HOME ASSIGNMENT SOLUTIONS GENERAL ORGANIC · PDF fileNo. of asymmetric or chiral carbon = 2; ... Therefore two forms are possible. 13. (D) 3 3 H | CH C ... The separation of a racemic

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SOLUTIONS

1. (B)

In this structure chiral carbon atom is present since it is optically active.

2. (D)

Chiral centre is present. Hence, it exists as optical isomers or enantiomorphs.

3. (C) C6H5CH2CH3 chiral centre is absent

4. (C) Lactic acid shows optical isomerism.

5. (A)

6. (A) molecular symmetry

Mesotartaric acid is optically inactive due to interval compensation i.e. the effect one half of the

molecule is neutralized by other.

7. (B) optical isomerism because chiral centre is present

8. (B) A chiral molecule has no superimposable mirror images

9. (B)

10. (D)

Number of enantiomers = 2n (n = asymmetric carbon atom) = 22 = 4.

11. (B) Glucose and fructose have similar molecular

Formula with difference of functional group, so they are functional isomers.

CHEMISTRY BOOK-7 HOME ASSIGNMENT SOLUTIONS GENERAL ORGANIC CHEMISTRY



12. (A)

13. (A)

14. (A)

15. (C)

16. (B)

17. (B) Structures are not mirror image of each other

which are nonsuperimposable so they are enantiomers.

18. (A) The product will be optically active because the

Chiral carbon atoms of both the acid and alcohol

retain their configuration.

19. (A)

20. (B) Ithas plane of symmetry.

21. (D)

It has 2 optically active carbon atoms and it has no place of symmetry. So, number of optical isomers

= 22 =4.

22. (A)

23. (D) 2,4- dibromopentane contains two similar

asymmetric carbon atoms and any compound with two or more asymmetric carbon atoms but also

having a plane of symmetry is called meso-compound.

24. B)

25. (B)

26. (A)

27. (A)



No. of asymmetric or chiral carbon = 2; Therefore no. of optical isomers = 22 2 4;n No. of

optically compound is optically inactive due to internal compensation.

28. (D)

29. (D)

30. (A)

It shows geometrical isomerism but does not show optical isomerism.

31. (A)

32. (C) 3o alcohol has no H-atom at (C–OH)

33. (C)

34. (A) It lacks -carbon.

35. (C)

36. (C)

Pentavalent carbon is generated which is not possible.

37. (B)

38. (B)

39. (C)

40. (A)

41. (D)

42. (A)



43. (A)

44. (C) SN1-reaction at 2o alkyl halide BE of (C–Br) <

(C–Cl). Thus (C–Br) is affected.

45. (C)

46. (A)

47. (A)

48. (B)

49. (A) LG Basicity

1

50. (B) Better leaving group increases rate of SN2 reaction.

51. (B) SN2 reactions proceed with inversion of configuration.

52. (B)

53. (B)

54. (D)

55. (B)

56. (A)

57. (A)

58. (C)

(1) As the size of substrate increases rate of reaction decreases

(2) As the stability of transition state increases rate of reaction increases

(3) Rate of reaction increases with better leaving groups.

(4) Electron donating groups increases rate of reaction.

59. (D) Rate of SN1 reaction depends on following factors:

(a) As the stability of cation increases rate of reaction increases

(b) Rate of reaction increases with better leaving groups.

(c) Electron donating group increases rate of reaction.

60. (D)

61. (A) (1, 2) Methyl shift leading to formation of 3o carbocation.

62. (A) -elimination

63. (D) Good L. G and stable carbocation

64. (B) E0 required good L.G.

65. (C)



66. (C) reassignment possible

67. (A)

68. (B)

69. (D) Addition of O3, on alkene followed by reduction with Zn/H2O

70. (B) Look for stable carbocation.

71. (A) More is the acidity of -hydrogen more will be reactivity of alcohol dehydration.

72. (C)

73. (B)

74. (C) T.S. stability

More is the acidity of -hydrogen more will be reactivity of alcohol for dehydration.

75. (A)

76. (C)

77. (B) Bulky base eliminates the L.G. to give Hoffmann product.

78. (B)

79. (B)

80. (B)

81. (B)

82. (C)

83. (C)

84. (B)

85. (C)

86. (A)

87. (C)

88. (A)

89. (A)

90. (D)



SOLUTIONS

1. (D) 0 016 0.50 0.35 2.8obs D

c

2. (B) A rapid umbrella type inversion rapidly converts the structure III to its enantiomer; hence the two

enantiomers are not separable.

3. (D) Compound gives Atropisomerism and is optically active (O.A) due to absence of plane of symmetry

(P.O.S) and (O.A) due to absence of plane of symmetry (P.O.S) and centre of symmetry (C.O.S).

4. (D)

5. (A)

6. (A)

7. (B)

8. (D)

9. (D)

10. (A)

11. (C) According to sequence Rule, the priority order are as under

2 5 3

41 2 3

,Cl C H CH H So in R configuration

12. (B)

3

*

2 3

|

|OH

CH

COOH C CH CH

One chiral centre. Therefore two forms are possible.

13. (D)

3

3

H

|

CH C* COOH

|

CH

asymmetriccentre

2-hydroxy propanoic acidlactic acid

14. (C)

3 3

*

|OH

CH CH CH CH CH Total possible isomers are four.

CHEMISTRY BOOK-7 OBJECTIVE QUESTIONS

SOLUTIONS GENERAL ORGANIC CHEMISTRY



cis – R

trans -R

cis- S

trans- S

15. (B)

16. (C) Order of SN2 reactivity is 1o> 2o> 3o

17. (B) In SN1 reaction, rate of substitution depends on the concentration of substrate. In case of SN1 reaction,

there is racemization. SN1reaction will be faster, if carbocation formed is more stable.

18. (B)

19. (D) Smaller the R group reactivity will be higher towards SN2 reaction.

Better will be leaving group, faster will be rate.

Relative reactivity F– Cl– Br– I–

1 200 10000 20000

20. (B)

21. (C)

22. (B) The reactivity order can be explained on the basis of C–X bond strength. Lower the bond strength,

higher is the reactivity.

23. (A) CH3 – Cl < CH3 – CH2 – Cl >

24. (A)

25. (A)

26. (D) Tertiary alkyl halides are the least reactive in SN2 reactions because bulky groups hinder the

approaching nucleophiles.

27. (C) In (a) Br is attached to vinylic carbon, hence SN reaction is not possible. In (b) and (d) elimination is

preferred (with rearrangement of carbocation) to substitution.

28. (B)

29. (B)

30. (B) In tertiary alkyl halides steric hindrance does not allow substitution by SN2 mechanism in which the

nucleophile attacks on the carbon atom and the reaction takes place in single step.

31. (B)

32. (D)

33. (A)

34. (C)

35. (A)

36. (A)

37. (D)

38. (D)

39. (C)

40. (B)

41. (D) It is an example of nucleophilic substitution reaction.



42. (C)

43. (B)

44. (A)

45. (C)

46. (C) The presence of an asymmetric C atom is not essential, e.g., allenes of the type (RRC = C = = CRR)

are optically active although they do not contain chiral c atoms.

47. (A) Since (A) and (B) are enantiomers, so specific rotation of B is –52o (because enantiomers have equal

and opposite specific rotation).

48. (A) Terminal groups are different. The number of asymmetric C atoms is four.

* * * *

2 2HOCH CH OH CH OH CH OH CH OH CH OH

49. (C)

obs

D lc

, where c is concentration in g/mL and l is length of the tube in dm (decimeter).

oo

D

1.216

1 0.075

50. (C)

2 5

*

3 3 7

C H|

H C C C H|H

(Total C = 7)

(optically active alkane)

51. (B) Although the meso compounds contain asymmetric C atoms, but they are optically inactive due to

superimposable mirror image or due to the presence of my element of symmetry.

52. (A) The separation of a racemic mixture is called resolution.

53. (B)

54. (B)

55. (C)

56. (C) The Fischer projection of (I) and (II) is wrong because the functional group is not at the top. On

rotating (I) and (II)by 180o, they are represented as shown.

Now compare I, II, III and IV

(I) and (III) Enantiomers



(II) and (IV) Identical or same

Therefore, the pairs of diastreomers are:

(I), (II); (I), (IV); and (II), (III).

Bu the answer is (C) (I, II)

57. (C) I and IV, both are symmetric molecules.

58. (D) For meso structure the 2R, 3S is identical with the 2S, 3R, it is not necessary to indicate the number

and thus R, S or S, R designation is also correct.

59. (B)

60. (D) An enantiomer can be converted into diastereomer by inverting two groups on the asymmetric carbon

atom.

61. (D) They do not have plane of symmetry (P.O.S)

62. (D) Compound gives Atropisomerism and is optically active (O.A) due to absence of plane of symmetry

(P.O.S) and (O.A) due to absence of plane of symmetry (P.O.S) and centre of symmetry (C.O.S).

63. (C) Structure I is charged to II by three (even number) interchange of groups, thus they should be enantiomers.

On the other hand, structure III is derived from I by two interchanges, thus they should be identical.

64. (D)

65. (C) Racemic mixture can be resolved by using optically active compounds.

66. (B)

67. (B)

68. (D)

69. (C)

70. (D)

71. (B)

72. (C) General order of SN2 reaction is 1o> 2o> 3o

73. (C)

74. (A)

75. (B)

76. (A)

77. (A) Inversion of configuration takes place in SN2 mechanism.

78. (B) In tertiary alkyl halides steric hindrance does not allow substitution by SN2 mechanism in which the

nucleophile attacks on the carbon atom and the reaction takes place in single step.

79. (C) Order of reactivity of alkyl halides towards SN1 reaction on the basis stability of intermediate

carbocations is 3o> 2o> 1o> CH3 – X.

80. (B)

81. (D) The reaction is intramolecular SN2 reaction.

82. (A) Order of reactivity of different halo compounds towards nucleophilic substitution reactions are :

allyl chloride > vinyl chloride > chlorobenezene.

83. (A)

84. (B)



85. (D) -carbonyl does not give SN1 reaction due to the presence of strong –I group on the carbon having

halogen.

86. (C)

87. (A)

88. (C)

89. (C)

90. (C)

91. (A)

92. (D)

93. (D)

94. (B)

95. (C)

96. (C)

97. (A)

98. (D)

99. (C)

100. (C)

101. (D)

102. (C) Due to inversion, reaction is SN2.

103. (C)

104. (D)

105. (C) Ethanolic KOH removes hydrogen and halogen atoms from a haloalkane, thus it is a dehydrohalogenating

reagent and gives the same type of reactions.



SOLUTIONS 1. (B)

2. (C)

3. (C) Primary and secondary alkyl chlorides are prepared from the respective alcohols by using HCl

gas and anhydrous ZnCl2 (Groove’s process).

Note :Tertiary alcohols are very reactive and hence they react readily with conc. HCl even in the

absence of ZnCl2.

4. (B) 6 5 2 6 5 2C H CH Cl KCN aq C H CH C N KCl

5. (A) alc. KOH

3 2 2 3 2CH CH CH Br CH CH CH

(A)

aq.KOH3 33 3

||

Propan-2-ol

CH CH CHCH CH CH

BrOH

6. (B)

7. (C)

8. (B) 22H /Ni

3 3 3 2 2Methyl chloride Methyl cyanide Ethyl amine (primary amine)

CH Cl KCN CH CN CH CH NH

9. (B) Straight chain alkyl halides have greater boiling point than their isomers.

10. (D) In polar aprotic solvent ionic compound NaCl precipitates out as product therefore substitution of

Cl takes place in preference of I .

11. (D) -

3

3

3|

CH O

3 2CH OH|

CH

H C C CH Br A

H

?

Alkyl halide is 1°.Keep in mind 1° halide give product by SN 2 / E – 2 mechanism and 1° halide

always gives substitution reaction except when strongly hindered base is used.

ex.: With

3|

( )

3|

3

CH

CH C O

CH

it gives mainly elimination.

The reaction involves carbocation intermediate.i.e.

3|

23|

(primary carbocation)

CH

CH C CH

H

but as it is a primary carbocation it will rearrange to give a tertiary carbocation, which completes the

reaction

3|

3|

3teritiary carbocation

CH

CH C

CH

CHEMISTRY BOOK-7 HOME ASSIGNMENT SOLUTIONS

HALOGEN CONTAINING ORGANIC COMPOUNDS

HBr



Stability of carbocation : 3° > 2° > 1° >3CH

It is because the stability of a charged system is increased by dispersal of the charge. The more stable

the carbocation, the faster it is formed.

N.B. – Rearrangement can be done in two ways.

12. (B) Aryl halides (e.g. C6H5Cl) do not hydrolysed by SN1 mechanism under ordinary conditions.

13. (C) .

: CN :

has ambidentate character because.

: CN :

has double attacking site

NaCN

Major MinorRCN + RNC.R I

14. (A) As a result of resonance, the carbon-chloride bond acquires some double bond character. Hence

vinyl

chloride does not undergo nucleophilic substitution reactions.

15. (B) Elimination of Br become easier due to the presence of carbonyl group.

16. (A) Chlorobenzene is less reactive than benzyl chloride

In chlorobenzene the lone pairs present on Cl atom get involved in resonance with electrons of

benzene due to which C-Cl bond acquires double bond character. Hence, reactivity decreases.

17. (B) 6 5 2 6 5 2C H CH Cl KCN C H CH C N KClaq

18. (A)

Ag – O – N = O is a covalent compound. Therefore, attack of nucleophile occurs through Nitrogen

atom. Hence, nitroethane is formed.

19. (B) They are polar but fail to form H-bonds with water.

20. (A)

21. (A)

22. (B) 3liq. NH

3 2 2 2 3 3196 KCH CH CHCl 2NaNH CH C CH 2NaCl 2NH

23. (A)



24. (C)

25. (B)

26. (A)

27. (C) Swartz reaction, 3 3CH Cl AgF CH F+AgCl

3 3CH Br AgF CH F AgBr

28. (A) If only aryl halide reacts with sodium in presence of ether, the reaction is called “Fittig” reaction.

29. (C) For the same alkyl group, the boiling points of alkyl halides decrease in the order:

RI > RBr > RCl > RF

This is because with the increase in size and mass of halogen atom, the magnitude of vander Waal’s

forces increases.

30. (D) /\

OH

RMgX HOH RH Mg

X

Here, 3 2 3RH CH CH CH

i.e., 2 2 3R CH CH CH

Hence the alkyl halide is propyl halide.

31. (B)

32. (B) According to Saytzeff’s rule, the major product will be that one which contains more number of

substituents around the double bond.

33. (C)


Cl– takes place in presence to I–.


Cl– takes place in presence to I–.

36. (A)

37. (A)

Order of dipole moment is : 3 2 2 3 4CH Cl CH Cl CHCl CCl

38. (B) Though polar, alkyl halides cannot form hydrogen bonds with water hence they are insoluble in

water.

39. (C) (CH3)3CCl> (CH3)2CHCH2Cl is not correct as boiling point of (CH3)3CCl is smaller than

(CH3)2CHCH2Cl. It is due to large surface area and shorter alkyl chain.

40. (A) ClCH2CH2CH2Br + KCN ClCH2CH2CH2CN + KBr

The order of reactivity is I Br C.l



41. (D) 3PI KCN Hydrolysis

3 3 3 3X Y Z

CH OH CH I CH CN CH COOH

42. (B)

43. (C) RX + Mg RMaX. Thus, Grignard reagent (RMgx) is formed by reaction of dry magnesium

(Mg) with alkyl halide (RX) in the presence of ether.

44. (D) C2H5Br + KCN C2H5CN+ KBr

Due to ionic nature, the attack by CN occurs through C atom and alkyl cyanide is formed.

45. (A)

46. (B) In haloarenes, carbon of C – X is sp2 hybridised while in haloalkanes it is sp3 hybridised.

47. (B)

48. (C)

NaOEt

2 2 3 EtOH

3

Cl CH CH CH CH|CH

2H

2 2 3 3 2 3catalyst

3 32-methylbutane

CH C CH CH CH CH CH CH

| |CH CH

49. (D)

It follows E2 mechanism

Hughes and Ingold proposed that biomolecular elimination reactions take place when the two

groups to be eliminated are transi.e. E2 reactions are stereo selectively trans.

50. (D) As C– I bond is weakest, (CH3)3C–I will undergo SN1 reaction most readily.

51. (B) Diethybromomethane is

5 4 3 2 1

2 2 33

Br|

H CH C CH CH CH

Its IUPAC name is 3-bromopentane.

52. (C)

3 3 3 2Chloromethane excess Methanamine

CH Cl NH CH NH HCl

Primary amine is obtained as a major product by taking large excess of ammonia.

53. (A) CH3CH2 – *

CH Br – CH3 2- Bromobutane

54. (B) The reactivity of alcohols towards halogen acids decreases in the other : 3o> 2o> 1o.

55. (C)

dry ether

3 3 3 3

X

CH CH CH Mg CH CH CH| |Br MgBr

2D O

3 3

Y

CH CH CH|OD

56. (A)

57. (D) 3 2 3 3

3 3

Br|

CH CH CH HBr CH C CH| |CH CH

2-bromo-2-methyl propane

58. (A)

59. (B)

60. (D)



61. (A)

As –CCl3 is a m-direction group.

62. (D) CHI3 gives a yellow ppt. of AgI.

63. (C)

64. (B)

Saytzeff's rule

3 2 3 3 3 2alc.

CH CH CH CH KOH CH CH CH CH KBr H O

65. (C) The correct order of nucleophilic substitution reactions

66. (B)

67. (D)

+2 4H O LiAlHKCN

3 3 3 3 2CH Br CH CN CH COOH CH CH OH(A) (B) (C)

68. (C) Swartz raction,

3 3CH Cl AgF CH F AgCl

3 3CH Br AgF CH F AgBr

69. (D) Its vapours are non-inflammable (i.e. do not catch fire).

Hence used as fire extinguishers under the name pyrene.

70. (C) Aromatic halides cannot be used in Williamson ether synthesis. However, if strong electron

withdrawing group at ortho and para-positions are present then synthesis may take place, e.g.,

71. (D) NBS

2 3 2 2CH CHCH CH CHCH Br

Allylic position AgCN

is affected 2 2CH CHCH NC

72. (C) KOH

3 3CH CHCl CH CHO

73. (D) Only 1o amine gives carbyl amine reaction.

74. (A)

75. (A)

76. (D)

77. (D)

78. (B)

79. (C) (a) CCl3CHO chloral

(b) CHCl3 chloroform

(C)



(D) CCl4carbontetrachloride

80. (C)

81. (B)

82. (B) When two molecules of alkyl halides (same or different) react with metallic sodium in the

presence of ether to form alkane, this reaction is called Wurtz reaction.

83. (D)

84. (B) CHCl3is stored in dark coloured bottles and properly stored to keep out air because in air because

in presence of air or light, it slowly oxidized to poisonous gas phosgene (carbonyl chloride).

CHCl3 +

Air or

2 2lightphosgene

1O COCl HCl

2

RNH2 + alc. KOH + CHCl3carbylamineRNC

85. (A)

86. (C) Chloroflourocarbon is used in air-conditioners and in domestic refrigerators for cooling purposes.

Its main drawback is this, it is responsible for ozone depletion.

87. (C)

88. (B) Haloform compounds with the formula CHX3, where X is a halogen atom.

Haloforms are trihalogen derivatives of methane.

Example: Chloroform CHCl3.

89. (C)

90. (B)

91. (D) On sulphonation of chlorobenzene, ortho and para chlorobenzene is formed because –Cl group is

para and ortho directing.



92. (B) This method is not applicable for the preparation of aryl halides because the C–O bond in phenol

has a partial double bond character and is difficult to break being stronger than a single bond.

93. (D) Chloropicrin is nitrochloroform. It is obtained by the nitration of chloroform with HNO3.

3HNO

3 2 3Chloroform Chloropicrin

HCCl O NCCl

Chloropicrin is a liquid, poisonous and used as an insecticide and a war gas.

94. (A)

95. (D) Para-di chlorobenzene has most symmetrical structure than others. It is found as crystalline

lattice form, therefore, it has highest melting point (52oC) due to symmetrical structure.

96. (D)

97. (A) –CH3 group is o, p-directing.

98. (B)

99. (A) The presence of an electron withdrawing group (–NO2) at ortho and para-positions increase the

reactivity of haloarenes.

100. (B) In haloarenes, carbon of C–X is sp2 hybridised while in haloalkanes it is sp3 hybridised.

101. (C) Chloroform gets oxidized by air in sunlight to poisonous phosgene gas.

3 2 22CHCl O 2COCl 2HClh

102. (B)

103. (B)

104. (D)

105. (B) During chlorination of benzene, anhydrous AlCl3, being a Lewis acid helps in generation of the

electrophile Cl+ by combining with the attacking reagent.

The electrophile Cl+ attacks the benzene ring in this reaction.

Documents

HOME ASSIGNMENT SOLUTIONS GENERAL ORGANIC · PDF fileNo. of asymmetric or chiral carbon = 2; ... Therefore two forms are possible. 13. (D) 3 3 H | CH C ... The separation of a racemic