethane. 41. Compound H is formed by the reaction of (c) It is prepared by the reaction of benzene O with benzoyl chloride in presence of (a) + PhMgBr anhydrous aluminium chloride. Ph H (d) It reacts with freshly prepared ammoniacal silver nitrate solution. O Read the passage given below and answer the questions (46 to 50): (b) + PhCH2MgBr Ph CH3 A compound (X) containing C, H and O is unreactive towards sodium. It also does not react with Schiff’s O reagent. On refluxing with an excess of hydroiodic acid, (X) yields only one organic product (Y). On (c) + PhCH2MgBr hydrolysis, (Y) yields a new compound (Z) which can Ph H be converted into (Y) by reaction with red phosphorous and iodine. The compound (Z) on oxidation with O Me potassium permanganate gives a carboxylic acid. The equivalent weight of this acid is 60. (d) + MgBr Ph H Ph