NEET Chemistry

Nomenclature Of Organic Compounds

Question 1. The IUPAC name of the compound (CH₃)₂CH-CH=CH-CHOH-CH₃ is

1. 5-methyl-hex-3-en-2-ol
2. 2-methyl-hex-3-en-5-ol
3. 2-hydroxy-5-nethyl-3-hexene
4. 5-hydroxy-2-methyl-3-hexene

Answer: 1. 5-methyl-hex-3-en-2-ol

Solution: Follow IUPAC rules.

Read And Learn More: NEET General Organic Chemistry Notes, Question And Answers

Question 2. Which of the following represents the condensed formula for pentanes?

1. CH₃(CH₂)₃CH₃
2. (CH₃)₃CCH₃
3. (CH₃)₂CHCH₂ CH₃
4. All of these

Answer: 4. All of these

Solution: In condensed structural formula, parenthesis is used for the identical groups of atoms.

Question 3. Which is the most stable carbocation?

1. isopropyl cation
2. Triphenylmethyl cation
3. Ethyl cation
4. n-propyl cation

Answer: 2. Triphenylmethyl cation

Solution: Aryl carbonium ions are more stable than the alkyl carbonium ions. The order of stability of carbocation is

Triphenyl methyl >Diphenyl methyl> Benzyl>Allyl>30>2°>1°> methyl carbocation.

Question 4. Which of the following is phenyl ethanoate?

Answer: 3

Solution: Esters are named by prefixing the name of the alkyl or aryl group (of OR’ part) before the name of the parent acid and changing the suffix “ic” acid to ate. Hence, the structure of phenyl ethanoate is

Question 5. IUPAC name of CH₂=CH—CH(CH₃)₂ is:

1. 1,1-dimethyl-2-propene
2. 3-methyl-1-butene
3. 2-vinyl propane
4. 1-isopropyl ethylene

Answer: 2. 3-methyl-1-butene

Solution: Follow IUPAC rules.

Question 6. In the structure, the number of carbon atoms are:

1. One primary, two secondary and one tertiary
2. Four primary, two tertiary and one secondary
3. One primary, one secondary, one tertiary and one quaternary
4. Five primary, one secondary, one tertiary and one quaternary

Answer: 4. Five primary, one secondary, one tertiary and one quaternary

Solution: A primary carbon is one which is joined to one carbon atom. A secondary carbon atom is joined to two carbon atoms and the tertiary carbon is attached to three carbon atoms. A quaternary carbon has all its four valencies attached to carbon atoms.

Question 7. IUPAC name of the following compound is

1. 3, 5-dimethylcyclohexene
2. 3, 5-dimethyl-1-cyclohexene
3. 1, 5-dimethyl-5-cyclohexene
4. 1, 3-dimethyl-5-cyclohexene

Answer: 1. 3, 5-methylcyclohexane

Solution: Its IUPAC name is 3,5-dimethylcyclohexene.

Question 8. IUPAC name ofis

1. N, N- dimethylethanolamine
2. N-methyl, N-ethylmethanamine
3. Dimethyl-ethylamine
4. None of the above

Answer: 1. N, N- dimethylethanolamine

Solution: Follow IUPAC rules.

Question 9. The IUPAC name of CH₃-CH=CH-C=CH is

1. Pent-3-en-1-yne
2. Pent-3-en-4-yne
3. Pent-2-en-4-yne
4. Pent-2-en-3-yne

Question 10.

1. 1-amino prop-2-enal
2. 3-amino prop-2-enal
3. 1-amino-2-formyl ethene
4. 3-amino-1-oxoprop-2-ene

Answer: 2. 3-amino prop-2-enal

Solution: Follow IUPAC rules.

Question 11. The IUPAC name of an unsymmetrical ether with the molecular formula of C₄H₁₀O is

1. Ethoxy propane
2. Methoxy ethane
3. Ethoxy ethane
4. Methoxy propane

Answer: 4. Methoxy propane

Solution: Only two unsymmetrical ethers are possible from the formula C₄H₁₀O viz

CH₃ CH₂ CH₃-O-CH₃

Question 12. IUPAC name of urea is:

1. Diamino ketone
2. 1-aminoethanamide
3. 1-aminomethanamide
4. amino acetamide

Question 13. IUPAC name of C₆ H₅ CN is:

1. Phenyl nitrile
2. Benzene nitrile
3. Benzyl nitrile
4. Phenyl cyanide

Answer: 2. Benzene nitrile

Solution: Follow IUPAC rules.

Question 14. The IUPAC name of the compound is

1. 2(carboxymethyl)-pentane-1,5-dioic acid
2. 3-carboxy hexane-1, 6-dioic acid
3. Butane-1, 2, 4-tricarboxylic acid
4. 4-carboxy hexane-1, 6-dioic acid

Answer: 2. 3-carboxy hexane-1, 6-dioic acid

Solution: IUPAC name of the above compound is 3-carboxyhexane-1, 6-dioic acid.

Question 15. IUPAC name of

1. Dimethyl amine
2. 2-amino propane
3. Isopropyl amine
4. 2-propanamide

Answer: 4. 2-propanamide

Solution: 2-propanamide

Question 16. The IUPAC name of

1. 1, 1-dimethyl-1, 3-butanediol
2. 2-methyl-2, 4-pentanediol
3. 4-methyl-2, 4-pentanediol
4. l, 3, 3-trimethyl-1, 3-propanediol

Answer: 2. 2-methyl-2, 4-pentanediol

Solution: Follow IUPAC rules.

Question 17. The IUPAC name of the compound is

1. 2,4-dimethylhexanone-3
2. 2,6-dimethylheptanone-4
3. 2,6-dimethylhexanone-4
4. 2,6-dimethylheptanone-5

Answer: 2. 2,6-dimethylheptanone-4

Solution: Follow IUPAC rules.

Question 18. Lactic acid is:

1. Propionic acid
2. ^-hydroxy propanoic acid
3. a-hydroxy propanoic acid
4. None of the above

Answer: 3. a-hydroxy propanoic acid

Solution: Lactic acid is also called as a-hydroxy propionic acid or 2-hydroxy propanoic acid.

Question 19. The compound which contains all the four 1°, 2°, 3° and 4° carbon atoms is

1. 2, 3-dimethyl pentane
2. 3-chloro-2, 3-dimethyl pentane
3. 2, 3, 4-trimethylpentane
4. 3, 3-dimethyl pentane

Answer: 2. 3-chloro-2, 3-dimethyl pentane

Solution: 3-chloro-2, 3-dimethyl pentane contains all the four 1°,2°, 3° and 4° carbon atoms.

Question 20. The name formic acid was given for HCOOH because it was prepared from:

1. Acetum
2. Ant
3. Wood
4. Oxalis plant

Answer: 2. Ant

Solution: Formic acid was obtained from ant (Formica in Greek). This is the trivial name for HCOOH.

Question 21. The IUPAC name of the compound having the formula CH= C — CH=CH₂ is:

1. 1-butene-3-yne
2. 3-butene-1-yne
3. 1-butyn-3-ene
4. But-1-yne-3-ene

Answer: 1. 1-butene-3-yne

Solution: Follow IUPAC nomenclature.

Question 22. The IUPAC name of the below compound is

1. 1- cyclohexane-2,4-dienylethanone
2. 3- cyclohexane-2,4-dienylethanone
3. 1- cyclohexane-3,5-dienylethanone
4. 3- cyclohexane-3,5-dienylethanone

Answer: 1. 1- cyclohexane-2,4-dienylethanone

Solution: Follow IUPAC nomenclature.

Question 23. The IUPAC name of the below compound is

1. 1-methyl-1-amino propane
2. 2-amino butane
3. 2-methyl-3-amino propane
4. None of the above

Answer: 2. 2-amino butane

Solution: Follow IUPAC rules.

Question 24. The correct IUPAC name of (C₂ H₅)₄ C is:

1. Tetraethyl methane
2. 2-ethyl pentane
3. 3,3-diethyl pentane
4. None of these

Answer: 3. 3,3-diethyl pentane

Solution: Follow IUPAC rules.

Question 25. Which IUPAC name is wrong?

Answer: 3

Solution: It is 3-methyl butan-2-ol.

Question 26. The structural formula of 2,2,3-trimethyl hexane is:

Answer: 3

Solution: Follow IUPAC rules.

Question 27. IUPAC name of acraldehyde is

1. But-3-en-1-al
2. Propenyl aldehyde
3. But-2-ene-1-al
4. Prop-2-en-1-al

Answer: 4. Prop-2-en-1-al

Solution: IUPAC name of acraldehyde

Question 28. The IUPAC name of the compound is

1. 2-chloro-4-methyl-2-pentene
2. 4-chloro-2-methyl-3-pentene
3. 4-methyl-2-chloro-2-pentene
4. 2-chloro-4, 4-dimethyl-2-butene

Answer: 1. 2-chloro-4-methyl-2-penton

Solution: Follow IUPAC rules.

Question 29. 2-methyl pent-3-ene is achiral because it has:

1. A centre of symmetry
2. A plane of symmetry
3. Symmetry at C₂ carbon
4. Both centre and a plane of symmetry

Answer: 4. Both centre and a plane of symmetry

Solution: If the structure of the compound is drawn then we find that a plane of symmetry is present in this compound. The two halves are equal hence the compound does not have a chiral point, in addition to this it has a Centre. The two sides of the compound on the different sides of the centre atom are similar. Both the reasons are present which signifies that the molecule is achiral.

Question 30. The IUPAC name of the compound is

1. 3,4-dimethyl-3-n-propyl nonane
2. 4-ethyl-4,5-dimethyl decane
3. 6,7-dimethyl-7-n-propyl nonane
4. 6,7-dimethyl-7-ethylene

Answer: 2. 4-ethyl-4,5-dimethyl decane

Solution: Follow IUPAC rules.

Question 31. Give the correct IUPAC name for

1. 2-ethoxy-5-chloroethane
2. l-chloro-4-ethoxy-4-methyl butane
3. 1-chloro-4-ethoxy pentane
4. Ethyl-1-chloropentylether

Answer: 1. 2-ethoxy-5-chloroethane

Solution: According to the IUPAC system, ether is named as alkoxy alkane. The larger alkyl group forms the parent chain while the lower alkyl group takes in ethereal oxygen and forms a part of the alkoxy group.

Question 32. The IUPAC name of the compound, CH₃ CH = CHC = CH is:

1. Pent-4-yn-2-ene
2. Pent-3-en-1-yne
3. Pent-2-en-4-yne
4. Pent-1-yn-3-ene

Answer: 2. Pent-3-en-1-yne

Solution: Follow IUPAC rules.

Question 33. The IUPAC name of C₆ H₅ COCl is

1. Benzoyl chloride
2. Benzene chloro ketone
3. Benzene carbonyl chloride
4. Chloro phenyl ketone

Answer: 1. Benzoyl chloride

Solution: Follow IUPAC rules.

Question 34. IUPAC name of the compound

1. 5- methyl-4-isopropyl-6, 6’diethyloctane
2. 3, 3-dimethyl, 3-ethyl-5- isopropyl octane
3. 3, 3-diethyl-4-methyl-5-(1,1-dimethyl) octane
4. 3, 3- diethyl-4-methyl-5-(1’-methylethyl) octane

Answer: 4. 3, 3- diethyl-4-methyl-5-(1’-methyl ethyl) octane

Solution: Select the longest possible carbon atom chain, number it and name the compound according to IUPAC rules.

Question 35. The IUPAC name of is

1. But-3-enoic acid
2. But-1-enoic acid
3. Pent-4-enoic acid
4. Prop-2-enoic acid

Question 36. The correct IUPAC name of the below molecule is

1. Isopropyl benzene
2. Cumene
3. Phenyl isopropanol
4. 2-phenyl propane

Answer: 4. 2-phenyl propane

Solution: Follow IUPAC rules.

Question 37. The IUPAC name of the compound is

1. 2-methyl pent-1-en-4-yne
2. 4-methyl pent-4-en-1-yne
3. 2-methyl pent-2-en-4-yne
4. 4-methyl pent-1-en-4-yne

Answer: 1. 2-methylpent-1-en-4-yne

Solution: Follow IUPAC rules.

Question 38. Freon-114 is an organic compound. k is chemically called 1,2dichlorotetrafluoroethane. Its correct structural formula is:

Answer: 3

Solution: Follow IUPAC rules.

Question 39. The IUPAC name of the following compound is

1. Propionic anhydride
2. Di propanoic anhydride
3. Ethoxy propanoic acid
4. Propanoic anhydride

Answer: 4. Propanoic anhydride

Solution: In the IUPAC system, anhydrides are named as alkanoic anhydride.

Question 40. IUPAC name of the compound is

1. 2-ethyl-3-methyl-hexa-l-en-4-yne
2. 5-ethyl-4-methyl-hexa-2-yn-5-ene
3. 3-methylene-4-methylhepta-5-yne
4. 5-methylene-5-ethyl-4-methylhepta-2-yne
5. Answer: 1. 2-ethyl-3-methyl-hexa-l-en-4-yne

Solution:-do-

Question 41. Bicyclo (1,1,0) butane is

Answer: 3

Its IUPAC name is bicyclo [1,1,0] butane.

Question 42. t-butyl alcohol is

1. 2-methyl propane-2-ol
2. 2-methyl propane-1-ol
3. 3-methyl butane-1-ol
4. 3-methyl butane-2-ol

Answer: 1. 2-methyl propan-2-ol

Solution: 2-methyl propan-2-ol is tert-butyl alcohol.

Question 43. IUPAC name of C₆ H₅ COCl is:

1. Benzal chloride
2. Benzenechloro ketone
3. Benzene carbonyl chloride
4. Chloro phenyl ketone

Answer: 3. Benzene carbonyl chloride

Solution: Follow IUPAC rules.

Question 44. The IUPAC name for tertiary butyl iodide is

1. 4-iodo butane
2. 2-iodo butane
3. 1-iodo-3-methyl propane
4. 2-iodo-2-methylpropane

Answer: 4. 2-iodo-2-methylpropane

Solution: Its IUPAC name is 2-iodo-2-methyl propane.

Question 45. The correct structure of dimethyl butane is:

1. CH₃CH₃—C=C—CH₂ CH₃
2. (CH₃)₃C —C —C≡CH
3. CH₃—C=CCH(CH₃)₂
4. 

Answer: 2. (CH₃)₃C —C —C≡CH

Question 46. Glyoxal is

1. CH₂OH-CH₂OH
2. CHO-CH₂OH
3. COOH-CO-COOH
4. CHO-CHO

Answer: 4. CHO-CHO

Solution:

Question 47. Give the IUPAC name for,

1. Ethyl-4- oxo heptanoate
2. Methyl-4- oxo heptanoate
3. Ethyl-4- oxo hexanoate
4. Methyl 4- oxo hexanoate

Answer: 4. Methyl 4- oxo hexanoate

Solution: Follow IUPAC rules.

Question 48. The IUPAC name of the compound,

1. 2-ethylbutanamide
2. 2-methylbutanamide
3. 1-amino-2-methylpropane
4. None of the above

Answer: 2. 2-methylbutanamide

Solution: Follow IUPAC rules.

Question 49. The IUPAC name of CH₃ COCH(CH₃)₂ is

1. Isopropyl methyl ketone
2. 2-methyl-3-butanone
3. 4-methyl isopropyl ketone
4. 3-methyl-2-butanone

Answer: 4. 3-methyl-2-butanone

Solution:

The keto functional group is given priority.

Question 50. The IUPAC name of the following compound is

1. Bicyclo [2,2,0] octane
2. Bicyclo [0,2,2] hexane
3. Bicyclo [2,1,1] hexane
4. Bicyclo [2,2,0] hexane

Answer: 4. Bicyclo [2,2,0] hexane

Solution: Hence, the correct IUPAC name is bicyclo [2,2,0] hexane.

Question 51. The IUPAC name of the compound,

1. 1,2-dimethyl-2-butenol
2. 3-methylpent-3-en-2-ol
3. 3,4-dimethyl-2-buten-4-ol
4. 2,3-dimethyl-3-pentenol

Answer: 2. 3-methyl pent-3-en-2-ol

Solution: Follow IUPAC rules.

Question 52. IUPAC name of (CH₃)₃ CCl is

1. n-butyl chloride
2. 3-chloro butane
3. 2-chloro 2-methyl propane
4. t-butyl chloride

Answer: 3. 2-chloro 2-methyl propane

Solution: Follow IUPAC rules.

Question 53. What is the correct IUPAC name of the compound?

1. 4-methoxy-2-nitrobenzaldehyde
2. 4-formyl-3-nitro anisole
3. 4-methoxy-6-nitrobenzaldehyde
4. 2-formyl-5-methoxy nitrobenzene

Answer: 1. 4-methoxy-2-nitrobenzaldehyde

Solution: Follow IUPAC rules.

Question 54. What is the correct IUPAC name of the compound?

1. 5-methyl hexanol
2. 2-methyl hexanol
3. 2-methylhex-3-enol
4. 4-methyl pent-2-enol

Answer: 4. 4-methyl pent-2-enol

Solution: —do—

Question 55. The IUPAC name of the compound is:

1. But-2-ene-2,3-diol
2. Pent-2-ene-2,3-diol
3. 2-methylbut-2-ene-2,3-diol
4. Hex-2-ene-2,3-diol

Answer: 2. Pent-2-ene-2,3-diol

Solution: Follow IUPAC rules.

Question 56. Which nomenclature in IUPAC is incorrect?

1. Pentyne-3
2. Pentyne-2
3. Hexyne-3
4. None of these

Answer: 1. Pentyne-3

Solution: CH₃ C = CCH₂ CH₃; It is always pentyne-2 and never pentyne-3.

Question 57. The IUPAC name of the compound CH₃ CONHBr is

1. 1-bromoacetate
2. Ethanoylbromide
3. N-Bromoethanamide
4. None of these

Answer: 3. N-Bromoethanamide

Solution: Follow IUPAC rules.

Question 58. The IUPAC name of the compound is:

1. 3-ethyl-4,4-dimethyl heptane
2. 1,1-diethyl-2,2-dimethyl pentane
3. 4,4-dimethyl-5,5-diethyl pentane
4. 5,5-diethyl-4, 4-dimethyl pentane

Answer: 1. 3-ethyl-4,4-dimethyl heptane

Solution: Follow IUPAC rules.

Question 59. The correct IUPAC name of the compound is:

1. 3-(1-ethyl propyl) hex-1-ene
2. 4-Ethyl-3-propyl hex-1-ene
3. 3-Ethyl-4-ethenyl heptane
4. 3-Ethyl-4-propyl hex-5-ene

Answer: 2. 4-Ethyl-3-propyl hex-1-ene

Solution: Follow IUPAC rules.

Question 60. The IUPAC name of the following compound is

1. 4-bromo-3-cynophenoal
2. 2-bromo-5-hydroxy benzonitrile
3. 2-cyano-4-hydroxybromobenzene
4. 6-bromo-3-hydroxy benzonitrile

Answer: 2. 2-bromo-5-hydroxy benzonitrile

Solution: The Cyano group has the highest priority therefore, parent name must be benzonitrile. Br occurs at 2- position, and hydroxyl at 3-position, hence the IUPAC name is 2-bromo-5-hydroxy benzonitrile.

Question 61. IUPAC name of the compound is

1. 2-chloromethyl-4-methyl-hexanal
2. 1-chloro-4-ethyl-2-pentanal
3. 1-chloro-4-methyl-2-hexanal
4. 1-chloro-2-aldo-4-methyl hexane

Answer: 1. 2-chloromethyl-4-methyl-hexanal

Solution: Follow IUPAC rules.

Question 62. Which one is not in the IUPAC system?

Answer: 2

Solution: It should be 4-ethyl-3-methyl heptane.

Question 63. The correct IUPAC name is

1. 1- cyclopropyl Cyclobutane
2. 1, 1-cyclobutane
3. 1- cyclobutane-1- cyclopropane
4. None of the above

Answer: 1. 1- cyclopropyl Cyclobutane

Solution: Follow IUPAC rules.

Question 64. The IUPAC name of the compound

1. 3, 3-dimethyl-1-hydroxy cyclohexane
2. 1, 1-dimethyl-3- hydroxy cyclohexane
3. 3, 3- dimethy-1- cyclohexanol
4. 1,1-dimethyl-3-cyclohexanol

Answer: 3. 3, 3- dimethyl-1- cyclohexanol

Solution: Carbon with – OH group is given C₁ thus it is 3, 3-dimethyl-1-cyclohexanol.

Question 65. The IUPAC name of the compound is

1. But-2-en-l-ol
2. l-hydroxy but-l-ene
3. 4-hydroxy butene-3
4. But-l-en-l-ol

Answer: 4. But-l-en-l-ol

Solution: Follow IUPAC rules.

Question 66. The name of (CH₃)₂ HC—O—CH₂—CH₂—CH₃ is:

1. Isopropyl propyl ether
2. Dipropyl ether
3. Di-isopropyl ether
4. Isopropyl propyl ketone

Answer: 1. Isopropyl propyl ether

Solution: Ether group (—O—) has propyl and isopropyl groups on its two sides.

Question 67. IUPAC name of, CH₃ CH(OH) CH₂ CH₂ COOH is:

1. 4-hydroxy pentanoic acid
2. 1-carboxy-3-butanoic acid
3. 1-carboxy-4-butanol
4. 4-carboxy-2-butanol

Answer: 1. 4-hydroxy pentanoic acid

Question 68. The prefix name of —SH group in IUPAC system is:

1. Mercapto
2. Thiol
3. Sulphide
4. None of these

Answer: 1. Mercapto

Solution: Follow IUPAC rules.

Question 69. The correct structure of 4-bromo-3-methyl-but-1-ene.

1. Br-CH=C(CH₃)₂
2. CH₂=CH-CH(CH₃)-CH₂ Br
3. CH₂=C(CH₃)CH₂ CH₂ Br
4. CH₃-C(CH₃)=CHCH₂-Br

Answer: 2. CH₂=CH-CH(CH₃)-CH₂ Br

Solution: Follow IUPAC rules.

Question 70. Which of the following represents neo-pentyl alcohol?

1. CH₃CH(CH₃)CH₂CH₂OH
2. (CH₃)₃C-CH₂OH
3. CH₃(CH₂)₃ OH
4. CH₃CH₂CH(OH)C₂H₅

Answer: 2. (CH₃)₃C-CH₂OH

Solution: (CH₃)₃C-CH₂OH is neo-pentyl alcohol.

Question 71. IUPAC name of the compound, ClCH₂ CH₂ COOH is:

1. 3-chloropropanoic acid
2. 2-chloropropanoic acid
3. 2-chloroethanoic acid
4. Chlorosuccinic acid

Answer: 1. 3-chloropropanoic acid

Solution: Follow IUPAC rules.

Question 72. Which of the following IUPAC names is correct?

1. 2-methyl-3-ethyl pentane
2. 2-ethyl-3-methyl pentane
3. 3-ethyl-2-methyl pentane
4. 3-methyl-2-ethyl pentane

Answer: 3. 3-ethyl-2-methyl pentane

Solution: -do-

Question 73. IUPAC name of, (C₂H₅)₂CHCH₂OH is:

1. 2-ethyl butanol-1
2. 2-methyl pentanol-1
3. 2-ethyl pentanol-1
4. 3-ethyl butanol-1

Answer: 1. 2-ethyl butanol-1

Solution: Follow IUPAC rules.

Question 74. 3-phenylpropanoid acid is IUPAC name of:

1. Mendaleic acid
2. Pyruvic acid
3. Succinic acid
4. Cinnamic acid

Answer: 4. Cinnamic acid

Solution: C₆ H₅—CH=CHCOOH is cinnamic acid.

Question 75. IUPAC name of the compound is:

1. 1,2,3-trieyanopropane
2. Propane-1,2,3-tricarbonitrile
3. 1,2,3-cyclopropane
4. Propane triarylamine

Answer: 2. Propane-1,2,3-tricarbonitrile

Solution: Follow IUPAC rules.

Question 76. IUPAC name of the compound is

1. 4-isopropyl, 6-methyl octane
2. 3-methyl, 5-(1-methyl ethyl) octane
3. 3-methyl, 5-isopropyl octane
4. 6-methyl, 4-(1-methyl ethyl) octane

Answer: 2. 3-methyl, 5-(1-methyl ethyl) octane

Solution: IUPAC name of compound.

Straight chain which contains a large number of side chains is taken as the parent chain and the counting starts from that side where the side chain is nearest.

Question 77. IUPAC name of the compound is:

1. 4-bromo-3-ethyl-1,4-pentadiene
2. 2-bromo-3-ethyl-1,4pentadiene
3. 2-bromo-3-ethyl-1-5-pentadiene
4. None of the above

Answer: 2. 2-bromo-3-ethyl-1,4pentadiene

Solution: The IUPAC name of this molecule is 2-bromo-3-ethyl-1, 4-pentadiene.

Question 78. IUPAC name of the compound is:

1. 4,5-dimethyl oct-4-ene
2. 3,4-dimethyl oct-5-ene
3. 4,5-dimethyl oct-5-ene
4. None of the above

Answer: 1. 4,5-dimethyl oct-4-ene

Solution: Follow IUPAC rules.

Question 79. The correct IUPAC name of the following compound is

1. 5, 6-dimethyl-8-methyl dec-6-ene
2. 6-butyl-5-ethyl-3-methyl oct-4-ene
3. 5, 6-diethyl-3-methyl dec-4-ene
4. 2, 4, 5-triethyl non-3-ene

Answer: 3. 5, 6-diethyl-3-methyl dec-4-ene

Question 80. Which of the following compounds has incorrect IUPAC nomenclature?

Answer: 4

Question 81. The IUPAC name of the compound is

1. 1,2,3-tricarboxylic-2,1-propane
2. 3-carboxy-3-hydroxy-1,5-pentanoic acid
3. 3-hydroxy-3-carboxy-1,5-pentanoic acid
4. None of the above

Answer: 2. 3-carboxy-3-hydroxy-1,5-pentanoic acid

Solution: Follow IUPAC rules.

Question 82. IUPAC name of is

1. Bicyclo (2,1,0) pentane
2. 1,2-cyclopropyl Cyclobutane
3. Cyclopentane (4,3) annulene
4. 1,2-methylene Cyclobutane

Answer: 1. Bicyclo (2,1,0) pentane

Solution: Follow IUPAC rules.

Question 83. The IUPAC name of neopentane is:

1. 2,2-dimethylpropane
2. 2-methylpropane
3. 2,2-dimethylbutane
4. 2-methyl butane

Answer: 1. 2,2-dimethylpropane

Solution: Follow IUPAC rules.

Question 84. The IUPAC name of compound shown below is

1. 2-bromo-6- chlorocyclohex-1-ene
2. 6-bromo-2-chlorocyclohexene
3. 3-bromo-1-chlorocyclohexene
4. 1-bromo-3-chlorocyclohexene

Answer: 3. 3-bromo-1-chlorocyclohexene

Solution:

Unsaturation (double bond) is given priority over halogen, then the lowest set of locants. So, the correct IUPAC name is 3-bromo-1-chlorocyclohexene.

Question 85. The IUPAC name of 

1. 4-hydroxy-1-methylpentanal
2. 4-hydroxy-4-methyl pent-2-en-1-al
3. 2-hydroxy-4-methyl pent-2-en-5-al
4. 2-hydroxy-3-methyl pent-2-en-5-al

Answer: 2. 4-hydroxy-4-methyl pent-2-en-1-al

Solution: Follow IUPAC rules.

Question 86. IUPAC’s name of is 

1. 3-chlorobutanol
2. 3-chlorobutanaldehyde
3. 3-chlorobutanol
4. 2-chlorobutanol

Answer: 3. 3-chlorobutanol

Solution:

The order is priority is -CHO > -Cl

Question 87. The IUPAC name of is 

1. 2-methyl-3-butanone
2. 3-methyl-butan-2-one
3. 3-methyl butanone
4. None of these

Answer: 2. 3-methyl-butane-2-one

Solution: Follow IUPAC rules.

Question 88. The IUPAC name of compound is

1. 2-ethyl prop-2-en-1-ol
2. 2-hydroxymethylbutan -1-ol
3. 2-methylenebutan-1-ol
4. 2-ethyl-3-hydroxypropyl-1-ene

Answer: 1. 2-ethyl prop-2-en-1-ol

Solution:

Question 89. Which nomenclature is not according to the IUPAC system?

Answer: 3

Solution: CH₂=CH-CH₂-Br is allyl bromide. Its IUPAC name is 3-bromopropene or 3-bromoprop-1-ene. Because here double bond is a functional group while halogen is a substituent. Double bond is given more priority and hence numbering starts from left.

Question 90. Acetonitrile is

1. CH₃CN
2. CH₃COCN
3. C₂H₅CN
4. C₆H₅CN

Answer: 1. CH₃CN

Solution: CH₃ C=N is known as acetonitrile or methyl cyanide.

Question 91. Isobutyl chloride is:

1. CH₃CH₂CH₂CH₂Cl
2. (CH₃)₂CHCH₂ Cl
3. CH₃CH₂CHClCH₃
4. (CH₃)₃C-Cl

Answer: 2. (CH₃)₂CHCH₂ Cl

Solution: It should contain (CH₃)₂CH- group to be named as iso.

Question 92. The given compound in IUPAC is called as

1. Diacetone
2. Acetone amine
3. Diacetone amine
4. 4-amino-4-methyl pentane-2-one

Answer: 4. 4-amino-4-methyl pentane-2-one

Solution: Follow IUPAC rules.

Question 93. The IUPAC name of compound is

1. 1-choloro-2,3-epoxypropane
2. 3-chloro-1,2-epoxypropane
3. 1-chloroethoxymethane
4. None of the above

Answer: 2. 3-chloro-1,2-epoxypropane

Solution: Follow IUPAC rules.

Question 94. Neo-heptyl alcohol is correctly represented as:

Answer: 3

Solution: Follow IUPAC rules.

Question 95. IUPAC name of the compound is CH₃ CH₂ C(Br)=CH-Cl

1. 2-bromo-1-chloro butene-1
2. 1-chloro-2-bromo butene-1
3. 3-chloro-2-bromo butene-2
4. None of the above

Answer: 1. 2-bromo-1-chloro butene-1

Solution: Follow IUPAC rules.

Question 96. The IUPAC name of CH₃-CH₂-CHO is

1. Propanal-1
2. 2-methyl butanal
3. Butanal-1
4. Pentanal-1

Answer: 1. Propanal-1

Solution: IUPAC name of CH₃ CH₂ CHO is propan-1-al.

Question 97. The IUPAC name of is

1. 4-hydroxy-1-methylpentanal
2. 4-hydroxy-2-methylpentanal
3. 3-hydroxy-2-methylpentanal
4. 3-hydroxy-3-methylpentanal

Answer: 2. 4-hydroxy-2-methylpentanal

Solution: Follow IUPAC rules.

Question 98. Select the correct statement:

1. The prefixes are written before the name of compound
2. The suffixes are written after the name of compound
3. The IUPAC name of a compound is always written as one word
4. All of the above

Answer: 4. All of the above

Solution: These are the IUPAC rules.

Question 99. Pick out the correct statement from the following and choose the correct answer from the codes given below

1. Hexa-1, 5-diene is a conjugated diene
2. Prop-1, 2-diene is conjugated diene
3. Hexa-1, 3-diene is a conjugated diene
4. Buta-1, 3-diene is an isolated diene
5. Prop-1, 2-diene is a cumulative diene

 

1. 1, 2
2. 2, 3
3. 4, 5
4. 2, 5

Answer: 3. 4, 5

Solution: In conjugated diene alternate single and double bonds are present while in cumulative diene, double bonds are present at adjacent positions.

Hence, statements 3 and 5 are correct.

Question 100. The IUPAC name for

1. 1,1-dimethyl-1,2-butanediol
2. 2-methyl-2,4-pentanediol
3. 4-methyl-2,4-pentanediol
4. 1,3,3-dimethyl-1,3-propanediol

Answer: 2. 2-methyl-2,4-pentanediol

Solution: -do-

Question 101. Formulae of phenyl carbinol and chloral are respectively:

1. C₆ H₅.CH₂ CH₂ OH and CHCl₂ CHO
2. C₆ H₅ CH₂ OH and CCl₃ CHO
3. C₆ H₅ OH and CH₂ Cl.CHO
4. C₆ H₅ CHO and CHCl₂ CHO

Answer: 2. C₆ H₅ CH₂ OH and CCl₃ CHO

Solution: Carbinol is trivial name for HCH₂ OH. Thus, C₆ H₅ CH₂ OH is phenyl carbinol and chloral is CCl₃ CHO.

Question 102. The correct name for the following hydrocarbon is

1. Tricycle [4.1.0] heptane
2. Bicyclo [5.2.1] heptane
3. Bicyclo [4.1.0] heptane
4. Bicyclo [4.1.0] hexane

Answer: 3. Bicyclo [4.1.0] heptane

Solution:

This compound contains 7 carbon atoms, so the corresponding alkane is heptane. Two bridges contain 4 and 1 carbon atom respectively and one bridge does not contain any carbon atom. So, the name of the compound is Bicyclo (4,1,0) heptane.

Question 103. The IUPAC name of the compound is:

1. 5-chloro-hex-2-ene
2. 2-chloro-hex-5-ene
3. l-chloro-1-methyl-pent-3-ene
4. 5-chloro-5-methyl-pent-2-ene

Answer: 1. 5-chloro-hex-2-ene

Solution: First the longest continuous chain of carbon atoms is selected. Number the chain from the side containing senior functional group (i.e., the group placed above in the seniority table).

Question 104. The IUPAC name of the compound is:

1. 1,1-diethyl1-2, 2-dimethyl pentane
2. 4,4-dimethyl-5,5-diethyl pentane
3. 5, 5-diethyl-4,4-dimethyl pentane
4. 3-ethyl-4,4-dimethyl heptane

Answer: 4. 3-ethyl-4,4-dimethyl heptane

Solution: -do-

Question 105. 2, 3-dimethyl hexane contains _______ tertiary ____ secondary and ______ primary carbon atoms, respectively

1. 2, 2, 4
2. 2, 4, 3
3. 4, 3, 2
4. 3, 2, 4

Answer: 1. 2, 2, 4

Solution: The structure of 2, 3-dimethyl hexane is

So, the number of tertiary carbon atoms = 2

The number of secondary carbon atoms = 2

The number of primary carbon atoms = 4

Question 106. IUPAC name of following compound is:

1. 2-cyclohexyl butane
2. 2-phenylbutazone
3. 3-cyclohexyl butane
4. 3-phenylbutazone

Answer: 2. 2-phenylbutazone

Solution: Follow IUPAC rules.

Question 107. 2-methyl-2-butene will be represented as:

Answer: 4

Question 108. Pick out the alkane which differs from the other members of the group

1. 2,2-dimethyl propane
2. Pentane
3. 2-methyl butane
4. 2, 2-dimethyl butane

Answer: 4. 2, 2-dimethyl butane

Solution: 2, 2-dimethyl butane is 6-carbon hydrocarbon (C₆ H₁₄)

Question 109. In IUPAC suffix name of —COX is:

1. Oyl halide
2. Halo carbonyl
3. Carbamoyl
4. None of these

Answer: 1. Oyl halide

Solution: Follow IUPAC rules.

Question 110. IUPAC name of following compound is:

1. 3-methyl cyclo-1-buten-2-ol
2. 4-methyl cyclo-2-buten-1-ol
3. 4-methyl cyclo-1-buten-3-ol
4. 2-methyl cyclo-3-buten-1-ol

Answer: 2. 4-methyl cyclo-2-buten-1-ol

Solution: Follow IUPAC rules.

Question 111. The correct IUPAC name of the acid is

1. Z-3-ethyl-4-methyl hex-3-en-1-oic acid
2. Z-3-ethyl-4-methyl hexanoic acid
3. Z-3, 4-diethyl pent-3-en-1-oic acid
4. E-3-ethyl-4-methylhex-4-en-1-oic acid

Answer: 3. Z-3, 4-diethyl pent-3-en-1-oic acid

Solution:

[The configuration of this compound is E because bulkier groups are present opposite of the double bond.]

Question 112. The IUPAC name of aldehyde is

1. Prop-2-en-1-al
2. Propenyl aldehyde
3. But-2-en-1-al
4. Propenal

Answer: 1. Prop-2-en-1-al

Solution: CH2=CH-CHO (Prop -2-en-1-al)

Question 113. Write the IUPAC name of the below compound.

1. 3-methoxy butane
2. 2-methoxy butane
3. 3-methyl-3-methoxy propane
4. Butoxy methane

Answer: 2. 2-methoxy butane

Solution: According to the IUPAC system ethers are named as alkoxy alkanes. The larger alkyl group forms the parent chain while the lower alkyl group is taken with the ethereal oxygen and forms a part of the alkoxy group.

Question 114. IUPAC name of the compound is

1. 4-butyl-2,5-hexadien-l-al
2. 5-vinyloct-3-en-l-al
3. 5-vinyloct-5-en-8-al
4. 3-butyl-1,4-hexadien-6-al

Answer: 1. 4-butyl-2,5-hexadien-l-al

Solution: Follow IUPAC rules.

Question 115. The IUPAC name of the compound is

1. Ethoxy propane
2. 1, 1-dimethyl ether
3. 2-ethoxy zso-propane
4. 2-ethoxy propane

Answer: 4. 2-ethoxy propane

Solution:

The above compound is an ether and its name is written as alkoxy alkane. Oxy is attached to the lower group. Hence, the IUPAC name of the above compound is 2-ethoxy propane.

Question 116. The IUPAC name of the compound

1. Ethane nitrile
2. Methane isonitrile
3. Ethane isonitrile
4. None of these

Answer: 2. Methane isonitrile

Solution: CH₃NC is methaneisonitrile.

Question 117. The structural formula of methyl aminomethane is:

1. (CH₃)₂ CHNH₂
2. (CH₃)₃ N
3. (CH₃)₂ NH
4. CH₃ NH₂

Answer: 3. (CH₃)₂ NH

Solution: IUPAC name is N-methyl methanamine.

Question 118. The IUPAC name of CH₃-CCH(CH₃)₂ is

1. 4-methyl-2-pentyne
2. 4, 4-dimethyl-2-butyne
3. methyl isopropyl acetylene
4. 2-methyl-4-pentyne

Answer: 1. 4-methyl-2-pentyne

Solution: Follow IUPAC rules.

Question 119. Which one of the following is the correct formula for dichlorodiphenyltrichloroethane?

Answer: 1

Solution: Follow IUPAC rules.

Question 120. IUPAC name of (CH₃)₂N-C₂H₅ is:

1. Dimethyl ethyl amine
2. Dimethyl aminomethane
3. Dimethyl aminoethane
4. N, N-dimethylethanolamine

Answer: 4. N, N-dimethylethanolamine

Question 121. 3-methyl penta-1,3-diene is:

1. CH₂=CH(CH₂)₂CH₃
2. CH₂=CHCH(CH₃)CH₂CH₃
3. CH₃CH=C(CH₃)CH=CH₂
4. CH₃-CH=CH(CH₃)₂

Answer: 3. CH₃CH=C(CH₃)CH=CH₂

Solution: Follow IUPAC rules.

Question 122. The IUPAC name of the compound is

1. 2-iodo-3-chloro-4-pentanoic acid
2. 4-oxo-3-chloro-2-iodo Pentanoic acid
3. 4-carboxy-4,3-chloro-2-butanone
4. 3-chloro-2-iodo-4-oxo-pentanoic acid

Answer: 4. 3-chloro-2-iodo-4-oxo-pentanoic acid

Solution: Report prefixes in alphabetical order.

Question 123. The IUPAC name for CH₃COCH(CH₃)₂ is:

1. 4-methyl isopropyl ketone
2. 3-methyl-2-butanone
3. Isopropyl methyl ketone
4. 2-methyl-3-butanone

Answer: 2. 3-methyl-2-butanone

Question 124. Give the IUPAC name of the compound

1. 1,1,3-trimethyl cyclo hex-2-ene
2. 1,3,3-trimethyl cyclo hex-l-ene
3. 1,1,5-trimethyl cyclo hex-5-ene
4. 2,6,6-trimethyl cyclo hex-l-ene

Answer: 2. 1,3,3-trimethyl cyclo hex-l-ene

Solution: Follow IUPAC rules.

Question 125. The name of the compound is:

1. (2Z,4Z)-2, 4-hexadiene
2. (2Z-, 4E)-2, 4-hexadiene
3. (2E, 4Z)-2, 4-hexadiene
4. (4E, 4Z)-2, 4-hexadiene

Answer: 1. (2Z,4Z)-2, 4-hexadiene

Solution: The name of the compound is (2Z,4Z)-2, 4-hexadiene.

Question 126. The IUPAC name of the compound is

1. 2-amino-3-hydroxy propanoic acid
2. l-hydroxy-2-amino propan-3-oic acid
3. l-amino-2-hydroxy propanoic acid
4. 3-hydroxy-2-amino propanoic acid

Answer: 1. 2-amino-3-hydroxy propanoic acid

Solution: Follow IUPAC rules.

Question 127. What is the formula of tertiary butyl alcohol?

1. CH₃-CH(CH₃)-CH₂-OH
2. CH₃-(CH₂)₂OH
3. CH₃-CH(OH)-CH₂-CH₃

Answer: 4. CH₃-CH(OH)-CH2-CH₃

Solution:

is the formula of tertiary butyl alcohol as in it – OH group is attached to tertiary carbon.

Question 128. The IUPAC name of the compound is

1. 2-ethoxy pentane
2. 4-ethoxy pentane
3. Pentyl-ethyl ether
4. 2-pentoxy ethane

Answer: 1. 2-ethoxy pentane

Solution: Follow IUPAC rules.

Question 129. The IUPAC name of the compound is

1. 2-ethyl-3-methyl butanol chloride
2. 2,3-dimethyl pentanol chloride
3. 3,4-dimethyl pentanol chloride
4. l-chloro-l-oxo-2,3-dimethyl pentane

Answer: 2. 2,3-dimethyl pentanol chloride

Question 130. The IUPAC name of the compound is

1. 1,2,3-trihydrosypropane
2. 3-hydroxy pentane-l,5-diol
3. 1,2,3-hydroxy-propane
4. Propane-1,2,3-triol

Answer: 4. Propane-1,2,3-triol

Solution: Follow IUPAC rules.

Question 131. The IUPAC name of is

1. 2-methyl-3-bromohexanal
2. 3-bromo-2-methylbutanal
3. 2-bromo-3-bromobutanal
4. 3-bromo-2-methylpentanal

Answer: 4. 3-bromo-2-methylpentanal

Solution: Follow IUPAC rules.

Question 132. The IUPAC name of

1. 4-propoxy pentane
2. Pentyl-propyl ether
3. 2-propoxy pentane
4. 2-pentoxy propane

Answer: 3. 2-propoxy pentane

Solution: Follow IUPAC rules.

Question 133. The systematic (IUPAC) name of the compound with the following structural formula shall be

1. 1-ethyl-2-methyl cyclohexene
2. 2-methyl-l-ethyl cyclohexene
3. 3-ethyl-2-methyl cyclohexene
4. 4-ethyl-3-methyl cyclohexene

Answer: 4. 4-ethyl-3-methyl cyclohexene

Solution: Follow IUPAC rules.

Question 134. Pyridine is:

1. An aromatic compound and a primary base
2. A heterocyclic amino compound and a tertiary base
3. An aromatic amino compound forms salts
4. A cyano derivative of benzene and a secondary base

Answer: 2. A heterocyclic amino compound and a tertiary base

Solution: Pyridine is a heterocyclic compound having six carbon-membered ring formed by C and N atoms

Question 135. The IUPAC name of (CH₃)₃C-CH=CH₂ is

1. 1, 1, 1-trimethyl-2-propene
2. 3, 3, 3-trimethyl-2-propene
3. 2, 2-dimethyl-3-butene
4. 3, 3-dimethyl-1-butene

Answer: 4. 3, 3-dimethyl-1-butene

Solution:

IUPAC name=3, 3-dimethyl-l-butene.

Question 136. The name of the compound is:

1. 2-pentanone
2. Pentanone-2
3. Pentan-2-one
4. All are correct

Answer: 4. All are correct

Question 137. Write the IUPAC name of

1. 3-methyl pentane-3-ol
2. 3-hydroxy hexane
3. 3-hydroxy-3-methyl pentane
4. All of the above

Answer: 1. 3-methyl pentane-3-ol

Fundamental Concepts In Organic Reaction Mechanism

Question 1. Chlorine in vinyl chloride is less reactive because?

1. sp²-hybridized carbon has more acidic character than sp3-hybridized carbon
2. C—Cl bond develops partial double bond character
3. Of resonance
4. All of the above are correct

Answer: 3. Of resonance

Solution: CH₂=CH−Cl−↔ CH₂−CH−Cl+CH₂=CH-Cl-↔ CH₂-CH-Cl+ is formed between C and Cl. Here it is less reactive due to resonance.

Read And Learn More: NEET General Organic Chemistry Notes, Question And Answers

Question 2. What information is provided by the reaction mechanism?

1. The bonds broken and formed
2. The reaction intermediates
3. The relative rates of discrete steps, especially the slowest one
4. All of the above

Answer: 4. All of the above

Solution: These are characteristics known from the mechanism of reaction.

Question 3. In which of the following ways does the hydride ion tend to function?

1. An electrophile
2. A nucleophile
3. A free radical
4. An acid

Answer: 2. A nucleophile

Solution: Hydride ions are formed when hydrogen accepts a proton, so it has a tendency to donate electrons. Since, hydride ion (H–) has a tendency to donate electrons, it functions as an nucleophile.

Question 4. Which of the following is the weakest base?

1. Ethyl amine
2. Ammonia
3. Dimethyl amine
4. Methyl amine

Answer:  2. Ammonia

Solution: Alkyl group (an electron releasing (+I group) increases electron density at N-atom, hence, basic nature is increased. In ammonia, no alkyl group is present, so it is least basic.

Question 5. An S_N2 reaction at an asymmetric carbon of a compound always gives

1. A mixture of diastereomers
2. A single stereoisomer
3. An enantiomer of the substrate
4. A product with opposite optical rotation

Answer: A single stereoisomer

Solution: In S_N2 reactions, the nucleophile attaches itself from the direction opposite to that of the nucleophile already present in the second step, the previous nucleophile is removed and a single stereoisomer is obtained

Question 6. The most common type of reaction in aromatic compounds is

1. Elimination reaction
2. Addition reaction
3. Electrophilic substitution reaction
4. Rearrangement reaction

Answer: 3. Electrophilic substitution reaction

Solution: Due to presence of delocalized π-electrons in the aromatic compounds, the electron density is maximum inside the ring. Therefore, aromatic compounds undergo electrophilic substitution reaction and resistance to addition reactions.

Question 7. Which behaves both as a nucleophile as well as an electrophile?

1. CH₃OH
2. CH₃NH₂
3. CH₃CN
4. CH₃Cl

Answer: 3. CH₃CN

Solution: CH₃NH₂ and CH₃OH are nucleophiles, CH₃Cl is an electrophile. But CH₃CN is a nucleophile due to the presence of a lone pair of electrons on N and is an electrophile due to the presence of a partial positive charge on C.

Question 8. (CH₃)₄N+ is neither an electrophile nor a nucleophile because it

1. Does not have electron pair for donation as well as cannot attract electron pair
2. Neither has electron pair available for donation nor can accommodate electrons since all shells of N are fully occupied
3. Can act as Lewis’s acid and base
4. None of the above

Answer:  2. Neither has electron pair available for donation nor can accommodate electrons since all shells of N are fully occupied

Solution: It’s a fact.

Question 9. If X is halogen the correct order for S_N2 reactivity is:

1. R₂CHX >R₃CX > RCH₂X
2. RCH₂X >R₃CX > RCH₂X
3. RCH₂X >R₂CHX >R₃X
4. R₃CX >R₂CHX > RCH₂X

Answer: 3. RCH₂X >R₂CHX >R₃X

Solution: Steric hindrance in tertiary halides gives rise to less reactivity for S_N2.

Question 10. Which of the following would react most readily with nucleophiles?

Answer: 3

Solution:

Nucleophiles always attack electron deficient sites. Presence of electron withdrawing groups such as NO₂,CHO etc decreases the electron density on the benzene nucleus, hence such groups activate the ring towards nucleophilic attack.

The presence of electron releasing groups such as R or “OR” increases the electron density, thus deactivating the nucleus towards nucleophilic attack. NO₂ group activates the ring more than Cl towards nucleophilic attack,hence reacts readily with nucleophile.

Question 11. Which does not have sp²-hybridised carbon atom?

1. Acetamide
2. Acetic acid
3. Acetonitrile
4. Acetone

Answer: 3. Acetonitrile

Solution: CH₃CN has sp³ and sp-hybridised carbon atoms.

Question 12. The basicity of aniline is less than that of cyclohexylamine. This is due to

1. +R effect of – NH₂group
2. -I effect of – NH₂ group
3. -R effect of –NH₂ group
4. Hyperconjugation effect

Answer: 1. +R effect of – NH₂ group

Solution: -NH₂ has +R effect, it donates electrons to the benzene ring. As a result, the lone pair of electrons on the N-atom gets delocalized over the benzene ring. As a result, the lone pair of electrons on the N-atom gets delocalized over the benzene ring and thus it is less readily available for protonation. Hence, aniline is a weaker base than cyclohexylamine.

Question 13. Which of the following is an elimination reaction?

Answer: 3

Solution:

Is an example of an elimination reaction.

Question 14. The function of AlCl₃ in Friedel-Crafts reaction is

1. To absorb HCl
2. To absorb water
3. To produce nucleophile
4. To produce electrophile

Answer: 4. To produce electrophile

Solution: The function of AlCl₃, in the Friedel-Craft reaction, is to produce electrophile, which later adds to the benzene nucleus

Question 15. The inductive effect

1. Implies the atom’s ability to cause bond polarization
2. Increases with increase of distance
3. Implies the transfer of lone pair of electrons from more electronegative atom to the lesser electronegative atom in a molecule
4. Implies the transfer of lone pair of electrons from lesser electronegative atom to the more electronegative atom in a molecule

Answer: 1. Implies the atom’s ability to cause bond polarization

Solution: It’s a fact.

Question 16. Among the following, the dissociation constant is highest for

1. C₆H₅OH
2. C₆H₅CH₂OH
3. CH₃-C≡CH
4. CH₃NH₃⁺ Cl^–

Answer: 3. CH₃-C≡CH

Solution: Dissociation of a proton from CH₃NH₃⁺Cl^– is very difficult due to I^– effect of Cl–and N⁺ while in C₆H₅OH due to the resonance stabilization of phenoxide ion protons are eliminated easily. Similarly, due to H-bonding in C₆H₅CH₂OH, it can be eliminated easily and in CH₃C≡CH  The proton is acidic in nature hence; it can be dissociated.

Question 17. In the following carbocation, H/CH₃  that is most likely to migrate to the positively charged carbon is

1. CH₃at C⎼4
2. H at C⎼4
3. CH₃ at C⎼2
4. Hat C⎼2

Answer: 4. Hat C⎼2

Solution: Due to the H–shift from C₂ to C₃ Driving force is conjugation from oxygen. Also, bulky groups hinder the hydride shift.

Question 18. The reaction  (Major) The correct statement (s) are

1. 2-butene is Saytzeff product
2. 1-butene is Hofmann (s) product
3. The elimination reaction follows Saytzeff rule
4. All of the above

Answer: 4. All of the above

Solution: The elimination takes place according to the Saytzeff rule. The most substituted alkene (butane-2) is called Saytzeff product whereas less substituted alkene (butane-1) is called Hofmann product

Question 19. Hyperconjugation involves overlap of the following orbitals

1. σ- σ
2. σ-ρ
3. p-p
4. π-π

Answer: 2. σ-ρ

Solution: Hyperconjugation arises due to the partial overlap of a sp³-s (a C-H bond) with the empty p-orbital of an adjacent positively charged carbon atom.

Question 20. The compound which reacts with HBr obeying Markownikoff’s rule is

Answer: 4.

Solution: Markownikoff’s rule is obeyed during addition of unsymmetrical addendum on unsymmetrical alkene.

Question 21. The stability of carbanions in the following; is in the order of:

Answer: 4.

Solution: RC≡C- is the most stable because C atom carrying the negative charge is sp hybridized (most electronegative C). Both

2 and 3 have negative charge on sp² hybridized C atom but 3 is less stable due to electron-releasing alkyl groups. 4  is the least stable as the negative charge is carried by sp³ hybridized C atom.

Question 22. CH₃Br + Nu– ⟶CH₃-Nu+Br– The decreasing order of the rate of the above reaction with nucleophiles (Nu–) 1 to 4 is [Nu– = (1) PhO–,(2) AcO–,(3) HO–,(4) CH₃O–]

1. 4>3>1>2
2. 4>3>2>1
3. 1>2>3>4
4. 2>4>3>1

Answer: 2. 4>3>2>1

Solution: C₆H₅O^–  possess less nucleophilicity due to the stabilized nature of phenoxide ions. CH₃OH is a weaker acid than CH₃COOH and thus CH₃O^– is a stronger base.

Acidic order: CH₃COOH >H₂O >CH₃OH

Question 23. Which of the following belongs to +I group?

1. –OH
2. –OCH₃
3. –COOH
4. –CH₃

Answer: 4. –COOH

Solution: It is exhibited when an electron releasing group is attached to the carbon chain. Example – Alkyl groups. “The more the number of alkyl groups, more is the +I effect.”

Question 24. The electrophile involved in the reaction is?

1. Di chloromethyl cation
2. Dichlorocarbene (:CCl₂)
3. Trichloromethyl anion (C ̅Cl₃)
4. Formyl cation

Answer: 2. Dichlorocarbene (:CCl₂)

Solution: It is the Reimer-Tiemann reaction. The electrophile formed is: CCl₂ (Dichlorocarbene)

Question 25. In the presence of peroxide, hydrogen chloride, and hydrogen iodide do not give anti-Markownikov’s addition to alkenes because

1. Both are highly ionic
2. One is oxidizing and the other is reducing
3. One of the steps are exothermic in both the cases
4. All the steps are exothermic in both the cases

Answer: One is oxidizing and the other is reducing

Solution: Follow mechanism of addition of HCl and HI in presence of peroxide. One of the chain propagation steps is endothermic in both cases.

Question 26. According to Cahn-Ingold-Prelog sequence rules, the correct order of priority for the given group is

1. –COOH > –CH₂OH > –OH > –CHO
2. –COOH > –CHO > –CH₂OH > –OH
3. –OH > –CH₂OH > –CHO > –COOH
4. –OH > –COOH > –CHO > –CH₂OH

Answer: 4. –COOH > –CHO > –CH₂OH > –OH

Solution: According to Cahn-Ingold-Prelog sequence rules, the priority of groups is decided by the atomic number of their atoms. When the atom (which is directly attached to the asymmetric carbon atom) of a group has higher atomic number, then the group gets higher priority. Groups with atoms of comparable atomic number having double or triple bond, have higher priority than those that have single bond. Hence, the order of priority of group is -OH>-COOH>-CHO>-CH₂OH

Question 27. Which of the following cannot undergo nucleophilic substitution under ordinary conditions?

1. Chlorobenzene
2. Tert-butylchloride
3. Isopropyl chloride
4. None of these

Answer: 1. Chlorobenzene

Solution: C-Cl bond is aryl chloride is stable due to delocalization of electrons by resonance. Also C-Cl bond possess a double bond character like vinyl chloride, hence SN reactions are not possible in chlorobenzene under ordinary conditions.

Question 28. Electromeric effect is

1. Permanent effect
2. Temporary effect
3. Resonance effect
4. Inductive effect

Answer: 2. Temporary effect

Solution: The electromagnetic effect occurs only in the presence of an attacking reagent. It operates in the molecules having multiple bonds. Since it exists only on the demand of attacking reagents, it is a temporary effect. example,

Question 29. The least active electrophile is

Answer: 3

Solution: In the given electrophile

Group is the same. So, only X affects their activity, i.e., we have to discuss activity due to

1. –OCH₃
2. –Cl
3. -N<_Me^Me
4. -S – CH₃

Since, amines are less active, therefore, electrophile (c) will be least active.

Question 30. Carbanions initiate

1. Addition reactions
2. Substitution reactions
3. Both (1) and (2)
4. None of these

Answer:  Both (1) and (2)

Solution: It is a fact.

Question 31. 2-hexyne gives trans-2-hexene on treatment with

1. Li/NH₃
2. Pd/BaSO₄
3. LiAlH₃₄
4. Pt/H₂

Answer: 1. Li/NH₃

Solution: 2-hexyne gives trans-2-hexene on treatment Li/NH₃

Question 32. The chief reaction product of reaction in between n-butane and bromine at 130°C is 

Answer: 2

Solution: 2°H is more reactive than 1°.

Question 33.  product.

Predominant product is

Answer: 2

Solution: Due to \(R_3 \stackrel{\oplus}{\mathrm{N}}-\) (e– withdrawing tendency) carbocation will appear farther to that (terminal).

Hence, the product is R₃N-CH₂-CH₂Br.

Question 34. Reaction of methyl bromide with aqueous sodium hydroxide involves

1. Racemisation
2. S_N1 mechanism
3. Retention of configuration
4. S_N2 mechanism

Answer: 4. S_N2 mechanism

Solution:

Since, the reaction rate depends upon the concentration of both reactant and nucleophile, it is a S_N2 reaction. It involves inversion of configuration.

Question 35. The reaction is fastest when X is

1. OCOR
2. OC₂H₅
3. NH₂
4. Cl

Answer: 4. Cl

Solution: The best-leaving group (poorest nucleophile) is Cl⊕, thus the fastest reaction is with Cl.

Question 36. Addition of Br₂ on cis-butene-2 gives

1. A racemic mixture of 2,3-dibromobutane
2. Meso form of 2,3-dibromo butane
3. Dextro form of 2,3-dibromobutane
4. Laevo form of 2,3-dibromobutane

Answer: 1. A racemic mixture of 2,3-dibromobutane

Solution: Follow mechanism of addition reaction.

Question 37. The order of stability of carbanions is

1. CH₃–>1°>2°>3°
2. 3°>2°>1°>CH₃^–
3. –3°>1°>2°>CH₃^–
4. –2°>3°>1°>CH₃^–

Answer: 1. CH₃–>1°>2°>3°

Solution: The positive inductive effect of CH₃ group on carbanions intensifies negative charge on C– centre and thus, 3° Carbanion is more reactive.

Question 38. In Cannizzaro’s reaction given below \(2 \mathrm{PhCHO} \stackrel{\mathrm{OH}^{\ominus}}{\longrightarrow} \mathrm{PhCH}_2 \mathrm{OH}+\mathrm{PhCO}_2^{\ominus}\)

The slowest step is:

1. The attack of ::OH^⊝ at the carboxyl group
2. The transfer of hydride to the carbonyl group
3. The abstraction of proton from the carboxylic group
4. The deprotonation of Ph CH₂OH

Answer: 2. The transfer of hydride to the carbonyl group

Solution: In Cannizzaro reaction the transfer of H〗 – to another carbonyl group is difficult and the slowest step. (Rate determining step or key step)

Question 39. Which of the following has the highest nucleophilicity?

1. F^–
2. OH^–
3. CH₃^–
4. NH₂^–

Answer: 3. CH₃^–

Solution: Stronger is an acid, weaker is its conjugate base or weaker is the nucleophile. The acidic character order is HF >H₂O >NH₃>CH₄.

Question 40. The following compound will undergo electrophilic substitution more readily than benzene

1. Nitrobenzene
2. Benzoic acid
3. Benzaldehyde
4. Phenol

Answer: 4. Phenol

Solution: During electrophilic substitution, electrophile attacks the double bond of the benzene ring. The aromatic compounds having electron donating groups undergo electrophilic substitution more easily due to the favorable effect of the electron donating group.

NO₂, COOH, and CHO groups are electron-withdrawing groups so they decrease the reactivity of organic compounds. -OH group is an electron-donating group, so it increases the electron density in the benzene ring and increases the rate of reaction

Question 41. The following reaction is an example of …. reaction. \(\mathrm{C}_2 \mathrm{H}_4 \mathrm{Br}_2 \stackrel{\text { Alc.KOH }}{\longrightarrow} \mathrm{C}_2 \mathrm{H}_2\)

1. Addition
2. Dehydrobromination
3. Substitution
4. Debromination

Answer: 2. Dehydrobromination

Solution:

This is a dehydrohalogenation reaction.

Question 42. The number of optically active products obtained from the complete ozonolysis of the given compound is

1. 0
2. 1
3. 2
4. 4

Answer: 1. 0

Solution: Ozonolysis of the compound may be given as:

Question 43. The isomeric mono substitution products theoretically possible for the structure, CH₂=HC-CH₂-CH₂ CH=CH₂ are∶

1. 3
2. 2
3. 4
4. 6

Answer: 1. 3

Solution: CHCl = CHCH₂CH₂CH = CH₂: CH₂= CClCH₂CH₂CH = CH₂

Question 44. The most stable carbanion is

1. CH^⊝₃
2. RCH₂^⊝
3. R₃C^⊝
4. ^⊝CH₂CHO

Answer: 4. ^⊝CH₂CHO

Solution: ^⊝CH ₂CHO is the most stable carbanion since it is stabilized by resonance

Question 45. Which of the following is the correct order of priority of groups in D-glyceraldehyde?

1. OH (1), CHO (2), CH₂OH(3) and H (4)
2. OH (1), CH₂OH (2), CHO (3) and H (4)
3. CH₂OH (1), CHO (2), OH (3) and H (4)
4. CHO (1), OH (2), CH₂OH(3) and H (4)

Answer: 1. OH (1), CHO (2), CH₂OH(3) and H (4)

Solution: The structure of D-glyceraldehyde is as

The priority of groups is decided by the following rules

1. Atom having higher atomic number gets higher priority
2. If the priority cannot be decided by rule 1 then the next atoms are considered for priority assignment.
3. Where there is a = bond or=bond both atoms are considered to be duplicated or triplicated ( has higher priority than –CH₂OH). Hence, the correct order of priority of groups in

D-glyceraldehyde is as: OH(1),CHO(2),CH₂OH (3)and H(4)

Question 46. Which of the following compounds are not arranged in order of decreasing reactivity towards electrophilic substitution?

1. Fluorobenzene > chlorobenzene > bromobenzene
2. Phenol>n-propyl benzene> benzoic acid>
3. Chlorotoluene >para-nitrotoluene>2-chloro-4-nitro toluene
4. Benzoic acid> phenol>n-propyl benzene

Answer: 4. Benzoic acid> phenol>n-propyl benzene

Solution:  -COOH group is a deactivating group

∴ Benzoic acid is less reactive towards electrophilic substitution.

So, benzoic acid> phenol>n-propyl benzene is not arranged correctly.

Question 47. During the elimination reactions, the hybrid state of carbon atoms involved in change shows:

1. sp³ to sp² nature
2. sp² to sp nature
3. No change in hybridized state
4. Either of the above

Answer: 4. Either of the above

Solution: CH₃CH₂X ⟶ CH₂= CH₂ (sp³ to sp²);

CH₂= CHX ⟶ CH ≡ CH (sp² to sp);

CH₂XCH₂CH₂X⟶∆ (No change).

Question 48. Which of the following is the strongest base?

1. Acetamide
2. Aniline
3. Methylamine
4. Dimethylamine

Answer: 4. Dimethylamine

Solution: Presence of methyl group on NH₃molecule increases the tendency of the N atom to lose electron pair. However, tertiary amines are less basic due to steric hindrance.

Question 49. One of the stable resonating forms of methyl vinyl ketone is

Answer: 2

Solution: The most stable one is that in which the positive and negative charges reside on the most electropositive and most electronegative atoms of the species respectively.

Question 50. The reagent showing addition on alkene against the Markownikoff’s rule

1. Br₂
2. H₂S
3. HF
4. HBr

Answer: 4. HBr

Solution: Kharasch effect involves addition of HBr.

Question 51. Among the following compounds, the most acidic is

1. p-nitrophenol
2. p-hydroxybenzoic acid
3. o-hydroxybenzoic acid
4. p-toluic acid

Answer: 3. o-hydroxybenzoic acid

Solution: A monosubstituted benzoic acid is stronger than a monosubstituted phenol as the former being a carboxylic acid. Among the given substituted benzoic acid, ortho-hydroxy acid is the strongest acid although – OH causes electron donation by resonance effect which tends to decrease acid strength. It is due to the very high stabilization of conjugate base by intramolecular

H-bond which outweighs the electron donating resonance effect of – OH.

The overall order of acid strength of the given four acids is ortho-hydroxy benzoic acid (pKa =2.98)> Toluic acid pk_a = 4.37)>p-hydroxybenzoic acid (pk_a = 4.58)>p-nitrophenol (pka = 7.15).

Question 52. The order of reactivities of the following alkyl halides for a S_N2 reaction is

1. RF > RCl > RBr > RI
2. RF > RBr > RCl > RI
3. RCl > RBr > RF > RI
4. RI > RBr > RCl > RF

Answer: 4.  RI > RBr > RCl > RF

Solution: The rate of reaction follows the order RI> RBr > RCl> RF; whether it obeys S_N1 or S_N2 mechanism due to steric hindrance of the alkyl group.

Question 53. Electrophiles are

1. Lewis bases
2. Lewis’s acids
3. Amphoteric
4. None of these

Answer: 2. Lewis’s acids

Solution: Electrophiles are electron pair acceptors.

Question 54. Product in the reaction is?

Answer: 2

Solution: In such cases where the migrating group is a cycloalkyl group, ring expansion may occur.

Question 55. Which one of the following is an intermediate in the reaction of benzene with CH₃Cl in the presence of anhydrous AlCl₃?

1. Cl⁺
2. CH₃^–
3. CH₃⁺
4. 

Answer: 3. CH₃⁺

Solution: CH₃⁺ acts as an intermediate in the given reaction (Friedel Craft’s alkylation). It is an example of electrophilic aromatic substitution. In this reaction CH₃⁺is electrophile.

Question 56.+I effect is shown by

1. -CH₃
2. -Br
3. -Cl
4. -NO₂

Answer: 1. -CH₃

Solution: +I effect is shown by –CH₃ while –I effect is shown by –Br,-Cl and-NO₂.

Question 57. Amongst the following which of the above are true for S_N2 reaction?

1. The rate of reaction is independent of the concentration of the nucleophile.

2. The nucleophile attacks the carbon atom on the side of the molecule opposite to the group being displaced.

3. The reaction proceeds with simultaneous bond formation and bond rupture.

1. 1,2
2. 1,3
3. 1,2,3
4. 2,3

Answer: 4. 2,3

Solution: Follow the characteristics of S_N2 mechanism.

Question 58. In the nucleophilic substitution reactions (SN2 or SN1), the reactivity of alkyl halides follows the sequence

1. R-I>R-Br>R-Cl>R-F
2. R-Cl>R-F>R-Br>R-I
3. R-F>R-Cl>R-Br>R-I
4. R-I>R-F>R-Cl>R-Br

Answer: 1. R-I>R-Br>R-Cl>R-F

Solution: The correct order of reactivity is RI>RBr>RCl>RF. It is due to the fact that the weaker the base, the better it will be for the leaving group. Hence, I– is the best-leaving group.

Question 59. Among the following compounds which can be dehydrated very easily is

Answer:

Solution: In the above reaction more stable carbocation is generated hence, the compound dehydrates very easily.

 Question 60. Which of the following is not a nucleophile?

1. BF₃
2. CN–
3. OH–
4. NH₃

Answer:  1. BF₃

Solution: All neutral covalent compounds in which the central atom has incomplete octet are electrophile. For example, BeCl₂,BH3,ZnCl₂,AlCl₃

Question 61. In the following reactions,

The major products (A) and (C) are respectively

Answer: 3

Solution: In the first reaction, dehydration is governed by the Saytzeff rule which gives a more substituted alkene product. Here, the secondary carbocation formed undergoes 1, 2 hydride shifts and gives more stable tertiary carbocation which further undergoes beta elimination to give the major product (A). Thus, 2- Methyl but-2-ene is the major product.

HBr in the absence of peroxide.

This reaction is governed by the Markovnikov rule according to which when an unsymmetrical alkene undergoes hydrohalogenation, the negative part of the addendum (adding molecule) gets attached to that doubly bonded C which possesses a lesser number of hydrogen atoms. when an unsymmetrical reagent. Thus, in the above case, 2-Methyl 2-bromobutane will be the major product.

Question 62. The rate of the reaction,

i.e., 4<3<2<1

The increasing order of speed of the above reaction is

1. 4,3,2,1
2. 1,2,3,4,
3. 1,4,3,2
4. 3,2,1,4

Answer: 1. 4,3,2,1

Solution: The rate of reaction is influenced by the hyperconjugation effect of group R. It depends on the electron-donating power of the alkyl group (R). The electron-releasing power of the R group depends on the number of hydrogen present in α carbon. The increasing order of speed with R group in the reaction is i.e., 4<3<2<1

Question 63. The organic chloro compound, which shows complete stereochemical inversionduring and S_N2 reaction, is

1. CH₃Cl
2. (C₂H₅)₂CHCl
3. (CH₃)₃CCl
4. (CH₃)₂CHCl

Answer: 1. CH₃Cl

Solution: S_N2 order: methyl >1°>2°>3°.

Question 64. Vinyl chloride undergoes

1. Only addition reactions
2. Only elimination reactions
3. Both (1) and (2)
4. Substitution reactions

Answer: 3. Both (1) and (2)

Solution: Vinyl chloride(CH₂=CHCl) undergoes addition and elimination reactions. Substitution’s reaction is shown by compounds having single bonds only.

Question 65. Which of the following is the strongest nucleophile?

1. Br^–
2. :OH^–
3. :CN^–
4. C₂H₅O^–

Answer: C₂H₅O^–

Solution: The order of nucleophilicity depends upon the nature of alkyl group ‘R’ on which a nucleophile to attack as well as on the nature of solvent. However, if these are same, then weaker is the acid, stronger is base, i.e., stronger is nucleophilicity. This acidic character is.

HI > HBr > HCl > HCN >H₂O > EtOH

Question 66. Isopropyl chloride undergoes hydrolysis by

1. S_N1 mechanism
2. S_N2 mechanisms
3. S_N1 and S_N2 mechanisms
4. Neither S_N1 nor S_N2 mechanism

Answer: 3. S_N1 and S_N2 mechanisms

Solution: Iso-propyl chloride is a 2° halide and 2° halides can undergo hydrolysis either by S_N1
or S_N2 mechanism depending upon the nature of the solvent used.

Question 67. Which of the following is not true for carbanions?

1. The carbon carrying the charge has eight valence electrons
2. They are formed by heterolytic fission
3. They are paramagnetic
4. The carbon carrying the charge is sp³ hybridised

Answer: 3. They are paramagnetic

Solution: Carbanions contain even number of valence electrons and thus, show diamagnetic behavior.

Question 68. Free radicals can undergo

1. Disproportionation to two species
2. Rearrangement to a more stable free radical
3. Decomposition to give another free radical
4. All of the above are correct

Answer:  4. All of the above are correct

Solution: These are the characteristics of free radicals.

Question 69. Due to the presence of an unpaired electron, free radicals are

1. Cations
2. Anions
3. Chemically inactive
4. Chemically reactive

Answer:  4. Chemically reactive

Solution: Free radicals have unpaired electrons but are neutrals and are reactive.

Question 70. Which represents nucleophilic aromatic substitution reaction?

1. Reaction of benzene with Cl₂ in sunlight
2. Benzyl bromide hydrolysis
3. Reaction of NaOH with dinitrofluorobenzene
4. Sulphonation of benzene

Answer: 3. Reaction of NaOH with dinitrofluorobenzene

Solution: Reaction of NaOH with dinitrofluorobenzene represents nucleophilic aromatic substitution reaction because the –NO₂ group is a deactivating group. They make benzene nucleus electrons deficient and facilitate the nucleophile to attack the ring.

Question 71. The S_N1 mechanism for substitution reaction by nucleophile is favored by

1. Low concentration of nucleophile
2. Weak nature of nucleophile
3. Polar solvent
4. All of the above

Answer: 4. All of the above

Solution: These are the characteristics of S_N1 mechanism.

Question 72. Which of the following is the most stable radical?

Answer: 4

Solution: The stability of alkyl free radicals can be explained by hyperconjugation and number of resonating structures due to the hyperconjugation.

Question 73. Which of the following types of reaction occurs when a substituent has got a double bond with an evenly distributed π electron cloud?

1. Electrophilic addition
2. Nucleophilic addition
3. Any of the (1) and (2)
4. None of the above

Answer: 1. Electrophilic addition

Solution: Such a condition is seen when π bond is formed between similar atoms

Question 74. The formation of acetylene from ethylene is an example of

1. Addition reaction
2. Substitution reaction
3. Elimination reaction
4. Condensation reaction

Answer:  3. Elimination reaction

Solution: CH₂=CH₂→-H₂CH≡CH Conversion of ethylene into acetylene is an example of an elimination reaction.

Question 75. Which of the following can act as a nucleophile?

1. BF₃
2. FeCl₃
3. ZnCl₂
4. C₂H₅MgBr

Answer: 4. C₂H₅MgBr

Solution: Grignard reagents can act as electrophile and nucleophile.

Question 76. Phenol is more acidic than

Answer:

Solution: Methoxy group, due to +I effect, increase electron density on OH- group, thus making it less acidic. Thus, o-methoxy phenol and acetylene are less than phenol. p-nitrophenol is more acidic than phenol.

Question 77. The formation of ethylene from acetylene is an example of

1. Elimination reaction
2. Substitutions reaction
3. Condensation reaction
4. Addition reaction

Answer: 4. Addition reaction

Solution: The formation of ethylene from acetylene is an example of additional reaction

Question 78. Which of the following compounds will be most reactive towards nucleophilic addition reaction?

1. CH₃COCH₂CH₂CH₂CH₃
2. CH₃CH₂COCH₂CH₂CH₃
3. CH₃CH₂CH₂CH₂CH₂CHO
4. CH₂ – CH₂– CO- CH-  CH₃– CH₃

Answer: 3. CH₃CH₂CH₂CH₂CH₂CHO

Solution:

The case with which a nucleophile attacks the carbonyl groups depends upon the electron-deficiency, i.e, the magnitude of the positive charge on the carbonyl carbon. Since, an alkyl group has an electron-donating inductive effect.

(+I effect),therefore, the greater the number of alkylgroups attached to the carbonyl groups greater is the electron-density on the carbonyl carbon and  hence, lower is its reactivity towards nucleophilic addition reactions.

Question 79. Among the following compounds (1-3) the correct order of reaction with electrophilic reagent is

1. 2>3>1
2. 3<1<2
3. 1>2>3
4. 1=2>3

Answer: 3. 1>2>3

Solution: Methoxy group is electron releasing group it increases electron density of benzene nucleus –NO₂ group is an electron withdrawing group, it decreases the electron density of the benzene nucleus. Thus, the order of reaction with electrophilic regent is

Question 80. Which of the following applies in the reaction CH₃CHBrCH₂CH₃→Alco.KOH?

 CH₃CH=CHCH₃ (Major product)

CH₂=CHCH₂CH₃ (Minor product)

1. Hofmann’s rule
2. Saytzeff rule
3. Kharasch effect
4. Markownikoff’s rule

Answer: 2. Saytzeff rule

Solution: This reaction is governed by Saytzeff rule. According to this rule the elimination of β-hydrogen atom take place from the carbon having the lesser number of H-atoms or in other words a stable alkene is formed. (More substituted alkene is more stable)

Question 81. Which of the following is most basic?

Answer: 4

Solution: Electron donors are bases. (Piperidine), hence, it is most basic.

Question 82. Arrange the following compounds in order of their decreasing reactivity with an electrophile, E^⊕.

1. Chlorobenzene,

2. 2,4-dinitrochlorobenzene,

3. p-nitrochlorobenzene

1. 3>2>1
2. 1>3>1
3. 1>3>2
4. 1>2>3

Answer: 3.1>3>2

Solution: Chlorobenzene has only one deactivating group, .e.,-Cl. In 2, 4-dinitrochlorobenzene three deactivating group, i.e., two –NO₂ and one –Cl are present and p-nitrochlorobenzene two deactivation groups, i.e., one NO₂ and one Cl is present. So, the order of reactivity is 1>3>2.

Question 83. Sulfur trioxide is

1. An electrophile
2. A nucleophile
3. A homolytic reagent
4. A base

Answer:  1. An electrophile

Solution: SO₃ can accept lone pair of electrons in the d-subshell.

Question 84. The chemical name of anisole is

1. Ethanoic acid
2. Methoxy benzene
3. Propanone
4. Acetone

Answer: 2. Methoxy benzene

Solution: It is an ether and the name of ether is given as alkoxy alkane. So, its name is methoxy benzene.

Question 85. Consider the following carbanions

The correct order of stability is

1. 1>2>3
2. 3>2>1
3. 2>3>1

Answer: 3. 2>3>1

Solution: -NO₂ group shows-M effect while CH₃O- group shows +M effect. (-Meffect stabilizesan anion). Hence, the order of stability is

Question 86. Which of the following shows S_N1 reaction most readily?

Answer: 2

Solution: S_N1 Reaction is most favourable for tertiary substances.

Question 87. Heterolysis of CH₃CH₂CH₃ results in the formation of

Answer: 3.

Solution: In heterolysis, the covalent bond is broken in such a way that one species (less electronegative) is deprived if its own electron, while the other species gain both the electron CH₃CH₂CH₃ → ^–CH₃+ ⁺C₂H₂

Question 88. The S_N1 reactivity of the following halides will be in the order

1 .(CH₃)₃CBr

2. (C₆H₅)₂CHBr

3. (C₆H₅)₂C(CH₃)Br

4.  (CH₃)₂CHBr

5. C₂H₅Br

1. 5>4>1>2>3
2. 2>1>3>5>4
3. 1>3>5>2>4
4. 3>2>1>4>5

Answer: 4. 3>2>1>4>5

Solution: S_N1(Unimolecular nucleophilic substitution reactions) Rate∝ (substrate)

Rate determining step in the formation of carbocation depends on the stability of the carbocation formed. The stability of carbocations follows the order

∵ Order of S_N1reactivity is

(C₆H₅)₂C(CH₃)Br>(C₆H₅)₂CHBr>(CH₃)₃CBr>(CH₃)₂CHBr>C₂H5Br

i.e., 3>2>1>4>5

Question 89. Which of the following acids has the smallest dissociation constant?

1. CH₃CHFCOOH
2. FCH₂CH₂COOH
3. BrCH₂CH₂COOH
4. CH₃CHBrCOOH

Answer: 3. BrCH₂CH₂COOH

Solution:  BrCH₂CH₂COOH is the weakest acid and has the lowest dissociation constant because.

i.e., of Br is lesser than F and is far away from –the COOH group.

Question 90. The reagent used in the dehydrohalogenation process is

1. Alcoholic KOH
2. NaNH₂
3. C₂H₅ONa
4. All of these

Answer: 4. All of these

Solution: All are used as dehydrohalogenating agent.

Question 91. Why is light necessary to bring in chlorination r

1. The dissociation of Cl₂ gives Cl° radical
2. The Cl₂ molecule absorbs light to show homolytic bond fission
3. The formation of Cl° free radical propagates the chain reaction
4. All of the above

Answer: 4. All of the above

Solution: Follow the mechanism of free radical substitution.

Question 92. The structure remaining after one H is removed from hydrocarbon is

1. Alkyl group
2. Alkenyl group
3. Alkynyl group
4. All of these

Answer: 4. All of these

Solution: Removal of H from alkane, alkene, and alkyne gives alkyl, alkenyl, and alkynyl groups respectively.

Question 93. Which of the following statements is not correct?

1. A  >C=C< group is made up of 4 σ -bonds and 2 π-bonds
2. A σ-bond is stronger than π-bond
3. A σ-bond can exist independently of π-bond
4. A double bond is stronger than a single bond

Answer: 1. A group is made up of 4 σ -bonds and 2 π-bonds

Solution:

Question 94. Stability of which intermediate is not governed by hyperconjugation?

1. Carbon cation
2. Carbon anion
3. Carbon-free radical
4. None of these

Answer: 2. Carbon anion

Solution: The stability of carbanion is not governed by hyperconjugation. Its stability depends on the +I or -I group.

Question 95. State the hybridization of carbon present in triplet carbene

1. Sp³
2. Sp²
3. Sp
4. None of these

Answer: 3. Sp

Solution: The state of hybridization of carbon in triplet carbene is sp.

Question 96. The decreasing order of nucleophilicity among the

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

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

Solution: If acid is weak, its conjugate base (nucleophile) is strong and vice versa.

Question 97. Following reaction, (CH₃)₃CBr+H₂O→ (CH₃)₃COH+HBr is an example of

1. Elimination reaction
2. Free radical substitution
3. Nucleophilic substitution
4. Electrophilic substitution

Answer: Nucleophilic substitution

Solution: (CH₃)₃CBr+H₂O→ (CH₃)₃C-OH+HBr Br is substituted by –OH–(nucleophile)S_N1 (Unimolecular nucleophilic substitution reaction)

Question 98. The correct sequence of steps involved in the mechanism of Cannizzaro’s reaction are

1. Nucleophilic attack, transfer of H^– and transfer of H⁺
2. Transfer of H^–, transfer of H+ and nucleophilic attack
3. Transfer if H⁺, nucleophilic attack and transfer of H^–
4. Electrophilic attack by OH^–, transfer of H+ and transfer of H

Answer: 1. Nucleophilic attack, transfer of H^– and transfer of H⁺

Solution: The Cannizzaro reaction is as

Methyl alcohol acetic acid

The mechanism of Cannizzaro reaction is as

Step 1 Attack of nucleophile OH– to the carbonyl carbon

 

Step 2 The transfer of hydride ions from anion (I) to the second molecule of aldehyde and finally rapid transfer of protons takes place.

Question 99. The arrangement of decreasing order of stability of Free radicals is.

Answer: 2

Solution: Follow the concept of hyperconjugation.

Question 100. Which chlorine atom is more electronegative in the following?

Answer: 4

Solution: More the number of hyper-conjugated structures, more will be electronegative chlorine atoms.

8-hyper conjugative structures

∵ 8-hyperconjugative structures are possible for (d)

∴ Chlorine in this is most electronegative.

Question 101. Arrange the following carbocations in order of stability

1. Benzy – 1

2. Allyl -2

3. Methyl -3

4. Vinyl – 4

1. 4>3>2>1
2. 1>2>3>4
3. 2>4>3>1
4. 3>2>1>4

Answer: 2. 1>2>3>4

Solution:

Methyl CH₃ CH₂= ⁺CH given options can be solved on the basis of conjugative and hyperconjugative structures

Question 102. The chain initiating species in the free radical chlorination of methane is?

1. Cl free radical
2. HCl
3. CH₃ radical
4. Methylene radical

Answer:  1. Cl free radical

Solution: Initiation step involves the splitting of a chlorine molecule which then forms two chlorine atoms; this process is initiated by ultraviolet radiation or sunlight. As we know, chlorine has one unpaired valence electron, which will act as a free radical.

Question 103. Among the following alkenes 

1. 1-butene

2. Cis-2-butene,

3. Trans-2-butene the decreasing order of stability is

1. 3 > 2 > 1
2. 3> 1 > 2
3. 1 > 2 > 3
4. 2 > 1 > 3

Answer: 1.3 > 2 > 1

Solution: Based on the heat of hydrogenation.

Question 104. The substitution reaction among the following is

Answer: 3

Solution:

1.  It is Diels Alder’s reaction (cyclo addition)
2. It is nucleophilic addition reaction
3. It is nucleophilic substitution reaction
4.  It is electrophilic addition reaction

Question 105. Which one of the following compounds will be most readily dehydrated?

Answer: 3

Solution: On the basis of stability of carbocation formed.

Question 106. Consider thiol anion (RS^⊝)and alkoxy anion(RS^⊝). Which of the following statement(s) is correct?

1. RS^⊝ is less basic and less nucleophilic than RO^⊝
2. RS^⊝ is less basic but more nucleophilic than RO^⊝
3. RS^⊝ is less basic and more nucleophilic than RO^⊝
4. RS^⊝ is more basic but less nucleophilic than RO^⊝

Answer: 2. RS^⊝ is less basic but more nucleophilic than RO^⊝

Solution: Nucleophilic strength increases down a column of the Periodic Table (in solvents that can have hydrogen bonds, such as water, alcohols, thiol alcohols).

Nucleophilic strength RO–<RS^–

Base strength RO^⊝ > RS^–

Thus, RO^⊝ is more nucleophilic but less basic than RO

Question 107. Which of the following alkyl halides is used as a methylating agent?

1. C₂H₅Cl
2. C₂H₅Br
3. C₂H₅I
4. CH₃I

Answer:   4. CH₃I

Solution: Methyl halides are methylating agents.

Question 108. The correct stability order for the following species as

1. 2>4>1>3
2. 1>2>3>4
3. 2>1>4>3
4. 1>3>2>4

Answer: 4. 1>3>2>4

Solution: 1>3>2>4

Question 109. Which of the following statement(s) is incorrect?

1. The rate of reaction increases with increase in water concentration in the hydrolysis of tertiary butyl bromide in methanol and water
2. The relative nucleophilicity in protic solvent is CN–>I–>OH>Br–>CI–>F–>H₂O
3. In S_N2 reactions, the order of reactivity of alkyl halides is in the order methyl > primary > secondary > tertiary
4. S_N2 reaction involves carbonium ions

Answer: 4. S_N2 reaction involves carbonium ions

Solution: S_N2 r reaction does not involve ion formation, these in fact involve the formation of transition state.

Question 110. Which of the following is an electrophile?

1. Na⁺
2. Li⁺
3. H⁺
4. Ca₂⁺

Answer: 3. H⁺

Solution: Positively charged species in which central atom has incomplete octet is called an electrophile, H+,X+,R+ are electrophile.

Question 111. The-Ieffect is shown by

1. –COOH
2. –CH₃
3. –CH₃CH₂
4. –CHR₂

Answer: 1. –COOH

Solution: –COOH is an electron withdrawing group.

Question 112. Identify the product in the given reaction: CH₃-CH=CH₂+NOCl ⟶ Product

1. CH₃CHCl.CH₂.NO
2. CH₃CH(NO).CH₂Cl
3. CH₃CH₂CH(Cl)(NO)
4. CH₂(NO).CH₂.CH₂Cl

Answer: CH₃CHCl.CH₂.NO

Solution: This reaction is an example of electrophilic addition reaction and its addition takes place according to Markownikoff’s rule. Negative part of the additive reagent adds to the less hydrogenated or more substituted carbon atom of the double bond of unsymmetrical alkene.

Question 113. Which of the following is least reactive in a nucleophilic substitution reaction?

1. (CH₃)₃C-Cl
2. CH₂=CHCl
3. CH₃CH₂Cl
4. CH₂=CHCH₂Cl

Answer: 2. CH₂=CHCl

Solution: Chlorine of vinyl chloride (CH₂=CHCl) is non-reactive (less reactive) towards nucleophile in nucleophilic substitution reaction because it shows the following resonating structure due to +M effect of –Cl atom.

In structure II, Cl-atom have positive charge and partial double bond character with C of vinyl group, so it is more tightly attracted towards the nucleus and it does not get replaced by nucleophile in SN– reaction.

Question 114. The bond that undergoes heterolytic cleavage most easily is

1. C–O
2. C–C
3. C–H
4. O–H

Answer: 4. O–H

Solution: Greater the difference in electronegativity of bonded atoms easier will be heterolytic cleavage.

Question 115. Arrange the following in order of increasing dipole moment

1. Toluene

2.  m-dichlorobenzene

3. o-dichlorobenzene

4. p-dichlorobenzene

1. 1 < 4 < 2 < 3
2. 4 < 1 < 2 < 3
3. 4 < 1 < 3 < 2
4. 4 < 2 < 1 < 3

Answer: 2. 4 < 1 < 2 < 3

Solution: In o-,m-,p- derivatives vectors are at 60°,120° and 180°. Thus, para has zero dipole moment. Alsoortho form has more dipole moment than metaform.

Question 116. Which of the following cannot show electromeric effect?

1. Alkenes
2. Ketones
3. Aldehydes
4. Ethers

Answer: 4. Ethers

Solution: Electromeric effect involves complete transfer of π-electron pair to more
electronegative atom on the need of attacking reagent.

Question 117. Which one is an elimination reaction?

1. CH₃CH₃+ Cl₂⟶ CH₃CHC₂l + HCl
2. CH₃Cl + KOH(aq.)⟶ CH₃OH + KCl
3. CH₂= CH2 + Br ⟶CH₂BrCH₂Br
4. C₂H5Br + KOH(alc.)⟶ C₂H₄+ KBr + H₂O

Answer: 4. C₂H5Br + KOH(alc.)⟶ C₂H₄+ KBr + H₂O

Solution: Elimination reactions involves the removal of a molecule (HBr here) from a substrate.

Question 118. In the following reaction sequence, the chain initiation step is?

Answer: 1

Solution: Chain initiation step involves the formation of free radicals only.

Question 119. The highest electrical conductivity of the following aqueous solutions is of

1. 0.1 M difluoroacetic acid
2. 0.1 M fluoroacetic acid
3. 0.1 M chloroacetic acid
4. 0.1 M acetic acid

Answer: 1. 0.1 M difluoroacetic acid

Solution: Fluoro group causes negative inductive effect increasing ionization, thus 0.1M difluoroacetic acid has highest electrical conductivity.

Question 120. The most stable carbocation is

Answer: 4

Solution: 4 is with maximum conjugative structures among them.

Question 121. Most stable carbonium ion is

Answer: 3

Solution: In the triphenyl methyl carbonium ion the π-electrons of all the three benzene rings are delocalised with the vacant p-orbitals of the central carbon atom. So, it is resonance stabilized. It is the most stable of all the carbonium ions given is stabilized by hyperconjugation, asecond order resonance.

Question 122. Which of the following resonating structures of 1-methoxy-1, 3-butadiene is least stable?

Answer: 3.

Solution: The octet of all atoms is complete in the structures a and b. The molecule in which all the atoms have completed octet is more stable than atoms which have incomplete octet. Larger the number of resonating structures, larger will be the stability, thus structures a and b are stable.

In structure (4), the electron deficient of positive charged carbon is duly compensated by one pair of electrons of adjacent oxygen atoms while such neighbor group support is not available in structure (3). Hence, structure (3) is least stable in comparison to structure (41).

Question 123. The reaction is an example of

1. Electrophilic addition
2. Electrophilic substitution
3. Nucleophilic substitution
4. Nucleophilic addition

Answer: 4. Nucleophilic addition

Solution: It is an example of a nucleophilic addition reaction.

Question 124. The reaction of sodium ethoxide with iodoethane to from diethyl is termed as

1. Electrophilic substitution
2. Nucleophilic substitution
3. Electrophilic addition
4. Radical substitution

Answer: 2. Nucleophilic substitution

Solution: When sodium ethoxide reacts with iodoethane, diethyl ether is obtained (Williamson’s synthesis) The mechanism of this reaction is as follows

Since, the reaction involves the substitution of a group by a nucleophile, it is an example of nucleophilic substitution reaction.

Question 125. Which group has the highest + Inductive effect?

1. CH₃–
2. CH₃CH₂–
3. (CH₂)₂CH-
4. (CH₃)₃C

Answer: 4. (CH₃)₃C-

Solution: The increasing order is: -CH₃<CH₃-CH₂– <(CH₃)₂CH-<(CH₃)₃C

Question 126. When SCN^– is added to an aqueous solution containing Fe(NO₃)₃,the complex ion produced is

1. [Fe(OH₂)2(SCN) ]2⁺
2. [Fe(OH2)5(SCN) ]2⁺
3. [Fe(OH2)8(SCN) ]2⁺
4. [Fe(OH2)(SCN)]6⁺

Answer: 2. [Fe(OH2)5(SCN) ]2⁺

Solution: On adding SCN^– to an aqueous solution of Fe(NO3)3, a blood red colour, due to formation of [Fe(H2O)5(SCN)]⁺ complex is obtained. This test is used for the detection of Fe3⁺ ion. SCN^– +Fe(NO3)3+ 5H2O→ [Fe(OH2)5(SCN)]2+ + 3NO3^– Blood red colour

Question 127. Which shows the easier electrophilic substitution in a ring?

1. N-acetyl aniline
2. C₆H₅NH₃Cl
3. Aniline
4. Nitrobenzene

Answer: 3. Aniline

Solution: Ortho and para-directing groups facilitate the ring for electrophilic substitution reaction. –NH₂ group increases the electron density in the ring, hence activates electrophilic substitution.

Question 128. Correct gradation of basic character

1. NH₃CH₃NH₂>NF₃
2. CH₃NH₂>NH₃>NF₃
3. NF₃>CH₃NH₂>NH₃
4. CH₃NH₂>NF₃>NH₃

Answer: 2. CH₃NH₂>NH₃>NF₃

Solution: In the gas phase, tertiary amines are more basic than secondary amines which are more basic than ammonia -Group present on the central atom decreases electron density, hence decreases basicity

CH₃NH₂>NH₃>NF₃

Question 129. A carbonium ion contains

1. A positively charged carbon center
2. A negatively charged carbon center
3. A carbon with an odd electron on it
4. None of the above

Answer: 1. A positively charged carbon center

Solution: It is a fact.

Question 130. In which of the following species the central carbon atom is negatively charged?

1. Carbonium ion
2. Carbanion
3. Carbocation
4. Free radicals

Answer: 2. Carbanion

Question 131. Alkaline hydrolysis of an ester (A) gives alcohol and salt

The correct statement about the reaction is

1. In alcohol configuration about chiral carbon atom is retained
2. In alcohol configuration about chiral carbon atom is inverted
3. Alcohol loses optical activity
4. All statements are incorrect

Answer: 1. In alcohol configuration about chiral carbon atom is retained

Solution: No bond around chiral carbon is broken and so configuration will be retained.

Question 132. Formic acid is a stronger acid than acetic acid. This can be explained using

1. +Meffect
2. -I effect
3. +I effect
4. -M effect

Answer: 2. -I effect

Solution: Electron withdrawing group has –I effect while electron donating group has +I effect. In CH₃COOH, the alkyl group (-CH₃) due to its greater +I effect increases the electron density on the oxygen atom of the O-H bond. Due to this the release of H⁺ ions in acetic acid will be more difficult as compared to the formic acid.

Question 133. Zero inductive effect is shown by? 

1. C₆H₅—
2. —H
3. CH₃
4. — Cl—

Answer: 4. — Cl—

Solution: Inductive effect of groups is measured with respect to H.

Question 134. The hydrolysis of alkyl halides by aqueous NaOH is best termed as

1. Electrophilic substitution reaction
2. Electrophilic addition reaction
3. Nucleophilic addition reaction
4. Nucleophilic substitution reaction

Answer: 4. Nucleophilic substitution reaction

Solution: \(R-X \stackrel{\mathrm{NaOH}}{\longrightarrow} R-\mathrm{OH}+\mathrm{NaX} R-X \stackrel{\mathrm{OH}^{-}}{\longrightarrow} R-\mathrm{OH}+X^{-}\)This is nucleophilic substitution.

Question 135. The intermediate during the addition of HCl to propene in the presence of peroxide

Answer: 2

Solution: Addition of HCl is not peroxide effect and it occurs via electrophilic addition.

Question 136. The structure which has positive charge on the oxygen atom

Answer: 1.

Solution:

Question 137. Addition of Br₂ on CH₂=CH₂ in presence of NaCl (aq.) gives?

1. CH₂Br.CH₂Br
2. CH₂Br.CH₂Cl
3. CH₂Br.CH₂OH
4. All of these

Answer: All of these

Solution: Once the carbocation is formed as an intermediate, the nucleophile Cl^– and OH^– present in solution also attach it in addition of Br^–

Question 138. The oxygen atom in phenol

1. Exhibits only inductive effect
2. Exhibits only resonance effect
3. Has more dominating resonance effect than inductive effect
4. Has more dominating inductive effect than the resonance effect

Answer: 3. Has more dominating resonance effect than inductive effect

Solution: The oxygen atom in phenol has more dominating resonance effect than inductive effect. Increase in charge separation decreases the stability of a resonating structure. Stability of resonating structure in decreasing order will be

1>2 ≡ 4>3

Question 139. The reaction intermediate produced, by homolytic cleavage of a bond is called

1. Carbene
2. Carbocation
3. Carbanion
4. Free radical

Answer: 4. Free radical

Solution: In homolytic cleavage, covalent bond is cleaved in such a way that each atom takes its shared electrons with itself and free radicals are formed.

Question 140. Among the following the least stable resonance structure is

Answer: 1

Solution: Two positive charges present at the adjacent place, elevates the energy, thus lowering the stability most.

Question 141. Carbocation can undergo

1. Loss of a proton
2. Addition to multiple bond
3. Combination with anions
4. All of the above

Answer: 4. All of the above

Solution: These are characteristics of carbocations.

Question 142. The electromeric effect in organic compounds is

1. Temporary effect
2. Permanent effect
3. Temporary-permanent effect
4. None of the above

Answer: 1. Temporary effect

Solution: It is raised on the need of attacking reagent, example,

Question 143. Dehydration of alcohol usually goes by

1. E₁ mechanism
2. E₂ mechanism
3. E₁ cb mechanism
4. S_N2 mechanism

Answer: 

Solution: Alcohols undergo dehydration usually by E₁ mechanism. This is because elimination is preferred in the case of tertiary alcohols, for example,

Question 144. Removal of a hydride ion from a methane molecule will give

1. Methyl radical
2. Carbonium ion
3. Carbanion
4. Methyl group

Answer: 2. Carbonium ion

Solution: CH₄⟶CH₃⁺+ + H^–; CH₃⁺ is methyl carbonium.

Question 145. The electrophile involved in the Sulphonation of benzene is

1. SO₃⁺
2. SO₄^2-
3. HSO₄^–
4. SO₃

Answer: 4. SO₃

Solution: The electrophile involved in the Sulphonation of benzene is SO₃ 2H₂SO₄→ SO₃+H₃O++HSO₄^–

Question 146. In the following carbocations, the stability order is

1. R-CH₂+ CH₃

2.

3.

4.

1. 3> 2 >4 > 1
2. 2 > 3 >4> 1
3. 3 > 2 > 1 > 4
4. 3> 4>2> 1

Answer: 1. 3> 2 >4 > 1

Solution: Cyclopropyl methyl carbocations are more stable than benzyl carbocations due to conjugation between bent orbitals of the cyclopropyl group.

Question 147. Which of the following undergoes nucleophilic substitution exclusively S_N1 mechanism?

1. Benzyl chloride
2. Isopropyl chloride
3. Chlorobenzene
4. Ethyl chloride

Answer: 1. Benzyl chloride

Solution: Benzyl carbonium is more stable due to resonance and thus, benzyl chloride is more reactive.

Question 148. Which of the following species is not electrophilic in nature?

1. \(\stackrel{\oplus}{C_1}\)
2. \(\mathrm{BH}_3\)
3. \(\mathrm{H}_3 \stackrel{\oplus}{O}\)
4. \(\stackrel{\oplus}{\mathrm{N}} \mathrm{O}_2\)

Answer: 3.

Solution: H₃⁺O cannot accept electron pairs.

Question 149. Which of the following reactions is an example of a nucleophilic substitution reaction?

1. RX + Mg ⟶ RMgX
2. RX+KOH ⟶ROH+KX
3. 2RX + 2Na ⟶ R —R + 2NaX
4. RX + H₂⟶ RH + HX

Answer: 2. RX+KOH ⟶ROH+KX

Solution: X–is replaced by OH–.

Question 150. Examine the following statements regarding the S_N2 reaction

1. The rate of reaction is independent of the concentration of nucleophile

2. The nucleophile attacks the carbon atom on the side of molecule opposite to the group being displaced

3. The reaction proceeds with simultaneous bond formation and rupture

4. Which of the above-written statements is correct?

1. 1, 2
2. 1, 3
3. 1, 2, 3
4. 2, 3

Answer: 4. 2, 3

Question 151. The number of different substitution products possible when ethane is allowed to react with bromine is sunlight are:

1. 9
2. 6
3. 8
4. 5

Answer: 1. 9

Solution: CH₃CH₂Cl; CH₃CHCl₂; CH₂ClCH₂Cl; CH₃CCl₃; CH₂ClCHCl₂; CH₂ClCCl₃; CHCl₂CHCl₂;CHCl₂CCl₃; CCl₃CCl₃

Question 152. Select the strongest bond amongst the following

Answer: 4.

Solution: Triple bond possesses maximum bond energy.

Question 153. Who proposed the concept of hyperconjugation?

1. Nathan and Baker
2. Mullikan
3. Kekule
4. Kolbe

Answer:  1. Nathan and Baker

Solution: It is a fact.

Question 154. Among the following compounds nitrobenzene, benzene, aniline and phenol, the strongest basic behavior in acid medium is exhibited by

1. Phenol
2. Aniline
3. Nitrobenzene
4. Benzene

Answer: 2. Aniline

Solution: Due to the presence of a lone pair on the “N” atom.

Question 155. Mesomeric effect involves delocalization of

1. Pi-electrons
2. Sigma electrons
3. Protons
4. None of these

Answer: 1. Pi-electrons

Solution: Mesomeric effect involves the complete transfer of π or lone pair of electrons to the adjacent atom or covalent bond. Hence, it involves the delocalization of pi (π) electrons.

Question 156. Consider the following reaction, Identify the structure of the major product ‘X’

Answer: 2

Solution: Br* is less reactive and more selective and thus formation of 3° free radicals will be the major product.

Question 157. The alkyl halide that undergoes S_N1 reaction more readily is

1. Ethyl bromide
2. Isopropyl bromide
3. Vinyl bromide
4. n-propyl bromide

Answer: 3. Vinyl bromide

Solution: S_N1 mechanism involves the formation of carbocation intermediate. Hence, the species which gives the most stable carbocation readily undergoes S_N1 mechanism. t-butyl bromide gives the most stable carbocation, i.e., 3° carbocation, so it readily undergoes S_N1 reaction.

Question 158. The homolytic fission of a hydrocarbon results in the formation of:

1. Carbonium ions
2. Free radicals
3. Carbanions
4. Carbenes

Answer: 2. Free radicals

Solution: Homolytic bond fission is one in which each entity involved in bond formation retains its electron involved in a shared pair of electrons to form free radicals.

Question 159. Which of the following intermediates have the complete octet around the carbon atom?

1. Carbonium ion
2. Carbanion
3. Free radical
4. Carbene

Answer: 2. Carbanion

Solution: Carbanion (CH₃^⊖) Here, the carbon atom carries a negative charge with a lone pair of electrons has eight electrons in the outermost orbit, and completes its octet.

Reactions in which carbanions are formed as intermediate are said to proceed by a “Carbanion mechanism”. Carbanion is sp³ hybridized, three sp³ hybrid orbitals form covalent bonds with three atoms while the fourth sp³ hybrid orbital has a non-bonding pair of electrons. It is pyramidal in shape as similar to NH_3.

Question 160. The kind of delocalization involving sigma bond is called

1. Inductive effect
2. Hyperconjugation effect
3. Electromeric effect
4. Mesomeric effect

Answer: 1. Inductive effect

Solution: The inductive effect is the permanent effect on σ-electrons. It involves the electron displacement along the chain of saturated carbon atoms due to the presence of a polar covalent bond at one end of the chain.

Question 161. An electrophilic reagent must have

1. A vacant orbital
2. An orbital containing one electron
3. An orbital containing two electrons
4. All completely filled atomic orbitals

Answer: 1. A vacant orbital

Solution: The reagent having an affinity for electrons is known as an electrophilic reagent. The electron-deficient species works as an electrophilic reagent. The electrophilic reagent as the name indicates loves the electron because it lacks electrons.

Question 162. The electrophile, E attacks the benzene ring to generate the intermediate σ complex. Of the following, which σ-complex is of lowest energy?

Answer: 2.

Solution: Structure B will be of lowest energy due to resonance stabilization of positive charge.

In all other three structures, the presence of electron-withdrawing NO₂ groups will destabilize the positive charge and hence they will have greater energy.

Question 163. Nucleophiles are:

1. Electron loving
2. Electron hating
3. Nucleus loving
4. Nucleus hating

Answer: 3. Nucleus loving

Solution: Nucleophiles are electron-rich species and can donate lone pairs of electrons to carbocation or any positive centre.

Question 164. Conversion of CH₄ to CH₃Cl is an example of which of the following reactions?

1. Electrophilic substitution
2. Free radical addition
3. Nucleophilic substitution
4. Free radical substitution

Answer: 4. Free radical substitution

Solution: This is an example of a free radical substitution reaction

Question 165. The S_N2 mechanism for, R-X+KOH(aq)⟶R-OH+KX follows with

1. 100% inversion
2. 50% inversion
3. 40% inversion
4. 30% inversion

Answer: 1. 100% inversion

Solution: The S_N2 mechanism always involves 100% inversion since nucleophile attacks from the back side of the leaving group.

⇒ \(R-X \stackrel{\mathrm{OH}^{-}}{\longrightarrow} \mathrm{H} \overline{\mathrm{O}}-\cdots \cdot R^{-\cdots-\cdots} X\)
↓
⇒ \(\mathrm{HO}-R+X^{-}:\)

Question 166. The reaction (CH₃)₃CBr →(H₂O) →(CH₃)₃C.OH is:

1. Elimination reaction
2. Free radical reaction
3. Substitution reaction
4. Displacement reaction

Answer: 3. Substitution reaction

Solution: Br is replaced by –OH.

Question 167. The stability of the carbocation decreases in the order

1. R₂CH+>R₃C+>RCH₂+>CH⁺
2. R₃C+>R₂CH+>RCH₂+>CH₃⁺
3. CH₃+>R₂CH+>RCH₂+>R₃C⁺
4. CH₃+>RCH₂+>R₂CH+>R₃C⁺

Answer: 2. R₃C+>R₂CH+>RCH₂+>CH₃⁺

Solution: Stability of alkyl carbocations can be explained by inductive effect and hyperconjugation. According to these two affects, the stability order is

Question 168. Most stable carbocation is formed during the heating of which of the following compounds with conc.H₂SO₄?

1. (CH₃)₃COH
2. C₆H₅CH₂OH
3. (CH₃)₂CHOH
4. CH₃CH(OH)CH₂CH₃

Answer: 2. C₆H₅CH₂OH

Solution: C₆H₅CH₂⁺ is stabilized by conjugation while intermediates of the rest of the compounds given are stabilized by hyperconjugation.

Question 169. Which of the chlorides is less reactive towards hydrolysis?

1. Vinyl chloride
2. Allyl chloride
3. Ethyl chloride
4. t-butyl chloride

Answer: 1. Vinyl chloride

Solution: Due to resonance, the partial double bond character is created on vinyl chloride. So, the chlorine atom is not replaced easily

Question 170. The shifting of electrons of a multiple bond under the influence of a reagent is called as

1. I-effect
2. E-effect
3. M-effect
4. T-effect

Answer:  2. E-effect

Solution: It is the definition of electromeric effect.

Question 171. Which of the following is an example of a substitution reaction?

4. None of the above

Answer:  1

Solution: Replacement of an atom or group by other atom or group is known as substitution reaction

Question 172. Which of the following aromatic acids is most acidic?

Answer: 2

Solution: Due to resonance; the carbonyl group of benzoic acid is coplanar with the ring. If the electron-withdrawing substituent (i.e.,-I showing) is present at the ortho position, it prevents the) coplanarity and thus, the resonance. Hence, it makes the acid stronger. Thus, among the given acids, ortho hydroxy benzene acid is the most acidic.

Question 173. Which of the following is most reactive towards electrophilic nitration?

1. Toluene
2. Benzene
3. Benzoic acid
4. Nitrobenzene

Answer: 1. Toluene

Solution: Due to +I effect of CH₃ in toulene, it is more reactive than benzene. Due to electron withdrawing nature of the –COOH group in benzoic acid and –NO₂ group in nitrobenzene, both benzoic acid and nitrobenzene are less reactive than benzene.

Question 174. The ion formed by the reaction of HNO₂ and H₂SO₄ is

1. Nitronium ion
2. Nitrosonium ion
3. Nitrite ion
4. Nitrate ion

Answer: 2. Nitrosonium ion

Solution: HNO₂ +H₂SO₄⟶NO++HSO₄^––+H₂O

Question 175. Which of the following cannot show S_N1 reaction?

Answer: 4

Solution: Primary and secondary alkyl halides give S_N2 reaction

Question 176. Which one of the following has the most nucleophilic nitrogen?

Answer:  1

Solution:

Nucleophiles are the species which have excess electrons. Among the given species, the lone pair of nitrogen of pyrrole is involved in the delocalization of the ring and, thus, are not available for donation. In aniline, the lone pair is involved in conjugation with the π-electrons of the ring while in pyridine, these are relatively free for donation. Thus, the nitrogen of pyridine is most nucleophilic.

(Phenyl and –COCH in toulene, it is more reactive than benzene. Due to electron withdrawing nature of the –CO both are electron withdrawing groups, thus decreasing the nucleophilicity of nitrogen).

Question 177. The most stable carbonium ion among the following is

Answer: 3

Solution: \(\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{HC}_6 \mathrm{H}_5\) is the most stable since the positive charge can be delocalized on both phenyl rings

Question 178. Which of the following is an electrophilic reagent?

1. RO^–
2. BF₃
3. NH₃
4.  

Answer: 2. BF₃

Solution: BF₃ is an electron-deficient compound.

Question 179. The formation of cyanohydrin from a ketone is an example of

1. Electrophilic addition
2. Nucleophilic addition
3. Electrophilic substitution
4. Nucleophilic substitution

Answer: 2. Nucleophilic addition

Solution: Ketone undergoes nucleophilic addition reaction because the nucleophilic end of reagent attack is first followed by the electrophilic end of the reagent.

Question 180. Which of the following orders is correct regarding the acidity of the carboxylic group?

1. CH₃CH₂CH(Cl)COOH>CH₃CH(Cl)CH₂COOH>ClCH2CH2CH₂COOH
2. CH₃CH₂CH(Cl)COOH<CH₃CH(Cl)CH₂COOH<ClCH2CH2CH₂COOH
3. CH₃CH₂CH(Cl)COOH>CH₃CH(Cl)CH₂COOH<ClCH2CH2CH₂COOH
4. CH₃CH₂CH(Cl)COOH<CH₃CH(Cl)CH₂COOH>ClCH2CH2CH₂COOH

Answer:  1. CH₃CH₂CH(Cl)COOH>CH₃CH(Cl)CH₂COOH>ClCH2CH2CH₂COOH

Solution: The presence of -I-showing group like Cl increases the acidic character of carboxylic acids and the acidity reduces with increase in the distance between – COOH and – /-showing group.

Question 181. Which step is the chain termination step in the following mechanism?
Answer: 

1. \(\mathrm{Cl}_2 \stackrel{h v}{\longrightarrow} \mathrm{Cl}^{\bullet}+\mathrm{Cl}^{\bullet}\)
2. \(\mathrm{Cl}^{\bullet}+\mathrm{CH}_4 \longrightarrow \stackrel{\bullet}{\mathrm{C}} \mathrm{H}_3+\mathrm{HCl}\)
3. \(\stackrel{\bullet}{\mathrm{C}} \mathrm{H}_3+\mathrm{Cl}_2 \longrightarrow \mathrm{CH}_3 \mathrm{Cl}+\mathrm{Cl}^{\bullet}\)
4. \(\mathrm{Cl}^{\bullet}+\dot{\mathrm{C}} \mathrm{H}_3 \longrightarrow \mathrm{CH}_3 \mathrm{Cl}\)

Answer: \(\mathrm{Cl}^{\bullet}+\dot{\mathrm{C}} \mathrm{H}_3 \longrightarrow \mathrm{CH}_3 \mathrm{Cl}\)

Solution: From the above mechanism it is clear that step 2nd is the chain propagation step because in this step regeneration of species takes place. So, the 3rd & 4th step is the termination step because after these steps no species are available for further reaction.

As two chlorine free radicals reacts together to form chlorine molecule and one chlorine free radical & one methane free radical react together to form chloromethane.

Question 182. Relative stabilities of the following carbocations will be in the order

1. \(\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_3\)

2. \(\mathrm{CH}_3 \stackrel{\oplus}{\mathrm{C}} \mathrm{H}_2\)

3. \(\stackrel{\oplus}{\mathrm{C}} \mathrm{H}_2 \mathrm{OCH}_3\)

1. 3>2>1
2. 3<2<1
3. 2>3>1

Answer:  1. 3>2>1

Solution: The dispersal of the charge stabilizes the carbocation. More the number of alkyl groups; the greater is the dispersal of positive charge and therefore, more the stability of carbocation, C₂H₅+>CH₃+,O-CH₃ is also an electron donating group, thus it will increase the stability of carbocation, hence, the correct order of stability is 3>2>1

Question 183. The addition of HI on double bonds of propene yields isopropyl iodide as a major product. It is because the addition proceeds through:

1. More stable carbocation
2. More stable carbanion
3. More stable free radical
4. Homolysis

Answer:  1. More stable carbocation

Solution: Formation of 2° carbocation, i.e.,

⇒ \(\mathrm{CH}_3 \mathrm{CH}=\mathrm{CH}_2 \stackrel{\stackrel{+\delta}{\mathrm{H}}-\stackrel{-}{\mathrm{H}}}{\longrightarrow} \mathrm{CH}_3 \stackrel{+}{\mathrm{C}} \mathrm{HCH}_3\)

Question 184.

1. The above reaction proceeds through
2. Free radical substitution
3. Nucleophilic substitution
4. Electrophilic substitution
5. None of the above

Answer:  1. The above reaction proceeds through

Solution: It is a free-radical substitution reaction, initiated by the formation of chlorine atoms Cl⋅ from Cl₂.

The initial step is hydrogen abstraction from toluene, and the most stable free radical is formed, C₆H₅CH₂, the one in which the radical center is in resonance with the aromatic ring.

Question 185. Arrange in order of increasing acidic strength.

1. X>Z>Y
2. Z<X>Y
3. X>Y>Z
4. Z>X>Y

Answer: 1. X>Z>Y

Solution: The pKa value of the carboxylic group is less than pKa of NH₃⁺ in amino acid and —NH₃⁺ will have comparatively less pKa than (Y) due to –I effect of the carboxylic group. We know that acidic strength in inversely proportional to pKa. Hence, the correct order of acidic strength is

Question 186. Heterolysis of the carbon-chlorine bond produces

1. Two free radicals
2. Two carbonium ions
3. Two carbanions
4. One cation and one anion

Answer: 4. One cation and one anion

Solution: Heterolysis involves the bond fission in a manner when either of the two atoms involved in bond fission retains the shared pair of electrons, producing positive and negative ions, example,

Question 187. Which of the following statements is correct?

1. Allyl carbonium ion ismore stable than propyl carbonium ion
2. Propyl carbonium ion (CH₂=CH–CH₃⁺ ) is more stable than the allyl carbonium ion
3. Both are equally stable
4. None of the above

Answer: 3. Allyl carbonium ion ismore stable than propyl carbonium ion

Solution: Allyl carbocations are more stable than the alkyl carbocations due to the resonance stabilization.

Question 188. The order of stability of the following carbanion is

1. 1>2>3>4
2. 1>3>2>4
3. 4>3>2>1
4. 3>4>1>2

Answer: 4. 3>4>1>2

Solution: I can have a maximum of 3 hyper conjugative structures. 2 has maximum of 5 hyper conjugative structures, 3 has 2 conjugative structures while 4 has 1 conjugative structure.

Question 189. Following reaction is,

1. S_N
2. S_E
3. El
4. EI-CB

Answer:  1. S_N

Solution: Diazonium salts are highly reactive. In the Sandmeyer reaction diazo group is replaced by chlorine or bromine in presence of CuCl or CuBr. (Substitution reaction)

⇒\(\mathrm{C}_6 \mathrm{H}_5 \stackrel{\oplus}{\mathrm{N}_2} \stackrel{\ominus}{\mathrm{Cl}} \stackrel{\mathrm{CuCl}}{\longrightarrow} \mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl}+\mathrm{N}_2\)

Question 190. Which one of the following is most reactive towards electrophilic attack?

Answer: 2.

Solution: O- and p-directing groups facilitate SE reactions whereas m-directing groups deactivate the benzene ring for SE reactions.

Question 191. Heterolysis of propane gives

1. Methyl and ethyl free radicals
2. Methylium cation and ethyl anion
3. Methyl anion and ethidium cation
4. Methylium and ethylium cations

Answer: 3. Methyl anion and ethidium cation

Solution: \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_3 \stackrel{\text { Hetrolysis }}{\longrightarrow} \overline{\mathrm{C}} \mathrm{H}_3+\mathrm{CH}_3 \stackrel{+}{\mathrm{C}} \mathrm{H}_2, \mathrm{CH}_3 \stackrel{+}{\mathrm{C}} \mathrm{H}_2\) due to dispersal of positive charge on ethylium ion on account of positive inductive effect. Thus, propane will not give\(\stackrel{+}{\mathrm{C}} \mathrm{H}_3 \text { and } \mathrm{CH}_3 \overline{\mathrm{C}} \mathrm{H}_2\)

Question 192. Which among the following compounds is most acidic?

Answer: 2.

Solution: Ortho nitrophenol is the most acidic because the electron withdrawing group increases acidic character due to –I effect of NO₂

Question 193. Hyperconjugation is

1. σ-π delocalisation
2. No bond resonance
3. σ-π odd electron
4. All of these

Answer: 4. All of these

Solution: It is a fact.

Question 194. Identify the reaction.

1. Substitution reaction
2. Elimination reaction
3. Rearrangement reaction
4. None of the above

Answer:  3. Rearrangement reaction

Solution: It is a fact.

Question 195. Dehydrogenation of ethanol to give ethanal is known as

1. Addition reaction
2. α-α elimination reaction
3. α-β elimination reaction
4. α-― elimination reaction

Answer: 2.  α-α elimination reaction

Solution: The reaction in which 2 atoms from a molecule are removed from the same atom is called α – α-elimination. It leads to the formation of an electron deficient reactive intermediate.

Question 196. The stability of a carbonium ion depends upon

1. The bond angle of the attached group
2. The substrate with which it reacts
3. The inductive effect and hyper-conjugative effect of the attached group
4. None of the above

Answer: 3. The inductive effect and hyper-conjugative effect of the attached group

Solution: It is a fact.

Question 197. Which of the following is an electrophile?

1. : CCl₂
2. CO₂
3. H₂O
4. NH₃

Answer:  1. : CCl₂

Solution: Electron deficient species or electron acceptor is electrophile. For example

⇒ \(\dot{\mathrm{C}} \mathrm{H}_3, \ddot{\mathrm{C}} \mathrm{H}_2, \dot{\mathrm{C}} X_2\)

Question 198. The reaction of phenol with chloroforms/sodium hydroxide to give o-hydroxy benzaldehyde involves the formation of

1. Dichlorocarbene
2. Trichloro carbene
3. Chlorine atoms
4. Chlorine molecules

Answer: 1. Dichlorocarbene

Solution: Phenol reacts with chloroform and NaOH to give o-hydroxy benzaldehyde or salicylaldehyde. In this reaction dichlorocarbene (∶CCl₂) electrophile is generated. This reaction is called the Reimer-Tiemann reaction.

⇒\(\mathrm{OH}^{-}+\mathrm{CHCl}_3 \longrightarrow \mathrm{HOH}+\underset{\text { unstable }}{: \mathrm{CCl}_3^{-}}\)

∶CCl^–₃ → Cl^– +∶CCl₂

Question 199. Carbanion can undergo:

1. Rearrangement
2. Combination with cation
3. Addition to a carbonyl group
4. All of the above are correct

Answer: 4. All of the above are correct

Solution: These are characteristics of carbanion.

Question 200. During the nitration of benzene, the attacking electrophile is

1. NO₃^–
2. NO₂^–
3. NO₂⁺
4. HNO₃

Answer: 3. NO₂⁺

Solution: During nitration of benzene the attacking electrophile is NO₂⁺. It is formed as follows by reaction between HNO₃ and H₂SO₄.

Question 201. The reaction which is not an example of nucleophilic substitution among the following is

Answer:  3

Solution: CH₃ -CH₂ -CH₂-Cl+alc.KOH⟶CH₃-CH=CH₂ It is an example of an elimination reaction.

Question 202. The Kolbe’s electrolysis proceeds via

1. Nucleophilic substitution mechanism
2. Electrophilic addition mechanism
3. Free radical mechanism
4. Electrophilic substitution reaction

Answer: 3.  Free radical mechanism

Solution: The Kolbe’s electrolysis proceeds via free radical mechanism. For example, when sodium propionate is electrolysed, n-butane, ethane and ethylene are obtained. The propionate ion discharges at the anode to form the free radicals.

Question 203. In a S_N2 substitution reaction of the type, which one of the following has the highest relative rate?

⇒ \(\mathrm{R}-\mathrm{Br}+\mathrm{Cl}^{-} \stackrel{\mathrm{DMF}}{\longrightarrow} R-\mathrm{Cl}+\mathrm{Br}^{+}\)

Answer:  3

Solution: For S_N2 mechanism, there should be the least steric hindrance.

Question 204. Which of the following sodium compound/compounds are formed when an organic compound containing both nitrogen and sulphur is fused with sodium?

1. Cyanide and sulfide
2. Thiocyanate
3. Sulphite and cyanide
4. Nitrate and sulfide

Answer: 2. Thiocyanate

Solution: Na reacts with C, N, and S to form NaCNS (sodium thiocyanate).

Question 205. Which one of the following explain, why propene undergoes electrophilic addition withHBr, but not with HCN?

1. Br–is better nucleophile than CN^–
2. HBr being better source of proton as it is stronger acid than HCN
3. HCN attacks preferentially via lone pair of nitrogen
4. The C-Br bond being stronger is formed easily as compared to C-CN bond

Answer: 2. HBr being better source of proton as it is stronger acid than HCN

Solution: HBr being a better source of proton. It gives a H⁺ and a Br^– ion

Thus, H⁺ attack the π bond of propene to form carbonium ion as

Question 206. Di-chloroacetic acid is a stronger acid than acetic acid. This is due to the occurrence of

1. Mesomeric effect
2. Hyperconjugation
3. Inductive effect
4. Steric effect

Answer: 3. Inductive effect

Solution: Di-chloro acetic acid due to presence of two electron with drawing chloro groups (-I showing group)is more acidic than acetic acid(+Ishowing-CH₃ group).

Question 207. The ease of nitration of the following three hydrocarbons follows the order

1. 2=3≈1
2. 2>3>1
3. 3>2>1
4. 1=3>2

Answer:  2. 2>3>1

Solution: Stability order of arenium ion 2>3>1

Question 208. Consider the following carbanions

The correct order of stability is

1. 1>2>3
2. 3>2>1
3. 2>3>1
4. 1>3>2

Answer: 3. 2>3>1

Solution: – NO₂ group shows –Meffect white CH₃O-group shows +Meffect (–M effect stabilizes an anion)

Question 209. Williamson’s synthesis involves

1. S_N1 mechanism
2. Nucleophilic addition
3. S_N2 mechanism
4. SE mechanism

Answer: 3. S_N2 mechanism

Solution: When sodium or potassium alkoxide is heated with an alkyl halide to give ether, the reaction is known as Williamson’s synthesis.

RONa+R’ X→ R-O-R’+NaX

This is an example of nucleophilic substitution reaction and follow S_N2 mechanism

Question 210. The reaction is

⇒ \(\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{Br} \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow}\left(\mathrm{CH}_3\right)_3 \mathrm{COH}\)

1. Elimination
2. Substitution
3. Free radical
4. Addition

Answer: 2. Substitution

Solution: The reaction \(\left(\mathrm{CH}_3\right)_3 \mathrm{C}-\mathrm{Br} \stackrel{\mathrm{H}_2 \mathrm{O}}{\longrightarrow}\left(\mathrm{CH}_3\right)_3 \mathrm{COH}\) is an example of a substitution reaction

Question 211. The reaction \(\mathrm{C}_2 \mathrm{H}_5 \mathrm{I}+\mathrm{KOH} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{KI}\)  is called

1. Hydroxylation substitution
2. Electrophilic substitution
3. Nucleophilic substitution
4. Dehydroiodination

Answer: 3. Nucleophilic substitution

Solution: Nucleophiles may be neutral or negatively charged, whereas substrates undergoing nucleophilic substitution may be neutral or positively charged.

⇒ \(\mathrm{C}_2 \mathrm{H}_5-\mathrm{I}+\mathrm{OH}^{-} \rightarrow \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}+\mathrm{I}^{-}\)

Question 212. In methanol solution, bromine reacts with ethylene to yield BrCH₂CH₂OCH₃ in addition to 1,2-dibromoethane because:

1. The intermediate carbocation may react with Br– or CH₃OH
2. The methyl alcohol solvates the bromine
3. The reaction follows Markownikoff’s rule
4. This is a free radical mechanism

Answer:  1. The intermediate carbocation may react with Br– or CH₃OH

Solution: In methyl alcohol solution, bromine reacts with ethylene to yield BrCH₂CH₂OCH₃ in addition to 1,2-dibromoethane because the intermediate bromonium ion formed initially may react with Br− or CH₃OH.

Question 213. In a compound electrophilic substitution has occurred. The substitute-Eare methyl –CH₂Cl,-CCl₃ and–CHCl₂. The correct increasing order towards electrophilic substitution is 

1. -CH₃<-CH₂Cl<-CHCl₂<-CCl₃
2. -CH₃<-CHCl₂<-CH₂Cl<-CCl₃
3. -CCl₃<-CH₂Cl<-CHCl₂<-CH₃
4. -CCl3<-CHCl₂<-CH₂Cl<-CH₃

Answer: 4. -CCl3<-CHCl₂<-CH₂Cl<-CH₃

Solution: Chlorine atoms are strongly electronegative (show negative inductive effect i.e., -I effect). They deactivate the ring towards an electrophilic reaction. The increasing order of substituent-E towards electrophilic substitution is -CCl₃<-CHCl₂<-CH₂Cl<-CH₃

Question 214. The addition of HBr on butene-2 in presence of peroxide follows the:

1. Electrophilic addition
2. Free radical addition
3. Nucleophilic addition
4. None of these

Answer: 2. Free radical addition

Solution: Follow the mechanism of the Kharasch effect.

Question 215. A neutral divalent carbon intermediate produced by the removal of two attached atoms is called

1. Free radical
2. Carbanion
3. Carbocation ion
4. Carbene

Answer: 4. Carbene

Solution: Follow carbenes.

Question 216. On exciting Cl₂ molecules by UV light, we get

1. Cl^•
2. Cl^+-
3. Cl^–
4. All of these

Answer: 1. Cl^•

Solution: Covalent bonds are cleaved in a homolytic way in the presence of UV light. It results in the formation of free radicals.

⇒ \(\mathrm{Cl} \stackrel{\bullet}{\mathrm{Cl}} \stackrel{\mathrm{UV}}{\longrightarrow} \dot{\mathrm{Cl}}+\dot{\mathrm{C}} \mathrm{l}\) (Chlorine Free Radicals)

Question 217. For the reaction,

1. Chloro benzene and carbon tetrachloride
2. Meta chloro benzotrichloride ortho,
3. Para chloro benzo trichloride
4. None of these

Answer: 

Question 218. Which of the following compounds is resistant to nucleophilic attack by hydroxyl ion?

1. Methyl acetate
2. Acetonitrile
3. Acetamide
4. Diethyl ether

Answer: 4. Diethyl ether

Solution: Diethyl ether is resistant to nucleophilic attack by hydroxyl ion.

Question 219. Among the following compounds (1-2) the correct order of reaction with the electrophile is

1. 2>3>1
2. 3<1<2
3. 1>2>3
4. 1≈2>3

Answer: 3. 1>2>3

Solution: Activating groups like –OCH₃,-OH etc activates the benzene ring towards electrophilic substitution while deactivating groups like-NO₂,-COOH etc. deactivates the benzene ring towards electrophilic substitution. Thus, order of reaction towards electrophile (of the given compounds) is as 1>2>3.

Question 220. In the given structure, which carbon atom is most electronegative?

1. 1
2. 2
3. 3
4. 4

Answer: 4. 4

Solution: Electronegativity of different hybrid and unhybrid orbitals in decreasing order is as follows s>sp>sp²>sp³>p

Question 221. S_N1 reaction on optically active substrates mainly gives

1. Retention in configuration
2. Inversion in configuration
3. Racemic product
4. No product

Answer: 

Solution: S_N1 mechanism gives rise to 50% inversion as it involves front seat as well as back seat substitution. This leads to racemic products.

Question 222. The stability order for carbocations given below are?

1. 1< 2 < 3
2. 3 < 2 < 1
3. 3< 1 < 2
4. 2 < 1 < 3

Answer: 1. 1< 2 < 3

Solution: Vinyl carbocations are more stable than primary carbocation but less stable than secondary carbocation.

Question 223. Which one of the following reactions is a condensation reaction?

1. HCHO ⟶para-formaldehyde
2. CH₃CHO⟶para-aldehyde
3. CH₃COCH₃⟶ mesityl oxide
4. CH₂= CH₂⟶ polyethylene

Answer: 3. CH₃COCH₃

Solution: Rest all are polymerization reactions.

Question 224. Inductive effect involves

1. Delocalization of σ-electrons
2. Displacement of σ-electrons
3. Delocalization of π-electrons
4. Displacement of π-electrons

Answer: 2. Displacement of σ-electrons

Solution: Inductive effect involves only displacement (and not delocalization) of σ-electrons.

Question 225. The most stable carbocation among the following is?

Answer: 1

Solution: Due to the property of resonance, extra stability is seen in 3° carbocation.

Question 226. Which group has the maximum-Inductive effect?

1. -NO₂
2. -CN
3. -COOH
4. -F

Answer: 1. -NO₂

Solution: The increasing order of inductive effect is: -F<-COOH<-CN<-NO₂.

Question 227. Which of the following statements is not characteristic of free radical chain reaction?

1. It gives a major product derived from most stable free radical
2. It is usually sensitive to change in solvent polarity
3. It proceeds in three main steps like initiation, propagation and termination
4. It may be initiated by UV light

Answer: 2. It is usually sensitive to change in solvent polarity

Solution: Free radical chain reaction is initiated by UV light. It proceeds in three main steps likeinitiation, propagation and termination. It gives major products derived from the most stable free radical.

Question 228. Reactivity towards nucleophilic addition reaction of

1.  HCHO

2. CH₃CHO

3. CH₃COCH₃ is

1. 2>3>1
2. 3>2>1
3. 1>2>3
4. 1>3>2

Answer: 3. 1>2>3

Solution: The nucleophilic addition reaction is the characteristic addition of carbonyl compounds. Reactivity order of carbonyl compounds is in the order.

This is due to an increase in the intensity of charge on carbon of the carbonyl group due to +I effect of alkyl groups.

Question 229. Which of the following is a nucleophilic addition reaction?

1. Hydrolysis of ethyl chloride by NaOH
2. Purification of acetaldehyde by NaHSO₃
3. Alkylation of anisole
4. Decarboxylation of acetic acid

Answer: 2. Purification of acetaldehyde by NaHSO₃

Solution: Sodium hydrogen sulphite adds to aldehydes and ketones to form crystalline bisulphite addition products. The product is water soluble and can be converted back to the original carbonyl compound by treating it with dilute mineral acid or alkali. Therefore, these are useful for separation and purification of aldehydes like acetaldehydes.

Question 230. When thiourea is heated with metallic sodium, the compound which can’t be formed is

1. NaCNS
2. NaCN
3. Na₂SO₄
4. Na₂S

Answer: 3. Na₂SO₄

Solution: The chemical formula of thiourea is NH₂CSNH₂ so here Na₂S, NaCN and NaCNS will be formed but not Na₂SO₄

Question 231. Which of the following statements is correct?

1. +Igroup stabilizes a carbocation
2. +Igroup stabilizes a carbanion
3. -I group stabilizes a carbocation
4. -I group stabilizes a free radical

Answer: 1. +Igroup stabilizes a carbocation

Solution: +I group stabilizes carbocation due to the dispersal of positive charge on the + I effect group also.

Question 232. The stability of 2,3-dimethyl but-2-ene is more than 2-butene. This can be explained in terms of?

1. Resonance
2. Hyperconjugation
3. Electromeric effect
4. Inductive effect

Answer: 2. Hyperconjugation

Solution: The former possesses 12α-H atoms whereas, later possesses six α-H atoms. More is the no. of α-H atom, more is the delocalization and more is the stability.

Question 233. Which of the following is singlet carbine?

Answer:  2

Solution: An intermediate neutral species having a divalent carbon atom with 6 valence electrons out of which two are present in the same orbital with opposite spins is called singlet carbene.

Question 234. The product of the reaction is

Answer:  2

Solution: When 3-methyl-2-pentene is reacted with hypochlorous acid, the product formed is 2- chloro-3-methyl pentanol-3. The reaction obeys Markovnikov’s addition and the hydroxide and chloro group is added across the double bond of two carbons.

Question 235. Alkyl halide can be converted into alkene by

1. Nucleophilic substitution reaction
2. Elimination reaction
3. Both nucleophilic substitution and elimination reaction
4. Rearrangement

Answer: 2. Elimination reaction

Solution: \(\left.\mathrm{R}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{X}+\mathrm{KOH} \text { (alc. }\right) \rightarrow \mathrm{R}-\mathrm{CH}=\mathrm{CH}_2+\mathrm{KX}+\mathrm{H}_2 \mathrm{O}\)

Alkyl halide undergo β-elimination to form alkene.

Question 236. Which step is the chain propagation step in the following mechanism?

Answer: 2

Solution: The chain propagation step involves the use of free radical and regeneration of another free radical.

Question 237. The addition reaction among the following is

Answer: 2

Solution: It is an example of an addition reaction.

Question 238. In this reaction, an asymmetric centre is generated. The acid obtained would be

⇒ \(\mathrm{CH}_3 \mathrm{CHO}+\mathrm{HCN} \rightarrow \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{CN} \stackrel{\text { H.OH }}{\longrightarrow} \mathrm{CH}_3 \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}\)

1. 50%D+50%L-isomer
2. 20%D+80%L-isomer
3. D-isomer
4. L-isomer

Answer: 1. 50%D+50%L-isomer

Solution: Lactic acid obtained in the given reaction is an optically active compound due to the presence of chiral C-atom. It exists as d and l forms whose ratio 1:1.

Question 239. The reaction is an example of

1. Nucleophilic substitution
2. Electrophilic addition
3. Elimination reaction
4. Nucleophilic addition

Answer: 2. Electrophilic addition

Solution: A hydrogen halide containing a highly polar H-Xbond can easily lose to the pi bond of an alkene. The result of the attack of H^⊕ is an intermediate carbocation, which quickly undergoesreaction with the negative halide ion (X–) to yield an alkyl halide

Question 240. How many structures of F are possible?

1. 2
2. 5
3. 6
4. 3

Answer:  4. 3

Question 241. Which one of the following does not show resonance?

1. Carbon dioxide
2. Benzene
3. Nitromethane
4. Propane

Answer: 4. Propane

Solution: Alkanes do not show resonance.

Question 242. The reagent used in dehalogenation process is

1. KOH alc.
2. Zn dust + alc.
3. Na
4. KOH(aq)

Answer: 2. Zn dust + alc.

Solution: Zn dust is used for dehalogenation

⇒ \(\mathrm{CH}_2 X . \mathrm{CH}_2 X \stackrel{\text { Zn dust }}{\longrightarrow} \mathrm{CH}_2=\mathrm{CH}_2\)

Question 243. Which alkyl halide is preferentially hydrolyzed by S_N1 mechanism?

1. (CH₃)₃C.Cl
2. CH₃CH₂CH₂Cl
3. CH₃CH₂Cl
4. CH₃ Cl

Answer: 1. (CH₃)₃C.Cl

Solution: Tertiary halide always favors S_N1 mechanism (as they give comparatively stable carbocation) white primary halide favors S_N2 mechanism.

Question 244. Which of the following has the most acidic hydrogen?

1. 3-hexanone
2. 2, 4-hexanedione
3. 2, 4-hexanedione
4. 2, 3-hexanedione

Answer: 2. 2, 4-hexanedione

Solution: When methylene group (-CH₂) is attached with two electron withdrawing groups (like,- CHO,>C=O,-COOH,-CN,-X,etc), its acidity will increase due to –I effect of the electron withdrawing groups.

Question 245. In the reaction, water is formed by the combination of?

1. Hydroxyl of acid with alcoholic hydroxyl hydrogen
2. Hydroxyl of alcohol with carboxylic hydrogen
3. Both the above changes
4. None of the above

Answer: 1. Hydroxyl of acid with alcoholic hydroxyl hydrogen

Solution: Follow the mechanism of esterification.

Question 246. The function of soda lime, a mixture of solid NaOH and solid CaO during the decarboxylation of carboxylic acids is

1. To increase the rate of reaction
2. To decrease the rate of reaction
3. To change the rate of reaction
4. None of the above

Answer: 2. To decrease the rate of reaction

Solution: CaO is added to NaOH to retard activity of NaOH, otherwise decarboxylation of acids will occur more violently.

⇒ \(R \mathrm{COONa} \stackrel{\mathrm{NaOH}+\mathrm{CaO}}{\longrightarrow} R-\mathrm{H}+\mathrm{Na}_2 \mathrm{CO}_3\)

Question 247. Example of chlorinolysis among the following is

Answer: 4.

Solution: Chlorinolysis involves the substitution reactions by chlorine.

Question 248. The correct order of nucleophilicity among the following is

1. 3, 2, 1, 4
2. 1, 2, 3, 4
3. 4, 3, 2, 1
4. 2, 3, 1, 4

Answer: 4. 2, 3, 1, 4

Solution: The nucleophilicity depends on the strength of the conjugate acid of the nucleophile. If the conjugate acid of the nucleophile is a weak acid, then the corresponding nucleophile will be stronger in nature.

Going by that fact, the conjugate acids of the above nucleophiles are acetic acid, methanol, hydrogen cyanide and toluene para sulphonic acid.

Among the following acids, the weakest acid is methanol. Hence its conjugate base, methoxy group will be the strongest nucleophile, while the weakest one will be the p-toluene sulphonate.

Question 249. The stability of

In the increasing order is

1. 3<1<4<2
2. 1<2<3<4
3. 4<3<2<1
4. 2<3<4<1

Answer: 1. 3<1<4<2

Solution: Can be solved on the basis of hyperconjugative structures

Question 250. Dehydrohalogenation of an alkyl halide is a/an

1. Nucleophilic substitution reaction
2. Elimination reaction
3. Both nucleophilic substitution and elimination reaction
4. Rearrangement

Answer: 2. Elimination reaction

Solution:

1.  In nucleophilic substitution reaction more powerful nucleophile replaces weaker nucleophile.
2. In rearrangement reaction atoms replace their position within the molecule.
3. In elimination reaction small molecules (example, H₂O, NH₃) are lost.

⇒ \(\mathrm{R}-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Cl}+\mathrm{KOH} \text { (alc.) } \stackrel{\Delta}{\longrightarrow} \mathrm{RCH}=\mathrm{CH}_2+\mathrm{KCl}+\mathrm{H}_2 \mathrm{O}\)

∵ KCl and H₂O molecules are lost during reaction.

∴ It is an elimination reaction.

Question 251. Electrophiles are

1. Electron loving species
2. Electron hating species
3. Nucleus loving reagents
4. Nucleus hating reagents

Answer: 1. Electron loving species

Solution: Electrophiles are electron deficient species which can share a lone pair of electrons with carbanion and are thus called Lewis’s acids.

Question 252. Correct order of nucleophilicity is

1. I^–>Br^–>Cl^–>F^–
2. F^–>Cl^–>Br^–>I^–
3. Cl^–>F^–>Br^–>I^–
4. I^–>Cl^–>Br^–>F^–

Answer: 1. I^–>Br^–>Cl^–>F^–

Solution: Nucleophilicity increases on going down in the group of the Periodic Table

⇒ \(\mathrm{I}^{\ominus}>\mathrm{Br}^{\ominus}>\mathrm{Cl}^{\ominus}>\mathrm{F}^{\Theta}\)

Question 253. Polarization of electron in acrolein may be written a

Answer:

Solution: Due to -R effect of –CHO group, oxygen carries –δ charge while the terminal carbon carries + δ,ie,

⇒ \(\begin{aligned}
& +\delta \\
& \mathrm{CH}
\end{aligned}=\mathrm{CH}-\mathrm{CH}=\mathrm{O}^{-\delta}\)

Question 254. Which of the following does not show electromeric effect?

1. Alkenes
2. Ethers
3. Aldehyde
4. Ketones

Answer: 2. Ethers

Solution: Electromeric effect implies complete transfer of π electrons in presence of a reagent.

Since, simple ethers do not contain a multiple bond, therefore, they do not show electromeric effect

Question 255. The chlorination of methane to give CCl₄ is an example of

1. Addition
2. Elimination
3. Substitution
4. Chain reaction

Answer: 4. Chain reaction

Solution: Halogenation of methane is a chain reaction and propagate through free radicals.

Question 256. An organic compound C₅H₁₁X on dehydrohalogenation gives pentene-2 only. What is the halide?

1. CH₃CH₂CHXCH₂CH₃
2. (CH₃)₂CHCHXCH₃
3. CH₃CH₂CH₂CHXCH₃
4. CH₃CH₂CH₂CH₂CH₂X

Answer: 1.  CH₃CH₂CHXCH₂CH₃

Solution: Follow Saytzeff rule for elimination. 3-halopentane will give only pentene-2.

Question 257. Which of the following is the correct order of decreasing S_N2 reactivity? (X=α halogen)

1. RCH₂X>R₃CX>R₂CHX
2. RCH₂X>R₂CHX>R₃CX
3. R₃CX>R₂CHX>RCH₂X
4. R₂CHX>R₃CX>R₂CH₂X

Answer:  2. RCH₂X>R₂CHX>R₃CX

Solution: S_N2 reactions are greatly controlled by steric factors. S_N2 reactivity decreases as bulkyness of alkyl group increases.

Question 258. Chloroacetic acid is a stronger acid than acetic acid. This can be explained using

1. -M effect
2. -I effect
3. +Meffect
4. +I effect

Answer:  2. -I effect

Solution: Cl is an electron-withdrawin (i.e.,-I showing) group. It withdraws electrons when attached to the carboxylic acid and decreases the electron density on the oxygen atom. This will facilitate the release of H⁺ by making O-H bond more polar and thus –Cl increases the acidity of acetic acid when attached at, α position because of –I effect.

Question 259. Anti-Markovnikov’s addition of HBr is not observed in

1. Propene
2. Butene-1
3. But-2-ene
4. Pent-2-ene

Answer: 3. But-2-ene

Solution: The rule is valid for unsymmetrical alkene.

Question 260. Consider the following carbocations,

1. 2<1<3<4
2. 2<3<1<4
3. 3<1<2<4
4. 4<3<1<2

Answer: 1. 2<1<3<4

Solution: Resonance and inductive effect decide the stability of carbocations.

∴ Correct order of stability is 2<1<3<4

Question 261. Which of the following orders is not correct regarding the –I effect of the substituents?

Answer: 3

Solution: -I power of groups in decreasing order with respect to the reference HNO₂>CHO>COOR>F>Cl>Br>I>OH>OR>NH₂

Question 262. The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

1. 3°<2°<1°
2. 3°>2°>1°
3. 3°<2°>1°
4. 3°>2°<1°

Answer: 2. 3°>2°>1°

Solution: Such dehydrohalogenation follow E2 mechanism. The driving force of such a reaction is the stability of alkene produced. Since, tertiary alkyl halide can give more substituted alkene, it reacts fastest followed by secondary and primary i.e.,3°>2°>1°.

Question 263. Which of the following species is paramagnetic?

1. A carbocation
2. A free radical
3. A carbanion ion
4. All of these

Answer:  2. A free radical

Solution: Free radicals have unpaired electrons.

Question 264. The fairly neutral character of CH3OH is changed to which of the following by adding sodium metal?

1. Acidic
2. Neutral
3. An electrophile
4. A nucleophile

Answer: 4. A nucleophile

Solution: 

⇒ \(\mathrm{CH}_3{ }^{-} \mathrm{O} \text { is nucleophile } ; \mathrm{CH}_3 \mathrm{OH}+\mathrm{Na} \rightarrow \mathrm{CH}_3 \mathrm{O}^{-} \mathrm{Na}^{+}+\frac{1}{2} \mathrm{H}_2\)

Question 265. Which of the following order is correct regarding the acidity of carboxylic acids?

1. Cl₃CCOOH>Cl₂CHCOOH>ClCH₂COOH
2. Cl₃CCOOH>Cl₂CHCOOH<ClCH₂COOH
3. Cl₃CCOOH<Cl₂CHCOOH>ClCH₂COOH
4. Cl₃CCOOH<Cl₂CHCOOH<ClCH₂COOH

Answer: 1. Cl₃CCOOH>Cl₂CHCOOH>ClCH₂COOH

Solution: As the –Igroup increases at the α-carbon, acidity increases.

Question 266. The total number of contributing structures showing hyperconjugation (Involving – C – H bonds) for the following carbocation is

1. Three
2. Five
3. Eight
4. Six

Answer: 4. Six

Solution: There are total 6α-H to sp² carbon and they all can participate in hyperconjugation.

Question 267. RX+I–→ R-I+X– is an example of … reaction.

1. Nucleophilic addition
2. Nucleophilic substitution
3. Electrophilic addition
4. Elimination

Answer: 2. Nucleophilic substitution

Solution: RX+I–→ R-I+X–

This reaction is an example of nucleophilic substitution.

Question 268. The correct order of increasing basicity of the given conjugate bases (R=CH₃) is

Answer: 4

Solution: In carboxylate ion, the negative charge is present on the oxygen, a most electronegative element here, thus it is resonance stabilized.

HC≡C–: Carbon is sp-hybridized so its electronegativity is increased higher relative to nitrogen. N̅H₂: Nitrogen is more electronegative than sp3- hybridized C-atom. From the above discussion, it is clear that the order of the stability of conjugated bases is as RCOO–>HC≡C–>N̅H₂>R–and higher is the stability of conjugated bases, lower will be basic character. Hence, the order of basic character is as

RCOO–<HC≡C–<N̅H₂<R–

Question 269. The reaction, CH₂= CHCHO → HX gives?

1. CH₃CHXCHO
2. CH₂XCHCHO
3. CH₂= CHCHX₂
4. None of these

Answer: 2.

Solution: The negative inductive effect of –CHO group plays a role to give anti Markownikoff’s addition.

Question 270. A solution of D (+)-2-chloro-2-phenylethane in toluene racemizes slowly in the presence of small amount of SbCl5 due to the formation of

1. Carbanion
2. Carbene
3. Free radical
4. Carbocation

Answer: 4. Carbocation

Solution: The solution of D(+)-2-chloro-2-phenyl ethane in toluene racemizes slowly in the presence of SbCl₂ due to the formation in carbocation.

Question 271. During the debromination of meso-dibromo-butane, the major compound formed is

1. n-butane
2. l-butene
3. Cis-2-butene
4. Trans-2-butene

Answer:  4. Trans-2-butene

Solution: Follow the mechanism of debromination.

Question 272. Reaction

1. Electrophilic substitution
2. Nucleophilic substitution
3. Electrophilic addition
4. Nucleophilic addition

Answer: 4. Nucleophilic addition

Solution: Carbonyl compounds show nucleophilic addition.

Question 273. Which one of the following gives a white precipitate with AgNO₃?

Answer: 2

Solution: A chloride linked with alkyl group is replaced with AgNO₃ and give white precipitate of AgCl.

Question 274. The + I.E. (inductive effect) is shown by?

1. CH₃
2. —OH
3. F
4. -C₆H₅

Answer: 1

Solution: —CH₃ is an electron-repelling group.

Question 275. Chlorobenzene is o, p-directed in electrophilic substitution reaction. The directing influence is explained by

1. +M of Ph
2. +I of Cl
3. +M of Cl
4. -I of Ph

Answer: 3. +M of Cl

Solution: Chlorobenzene is o, p directing in electrophilic substitution reaction. The directing influence is explained by +M of Cl atom

Question 276. List the following alkoxide nucleophile in decreasing order of their SN2 reactivity

1. Me₃CO–

2. MaO–

3. MeCH₂O–

4. Me₂CHO–

1. 2>3>5>4>1
2. 5>3>2>1>4
3. 1>5>2>3>4
4. 3>5>1>2>3

Answer: 1. 2>3>5>4>1

Solution: Epoxide is an ambident substrate for nucleophilic substitution reactions. In protonated epoxide carbon-2 and carbon-3 both acquire some positive charge due to the highly electronegative atom.

Question 277. Due to the presence of an unpaired electron free radicals are

1. Cations
2. Anions
3. Chemically inactive
4. Chemically reactive

Answer: 4. Chemically reactive

Solution: Free radicals have unpaired electrons, but are neutrals and are reactive.

⇒ \(\stackrel{\bullet}{\mathrm{C}} \mathrm{H}_3+\stackrel{\bullet}{\mathrm{C}} \mathrm{H}_3 \longrightarrow \mathrm{CH}_3-\mathrm{CH}_3\)

Question 278. Which one of the following carbanions is the least stable?

1. CH₃CH₂–
2. HC≡C–
3. (C₆H₅)₃C–
4. (CH₃)₃C–

Answer: 4. (CH₃)₃C–

Solution: An organic ion with a pair of available electrons and a negative charge on the central carbon atom is called a carbanion. Electron attracting group – CN, >C= Oincreases stability, and electron releasing group (-CH₃ etc) decreases the stability of carbanion. In (CH₃)₃C–, three –CH₃ groups (electron releasing group) are present, so it is least stable.

Question 279. The compound which gives the most stable carbonium ion on dehydration is

1. CH₃CH(CH₃)CH₂OH
2. (CH₃)₃COH
3. CH₂=CHCH₂CH₂OH
4. CH₃CHOHCH2- CH₃

Answer: 2. (CH₃)₃COH

Solution: Increasing order of stability of carbocation. 1°carbocation <2° carbocation<3°carbocation

Question 280. Most stable carbonium ion is

1. \(\stackrel{+}{\mathrm{C}_2} \mathrm{H}_5\)
2. \(\left(\mathrm{C}_6 \mathrm{H}_5\right)_3 \stackrel{+}{\mathrm{C}}\)
3. \(\left(\mathrm{CH}_3\right)_3 \stackrel{+}{\mathrm{C}}\)
4. \(\mathrm{C}_6 \mathrm{H}_5 \stackrel{+}{\mathrm{C}} \mathrm{H}_2\)

Answer: 3. \(\left(\mathrm{CH}_3\right)_3 \stackrel{+}{\mathrm{C}}\)

Solution: is resonance stabilized? In the triphenyl methyl carbonium ion, the π-electrons of all the three benzene rings are delocalised with the vacant p- orbital of central carbon atom. So, it is resonance stabilized. Therefore, it is the most stable of the given carbonium ions.

More the number of resonatic structures, the more will be the stability.

Question 281. Which of the following species does not exert a resonance effect? 

1. C₆H₅NH₂
2. C₆H₅OH
3. C₆H₅Cl

Answer:

Solution: Among the given species does not exert a resonance effect.

Structure 2 is not possible because in it, nitrogen contains 10 valence electrons.

Question 282. The compound can be distinguished by?

1. Chlorinated products
2. Products formed by addition of bromine
3. Reaction with H₂/Ni
4. None of the above

Answer: 2. Products formed by addition of bromine

Solution: Addition of Br₂ gives altogether different products units cis and trans butene-2.

Question 283. When two halogen atoms are attached to two adjacent carbon atoms, the dihaloalkane is called?

1. Alkylidene dihalide
2. Alkane dihalide
3. Alkylene dihalide
4. Alkyl halide

Answer: 3. Alkylene dihalide

Solution: Vicinal or alkylene dihalides.

Question 284. Reactions involving heterolytic fission are said to proceed via :

1. Ionic mechanism
2. Polar mechanism
3. Both (1) and (2)
4. None of these

Answer: 3. Both (1) and (2)

Solution: Heterolytic bond fission gives rise to formation of ions.

Question 285. Select the organic compound which was prepared for the first time in laboratory from its elements

1. Urea
2. CH₃COOH
3. C₂H₅OH
4. None of these

Answer: 2.

Solution: Just after few years when Wohler prepared urea from KCNO and (NH₄)2SO₄, Kolbe prepared acetic acid in the laboratory from its element and gave the final blow to Vital force theory.

Question 286. Which of the following is the most stable cation?

1. \(\mathrm{F}_3 \mathrm{C}-\mathrm{CH}_2^{\oplus}\)
2. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH}^{\oplus}\)
3. \(\mathrm{CH}_3^{\oplus}\)
4. \(\mathrm{CF}_3^{\oplus}\)

Answer: 2. \(\left(\mathrm{CH}_3\right)_2 \mathrm{CH}^{\oplus}\)

Solution: Due to the presence of methyl group positive inductive effect increases and the stability of carbocation also increases. The stability order of carbocation is Tertiary > Secondary> Primary

Question 287. Which of the following has the highest degree of coordination bond?

1. CH₃OH
2. AlCl₃
3. 
4. BF₃O(Et)₂

Answer: 4. BF₃O(Et)₂

Solution: Three coordinate bonds on O atom.

Question 288. Which of the following is an electrophile?

1. H₂O
2. SO₃
3. NH₃
4. ROR

Answer: 2. SO₃

Solution: The species which are electron deficient and accept a pair of electrons are called electrophile. Hence, SO₃ is an electrophile as it contains an electron deficient centre. While H₂ O,NH₃ and R-O-R are nucleophiles.

Question 289. A molecule is R₃C—H. If H is replaced by Z(R₃C—Z) and on doing so electron density on R₃—C part increases, then Z is

1. Electron attracting group
2. Electron withdrawing group
3. Electron repelling group
4. Either of the above

Answer: 3. Electron repelling group

Solution: The Z repels electrons and thus, electron density increases on R3C part.

Question 290. Which of the following statements is incorrect?

1. S_N2 reaction proceeds with inversion
2. S_N1 reaction proceeds with racemisation
3. S_N2 reaction involves transition state
4. In transition state, one end carries δ+, and another end carriesδ–charge

Answer: 4. In transition state, one end carries δ+, and another end carriesδ–charge

Solution: S_N2 reaction proceeds with inversion and a transition state is formed which does not carry any charge.

Question 291. Which of the following is an example of an elimination reaction?

1. Chlorination of methane
2. Dehydration of ethanol
3. Nitration of benzene
4. Hydroxylation of ethylene

Answer: 2. Dehydration of ethanol

Solution: Chlorination of methane is a free radical substitution reaction.
Dehydration of ethanol is an elimination reaction.

Nitration of benzene is an electrophilic substitution reaction. Hydroxylation of ethylene is a redox reaction.

Question 292. C₃H₅Cl+aq.NaOH→ C₂H₅OH+NaCl; this reaction is

1. Electrophilic substitution of 1 order
2. Electrophilic substitution of 2 order
3. Nucleophilic substitution of 1 order
4. Nucleophilic substitution of 2 order

Answer: 4. Nucleophilic substitution of 2 order

Solution: The given reaction can be represented as

NaOH→ Na++OH–

Since in this reaction, a nucleophile replaces the other group, it is an example of nucleophilic substitution reaction. The mechanism shows that the rate depends on the concentration of both alkyl halide and nucleophile. So, it is an example of

S_N2(nucleophilic substitution of 2 order reaction.

Question 293. Which of the following statements (s) is (are) not true?

1. Carbanions and carbonium ions, usually exist in ion pairs or else solvated
2. Acidity increases and basicity decreases in going from left to right across a row of Periodic CH₄<NH₃<H₂O<HF(acidity), CH₃–>NH₂–>OH–>F– (basicity)
3. RCOOH like RCOR reacts with H₂ NOH to give an oxime
4. Decreasing order of ionizing power of solvents is CF₃COOH>HCOOH>H₂O>CH₃COOH>CH₃OH>C₂H5OH>(CH₃)₂SO>CH₃CN

Answer:  3. RCOOH like RCOR reacts with H₂ NOH to give an oxime

Solution: Aldehydes and ketones combine with a variety of compounds of the Z-NH₂ to formoxime

Question 294. Which of the following is arranged according to the nature indicated?

1. Electrophile
2. Electrophile
3. Electrophile –CH₃OH,N₃–. Nucleophile –NO₂⁺,Br⁺
4. Electrophile –Br⁺,N₃^–,Nucleophile –CH₃OH,

Answer: 1. Electrophile

Solution: Electrophiles are the species having a tendency to accept a pair of electrons, example.,NO₂+,Br+ etc. Nucleophiles are the species having a tendency to donate a pair of electrons. example,CH₃OH.N₃–

Question 295. Which of the following belongs to –I group?

1. -C₆H₅
2. -CH₃
3. -CH₂CH₃
4. -C(CH₃)₃

Answer: 1. -C₆H₅

Solution: C₆H₅ – (phenyl) group has – I effect group

Question 296. Arrange the carbanions, in order of their decreasing stability (CH₃)₃C̅,C̅Cl₃,(CH₃)₂C̅H,C₆H₅C̅H₂

1. C₆H5C̅H₂>C̅Cl₃>(CH₃)₂C̅>(CH₃)₂C̅H
2. (CH₃)₂C̅H>C̅Cl₃>C₆H₅CH₂>(CH₃)₃C̅
3. C̅Cl₃>C₆H₅C̅H₂>(CH₃)₂C̅H>(CH₃)₃C̅
4. (CH₃)₃C̅>(CH₃)₂C̅H>C̅H₂>C̅Cl₃

Answer:

Solution: −I effect [e− withdrawing] exerting groups stabilize carbanion by the dispersal of their negative charge while +I effect exerting [e− releasing] groups destabilize the carbanion by increasing electron density on them.

On the other hand, resonance stabilized carbanion is stable due to the involvement of their lone pair of electrons with the delocalization of π-electrons of the attached phenyl group

Question 297. Which one of the nitrogen atoms in the below molecule is the most nucleophilic?

1. 3
2. 1
3. 1 A
4. 2 three N atoms

Answer: 2.1

Solution: When the nucleophilic site is the same, nucleophilicity parallels basicity. It means more the basic the nucleophile, stronger is the nucleophile. \(\mathrm{H}_2 \ddot{\mathrm{N}}(\mathrm{I})\) is the most nucleophilic

Furthermore the NH₂ group is away from the –C-group and is not involved in resonance. Hence, its lone pair is readily available.

Question 298. The increasing order of positive I-effect shown by H,CH3,C2H5 and C3H7 is

1. H <CH₃<C₂H₅<C₃H₇
2. H>CH₃<C₂H₅>C₃H₇
3. H <C₂H₅<CH₃<C₃H₇
4. None of the above

Answer: 1. H <CH₃<C₂H₅<C₃H₇

Solution: Follow inductive effect.

Question 299. Which of the following reactions proceeds via secondary free radical?

Answer: 2

Solution: 1-Propane undergoes bromination in the presence of UV light through a secondary free radical mechanism as given below.

Question 300. Which of the following will be easily nitrated?

Answer: 1

Solution: Nitration of aromatic compounds takes place by an electrophile. The electrophile will be more attracted towards electron rich positions in the benzene ring. Hence, electron donating groups will be easily nitrated.

Toluene will be most easily nitrated among these compounds due to presence of electron donating group (i.e.,CH₃). Nitrobenzene will be most slowly nitrated due to the presence of electron withdrawing group (i.e.,NO₂). CH₃NO₂ will be formed by free radical substitution of CH₄

Question 301. In E₂ elimination, some compounds follow Hofmann’s rule which means:

1. The double bond goes to the most substituted carbon
2. The compound is resistant to elimination
3. No double bond is formed
4. The double bond goes mainly towards the least substituted carbon

Answer: 1. The double bond goes mainly towards the least substituted carbon

Solution: Follow elimination rules.

Question 302. t-butyl chloride reacts with OH– by S_N1 mechanism and rate ∝[t-buty1 chloride].

1. One of the reasons for this is that
2. Stereochemical inversion takes place t- buty1 carbocation is first formed which is more stable
3. The product t-butyl alcohol is more stable
4. The intermediate t-butyl carbocation is stabilized by solvation

Answer: 2. Stereochemical inversion takes place t- buty1 carbocation is first formed which is more stable

Solution:

Rate ∝[t-butyl chloride]

Tertiary butyl carbocation is first formed which is more stable

Question 303. Resonance in benzene is accompanied by delocalization of π-electrons. Each π- electron is attached with:

1. 4 carbon
2. 2 carbon
3. 3 carbon
4. 6 carbons

Answer: 4. 6 carbons

Solution: Each π-electron is delocalized over six carbon atoms in ring.

Question 304. Which of the following is not a nucleophile?

1. BF₃
2. NH₃
3. CN^–
4. OH^–

Answer: 1. BF₃

Solution: Electron donors having lone pair of electrons are nucleophile. BF₃ is not nucleophile because it does not have lone pair of electrons. It is in fact Lewis’s acid because it accepts pair of electrons. NH₃,CN^– and OH^– all have lone pair of electrons, so they are nucleophiles.

Question 305. Among the following the strongest nucleophile is

1. C₂H₅SH
2. CH₃COO^–
3. CH₃NH₂
4. NCCH₂^–

Answer: 1. C₂H₅SH

Solution: Nucleophiles are those substances which can donate a pair of electrons. They can be neutral or negatively charged. The nucleophilic power depends on the tendency of species to donate the electrons. Due to the presence of +Ieffect it increases. Hence, higher the +I effect, higher the nucleophilic power. The +I effect of ethyl is greater than +I effect of methyl group

Question 306. Which of the following is the most stable carbocation?

1. \(\stackrel{+}{\mathrm{C}} \mathrm{H}_3\)
2. R\(\stackrel{+}{\mathrm{C}} \mathrm{H}_2\)
3. \(R_2 \stackrel{+}{\mathrm{C}} \mathrm{H}\)
4. \(R_3 \stackrel{+}{\mathrm{C}}\)

Answer: 4. \(R_3 \stackrel{+}{\mathrm{C}}\)

Solution: In case of alkyl carbocations as the number of R group decreases stability decreases. Thus, the correct order of stability of carbocation is \(R_3 \stackrel{+}{\mathrm{C}}\)>\(R_2 \stackrel{+}{\mathrm{C}} \mathrm{H}\)>R\(\stackrel{+}{\mathrm{C}} \mathrm{H}_2\)>\(\stackrel{+}{\mathrm{C}} \mathrm{H}_3\)

Question 307. The appropriate reagent for the following transformation is

1. Zn (Hg),
2. HCl
3. NH₂NH₂,OH–H₂/Ni
4. NaBH₄

Answer: 2.

Solution: Both Wolf-Kishner and Clemmensen reduction are used to convert The latter is not suitable as it will also attack –OH group of rings.

Question 308. The correct order for homolytic bond dissociation energies. (∆Hin kcal/mol) for CH₄ (1),C₂H₆ (2) and CH₃Br(3), under identical experimental conditions

1. 3>2>1
2. 2>3>1
3. 3>1>2
4. 1>2>3

Answer: 2. 2>3>1

Solution: The order of homolytic bond dissociation energies of CH₄, C₂H₆ and CH₃Br is as CH₄> C₂H₆> CH₃Br ∆H (kcal/mol) 105> 100 > 70

Question 309. Which one is least reactive in a nucleophile substitution reaction?

1. CH₃CH₂Cl
2. CH₂=CHCH₂Cl
3. CH₂=CHCl
4. (CH₃)₃CCl

Answer: 3. CH₂=CHCl

Solution: Vinyl chloride is the least reactive for SN reaction due to resonance

Question 310. Homolytic fission of C—C bond in ethane gives an intermediate in which carbon is …. hybridized.

1. Sp³
2. Sp²
3. Sp
4. Sp²d

Answer: 2. Sp²

Solution: Homolytic fission of the C-C bond in ethane gives free radicals. In a free radical the central carbon atom is sp² sp²-hybridised.

Question 311. Which of the following orders regarding relative stability of free radicals is correct?

1. 3°<2°<1°
2. 3°>2°>1°
3. 1°<2°>3°
4. 3°>2°<1°

Answer: 2. 3°>2°>1°

Solution: Free radicals are electron-deficient compounds. Alkyl groups are electron donor groups and they increase the stability of free radicals.

∴ The more the number of alkyl groups, the more will be stability of free radicals.

∴ 3°>2°>1° is the correct order of stability of free radicals.

Question 312. Which of the following is most reactive towards nucleophilic substitution reaction?

1. CH₂=CH-Cl
2. C₆H₅Cl
3. C₆H₅CH₂Cl
4. ClCH₂-CH=CH₂

Answer: 2. C₆H₅Cl

Solution: During the nucleophilic substitution weaker nucleophile is replaced by a stronger nucleophile. The compound having a C-Cl bond which can be most easily broken will be the most reactive towards nucleophilic substitution reaction. In vinyl chloride CH₂=CH-Cl and chlorobenzene C₆H₅Cl the C-Cl bond has partial double bond character due to resonance.

∴ They do not give nucleophilic substitution reactions easily

Benzyl chloride, gives nucleophilic substitution easily because they carbocation formed is stabilized due to resonance.

Question 313. Correct order of stability is

1. HC≡C̅>CH₂=C̅H>CH₃-C̅H₂
2. CH₃-C̅H₂>CH₂=C̅H>CH ≡C̅
3. CH₃-C̅H₂>CH ≡CH≅CH₂=C̅H
4. All are equally stable

Answer: 2. CH₃-C̅H₂>CH₂=C̅H>CH ≡C̅

Solution: Stability of alkyl carbanion  and magnitude of negative charge ∝+I power of the group. Hence, acetylenic carbanion is more stable than vinylic carbanion which is more stable than alkyl carbanion

Question 314. During the addition of bromine on ethene, the first species formed is

Answer: 1

Solution: Addition of Br₂ on ethene follows electrophilic addition Intermediate is cyclic bromonium ion

Question 315. For all practical purposes, the influence of the inductive effect is neglected after

1. 2nd carbon atom
2. 1st carbon atom
3. 3rd carbon atom
4. None of these

Answer: 1. 2nd carbon atom

Solution: Follow inductive effect.

Question 316. Which of the aldehyde is the most reactive?

1. C₆H₅-CHO
2. CH₃CHO
3. HCHO
4. All the equally reactive

Answer: 3. HCHO

Solution: Among carbonyl compounds, reactivity decreases with increase in the alkyl groups as alkyl groups (+I effects) decrease the positive character on the C-atom. Thus, the correct order of reactivity is HCHO>CH₃CHO>C₆H₅CHO

Question 317. The stabilization due to resonance is maximum in:

1. Cyclohexane
2. Cyclohexene
3. 1 ,3-cyclohexadiene
4. 1,3,5-cyclohexatriene

Answer: 4. 1,3,5-cyclohexatriene

Solution: C₆H₆ has more canonical forms.

Question 318. In which of the following molecules, the resonance effect is not present?

Answer: 2

Solution: If positive charge is present on the nitrogen, then the positive charge will not be in conjugation to the ring because in this case nitrogen will become pentavalent

Question 319. Hydride ion transfer takes place in

1. Frankland method
2. Wurtz reaction
3. Cannizzaro’s reaction
4. Wolff-Kishner reduction

Answer: 3. Cannizzaro’s reaction

Solution: Cannizzaro reaction involves oxidation as well as reduction of aldehydes having lack of α-H atom. The mechanism of this reaction is as

1. Attack of OH^– on carbonyl carbon

 

Question 320. The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system of nomenclature is

1. -COOH,-SO₃H,-CONH₂,-CHO
2. -SO₃H,-COOH,-CONH₂,-CHO
3. -CHO,-COOH,SO₃H,-CONH₂
4. -CONH₂,-CHO,-SO₃H,-COOH

Answer: 1. -COOH,-SO₃H,-CONH₂,-CHO

Solution: The correct decreasing order of priority for the functional groups of organic compounds in the IUPAC system is -COOH>SO₃H>-COOR>COCl>-CONH₂>-CN>-CH=O

Question 321. Which one is a nucleophilic substitution reaction among the following?

Answer: 4

Solution: Nucleophile (-NH₃) replaces other nucleophile (-Br) in the reaction.

Question 322. LiAlH₄ is used as

1. Oxidizing agent
2. Reducing agent
3. A mordant
4. A water softener

Answer: 2. Reducing agent

Solution: It is a strong reducing agent.

Question 323.

1. \(\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{Br} \stackrel{\text { LAH }}{\longrightarrow} \mathrm{C}_2 \mathrm{H}_6\) and LAH

2. (CH₃)₃CBr → Alkene, the reason for this is

1. S_N2(2) E₁ mechanism
2. S_N1, (2) E₂ mechanism
3. S_N1, (2) E₁ mechanism
4. S_N2, (2) E₂  mechanism

Answer: 1. S_N2(2) E₁ mechanism

Solution:

Question 324. Which of the following orders is correct regarding the –I effect of the substituents?

1. -NR₂>-OR>-F
2. -NR₂<-OR<-F
3. -NR₂>-OR<-F
4. -NH₂<-OR>-F

Answer:  2. -NR₂<-OR<-F

Solution: Correct order of -I effect is −NR2<−OR<−F.

Question 325. CH₃CH₂Cl undergoes homolytic fission, produces

Answer: 1

Solution: In homolysis, the covalent bond is broken in such a way that each resulting species known as free radical.

⇒ \(\mathrm{CH}_3 \mathrm{CH}_2-\mathrm{Cl} \underset{\text { fission }}{\stackrel{\text { Homolytic }}{\longrightarrow}} \mathrm{CH}_3 \stackrel{\dot{\mathrm{C}} \mathrm{H}_2}{ }+{ }^{\bullet} \mathrm{Cl}\)

Question 326. Nitration of benzene is

1. Electrophilic substitution
2. Electrophilic addition
3. Nucleophilic substitution
4. Nucleophilic addition

Answer: Electrophilic substitution

During nitration, benzene ring is attacked by NO₂⁺ and hydrogen of benzene ring is replaced by NO₂ group.

∴ Nitration of benzene is an electrophilic substitution because NO₂⁺ is an electrophile.

Question 327. Bromination of alkanes involves

1. Carbanions
2. Carbocations
3. Carbenes
4. Free radicals

Answer: 4. Free radicals

Solution: Bromination of alkanes in the presence of sunlight involves the formation of free radical, example,

⇒ \(\mathrm{CH}_4 \underset{\mathrm{hv}}{\stackrel{\mathrm{Br}_2}{\longrightarrow}} \mathrm{CH}_3 \mathrm{Br}\)

Mechanism: 

Question 328. Conversion of chlorobenzene to phenol involves

1. Electrophilic substitution
2. Nucleophilic substitution
3. Free radical substitution
4. Electrophilic addition

Answer: 2. Nucleophilic substitution

Solution:

In this process one group is replaced by other, hence, it is a substitution process and both the leaving and attacking groups are nucleophilic, therefore it is an example of a nucleophilic substitution reaction.

Question 329. Which of the following is most reactive towards elimination reaction?

1. RCOO^–
2. CN^–
3. NO3^–
4. RO^–

Answer: 4. RO^–

Solution: With the increasing basicity of the added base, the rates of the elimination reactions have been found to increase. Thus, RO– is most reactive

Question 330. Dehydrohalogenation in the presence of OH^– is correctly represented by

Answer:  2

Solution: The dehydrohalogenation in presence of OH^– is correctly represented by In this mechanism the base OH– removes a proton from the β carbon.

Question 331. The most unlikely representation of resonance structures of p-nitro phenoxide ions

Answer: 2

Solution: “N” is pentavalent which is not possible.

Question 332. The enol form of acetone after treatment with D₂O gives

Answer: 2

Solution: The enol form of acetone on treatment with D₂O undergoes enolization, deutration (addition of D₂O) and dehydration (removal of H₂O). The repeated enolization, deutration and dehydration ultimately gives CD₃COCD₃.

Question 333. Which of the following is free radical?

1. Cl⁺
2. Cl^–
3. Cl
4. • NO₂

Answer:  3. Cl

Solution: Free radicals are represented by putting a dot on the entity.

Question 334. The effect involving the complete transfer of a shared pair of electrons to one of the atoms joined by a multiple bond at the requirement of attacking reagent is called

1. Inductive effect
2. Mesomeric effect
3. Electromeric effect
4. None of these

Answer:  3. Electromeric effect

Solution: The definition of electromeric effect.

Question 335.

In  Electrophilic substitution occurs at

1. Ortho⁄paraat first ring
2. Meta at first ring
3. Ortho⁄para at second ring
4. Meta at second ring

Answer: 3. Ortho⁄para at second ring

Solution: Second ring is in conjugation with lone pair of oxygen

Question 336. The increasing order of stability of the following free radicals are

Answer:  1

Solution: Free radicals’ stability

Question 337. Predict the nature of principal product in the reaction,

1. BrCH₂CH₂CH₂CH₂Br+KOH (alc.)⟶Product1,3-butadiene
2. Cyclobutane
3. BrCH₂CH₂CH = CH₂
4. None of these

Answer: 1. BrCH₂CH₂CH₂CH₂Br+KOH (alc.)⟶Product1,3-butadiene

Solution: Follow the elimination of HBr from two ends.

Question 338. Which of the following requires radical intermediate?

Answer: 3

Solution:

⇒ \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2+\mathrm{HBr} \underset{\text { peroxide }}{\stackrel{\text { Organic }}{\longrightarrow}} \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{Br}\)  Requires radical intermediate.

Question 339. The following reaction is an example of \(>\mathrm{C}=\mathrm{O}+\mathrm{H}_2 \mathrm{NOH} \rightarrow>\mathrm{C}=\mathrm{NOH}+\mathrm{H}_2 \mathrm{O}\)

1. Substitution
2. Elimination
3. Addition
4. Addition elimination

Answer: 4. Addition elimination

Solution: In the reaction Both addition and elimination take place simultaneously. Thus, the reaction is an addition elimination

Question 340. In the reaction of phenol with chloroform and aqueous solution of NaOH at 70°C, the electrophile attacking the ring is

1. CHCl₃
2. CHCl₂
3. ∶CCl₂
4. COCl₂

Answer: 3. ∶CCl₂

Solution: When phenol reacts with chloroform and aqueous NaOH solution, it gives salicylaldehyde. CHCl₃+OH–⇋H₂O+CCl₃–

Question 341. In electrophilic aromatic substitution reaction, the nitro group is meta-directing because it

1. Decreases electron density at ortho and para positions
2. Decreases electron density at the meta position
3. Increases electron density at meta position
4. Increases electron density at ortho and para positions

Answer: 3. Increases electron density at meta position

Solution: When nitro group is present in the benzene nucleus, it withdraws electrons from o and p-positions. Thus, the electron density at the o and p-positions decrease. m-positions become positions of comparatively higher electron density and therefore, electrophilic attack occurs at mpositions.

Question 342. Which of the following compounds yields most stable carbanion after rupture of the bond?

Answer:  2

Question 343. The equation of Benzon 

1. C₂H₅OH
2. C₆ H₆
3. CH₄
4. CHO

Answer: 2. C₆ H₆

Solution: 

Question 344. Dehydration of alcohol is an example of which type of reaction?

1. Substitution
2. Elimination
3. Addition
4. Rearrangement

Answer: 2. Elimination

Solution: Dehydration of alcohol involves the loss of two atoms or groups from the adjacent carbon atoms; hence it is an example of a β-elimination reaction.

Question 345. The halogen compound which most readily undergoes nucleophilic substitutions is

1. CH₂=CHCl
2. CH₃CH=CHCl
3. CH₂=CHC(Cl)=CH₂
4. CH₂=CHCH₂Cl

Answer: 4. CH₂=CHCH₂Cl

Solution: CH₂=CH.CH₂Cl compound undergoes nucleophilic substitution most readily.

Question 346. Reactivity towards nucleophilic addition reaction of is

1. HCHO

2. CH₃CHO

3. CH₃COCH₃

1. 2>3>1
2. 3>2>1
3. 1>2>3
4. 1>2<3

Answer: 3. 1>2>3

Solution: Aldehydes and ketones readily undergo nucleophilic addition reaction. The order of reactivity is as the +I effect of alkyl group increases

Question 347. H₂C=O behaves as:

1. Nucleophile
2. Electrophile
3. Both (1) and (2)
4. None of these

Answer:  3. Both (1) and (2)

Solution: O is more electronegative than C, therefore, C carries a small positive charge and, O carry a small negative charge. In other words, it acts as an electrophile due to the presence of a partial positive charge on C and act as a nucleophile due to the presence of a partial negative charge on O.

Question 348. The strongest base among the following is?

1. NH₄⁺
2. :NH₃
3. :NH₂^–
4. :OH^–

Answer:  3. :NH₂^–

Solution: Least stable be the species ion, more basic be the given species. As NH₂− (amide ion) is the least stable species ion, it is the most basic species.

Question 349. In hyperconjugation, the atom involved is

1. β-H atom
2. α-H atom
3. γ-H atom
4. All of these

Answer: 2. α-H atom

Solution: —do—

Question 350. Addition of Br₂ on trans-butene-2 gives:

1. A racemic mixture of 2,3-dibromobutane
2. Meso form of 2,3-dibromobutane
3. Dextro form of 2,3-dibromobutane
4. Laevo form of 2,3-dibromobutane

Answer: 2. Meso form of 2,3-dibromobutane

Solution: Follow the mechanism of addition reactions.

Question 351. Which one of the following compounds is most reactive towards nucleophilic addition?

1. CH₃CHO
2. PhCOCH₃
3. PhCOPh
4. CH₃COCH₃

Answer: 1. CH₃CHO

Solution:

Carbonyl compounds undergoes nucleophilic addition reaction.

If group or atom attached with carbonyl carbon shows negative inductive effect, then it decreases electron density on carbonyl carbon and facilitate the attack of nucleophile, hence reactivity of the carbonyl compound increases.

The aromatic aldehydes and ketones are less reactive than their aliphatic analogues due to +R effect of benzene ring. The increasing order of the nucleophilic addition reaction in the following compounds will be.

CH₃CHO>CH₃COCH₃>PhCOCH₃>PhCOPh

Question 352. In which of the reactions, addition takes place according to Markownikoff’s rule?

1. CH₃CH=CHCH₃+Br→
2. CH₂=CH₂+HBr→
3. CH₃CH=CH₂+HBr→
4. CH₃CH=CH₂+Br₂→

Answer: 3. CH₃CH=CH₂+HBr→

Solution: Markownikoff’s rule is for the addition of an unsymmetrical additive on an unsymmetrical alkene.

Question 353. S_N1 mechanism for the reaction, R-X+KOH ⟶ROH+KX follows \(R-X \rightarrow R^{+}+X^{-} \stackrel{\mathrm{OH}^{-}}{\rightarrow} \quad \mathrm{R}-\mathrm{OH}\)

1. Carbocation mechanism
2. Carbanion mechanism
3. Free radical mechanism
4. Either of the above

Answer: 1. Carbocation mechanism

Solution: \(R-X \rightarrow R^{+}+X^{-} \stackrel{\mathrm{OH}^{-}}{\rightarrow} \quad \mathrm{R}-\mathrm{OH}\)

Question 354. “The negative part of the addendum adds on the carbon atom joined to the least number of hydrogen atoms.” This statement is called?

1. Markownikoff’s rule
2. Peroxide effect
3. Baeyer’s strain theory
4. Thiele’s theory

Answer: 1. Markownikoff’s rule

Solution: It is Markownikoff’s rule.

Question 355. The following reaction is described as

1. S_E2
2. S_N2
3. S_N1
4. S_N0

Answer: 2. S_N2

Solution: In this reaction inversion takes place. Hence, it is an example of S_N2 reaction. In this mechanism the attack of OH^– ions take place from the back side while the Br– ion leaves from the front side

Question 356. A carbonium ion is formed when a covalent bond between the two atoms in an organic compound undergoes?

1. Homolysis
2. Heterolysis
3. Cracking
4. Pyrolysis

Answer: 2. Heterolysis

Solution: Heterolytic bond fission produces positive and negative ions.

Question 357. Grignard reagent adds to

Answer: 2

Solution: Grignard reagent reacts with >C=O, -C≡N,>C=S as follows

Question 358. What will be the compound if two valencies of the carbonyl group are satisfied by two alkyl groups?

1. Aldehyde
2. Ketone
3. Acid
4. Acidic anhydride

Answer: 2. Ketone

Solution:  – do –

Question 359. Among the following anions

1. CH₃^–

2. NH₂^–

3. OH^–

4. F– the order of basicity is?

1. 1>2>3>4
2. 2>2>3>4
3. 3>2>1>4
4. 3>1>2>4

Answer:  1. 1>2>3>4

Solution: Stronger is the acid, weaker is its conjugate base or weaker is its nucleophilicity. The acidic order HF >H₂O >NH₃>CH₄

Question 360. Intermediate product formed in the acid catalyzed dehydration of n-propyl alcohol is?

1. CH-CH₂-CH₃
2. CH₃-CH=CH₂
3. \(\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3\)
4. \(\mathrm{CH}_3-\mathrm{CH}_2-\stackrel{+}{\mathrm{C}} \mathrm{H}_2\)

Answer: 3. \(\mathrm{CH}_3-\stackrel{+}{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3\)

Solution: 2° carbocation is more stable.

Question 361. In the compound, electrophilic substitution occurs at 

1. Ortho/para position at ring
2. Meta position at the ring
3. Ortho/para position at the ring
4. Meta position at ring

Answer: 3. Ortho/para position at ring

Solution: Electrophilic substitution reaction takes place in compounds in which π-electrons are highly delocalised. The electrophile attacks the region of high electron density; therefore, electrophilic substitution occurs at ortho/para position at ring 2.

Assertion – Reasoning Type

Each question contains Statement 1 (Assertion) and Statement 2 (Reason). Each question has the 4 choices (1), (2), (3), and (4) out of which Only one is correct.

1. Statement 1 is True; Statement 2 is True; Statement 2 is correct explanation for Statement 1
2. Statement 1 is True; Statement 2 is True Statement 2 is not the correct explanation for Statement 1
3. Statement 1 is True, Statement 2 is False
4. Statement 1 is False, Statement 2 is True

Question 362.

• Statement 1: Electrophiles are electron rich in nature
• Statement 2: H₃O+,BF₃ and AlCl₃ are electrophile and can accept electron pair

Answer: 4. Statement 1 is False, Statement 2 is True

Solution: Electrophiles are electron deficient while nucleophiles are electron rich in nature, ie, electrophile can accept an electron pair while nucleophile donates an electron pair

Question 363.

• Statement 1: Dehydration of alcohol is an example of an elimination reaction
• Statement 2: When H₂SO₄or H₃PO4(concentrated) are used as dehydrating agent, the mechanism is E1

Answer: 2. Statement 1 is True; Statement 2 is True Statement 2 is not the correct explanation for Statement 1

Solution: Alcohols leading to conjugated alkenes are more easily dehydrated than the alcohols leading to non-conjugated alkenes

Question 364.

• Statement 1: The order of stability of carbocation are \(R_3 \mathrm{C}^{+}>R_2 \stackrel{+}{\mathrm{C}} \mathrm{H}>R_{\mathrm{R}}^{+} \mathrm{H}_2>\stackrel{+}{\mathrm{C}^{-}} \mathrm{H}_3\)
• Statement 2: The stability of carbocations is influenced by both resonance and inductive effects

Answer: 3. Statement 1 is True, Statement 2 is False

Solution: The stability of carbocation is explained on the basis of hyperconjugation and inductive effect hence the stability order of carbocation is 3°>2°>1°>Methyl carbocation

Question 365. 

• Statement 1: 
• Statement 2: A compound in which the positive and negative charges reside on the most electropositive and most electronegative atoms of the species respectively is more stable

Answer:  1.  Statement 1 is True; Statement 2 is True; Statement 2 is the correct explanation for Statement 1

Solution: Both structures are resonating structures of formic acid In negative charge is on oxygen but in negative charge is on carbon therefore (I) will be more stable than the FQ

Isomerism

Question 1. Increasing order of stability among the three main conformations i.e., Eclipse, Anti, Gauche. of 2-fluoroethanol is

1. Eclipse, Gauche, Anti
2. Gauche, Eclipse, Anti
3. Eclipse, Anti, Gauche
4. Anti, Gauche, Eclipse

Answer: 3. Eclipse, Anti, Gauche

Solution: HO – CH₂– CH₂ – F

Gauche conformation is comparatively more stable due to the hydrogen linkage in between F and H at O-atom. hence order is Eclipse, Anti staggere, Gauche.

Read And Learn More: NEET General Organic Chemistry Notes, Question And Answers

Question 2. How many isomers will C₃H₆ have?

1. 1
2. 2
3. Zero
4. 4

Answer: 2. 2

Solution: One propene and one cyclopropane.

Question 3. The molecular formula of diphenylmethane is C₁₃H₁₂

How many structural isomers are possible when one of the hydrogens is replaced by a chlorine atom?

1. 6
2. 4
3. 8
4. 7

Answer: 2. 4

Solution:

Question 4. In butane, which of the following forms has the lowest energy?

1. Gauche form
2. Eclipsed form
3. Staggered form
4. None of these

Answer: 3. Staggered form

Solution:  The staggered form has lower energy than the eclipsed form because the repulsive interaction between the H-atoms attached to two carbon atoms are minimal due to the maximum distance between them.

Question 5. Which of the following does not exhibit tautomerism?

Answer: 1

Solution: The compounds that contain the active methylene group at the adjacent position of the carboxyl group show tautomerism.

This compound does not contain an active methylene group, and hence does not exhibit tautomerism. Moreover, this compound is highly stable due to extensive cross-conjugation.

Question 6. The structure exhibits______________________ isomerism

1. Geometrical isomerism
2. Optical isomerism
3. Geometrical and optical isomerism
4. Tautomerism

Answer: 2. Optical Isomerism

Solution: One asymmetric carbon atom is present.

Question 7. Propanal and propanone are

1. Functional isomers
2. Enantiomers
3. Chain isomers
4. Structural isomers

Answer: 1. Functional isomers

Solution: Propanal and propanone are functional isomers

Question 8. Maleic and fumaric acids are :

1. Tautomers
2. Geometrical isomers
3. Chain isomers
4. Functional isomers

Answer: 2. Geometrical Isomers

Solution: Maleic and fumaric acids are geometrical isomers.

Question 9. Tautomerism is exhibited by

Answer: 1. 1

Solution: Tautomerism is a functional isomerism in which the isomers are readily interchangeable and they maintain a dynamic equilibrium with each other.

Question 10. An optically active compound is:

1. 1-bromobutane
2. 2-bromobutane
3. 1-bromo-2-methylpropane
4. 2-bromo-2-methylpropane

Answer: 2. 2-bromobutane

Solution: 2-bromobutane has asymmetric carbon atoms.

Question 11. The number of isomers possible for the compound with the molecular formula C₂BrClFI is

1. 3
2. 4
3. 5
4. 6

Answer:  4. 6

Solution: Molecular formula C₂BrClFl six isomers are possible.

Question 12. Which is incorrect about enantiomorphs?

1. They rotate the plane of polarized light in different directions
2. They have mostly identical physical properties
3. They have the same configuration
4. They have different biological properties

Answer: 3. They have the same configuration

Solution: Enantiomers differ in their configuration R or S. at the stereogenic centre.

Question 13. The total number of alkenes possible by dehydrogenation of 3-bromo-3- cyclopentyl hexane using alcoholic KOH is

1. 1
2. 3
3. 5
4. 7

Answer:  3. 5

Solution: The substrate has three different types of B-H, therefore, first, three structural isomers of alkenes are expected as

The last two alkenes 2 and 3 are also capable of showing geometrical isomerism hence two geometrical isomers for each of them will be counted giving a total of five isomers.

Question 14. The following compounds differ in

1. Configuration
2. Conformation
3. Structure
4. Chirality

Answer: 3. Structure

Solution: 

Since in the above structures, the positions of CI is different, these are position isomers, which is a type of structural isomerism.

Question 15. The correct statement about the compounds A, B and C is

1. A and B are identical
2. A and B are diastereomers
3. A and C are enantiomers
4. A and B are enantiomers

Answer: 4. A and B are enantiomers

Solution:

1. Enantiomers are pairs of optical isomers that are related as non-superimposable mirror images of each other.

2. Diastereomers are pairs of optical isomers that cannot be related as non-superimposable mirror images of each other.

∴ The only correct statement about given structures is that A. and B. are enantiomers.

Question 16. Isomerism among compounds due to the migration of a proton is known as

1. Geometrical
2. Optical
3. Tautomerism
4. Position

Answer: 3. Tautomerism

Solution: Definition of tautomerism.

Question 17. What is the number of possible optical isomers of glucose?

1. 3
2. 4
3. 12
4. 1

Answer:  4. 1

Solution: Glucose contains four chiral carbon atoms hence number of possible optical isomers is 2⁴ 

2⁴ = 2×2×2×2

= 16.

Question 18. Benz aldoxime exists in how many forms?

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Solution: Benz aldoxime can exist in two geometrical isomeric forms known cis and trans.

Question 19. In the following the most stable conformation of n-butane is?

Answer: 2. 2

Solution:

1. The anti-staggered conformation of n-butane is more stable than gauche-staggered and eclipsed conformations of n-butane.

2. In anti-staggered n-butane, the methyl groups are placed at a dihedral angle of 1800, and the steric hindrance is minimal in anti-form than in gauche form.

Question 20. Which one of the following monoenes does not exhibit geometric isomerism?

1. C₄H₈
2. C₃H₆
3. C₅H₁₀
4. C₈H₁₆

Answer: 2.  C₃H₆

Solution: The main conditions for exhibiting geometrical isomerism are

1. Presence of double bond.
2. Presence of different groups on the same double bonded carbon.
3. Presence of at least one similar group on adjacent double bonded carbon atoms.

does not exhibit geometric isomerism due to the presence of same group on double-bonded atom (C₁ )

Question 21. The number of optical isomers of CH₃CH(OH)CH(OH)CHO is

1. Zero
2. 2
3. 3
4. 4

Answer: 4. 4

Solution: Possible number of optical isomers = 2ⁿ

= 2²

= 4

Question 22. The isomeric cis-2-butene and trans-2-butene can be distinguished on the basis of their:

1. Their physical nature
2. Their reduction products
3. The products they give on ozonolysis
4. The products they give in addition to bromine

Answer:

Solution: The two forms give different products on the addition of Br₂ cis butene gives a racemic mixture whereas trans butene gives meso form of 2,3,3-dibromo butene.

Answer: 3.

Question 23. n- pentane, iso-pentane, and neo– pentane are examples for isomers of the type

1. Geometrical
2. Optical
3. Chain
4. Positional

Answer:  3. Chain

Solution: The compounds which differ in the nature of carbon chain are called chain isomers, for example,

Question 24. The number of isomers in C₄H₁₀O  are

1. 7
2. 8
3. 6
4. 5

Answer: 1. 7

Solution: There are seven isomers in C₄H₁₀O. Out of these seven isomers, three are of ether. four are of alcohol

Question 25. In a lactic acid molecule, the methyl group, a hydroxyl group, a carboxylic acid group and a hydrogen atom are attached to a central carbon atom showing optical isomerism due to the molecular geometry at the

1. The carbon atom of the methyl group
2. Carbon atom of the carboxylic acid group
3. Central carbon atom
4. Oxygen of the hydroxyl group

Answer: 3. Central carbon atom

Solution: The central carbon atom is chiral carbon.

Question 26. Geometrical isomerism is possible in case of

1. Pentene-2
2. Propane
3. Pentane
4. Ethene

Answer: 1. Pentene – 2

Solution:  Pentene-2 exhibits cis and frans-isomerism.

Question 27. Isopentane can form four isomeric mono bromo derivatives. How many of them are optically active?

1. 1
2. 2
3. 3
4. None of these

Answer: 2. 2

Solution:

Question 28. A cyclic stereoisomer having the molecular formula C₄H₇Cl is classified and tabulated. Find out the correct set of numbers.

Answer: 1.

Solution: The acyclic stereoisomers of C₄H₇Cl. are

1. Number of optical isomers = 2ⁿ = 2¹ = 2
2. Hence, the total number of geometrical isomers = 6
3. Total number of optical isomers =2.

Question 29. The optical isomers, which are not enantiomers are called

1. Conformer
2. Diastereomer
3. Mirror images
4. None of these

Answer: 2. Diastereomer

Solution: Diastereoisomers are a pair of optical isomers that cannot be related as non-superimposable mirror images of each Other.

Question 30. A racemic mixture is a mixture of:

1. Meso and its isomers
2. d-and its l-isomers in equal proportions
3. d-and its l-isomers in different proportions
4. Meso and d-isomers

Answer: 2. d-and its l-isomers in equal proportions

Solution: A mixture of 50-50% of d and it’s form is called a racemic mixture.

Question 31. 2-methyl pent-3-enoic acid shows:

1. Optical isomerism
2. Geometrical isomerism
3. Both 1 and 2
4. None of these

Answer: 3. Both 1 and 2

Solution: The acid exists in cis and trans forms:

Also, it has an asymmetric carbon atom

Question 32. The number of optical enantiomorphs of tartaric acid is

1. 3
2. 2
3. 4
4. 1

Answer: 2. 2

Solution: a = 2n-1; where n is no. of asymmetric carbon; when a molecule possesses symmetry.

Question 33. The optically active alkane with the lowest molecular weight is

Answer: 3

Solution: Only this is optically active due to the central carbon being asymmetric.

Question 34. Which of the following structures permits cis-trans isomerism?

1. X₂C = CY₂
2. XYC = CZ₂
3. X₂C = CXY
4. XYC = CXY

Solution: Molecules with two similar groups attached on either of the doubly bonded carbons do not show geometrical isomerism.

Question 35. Methoxy methane and ethanol are

1. Position isomers
2. Chain isomers
3. Functional isomers
4. Optical isomers

Answer: 3. Functional isomers

Solution:

CH₃OCH₃ – Methoxy methane ₂₃₅

C ₂H₅OH₆ – Ethanol ether. alcohol. functional group

(C ₂H6O) (C₂H₆O)molecular formula

In methoxy methane and ethanol both molecular formulas are the same but the functional groups are different, so they are functional isomers.

Question 36. The number of structural and configurational isomers of a bromo compound, C5HgBr, formed by the addition of HBr to 2-pentyne respectively.

1. 1 and 2
2. 2 and 4
3. 4 and 2
4. 2 and 1

Answer:  2 and 4

Solution: Draw the isomers.

Question 37. A compound contains 2 dissimilar asymmetric carbon atoms. The number of optically active isomers is:

1. 2
2. 3
3. 4
4. 5

Answer: 3. 4

Solution: a = 2; where n is no. of dissimilar asymmetric carbon atoms and a is no. of optically active isomers.

Question 38. Stereoisomers are geometrical or optical. which are neither superimposable nor mirror image to each other are called?

1. Enantiomers
2. Mesomers
3. Tautomers
4. Diastereomers

Answer: 4. Diastereomers

Solution: It is the definition of Diastereoisomers.

Question 39. α-D- (+)-glucose and P-D- (+)-glucose are?

1. Enantiomers
2. Conformers
3. Epimers
4. Anomers

Answer: 4. Anomers

Solution:  Alpha D glucose and beta D glucose are examples of anomers. Alpha D glucose can be written as a-D (+)-glucose, whereas beta D glucose can be represented as P-D (+)-glucose.

Question 40. Isomers of propionic acid are

1. HCOOC₂H₅  and CH₃COOCH₃
2. HCOOC₂H5  and C₃H₇COOH
3. CH₃COOCH₃  and  C₃H₇OH
4. C₃H₇OH and CH₃COCH₃

Answer:  2. HCOOC₂H₅and CH₃COOCH₃

Solution: Isomers of propionic acid are as

Question 41. Identify, which of the below does not possess any element of symmetry.

1. (+) and (-) tartaric acid
2. Carbon tetrachloride
3. Methane
4. Meso-tartaric acid

Answer: 1. (+) and (-) tartaric acid

Solution: (+) and (-) tartaric acid does not possess any element of symmetry.

Question 42. Cis-trans, isomers generally

1. Contain an asymmetric carbon atom
2. Rotate the plane of polarized light
3. Are enantiomorphs
4. Contains a double-bonded carbon atom

Answer: 4. Contains a double-bonded carbon atom

Solution: Cis-trans isomers generally contain double-bonded carbon atoms.

Question 43. How many primary amines are possible with the formula C₄H₁₁N?

1. 1
2. 2
3. 3
4. 4

Answer:  4. 4

Solution: The possible primary amine with the formula C₄H₁₁N are

Question 44. For which of the following parameters the structural isomers C₂H₅OH and CH₃OCH₃ would be expected to have the same values? Assume ideal behaviour.

1. Heat of evaporation
2. Vapour pressure at the same temperature
3. Boiling points
4. Gaseous densities at the same temperature and pressure

Answer: 4. Gaseous densities at the same temperature and pressure

Solution: In CH₃CH₂OH, there is intermolecular H-bonding, while it is absent in isomeric ether CH₃OCH₃

1. Larger heat is required to vaporise CH₃CH₂OH as compared to CH₃,OCH₃, thus 1  is incorrect.

2. CH₃CH₂OH is less volatile than CH₃OCH₃, thus vapour pressures are different, thus 2 is incorrect.

3. Boiling point of CH₃CH₂OH>CH₃OCH₃, thus c. is incorrect.

Density = \(\frac{\text { mass }}{\text { volume }}\) due to ideal behaviour at a given temperature and pressure volume and molar J volume or  mass are the same.

Hence, they have the same vapour density.

Question 45. A compound having a molecular formula C₄H₁₀ can exhibit.

1. Metamerism
2. Functional isomerism
3. Positional isomerism
4. All of these

Answer: 4. All of these

Solution: Alcohols show position isomerism; Ethers show metamerism; Alcohols and ethers show functional isomerism.

Question 46. Which one of the follow 2, 3-dibromopentane

1. 3, 3-dibromopentane
2. 4-bromo-2-pentanol
3. 3-bromo-2-pentanol
4. 2, 3-dibromopentane

Answer: 3-bromo-2-pentanol

Solution: Those compounds which contain two or more asymmetric carbon atoms but are optically inactive due to the presence of plane of symmetry, are called meso compounds. Meso compounds are optically inactive due to internal compensation.

Out of the given compounds only 2, 4-dibromopentane have a plane of symmetry, so it is a meso compound.

Question 47. Ethylene dichloride and ethylidene chloride are isomeric compounds. The false statement about these isomers is that
they

1. React with alcoholic potash and give the same product
2. Are position isomers
3. Contain the same percentage of chlorine
4. Are both hydrolysed to the same product

Answer:  4. Are both hydrolysed to the same product

Question 48. Which of the following acids does not exhibit optical isomerism?

1. Lactic acid
2. Tartaric acid
3. Maleic acid
4. α -amino acids

Answer: 3. Maleic acid

Solution:

It has no asymmetric carbon; however, it shows geometrical isomerism.

Question 49. Ethyl acetoacetate shows which type of isomerism

1. Chain
2. Optical
3. Metamerism
4. Tautomerism

Answer: 4. Tautomerism

Solution: Ethyl acetoacetate shows tautomerism.

Question 50. The number of isomeric hexanes is?

1. 5
2. 2
3. 3
4. 4

Answer: 1. 5

Question 51. Which among the following statements is correct with respect to the optical isomers?

1. Enantiomers are non-superimposable mirror images.
2. Diastereomers are superimposable mirror images.
3. Enantiomers are superimposable mirror images.
4. Meso forms have no plane of symmetry.

Answer: 1. Enantiomers are non-superimposable mirror images

Solution: Enantiomers are non-superimposable mirror images, for example, lactic acid

Diastereomers are non-superimposable and are not mirror images of each other. Moreover, the meso form has a plane of symmetry.

Question 52. Hydrogen cyanide and hydrogen isocyanide are:

1. Functional isomers
2. Positional isomers
3. Metamers
4. Chain isomers

Answers: 1. Functional isomers

Question 53. 2-pentanone and 3-methyl-2-butanone are a pair of isomers.

1. Functional
2. Chain
3. Positional
4. Stereo

Answer:  2. Chain

Solution: 2-pentanone and 3-methyl-2-butanone are chain isomers because they differ in carbon skeleton.

Question 54. Which of the following is a dynamic isomerism?

1. Metamerism
2. Geometrical isomerism
3. Tautomerism
4. Coordinate isomerism

Answer: 3. Tautomerism

Solution: Tautomerism is a dynamic isomerism because two forms keto and enol. of substance cannot be separated; they are in dynamic equilibrium with each other.

Question 55. Cyclobutane and butene-1 are?

1. Chain isomers
2. Position isomers
3. Ring-chain isomers
4. Metamers

Answer: 3. Ring-chain isomers

Solution: Both have different mode of linkage, i.e., chain and ring.

Question 56. Among the following which is the one that does not exhibit functional group isomerism.

1. C₂H₆O
2. C₃H₈P
3. C₄H₁₀
4. C₄H₁₀O

Answer: 3. C₄H₁₀

Solution: Alkanes never show functional isomerism, metamerism, tautomerism and geometrical isomerism.

Question 57. Vinyl alcohol and acetaldehyde are

1. Geometrical isomers
2. Keto-enol tautomers
3. Chain isomers
4. None of these

Answer: 2. Keto-enol tautomers

Solution: Isomers differ in the arrangement of atoms but exist in dynamic equilibrium with each other shows the tautomerism. Acetaldehyde and vinyl alcohol are keto-enol tautomers.

Question 58. The structures that do not actually exist are known as

1. Tautomers
2. Conformational isomers
3. Canonical structures
4. Optical isomers

Answer: 3. Canonical structures

Solution: Canonical structures proposed in resonance are not the real structure of compounds. The compound showing resonance has a definite structure which can however not be drawn on paper.

Question 59. Chiral molecules are those which:

1. Are not superimposable on their mirror images
2. Are superimposable on their mirror images
3. Show geometrical isomerism
4. Are unstable molecules

Answer: 1. Are not superimposable on their mirror images

Solution: Chiral molecules should not contain any kind of symmetry.

Question 60. d-tartaric acid and l-tartaric acid are which type of isomer?

1. Structural isomers
2. Diastereoisomers
3. Tautomers
4. Enantiomers

Answer: 4. Enantiomers

Solution: d and l configurations are non- superimposable on each other and are mirror images, so they are termed as enantiomers. Enantiomers are compounds that are mirror images but are non-super impossible on each other. The structures that are mirror images are called enantiomorphs.

Question 61. Which of the following shows geometrical isomerism?

1. C₂H₅Br
2. (CH₂) (COOH)₂
3. (CH)₂(COOH)₂
4. C₂H₆

Answer: 3. (CH)₂(COOH)₂

Solution: The compounds must fulfil two conditions to show geometrical isomerism.

1. The compound should have at least one C=C.
2. The two groups attached to the same carbon must be different.
3. Out of given choices only (3). fulfils both conditions andshows geometrical isomerism.

Question 62. α-D-glucose and β-D-glucose have a specific rotation of +112° and +19° respectively. In aqueous solution the  rotation becomes +52°. This process is called as?

1. Inversion
2. Racemization
3. Mutarotation
4. Enolization

Answer: 3. Mutarotation

Solution: The conversion of a-glucose to p-glucose is called mutarotation.

Question 63. Which of the following pairs of carbon skeletons is an example of isomerism?

Answer: 1. 1

Solution: In the rest of them, all carbon chains are the same.

Question 64. Which of the following is a chiral compound?

1. Hexane
2. n-butane
3. Methane
4. 2,3,4, trimethyl hexane

Answer: 2,3,4, trimethyl hexane

Solution: In hexane, all the C-atoms are symmetric, so no carbon atom is chiral.

In n-butane all C-atom present are symmetric, hence it is achiral In methane, all groups attached are the same, hence it is also achiral.

In 2,3,4-trimethyl hexane, there are two chiral centres making the compound asymmetric.

Question 65. Geometrical isomerism is shown by:

Answer: 2

Solution: A molecule having doubly bonded carbon atoms shows geometrical isomerism only if both the doubly bonded carbon has altogether different group, i.e., _baC =C_ab or _abC = C_ac or _dcC=C_ab.

Question 66. How many optically active forms are possible for a compound of the formula,

CHO.CHOH.CHOH.CHOH.CH₂OH?

1. 2
2. 4
3. 3
4. 8

Answer: 4. 8

Solution: a = 2ⁿ; n = 3.

Question 67. The lactic acid molecule has

1. One chiral carbon atom
2. Two chiral carbon atoms
3. No chiral carbon atom
4. Asymmetric molecule

Answer: 1. One chiral carbon atom

Solution: A chiral carbon atom has all four different groups attached to it.

∴ It has one asymmetric or chiral carbon atom.

Question 68. Which of the following is an optically active compound?

1. Lactic acid
2. Chloro acetic acid
3. Meso-tartaric acid
4. Acetic acid

Answer: 1. Lactic acid

Solution: One asymmetric carbon atom is present in a lactic acid molecule. Hence, it is an optically active compound.

Question 69. Geometrical isomerism is attributed to?
Answer:

1. By restricted rotation around C = C bond
2. By the presence of one asymmetric carbon atom
3. Due to different groups attached to the same functional group
4. By swing of hydrogen atom between two polyvalent atoms

Answer: 1. By restricted rotation around C = C bond

Solution: It is a fact.

Question 70. Among the following four structures I to IV it is true that,

1. All four are chiral compounds
2. Only 1 and 2 are chiral compounds
3. Only 3 is a chiral compound
4. Only 2 and 4 are chiral compounds

Answer:  2. Only 1 and 2 are chiral compounds

Solution: Chiral compounds which have one chiral centre. All four atoms or groups attached to carbon are different.

Question 71. The number of chiral centres in +. -glucose

1. 4
2. 3
3. 2
4. 1

Answer: 1. 4

Solution: A carbon atom which is attached by four different groups is called the chiral centre of asymmetric carbon atoms. +.-glucose has four chiral centres.

Question 72. Geometrical isomerism is possible in

1. Acetone-oxime
2. Isobutene
3. Acetophenone-oxime
4. Benzophenone-oxime

Answer: 3. Acetophenone-oxime

Solution: Acetophenone oxime can show geometrical isomerism.

Question 73. Which of the following oxime can show geometrical isomerism?

Answer: 2

Solution: Due to asymmetric carbon atoms.

Question 74. Lactic acid, CH₃CH(OH)COOH  molecule shows

1. Geometrical isomerism
2. Metamerism
3. Optical isomerism
4. Tautomerism

Answer: 3. Optical isomerism

Solution: It shows optical isomerism due to the presence of asymmetric carbon atoms.

Question 75. The number of structural isomers possible for an organic compound with molecular formula C₅H₁₂

1. 5
2. 3
3. 4
4. 2

Answer: 2. 3

Solution: These are isopentane, neopentane and n-pentane.

Question 76. Which is a chiral molecule?

1. CH₃Cl
2. CH₂Cl₂
3. CHBr₃
4. CHClBrI

Answer: 4. CHClBrI

Solution: Carbon is asymmetric as all its valencies are attached to different groups.

Question 77. How many isomers are possible for the alkane C₄H₁₀?

1. 3
2. 5
3. 2
4. 4

Answer: 3.

Solution: Two isomers

Question 78. Which of the following compounds exhibits rotamers?

1. 2-butene
2. Maleic acid
3. Butane
4. Fumaric acid

Answer: 2-butene

Solution: 2-butene exhibit rotamers. Rotamers are the isomers formed by restricted rotation.

Question 79. Nitroethane can exhibit one of the following kinds of isomerism

1. Metamerism
2. Optical activity
3. Tautomerism
4. Position isomerism

Answer: 3. Tautomerism

Solution: Nitroalkanes exhibit tautomerism. In it, a-H-atom is labile and forms nitrolic acid.

Question 80. CH₃-CHO-CH₂-CH₃ has a chiral centre. Which one of the following represent its R-configuration?

Answer: 2

Solution: Follow priority rules.

Question 81. Which of the following molecules is achiral?

Answer: 4

Solution: A molecule having an asymmetric carbon atom and is not superimposable on its mirror image then it is chiral while if it is superimposable on its mirror image, it is achiral.

Question 82. Which of the following statements is necessarily true in the case of isomeric organic compounds?

1. They are hydrocarbons
2. They are optically active
3. They yield the same products on complete combustion
4. They have same melting or boiling points

Answer: 3. They yield the same products on complete combustion

Solution: Due to the same molecular formula.

Question 83. What kind of isomerism is possible for 1-chloro-2-nitroethene?

1. Functional group isomerism
2. Position isomerism
3. E/Z isomerism
4. Optical isomerism

Answer: 3. E/Z isomerism

Solution: The structure of 1-chloro-2-nitroethene is as

Question 84. On monochlorination of n-pentane, the number of isomers formed is

1. 4
2. 3
3. 2
4. 1

Answer: 2. 3

Solution: CH₃CH₂CHClCH₂CH₃; CH₃CHClCH₂CH₂CH₃; CH₂ClCH₂CH₂CHCH₃.

Question 85. One of the following compounds exhibits geometrical isomerism

1. CH₃CH₂CH₂CH₃
2. CH₃-HCCH₃.-HC.CH₃-CH₃
3. CH₃-HCCH₃.-CH₃
4. CH₃CH = CH-CH₃

Answer: 4. CH₃CH = CH-CH₃

Solution: Geometrical isomers of CH₃CH=CH-CH₃ are

Question 86. Which types of isomerism is shown by 2, 3-dichlorobutane?

1. Structural
2. Geometric
3. Optical
4. Diastereomer

Answer: 3. Optical

Question 87. Optically active compound among the following is:

1. 2-ethylbutanol
2. n-butanol
3. 2,2-dimethylbutanol
4. 2-methylbutanol-1

Answer: 2-methylbutanol-1

Solution: Chiral carbon is present. Hence it is optically active.

Question 88. Which of the following is a pair of functional isomers?

1. CH₃COCH₃,CH₃CHO
2. C₂H₅CO₂H,CH₃CO₂CH₃
3. C₂H₅CO₂H,CHCO₂C₂H₅
4. CH₃CO₂H,CH₃CHO

Answer: 2. C₂H₅CO₂H,CH₃CO₂CH₃

Solution: Carboxylic acid and esters show functional group isomerism. When two compounds have the same molecular formula but different functional groups, then functional isomerism arises.
for example., C₂H₅COOH and CH₃COOCH₃

Question 89. Diastereomers can be separated by?

1. Fractional distillation
2. Simple disillation
3. Electrophoresis
4. All of these

Answer: 1. Fractional distillation

Solution: It is a fact.

Question 90. Geometrical isomerism is not shown by

1. 1, 1-dichloro-1-pentene
2. 1,2-dichloro-1-pentene
3. 1, 3-dichloro-2-pentene
4. 1, 4-dichloro-2-pentene

Answer: 1, 1-dichloro-1-pentene

Solution: 1, 1-dichloro-1-pentene does not exhibit geometrical isomerism.

Question 91. n- pentane and 2-methyl butane are a pair of

1. Enantiomers
2. Stereoisomers
3. Diastereomers
4. Constitutional isomers

Answer: 4. Constitutional isomers

Solution: n-pentane and 2-methyl butane are constitutional isomers or chain isomers or skeletal isomers.

Question 92. A racemic mixture is formed by mixing two

1. Isomeric compounds
2. Chiral compounds
3. meso compounds
4. Enantiomers with chiral carbon

Answer: 2. Chiral compounds

Solution: Racemic mixture is formed by mixing two chiral compounds.

Question 93. CH₅≡N and C₆H₅N≡C exhibit which type of isomerism?

1. Position
2. Functional
3. Metamerism
4. Dextro isomerism

Answer:  2. Functional

Solution: CH₅≡N and C₆H₅N≡C are functional isomers.

Question 94. The production of an optically active compound from a symmetric molecule without resolution is termed as:

1. Walden inversion
2. Partial racemisation
3. Asymmetric synthesis
4. BVartial resolution

Answer: 3. Asymmetric synthesis

Solution: It is the definition of asymmetric synthesis.

Question 95. Metamers of ethyl propionate are

1. C₄H₉COOH and HCOOC₄H₉
2. C₄H₉COOH and C₇3COOC3H₇
3. CH₃COOCH₇ and CH₄COOC3H₇
4. CH₃COOC₃H₇and CH₇COOCH₇

Answer:  4. CH₃COOC₃H₇and CH₇COOCH₇

Solution: Metamers of ethyl propionate are as CH₃COOC₃H₇,C₃H₇COOCH₃

Question 96. Geometrical isomerism is not shown by which of the following? 

Answer: 3. CH₂ =C(Cl)CH₃

Solution: A molecule having doubly bonded carbon atoms shows geometrical isomerism only if both the doubly bonded carbon has altogether different group, i.e., _baC ≡C_ab or C_ac or _abC= _dcC=C_ab

Question 97. The number of optically active isomers of tartaric acid are

1. 1
2. 3
3. 4
4. 2

Answer: 4. 2

Solution: Total number of optical isomers = (2)ⁿ Where n = number of asymmetric carbon atoms.

=(2)²ⁿ= 4

Out of these four optical isomers, two are meso structures which are optically inactive.

∴ Only two structures d and l tartaric acid are optically active.

Question 98. The number of 1°,2° and 3° carbon atoms present in isopentane are respectively.

1. 3, 2, 1
2. 2, 3, 1
3. 3, 1, 1
4. 2, 2,1

Answer: 3. 3,1,1

Solution: (CH₃)₂CHCH₂CH₃ is isopentane.

Question 99. How many optically active stereomers are possible for butan-2, 3-diol?

1. 1
2. 2
3. 3
4. 4

Answer: 2. 2

Solution:

Where C* = Asymmetric C atom

It is a symmetrical molecule, so the number of optically active stereomers= 2n-1.

n = number of asymmetric C atom.

= 2^(2-1).

= 2^(-1)

= 2

Question 100. Which among the following statements is correct with respect to the optical isomers?

1. Enantiomers are non-superimposable mirror images
2. Diastereomers are superimposable mirror images
3. Enantiomers are superimposable mirror images
4. Meso forms have no plane of symmetry

Answer: 1. Enantiomers are non-superimposable mirror images

Solution: Enantiomers are non-superimposable mirror images. example, lactic  acid

Question 101. How many structural isomers are possible for C₄H₉CI?

1. 2
2. 4
3. 8
4. 10

Answer: 2. 4

Question 102. The number of isomeric alkenes with molecular formula C₆H₁₂ are

1. 8
2. 10
3. 11
4. 13

Answer: 4. 13

Solution: The number of isomeric alkenes with molecular formula C₆H₁₂ are 13.

Question 103. Tautomerism is not exhibited by

Answer: 1

Solution: For keto-enol isomerism a compound should have at least one a-hydrogen atom with respect to ketone group or in other words for tautomerism presence of the a-hydrogen atom is essential.

Does not exhibit tautomerism due to the absence of a-hydrogn atom.

Question 104. The compound which exhibits optical isomerism among the following is.

1. CH₃CHOHCH₃
2. (CH₃) ₂CHCH₂CH₃
3. CH₃CHClCH₂CH₃
4. CH₃CCl₂CH₂CH₃

Answer: 3. CH₃CHClCH₂CH₃

Solution: Due to asymmetric carbon atoms in it.

Question 105. Which is true about the following?

1. Only 3 is a chiral compound
2. Only 2 and 4 are chiral compounds
3. All four are chiral compounds
4. Only 1 and 2 are chiral compounds

Answer: 4. Only 1 and 2 are chiral compounds

Solution: The central carbon in 1 and 2 is asymmetric.

Question 106. Racemic compound has

1. An equimolar mixture of enantiomers
2. 1:1 mixture of enantiomer and diastereomer
3. 1:1 mixture of diastereomers
4. 1:2 mixture of enantiomers

Answer: 1. An  equimolar mixture of enantiomers

Solution: An equimolar mixture of the enantiomers dextro or laevo forms. is called a racemic mixture. It is represented as dl-form or ± form and is optically inactive due to the external compensation. The separation of racemic mixture into d- and l-forms is called as resolution.

Question 107. The number of stereoisomers obtained by bromination of trans-2-butene is?

1. 1
2. 2
3. 3
4. 4

Answer: 1. 1

Solution: Meso 2,3-dibromo butane is the product obtained upon bromination.

Question 108. Lactic acid shows optical activity in which state?

1. Solution state
2. Liquid state
3. Crystalline state
4. In all states

Answer: 4. In all states

Solution: Asymmetry is present in all the states.

Question 109. Which one of the starred carbons is the asymmetric one?

Answer: 1

Solution: It has altogether different groups.

Question 110. The number of isomers possible for the aromatic compound with the formula C₇H₈O? 

1. 2
2. 3
3. 4
4. 5

Answer: 4. 5

Solution: For the compound having formula C₇H₈O, there are 5 aromatic isomers.

1. Benzyl alcohol
2. o-methyl phenol
3. m-methyl phenol
4. p-methyl phenol
5. Anisole

Question 111. The number of isomers of the compound with molecular formula C₂H₂Br₂ is

1. 4
2. 3
3. 5
4. 2

Answer: 2. 3

Solution: The structure of isomers from C₂H₂Br₂ are CH₂=CBr₂

Question 112. The number of possible enantiomeric pairs that can be produced during monochlorination of 2-methyl butane

1. 2
2. 3
3. 4
4. 1

Answer: 1.  2

Solution: Two enantiomeric pairs are produced- If monochlorination occurs in no.1 carbon and no.3 carbon i.e., 1-Chloro-2-methyl butane and 2-Chloro-3-methylbutane.

Question 113. An alkane forms isomers if the number of carbon atoms are

1. ≥1
2. ≥2
3. ≥3
4. ≥4

Answer: 4. ≥4

Solution: Butane and isobutane and all higher alkanes show isomerism.

Question 114. Ethoxy ethane and methoxy propane are

1. Geometrical isomers
2. Optical isomers
3. Functional group isomers
4. Metamers

Answer: 4. Metamers

Solution: Ethers show metamerism.

Question 115. How many chiral compounds are possible on monochlorination of 2-methyl butane?

1. 2
2. 4
3. 6
4. 8

Answer: 2. 4

Solution:

Thus, out of four isomers only two have chiral carbon. Each has two isomers.

Question 116. Optical isomerism is shown by

1. Propanol-2
2. Butanol-2
3. Ethanol
4. Methanol

Answer: 2. Butanol-2

Solution: Optical isomerism is shown by an asymmetric carbon atom which has a carbon atom

Question 117. Which one of the following compounds will show optical isomerism?

1. (CH₃) ₂-CH-CH₂-CH₃
2. CH₃-CHOH-CH₃
3. CH₃-CHCl-CH₂-CH₃
4. CH₃-CCl₂-CH₂-CH₃

Answer: 3. CH₃-CHCl-CH₂-CH₃

Solution: Compound CH₃-CHCl-CH₂-CH₃ shows optical isomerism due to the presence of chiral carbon atoms.

Question 118. A molecule of urea can show

1. Chain isomerism
2. Position isomerism
3. Geometrical isomerism
4. Tautomerism

Answer:  4. Tautomerism

Solution: Urea shows tautomerism as

Question 119. If there is no rotation of plane polarised light by a compound in a specific solvent, though to be chiral, it means that:

1. It is certainly meso
2. It is a racemic mixture
3. It is certainly not chiral
4. None of the above

Answer: 1. It is certainly meso

Solution: Meso form is optically inactive.

Question 120. Which of the following does not show stereoisomerism?

Answer: 3

Solution: A, B and C will show stereoisomerism.

Question 121. An enantiomerically pure acid is treated with a racemic mixture of an alcohol having one chiral carbon. The ester formed will be

1. Optically active mixture
2. Pure enantiomer
3. Meso compound
4. Racemic mixture

Answer: 1. Optically active mixture

Solution:

• When optically active acid reacts with a racemic mixture of an alcohol, it forms two types of isomeric esters.
• In each, the configuration of the chiral centre of acid will remain thesame.
• So, the mixture will be optically active.

Question 122. How many chiral carbon atoms are present in 2, 3, 4- trichloro pentane?

1. 4
2. 1
3. 2
4. 3

Answer: 3. 2

Question 123. The type of isomerism observed in urea molecules is

1. Chain
2. Position
3. Geometrical
4. Functional

Answer: 4. Functional

Solution: NH₄CNO is a functional isomer of urea.

Question 124. The number of possible isomers for glucose are

1. 10
2. 14
3. 16
4. 20

Answer: 3. 16

Solution: Glucose has four dissimilar asymmetric carbon atoms; a = 24.

Question 125. The number of geometrical isomers in the case of a compound with the structure, CH₃ — CH=CH—CH=CH—C₂H₅ are

1. Four
2. Three
3. Two
4. Five

Answer: 1. Four

Solution: Two pairs of cis and transforms.

Question 126. The structures,CH₃—CH(NH₂)—CH₂—CH₂CH₃ and CH₃—CH₂—CH(NH₂)—CH₂CH₃ represent:

1. Chain isomers
2. Position isomers
3. Stereoisomers
4. Members

Answer: 2. Position isomers

Solution: 2-aminopentane and 3-aminopentane; Position is different.

Question 127. The total number of cyclic isomers possible for a hydrocarbon with molecular formula C₄H₆ is

1. 1
2. 3
3. 5
4. 7

Answer:  3. 5

Solution: C₄H₆ can have five cyclic isomers.

Question 128. Which of the following compounds exhibits geometrical isomerism?

1. C₂H₅Br
2. (CH)₂(COOH)₂
3. CH₃CHO(CH₂)₂
4. (COOH)₂

Answer: 2. (CH)₂(COOH)₂

Solution: (CH)₂(COOH)₂ actually represents HOOC−CH=CH−COOH which shows geometrical isomerism.

Question 129. The maximum number of alkene isomers for an alkene with molecular formula C₄H₈ are

1. 2
2. 3
3. 6
4. 5

Answer: 3. 6

Solution: The maximum number of alkene isomers for an alkene with molecular formula C₄H₈  are 6 namely, 1- butene, isobutylene, Cyclobutane, methyl cyclopropane, cis 2-butene, trans 2-butene.

Question 130. In fructose, the possible optical isomers are

1. 12
2. 8
3. 16
4. 4

Answer: 2. 8

Solution: Fructose has three chiral carbon atoms, hence the number of optical isomerism =2³

= 2×2×2

= 8

Question 131. The number of optical isomers of pent-3-en-2-ol.

1. 2
2. 4
3. 8
4. 16

Answer: 1. 2

Solution: CH₃CH = CH₂CHOHCH₃ has one asymmetric carbon.

Question 132. Which of the following acids shows stereoisomerism?

1. Oxalic acid
2. Tartaric acid
3. Acetic acid
4. Formic acid

Answer: Tartaric acid

Solution: In the case of 2,3-dihydroxybutanedioic acid, known as tartaric acid, the two chiral centres have the same four substituents and are equivalent. As a result, two of the four possible stereoisomers of this compound are identical due to a plane of symmetry, so there are only three stereoisomeric tartaric acids.

Question 133. How many types of functional groups can be present in an amine with the formula C₃H₉N?

1. 1
2. 2
3. 3
4. 4

Answer: 4. 4

Solution: 4 namely, propylamine, isopropyl amine, N-methyl ethylamine, and trimethylamine.

Question 134. The maximum number of stereoisomers possible for 3-hydroxy-2-methyl butanoic acid are

1. 1
2. 2
3. 3
4. 4

Answer: 4. 4

Solution:

Has two asymmetric carbon atoms and the molecule has no symmetry. Thus, the number of optical isomers = 2ⁿ

= 2²

= 4.

Question 135. An important chemical method to resolve a racemic mixture makes use of the formation of?

1. Mesocompound
2. Enantiomer
3. Racemers
4. Diastereoisomers

Answer: 4. Diastereoisomers

Solution: The resolution of the racemic mixture involves the formation of Diastereoisomers.

Question 136. Which one of the following pairs represents stereoisomerism?

1. Structural and geometrical isomerism
2. Linkage and geometrical isomerism
3. Chain and rotational isomerism
4. Optical and geometrical isomerism

Answer: 4. Optical and geometrical isomerism

Solution: Stereoisomerism is of two types, geometrical and optical.

Question 137. How many asymmetric carbon atoms are present in?

1. 1, 2-methylcyclohexane

2.  3-methyl cyclopentane and

3.  3-methylcyclohexane?

1. Two, one, one
2. One, one, one
3. Two, none, two
4. Two, none One

Answer: 1. Two, one, one

Solution: They contain two,, asymmetric carbon atoms respectively.

Question 138. Which class of compounds can exhibit geometrical isomerism?

1. C₆H₅CH = NOH
2. CH₃CH = CHCH₃
3. HOOCCH- CH₂– CHCOOH
4. All of the above

Answer: 4. All of the above

Solution: All of them show geometrical isomerism.

Question 139. Which type of isomerism is most common among the ethers?

1. Metamerism
2. Functional
3. Chain
4. Position

Answer: 1. Metamerism

Solution: Ethers show metamerism. Metamerism arises when a polyvalent functional group (For example, -O-,>C=O., etc.) is attached to different alkyl groups but the molecular formula remains the same for example, C₂H₅-O- C₂H₅ and CH₃-O-C₃H₇

Question 140. How many chiral isomers can be drawn from 2-bromo, 3-chloro butane?

1. 2
2. 3
3. 4
4. 5

Answer: 3. 4

Solution: 

∴ Number of asymmetric carbon atoms=2

∴ Number of chiral isomers = 2ⁿ

= 2²

= 4

Question 141. Identify the compound that exhibits tautomerism

1. 2-butene
2. Lactic acid
3. 2-pentanone
4. Phenol

Answer: 3. 2-pentanone

Solution:

Question 142. Which of the following may exist in enantiomorphs?

Answer: 4

Solution: Due to the presence of asymmetric carbon atoms.

Question 143. The two optical isomers given below, namely are

1. Enantiomers
2. Geometrical isomers
3. Diastereomers
4. Structural isomers

Answer: 3. Diastereomers

Solution: Diastereomers are a type of stereoisomer. Diastereomers are defined as non-mirror image, non-identical stereoisomers. Hence, they occur when two or more stereoisomers of a compound have different configurations at one or more of the equivalent stereo centres and are not mirror images of each other.

Question 144. CH₃CH₂OH and CH₃OCH₃are the examples of

1. Chain isomerism
2. Functional isomerism
3. Position isomerism
4. Metamerism

Answer: 2. Functional isomerism

Solution: CH₃CH₂OH and CH₃OCH₃ have different functional groups. (ie., -OH in alcohol and –O-in ether), hence they are examples of functional isomerism.

Question 145. Which is optically active?

1. Isobutyric acid
2. β-chloropropionic acid
3. Propionic acid
4. α-chloropropionic acid

Answer: 4. α-chloropropionic acid

Solution: CH₃CHClCOOH contains an asymmetric carbon atom.

Question 146. Compounds that rotate plane polarised light in a clockwise direction are known as?

1. Dextrorotatory
2. Laevorotatory
3. Optically inactive compounds
4. Racemic

Answer: Dextrorotatory

Solution: A characteristic of dextrorotatory.

Question 147. Maleic acid and fumaric acid are

1. Position isomers
2. Geometric isomers
3. Enantiomers
4. Functional isomers

Answer: 2. Geometric isomers

Solution: The structures of maleic and fumaric acids are given below

The structures of fumaric and maleic acid suggest that they are geometrical isomers because they have the same molecular formula but different spatial arrangements of atoms around a double bond.

Question 148. The total number of isomeric carbocations possible for the formula C₄H₉⁺

1. 3
2. 4
3. 2
4. 5

Answer: 2. 4

Solution: The possible carbocations are (CH₃)₃C⁺, (CH₃)₃CHC⁺H₂,CH₃CH₂CH₂C⁺H₂ and CH₃CH₂C⁺HCH₃

Question 149. Glucose and fructose are

1. Chain isomers
2. Position isomers
3. Functional isomers
4. Optical isomers

Answer: 3. Functional isomers

Solution: Glucose has an aldehyde group and a fructose keto group. The general formula for both is C₆H₁₂O₆.

Question 150. The compound isomeric with acetone is

1. Propionaldehyde
2. Propionic acid
3. Ethoxy ethane
4. None of these

Answer: 2. Propionaldehyde

Solution: Both have the same molecular formula.

Question 151. How many structural formulae are possible for C₅H₁₁Cl?

1. 6
2. 8
3. 10
4. 12

Answer: 2. 8

Solution: 1-chloropentane, 2-chloropentane, 3-chloropentane,1-chloro-3-methylbutane, 1-chloro-2-methylbutane, 1-chloro-2,2 dimethylpropane,2-chloro3-methylbutane, 2-chloro-2-methylbutane

Question 152. Total number of isomeric aldehydes and ketones that can exist with the molecular formula C₅H₁₀O :

1. 5
2. 8
3. 6
4. 7

Answer: 4. 7

Solution: Draw all possible isomers.

Question 153. Isomerism exhibited by acetic acid and methyl formate is

1. Functional
2. Chain
3. Geometrical
4. Central

Answer: 1. Functional

Solution: Acid has —COOH group whereas, ester has —COOR group.

Question 154. The number of isomeric structures for C₂H₇N would be?

1. 4
2. 3
3. 2
4. 1

Answer: 3. 2

Solution: CH₃CH₂NH₂and CH₃NHCH₃.

Question 155. Which of the following will have a meso-isomer also?

1. 2-chloroquine
2. 2, 3-dichlorobutane
3. 2, 3-dichloromethane
4. 2-hydroxy propanoic acid

Answer: 2. 2, 3-dichlorobutane

Solution:

One asymmetric carbon atom, forms d, and l-optical and isomers.

Two asymmetric carbon atoms, forms d, l and meso forms

Meso due to internal compensation

Two asymmetric carbon atoms but does not have symmetry. Hence, the meso form is not formed.

Question 156. 

Compound can exhibit

1. Geometrical isomerism
2. Tautomerism
3. Optical isomerism
4. Geometrical and optical isomerism

Answer: 3. Optical isomerism

Solution: A compound could be optically active only when it contains, at least one asymmetric carbon atom or a chiral centre.

Question 157. Select R-isomers from the following

1. 1 and 3
2. 2, 4 and 5
3. 1,2, and 3
4. 2 and 3

Answer: 3. 1,2, and 3

Solution: Draw the orientations.

Question 158. A similarity between optical and geometrical isomerism is that

1. Each forms an equal number of isomers for a given compound
2. If in a compound, one is present then so is the other
3. Both are included in stereoisomerism
4. They have no similarity

Answer:  3. Both are included in stereoisomerism

Solution: Both geometrical and optical isomerism are included in stereoisomerism.

Question 159. Acetone and propen-2-ol are?

1. Positional isomers
2. Keto-enol tautomers
3. Geometrical isomers
4. Chain isomers

Answer: 2. Keto-enol tautomers

Solution: Acetone has a ketone functional group and propenal have an aldehyde functional group therefore, these are functional isomers.

Question 160. How many isomers are possible for the compound having a molecular formula C₃H₅Br₃?

1. 5
2. 4
3. 6
4. 8

Answer: 1. 5

Solution: Draw position and chain isomers.

Question 161. Which one of the following shows functional isomerism?

1. C₂H₄
2. C₃H₆
3. C₂H₅OH
4. CH₂Cl₂

Answer: C₂H₅OH

Solution: Ethyl alcohol shows functional isomerism with dimethyl ether.

C₂H₅OH – Alcohol

CH₃-O-CH₃ – Ether 

Question 162. How many isomers of C₅H₁₁OH will be primary alcohols?

1. 5
2. 4
3. 3
4. 2

Answer: 2. 4

Solution: A total of 8 structural isomers of pentanol i.e., 1-pentanol, 2-pentanol, 3-pentanol, 2-methylbutan-1-ol, 3-methylbutan-1-ol, 2-methylbutan-2-ol, 2-methylbutan-3-ol and 2,2-dimethylpropanol are possible. Out of these 8, 4 are primary alcohols.

Question 163. The molecular formula of a saturated compound is C₂H₄Br₂. This formula permits the existence of which isomers?

1. Functional isomers
2. Optical isomers
3. Positional isomers
4. Cis-trans isomers

Answer: Positional isomers

Solution: 1,1-dibromoethane and 1,2-dibromoethane.

Question 164. C₆H₁₂ on the addition of HBr in the presence and in the absence of peroxide gives _______product.

1. Hexene-3
2. 2,3-dimethyl butane-2
3. Symmetrical alkene
4. All of these

Answer: 4. All of these

Solution: There are two symmetrical hexenes as given in (1) and (2).

Question 165. Buta-1,3-diene and But-2-yne are:

1. Position isomers
2. Functional isomers
3. Chain isomers
4. Tautomers

Answer: 2. Functional isomers

Solution: But-2-yen and Buta -1, 3-diene both have the same molecular formula of C₄H₆. In both of them, the functional group is different but the formula is the same. – Hence, they show functional isomerism.

Question 166. Which of the following compounds are optically active?

1. (CH₃)₂CHCH₂OH
2. CH₃CH₂OH
3. CCl₂F₂ CH₃
4. CHOHC₂H₅

Answer: 4. CHOHC₂H₅

Solution: CH₃CHOHC₂H₅ is optically active because it has chiral C*-atom

Question 167. Compounds whose molecules are superimposable on their mirror images even though they contain asymmetric carbon atoms or chiral centres are known as

1. Enantiomers
2. Racemers
3. Mesomers
4. Conformers

Answer: 3. Mesomers

Solution: Mesoforms are optically inactive as they are superimposable to their mirror images.

Question 168. Position isomerism is shown by

1. O-nitrophenol and p-nitrophenol
2. Dimethyl ether and ethanol
3. Pentan-2-one and pentan-3-one
4. Acetaldehyde and acetone

Answer: O-nitrophenol and p-nitrophenol

Solution: O-,m-,p- isomers are position isomers.

Question 169. The number of stereoisomers possible for a compound of the molecular formula

CH₃-CH=CH-CH(OH)-Me is

1. 3
2. 2
3. 4
4. 6

Answer:

Solution: There are four stereoisomers cis-R cis-S trans-R trans-S

Question 170. Which of the following will exhibit cis-trans isomerism?

1. CH₂Br–CH₂Br
2. CBr₃–CH₃
3. CHBr=CHBr
4. CBr₂=CH₂

Answer: 3. CHBr=CHBr

Solution: Due to the restricted rotation of a double bond, the alkene shows geometrical isomerism because the relative position of atoms or groups attached to the carbon atoms of the double bond gets fixed.

If the same groups or atoms are attached to the double bond bearing carbon, then alkene doesn’t show geometrical isomerism.

Question 171. The total number of cyclic structural as well as stereoisomers possible for a compound with the molecular formula C₅H10 is

1. 2
2. 4
3. 6
4. 7

Answer: 3. 6

Solution: The total number of cyclic isomers is six as shown below.

Question 172. The maximum number of possible optical isomers in 1-bromo-2-methylcyclohexane is

1. 4
2. 2
3. 8
4. 16

Answer: 1. 4

Solution: The molecule contains two carbons. The number of optical isomers is given by 2n, where n=No. of chiral carbons.

∴ Optical isomers 2² = 4.

Question 173. How many stereoisomers does this molecule have? CH₃CH=CHCH₂CHBrCH₃

1. 6
2. 8
3. 4
4. 2

Answer: 4. 2

Solution:

It has one chiral centre (two enantiomers) and two geometrical isomers cis⎼d, trans⎼d, cis⎼ and trans⎼l.

Question 174. How many optically active stereoisomers are possible for butane-2, 3-diol?

1. 0
2. 1
3. 2
4. 3

Answer: 4. 3

Solution: The structure of butane-2, 3-idol is as

∵ Optical isomers in compounds have similar asymmetric carbon atoms, which are even in number =2n-1

Here, n = 2

∴ Total number of optically active stereoisomers =

= 2(2)-1

= 4- 1=  3

Question 175. The number of isomeric pentyl alcohols are

1. Two
2. Four
3. Six
4. Eight

Answer:  4. Eight

Solution: n-pentanol, 2-pentanol, 3-pentanol, 2-methylbutanol, 2-methylbutan-2-ol, 3-methylbutanol, 2, 2-dimethypropanol, and 3-methylbutan-2-ol (8 isomers)

Question 176. Which of the following is most likely to show optical isomerism?

Answer: 2

Solution: Optical isomerism is shown by compounds which have one or more chiral carbon atoms.

∵ It has asymmetric or chiral carbon atoms,

∴ It shows optical isomerism.

Question 177. Which type of isomerism is shown by propanal and propanone?

1. Functional group
2. Metamerism
3. Tautomerism
4. Chain isomerism

Answer: 1. Functional group

Solution: When two compounds have similar molecular formulas but differ in the functional group then the isomerism is called functional group isomerism i.e.,

Question 178. The number of isomers possible for C4H8O is

1. 3
2. 4
3. 5
4. 6

Answer: 4. 6

Solution: There are six isomers possible for the compounds having molecular formula C₄H₈O, which are as follows

Question 179. Which of the following statements are correct?

1. Desmotropism is another name for tautomerism
2. Allyl carbocation is less stable than isopropyl carbocation
3. -I effect is exhibited by -NH₃⁺
4. The formula CH₂Cl₂ is non-polar

Answer: 1. Desmotropism is another name for tautomerism

Solution: Desmo (bond), tropism (turn). Thus, desmotropism, i.e., isomerism arised due to the turning of bond was the name given to tautomerism.

Question 180. Which of the following structures are superimposable?

1. 1 and 2
2. 2 and 3
3. 1 and 4
4. 1 and 3

Answer: 4. 1 and 3

Solution: (1) and (3) are enantiomeric forms of each other.

Question 181. The alkene that exhibits geometrical isomerism is

1. Propene
2. 2-methylpropene
3. 2-butene
4. 2-methyl-2-butene

Answer: 2 -butene

Solution: 2-Butene may exist as cis and trans isomers. The cis-isomer has the two methyl groups on the same side and the trans-isomer has the two methyl groups on opposite sides. Due to restricted rotation around the double bond, it exhibits geometrical isomerism.

Question 182. The isomers which are interconverted through rotation around a single bond are

1. Conformers
2. Diastereomers
3. Enantiomers
4. Position isomers

Answer: 1. Conformers

Solution: The isomerism which arises due to rotation about a C-C is called conformational isomerism and the isomers are called conformational isomers or rotational isomers or conformers.

Question 183. The total number of acyclic isomers including the stereoisomers with the molecular formula C₄H₇Cl

1. 11
2. 12
3. 9
4. 10

Answer: 2. 12

Question 184. n-pentane and neopentane exhibit

1. Functional isomers
2. Geometrical isomers
3. Chain isomers
4. Position isomers

Answer: 3. Chain isomers

Solution: N-pentane and isopentane or 2-methylbutane are chain isomers since both have different hydrocarbon chains.

Question 185. Geometrical isomerism is shown by

1. – C –C –
2. >C=C<
3. C≡C
4. None of these

Answer: 2. >C=C<

Solution: Geometrical isomerism is shown by >C=C< only when identical groups are not present on the double bonded carbon atoms.

Question 186. Example of geometrical isomerism is

1. 2-butanol
2. 2-butene
3. Butanal
4. 2-butyne

Answer: 2. 2-butene

Solution: 2-Butene may exist as cis and trans isomers. The cis-isomer has the two methyl groups on the same side and the trans-isomer has the two methyl groups on opposite sides. Due to restricted rotation around the double bond, it exhibits geometrical isomerism.

Question 187. Two crystalline forms of a substance, one being a mirror image of the other are called?

1. Pentane
2. Chain isomers
3. Stereoisomers
4. Functional isomers

Answer: 3. Stereoisomers

Solution: The mirror-image isomerism is a class of stereoisomerism and is included in optical

isomerism.

Question 188. Which pair represents chain isomers?

1. CH₃CHCl₂ and ClCH₂CH₂Cl
2. Propyl alcohol and isopropyl alcohol
3. 2-methyl butane and neopentane
4. Diethyl ether and dipropyl ether

Answer: 3. 2-methyl butane and neopentane

Solution: Note that propyl (propan-l-ol) and isopropyl alcohol (propan-2-ol) are position isomers.

Question 189. Different structures generated due to rotation about, the C – C axis, of an organic molecule, are examples of

1. Geometrical isomerism
2. Conformational isomerism
3. Optical isomerism
4. Structural isomerism

Answer: 2. Conformational isomerism

Solution: The different arrangement of atoms in space that results from the carbon-carbon single bond free rotation by 360° are called conformations or conformational isomers and this phenomenon is called conformational isomerism.

Question 190. Which of the following compounds is optically active?

1. 1 – butanol
2. Isopropyl alcohol
3. Acetaldehyde
4. 2-butanol

Answer: 4. 2-butanol

Solution: 2-butanol is optically active as it contains a chiral carbon atom.

Question 191. Which of the following will have meso isomers also?

1. 2-hydroxy propanoic acid
2. 2,3-dichlorobutane
3. 2,3-dichloromethane
4. 2-chloroquine

Answer: 2. 2,3-dichlorobutane

Solution: 2,3-Dichlorobutane has a meso-isomer due to the presence of the plane of symmetry.

Question 192. The isomerism which exists between CH₃CHCl₂ and CH₂ClCH₂Cl is

1. Chain
2. Functional
3. Positional
4. Metamerism

Answer: 3. Positional

Solution: The positions of Cl are different.

Question 193. Alkyl cyanide R-C≡N and alkyl isocyanides R–N≡C are

1. Tautomers
2. Metamers
3. Functional isomers
4. Geometrical isomers

Answer: Functional isomers

Solution: Both have different functional groups, i.e., –CN and –NC.

Question 194. Out of the following, the alkene that exhibits optical isomerism is

1. 3-methyl-2-pentene
2. 4-methyl-1-pentene
3. 3-methyl-1-pentene
4. 2-methyl-2-pentene

Answer: 3-methyl-1-pentene

Solution: For a compound to show optical isomerism, the presence of chiral carbon atoms is a necessary condition. H₂C=HC−H|C∗|CH₃  −CH₂−CH₃ 3-methyl-1-pentene

Question 195. Which of the following compounds will show metamerism?

1. CH₃– CO-C₂H₅
2. C2H5-S-C₂H₅
3. CH₃-O-CH₃
4. CH₃-O-C₂H₅

Answer: 3. CH₃-O-CH₃

Solution: Compounds having bivalent functional groups (like C=O, –O–, –S – etc) with at least 4 carbon atoms (in the case of ether and thioether) or at least 5 carbon atoms (in the case of ketones) exhibit metamerism. Hence, C₂H₅-S-C₂H₅ will show metamerism.

Question 196. Which of the following compounds exhibit stereoisomerism?

1. 3-methyl butyne –1
2. 2-methyl butene –1
3. 2-methyl butanoic acid
4. 3-methyl butanoic acid

Answer: 3. 2-methyl butanoic acid

Solution: 2-methyl butanoic acid exhibits stereoisomerism.

It shows optical isomerism because it contains an asymmetric carbon atom.

Question 197. Who proposed the tetrahedral mirror image structures to a pair of enantiomers?

1. Kekule
2. Wohler
3. Van’t Hoff
4. None of these

Answer: 3. Van’t Hoff

Solution: It is a fact.

Question 198. Geometrical isomers differ in

1. Position of functional groups
2. Position of atoms
3. Spatial arrangement of atoms
4. Length of the carbon chain

Answer: 3. Spatial arrangement of atoms

Solution: It is a fact.

Question 199. The number of isomeric alkanes having the molecular formula C₂H₁₂ is

1. Three
2. Five
3. Nine
4. Thirty-two

Answer: 1. Three

Solution: The isomers alkanes having the molecular formula C₂H₁₂ are as

Question 200. Two crystalline forms of a substance, one being a mirror image of the other are called

1. Pentane
2. Chain isomers
3. Stereoisomers
4. Functional isomers

Answer: 3. Stereoisomers

Solution: is a form of isomerism in which molecules have the same molecular formula and sequence of bonded atoms, but differ in the three-dimensional orientations of their atoms in space.

Question 201. A molecule having three different chiral carbon atoms, how many stereoisomers will it have?

1. 8
2. 3
3. 9
4. 6

Answer: 1. 8

Solution: The number of stereoisomers=2’

(Here, n=chiral carbon atom) Thus, the number of stereoisomers

= 2³

= 2×2×2

= 8

The mirror-image isomerism is a class of stereoisomerism and is included in optical isomerism.