Unit-II - Organic Chemistry - [Solutions]

Aakash IIT-JEE - Regd. Office : Aakash Tower, Plot No. 4, Sector-11, Dwarka, New Delhi-75 Ph.: 45543147/8 Fax : 25084124(133)

Section A : Straight Objective Type1. Answer (3)

Due to ortho effect,

COOHOH

is most acidic.

2. 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

=

22 HCCHCH resonance stabilised

=

HCCH2 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)

The Fisher projection of the compound is HO H

CH3 CH3

OHH

and it has R & R configuration on both

asymmetric C-atom.8. Answer (1)

H C C C C HH

H

H

HH H

p orbitals of carbon being restricted rotation

Organic Chemistry UNIT 2


Success Magnet (Solutions) Organic Chemistry

9. Answer (3)COOHHOHCHOHCCH3

No. of chiral carbon atoms = 2No. of optical isomers = 2n = 22 = 4

10. Answer (2)

123

4

56 7 Cl

3

3

3

7

Positive charge due to delocalisation of electrons is present 1, 3, 5 and 7 carbon.11. Answer (1)

Bridge head carbocation is not possible.12. Answer (2)

OH

ring expansion

H+

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)

HNO3 decomposes Na2S and NaCN present in Lassaigne extract becuase otherwise they will giveprecipitation with AgNO3.


Organic Chemistry Success Magnet (Solutions)

18. Answer (3)

Stabilised due to aromatic nature and is resonance stabilised.

19. Answer (3)

CH = CH O H CH CHO2

O

O

OH

OH

NH NH2

:

20. Answer (4)Pyrolle is least basic as the lone pair is contributing to aromaticity of the molecule.

In pyridine N

N is sp2 hybridised.

NH

is most basic, therefore 6 member ring is least strained.

21. Answer (2)

CH O CH3 2

: stabilised due to lone pair of O.

22. Answer (2)

B C cyclic delocalisation of 6 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 bridgeheadcarbon is not feasible.

26. Answer (2)2 has (4n + 2) electrons and planar ring.



27. Answer (4)

CH OH2

CH OH2

OHOHHHO

H

H S

S CH OH2

CH OH2

OHOH

HHOH

H S

S

28. Answer (1)

lCBr+

, 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 %88.131006.7593

8.3251163110031

=++

=

+++

=

35. Answer (1)

Cl C CH CH3

Cl

Cl

< Cl C CH2CH2

Cl

Cl

(more stable carbocation)

36. Answer (4)H3O+ gives A i.e. Markownikovs addition, through carbocation formation. B formed by hydroborationoxidation that gives anti-Markownikovs product.

37. Answer (2)

H+

H

Hydride transfer

H



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 C H Br2 5C H2 5

42. Answer (2)

C C C CCH3

CH3

H

H

C CH3H

H2 CH = CH CCH

CH CH3CH3

(optically inactive)H

CH3

Triple bond is more sensitive for addition H2 than double bond.43. Answer (3)

CHOCHHCCaCCCaO 3SOHHg

22OH

242

22

++

44. Answer (2)

C = CH

H C3 H

CH3 CH N2 2: CH2 C C

H

H C3 H

CH3

45. Answer (2)Na give electron resulting the formation of free radical.

46. Answer (2)

COONaSodalime

COCl

Anhyd. AlCl3

CO

A B (ketone)47. Answer (1)

Cl AlCl3

48. Answer (2)

CH CH C CH3 2 H O2Hg , H2+ +CH CH C = CH3 2 2

OHCH CH C CH3 2 3

O



49. Answer (2)KMnO4 (H+) cold gives syn hydroxylation.

50. Answer (3)

ClCl,

Cl

Cl,

Cl

Cl, Cl

Cl

51. Answer (4)Peracids give epoxy formation.

52. Answer (1)Peroxy acid form epoxy alkane.

53. Answer (2)

CH C CNa3 C H Br2 5 CH C H CH3 2 3 C C54. Answer (2)

OCHOO3

55. Answer (2)Thiophene undergoes electrophilic substitution at 2nd position.

56. Answer (2)

(CH ) C == CH3 2 2 H+ (CH ) C CH3 2 3 (CH ) C == CH3 2 2 (CH ) C CH3 2 3

CH C (CH2 3)2H+

(CH ) C 3 3 CH3CH == C

CH 3CH 3

57. Answer (2)CH = CH CH = CH2 2 H21, 4 addition CH CH = CH CH3 3

A

O /H3 2O1, 4 addition 2CH COOH3

B(Zinc is not present hence CH3CHO first formed oxidised to CH3COOH)

58. Answer (3)

CH CH CH == CH3 2 H+

CH CH CH CH3 3CH3

1-2 H shift CH C CH CH3 2 3CH3

Br CH C CH CH3 2 3CH3

Br

(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.



61. Answer (3)Carbocation formation and ring expansion takes place.

62. Answer (1)Due to high electronegativity of carbon.

63. Answer (3)

CH C C CH3 2 CH3O3 CH C C C H3 52

O O

Li AlH4

CH C C C H3 52

OH OH

H HChiral centre = 2Stereoisomer = 4

64. Answer (4)Free radical substitution takes place in presence of sunlight.

65. Answer (3)

COCl

NO2

Anhyd.AlCl3

CO

NO2

Given product.

66. Answer (1)

1 Kolbes electrolytic reaction.

67. Answer (4)

33CH

3NaNH

3 CHCCCHNaCCCHHCCCH 32

68. Answer (3)Williamsons 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.



74. Answer (2)

Cl

Cl

NH NH

2

3 ClBenzyne

NH2

Cl

NH2H H2 N NH2 Cl

NH2

.

75. Answer (4)

C H Br2 5

aq KOH

alc

C H OH(1)

2 5

CH = CH2 2 (2)

C2H5OH and CH3OCH3 are isomers

22H

52 CHCHOHHC =+

.

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

F

NO2

NO2

CH3+ OH

F

NO2

NO2

CH3

OH

F

NO2

O2N

CH3

OH

79. Answer (3)Iodoform test.

80. Answer (1) NO2 group show I & R effect.

81. Answer (4)

+ I2

I

+ HI .



82. Answer (3)SN1 reaction carbocation formation.

83. Answer (2)

Br

BrAgOH Br Br

OH

.

84. Answer (4)Depends on formation of free radical

Br(1 eq)

2

h

Isomeric monobrominated product

Br2h

85. Answer (4)Carbene formation then ring expansion takes place.

86. Answer (1)Due to presence of benzylic group

CH Cl2

Cl

NaOH CH OH2

Cl

87. Answer (3)

CH CH SH3 2CH O

CH OH33

CH CH S3 2CH CH22 C

O

CH CH SCH CH O23 2 2H O/H2

+

CH CH SCH CH OH23 2 2 .

88. Answer (3)SN1 involves cyclic formation and

R OS = OCl Cl

S = OO

R+

R Cl + SO2

Where R = CDH

CH3.



89. Answer (4)

OH

OH

O H

OH

OH

O

Br

90. Answer (4)

CH CH = C3H

COOHBr Cl+

CH CH C COOH3Cl Br

H

4 stereoisomerstherefore2 chiral carbon

91. Answer (2)Chlorination takes place at more probable free radical site.

92. Answer (1)

CH C CH3 2O

H OH2

18

+H3C C CH2

OH

CH3

H O218

Stable carbocation

C C CH2H OH3

CH3

O18HH

C C CH2H OH3

CH3

O18

H

93. Answer (1)

C)CH( 33 Stable carbocation.94. Answer (3)

N

OH

O

O less stronger than

N OO

OH

Intramolecular H-bonding.



95. Answer (1)

C

C

O

O

O + 2

OH

H

C

CO

O

OHOH

+ H O2

(phenolphthalein)

96. Answer (1)COOH

+ NaHCO3

COONa

+ CO + H O2 2

Phenol does not react with NaHCO3.97. Answer (3)

R O R + HI R OH + RI 3AgNO AgI ppt.

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

pKa acid of Strength1

101. Answer (3)E1 Saytzeff oriented product.

102. Answer (3)Ring expansion

CH OH2H /+ CH2

HH+

103. Answer (3)Both NaBH4 and LiAlH4 reduces acid chloride to alcohols.

104. Answer (3)

OH

NO2 is weaker acid than H2CO3.



105. Answer (3)OH

+ FeCl3O Fe O

O+ HCl

(ferric phenoxide)or (C H O) Fe6 5 3

[Fe(C H O) (H O) ]ClSoluble complex

6 5 3 2 3 3

H O2

106. Answer (3)

CH3 KMnO4CH3

OH

OHH

32

CrOCH COOH

3

3

CH3OH

O

107. Answer (3)THF is good solvent for Grignand reagent.

108. Answer (2)CH OH2CH OH2

PCl5 CH Cl2CH Cl2

+ POCl3

109. Answer (4)

OHH+

FTrans (anti addition)

BrCCl

2

4Mesoform

(1)

cis

Br

anti addition2

Racemicmixture

d, forml

Mesoform + d, = 3 stereoisomerl1 2

110. Answer (1)Ether have low boiling point, more volatile whereas alcohols due to presence of intermolecular H bonding havehigh boiling point therefore less volatile.

CH OCHCH3 3

CH3

HIS 2N

CH I + HO CH CH3 3

CH3

NaOI

CH COONa + CHI3 3(yellow ppt)



111. Answer (2)

CH3

O

H+

OH

CH OH2

112. Answer (1)OC OHC O

OH+

HO CH

HO CH2

2

A

O

O O

O

HO CH

HO CH

2

2

H SO2 4

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 secondaryalcohol produces ketone.

115. Answer (2)

CHClKOH

3

OH OH

CHO

ReimerTiemannreaction

Cannizzaroreaction

50%KOH

OH

COOK+

OH

CH OH2

.

116. Answer (1)

C H CO H6 5 3 OHBr

OH

Br(A) Trans 2-Bromocyclohexanol

117. Answer (4)

Compound -1 CH = CH O 2 CHCH 3

CH 3

H O

H2

+CH CH3 2 3OH + CH C OH

CH3

H

Compound-2 CH = C O 2 C H2 5H O

H2

+CH C OH + C H OH3 2 5

CH3

H

CH3

Both 1 & 2 gives same product.



118. Answer (1)C H CH 6 5 2 CH OH2 CH2

LiAlH(reduce both C = C andC = O bond)

4

C H CH =6 5 CH CH = Ocrotonaldehyde

NaBH4(does not reduce

C = C bond

C H CH =6 5 CH CH OH2(Y)

(X)

119. Answer (4)CH3OH < CH3 O CH3 < C6H5OH.

120. Answer (3)Jones reagent oxidies 1 alcohol to aldehyde

CH OH2 Jone'sreagent

CHO

In choice (4), tertiary alcohol is resistant to oxidation at room temperature.

R X + ORaN+

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 + [(CH ) CO] Al3 3 3 > C = O

125. Answer (1)O CHO KOH

50%+

O COOH O CH OH2

no hydrogen atom

Undergo Cannizzaro



126. Answer (2)

CH CH = CH 3 CH CH3H CrO in aq

acetone (Jones reagent)2 4

OH

CH 3 CH = CH C CH3

O

127. Answer (2)

CHO(CHOH) CH OH

4

2

5HIO4 5HCOOH + HCHO

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 H.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 resonancestabilize therefore it give substitution at temperature and pressure.

131. Answer (4)

HO CH CH2NH3

COOH is most acidic so pH is low.

132. Answer (3)

C = O + H+

HO

RR C = O H

OH

R C OH

OH

133. Answer (3)

CH CH CH C N3 2 2CH MgI3 CH CH CH C = NMgI3 2 2

H O3+

CH CH CH C = O3 2 2

CH3

+ Mg IOH

CH3



134. Answer (4)

CH C Cl3H S2

[H S H]CH C SH + HCl3

O O

135. Answer (4)

CH C O 3 CH3 + CH Mg Br3

O

CH C O 3 CH3

MgBr

CH3

H O2 CH C + 3 MgBr

CH3

O

OH+ CH OH3

(1) CH MgBr(2) H O

3

3+

CH C 3 CH3

CH3

OH

O

136. Answer (2)

H O CH 2 C OH

OSOCl2 Cl CH 2 C Cl

OCH OH3 Cl CH 2 C OCH3

O

137. Answer (3)

CH3 CH2 C NH CH3

O

It should be noted that acid will be neutralised with amine.138. Answer (2)

keto acid undergo decarboxylation fast due to formation of resonance stabilised carbanion.

PhCOCH2COOH

PhCOCH3.

138(a). Answer (3) IIT-JEE 2008

Ph

O O

OH

CO2 Ph

CCH3

OI + NaOH 2

PhC

O

ONa + CHI3

GE-keto acid

iodoform reaction

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.



140. Answer (2)

O

O

+ (H Br)

O H

O

Br

OH

OBrH

OHBr

OH

H+

141. Answer (1)Baeyers villiger oxidation.

142. Answer (3)

O

NH OH

H2

+

NOH

Na/C H OH2 5

NH2

H CH3

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-Markownikovs product)

146. Answer (4)Pinacol - pinacolone rearrangement.

147. Answer (1)

O O

O

Stable aromatic

O

Stable aromatic

(1)

(2)

(3)

Anti - aromatic



148. Answer (4)

NBS Mgdry ether

Br MgBr

CH CN3

H C C = N MgBr3

H O2

H C C = O3

149. Answer (1)Br

Red P2CH CH COOH3 2 CH CHCOOH3

Br

NH3 CH CHCOOH3

NH2

150. Answer (1)

NH2

+ Cl C C Cl

HO

H

H N C CH Cl2

O

AlCl3

N

CH2

H

C = O

H

orON

151. Answer (1)Coupling occurs prefenertially in the para position to the hydroxyl group. But it this position is blocked thencoupling occurs at ortho position.

152. Answer (3)

NO2Br /FeBr2 3

NO2NaNO

HCl2

N2Cl

Br 280 K(X)Br

(Z)

Cu powderCl

Br(W)MgEther

Cl

MgBr

C H COCHH O

6 5 3

3+

Cl

C H6 5OH

CH3

(F) (A)



153. Answer (2)

NH2NaNO

HCl2

N2ClOH

phenanthrene

154. Answer (1)

KCN

OH Cl OH CN

SOCl2

Cl CN

H /Pt2

Cl CH2NH2

NaOHHCl N

155. Answer (2)

NH3

O OH O NH2A

KOHBr2

NH2B

CHClKOH

3

NCC

H O3+

NH2

+ HCOOH

DE

(B & D are same]

156. Answer (1)

NaNO /HCl0 5C

2

NO2

F

NH2

NMe2

N Cl2

NMe2

H /Ni2

NH2

NMe2

157. Answer (2)Hoffmann elimination takes place, resulting the formation of less substituted alkene.

158. Answer (1)

CH C H3

O

gives iodoform (haloform) test.159. Answer (3)

CH CH COOH3 2SOCl2 CH CH COCl3 2

(CH CH CH ) NH3 2 2 2 CH CH 3 2 C N

OC H3 7

C H3 7

LiAlH4 (C H ) N3 7 3A CB



160. Answer (2)Less substituted alkene is more stable

N CH3

CH3CH3

OH H O2 N

CH2

CH3CH3

+ H O2

161. Answer (3)

C NH + O = C C CH 2 3CH3

H3C H CH3

H H

C N = C C CH3CH3

H3C H CH3

H H

CH C N CH CH(CH )3 3 22H

CH3 H

H /Pt2

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)

O + H +OH

NH

OH

NH

OH2NH

N



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 Hinsbergreagent.

168. Answer (2)

CH I3

N H

N CH3CH3

I

AgOH

NCH3

CH3I

AgOH

N CH3CH3

OHN

CH3

CH3

CH I excess3

CH3

169. Answer (2)

N H O2 2

+ NH OH2

170. Answer (1)

C

O

O

O

+ NH3C

O

C

O

OH

NH2

171. Answer (3)

O

C NH2H /Pd.

high pressure2 CH NH2 2

H /Pd.2 CH NH2 2

H2 reduces CONH2 and then benzene ring.



172. Answer (3)

22OH CORNHOCNR 2 + == .

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)

NH2O

CH3

NH3 OH

177. Answer (1)

O

O

HNN

H

OHH

O

OH

O

O

NN

H

OHH

O

Anion (conjugate pair) stabilize by resonance.178. Answer (1)

323OH

223LiAlH

3 HNCHCHNHCHCHNCCH 34

+

.

179. Answer (2)

C = O + H N NH C NH 2 2 2

OCH3

Hsemi carbazide

Pyridine

C = N . NH C NH2H3C

H

O

Semi carbazone

NH2 of

O

C NH2 does not undergo condensation due to its resonance stabilization.

O

C NH2

O

C = NH2



180. Answer (3)1 amine oxidise to nitro group and 2 alcohol oxidise to ketone with cold KMnO4.

181. Answer (4)NH OH

H2

+ O H

Et OH

OHH2NOH OH

N OH

H H

N OH

H

N O H

O

H O2

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)

CH CH3 2 C C = O

H

CH3

H

+ H NCH2 3

CH CH3 2 C C = N CH3H

CH3 H

H /Pt.2

CH CH3 2 C C N CH3H

CH3 H 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)

n CH2 = CCH3

CH3C CH C CH2 2

CH3

CH3

CH3

CH3



189. Answer (4)

Teflon FCF

FCF n

is fully fluorinated polymer.

190. Answer (3)

Amino acid are amphoteric in nature. The acidic properties are due to amino group

3HN .

191. Answer (1)Alanine is NH CH COOH2

CH3

NH CH COOH3

CH3(At low pH)i.e. acidic pH NH CH COO2

CH3(At high pH)i.e. basic pH

192. Answer (1)During vulcanization of natural rubber S S crosslinks are formed which make it more elastic.

193. Answer (3)

n CH = CH22343 K 373 K, 6 7 atm

Ziegler Natta catalyst[TiCl + (C H ) Al]4 2 5 3

( CH CH )2 2 n

194. Answer (3)The -helix is known as 3.613 helix since each turn of the helix has approximately 3.6 amino acids and 13membered 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)

Proline is heterocyclic amino acid. H N H

COOH

amino acid have a primary amino group except proline.198. Answer (3)

Acrilan is PAN (Poly acrybonitrile).199. Answer (4)

Caprolactum is monomer of Nylon-6 or Perlon N

HO



200. Answer (1)

Bakelite is made up of

OH

(phenol) and HCHO (formaldehyde)

OH

+ HCHO

OHH C2 CH2

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 OHCH OH2

204. Answer (1)Fat 130 ATP, carbohydrate 38 ATP, protein 5 ATP.

205. Answer (1)

HCH OH2

OHOH

H

OH

H O

OH

H D glucopyranose

206. Answer (2)

CHO

CHOH

(CHOH)3CH OH2

C H NNH6 5 2H O2

CH = NNHC H6 5CHOH

(CHOH)3CH OH2

C H NNH6 5 2C H NH , NH6 5 2 3

CH = NNHC H6 5C = O C H NNH6 5 2

H O2

CH = NNHC H6 5C = NNHC H6 5

CH OH2

(CHOH)3

glucosozoneCH OH2

(CHOH)3

207. Answer (2)Only phospholipids form bilayer memberane.



208. Answer (2)

C H O + H O12 12 11 2H+

invertase D(+)-glucose + D()-fructoseC H O6 12 6 C H O6 12 6

( ) = 52.7 D ( ) = 92.4 Dsucrose( ) = +66.5 D

invert ( ) = 19.9 D209. Answer (3)

C12H22O11 + 18[O] conc. HNO3 COOH

COOHoxalic acid

+ 5H O2

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)

C H O + H O12 22 11 2( ) = +66.5 D

H+

invertase C H O6 12 6D(+)-glucose( ) = 52.7 D

+ C H O6 12 6D()-fructose( ) = 92.4 D

invert sugar ( ) = 19.9 D213. 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 osazoneonly C1 and C2 are involved.

215. Answer (4)-D-glucose and D-glucose are anomers.

216. Answer (3)

C N

O H peptide bond of protein form Hbond.

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.



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 3HN is acidic group.223. Answer (3)

Amide linkage ( CONH2)224. Answer (4)

5 Chiral carbon present in Dglucose H

CH OH2OH

OH

H

HO

H O

OH

HH*

*

*

*

*

225. Answer (3)

H C = O

(CHOH)4CH OH2

HCN H C OH

(CHOH)4CH OH2

H O3+

COOH(CHOH) CH OH

5

2

HI

COOH(CH ) CH

2 5

3

Heptanoic acid

CN

226. Answer (3)

CH OH

C= O

CH OH

2

2

No chiral carbon

No stereoisomers.

227. Answer (2)Ruff degradation.



Section - B : Multiple Choice Questions1. Answer (1, 2)

In cis form same groups are present on the same side of double bond.

2. Answer (1, 2, 3, 4)

In +

and

+

aromaticity is maintained in canonical forms, while benzyl carbocation looses its

aromaticity in canonical forms.

(CH3)3+

C 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 oftrans 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 giveBeleistein test.

5. Answer (1, 3)

COOHOH

is more acidic than HCOOH due to ortho effect while

COOH

and

COOH

CH3

are less acidic

than HCOOH.

6. Answer (1, 2)

OH

C O

C

C

C

O

O

H

H

C O

C

C

O

H

HOOC H

,

Conjugate baseof maleic acid

Conjugate baseof fumaric acid

(stability)



7. Answer (1, 3)

F SbF5

F

Aromatic systemsSbF5

8. 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) makeanother enantiomeric pair.

13. Answer (1, 2)

H

COOH

CH3

H N2

OHH

S

R23

OH

COOH

H

OHH

S

R

23

COOH14. Answer (2, 3, 4)

(1) molecule contains 3 (1 H), 8 (2 hydrogens) and 3 (3 hydrogens).15. Answer (1, 3, 4)

OO 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 + CH MgI CH + Mg3 4I

OR0.37

xx = 74

112 cc22400

CH C CH OH3 2

H

CH3

H+

CH C = CH3 2CH3

O3 CH C = O + HCHO3CH3



18. Answer (1, 3)A Presence of benzylic H

B Absence of benzylic H

19. Answer (1, 3)

C5H12 + Cl2 H C C CH Cl3 2

CH3

CH3

NaWurtz reaction H C C CH CH C CH3 2 2 3

CH3

CH3

CH3

CH3(B) (C)

20. Answer (1, 2)

O

O OO

CHOC

O

H

C

O

H CHO

O3HOH

OH

OO

H

21. Answer (1, 2, 3)Wurtz reaction is a type of free radical reaction.

C2H5 Cl2NaCl ++ C2H5 etherdry 2NaClHCbutane104+

Ethyl free radical can disproportionate to give ethane and ethene.

Ethane62

Ethene4252 HCHCHC2 +

22. Answer (1, 2, 4)

On applying selectivityreactivity principle, only CH CH CH CH3 3CH3

Cl has about 35% in the mixture. Others

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, whilebut-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 reductiveozonolysis, 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 ofHCO3H, anti addition occurs.



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 brominationBr ortho para directing

28. Answer (1, 2)

NBS

Br

(A)29. Answer (2, 3)

Due to presence of unsaturation.30. Answer (1, 2, 4)

Clanhyd AlCl3 hydride shift

more stablecarbocation

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 productromin

73major

273 ONOHCNOHC +

due to ambident nucleophilic nature of nitro group.

34. Answer (1, 3)

LiAlH4

CH2

H C3 Br

CH3

CH3(A)

LiAlH4

CH3H C2

Br CH3H C3

(B)



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

BrBr

HNO3

Br

BrBr

NO2

Highest melting point 1, 3, 5 tribromobenzne.

38. Answer (2, 4)

(2) AgBrROROAgBrRdry

2 ++

(4) CH3CH2OH K413SOH 42

CH3 CH2 O CH2 CH3

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) cant give an alkene

41. Answer (2, 3)3 alkyl halide mainly gives SN1 mechanism. Polar protic solvent favours SN1 mechanism while polar aproticsolvent 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)CH3CH2CH2OH + SOCl2 Pyridine CH3CH2CH2Cl + SO2 + HClCH3CH2CH2OH + 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



46. Answer (1, 2)

OH

H+

H O2+

major minor product47. Answer (3, 4)

(2) RNH2 + CHCl3 + alc.KOH RNC + KCl + H2O

(4)

NH2

+ CHCl3 + alc.KOH

NC

+ 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)

C N

O H

is o, p directing due to N

H

group.

50. Answer (1, 3)Only 1 amine gives carbyl amine test.

51. Answer (1, 2)cis and trans isomers C = C PhH

PhH & C = C

PhHPh H by elimination.

52. Answer (1, 2, 3)

CH Br2

H C5 2

1

2

34

Br

CH2

HOHH+

+CH OH2

H C5 2

CH2

H C5 2

CH2

H C5 2

HOHH+

CH2

H C5 2

OHCH2

H C5 2OH

(1)

(2)

(3)

HOHH+

1

H C5 2



53. Answer (2, 3, 4)Br

C H ONa2 5

54. Answer (2, 3, 4)

OH

HI/P

55. Answer (1, 2)Water favours the formation of a polar compound.

56. Answer (1, 4)

Secondary alcohol containing CH

OH

CH3 linkage can give all given three reactions.

57. Answer (2, 4)

O OH+16

16

H

+

OH18 H O2

18

+ OH16

+

58. Answer (1, 2, 3, 4)In (1), (2) and (3) intramolecular hydrogen bonding is present.

59. Answer (1, 2, 3)

Iodoform test is given by alcohols having the group R CH CH3

OH

. Thus, all the three except (4) give this test.

60. Answer (1, 3)

CH CH C CH3 2 (i) NaNH2(ii) C H Br2 5 CH CH C C C H3 2 2 5

H2Pd/BaSO4

C = CC H2 5

H

H5C2

H

alk.KMnO4

C H2 5

C H2 5

OHOH

HH

(X)

(Y)

optically inactive (Z)

cis alkene

meso compound



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 C CH

Cl OH

Cl OH

and F C CH

F OH

F OH

No hydrogen bonding occurs with Br and I due to large size and lesser electronegativity.

63. Answer (1, 3)

C H CHCH6 5 3

OH

and CH CH CH3 3

OH

gives iodoform test due to presence of CH3 group adjacent to OH group.

64. Answer (1, 3)In option (1) and (3), most stable carbocation is formed as an intermediate.

65. Answer (1, 2)

OHNO2

and

OHNO2

NO2

cant give effervescences with NaHCO3 but

OHNO2

NO2

O N2 can

give effervescences with NaHCO3 due to more acidic nature.

66. Answer (2, 3)

HO C CH3

O

and H N C CH2 3

O

cant 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 carboxylicacid is 1>2>3. Other statements are correct.

69. Answer (1, 2, 4)

OH

NO2 NO2

NO2

,

COOH

,

SO H3

are stronger acid than H2CO3 acid.



70. Answer (2, 4)

D OH

HHCH = CH CH CH3

OHK Cr O2 2 7

H+

D O

H C OH

O

+ HO C C CH3SOCl2

Cl C C CH3

OO

+

D

H C Cl

O

(Y)(X)

OO

O

71. Answer (1, 2)

O OCH3

H+

H O2 HOOH OH

+ CH OH3H O2

OOH H

72. Answer (1, 3)(3) does not contain hydrogen and (1) contains vinylic hydrogen which cannot be easily removed.

73. Answer (3, 4)

O

O

OHconc.

COOOH

Cl

COOC H2 5

CCl3

Br

aq. KOHaq. KOH

74. Answer (1, 2, 3, 4)

R RCHO COOAgNO /NH OH3 4

H C OH

OAgNO /NH OH3 4 HCO3

NHOH NO

AgNO /NH OH3 4

(R can be CH3 C6H5)



75. Answer (1, 3, 4)CH

CH MgBr2

MgBr

CH2 CH2 OMgBr

OHB H2 6

HCHO

CH2

H2O /OH2

+O

H O /OH2 2

76. Answer (1, 2, 3, 4)O

O

O Phenol Phenolpthaline

ResorcinolCatachol

Nitration

FluorisienAlizarin

2,4 D.N.P.

77. Answer (2, 3)OMDM gives MarkownikovsHBO gives Anti-Markownikovs productNo, carbocation formation No, rearrangement takes place

78. Answer (1, 2, 3)

(1) H C CH CH CHO3 2

CH3NaOH CH C C C CHO3

CH3 H H

C H

CH3H3COHH

aldol

(2) H C CH CH C = O + CH MgBr3 2 3

CH3 HH O3

+

CH CH CH C CH3 2 3

CH3 OH

H(3) 3-pentanone and 3 methyl butanol have same molecular formula(4) On Wolff Kishner reduction it gives 2 methyl butane.



79. Answer (1, 2, 3)

(1) PhCH MgBr + CH C H2 3 CH C CH Ph3 2OH

H

O

(2) CH CH CH + PhLi2 3 H O3+

OPhCH C CH2 3

OH

H

(3) Ph C C H + CH MgBr3

OH

H

H O3+

PhCH C CH2 3

OH

H80. Answer (1, 2, 4)

(1)

CH3

(2)

(4) CH3

CH381. Answer (2, 3, 4)

(2) NaBH4 reduces carbonyl to 2 alcohol.(4) Silver mirror test given by aldehydes.

82. Answer (1, 2)N Cl2

Cu/KCN

CN

H O3+

COOH

NaOHCaO

(X) (Y) (Z)83. Answer (1, 3)

(1) H C CH3

OH

(O)KMnO4

C CH3

O

(3) + CH COCl3AnhyAlCl3

COCH3



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 methylketone.

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) )B(

23)A(

223 COOHCHCHOHCHCHCH

(4) CH3CH2COOH93. Answer (1, 2, 3)

CH3Cl

CH3

ClCH3

Cl

KNH2

KNH2/

CH3 CH3

NH2

CH3 CH3NH2



94. Answer (1, 4)NO2 NO

Fe/H O(v)2NH2

OH

strong acidelectrolysis

95. Answer (1, 4)

R RNH2 OHHNO2

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 (CH ) NH CO (CH ) CO ]2 6 2 4 n

PMMA = CH C 2

CH3

COOCH3 n99. 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)

CH CH + HCN Ba+2 polymerisation (PAN) CH CH 2

CN n



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 ComprehensionC1. 1. Answer (4)

HH

sp3 hybridised

sp3 hybridised

So non planar (so not aromatic).

(I)

4 electrons (so not aromatic).

2. Answer (4)

NH

E+

NH

NH

HE

E+

NH

NH

HE

+ E+

NH

NH

HE



3. Answer (2)

O+

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 withthe presence of electron releasing group.

C3. 1. Answer (1)

H H

O OH

Aromatic, hence very stable.2. Answer (2)

In (1), (3) & (4) the group attached to CO, involves, CO

in resonance.

O O

Cl

O O

Cl3. Answer (4)

O O O OH

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 2 amine is more basic than 3 & then comes 1.

3. Answer (4)For iodine + R is least.

C5. 1. Answer (1)

++= cos2222

1



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

no. of hyperconjugative structures = 3 2 1Alkyl 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 H+ methylshift

C7. 1. Answer (4)[O]

O , H O3 2 2COOHCOOH

2. Answer (3)

CH = CH2 2MCPBA

OCH MgI3

OMgI

CH3H O3

+

OH(A) (B)

3. Answer (4)

CH CHO + 3

O+ HCHO

C8. 1. Answer (1)Diels alder reaction involves (4 + 2) cycloaddition.

2. Answer (2)

HH

C = C C = CH H

H

H

H+ H

H

Br

Br

Br

BBr



3. Answer (3)

+

B E

O /H O3 2 COOHCOOH

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 attacksthe 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 aresame. 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 conjugateddiene 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.

HCH3

OHCH3

3. Answer (4)OsO4 Syn hydroxylationH2 Syn additionKMnO4 Syn hydroxylationBr2/CCl4 Anti addition

C12. 1. Answer (2)

ClCl

(d) + ( )l

Cl

(d) + ( )l

Cl



2. Answer (2)

Cl probability = 1 3 = 3

Clprobability = 2 3.8 = 7.6

Clprobability = 1 5 = 5

Clprobability = 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, NH2is a strong base and hence very weak leaving group, so cannot undergo SN reaction

C14. 1. Answer (4)

OCH3 is a better base.

2. Answer (1)SN1

3. Answer (3)E1

C15. 1. Answer (3)

( ) n 56n 42M(w) vesderivati acetyl of .tW(w) alkali used of .tW

=+

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.



C16.

OH

NaOH

ONa

(i) CO4 - 7 atm P

(ii) H

2

+

OH

COOH

A B

NaOHCaO

OH

CBr2(CH CO) O3 2

O COCH3

COOH

NaOH OH

COONa + CH COOH3

OHBrBr

Br D

G H

E

1. Answer (2)2. Answer (1)3. Answer (3)

C17. 1. Answer (2)

BrMg

ether MgBrCH CHO3 CHCH3

OH

H+

CHCH3Ring

expansionBr

CH3

Br

2. Answer (2)

CC H3 7

H3C

OH

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)

Compound B is C7H6O3

O H

C = OOH

H-bond



2. Answer (1)

OH

COOH

OCOCH3COOHCH COCl3

H+

PhOHH+

(C)

OHCOOPh

(D)

(aspirin)

(salol)

3. Answer (2)Compound D is salol.

C19. 1. Answer (3)

CH3

OH

B

O CH3

A =

2. Answer (3)O COCH3

YCOOH

, Aspirin used as antipyretic.

3. Answer (1)

CH3

OH O CH3

& C H O7 8

C20. 1. Answer (1)O

CH3Carbanion becomes aromatic.

2. Answer (1)O

H C2 H

O

H C2 HCarbanion stabilize by resonance.



3. Answer (1)

H C C OH3

H

NaOI

COONa

+ CHI3

C21. 1. Answer (3)NH2

HNO

H SO3

2 4

NH2

NO2

+

NH2

NO2M

(47%)P

(51%)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 ofortho substitution is steric hinderance.

C22. 1. Answer (3)

I

Ag O2

N(CH )3 3

OH

+ NMe + H O3 2NMe3

2. Answer (2)

CH NH2 23CH I3

CH N2

Me

Me

Me

I

Ag O2

+ Me N 3 + H O2

3. Answer (4)

N Me

nBu Et

OH CH = CH22



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)

CH NH2 2A B

O

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)Hbonding in Nylon.

3. Answer (4)It is prepared by condensation of adipic acid and hexamethylene diamine.

C26. 1. Answer (1)

C N

O H H bonding

2. Answer (2)Teflon is polymer of CF2 = CF2.

3. Answer (3)Polyacrylonitrile

H HC = CH CN

H H( C C ) H CN

n

C27. 1. Answer (2)

62

8.703.30pH =+=

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 2NH2 group, So pH must be greater than 7.



C28. 1. Answer (2)Shortening of carbon chain of one carbon.

2. Answer (3)Name of reaction Ruff degradation.

3. Answer (4)(A) C = NOH, (B) Pentyacetate(C) Cyanide, (D) Aldopentose

C29. Starch(C H O )6 10 5 n

200 250C

(C H O )6 10 6 n

H O2 C H O12 22 11H O2 C H O6 12 6

conc.H SO2 4

(A)Dextrin

(C) (D)(E)

Black

1. Answer (2)2. Answer (1)3. Answer (1)

C30. 1. Answer (3)Elimination dominates over substitution.

2. Answer (4)A C3H6B CH CH CH3 3

BrC CH CH CH3 3

OH3. Answer (3)

Substitution product

Cl OH OH

C31. 1. Answer (1)

R C R

OH

OH

(O)R C R

O

2. Answer (4)All of these contain Cr+6

PCCN

HCrO3Cl

3. Answer (3)nfactor = 6



C32. 1. Answer (3)

C H C C H + MgX2 5 2 5 C H2 5

OH O3

+

C H C C H2 5 2 5

OH

C H2 5

H O2

C H C = C CH2 5 3C H2 5

C H C = O + CH C H2 5 3

O

C H2 5

H

2. Answer (3)Less hindered more reactive

3. Answer (3)C = O C

sp2 sp3

C33. 1. Answer (3)CH3

H SO2 4

CH3

SO H3 PCl5

CH3

SO Cl2 NH3

CH3

SO NH2 2

COOH

SO NH2 2

CO NH

SO2

CH3

SO H ( P)3

H SO2 4

2. Answer (1)CH3 CH3 CH3 CH3

SO H3

PCl5

SO Cl2

NH3

SO NH2 2

(i) ClOH(ii) NaOCl

SO NNa2+

Cl

3. Answer (4)

CH3

SO H3

PCl5

CH3

SO Cl2

C H OH6 5

CH3

SO OC H2 6 5ester

H O218

CH3

SO OH218

+

OH



C34. 1. Answer (2)Rearrangement of carbocation.

2. Answer (3)Intermediate 2 carbocation of AIntermediate 1 carbocation of B

3. Answer (3)

OHCH NH2 2

HNOHN

2

2

+

OH

CH2ring expansion

OH OH O

C35. 1. Answer (2)

CH C CH2 3

O

CH C H2&

O

2 carbanion 4 product2. Answer (4)

Carbanion stabilize by resonance.3. Answer (4)

(1) & (2) option no H atom(3) option give chloroform reaction.

Section - D : Assertion - Reason Type1. Answer (1)

That acid is easily decarboxylate which produce most stable carbanion.

2. Answer (1)Due to I effect producing group.

3. Answer (1)

=+

Major23

Minor23 CHOCHHCOCH

Stabilize by resonance, the second structure has octets on all atoms and an additional bond.

4. Answer (3)

HONO + H SO2 2 4 H O NO + HSO2 4

Hbase acid

5. Answer (2)

HOX + H + H O X (acid) H

H O + X2electrophile

6. Answer (2)The bond energy of allylic carbon hydrogen bond in propene is less than the bond energy of benzylic carbon-hydrogen bond in toluene.



7. 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

NH2

H O/H or OH2+

OH

NH2

(1-5 migration of H atom).

11. Answer (3)1 is R, II S.

12. Answer (1)

C H

ClH

O

Cl OH

H-bonding

Cl C

13. Answer (2)

No. of isomers 21n

n 22

.

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 Cl C3 is less basic than F C3 because fluorine can dispersecharge only by an inductive effect while Cl disperse charge by inductive effect as well as p p bondingdelocalisation.

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 = 75.021085.109

=

(negligible).



20. Answer (1)

CH C3CH3O CH3

CH C3

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 inpyrrole the non-bonding pair on nitrogen is part of aromatic rextet.

23. Answer (1)Number of hyperconjugative structure 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)O H

H

O H

HH

H

H-bonding

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)

C C

Br+

Intermediate ionbromonium ion

30. Answer (2)Correct R Mesotartaric acid has molecular symmetry.

31. Answer (4)

CH C C H3 H C C ==== C H3 CH C == C H3Hg+2 Hg Hg+

OH H

Hg+2CH C == CH3 2

OHOH H (enol)



32. Answer (2)(II) Does not react with NH3 since double bond is not polarised by an electron withdrawing group.

(I) Form product (III) Via H N C C C == C3 +CH3 H

CH3 H

O

CH3

intermediate.

33. Answer (3)Alkyne gives carbonyl compounds

R C C H (i) H O(BH )22 3THF(ii) H O OH2 2/

R CH C H2O

34. Answer (4)It involves formations of vinyl carbocation, which is unstable reaction does not takes place.

35. Answer (1)

OH

ringexpansion

H+ H+

36. Answer (3)Order of boiling pointStraight 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.



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 ofhalogen 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

H

Benzene(aromatic)

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 p-p 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.



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 3 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 hydrogen from CH3NO2.

70. Answer (4)2, 2 dimethyl propanoic acid has no 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)

NH2C

NH2

NH

H2NC

NH2

NH2H O2 (is stabilised by resonance)

79. Answer (4)In alkaline medium with ZnNaOH, nitrobenzene gives hydroazobenzene.



80. Answer (1)CCl 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 isdelocalised.

R C NH2

O

R C = NH2

O +

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 cleavescarboxy - 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 differentpolymer chains to give a three dimensional network solid.

88. Answer (1)

Carbocation formed from styrene )CHHCHC( 356 +

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 Tollens reagent.

94. Answer (3)Cn(H2O)n General formula.



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)-D-glucose and -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 bothboubly 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)

(A)

Aromatic and resonance stabilised (4n + 2 = 6)

(B) CH C C = C3CH3

CH3

CH3

H

+1 groupresonance

CH C = C C3

CH3 CH3

CH3H

(C) Aromatic and resonance stabilised (4n + 2 = 6)

(D)N O

OCH2

Aromatic resonance and I effect of NO2 group.



4. Answer - A(p, r), B(p, q), C(s), D(r)Friedel Crafts reaction

CH Cl3anhyd. AlCl3 CH3

CH3

Reimer Tiemann reactionOH

CHCl , KOH3

OHCHO

salicylaldehydeIntermediate Carbene, an electrophile attack on electron rich ringAldol condensation

CH C H3

OOH CH C H2

O

carbanion

CH C H3

OCH C H2

O

CH CH CH CHO3 2

OH

Acid catalysed hydration Carbocation is intermediate

5. Answer - A(r), B(p), C(q), D(q, s)Chlorination in presence of h 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)

(A)H

Very acidic hydrogen so reacts with base it is allylic hydrogen which will be subslituted by

Cl2/h. Presence of double bond will make it react with Br2 water.

(B) Acidic hydrogen HOCHH 2

(C) Undergoes nitration with (HNO3 + H2SO4) as well as free radical addition with Cl2/h.

(D) free radical substitution with Cl2/h.7. Answer - A(s), B(q, r, s), C(p, s), D(q, r, s)

(A) Cis elimination of meso compound produces trans isomer



(B) Elimination is possible from non-stereo centre. So, is a stereospecific reaction and a stereoselectivereactions and undergoes antielimination.

BrH

C

Br H

C

HH

H

HHH

(C) Carbocation would be formed, reactant is optically active product will show geometrical isomer hence thereaction 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 whileterminal 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) CH4 + Cl2 h 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.

(D)O

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) CH3 C C H HBO

OH2 CH3 CH = CH OH ationtautomeris CH3 CH2CHO

(B) CH3 C CH OMDM CH C = CH3 2OH

ationtautomeris CH C CH3 3

O

(C) Oxymercuration demercuration involves addition of H2O giving Markownikovs product.(D) Oxo process.



13. Answer - A(s), B(p), C(r), D(q)

Cl % = %54.201002.29

610058.34316

16==

+++

Cl % = %12.172.29

1005=

Cl % = %0.261002.29

6.7=

Cl % = %27.101002.293

=

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)

(A) C CH3OH

HH+ CH CH3

CH

= CH2

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 h 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 arepresent then elimination through E1CB.



19. Answer - A(q), B(s), C(p), D(r)

(A)Cl

reacts with aqueous KOH gives SN2 reaction.

(B)Cl

reacts with alcoholic KOH gives E2 reaction and converted into alkene.

(C)Cl

reacts with H2O gives SN1 reaction and form tertiary alcohol.

(D)OH

reacts with H+ and on heating gives E1 reaction due to formation of tertiary carbocation.

20. Answer - A(r), B(p, r, s),

C(q, r), D(p, r)(D) Reactant molecule undergoes anti-elimination but as the product cant exhibit stereoisomerism thats 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)

(A)OCH3

OCH3

OCH3

3HI OH

OH

OH

+ 3CH I3

So, product reacts with Na and CH3I is one of the product.

(B)OCH3

OCH3

OCH36HI

II

I+ 3CH I + 3H O3 2

(C) OPh

OPh

OPh

3HI

II

I3PhOH +



(D)CH OH2CHOHCH OH2

5HICH3CH ICH3 2 Alkyl halide

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)

(A)O

+ C = OH

HOH

O

OH

(B)O

+C H

O

OH

O

CH

(C) C = OH

H+ CH C H3

OHO

CH C CH C H3 2

H O

OH

(D) CO

H+ CH C H3

C = C C H

H

H OO

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 andReimer 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



(C)OH

PCC

O

OH Zn/H O2+ O3

OH

OO

OHHBO

OH

OH

(D) O3Zn/H O2OO

29. Answer - A(q, r), B(p, r), C(s), D(p, s)

CH3 C

O

, gives haloform test.CHO gives positive Tollens 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/, 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 H.Aldol condensation is responded by carbonyl compounds containing acidic hydrogen. Refomatsky reaction isresponded by aldehydes only Tollens reagent can oxidise aldehydes only.

32. Answer - A(p, q, r, s), B(p, r), C(p, q, r, s), D(r)-keto acids can be decarboxylated by heating.

33. Answer - A(r), B(s), C(p), D(q)

(A)

COCl

H O3+

COOH

(B)COCl

NH3

C NH2

(Amide)

O

(C)H OH2

CH3COOH ester

(D)

COCl

RCOOH Anhydride



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)RMgX 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 Tollens reagent37. 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 heteroaromatic and 4 N is aliphatic.

39. Answer - A(q), B(p, q, s), C(p, r), D(p, s)

(A) NCl

OHNOH

H+N

O H O

Ag+

N

+

(B)

NH2

HNO2

N2

OH

2 alcohol gives blue colourwith Victor Meyer

HClring expansion

(C) No ring expansion since CH2

is stable than

CH2OH gives red colour with

Victor Meyer.

(D)O N2 NH2 O N2 OH

gives Victor Meyer test



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 -amino acids and carboxylic acid having

amide linkage.

(B) Nylon66 Condensation polymer having 2 monomeric units forming N C O

H

bond.

(C) Buna N It is a copolymer of butadiene and vinyl cyanide.(D) PHBV Poly hydroxy butyrate CO hydroxy valerate is a copolymer of 3-hydroxy butanoic acid

and 3-hydroxy pentanoic acid in which the monomer are connected by ester linkages.

CH CH CH COOH + CH CH CH CH COOH3 2 3 2 2

OH

OH

PHBV

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 -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 and 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.



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 -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 configurationafter 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. (i)

O

I Highly strainedring

O

The positive charge is resonance stabilised

O

O

II

O

C

Greater charge separation in I assigns it a higher dipole moment than II.

(ii)NH

NH

In N H , no such resonance is possible. Hence, electron density is more over nitrogen atom.



(iii)I

is aromatic as it involves cyclic delocalization and follows Huckels (4n + 2) electron rule.

II is not aromatic and hence is less stable.

2. (i)O

CH2H3C

OCH3H3C

OOH

Both OC and OH groups are in trans position and so no hydrogen bonding exists and can easily

tautomerise and it is more polar

OH

OH C3 CH3 CH3H C3

O O

(I) (II)Here, intramolecular hydrogen bonding exists and is less polar. Thus compound (I) is more acidicthan (II).

(ii)N O

is basic but NH

Ois not because in the first case, the lone pair on nitrogen do not

participate in resonance with O since it will generate a double bond on bridge head position while

in the second case it does participate in resonance, decreasing its basicity.

NH

O

NH

O+

(iii) OOH

is more acidic than O OH

because its conjugate base is more stabilised.

OO

O

O

O

O

O

OO

O

3. (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 phenylrings are not in the same plane. One of the ring rotates about CC bond axis because of bulkysubstituents at ortho o positions of two adjacent phenyl rings and the two rings are perpendicular toeach other. So (ii) is optically active.

(b) Compound (i) is optically active because there is no plane of symmetry which can cut the molecule intotwo 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.



4. (i) N OHN

O

(ii) Ph C C CH3

O

O

Ph C C === CH2

O O

H

(iii)O OH O

O

(iv) N == OHO N OHO

(v)OO

O

OHHO

OH

(vi)O OH

(vii) NH

N

(viii) CH NO3 2 CH N2 == OH

O

(ix) CO

CH3CH3

H3C CH3 C

OH

CH3CH3

H3C CH3

(x) H C3 CH3O

H C3 CH3OH

(xi) C = OH C3

H C3C OH

H C2

H C3

(xii)CH3CH3

H3CO

H3C has no -H atom; hence keto-enol tautomerism is not possible.



5. (i)

OHBr Br

Br

(ii)OH

(iii)Cl CH3

N

O

H

(iv)

OEtOEt

NO2NO2

Cl

6.

Documents

Unit-II - Organic Chemistry - [Solutions]