Unit-II - Organic Chemistry - [Solutions]
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UNIT
Organic Chemistry
2
Section A : Straight Objective Type 1.
Answer (3)
COOH OH Due to ortho effect, 2.
is most acidic.
Answer (3) On the abstraction of H(d), allyl free radical as well as tertiary free radical is formed.
3.
Answer (3) Electron withdrawing group increases rate for SN reactions.
4.
Answer (2)
⊕
⇒
Aromatic + resonance stabilised
⊕
CH2 ⇒
resonance stabilised ⊕
CH2 = CH − C H2
⇒
resonance stabilised
⊕
CH2 = C H ⇒ least stable due to presence of positive charge on sp hybridised carbon. 5.
Answer (4) Number of stereoisomers = 2n (n = number of asymmetric carbon atoms + number of double bonds)
6.
Answer (1) Meso tartaric acid is optically inactive due to presence of plane of symmetry.
7.
Answer (1)
CH3
CH3
The Fisher projection of the compound is HO
H and it has R & R configuration on both H
OH
asymmetric C-atom. 8.
Answer (1)
p orbitals of carbon being restricted rotation H H—C—C—C—C—H H H H
H
H
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Organic Chemistry
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9.
Answer (3)
CH3 − C HOH − C HOH − COOH ∗
∗
No. of chiral carbon atoms = 2 No. of optical isomers = 2n = 22 = 4 10. Answer (2) 6
7
Cl ⊕
4
1 3
⊕7
5
⊕ ⊕3
2
3
3
Positive charge due to delocalisation of π electrons is present 1, 3, 5 and 7 carbon. 11. Answer (1) Bridge head carbocation is not possible. 12. Answer (2)
OH
⊕ +
H
ring expansion ⊕
⊕
13. Answer (1) N2 is good leaving group, resulting in the formation of carbene. 14. Answer (4) –OCH3 group is +R group which would decrease the magnitude of ‘+’ charge. 15. Answer (1)
H
H OH
Conjugate base is resonance stabilised and is aromatic. 16. Answer (2) Compounds containing N and S respond to this test. 17. Answer (2) HNO 3 decomposes Na2 S and NaCN present in Lassaigne extract becuase otherwise they will give precipitation with AgNO3. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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18. Answer (3)
⇒
Stabilised due to aromatic nature and is resonance stabilised.
:
19. Answer (3)
CH = CH – O H
CH2 – CHO
O
OH
O
OH
NH
NH2
20. Answer (4) Pyrolle is least basic as the lone pair is contributing to aromaticity of the molecule. N is sp2 hybridised.
In pyridine N
∴
is most basic, therefore 6 member ring is least strained.
N H 21. Answer (2) :
⊕
CH 3 – O – CH 2 stabilised due to lone pair of O.
22. Answer (2) B – C cyclic delocalisation of 6π electrons.
23. Answer (2) Br – is a good leaving group and carbonyl site has electron deficient carbon. 24. Answer (1) Alkenes undergo electrophilic addition reaction. Carbonyl undergo electrophilic addition reaction. 25. Answer (4)
sp2 hybridised carbon cannot be readily attacked by nucleophile from back side and attack at bridgehead carbon is not feasible. 26. Answer (2) 2 has (4n + 2)π electrons and planar ring. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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27. Answer (4)
S
CH2OH H
OH
H
OH
HO
S
CH2OH
OH S
H
S
H
H
OH
HO
H
CH2OH
CH2OH
28. Answer (1) δ+
δ−
Br − C l , so Cl– will add at most stable carbocationic site.
29. Answer (2) Ketal formation. 30. Answer (4) 1 , 2 and 3 has plane of symmetry.
31. Answer (3) In ortho substituted amines, nitrogen moves out of the plane. 32. Answer (1) 1 and 2 are resonance stabilised but 2 is less stabilised since it has electron withdrawing nitro group.
33. Answer (2) Aliphatic amine is more basic and lone pair e– of pyrolle participate in resonance. 34. Answer (4) Percentage of 1 chloro-3-methyl butane =
1× 3 × 100 3 = × 100 = 13.88% 1× 3 + 6 × 1 + 1× 5 + 2 × 3.8 9 + 5 + 7.6
35. Answer (1)
Cl
Cl
⊕
⊕
Cl — C — CH — CH3 < Cl — C — CH2 — CH2 (more stable carbocation) Cl
Cl
36. Answer (4) H3O+ gives A i.e. Markownikov’s addition, through carbocation formation. B formed by hydroboration oxidation that gives anti-Markownikov’s product. 37. Answer (2)
H+
⊕
H ⊕
Hydride transfer H
⊕
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38. Answer (2) –OH group of molecule will behave as a basic site in presence of HCl. 39. Answer (2) Gauche form is more stable due to presence of intramolecular hydrogen bonding. 40. Answer (2) Lemeiux reagent will add OH at each site after cleavage of every bond. 41. Answer (3)
Br Br
2NaNH2
NaNH2
Na
C2H5Br
C2H5
42. Answer (2)
CH3
CH3
H
H2
C—C—C—C CH3
CH = CH – C CH3
C – CH3
H H
H
CH
(optically inactive)
CH – CH3
Triple bond is more sensitive for addition H2 than double bond. 43. Answer (3) 2+
H2O Hg Δ CaO + C ⎯⎯→ CaC2 ⎯⎯ ⎯→ C2H2 ⎯⎯ ⎯→ CH3 CHO H2SO 4
44. Answer (2) CH3
H C=C
H
H 3C
CH3
H
CH2N2 : CH2
C–C H 3C
H
45. Answer (2) ‘Na’ give electron resulting the formation of free radical. 46. Answer (2)
CO
COONa COCl
Sodalime Δ
Anhyd. AlCl3
A
B (ketone)
47. Answer (1) AlCl3
Cl
⊕
⊕
48. Answer (2)
OH CH3CH2 – C ≡ CH
H2O 2+
+
Hg , H
O
CH3CH2 – C = CH2
CH3 – CH2 – C – CH3
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49. Answer (2) KMnO4 (H+) cold gives syn hydroxylation. 50. Answer (3)
Cl Cl
Cl
Cl ,
,
Cl
Cl
Cl Cl
,
51. Answer (4) Peracids give epoxy formation. 52. Answer (1) Peroxy acid form epoxy alkane. 53. Answer (2) CH3 – C ≡ CNa 54. Answer (2)
O3
C2H5Br
CH3 – C ≡ C – C H2 – CH3
O CHO
55. Answer (2) Thiophene undergoes electrophilic substitution at 2nd position. 56. Answer (2) +
(CH3)2 C == CH2
H
(CH3)2 – C – CH3
(CH3)2 C == CH2
(CH3)2 – C – CH3 CH2 – C (CH3)2 +
–H
(CH3)3 C – CH3 CH == C
CH3 CH3
57. Answer (2) H2 1, 4 addition
CH2 = CH – CH = CH2
CH3 – CH = CH – CH3
O3/H2O 1, 4 addition
A
2CH3COOH B
(Zinc is not present hence CH3CHO first formed oxidised to CH3COOH) 58. Answer (3) Br +
CH3 – CH – CH == CH2
H
CH3 – CH – CH – CH3
1-2 H¯ shift
CH3 – C – CH2 – CH3 CH3
CH3
Br¯
CH3 – C – CH2 – CH3 CH3 (3° more stable than 2°)
59. Answer (3) Triple bond is more reactive than double bond for hydrogenation. 60. Answer (1) Chlorine free radical is prepared in presence of ultraviolet light. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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61. Answer (3) Carbocation formation and ring expansion takes place. 62. Answer (1) Due to high electronegativity of carbon. 63. Answer (3)
O O CH3 – C ≡ C – CH2CH3
O3
CH3 – C – C – C2H5 Li AlH4
OH OH CH3 – C – C – C2H5 H
H
Chiral centre = 2 Stereoisomer = 4
64. Answer (4) Free radical substitution takes place in presence of sunlight. 65. Answer (3) ⊕
COCl
CO Anhyd. AlCl3
NO2
Given product. NO2
66. Answer (1) 1 → Kolbe’s electrolytic reaction.
67. Answer (4) Θ
Θ ⊕
CH
NaNH2 3 CH3 − C ≡ C H ⎯⎯ ⎯ ⎯→ CH3 − C ≡ C Na ⎯⎯⎯ → CH3 − C ≡ C − CH3
68. Answer (3) Williamson’s synthesis. 69. Answer (4) Benzyl fluoride form soluble AgF and alkyl bromide give yellow ppt of AgBr. Vinyl chloride does not react. 70. Answer (3) Markownikov product. 71. Answer (1) Retention is dominating over inversion. 72. Answer (3) Saytzeff product. 73. Answer (1) Elimination of HCl resulting the formation of cyclic ether. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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74. Answer (2)
NH2
NH2
Cl NH2 Cl
NH2
– NH3
HNH2
Cl
Cl
– NH2
Cl
.
Benzyne
75. Answer (4)
aq KOH C2H5Br alc
C2H5OH (1) CH2 = CH2 (2)
C2H5OH and CH3OCH3 are isomers +
H C2H5OH ⎯⎯→ ⎯ CH2 = CH2 .
76. Answer (3) – NO2 group is electron withdrawing it show –I & –R effect. More over it stabilize anion more than other group. 77. Answer (1) Inversion product. 78. Answer (1) – NO2 at ‘o’ and ‘p’ position stabilize anion
NO2
F
NO2
F OH
–
+ OH CH3
CH3
NO2
NO2
O2N –δ
F OH –δ –δ CH3 NO2
79. Answer (3) Iodoform test. 80. Answer (1) – NO2 group show – I & – R effect. 81. Answer (4)
I + I2
+ HI .
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82. Answer (3) SN1 reaction carbocation formation. 83. Answer (2)
OH Br AgOH
Br
Br
.
Br 84. Answer (4) Depends on formation of free radical Br2 (1 eq) hν Δ
Isomeric monobrominated product
Br2 hν
85. Answer (4) Carbene formation then ring expansion takes place. 86. Answer (1) Due to presence of benzylic group
CH2Cl
CH2OH
NaOH
Cl
Cl
87. Answer (3)
O CH3O
CH3CH2SH
– CH3OH
CH3CH2S
CH2 – CH2 C
+
CH3CH2–S–CH2–CH2O
H2O/H
CH3CH2–S–CH2–CH2OH .
88. Answer (3) SN1 → involves cyclic formation and
R–O
+δ R
S=O
–δ O S=O
R – Cl + SO2
Cl
Cl
CH3 Where R =
.
C H
D
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89. Answer (4)
Br –
OH OH
O
O H
H
O
O H
90. Answer (4) +
H CH3 – CH = C
–
Br Cl
COOH
H CH3 – CH – C – COOH
Cl Br therefore 4 stereoisomers
2 chiral carbon 91. Answer (2)
Chlorination takes place at more probable free radical site. 92. Answer (1)
CH3
18
CH3 – C – CH2 O
H2O +
H3C – C – CH2
H
Stable carbocation
OH 18
H2 O CH3
CH3
CH3 – C – CH2OH
CH3 – C – CH2OH
18
O
O 18
H
H
H
93. Answer (1)
(CH3 )3 C⊕ → Stable carbocation. 94. Answer (3)
H
H
O
O
O N
less stronger than
O N
O
O Intramolecular H-bonding. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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95. Answer (1)
O
O
OH
C
C
Δ
O + 2
+ H2O
O
C
C H
O
OH
OH
(phenolphthalein) 96. Answer (1)
COOH
COONa
+ NaHCO3
+ CO2 + H2O
Phenol does not react with NaHCO3. 97. Answer (3) AgNO 3 → AgI ppt. R – O – R′ + HI → R – OH + RI ⎯⎯ ⎯⎯
98. Answer (1) Depends on electronegativity of elements and stability of conjugate pair. 99. Answer (3) Due to presence of intramolecular H bonding in (a). 100. Answer (4) Depends on pKa value
1 pKa ∝ Strength of acid 101. Answer (3) E1 Saytzeff oriented product. 102. Answer (3) Ring expansion +
CH2OH
H /Δ
CH2
+
–H H
103. Answer (3) Both NaBH4 and LiAlH4 reduces acid chloride to alcohols. 104. Answer (3)
OH NO2 is weaker acid than H CO . 2 3
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105. Answer (3)
OH O — Fe — O
+ FeCl3
+ HCl
O
(ferric phenoxide) or (C6H5O)3Fe H2O [Fe(C6H5O)3(H2O)3]Cl3 Soluble complex 106. Answer (3)
CH3
CH3 KMnO4
3° 2°
CH3 CrO3
OH CH3COOH H
OH O
OH 107. Answer (3) THF is good solvent for Grignand reagent. 108. Answer (2) CH2OH
CH 2Cl
PCl5
CH2OH
+ POCl3
CH 2Cl
109. Answer (4) +
Br2
H
Mesoform (1)
CCl4
OH
Trans (anti addition)
F
Racemic mixture d, l form
Br2 cis
anti addition
Mesoform + d, l = 3 stereoisomer 1 2 110. Answer (1) Ether have low boiling point, more volatile whereas alcohols due to presence of intermolecular H bonding have high boiling point therefore less volatile.
CH3 CH3–O–CH–CH3 CH3
HI SN2
CH3I + HO – CH– CH3 NaOI CH3COONa + CHI3 (yellow ppt)
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111. Answer (2)
CH2OH
O +
H
CH3 OH 112. Answer (1) O
C – OH C – OH
+
HO – CH2 HO – CH2
O
A
HO – CH2 CH2
H2SO4 Δ
HO
O
O
O
O
O O
113. Answer (2) A → Base abstract acidic H. B → Carbocation formation due to H+ ion. 114. Answer (1) B is primary alcohol and C is secondary alcohol. Primary alcohol produces carboxylic acid and secondary alcohol produces ketone. 115. Answer (2)
OH
OH CHCl3
OH CHO
KOH
OH
50%KOH
COOK
+
CH2OH
Cannizzaro reaction
Reimer Tiemann reaction
.
116. Answer (1)
C6H5CO3H
O
OH
HBr
Br (A)
Trans 2-Bromocyclohexanol
117. Answer (4)
CH3 Compound -1
CH2 = CH – O – CH
+
CH3
CH3 Compound-2
CH2 = C – O – C2H5
H2O H
+
CH3CH2OH + CH3 – C – OH H
CH3
H2O H
CH3
CH3 – C – OH + C2H5OH H
Both 1 & 2 gives same product. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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118. Answer (1)
C6H5 – CH2 – CH2 – CH2 – OH (X) LiAlH4 (reduce both C = C and C = O bond) NaBH4
C6H5 – CH = CH – CH = O
(does not reduce C = C bond
crotonaldehyde
C6H5 – CH = CH – CH2OH (Y)
119. Answer (4) CH3OH < CH3 – O – CH3 < C6H5OH. 120. Answer (3) Jone’s reagent oxidies 1° alcohol to aldehyde
CH2OH
CHO
Jone's reagent
In choice (4), tertiary alcohol is resistant to oxidation at room temperature. +
R – X + N a OR → R – O – R + NaX. 121. Answer (2) Order of boiling point of isomeric alcohols
3° > 2° > 1°.
122. Answer (2)
O
O +
O
O
O
O
123. Answer (4) Cross Cannizzaro reaction HCHO is always oxidised to HCOOH other part reduced to alcohols. 124. Answer (4)
C – OH + [(CH3)3CO]3 Al
>C=O
125. Answer (1)
O
CHO
KOH 50%
O
O
COOH
CH2OH
+
no ‘α’ hydrogen atom ∴ Undergo Cannizzaro Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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126. Answer (2)
O
OH H2CrO4 in aq
CH3 – CH = CH – CH – CH3
acetone (Jones reagent)
CH3 – CH = CH – C – CH3
127. Answer (2)
CHO (CHOH)4
5HIO4
5HCOOH + HCHO
CH2OH 128. Answer (2) (A) Br is good leaving group and carbocation stabilize by allylic resonance. (C) Nucleophilic attack fastest at
CO due to presence of αH.
129. Answer (1) NaBH4 reduces only carbonyl compounds. 130. Answer (2) Only acid bromide and alkyl bromide undergo hydrolysis with alkali Br attached with benzene ring is resonance stabilize therefore it give substitution at temperature and pressure. 131. Answer (4)
COOH is most acidic so pH is low.
HO – CH2 – CH NH3 132. Answer (3)
R +
C=O + H
R–C=O–H
HO
OH
R – C – OH OH 133. Answer (3)
CH3 CH3–CH2–CH2 – C
N
CH3MgI
CH3 – CH2 – CH2 C = NMgI +
H3O
CH3 CH3 – CH2 – CH2 – C = O + Mg
I OH
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134. Answer (4)
O
O H2S
CH3 – C – Cl
CH3 – C – SH + HCl
[H – S – H]
135. Answer (4)
O CH3 – C – O – CH3 + CH3 Mg Br O
O MgBr H2O
CH3 – C – O – CH3
CH3 – C
CH3
+ Mg–Br + CH3OH OH
CH3 (1) CH3MgBr (2) H3O
+
OH CH3 – C – CH3 CH3 136. Answer (2)
O
O
O
H – O – CH2 – C – OH
SOCl2
CH3OH
Cl – CH2 – C – Cl
Cl – CH2 – C – O–CH3
137. Answer (3)
O CH3 CH2 C
NH
CH3
It should be noted that acid will be neutralised with amine. 138. Answer (2) β keto acid undergo decarboxylation fast due to formation of resonance stabilised carbanion. Δ PhCOCH2COOH ⎯⎯→ PhCOCH3.
138(a). Answer (3)
O Ph
IIT-JEE 2008
O
O ∗
β-keto acid
Δ
OH
–CO2
C Ph
O I2 + NaOH
∗ CH3
iodoform reaction
C Ph
∗ ONa + CHI3
E
G
139. Answer (4) HIO4 can oxidise only two adjascent oxygen containing carbon atoms. Aldehydic and 2° alcohol oxidise to formic acid whereas 1° alcohol to aldehyde by periodic acid. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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140. Answer (2)
O
O–H +
+ (H – Br)
–
H
O
OH
OH Br
H Br O
Br
O
OH
141. Answer (1) Baeyers villiger oxidation. 142. Answer (3)
O
NH2
NOH NH2OH
Na/C2H5OH
+
H
CH3
H 143. Answer (2)
HVZ followed by elimination. 144. Answer (3) Ethylene glycol is protecting group for carbonyl. 145. Answer (3) (A) NaBH4 reduces cabonyl group. (B) HBO changes alkane to alcohol. (Anti-Markownikov’s product) 146. Answer (4) Pinacol - pinacolone rearrangement. 147. Answer (1)
O
O
(1) O (2)
Anti - aromatic
Stable aromatic O
(3)
Stable aromatic
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148. Answer (4)
MgBr
Br Mg dry ether
NBS
CH3CN H3C – C = O
H3C – C = N – MgBr H2O
149. Answer (1)
Br2
CH3CH2COOH
Red P
NH3
CH3–CH–COOH
CH3–CH–COOH
Br
NH2
150. Answer (1)
O NH2
O
H – N – C – CH2Cl
H
+ Cl – C – C – Cl H
AlCl3 H N O or
C=O
N
CH2
H 151. Answer (1) Coupling occurs prefenertially in the para position to the hydroxyl group. But it this position is blocked then coupling occurs at ortho position. 152. Answer (3)
NO2
NO2
N2Cl
Br2/FeBr3
NaNO2 Br (X)
Cl Cu powder
HCl 280 K
Br
Br
(W)
(Z)
Mg Ether
Cl
Cl CH3 C6H5
C6H5COCH3 +
H3O
OH (F)
MgBr (A)
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153. Answer (2)
NaNO2 NH2
–
OH
HCl
N2Cl phenanthrene
154. Answer (1)
KCN
SOCl2
NaOH
H2/Pt
–HCl OH CN
OH Cl
Cl
CN
Cl
CH2
N
NH2 155. Answer (2)
NH3
KOH Br2
Δ O
OH
O
NH2 A
CHCl3 KOH NH2
NC C
B
+
H3O
(B & D → are same]
+ HCOOH NH2
E
D
156. Answer (1)
F
NMe2
NMe2 NaNO2/HCl 0 – 5°C
NO2
NH2
NMe2 H2/Ni
N2Cl
NH2
157. Answer (2) Hoffmann elimination takes place, resulting the formation of less substituted alkene. 158. Answer (1)
O CH3 – C – H gives iodoform (haloform) test. 159. Answer (3)
CH3CH2COOH
O SOCl2
CH3CH2COCl A
(CH3CH2CH2)2NH
CH3 – CH2 – C – N B
C3H7 C3H7
LiAlH4
(C3H7)3N C
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160. Answer (2) Less substituted alkene is more stable
CH3 OH—
N CH3
Δ
CH2 + H2O
N
– H2O
CH3
CH3
CH3
161. Answer (3)
H
CH3
H
C – NH2 + O = C – C – CH3 H3C
CH3
H
H
CH3
H
C – N = C – C – CH3 H3C
CH3
H
H2/Pt H CH3 – C – N – CH2 – CH(CH3)2 CH3 H 162. Answer (3) III - 2 lone pair of electrons present on 2 N atom. IV - Presence of – SO2NH2 group. 163. Answer (4) Due to presence of +I effect of methyl group. 164. Answer (2) Formation of quaternary ammonium salt. 165. Answer (1)
OH +
OH
N
O + H
N
H
H
OH2 N H N
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166. Answer (3) Only benadryl is tertiary amine Choline
- quaternary ammonium salt
Benzedrine - primary amine Coniine
- secondary amine.
167. Answer (2) 2° amine react with Hinsberg reagent which forms insoluble material which does not soluble in NaOH. 1° amine react with Hinsberg reagent and finally form soluble complex. 3° amine does not react with Hinsberg reagent. 168. Answer (2)
N–H CH3I
AgOH Δ
CH3 N – CH3 – I
CH3 N
I
CH3 CH3
AgOH
CH3I excess
CH3 N – CH3 – OH
CH3
Δ
N CH3
169. Answer (2) H 2 O2
—N
+ NH2OH
Δ
170. Answer (1)
O
O
C O
+ NH3 C
C
OH NH2
O
O 171. Answer (3)
O C
NH2
H2/Pd. high pressure
CH2NH2
H2/Pd.
CH2NH2
H2 reduces – CONH2 and then benzene ring. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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172. Answer (3) H2O R − N = C = O ⎯⎯ ⎯→ RNH2 + CO2 .
173. Answer (2) Nitroglycerene is formed by reaction of glycol (alcohol) and nitric acid (acid). 174. Answer (4) Secondary amine has + I effect. 175. Answer (2) NaBH4 reduces only carbonyl compound where as LiAlH4 reduces carbonyl and cyanide group. 176. Answer (3)
NH2
O
NH3
OH
CH3 177. Answer (1)
H
O
N
H–N
O
–
N
N
OH
H H
O
H
O
O H H
O
O
O
Anion (conjugate pair) stabilize by resonance. 178. Answer (1) ⊕
+
LiAlH4 H3O CH3 − C ≡ N ⎯⎯ ⎯ ⎯→ CH3 − CH2 − NH2 ⎯⎯ ⎯→ CH3 − CH2 NH3 .
179. Answer (2)
O CH3
C = O + H2 N – NH2 – C – NH2
H Pyridine
semi carbazide
O H3C
C = N . NH – C – NH2
H
Semi carbazone O
– NH2 of – C – NH2 does not undergo condensation due to its resonance stabilization. O – C – NH2
O – C = NH 2
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180. Answer (3) 1° amine oxidise to nitro group and 2° alcohol oxidise to ketone with cold KMnO4. 181. Answer (4)
O
NH2OH
O–H
+
H Et OH
OH
OH
H2N–OH
N – OH H
H – H2O
N–O–H
N – OH H
182. Answer (2) NaOH react with carboxylic acid. 183. Answer (4) 3° amine cannot form amine oxide. 184. Answer (3) Due to stability of carbocation. 185. Answer (2)
CH3 CH3 – CH2 – C – C = O + H2NCH3 H
H CH3 H
CH3 – CH2 – C – C = N – CH3 H H2/Pt. CH3 H
H
CH3 – CH2 – C – C – N – CH3 H
H
186. Answer (3) Nylon has strongest intermolecular forces (i.e. hydrogen bonding) out of these. 187. Answer (1) Artificial silk is a polysaccharide while the other three are polyamides. 188. Answer (1)
CH3 n CH2 = C
CH3
CH3
C – CH2 – C – CH2 CH3
CH3
CH3
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189. Answer (4)
F Teflon
F
C C F
F
is fully fluorinated polymer. n
190. Answer (3) ⎛ ⊕ ⎞ Amino acid are amphoteric in nature. The acidic properties are due to amino group ⎜⎜ – N H3 ⎟⎟ . ⎝ ⎠
191. Answer (1) Alanine is NH 2 – CH – COOH CH3
CH3 (At low pH) i.e. acidic pH NH3 – CH – COOH
CH3 (At high pH) – i.e. basic pH NH2 – CH – COO
192. Answer (1) During vulcanization of natural rubber S – S crosslinks are formed which make it more elastic. 193. Answer (3) n CH2 = CH2
343 K – 373 K, 6 – 7 atm Ziegler – Natta catalyst
( – CH2 – CH2 – )n
[TiCl4 + (C2H5 )3Al]
194. Answer (3) The α-helix is known as 3.613 helix since each turn of the helix has approximately 3.6 amino acids and 13 membered ring is formed by hydrogen bonding. 195. Answer (4) Both insulin and carboxy peptidase contain Zinc. 196. Answer (3) Cashmilon is polyacrylonitrile. 197. Answer (4)
COOH Proline is heterocyclic amino acid. H – N
H
α amino acid have a primary amino group except proline. 198. Answer (3) Acrilan is PAN (Poly acrybonitrile). 199. Answer (4)
H N
O
Caprolactum is monomer of Nylon-6 or Perlon Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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200. Answer (1)
OH Bakelite is made up of
OH
(phenol) and HCHO (formaldehyde)
–
H2C
OH
CH2
–
+ HCHO CH2
–
201. Answer (1) Methoxy group is attached with carbonyl oxygen. 202. Answer (2) There is a hydroxyl group and an ether attached to the same carbon forming a hemiacetal. 203. Answer (1) OH group present on right side
H
OH CH2OH
204. Answer (1) Fat – 130 ATP, carbohydrate – 38 ATP, protein – 5 ATP. 205. Answer (1)
CH2OH O H α H OH
H OH OH H
α – D – glucopyranose
OH
206. Answer (2)
CHO CHOH
CH = NNHC6H5 C6H5NNH2 –H2O
CHOH
C6H5NNH2 –C6H5NH2, – NH3
CH = NNHC6H5 C=O
(CHOH)3
(CHOH)3
(CHOH)3
CH2OH
CH2OH
CH2OH
C6H5NNH2 –H2O
CH = NNHC6H5 C = NNHC6H5 (CHOH)3 CH2OH glucosozone
207. Answer (2) Only phospholipids form bilayer memberane. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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208. Answer (2) +
C12H12O11 + H2O sucrose (α )D = +66.5°
H invertase
C6H12O6 C6H12O6 D(+)-glucose + D(–)-fructose (α )D = 52.7 (α )D = –92.4° invert (α )D = –19.9°
209. Answer (3) C12H22O11 + 18[O]
conc. HNO3
COOH + 5H2O COOH oxalic acid
210. Answer (2) The excess glucose present in the blood is transported in the tissue where insulin converts it into glycogen. The deficiency of insulin causes diabetis. 211. Answer (2) Proteins give ninhydrin and molisch test. 212. Answer (3) +
C12H22O11 + H2O (α )D = +66.5°
H invertase
C6H12O6 + C6H12O6 D(+)-glucose D(–)-fructose (α )D = 52.7° (α )D = –92.4° invert sugar (α )D = –19.9°
213. Answer (1) Glucose and fructose are functional isomers. 214. Answer (1) Glucose and fructose have similar configuration on C3, C4 and C5 so they form same osazone. In osazone only C1 and C2 are involved. 215. Answer (4) α-D-glucose and β–D-glucose are anomers. 216. Answer (3) O
H
–C–N–
peptide bond of protein form H–bond.
217. Answer (2) Stearic acid, Palmatic acid and Oleic acid are higher fatty acids. Their Na salts are used as soaps. 218. Answer (2) Chemical name of vitamin E is tocopherol. 219. Answer (3) Purines contain two rings in which each ring contains 2 nitrogen atoms. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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220. Answer (4) Denaturation of protein converts quaternary, tertiary and secondary structure to primary structure. 221. Answer (2) Invertase for carbohydrates, urease for urea and trypsin for protein. 222. Answer (4) ⊕
pH = 1 (acidic). It accepts a proton and exist as cation – NH3 is acidic group. 223. Answer (3) Amide linkage (– CONH2) 224. Answer (4)
CH2OH H 5 Chiral carbon present in β–D–glucose
* HO
O
* H OH *
H *
H
OH
OH * H
225. Answer (3)
CN H–C=O
HCN
+
H – C – OH
(CHOH)4 CH2OH
(CHOH)4 CH2OH
H3O
COOH (CHOH)5 CH2OH HI COOH (CH2)5 CH3 Heptanoic acid
226. Answer (3)
CH2OH C= O CH2OH No chiral carbon ∴ No stereoisomers. 227. Answer (2) Ruff degradation. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Section - B : Multiple Choice Questions 1.
Answer (1, 2) In cis form same groups are present on the same side of double bond.
2.
Answer (1, 2, 3, 4) +
+
and
In
aromaticity is maintained in canonical forms, while benzyl carbocation looses its
aromaticity in canonical forms. +
(CH3)3C is more stable than benzyl carbocation due to 9 hyperconjugative bonds. +
CH2
is more stable than benzyl carbocation due to bent p-atomic orbital.
3.
Answer (2, 3) If one + I and one – I effect group is present on both doubly bonded carbon atom then the dipole moment of trans form is more than cis form.
4.
Answer (1, 2, 4) Beleistein test is not given by those compound in which F atom is present. Urea and thiourea also give Beleistein test.
5.
Answer (1, 3)
COOH
COOH
COOH OH is more acidic than HCOOH due to ortho effect while
and
are less acidic
CH3 than HCOOH. 6.
Answer (1, 2)
O O H H
C C C
–
O
H ,
OH C
HOOC
C C
C
–
O
(stability)
H
O Conjugate base of maleic acid
Conjugate base of fumaric acid
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7.
Success Magnet (Solutions)
Answer (1, 3) SbF5
F
⊕
F ⊕ SbF5
8.
Aromatic systems
Answer (1, 2) Triple bond is more prone to hydrogenation.
9.
Answer (1, 2, 4) Cyclooctatetraene has no planar structure.
10. Answer (1, 2, 3, 4) C4H10O can form 4 alcohols & 3 ethers which can exhibit chain, position, functional & metamerism. 11. Answer (2, 4) (2) and (4) contain chiral carbon so are optically active. 12. Answer (1, 2, 3) A, B, C, D are isomers of same compounds where (A) and (B) make, enantoimeric pair and (C), (D) make another enantiomeric pair. 13. Answer (1, 2)
COOH H2N
H
2 3
H
COOH
S H
R
OH S
2 3
H
OH
CH3
R
OH
COOH
14. Answer (2, 3, 4) (1) molecule contains 3 (1° H), 8 (2° hydrogens) and 3 (3° hydrogens). 15. Answer (1, 3, 4)
O
O has no acidic hydrogen to participate in enol formation.
16. Answer (1, 2, 3) Bromination takes place at most stable free radical. 17. Answer (1, 2, 3)
ROH + CH3MgI → CH4 + Mg 0.37 x x = 74
112 cc 22400
I O
R
H CH3 – C – CH2OH CH3
Δ +
H
CH3 – C = CH2
O3
CH3 – C = O + HCHO
CH3
CH3
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18. Answer (1, 3) A → Presence of benzylic H B → Absence of benzylic H 19. Answer (1, 3)
CH3
CH3
C5H12 + Cl2 → H3C – C – CH2Cl
Na Wurtz reaction
CH3
H3C – C – CH2 –CH2 – C – CH3
CH3 (B)
CH3
CH3 (C)
20. Answer (1, 2)
O O
C
H
CHO
≡ H O O O3
C
CHO
O
O
H
H O
H O
H
O
O
21. Answer (1, 2, 3) Wurtz reaction is a type of free radical reaction. dry ether C2H5 Cl + 2Na + Cl C2H5 ⎯⎯ ⎯⎯→ C 4H10 + 2NaCl
butane
Ethyl free radical can disproportionate to give ethane and ethene. •
2 C 2 H5 ⎯⎯→ C 2H4 + C 2H6 Ethene
Ethane
22. Answer (1, 2, 4)
CH3 On applying selectivity–reactivity principle, only CH3 – CH – CH – CH3 has about 35% in the mixture. Others
Cl
have less than 35% in the mixture. 23. Answer (1, 3, 4)
2-methyl 2-butene is more stable than but-2-ene due to presence of 9 hyperconjugative bonds, while but-2-ene contains 6 hyperconjugative bonds. Buta-1, 3-diene and 2-methyl buta-1, 3-diene are more stable than but-2-ene due to conjugation. 24. Answer (1, 2, 4) In all molecule, one H2 molecule is added but only (1), (2) and (4) can give symmetrical diketone on reductive ozonolysis, while (3) give only ketone. 25. Answer (1, 3, 4) In presence of 1% alkaline KMnO4, C6H5CO3H and OsO4/OH–, syn addition occurs while in presence of HCO3H, anti addition occurs. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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26. Answer (1, 3) A by E2 elimination, B by E1 elimination. Alc. KOH will undergo dehydrohalogenation by anti elimination. 27. Answer (1, 4) NBS → characteristic reagent for allylic bromination Br → ortho para directing 28. Answer (1, 2)
Br NBS
(A) 29. Answer (2, 3) Due to presence of unsaturation. 30. Answer (1, 2, 4)
Cl
⊕ anhyd AlCl3
hydride shift
⊕
more stable carbocation
31. Answer (2, 3) → will give only 1 mono halo substituted derivative.
→ will give 3 mono halo substituted derivative. 32. Answer (1, 3) SO2Cl2 and Cl2 replaces benzylic H. 33. Answer (2, 3) C3H7Br + AgNO2 → C3H7NO 2 + C3H7 ONO major
minor product
due to ambident nucleophilic nature of nitro group. 34. Answer (1, 3)
CH2
CH3 LiAlH4
H3C
Br
CH3 (A) CH3
CH3 Br H2C
LiAlH4
H3C
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35. Answer (1, 2, 3, 4) All are feasible reaction (1), (2) substitution as well as elimination (3) Ring expansion (4) Ring contraction 36. Answer (1, 2, 3, 4) 1-4 addition dominates at high temperature and with excess of HBr both double bond give addition product. 1-2 addition dominates at low temperature 37. Answer (1, 2, 4)
Br
Br NO2
HNO3 Δ
Br
Br
Br
Br
Highest melting point 1, 3, 5 tribromobenzne. 38. Answer (2, 4) (2) R − Br + Ag 2O → R − O − R + AgBr dry
H2SO4 ⎯⎯→ CH3 – CH2 – O – CH2 – CH3 (4) CH3CH2OH ⎯⎯ 413 K
39. Answer (1, 2, 3) In (1), (2) and (3) options, stable carbocation is formed as an intermediate, so, these give SN1 mechanism, while CH3Cl mainly gives SN2 mechanism because it is 1° alkyl halide and produce less stable carbocation. 40. Answer (1, 3) (1) and (3) on E2 elimination give an alkene but (2) and (4) can’t give an alkene 41. Answer (2, 3) 3° alkyl halide mainly gives SN1 mechanism. Polar protic solvent favours SN1 mechanism while polar aprotic solvent favours SN2 mechanism. 42. Answer (1, 3, 4) In (2) option, alkene is formed as major product while in other options, ether is formed as major product. 43. Answer (3, 4) Pyridine CH3CH2CH2OH + SOCl2 ⎯⎯ ⎯⎯→ CH3CH2CH2Cl + SO2↑ + HCl↑
CH3CH2CH2OH + PCl5 ⎯⎯→ CH3CH2CH2Cl + POCl3 + HCl 44. Answer (1, 2, 3) (1) Yellow ppt of AgI (2), (3) Iodoform CHI3 formation 45. Answer (1, 3) (1) Friedel Craft alkylation (3) Friedel Craft decarbonylation Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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46. Answer (1, 2) +
H H2O
+
OH major
minor product
47. Answer (3, 4) (2) RNH2 + CHCl3 + alc.KOH → RNC + KCl + H2O
NC
NH2 + CHCl3 + alc.KOH →
(4)
+ KCl + H2O
48. Answer (1, 2, 4) 1, 2, 4 gives Hoffmann product (1) Hindered base (2) EICB (stable conjugate base) (4) Hoffmann elimination 49. Answer (2, 3)
O
H
C –– N
H
is o, p directing due to – N – group.
50. Answer (1, 3) Only 1° amine gives carbyl amine test. 51. Answer (1, 2) cis and trans isomers
Ph H
C=C
H Ph & Ph H
C=C
Ph by elimination. H
52. Answer (1, 2, 3)
CH2 – Br 1
2 + CH2 –
–Br
H5C2
4
1 CH2 – OH
CH2
HOH + –H
⊕
⊕
3 H5C2
CH2
H5C2
H5C2 (2)
H5C2 +
–H
HOH
+
–H
CH2
HOH
CH2 OH
H5C2 OH (1)
H5C2 (3)
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53. Answer (2, 3, 4)
Br C2H5ONa
54. Answer (2, 3, 4) HI/P
OH 55. Answer (1, 2) Water favours the formation of a polar compound. 56. Answer (1, 4)
OH Secondary alcohol containing
CH3 linkage can give all given three reactions.
CH
57. Answer (2, 4) 16
O
H
+
+
16
O
+
16
+
OH
H 18
OH
18
H2O
58. Answer (1, 2, 3, 4) In (1), (2) and (3) intramolecular hydrogen bonding is present. 59. Answer (1, 2, 3) OH
Iodoform test is given by alcohols having the group R – CH – CH3 . Thus, all the three except (4) give this test. 60. Answer (1, 3)
(i) NaNH2 H2 CH3 – CH2 – C ≡ CH CH3 – CH2 – C ≡ C – C2H5 Pd/BaSO (ii) C2H5Br 4 (X)
H5C2
C2H5 C =C
H
H cis alkene (Y)
KMnO4 alk.
C2H5 H
OH
H
OH C2H5
meso compound optically inactive (Z) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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61. Answer (1, 3) CH3COCl and (CH3CO)2O can be used. 62. Answer (3, 4) Both (3) and (4) are stable due to intramolecular hydrogen bond formation
Cl
H
O
F
H
O
Cl – C —— CH and F – C —— CH Cl
H
O
F
H
O
No hydrogen bonding occurs with Br and I due to large size and lesser electronegativity. 63. Answer (1, 3)
C6H5CHCH3 and CH3CH – CH3 gives iodoform test due to presence of –CH3 group adjacent to –OH group.
OH
OH 64. Answer (1, 3)
In option (1) and (3), most stable carbocation is formed as an intermediate. 65. Answer (1, 2)
OH
OH
OH
NO2
NO2 and
NO2
O2N can’t give effervescences with NaHCO3 but
NO2
can
NO2
give effervescences with NaHCO3 due to more acidic nature. 66. Answer (2, 3)
O
O
HO – C – CH3 and H2N – C – CH3 can’t give iodoform on warming with NaOH and iodine due to resonance. 67. Answer (1, 2, 4) Pyroligneous acid cotnains 9 – 10% CH3COOH, 2– 4% CH3OH, 1– 2% acetone and rest is water. 68. Answer (1, 2) In esterification, H+ of alcohol reacts with OH– of carboxylic acid to form H2O and the reactivity of carboxylic acid is 1°>2°>3°. Other statements are correct. 69. Answer (1, 2, 4)
COOH
OH NO2 ,
NO2
,
SO3H are stronger acid than H2CO3 acid.
NO2 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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70. Answer (2, 4)
O CH = CH – CH – CH3
H H
K2Cr2O7
OH D
O
+ HO – C – C – CH3
+
H
O
D
OH
O
C – OH
H
SOCl2
O O
O
C – Cl
H
+ Cl – C – C – CH3 (Y)
O
D (X) 71. Answer (1, 2) +
OCH3
O
H H2O
+ CH3OH OH
HO OH
–H2O
O OH
H
72. Answer (1, 3) (3) does not contain α hydrogen and (1) contains vinylic hydrogen which cannot be easily removed. 73. Answer (3, 4) –
O
COO OH
–
OH conc.
O Cl CCl3
aq. KOH
aq. KOH
COOC2H5 Br
74. Answer (1, 2, 3, 4) R
CHO
AgNO3/NH4OH
R
–
COO
O H
C
OH
AgNO3/NH4OH
NHOH
HCO 3
–
NO
AgNO3/NH4OH
(R can be CH3 – C6H5) Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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75. Answer (1, 3, 4)
CH
CH2
OH
B2H6
–
H2O2/OH
–
H2O2/OH
CH2MgBr
CH2
CH2
OMgBr
HCHO
MgBr +
O
76. Answer (1, 2, 3, 4)
O O
Phenol
Phenolpthaline
O Resorcinol Catachol Nitration
Fluorisien Alizarin 2,4 – D.N.P.
77. Answer (2, 3) OMDM → gives Markownikov’s HBO → gives Anti-Markownikov’s product No, carbocation formation ∴ No, rearrangement takes place 78. Answer (1, 2, 3)
CH3
CH3 H
(1) H3C – CH – CH2 – CHO
NaOH
H
CH3 – C — C — C – CHO OH C – H
H
H3C CH 3 aldol
CH3
H
CH3
OH
+
(2) H3C – CH – CH2 – C = O + CH3MgBr
H3O
CH3 – CH – CH2 – C – CH3 H
(3) 3-pentanone and 3 methyl butanol have same molecular formula (4) On Wolff Kishner reduction it gives 2 methyl butane. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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79. Answer (1, 2, 3)
O
OH
(1) PhCH2MgBr + CH3 – C – H
CH3 – C – CH2 – Ph H OH
O +
(2) CH2 – CH – CH3 + PhLi
H3O
PhCH2 – C – CH3 H
H
O
OH +
(3) Ph – C – C – H + CH3MgBr
H3O
PhCH2 – C – CH3 H
H 80. Answer (1, 2, 4) CH3 (1)
(2)
CH3 (4)
CH3 81. Answer (2, 3, 4) (2) NaBH4 reduces carbonyl to 2° alcohol. (4) Silver mirror test given by aldehydes. 82. Answer (1, 2)
N2Cl
CN
COOH +
H3O
Cu/KCN
(X)
NaOH CaO
(Y)
(Z)
83. Answer (1, 3)
OH
O
(1) H – C – CH3
C – CH3 (O) KMnO4
COCH3 (3)
+ CH3COCl
Anhy AlCl3
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84. Answer (2, 3, 4) CH3CHO is less reactive than HCHO but more reactive than ketones. 85. Answer (2, 3, 4) Ethyl alcohol is less acidic than phenol. 86. Answer (1, 3) Sulphur and quinoline
N 87. Answer (1, 2, 3) In (4) 2° alcohol is formed. 88. Answer (1, 2, 3) NaHSO3 is used to separate ketone and aliphatic methyl ketone as well as aldehyde and aliphatic methyl ketone. 89. Answer (1, 2, 3, 4) In (3) – COOH group is reduced in preference to
C = O group.
90. Answer (1, 2, 3) Active methylene compounds are used in their reaction. 91. Answer (1, 2, 3, 4) (1) Cyanohydrin formation (2) Aldehydes give silver mirror (3) Cross aldol (4) Condensation with ammonia derivative 92. Answer (1, 2, 3, 4) (1) C2H5COOH (2) CH3CH2COOH (3) CH3 − CH2 − CH2OH ⎯⎯→ CH3 CH2COOH (A)
(B )
(4) CH3CH2COOH 93. Answer (1, 2, 3)
CH3 Cl CH3
CH3
NH2
KNH2
CH3
Δ
Cl CH3
CH3
CH3
KNH2/Δ
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94. Answer (1, 4)
NO2
NO Fe/H2O(v)
NH2 strong acid electrolysis
OH 95. Answer (1, 4)
R
NH2
HNO2
R
OH
But exceptionally methylamine forms ether. 96. Answer (1, 2, 3, 4) In all given compounds lone pair of N is resonance stabilized. 97. Answer (1, 2, 4) 3° amine has no H atom on N. 98. Answer (1, 3) Nylon66 =
[ NH – (CH2)6 – NH – CO – (CH2)4CO ]n
CH3 PMMA = — CH2 – C –
COOCH3
n
99. Answer (1, 2, 3, 4) 100. Answer (1, 2, 3) Fact 101. Answer (2, 4) Fact 102. Answer (1, 2, 4) 103. Answer (1, 3, 4 ) 104. Answer (2, 3, 4) Sucrose is a disaccaharides of glucose and fructose. 105. Answer (1, 3, 4) Nylon-66 is a copolymer monomers are hexamethylenediamine and adipic acid. 106. Answer (1, 2, 4) Natural silk is a polyamide. 107. Answer (1, 2, 4) +2
CH ≡ CH + HCN
Ba
polymerisation
(PAN) — CH2 – CH — CN
n
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108. Answer (3, 4) Fructose is equilibrium with glucose in alkaline solution and hence gives positive test of aldehydic group. 109. Answer (2, 4) At pH > 7, H+ from carboxylic acid will be lost. 110. Answer (1, 2, 3) Glycine is only amino acid which is optically inactive, since then is no asymmetric centre. 111. Answer (1, 2, 3, 4) 112. Answer (2, 3, 4) Guanine is purine. 113. Answer (3, 4) Remini and Schyrver are tests for formaldehyde. 114. Answer (3, 4) 115. Answer (1, 2, 3, 4)
Section - C : Linked Comprehension C1. 1. Answer (4) 3
sp hybridised H H
So non planar (so not aromatic). 3
sp hybridised (I) 4 π electrons (so not aromatic). 2. Answer (4)
N
N
H
H +
E
H E N H +
E N
N
H
H
H E
H E
+
+ E N
N
H
H
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Organic Chemistry
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3. Answer (2)
O δ– δ+
O– +
⇒
anti aromatic
C2. 1. Answer (4) + 1 group stabilises free radical. 2. Answer (2) Allyl carbocation is most stable. 3. Answer (1) Stability of carbanion increases with the presence of electron withdrawing group, whereas decreases with the presence of electron releasing group. C3. 1. Answer (1)
O
H
OH
H
Aromatic, hence very stable. 2. Answer (2)
O
In (1), (3) & (4) the group attached to CO, involves,
O
O
O
C
in resonance.
O
Cl
Cl
3. Answer (4) H O
O
O
O
stabilised due to intramolecular hydrogen bonding. C4. 1. Answer (2) Electron releasing group decreases acidic nature hence highest pKa. 2. Answer (2) In aqueous solution 2º amine is more basic than 3º & then comes 1º. 3. Answer (4) For iodine + R is least. C5. 1. Answer (1) μ = μ12 + μ 2 2 + 2 cos θ Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Organic Chemistry
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2. Answer (2) Symmetrical molecules have zero dipole moment. 3. Answer (1) CCl4 has zero dipole moment. As the number of chloride atoms increase dipole moment decreases. C6. 1. Answer (1)
CH2
2 1 no. of hyperconjugative structures = 3 Alkyl group make the ring electron rich by their tendency to make ring electron rich by hyperconjugation. 2. Answer (2) (3) is most stabilised by higher + 1 effect of isopropyl group. 3. Answer (3) CH = CH2
+
methyl
H
shift C7. 1. Answer (4)
[O]
COOH
O3, H2O2
COOH
2. Answer (3)
CH3 CH2 = CH2
+
CH3MgI
MCPBA
H3O
O OH
OMgI (A)
(B)
3. Answer (4)
O + HCHO
CH3CHO +
C8. 1. Answer (1) Diel’s alder reaction involves (4 + 2) cycloaddition. 2. Answer (2)
H H H
H
C=C–C=C B
H
+
H
H
H H
–
Br
–
Br
Br
Br Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Organic Chemistry
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3. Answer (3)
O3/H2O
+ B
COOH COOH
E
C9. 1. Answer (1) a is most substituted double bond, so most stable.
2. Answer (3) The structure has –OH group on most substituted carbons of the double bonds because the water attacks the carbocation that forms. The most stable carbocation is the one on the most substituted carbon. 3. Answer (1) All of the double bonds have two substituents. If on the one end of the bond both the substituents are same. Consequently there is no possibility for geometric isomers. C10. 1. Answer (4) Alkenes are more reactive towards electrophilic addition reactions but when product formed is conjugated diene alkynes give this reaction first. 2. Answer (1) Alkenes are more reactive for electrophilic addition reaction. 3. Answer (1) Alkenes are more reactive for hydrogenation. C11. 1. Answer (4) Cis alkene + Anti addition ⇒ Racemic mixture. 2. Answer (4) Product produced has no chance of having plane of symmetry.
H CH3
OH CH3
3. Answer (4) OsO4
⇒ Syn hydroxylation
H2
→ Syn addition
KMnO4
→ Syn hydroxylation
Br2/CCl4
→ Anti addition
C12. 1. Answer (2)
Cl Cl
Cl Cl (d) + (l)
(d) + (l)
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2. Answer (2)
Cl
probability = 1 × 3 = 3
probability = 2 × 3.8 = 7.6
Cl
probability = 1 × 5 = 5
Cl
Cl probability = 6 × 1 = 6 3. Answer (3) Bromination takes place at most stable free radical and free radical at bridge head carbon is least stable. C13. 1. Answer (3) Alkoxide ion is a better base and so will favour E2. 2. Answer (3) Same. 3. Answer (2) Vinyl halide and aryl halide have partial double bond character between carbon and halogen. in aniline, NH2 is a strong base and hence very weak leaving group, so cannot undergo SN reaction C14. 1. Answer (4) OCH 3 is a better base.
2. Answer (1) SN1 3. Answer (3) E1 C15. 1. Answer (3)
W t. of used alkali (w) × (M + 42 n) = 56 n t W . of acetyl derivatives (w) M → molecular mass of alcohol. 2. Answer (1) Secondary alcohol (glycol) gives formic acid. 3. Answer (1) Two 1° alcohol gives formaldehyde and C = O changes to CO2. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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OH
ONa (i) CO2
NaOH
C16.
OH
OH COOH NaOH CaO
4 - 7 atm P +
(ii) H
A
B
C Br2
(CH3CO)2O
OH
O – COCH3 COOH
Br
E OH
Br
Br
D
NaOH Δ COONa + CH3COOH G H
1. Answer (2) 2. Answer (1) 3. Answer (3) C17. 1. Answer (2)
OH Br
Mg ether
CH3CHO
CHCH3
MgBr +
H Br
Ring expansion Br
CH3
CH–CH3
2. Answer (2) OH
C 3H 7 C H 3C
Ph
There would be a racemic mixture since carbonyl group is planar and can attack from both side. 3. Answer (3) Alkyl group (carbanion) attack on carbon atom of carbonyl group. C18. 1. Answer (2)
O–H H-bond
Compound B is C7H6O3
C=O OH Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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2. Answer (1)
OCOCH3 CH3COCl
COOH
+
H
OH
(C) (aspirin) COOH
OH PhOH
COOPh
+
H
(D) (salol) 3. Answer (2) Compound D is salol. C19. 1. Answer (3)
OH
O – CH3
B
A = CH3
2. Answer (3)
O – COCH3 COOH , Aspirin used as antipyretic.
Y 3. Answer (1)
OH
O – CH3 C7H8O
& CH3 C20. 1. Answer (1)
O CH3 Carbanion becomes aromatic. 2. Answer (1)
O H2C
O H
H2C
H
Carbanion stabilize by resonance. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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3. Answer (1)
H COONa
H3C – C – OH NaOI
+
CHI3
C21. 1. Answer (3)
NH2
NH2
NH2
HNO3 +
H2SO4
NO2 NO2 P (51%)
M (47%)
Metal is formed due to formation of anilinium ion as intermediate because of basic nature of aniline. 2. Answer (2) The carbonyl group in electron withdrawing making the amide less basic than the amine. 3. Answer (2) The substitution is para, so the amide must be ortho-para directing. The best explanation for the lack of ortho substitution is steric hinderance. C22. 1. Answer (3)
Δ
Ag2O
+ NMe3 + H2O
—
OH
NMe3I—
N (CH3)3
2. Answer (2)
Me 3CH3I CH2NH2
CH2 – N
I–
Me Me
Ag2O
+ Me3N + H2O 3. Answer (4)
–
N – Me OH nBu
Δ
CH2 = CH2
Et
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C23. 1. Answer (1) Due to ortho effect. 2. Answer (2) Conjugate base stabilise more than other site. 3. Answer (2) Conjugate pair stabilise by resonance. C24. 1. Answer (3)
O CH2NH2
A
B
due to ring expansion.
2. Answer (2) Phenyl group migrated over to N atom. 3. Answer (2) R – N = C = O is common in Schmidt and Hoffmann reaction. C25. 1. Answer (3) The lone pair of electrons on the nitrogen of the amine attacks the electrophillic carbon of acid. 2. Answer (1) H–bonding in Nylon. 3. Answer (4) It is prepared by condensation of adipic acid and hexamethylene diamine. C26. 1. Answer (1)
O H H bonding
–C–N– 2. Answer (2) Teflon is polymer of CF2 = CF2. 3. Answer (3) Polyacrylonitrile
H
H
C=C H
CN
H
H
( C – C )n H
CN
C27. 1. Answer (2) pH =
3.30 + 8.70 =6 2
2. Answer (4) Amino acids show lowest solubility at isoelectric point since there is highest concentration of the dipolar ion. 3. Answer (4) Since it is basic amino acid with 2–NH2 group, So pH must be greater than 7. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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C28. 1. Answer (2) Shortening of carbon chain of one carbon. 2. Answer (3) Name of reaction → Ruff degradation. 3. Answer (4) (A) →
(B) → Pentyacetate
C = NOH ,
(C) → Cyanide, C29.
Starch (C6H10O5)n
(D) → Aldopentose
H 2O
C12H22O11
H 2O
(C)
C6H12O6 (D)
conc. H2SO4
(E) Black
200 – 250°C
(C6H10O6)n (A) Dextrin 1. Answer (2) 2. Answer (1) 3. Answer (1) C30. 1. Answer (3) Elimination dominates over substitution. 2. Answer (4) A → C3H6 B → CH3 – CH – CH3 Br
C → CH3 – CH – CH3
OH 3. Answer (3) Substitution product
Cl
OH
–
OH
C31. 1. Answer (1) OH
R–C–R
O (O)
R–C–R
OH 2. Answer (4) All of these contain Cr+6
PCC N 3. Answer (3) nfactor = 6
CrO3 Cl
H
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C32. 1. Answer (3)
O
OH +
C2H5 – C – C2H5 + C2H5MgX
H3O
C2H5 – C – C2H5
H Δ –H2O
C2H5 – C = C – CH3
C2H5
C2H5 O C2H5 – C = O + CH3 – C – H C2H5
2. Answer (3) Less hindered more reactive 3. Answer (3) C=O sp
C
2
sp
3
C33. 1. Answer (3)
CH3
CH3
CH3 SO3H
H2SO4
CH3 SO2Cl
PCl5
SO2NH2
NH3
CH3 COOH H2SO4
SO2NH2 SO3H (‘P’)
Δ
CO — NH SO2 2. Answer (1)
CH3
CH3
CH3
PCl5
SO3H
NH3
SO2Cl
CH3 (i) ClOH (ii) NaOCl
Cl +
SO2NH2
SO2NNa
3. Answer (4)
CH3
CH3
CH3 PCl5
C6H5OH
CH3
OH
18
H2O
+ 18
SO3H
SO2Cl
SO2 – OC6H5 ester
SO2OH
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C34. 1. Answer (2) Rearrangement of carbocation. 2. Answer (3) Intermediate 2° carbocation of A Intermediate 1° carbocation of B 3. Answer (3)
OH
OH HNO2
OH
ring expansion
H –N2
C35. 1. Answer (2) O
O
CH2
+
CH2NH2
OH
O
CH2 – C – CH3
CH2 – C – H
&
∴ 4 product 2 carbanion 2. Answer (4) Carbanion stabilize by resonance. 3. Answer (4) (1) & (2) option no α – H atom (3) option give chloroform reaction.
Section - D : Assertion - Reason Type 1.
Answer (1) That acid is easily decarboxylate which produce most stable carbanion.
2.
Answer (1) Due to –I effect producing group.
3.
Answer (1)
⎤ ⎡ ⊕ + ⎢CH − O − C H ↔ CH − O = CH ⎥ 2 3 2⎥ ⎢ 3 Minor •• ⎢⎣ Major ⎦⎥ Stabilize by resonance, the second structure has octets on all atoms and an additional bond. 4.
Answer (3)
HONO2 + H2SO4 base 5.
acid
H
Answer (2)
HOX + H (acid)
6.
⊕ – H – O – NO2 + HSO4
+
H–O—X H
H2 O + X electrophile
Answer (2) The bond energy of allylic carbon hydrogen bond in propene is less than the bond energy of benzylic carbonhydrogen bond in toluene.
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7.
Success Magnet (Solutions)
Answer (3) Hyperconjugation increases stability of carbocation and free radical, not carbanion.
8.
Answer (1) N does not have vacant d-orbital for expansion of octet. Maximum co-valency of N is 4.
9.
Answer (3) Tartaric acid isomers = 3. d, l and meso form.
10. Answer (2)
O
OH +
–
H2O/H or OH
(1-5 migration of H atom) .
NH2
NH2
11. Answer (3) 1 is R, II → S. 12. Answer (1)
H-bonding Cl
H
Cl – C
O C–H
Cl
O H
13. Answer (2) No. of isomers → 2n − 2
n −1 2 .
14. Answer (3) Double bond generating geometrical isomers and chiral centre give optical isomers. 15. Answer (2) Unimolecular
elimination.
16. Answer (4) A → CHCl3 is more acidic than CHF3 because Cl3C is less basic than F 3C because fluorine can disperse charge only by an inductive effect while Cl disperse charge by inductive effect as well as pπ – pπ bonding delocalisation. 17. Answer (1) Due to steric crowding electrons pair does not undergo resonance. 18. Answer (4) Triplet carbene is more stable than singlet carbene. 19. Answer (1) Angle strain of cyclopentane =
109.5° − 108 = 0.75 º (negligible). 2
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20. Answer (1)
CH3 CH3
CH3 – C ⊕
CH3 – C
O – CH3
O – CH3 ⊕
(resonance)
21. Answer (2) Vinyl chloride does not form vinyl carbocation with anhy. AlCl3. 22. Answer (3) Pyridine is less reactive than benzene towards EAS creating at the position due to - I effect of the N-atom while in pyrrole the non-bonding pair on nitrogen is part of aromatic rextet. 23. Answer (1) Number of hyperconjugative structure α stability. 24. Answer (3) Acidic hydrogen is present in 1-alkynes but not in alkyne-2. 25. Answer (2) Due to (1 – 4) position of H which causes hinderance in boat form. 26. Answer (1)
H-bonding
O–H H
O–H
H
H H
27. Answer (4) Ethylene is more reactive than acetylene towards electrophilic addition reaction. 28. Answer (4) Rate of nitration of C6H6 = rate of nitration of C6D6. 29. Answer (1) +
Br Intermediate
ion
C–C bromonium ion
30. Answer (2) Correct R → Mesotartaric acid has molecular symmetry. 31. Answer (4) Hg
+2
CH3 – C ≡ C – H
Hg
H3C – C ==== C – H
Hg
+2
CH3 – C == C – H O
O H
+
H
H
–Hg
CH3 – C == CH2 OH
H
(enol)
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32. Answer (2) (II) Does not react with NH3 since double bond is not polarised by an electron withdrawing group.
CH3 H (I) Form product
O
+
(III) Via H3N – C – C – C == C
CH3 H
intermediate.
CH3
33. Answer (3) Alkyne gives carbonyl compounds R–C≡ C–H
(i) H2O(BH3)2 THF (ii) H2O2/OH¯
R – CH2 – C – H O
34. Answer (4) It involves formations of vinyl carbocation, which is unstable ∴ reaction does not takes place. 35. Answer (1)
OH ring expansion +
+
–H
–H
36. Answer (3) Order of boiling point Straight chain > branched chain of isomeric hydrocarbon. 37. Answer (1) Removal of Cl is easy due to presence of NO2 group. 38. Answer (4) Anti-Markownikov product. 39. Answer (4) Chloroform is heavier than water. 40. Answer (1) Racemic mixture is obtained due to walden inversion. 41. Answer (4) C2H5Br + AgCN → C2H5CN (major product) 42. Answer (2) Aniline behaves as Lewis base for anhydrous AlCl3. 43. Answer (3) Addition of HBr to 2-pentene give meso product. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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44. Answer (3) 2° carbocation converts to 3° carbocation. 45. Answer (1) HCN is weak acid where as HI is strong. 46. Answer (3) Presence of electron withdrawing group at ortho position of Aryl halide increases nucleophilic substitution. 47. Answer (3) Correct reason :- In aryl halides electron density at the ring decreases due to electron withdrawing effect of halogen atom. 48. Answer (1) +
R—O — R + H
R — OH — R
49. Answer (1) Phenoxide ion stabilized by resonance. 50. Answer (2) Correct reason :- Formation of stable intermediate benzyl carbocation. 51. Answer (3) Correct reason :- Intramolecular H-bonding 52. Answer (1)
OH
Benzene (aromatic)
H 53. Answer (2) Correct reason :- OH group of salicyladehyde is less reactive due to presence of intramolecular H-bonding. 54. Answer (3) Correct reason :- Rearrangement of carbocation form stable saytzeff product. 55. Answer (2) Intramolecular aldol reaction. 56. Answer (1) +I effect of –CH3 group decrease dipole moment in ketone. 57. Answer (4) It forms 1 nitro 2 propanol. 58. Answer (1) Cl is good leaving group than –NH2 group. 59. Answer (4) p-chloro benzoic is more acidic than p-fluoro benzoic acid due to more effective pπ-pπ bonding. 60. Answer (4) Guanidine is more basic due to stability of its conjugate pair. 61. Answer (2) In strongly acidic medium protonation of hydoxyl amine takes place. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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62. Answer (4) Chloro ethanoic acid has lowest value of pKa and stronger acid. 63. Answer (3) Cross Cannizzaro is redox reaction. 64. Answer (2) Cl is good leaving group. 65. Answer (4) +I effect of methyl group. 66. Answer (3) It has 3α H atom. 67. Answer (1) HCHO has 2H for hydride shift and benzaldehyde is resonance stabilised. 68. Answer (4) NaBH4 reduces only carbonyl group, not carbon-carbon multiple bond. 69. Answer (3) OH– abstracts the α hydrogen from CH3NO2. 70. Answer (4) 2, 2 dimethyl propanoic acid has no α hydrogen ∴ does not gives HVZ reaction. 71. Answer (1) Zn-Hg/conc. HCl form cyclo alkene. 72. Answer (1) Umbrella effect. 73. Answer (3) Pyrrole is a weak base. It is aromatic because the non-bonding electrons on nitrogen are located in a p-orbital, where they contribute to aromatic sextet. 74. Answer (2) Due to hindrance. 75. Answer (1) +I effect of alkyl group increase electron density of N-atom. 76. Answer (4) Gabriel phthalimide is used for preparation of 1° amine in pure state. 77. Answer (2) The intermediate is stabilized by delocalization of negative charge on to the electronegative ion. 78. Answer (2)
NH C NH2
NH2
H2O
NH2
(is stabilised by resonance)
C H2N
NH2
79. Answer (4) In alkaline medium with Zn–NaOH, nitrobenzene gives hydroazobenzene. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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80. Answer (1) C–Cl bond in chlorobenzene is resonance stabilised, it has partial double bond character. 81. Answer (2) Amides are less basic than primary amines because lone pair of electrons on nitrogen atom in amides is delocalised. O
O–
R – C – NH2
+ R – C = NH2
82. Answer (4) Carbylamine reaction is applicable to primary amines. 83. Answer (1)
+ N
N
similarly on the three rings.
84. Answer (1) In acidic medium, aniline is converted to anilinium ion which does not couple. 85. Answer (1) Carboxypeptidase is an exopeptidase as it breaks the peptide chain at terminal ends. Carboxy peptidase cleaves carboxy - terminal amino acids having aromatic or branched aliphatic side chains. 86. Answer (3) Nylon is a polyamide of hexamethylene diamine and adipic acid. 87. Answer (1) Bakelite becomes hard on heating, hardening is due to formation of extensive cross-links between different polymer chains to give a three dimensional network solid. 88. Answer (1) +
Carbocation formed from styrene (C 6H5 − C H − CH3 ) is more stable than that formed from propene. 89. Answer (4) Glycine does not contain a chiral centre. 90. Answer (3) Teflon is fully fluoroninated polymer. 91. Answer (4) Nylon is a monomer of hexamethylene diammine and adipic acid. 92. Answer (3) Polybutadiene is chain growth polymer. 93. Answer (3) Both glucose and fructose reduces Tollen’s reagent. 94. Answer (3) Cn(H2O)n → General formula. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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95. Answer (1) On hydrolysis sucrose give unequal amount of glucose and fructose which changes sign of rotation. 96. Answer (2) Enzymes are generally made up of proteins. 97. Answer (2) Fructose is monosaccharide carbohydrate which do not undergo hydrolysis. 98. Answer (2) Existence of both cationic (NH3+) species and anionic species (COO–). 99. Answer (2) Starch is polymer of glucose. 100. Answer (4) α-D-glucose and β-D-glucose are anomers.
Section - E : Matrix-Match Type 1. Answer - A(p, q, s), B(p, q, r, s), C(p, q, r, s), D(p, q, r, s)
OH can react with Na, NaOH and NaNH2 but not with NaHCO3. While others react with Na, NaOH, NaNH2 and NaHCO3. 2. Answer - A(p, r, s), B(q, r), C(q, r), D(p, r, s) If two (+I) or two (–I) effect groups are present on doubly bonded carbon atom then dipole moment (cis > trans), Melting point (trans > cis) and boiling point (cis > trans). If one (+I) and on (–I) effect group is present on both boubly bonded carbon atom then dipole moment (trans > cis) and melting point (trans > cis) but boiling point (cis > trans). 3. Answer - A(q, r), B(p, q), C(r, q), D(q, r, s)
⊕ → Aromatic and resonance stabilised (4n + 2 = 6)
(A)
(B)
CH3 ⊕ CH3 – C – C = C
CH3 resonance
H +1 group
CH3 CH3 – C = C – C ⊕ H
CH3 CH3
→ Aromatic and resonance stabilised (4n + 2 = 6)
(C)
CH2 N (D)
CH3
O O → Aromatic resonance and –I effect of NO group. 2
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4. Answer - A(p, r), B(p, q), C(s), D(r) Friedel Crafts reaction
CH3Cl
CH3
⊕ CH3
anhyd. AlCl3
Reimer Tiemann reaction
OH
OH CHO
CHCl3, KOH
salicylaldehyde Intermediate → Carbene, an electrophile attack on electron rich ring Aldol condensation
O
O
CH3 – C – H
OH
CH2 – C – H carbanion O
O CH3 – C – H
CH2 – C – H
OH CH3 – CH – CH2 – CHO
Acid catalysed hydration → Carbocation is intermediate 5. Answer - A(r), B(p), C(q), D(q, s) Chlorination in presence of hν is a free radical reaction. Bromination of alkene proceeds through cyclic transition state and undergoes anti addition. Hydration involves carbocationic mechanism, carbocation formed in this case will not undergo rearrangement. Elimination (E1) proceeds through carbocation, which has tendency of rearrangement. 6. Answer - A(q, r, s), B(q), C(p, s), D(s)
H (A)
→ Very acidic hydrogen so reacts with base it is allylic hydrogen which will be subslituted by Cl2/hν. Presence of double bond will make it react with Br2 water.
(B)
(C) (D)
••
Acidic hydrogen H − CH2 − O − H → Undergoes nitration with (HNO3 + H2SO4) as well as free radical addition with Cl2/hν. → free radical substitution with Cl2/hν.
7. Answer - A(s), B(q, r, s), C(p, s), D(q, r, s) (A)
Cis elimination of meso compound produces trans isomer
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(B)
Success Magnet (Solutions)
Elimination is possible from non-stereo centre. So, is a stereospecific reaction and a stereoselective reactions and undergoes antielimination.
H H H
Br
Br
H
H (C)
H
C
C
H
H Carbocation would be formed, reactant is optically active product will show geometrical isomer hence the reaction is stereoselective.
8. Answer - A(p, q, r), B(p, s), C(p, q, r), D(p, q, r) (A)
Dehydration (elimination of H2O) proceeds through carbocationic intermediate which will undergo rearrangement.
(B)
Elimination of HF from fluoroalkanes yields Hoffmann product
(C)
Same as (1)
(D)
Elimination of HCl proceeds through carbocation intermediate.
9. Answer - A(p, q, s), B(p, r, s), C(p, q, s), D(p, r, s) Non terminal alkyne can form trans-alkene with Na/Liq.NH3 and does not react with ammonical AgNO3 while terminal alkynes do not form trans-alkene with Na/Liq.NH3 but reacts with ammonical AgNO3. 10. Answer - A(p, q, r, s), B(r), C(p, q, r), D(r) (A)
hν CH4 + Cl2 ⎯⎯ → CH3Cl + CH2Cl2 + CHCl3 + CCl4.
(B)
CHCl3 form carbene (:CCl2) with KOH.
(C)
CH3Cl, CH2Cl2 and CHCl3 have dipole moment but CCl4 has zero dipole moment.
O (D)
reacts with Cl2 and NaOH to give haloform reaction and produce chloroform.
11. Answer - A(p, r, s), B(p, s), C(q), D(p) (A)
Dehydration in (A) will follow E1 mechanism, would be brought by H+, reaction will proceed through carbocation, which can undergo rearrangement.
(B)
Same as (A) but carbocation will not undergo rearrangement, since it is benzylic carbon.
(C)
Follows E2 mechanism.
(D)
E1 mechanism follows carbocationic rearrangement.
12. Answer - A(p, s), B(p, r), C(r), D(q) (A)
H2O ⎯ ⎯ ⎯⎯ ⎯→ CH – CH CHO CH3 – C ≡ C – H ⎯⎯ ⎯→ CH3 – CH = CH – OH ←⎯ tautomerisation 3 2
(B)
tautomeris ation CH3 – C ≡ CH ⎯⎯ ⎯ ⎯ ⎯ ⎯⎯ ⎯→ CH3 – C – CH3 ⎯→ CH3 – C = CH2 ←⎯
(C)
Oxymercuration demercuration involves addition of H2O giving Markownikov’s product.
(D)
Oxo process.
HBO
OH
O
OMDM
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13. Answer - A(s), B(p), C(r), D(q)
Cl
%=
%=
Cl
6 ×1 6 × 100 = × 100 = 20.54% 6 × 1 + 3 + 4 × 3 .8 + 5 29.2
5 × 100 = 17.12% 29.2
%=
Cl
7.6 × 100 = 26.0% 29.2
3 Cl % = 29.2 × 100 = 10.27%
14. Answer - A(p, r), B(p, r), C(q, s), D(s) (A)
Alc. KOH will bring elimination, reaction but the carbocation formed is stable so reaction will proceed through E1.
(B)
Same.
(C)
Sterically less hindered alkyl halide so will undergo E2 mechanism.
(D)
E1CB.
15. Answer - A(p, r, s), B(p), C(q), D(r, s) Carbocation is stabilised by electron releasing group whereas it is destabilised by electron withdrawing group. 16. Answer - A(p, q, s), B(q), C(p, r, s), D(p, q, s)
OH (A)
C – CH3
+
H
⊕ CH – CH3
CH = CH2
H Elimination by E1 mechanism. (B)
2° alkyl halides are more prone to elimination in presence of sterically hindered base.
(C)
Carbocation formed is aromatic so reaction proceeds through SN1.
(D)
3° alcohol will undergo E1 mechanism in presence of H+.
17. Answer - A(q, r), B(q, r), C(p), D(s) (A)
Friedel Craft alkylation and a electrophilic substitution.
(B)
Friedel Craft acylation and a electrophilic substitution.
(C)
Only possibility for chlorobenzene is SN with KOH, though that also is not feasible.
(D)
Chlorination in presence of hν is free radical reaction.
18. Answer - A(r), B(q), C(p), D(s) Tertiary alkyl halide gives E1 reaction while secondary alkyl halide gives E2 reaction. If two basic groups are present then elimination through E1CB. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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19. Answer - A(q), B(s), C(p), D(r)
Cl (A)
reacts with aqueous KOH gives SN2 reaction.
Cl (B)
reacts with alcoholic KOH gives E2 reaction and converted into alkene.
Cl (C)
reacts with H2O gives SN1 reaction and form tertiary alcohol.
OH reacts with H+ and on heating gives E1 reaction due to formation of tertiary carbocation.
(D)
20. Answer - A(r), B(p, r, s), C(q, r), D(p, r) (D)
Reactant molecule undergoes anti-elimination but as the product can’t exhibit stereoisomerism that’s why reaction is not stereospecific.
21. Answer - A(p), B(s), C(q), D(q, r) (A)
Picric acid give CO2 with NaHCO3.
(B)
Tertiary alcohol gives white turbidity within few seconds.
(C)
2° alcohol gives white turbidity after 8 – 10 minutes. Ethanol gives iodoform test.
(D)
Ethanol give iodoform test and evolve H2 with sodium metal.
22. Answer - A(p, q, r), B(p, r), C(p, s), D(p, r) (A)
Stable carbocation intermediate due to rearrangement and stable Saytzeff product.
(B)
3° stable carbocation.
(C)
E2 elimination.
(D)
Unimolecular elimination.
23. Answer - A(q, s), B(p, r, s), C(q), D(p, r)
OCH3 OCH3
(A)
OH OH
3HI
+ 3CH3I OH
OCH3
So, product reacts with Na and CH3I is one of the product.
I
OCH3 OCH3
(B)
6HI
I + 3CH3I + 3H2O I
OCH3
I
OPh (C)
OPh OPh
3HI
I 3PhOH + I
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CH3
CH2OH (D)
5HI
CHOH CH2OH
CH – I 2º Alkyl halide
CH3
24. Answer - A(q), B(r), C(p, r), D(p) (A)
In aldol condensation, carbanion is formed as an intermediate.
(B)
In the formation of Grignard reagent, free redical is formed as an intermediate.
(C)
In the Ist step free radical is formed as an intermediate while in IInd step carbocation is formed as an intermediate.
(D)
In dehydration, carbocation is formed as an intermediate.
25. Answer - A(p, q), B(p, r), C(q, s), D(r, s)
O +
(A)
H
O
–
C=O
OH
OH
H
O O (B)
C–H
+
O
–
OH Δ
CH
O
H (C)
C = O + CH3 – C – H H
O
CH3 – C – CH2 – C – H OH
O (D)
H –
OH
C
H
O + CH3 – C – H
H
O
C=C–C–H H
26. Answer - A(p, q, r, s), B(p, s), C(q, s), D(q, s) Phenol gives alkoxy benzene with Friedel Craft reaction, chloroform gives carbyl amine, phosgene gas and Reimer Tiemann reaction. 27. Answer - A(r), B(s), C(p), D(q) Based on data and acidic strength. 28. Answer - A(q, s, r), B(q, r, s), C(p, q, r, s), D(q, r, s) (A)
A gives ketone with H3O+ and HgSO4 and by HBO also undergo ozonolysis to form bicarbonyl compounds
(B)
Alkene give ozonolysis and HBO reaction
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O
OH (C)
PCC
OH + O3
OH
Zn/H2O
O O OH
OH
HBO
OH O3 Zn/H2O
(D)
O
O
29. Answer - A(q, r), B(p, r), C(s), D(p, s)
O CH3
C
, gives haloform test.
—CHO gives positive Tollen’s test. —OH and —COOH, gives NaHCO3 test. 30. Answer - A(p, q, r, s), B(p, q, r, s), C(q, r), D(p, s) Acid with CaO/Δ, acyl chloride with (CH3)2Cd and alkyl cyanide with Grignard salt give ketone. 31. Answer - A(r), B(p, q, r), C(r, s), D(r, s) Cannizzaro reaction is responded by aldehydes containing no αH. Aldol condensation is responded by carbonyl compounds containing acidic hydrogen. Refomatsky reaction is responded by aldehydes only Tollen’s reagent can oxidise aldehydes only. 32. Answer - A(p, q, r, s), B(p, r), C(p, q, r, s), D(r) β-keto acids can be decarboxylated by heating. 33. Answer - A(r), B(s), C(p), D(q) COCl COOH +
H3O
(A)
O COCl
C – NH2 NH3
(B)
(Amide) H2OH CH3COOH
(C)
ester
COCl (D)
RCOOH
Anhydride
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34. Answer - A(r, s), B(p, r), C(p, q), D(r, s) (A)
Hofmann bromamide
(B)
Mannic condensation
(C)
Aldol condensation
(D)
Schemidt reaction
35. Answer - A(p, q, r, s), B(r, s), C(s), D(p, q, r, s) R—MgX reacts with acidic hydrogen. 36. Answer - A(p, r), B(p, q), C(p, q, r), D(s) (A)
2° amine gives insoluble material with Hinsberg and yellow oily layer of p-nitrosoamine.
(B)
1° amine gives Hinsberg test and alcohol give red colour with Victor Meyer.
(C)
1° amine → Hinsberg –NO2 → Red colour with Victor Meyer.
(D)
Only aldehyde give silver mirror with Tollen’s reagent
37. Answer - A(p, q), B(q), C(s), D(p, r, s) Amide can give amine with reduction as well as Br2/KOH 38. Answer - A(p, r), B(q, r), C(p, s), D(q, r) Only 3 N is present in benzene ring with sp2 hybridisation whereas 1 and 2 are non-benzenoid hetero aromatic and 4 N is aliphatic. 39. Answer - A(q), B(p, q, s), C(p, r), D(p, s) +
Cl (A)
Ag
N
⊕ N
OH
OH
N
+
–H
O–H + ⊕ N2
NH2
(C)
O
⊕ ring expansion
HNO2 HCl
(B)
N
⊕
⊕ CH2 is stable than
No ring expansion since
OH 2° alcohol gives blue colour with Victor Meyer
⊕ CH2OH gives red colour with
Victor Meyer.
O2N
NH2
O2N
OH
(D)
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40. Answer - A(p, q, r, s), B(p, r, s), C(r), D(q, r, s) (A)
Protein → Polymer of amino acid formed by condensation of α-amino acids and carboxylic acid having amide linkage.
O (B)
Nylon66 → Condensation polymer having 2 monomeric units forming – N – C – bond.
H (C)
Buna N → It is a copolymer of butadiene and vinyl cyanide.
(D)
PHBV → Poly hydroxy butyrate – CO – β – hydroxy valerate is a copolymer of 3-hydroxy butanoic acid and 3-hydroxy pentanoic acid in which the monomer are connected by ester linkages.
OH CH3 – CH – CH2 – COOH + CH3 – CH2 – CH – CH2 – COOH
PHBV
OH 41. Answer - A(p), B(p), C(q, s), D(q, r) (A)
Buna-S is a co-polymer of butadiene and styrene. It is an addition polymer.
(B)
Polythene is an additional polymer and is obtained by polymerizing ethylene.
(C)
Nylon 6, 6 is obtained by condensation polymerization of hexamethylene diamine and adipic acid.
(D)
Terylene is also called a polyester as it contain ester group. It is a condensation polymer.
42. Answer - A(q, p), B(q), C(r, s), D(s) (A)
Ethylene glycol is a monomor used in formation of Terylene and Glyptal.
(B)
Terephthalic acid is a monomer of Terylene.
(C)
Formaldehyde is a monomer used in formation of Bakelite as well as malmac.
(D)
Phenol is a monomer used in formation of Bakelite.
43. Answer - A(q), B(q), C(q, s), D(p, r, s) (A)
Nylon 6 → Formed by condensation of monomer.
(B)
Glyptal → It is a polyester formed by ethylene glycol and phthalic acid.
(C)
Nylon 2, 6 → It is a biodegradable, condensation polymer.
(D)
Cellulose → A polymer of β-glucose units joined together by a glycosidic linkage is natural occur polymer and is biodegradable.
44. Answer - A(p, q, s), B(p, r, s), C(s), D(s) (A)
Glucose is monosaccharides with 5 chiral carbon α and β –D– glucose differ in C1 configuration.
(B)
Glucose and mannose has same molecular formula C6H12O6 but differ in configuration at 1-carbon only.
(C)
Glucose and fructose are monosaccharides.
(D)
Ribose and glucose are monosaccharides.
45. Answer - A(p, q, r, s), B(p, q, s), C(p, s), D(s) (A)
Glucose and mannose are monosaccharide both reducing sugar and are C2-epimers.
(B)
Mannose and galactose are monosaccharides, reducing sugar.
(C)
Glucose and fructose are monosaccharides and reducing sugar.
(D)
Lactose and maltose are reducing sugar.
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46. Answer - A(p), B(p, q), C(r, s), D(p, q) (A)
Cellulose is polymer composed of D-glucose units which are joined by β-glycosidic linkages.
(B)
Proteins are nitrogeneous polymeric substance.
(C)
Lipids are esters of long chain fatty acids and alcohols and richest source of energy stored in the living bodies.
(D)
Nucleic acids are a group of high molecular mass biomolecules which are present in all living cells in format nucleoproteins. Nucleoproteins are made up of proteins and natural polymers.
47. Answer - A(q, r, s), B(p, r, s), C(q, r, s), D(q) Maltose, sucrose and lactose are disaccharides having glycosidic linkage and sucrose invert its configuration after hydrolysis. Fructose is reducing monosachharide sugar. 48. Answer - A(q, r), B(p, q), C(q, r), D(q, s) Glycine and alanine are amino acid form Zwitter ion. Protein has amide linkage form hydrogen bonds. DNA → Nucleotide containing deoxy ribose sugar.
Section - F : Subjective Type
1.
O
O
I Highly strained ring
The positive charge is resonance stabilised
(i)
O
O
O C
II
Greater charge separation in I assigns it a higher dipole moment than II.
(ii)
In
N
N
H
H N – H , no such resonance is possible. Hence, electron density is more over nitrogen atom.
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is aromatic as it involves cyclic delocalization and follows Huckel’s (4n + 2) π electron rule.
(iii)
I is not aromatic and hence is less stable.
II O
O
2.
(i)
CH2
H3C
CH3
H3C O
OH
O Both
and –OH groups are in trans position and so no hydrogen bonding exists and can easily
C
tautomerise and it is more polar
O
H
O
O
H3C
O
H3C
CH3
CH3
(I)
(II)
Here, intramolecular hydrogen bonding exists and is less polar. Thus compound (I) is more acidic than (II).
O (ii)
N
is basic but
O
is not because in the first case, the lone pair on nitrogen do not
NH
O
participate in resonance with
since it will generate a double bond on bridge head position while
in the second case it does participate in resonance, decreasing its basicity. –
O
O +
NH
(iii)
NH
O is more acidic than
because its conjugate base is more stabilised.
O
OH
O
OH
O
O
O
–
O 3.
O
O
O
O¯ O
(a)
In compound (i) and (iii), there is plane of symmetry passing through the compounds (molecular plane). Therefore, they are optically inactive. Compound (ii) does not have plane of symmetry as the two phenyl rings are not in the same plane. One of the ring rotates about C–C bond axis because of bulky substituents at ortho o′ positions of two adjacent phenyl rings and the two rings are perpendicular to each other. So (ii) is optically active.
(b)
Compound (i) is optically active because there is no plane of symmetry which can cut the molecule into two equal halves. Compound (ii) also does not have plane of symmetry so, it is optically active. Compound (iii) is optically inactive because of the presence of centre of symmetry.
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4.
N
N
(i)
O
OH
H
O (ii)
C
Ph
O
CH3
C
O
Ph — C — C === CH2
O
(iii)
O (iv)
OH
O
O
HO
N == O
O
O
HO
O
N — OH
OH
(v)
O
OH
(vi)
O
OH
H (vii)
N
N
OH (viii)
CH3NO2
CH2 == N O O
(ix)
H3C
C CH3
OH H3C
CH3 CH3
CH3
O (x)
H3C
(xi)
CH3 CH3
OH H3C
CH3
H 3C
C
CH3
H 2C C=O
H 3C
C – OH H 3C
O (xii)
H3C H3C
CH3 CH3 has no α-H atom; hence keto-enol tautomerism is not possible.
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OH Br 5.
Br
OH
(i)
(ii)
Br Cl (iii)
CH3
OEt O
Cl
(iv)
N H
NO2
CH3 6.
(i)
OEt
NO2
CH3
(CH2)2CH – CH == CH2
(CH2)2CH – CH – CH3 1, 2 hydride shift
H
CH3 (CH2)2 – C – CH2 – CH3 –H
(ii)
H
CH == CH2
CH – CH3
H
,
CH3
CH3 –H
7.
CH == CH2
CH3
H
(i) Yes, each ring atom contributes a p-orbital perpendicular to the ring, and there are 10[4(2) + 2] electrons. Azulene is a non-benzoid aromatic compound. (ii) Addition of an electrophile to either starred carbon leads to an intermediate in which the seven membered ring is rendered aromatic
H
H
E
+
E
==
E
+
(iii) Attachment of a nucleophile to any one of the three starred carbons. Show here would render the five membered ring aromatic for e.g. –
Nu
–
== H Nu
H Nu
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8.
Na HC ≡ CH ⎯⎯→ ⎯ HC ≡ C ¯ Na +
.......(1)
H2 HCl HC ≡ CH ⎯⎯→ ⎯ CH2 = CH2 ⎯⎯ ⎯→ CH3 − CH2Cl
.......(2)
Combine (1) and (2) +
HC ≡ C¯ Na + CH3CH2Cl
HC ≡ C – CH2 – CH3 i) Na ii) CH3CH2Cl
H3CH2CC ≡ CCH2CH3 Lindlar catalyst / H2
H3C – CH2 H
Na / liq NH3
CH2 – CH3
C == C
H H3CH2C
H
(Z) -3-hexene
CH2
H2C
H
(E) -3-hexene H3C
9.
CH2CH3
C == C
CH2
CH3
CH2 CH
HCl (excess)
Cl
CH Cl
A
Mg/ether
OMgCl H3C – C – OC2H5
O
CH H3C
H3C
MgCl CH
CH2
CH3
CH2 CH
CH3C – OC2H5
CH
MgCl
CH3
MgCl B
H3C
H3C
10.
O C CH CH2
H3C
CH2
H3C
+
δ MgCl CH δ− Br2
H3C
CH3
Br H3C
OMgCl
CH3
H
+
H3C
OH
CH3
H3C
H3C
C
CH3 CH2Br
Isobutene H
H3C H3C
CH3 Isobutane
Q
160 g Br2 is needed for 56 g of isobutene
∴
20 g Br2 is needed for 56 × 20 = 7 g of isobutene 160
Br2
No addition reaction
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Thus, Isobutene = 7 g, Isobutane = 3 g Now, 7 gm isobutene is converted into isobutane by reduction So, weight of isobutane =
58 × 7 = 7.25 g 56
∴ Isobutane = 7.25 + 3 = 10.25 g
Now, C 4H10 + Br2 ⎯⎯→ C 4H9Br + HBr [Isobutane reacts with Br2 to give monobromo product in light] So by reaction 58 g of C4H10 gives product = 137 g C4H9Br So (7.25 + 3) g of C4H10 gives product =
137 × 10.25 g 58
= 24.21 g C4H9Br CH3
11. A = H3C
Br
CH3
CH3
B = H3C
CH3 Br
HO
CH3 C = H 3C
CH3
D = H3C
CH3
CH3 OH
CH3 E = CH3COOH
COOH
F = H3C
12. A = H3C
CH3
B = H3C
CH3
CH3
CH3
OH C = H3C
CH3
D = H3C
CH3
CH3
CH3 E = HOOC
O
F = CH3COOH
O G = H3C CH3 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Organic Chemistry
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13. PhCH=CH2
Br2
Ph – CH – CH2 – Br
base
Ph – CH == CHBr + Ph – C == CH2
Br
(ii)
R–C=CH2
CH2I2 Zn / Cu
E+Z R CH3
H2 PtO2
base
PhC≡CH
Br
R–C(CH3)3
CH3
(iii)
(i) Base (i) Base HC≡CH (ii) CH I H3C – C ≡ C – H (ii) CH I H3C – C ≡ C – CH3 3
3
H 3C CH2 == CH2
(iv) 14. (i)
H3C
(i) BF3 – THF (ii) Br2
Δ
H2 Na, Liq. NH3
H 3C C == C
H
H CH3
KMnO4 OH¯
product
Product
This mechanism combines electrophilic addition to the non-aromatic π-bond, and electrophilic aromatic substitution. + H CH3 CH3
+ +
CH3
CH3 +
+
–H
H
(ii) This reaction involves a "tandem" electrophilic aromatic substitution. +
NO2
NO2
H
H
NO2 H
+
–H+
H
NO2
H +
H
H
The last structure shown here is a 1, 4-cyclohexadiene derivative, which we know will be very easily oxidised by HNO3 (which is a strong oxidising agent, as well as being strong acid and a source of NO2+). One possible mechanism for the necessary oxidation might involve extraction of the methine hydride by NO2+, giving carbocation 'a', which could deprotonate to aromatize the second ring. H
NO2
NO2
H
+
H
+
–H
NO2
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(iii) The first step is Markownikov protonation of the aromatic double bond to give the 2° carbocation. Then, either the Ph or CH3 migrates, followed by deprotonation.
CH3
CH3 H – OTs
H3C – C – CH == CH2
+
H3C – C – CH – CH3
Ph
Ph CH3
+
CH3
H3C – C – CH – CH3
H3C – C – CH – CH3 +
Ph –H+
H3C
Ph –H+ CH3
C == C Ph 15. CH3CH2CH2CH2Br
CH3
alc. KOH –HBr
H3C
Ph
(B) Br2
CH3CH2HC = CH2
CH3CH2CH(Br)CH2Br
(B)
(C)
CH 3CH 2CH(Br)CH 2Br
NaNH2 –2HBr
CH 3CH 2C ≡ CH
(C)
(D)
CH3CH2C ≡ CH + Ag2O Tollen's But-1-yne reagent
nBuCl
CH3 C == C
CH3CH2CH = CH2
(A)
16. (i)
H3C ==
NaNO2 EtOH
CH3CH2C ≡ CAg White ppt
nBu – NO2 NaNO2
ClCH 2OCH2CH3 EtOH
(ii) ONO – CH2OCH2CH3 + EtO – CH2OCH2CH3 17. A = C6H5CH2CH2Br B = C6H5CH(Br)CH3 C = C6H5CH = CH2 D = C6H5CH2CH2OH D and E are isomeric alcohols E = C6H5CH(OH)CH3 Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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18. (i) In aqueous solution, the given alkyl bromide ionizes to give corresponding resonance stabilised tertiary benzylic carbocation and Br–. This carbocation is attacked by ethanol follow by abstraction of proton by Br– to produce corresponding ether and HBr. Due to HBr, the resultant solution becomes acidic. But the aryl bromide cannot ionise even in aqueous solution due to partial double character bond between carbon and bromine as a result of resonance. Therefore no reaction will take place. (ii) In the first aryl fluoride, attack of nucleophile (OH–) is favoured by the presence of a strong electron withdrawing group (–NO2) at para position. On the other hand in second aryl fluoride, attack of nucleophile (OH–) is opposed by the presence of an electron donating methyl group at para position.
HBr
19.
Br
HO – CH2 – CH2CH2CH2 – CH2 – Br
HBr
Br – CH2CH2CH2CH2CH2 – Br
O⊕
O
H
CH3 20. (i)
CH3
CH3 – CH – CH – CH3
HBr
–H2O
CH3 – CH – CH – CH3
OH
OH ⊕ 2 CH3
CH3 Br
CH3 – CH2 – C – CH3
–
–
–H
CH3 – CH2 – C – CH3 ⊕
CH3 + CH3CH – C – CH3 H
Br
(ii) CH3 – CH2 – CH – OH
⊕ CH3 – CH2 – CH – OH2
HBr
CH3
21. (i)
–H2O
⊕ CH3 – CH2 – CH
CH3
CH2 – CH3
NBS in CCl4
Br
CH3
CH – CH3
alc. KOH
CH3 – CH2 – CH – Br CH3
CH = CH2
Br (ii) CH3CH2CH = CH2
HBr in H2O2
NH3
CH3CH2CH2CH2 – Br –HBr
CH2
CH2 +
+
(iii) HOCH2CH2CH = CH2
H
O
⊕ CH2
H
(iv)
CH2
Cl
Cl2/hν
CH3CH2CH2CH2 – NH2
CH3O
–H
O
OCH3
–
–Cl
(v) CH3CH2CH = CH2
NBS in CCl 4
CH3CHCH = CH2
alc. KOH
CH2 = CH – CH = CH2
Br Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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22. (i) Decreases (ii) Decreases (iii) Increases
CH3 23.
COO – CH – CH3 OH
NaOH, Δ
COONa + CH3 – CH – CH3
(A)
OH
(A)
OH
COOH OH (B)
OH NaOH Δ
CH3 – CH – CH3
CH3COONa + CH3I3 Yellow ppt.
(In NaOH)
14
24. (i)
MeO¯ + H2 C —— CH – CH2 – Cl
SN2
14
CH3O – CH2 – CH – CH2 – Cl O¯
O 14
MeO¯ – CH2 – CH — CH2 + Cl¯ O (ii) Place an O atom between the C's bonded to I's (a) (CH3)3COCH3CH2CH3 (b) cyclohexyl and methyl iodides and (c) The presence of two I's in the same product indicates a cyclic ether, i.e. tetrahydrophyran (iii) In addition to the amount of electron density on the nucleophilic site, steric affects also influence basicities. The greater the steric hinderance encountered in the formation of coordinate bonds, the weaker is the Lewis basicity. In tetrahydrofuran, the R group (the sides of the ring) are tied back leaving a very exposed O atom free to serve as a basic site. The 2° R group in di-isopropyl ether furnish more steric hinderance than do the 1° ethyl groups in ethyl ether. The order of decreasing Lewis basicity is thus tetrahydrofuran > diethyl ether > di-i-propyl ether.
25. [X] is
CH2 – CH2 – CH2 – C ≡ CH Pent-4-yn-1-ol
OH
CH2· CH2 · CH2 · C ≡ CH + 2CH3MgBr → 2CH4 OH
[X]
Q 84 g [X] gives 2×22.4 lt. CH4
0.42 g [X] will give
2 × 22.4 × 0.42 = 224 ml CH4 84
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HO
HO OH¯ Δ
C6H3 – CH2 – CH == CH2
26.
H3CO
C6H3CH == CH – CH3 H3CO
(A)
(C)
C10H12O2
C10H12O2
CH3O
H3CO
OH¯ Δ
C6H3 – CH2 – CH == CH2 CH3O
C6H3CH == CH – CH3 H3CO
(B)
(D)
C11H14O2
C11H14O2
OH
OH
OH
OH 27. (A)
(B)
C – CH2Cl
CH(OH)CH2NHCH3
O
OH OH (C)
C – CH2NHCH3 O H 3C
28. (i) (ii)
A=
OH HgOAc H
H 3C
OH
B=
A = CH3CH2CH2OH B = CH3OH
OH (iii)
A = PhMgBr
B = Ph
CH3
(iv)
H 3C — C — OH + C 2H 5OH CH3
(v)
RCO2Et + MgBr
MgBr R
R OH
followed by +
H3O
R H2/Pd
–H2O
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(vi)
No reaction
(vii)
6 HCOOH
OH +
29. (i)
H + –H3O
HO
ring expansion
HO O
O—H +
–H
OH (ii)
+ OH – H
OH
O H
O
H3C
OH
H3C
+ OH2
H
CH3
–H2O
(iii)
+
–H
CH3
Me
Ph
Me
Ph
Me
Ph
+
(iv) Me
H
Ph OH
Me
OH
Ph OH
OH
Me Ph C O
OH
Me
Me Ph Ph
Ph
+
–H
Me
H— O
H
+
(v)
Ph
Me
–H2O
H
+ OH
OH OH
OH2
+
O
–H
O– H
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O
O –
30. (i)
CH3–C–CH2–CH2–C–CH3 O
dil OH 100°C
H2C
O
C
C CH2
C
CH2
C O
O
H3C
CH2
CH2
CH2
CH3 H2O
O C CH C
O CH2
C
Δ, –H2O
CH2
CH2
CH2
CH2
C
CH3
HO
CH3
CH2
–
dil OH
CH3–C–CH2–CH2–CH2–CH2–C–CH3
(ii)
O
O
H3C – C – CH O H3C
CH2 CH2
C O
O H3C – C H3C
Δ
–
H2O(–OH ) CH3CO
H3CCO
–H2O
H3C
H3C
–
OH
O
CH2
–
(iii)
H3C–C–CH2CH2CH2CH2CH2–C–CH3 O
dil OH
CH2
H3C – C –CH O H3C
O
C
CH2 H2C
O
O
O
H3C – C
H3C – C Δ – H2O H3C
H3C
O H2O
H3C – C
–
–(OH )
H3C –
OH
O
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Organic Chemistry
Success Magnet (Solutions) O –
(iv)
dil OH
CH3–C–CH2–C–CH3 O
CH3–C–CH–C–CH3
O
O
O
CH3–C–CH2–C–CH3
O
O O H3C – C
O
CH3
O
CH
–
OH
CH2
H3C – C
C=O
O
CH3 C
dil. CH2
H3C – C
CH2 C
O
C CH
CH2
C
H3C – C
CH3
H3C – C
C
CH
H3C
O
O
C=O
O
H3C
CH2
O
O O
O
H3C – C
H3C – C
CH3 H3C – C HO C CH H3C C CH2 OH CH2 C
CH3
C=C
C=C Tautomerize
H3C – C
H3C – C
CH CH
CH3
31.
+
N H3C
(C)
O
O
and OH
N
+ H3C
Ph OH
CH3
NH
NH
H3C D
KOH
syn
(oxime)
Ph
H3C
OH
H3C
(B)
N
N
Anti
(A)
Ph
H5C6
OH
NH2OH, HCl
NH2
C OH
O
H5C6
CH CH
C
O
O
CH3
Ph
(E)
(D)
CH3 – COOK + (F)
NH2 CH3COCl (F)
N
O
H
,
E
KOH
CH3CN + PhCOOH White solid
CH3
(G)
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32. (i)
O O
O CH3 O
(ii)
CH3
CH3 Br2 NaOH
Br
COONa +
Br Br
+
H
O
O CH3 COOH
CH3
– CO2 Δ
H
Easy decarboxylation
SOCl2
(iii)
AlCl3
Zn-Hg HCl Conc. HCl
COOH
COCl
O (A)
O
(B)
O OH
(iv)
OH
+
O
CHO
OH +
NaBH4 CH3OH
OH
1 eq. H
(C)
H3O
O
CH
CHO
CH O
O
(A)
(B)
(C)
O O
O
PhMgBr
33.
Ph
Ph – C
C=O CH2
O H2O
CH2
H–OH Ph – C
..O
..
O
CH2
C – OH CH2
34. (i) Number of products are three
(ii) CH3 — C = CH2 | CH3
H/HOCl Cl + + OH –
δ+ CH3 — C —— CH2 | CH3 Cl δ +
OH
–
OH | CH3 — C — CH2 | | CH3 Cl
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35. (i)
A =
(ii)
C =
Success Magnet (Solutions)
N
CO
CH3
N
B =
OH
D = O
O (iii)
S
E =
CH2CH3
NH
C2H5
S
F =
N
C2H5
O
O O S
G =
N
SO3H
H =
O (iv)
I = (C2H5)2NH2 J = (CH3NH)2C
S
N(CH3)2 36. (i)
CH3OH + N(CH3)2
(ii)
CH2 = CH – CH3 + (major)
(iii)
CH 2
CH2 + N(CH3)2CH2CH2CH3
(iv)
CH2 + N(CH3)3
CH3
CH3 CH3 37. X = CH3
Y = CH3
CH
CH NH2 ∗ (optically active)
C
CH2CH3
OH (Optically inactive)
CH2OH 38. CH2
CH2Cl CH2 CH2Cl (B)
CH2
Conc. HNO3 Conc. H2SO4
NH
CH2CN
CH2CH2NH2
(C)
NO2 39. (i)
CH2CH2NH2
CH2
CH2OH (A)
CH2CN
(D)
O
NH2 (1) Sn/HCl — (2) OH
(E)
C Cl – C –
NH
C
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CH3 (ii)
N–H
excess CH3I
CH3 Moist Ag2O
⊕N
I
–
Δ
N
–AgI
⊕
CH3
N(CH3)2
–H2O
CH3
H O
O–H
O–H
HO
+
40. (i)
H
N(CH3)2
N(CH3)2
– H+
(CH3)2NH
H
HO
H–O
N(CH3)2
H+
– H2O
N(CH3)2
N(CH3)2 H – BH2CN – BH2CN
H2 / Pt
(ii)
1 equivalent
C≡N
O
O
O
CH2NH2
O
CH2
H
NH2
IMPE
N
N H
H
H
1 equivalent
CH = NH
–H2O
N
+
H2 / Pt
O H
H
N OH H
O H
H
+
–H
H
CH2 NH2
H2/P2
N
N H
41. (i) Buna-S – — CH2 – CH = CH – CH2 – CH – CH2 — , monomers ⇒ styrene, 1, 3-butadiene
n
(ii) PHBV – — O – CH – CH2 – C – O — , R = CH3, C2H5, monomers ⇒ 3-Hydroxy butanoic acid R
O
n
3-Hydroxy pentanoic acid Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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Organic Chemistry
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H O
(iii) Nylon 6
C
O
N (CH2)5
NH
caprolactum (monomers)
n
Cl (iv) Neoprene – — CH2 – C = CH – CH2 — , monomers = chloroprene n
(v) Glyptal – — OCH2 – CH2OOC
COO — n
(i) Buna-S – Polymer of butadiene Styrene – Chain growth. (ii) PHBV (Poly β-hydroxy butyrate – Co-β-hydroxy valerate) Monomen – 3 hydroxy butanoic acid and 3-hydroxy pentanoic, step growth polymer. (iii) Nylon 6 – Caprolactum step growth. (iv) Neoprene – 2 chloro 1-3 butadiene, chain growth polymer. (v) Glyptal – Phthalic acid and ethylne glycol step growth polymer. 42. Let the fraction of glucose present as the α anomer [(α) = +112.2º] is ‘a’, the fraction present as the β anomer [(α) = + 18.7º] is ‘b’ and the rotation of the mixture is + 52.6º, We have a(+ 112.2º) + b(+ 18.7º) = + 52.6º There is very little of the open-chain form present, so the fraction present as the α anomer (a) plus the fraction present as the β anomer(b) should account for all the glucose. a + b = 1 or b = 1 – a Substituting (1 – a) for b in the first equal, we have a(112.2º) + (1 – a)(18.7º) = 52.6º Solving this equation for a we have a = 0.36 or 36%. Thus b must be (1 – 0.36) = 0.64 or 64% α anomer = 36% β anomer = 64% The anomeric hydroxyl group is axial in α anomer and equitorial in the β anomer. So it is obivious that more stable β anomer predominates. Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124
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43. (i)
P = COOH
(ii)
Q = COOH
(iii)
R = COOH
(CHOH)4
(CHOH)4
CHOH
+ rearrangement
CH2OH gluconic acid
COOH gluconic acid
(CHOH)3
+ Ag (silver mirror)
CH2OH gluconic acid 44. The isoelectric point (pI) is the pH at which the amino acid exists only as a dipolar ion with net charge zero. At isoelectric point, for a neutral amino acid, pI =
(pK a + pK a ) 1
2
2 The dissociation of cationic form of valine can be represented as –
CO2H CHNH3
–
CO2
(pKa1) – OH
CHNH3
H+
(pKa2) – OH
CO2
CHNH2
H+
CH(CH3)2
CH(CH3)2
Net charge (+)
CH(CH3)2
(0)
(–1)
The species with zero net charge exists between species with (+ 1) and (–1) net charges.
pI =
(pK a + pK a ) 1
2
2
=
9.69 + 2.31 =6 2
When the pH of the solution equals to pI, the structure of valine is CO2–
CHNH3 CH(CH3)2 When the pH of the solution is two, the structure of valine is CO2H
CHNH3 CH(CH3)2 When the pH of the solution is 12, the structure of valine is CO2–
CHNH3 CH(CH3)2 45. α-glucopyranose is the cis-1, 2-diol and the β-anomer is the trans-1, 2-diol. Since the former can form a cyclic ester with periodic acid and the latter cannot, the former is oxidised more rapidly than the latter. 46. Only C-1 and C-2 are involved in osazone formation. Hence, aldohexoses and ketohexoses, which have the same configuration at C-3, C-4 and C-5 give the same osazone. CHO CHO CH2OH
H HO
OH
HO
H
H
HO
H
CO HO
H
H
OH
H
OH
H
OH
H
OH
H
OH
H
OH
CH2OH D-glucose
CH2OH
CH2OH
D-mannose
D-fructose
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47.
Success Magnet (Solutions)
OMe H
H
MeO CH2OMe
CH2OMe
H
H H
MeO
MeO O
H
H CH2OMe
OH H H 2, 3, 4, 6-tetra-o-methyl-D-glucose
H
+ H
HO
H
MeO
MeO
OMe
CH2OMe OMe
O
H
H
H MeO
OMe
OH H H 2, 3, 4, 6-tri-o-methyl-D-glucose
48. A ⇒ H2C = CH2 B ⇒ HO – CH2 – CH2 – OH C ⇒ HOOC
O O O
O
O
OH O
49. (a) (i) Sucrose has no free anomeric OH. (ii) D-glucose and D-fructose are linked by their anomeric OH’s, the α of one with β OH of ether (hydrolysis by emulsion). (iii) The glucose unit is a pyranoside because the C5 OH is unmethylated. (b) Fructose ring size and glycosidic linkage (actual linkage is α to glucose and β to fructose). (c) The sweeter form is fructopyranose. The high temperature causes a shift in the pyranose equilibrium towards the less sweet furanose. 50. (A) Contains 92.3% C & 7.3% H. Thus it is CH ≡ CH (B) Contains 90% C & 10% H, thus it is CH3 – C ≡ CH CH3
CH3
C CH
CH
CH
CH
CH3 CH3COCl
Red hot tube
AlCl3
COCH3 (D) CH3
(C)
CH3
CH3 SOCl2
CH3 I2/NaOH
CH
CH3 NH3
furanose
+ CHI3 COOH (E)
Br2 NaOH
COOH
COCl
CONH2
NH2
(F)
(G)
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51. Since the last compound is p-dibromobenzene, hence the starting compound must be o-toluedine, because it gives carbylamine reaction and can be diazotised
CH3
CH3 NH2
CH3 NHCOCH3
(CH3CO)2O
(A)
Br2 Fe Powder
CH3 NHCOCH3
NH2
H2O/H+
Br
Br
(B)
(C)
(D) NaNO2 / HCl 0–5ºC
Br
CH3
(i) Oxidation
CH3 +
Br
(ii) NaOH/CaO
N2 Cl
CuBr HBr
Br
—
Br
(E)
Br p-dibromobenzene
52. Since compound (A) on heating with KOH eliminates HBr, hence Br atom is in side chain not in the ring. Thus (A) is N-bromobenzamide which gives unstable species (B) which changes into phenyl isocyanate (C). The isocyanate on heating with KOH gives aniline, which is diazotised and the product couples with B-naphthol to give orange red dye.
O
O
C–NHBr
.. C–N
..
KOH
NH2
2KOH / Δ
N=C=O
Rearrangement
Phenyl isocyanate –K2CO3
Δ, –KBr, H2O
Aniline
(A)
(B)
(C)
(D) NaNO2 + HCl 0 – 5ºC
N=N–
OH
N2Cl
OH NaOH Coupling reaction
(Orange red dye)
(E)
53. The aromatic amine (A) if considered as 2-methyl pyridine, then all foregoing reactions can be explained as
N
N
CH3
CH3
⊕
(ii) Ag2O moist
H (A)
Δ
(i) CH3I
Ni / H2
N CH3
(B)
+
under reduced pressure
CH3 CH3{OH–
N CH3
(C)
CH3 CH3
(D)
CH2
N CH3
CH3 (E)
Quarternary ammonium hydroxide
(i) CH3I
N CH3 (D)
CH3 CH3
(i) CH3I
(ii) Ag2O, moist (iii) Δ
CH2
N CH3 (E)
(ii) Ag2O, moist (iii) Δ
CH2
CH3 (F)
(G)
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Organic Chemistry
Success Magnet (Solutions)
54. Ozonolysis of A′ gives 4-ethoxy-3-hydroxy benzaldehyde (C).
CHO A′
(i) O3
+ CH3 – CH = O
(ii) Zn/H2O
OH OC2H5
The A′ is obtained by heating, A with strong alkali i.e. the isomerisation of A into A′. This isomerisation involves the migration of double bond from terminal position to adjacent carbon. This isomerisation suggests the IInd part of the ozonolysis product is acetaldehyde. From these two ozonolysis product, the structure of A can be formulated as
OC2H5 OH
OC2H5 OH
strong alkali Δ CH2—CH = CH2
CH = CH – CH3 (A′)
(A)
By the structure (A) other structures can be formulated by the following reaction.
OC2H 5 OH
(CH 3)2 SO4
OC2H 5 OCH 3
NaOH CH 2–CH=CH 2 (A) reflux Δ Strong alkali
HI
CH 2–CH=CH 2 (B)
OH OH + C2H 5I CH 2–CH=CH 2
OC2H 5 OH
CH=CH–CH3 (A') (I) O3 (II) Zn/ H O 2 OC2H5 OH
CHO (C)
Strong alkali Δ OC2H 5 OCH 3
CH=CH–CH3 (B') (I) O3 (II) Zn/ H O 2 OC2H 5 OCH 3
CHO (D)
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Organic Chemistry
Success Magnet (Solutions)
55.
Since (A) has degree of unsaturation 2. The ozonolysis of A gives (B) having same number of carbon atom. Thus (A) must have one ring and one double bond. The only structure of A i.e. cyclopropene is possible containing one double bond and one ring. On the basis of (A) other structure may be formulated as
CH 2
CHO
(I) O 3
CH = CH (A)
Δ
CH 2
(II) Zn / H 2O
Na / C2H 5OH
CHO (B)
CH 2Cl
CH 2OH
PCl 5
CH 2
CH 2
CH 2Cl CH 2OH (C) C is 1º-alcohol hence gives red colour with leric ammonium nitrate & diacetyl derivative CH 2 – CH 2 – NH 2 CH 2
CH 2 – CN Reduction LiAlH4
CH 2
KCN CH 2 – CN
CH 2 – CH 2 – NH 2
(D)
(E)
2HCl
CH 2 – CH 2 NH 2 HCl
Δ
CH 2
H 2SO4
N | H (F)
CH 2 – CH 2 NH 2 HCl (E) ⊕ N
CH 3I N (G) Pyridine
CH 3 I–
Pd/C Δ
Quaternary salt
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