Here below some basic MCQ’s about “Amines and Nitro Compounds” with answer which is well explained for exam practice. Let’s check one by one which is given below.
1. Which of the following is the most basic amine?
a) Aniline
b) Methylamine
c) Diphenylamine
d) p-Nitroaniline
Answer: b) Methylamine
Explanation: Methylamine (CH₃NH₂) is a simple aliphatic amine, and aliphatic amines are generally more basic than aromatic amines because the lone pair on nitrogen is more readily available for protonation. In aniline, for example, the lone pair is delocalized into the benzene ring, reducing its basicity. Nitro groups (in p-Nitroaniline) also reduce the basicity by withdrawing electron density.
2. Which of the following will not undergo the Hoffmann bromamide reaction?
a) Methylamine
b) Benzylamine
c) Acetamide
d) Benzenesulfonamide
Answer: d) Benzenesulfonamide
Explanation: The Hoffmann bromamide reaction involves the conversion of an amide to a primary amine. Benzenesulfonamide is not an amide; it is a sulfonamide and thus will not undergo the Hoffmann reaction.
3. Which of the following reagents is used to distinguish between primary, secondary, and tertiary amines?
a) Carbylamine test
b) Lucas test
c) Hinsberg’s reagent
d) Tollens’ reagent
Answer: c) Hinsberg’s reagent
Explanation: Hinsberg’s reagent (benzenesulfonyl chloride, C₆H₅SO₂Cl) is used to distinguish between primary, secondary, and tertiary amines. Primary amines form a sulfonamide that is soluble in alkali, secondary amines form a sulfonamide that is insoluble in alkali, and tertiary amines do not react with Hinsberg’s reagent.
4. The nitro group (-NO₂) is a:
a) Strongly electron-donating group
b) Weakly electron-donating group
c) Weakly electron-withdrawing group
d) Strongly electron-withdrawing group
Answer: d) Strongly electron-withdrawing group
Explanation: The nitro group (-NO₂) is a strongly electron-withdrawing group due to the high electronegativity of nitrogen and oxygen, and the resonance effect, where the negative charge is delocalized over oxygen atoms. This reduces electron density on the rest of the molecule.
5. Aniline reacts with bromine water to give:
a) Mono-bromoaniline
b) Di-bromoaniline
c) Tri-bromoaniline
d) Tetra-bromoaniline
Answer: c) Tri-bromoaniline
Explanation: Aniline, due to the strongly activating nature of the amine group, reacts with bromine water to give 2,4,6-tribromoaniline, where bromine is substituted at the ortho and para positions of the benzene ring.
6. Which of the following compounds is formed when nitrobenzene is reduced with tin and hydrochloric acid?
a) Aniline
b) Benzylamine
c) Phenol
d) Benzene
Answer: a) Aniline
Explanation: Nitrobenzene (C₆H₅NO₂) is reduced to aniline (C₆H₅NH₂) when treated with reducing agents like tin (Sn) and hydrochloric acid (HCl). This is a standard reduction of the nitro group to the amino group.
7. Which of the following compounds can be prepared by the Gabriel phthalimide synthesis?
a) Methylamine
b) Aniline
c) Ethylamine
d) Both a and c
Answer: d) Both a and c
Explanation: Gabriel phthalimide synthesis is used to prepare primary aliphatic amines such as methylamine (CH₃NH₂) and ethylamine (C₂H₅NH₂). Aromatic amines like aniline cannot be synthesized by this method.
8. Which of the following is a method for converting an amide to an amine?
a) Hoffmann bromamide reaction
b) Gabriel synthesis
c) Sandmeyer reaction
d) Wurtz reaction
Answer: a) Hoffmann bromamide reaction
Explanation: The Hoffmann bromamide reaction converts an amide (RCONH₂) to a primary amine (RNH₂) by treating it with bromine (Br₂) and sodium hydroxide (NaOH).
9. The reduction of nitrobenzene in acidic medium produces:
a) Phenol
b) Aniline
c) Hydrazobenzene
d) Nitroethane
Answer: b) Aniline
Explanation: Reduction of nitrobenzene (C₆H₅NO₂) in an acidic medium (e.g., using Sn/HCl) results in the formation of aniline (C₆H₅NH₂), a primary aromatic amine.
10. Which of the following is NOT true about nitro compounds?
a) Nitro compounds are highly polar
b) Nitro compounds are generally insoluble in water
c) Nitro compounds show strong hydrogen bonding
d) Nitro compounds undergo reduction easily
Answer: c) Nitro compounds show strong hydrogen bonding
Explanation: Nitro compounds are polar due to the presence of the nitro group, but they do not exhibit strong hydrogen bonding because they lack a hydrogen atom attached to oxygen or nitrogen that could participate in hydrogen bonding. They are often insoluble in water and can undergo reduction reactions.
11. Which of the following reactions is used to prepare aniline from benzene?
a) Sandmeyer reaction
b) Gattermann reaction
c) Nitration followed by reduction
d) Friedel-Crafts alkylation
Answer: c) Nitration followed by reduction
Explanation: Aniline is prepared from benzene by first nitrating benzene to form nitrobenzene, followed by reduction of nitrobenzene to aniline using reducing agents like tin and hydrochloric acid.
12. In the reaction of aniline with acetic anhydride, the product formed is:
a) Acetanilide
b) Nitroaniline
c) Benzanilide
d) Aniline acetate
Answer: a) Acetanilide
Explanation: When aniline reacts with acetic anhydride (CH₃CO)₂O, it forms acetanilide (C₆H₅NHCOCH₃) via acetylation of the amino group. This reaction protects the amine group from further reactions.
13. What is the major product when nitrobenzene undergoes catalytic hydrogenation?
a) Aniline
b) Phenol
c) Benzylamine
d) Azobenzene
Answer: a) Aniline
Explanation: Catalytic hydrogenation of nitrobenzene (C₆H₅NO₂) with hydrogen (H₂) and a catalyst like palladium or platinum reduces the nitro group (-NO₂) to an amino group (-NH₂), forming aniline (C₆H₅NH₂).
14. Which of the following will not give a positive carbylamine test?
a) Aniline
b) Methylamine
c) Diethylamine
d) Ethylamine
Answer: c) Diethylamine
Explanation: The carbylamine test is used to detect primary amines. Only primary amines react with chloroform (CHCl₃) and alcoholic potassium hydroxide (KOH) to form isocyanides (foul-smelling compounds). Diethylamine (a secondary amine) will not give a positive result.
15. Reduction of nitrobenzene with zinc and ammonium chloride gives:
a) Aniline
b) Phenylhydroxylamine
c) Hydrazobenzene
d) Azobenzene
Answer: b) Phenylhydroxylamine
Explanation: Partial reduction of nitrobenzene with zinc dust and ammonium chloride (NH₄Cl) in neutral conditions yields phenylhydroxylamine (C₆H₅NHOH). Further reduction can produce aniline.
16. The reaction of methylamine with nitrous acid results in the formation of:
a) Methyl alcohol
b) Ethyl alcohol
c) Methanol
d) Nitromethane
Answer: a) Methyl alcohol
Explanation: When methylamine (CH₃NH₂) reacts with nitrous acid (HNO₂), it undergoes diazotization to form a diazonium salt, which then decomposes to form methyl alcohol (CH₃OH), nitrogen gas (N₂), and water.
17. Which of the following is a nucleophilic substitution reaction?
a) Conversion of aniline to diazonium salt
b) Conversion of nitrobenzene to aniline
c) Hoffmann bromamide reaction
d) Gabriel phthalimide reaction
Answer: d) Gabriel phthalimide reaction
Explanation: The Gabriel phthalimide reaction involves the nucleophilic substitution of a halide by the phthalimide anion to form a primary amine.
18. Which of the following reagents is used for the diazotization of aniline?
a) H₂SO₄
b) NaOH
c) HCl + NaNO₂
d) CH₃COOH
Answer: c) HCl + NaNO₂
Explanation: Diazotization of aniline is carried out by treating it with sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at low temperatures (0-5°C). This forms the diazonium salt, which is used in various synthetic reactions.
19. When p-nitroaniline is treated with bromine water, the major product is:
a) Mono-bromoaniline
b) Di-bromoaniline
c) 2,4,6-Tribromoaniline
d) 3,5-Dibromo-4-nitroaniline
Answer: d) 3,5-Dibromo-4-nitroaniline
Explanation: In p-nitroaniline, the -NO₂ group is a strong electron-withdrawing group that deactivates the benzene ring toward electrophilic substitution, but the -NH₂ group directs bromine to the ortho and para positions relative to it. The product is 3,5-dibromo-4-nitroaniline.
20. Which of the following does not involve the formation of a diazonium salt?
a) Sandmeyer reaction
b) Gattermann reaction
c) Hoffmann bromamide reaction
d) Coupling reaction
Answer: c) Hoffmann bromamide reaction
Explanation: The Hoffmann bromamide reaction does not involve diazonium salts. It is used to convert amides into amines by using bromine and a strong base.
21. Aniline when treated with excess bromine water gives:
a) 2,4,6-Tribromoaniline
b) p-Bromoaniline
c) m-Bromoaniline
d) o-Bromoaniline
Answer: a) 2,4,6-Tribromoaniline
Explanation: Aniline is highly reactive towards bromine due to the strong activating nature of the -NH₂ group. When treated with bromine water, it forms 2,4,6-tribromoaniline by substitution at the ortho and para positions.
22. The major product obtained when nitrobenzene is reduced using iron and HCl is:
a) Aniline
b) Hydrazobenzene
c) Phenylhydroxylamine
d) Benzylamine
Answer: a) Aniline
Explanation: Reduction of nitrobenzene (C₆H₅NO₂) using iron and hydrochloric acid gives aniline (C₆H₅NH₂) as the major product. This is a common method of reducing the nitro group to an amino group.
23. The Sandmeyer reaction is used for the synthesis of:
a) Halogenated aromatic compounds
b) Aliphatic amines
c) Nitro compounds
d) Alcohols
Answer: a) Halogenated aromatic compounds
Explanation: The Sandmeyer reaction involves the replacement of the diazonium group in an aromatic diazonium salt with a halogen (Cl, Br, or I) using copper salts (CuCl or CuBr). It is commonly used to synthesize halogenated aromatic compounds.
24. Which of the following statements is correct about nitro compounds?
a) Nitro compounds are basic
b) Nitro compounds are acidic
c) Nitro compounds are neutral
d) Nitro compounds show neither basic nor acidic character
Answer: b) Nitro compounds are acidic
Explanation: Nitro compounds, especially nitroalkanes, have an acidic hydrogen at the α-carbon due to the electron-withdrawing nature of the nitro group, which stabilizes the conjugate base.
25. Which of the following amines can be synthesized using the Hoffmann degradation?
a) Primary amines
b) Secondary amines
c) Tertiary amines
d) Quaternary amines
Answer: a) Primary amines
Explanation: The Hoffmann degradation (also known as Hoffmann bromamide reaction) converts an amide to a primary amine with one fewer carbon atom. Secondary, tertiary, and quaternary amines cannot be synthesized by this method.
26. Which of the following is true about the basicity of aromatic amines?
a) Aromatic amines are more basic than aliphatic amines
b) Aromatic amines are less basic than aliphatic amines
c) Aromatic amines do not act as bases
d) Aromatic amines have the same basicity as aliphatic amines
Answer: b) Aromatic amines are less basic than aliphatic amines
Explanation: Aromatic amines (like aniline) are less basic than aliphatic amines because the lone pair of electrons on the nitrogen atom is delocalized into the aromatic ring, making it less available for protonation.
27. Which of the following reactions will give nitrobenzene as a product?
a) Nitration of benzene
b) Reduction of aniline
c) Oxidation of phenol
d) Friedel-Crafts acylation
Answer: a) Nitration of benzene
Explanation: Nitrobenzene is formed by the nitration of benzene using a mixture of concentrated nitric acid (HNO₃) and sulfuric acid (H₂SO₄). This introduces a nitro group (-NO₂) onto the benzene ring.
28. Which of the following reactions involves the formation of a free radical?
a) Sandmeyer reaction
b) Hoffmann bromamide reaction
c) Diazotization
d) Acetylation of aniline
Answer: a) Sandmeyer reaction
Explanation: The Sandmeyer reaction involves the formation of free radicals when the diazonium salt is reacted with copper(I) halide (CuCl or CuBr), which replaces the diazonium group with a halogen.
29. Which of the following amines does not undergo diazotization?
a) Aniline
b) Methylamine
c) Phenylamine
d) Benzylamine
Answer: d) Benzylamine
Explanation: Benzylamine is a primary aliphatic amine, and only aromatic primary amines (like aniline) undergo diazotization. Aliphatic amines generally decompose when treated with nitrous acid, forming alcohols.
30. The reduction of nitrobenzene under strongly acidic conditions gives:
a) Phenylhydroxylamine
b) Hydrazobenzene
c) Azobenzene
d) Aniline
Answer: d) Aniline
Explanation: Nitrobenzene (C₆H₅NO₂) is reduced to aniline (C₆H₅NH₂) under strongly acidic conditions using a reducing agent such as tin and hydrochloric acid (Sn/HCl).
31. Which of the following compounds will give a positive Hinsberg test?
a) Ethylamine
b) Diethylamine
c) Triethylamine
d) Methylamine
Answer: a) Ethylamine
Explanation: The Hinsberg test distinguishes primary, secondary, and tertiary amines. Ethylamine, a primary amine, reacts with Hinsberg reagent to form a sulfonamide soluble in alkali, giving a positive test. Secondary and tertiary amines behave differently.
32. What is the product when acetamide is treated with Br₂ and NaOH (Hoffmann bromamide reaction)?
a) Methylamine
b) Ethylamine
c) Aniline
d) Acetone
Answer: a) Methylamine
Explanation: The Hoffmann bromamide reaction converts an amide (acetamide, CH₃CONH₂) to a primary amine with one less carbon atom. In this case, the product is methylamine (CH₃NH₂).
33. Aniline when treated with excess acetic anhydride produces:
a) Acetone
b) Acetanilide
c) Anilinium ion
d) N,N-Diacetylaniline
Answer: b) Acetanilide
Explanation: When aniline is treated with acetic anhydride, it undergoes acetylation to form acetanilide. This reaction protects the amine group and prevents further reactions on the nitrogen atom.
34. Which of the following does not give a positive nitrous acid test?
a) Aniline
b) Methylamine
c) Diethylamine
d) Benzylamine
Answer: c) Diethylamine
Explanation: The nitrous acid test is used to distinguish between primary, secondary, and tertiary amines. Diethylamine (a secondary amine) does not form diazonium salts with nitrous acid and gives a negative result in this test.
35. Which of the following reagents can be used to reduce nitro compounds to amines?
a) Fe/HCl
b) NaOH
c) KMnO₄
d) H₂SO₄
Answer: a) Fe/HCl
Explanation: Nitro compounds are typically reduced to amines using reducing agents like iron filings and hydrochloric acid (Fe/HCl) or other catalytic hydrogenation processes.
36. Which of the following compounds is not prepared by the reduction of a nitro compound?
a) Aniline
b) Phenylhydroxylamine
c) Azobenzene
d) Benzene
Answer: d) Benzene
Explanation: Benzene is not prepared by the reduction of a nitro compound. Compounds like aniline and phenylhydroxylamine are produced from the reduction of nitrobenzene.
37. When aniline reacts with cold HNO₂ in the presence of HCl, the product is:
a) Acetanilide
b) Benzenediazonium chloride
c) Phenol
d) Benzylamine
Answer: b) Benzenediazonium chloride
Explanation: Aniline reacts with cold nitrous acid (generated in situ from NaNO₂ and HCl) to form benzenediazonium chloride (C₆H₅N₂⁺Cl⁻), which is an important intermediate in synthetic organic chemistry.
38. What is the major product when aniline reacts with acetic anhydride?
a) Nitroaniline
b) Acetanilide
c) Benzanilide
d) Acetone
Answer: b) Acetanilide
Explanation: Acetic anhydride reacts with aniline to form acetanilide by acetylating the amino group (-NH₂). This reaction is used to protect the amine group in subsequent reactions.
39. Which of the following nitro compounds is most reactive towards nucleophilic substitution?
a) Nitrobenzene
b) 2,4-Dinitrochlorobenzene
c) m-Nitroaniline
d) p-Nitrotoluene
Answer: b) 2,4-Dinitrochlorobenzene
Explanation: 2,4-Dinitrochlorobenzene is highly reactive towards nucleophilic substitution because the two nitro groups strongly withdraw electron density from the ring, making the carbon bearing the chlorine more susceptible to nucleophilic attack.
40. Which of the following reagents converts aniline into a diazonium salt?
a) HCl and NaNO₂
b) Br₂ and Fe
c) KMnO₄ and NaOH
d) CH₃COOH and NaOH
Answer: a) HCl and NaNO₂
Explanation: Aniline is converted into benzenediazonium chloride by treating it with a mixture of sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at low temperatures (0-5°C).
41. Which of the following is the most basic compound?
a) Aniline
b) Ethylamine
c) Ammonia
d) Diphenylamine
Answer: b) Ethylamine
Explanation: Ethylamine (a primary aliphatic amine) is more basic than aromatic amines (like aniline or diphenylamine) because the electron-donating alkyl group increases the electron density on the nitrogen, making it more readily protonated.
42. Which of the following amines reacts with nitrous acid to form a stable diazonium salt?
a) Aniline
b) Benzylamine
c) Methylamine
d) Ethylamine
Answer: a) Aniline
Explanation: Aniline (an aromatic primary amine) reacts with nitrous acid to form a stable diazonium salt. Aliphatic amines like benzylamine and methylamine form unstable diazonium salts that decompose to alcohols.
43. Which of the following compounds will undergo the Hoffmann rearrangement?
a) Benzamide
b) Benzylamine
c) Benzene
d) Benzaldehyde
Answer: a) Benzamide
Explanation: The Hoffmann rearrangement (or Hoffmann bromamide reaction) involves the conversion of an amide (such as benzamide) to a primary amine with the loss of one carbon atom.
44. Which of the following is not a characteristic of nitro compounds?
a) They are generally explosive
b) They are basic
c) They are electron-withdrawing
d) They can undergo reduction to amines
Answer: b) They are basic
Explanation: Nitro compounds are not basic; in fact, they are typically acidic due to the electron-withdrawing nature of the nitro group (-NO₂).
45. What is the product when nitrobenzene is reduced with LiAlH₄?
a) Aniline
b) Phenol
c) Benzene
d) Nitroethane
Answer: a) Aniline
Explanation: Reduction of nitrobenzene with lithium aluminum hydride (LiAlH₄) results in the formation of aniline (C₆H₅NH₂) by reducing the nitro group (-NO₂) to an amino group (-NH₂).
46. Which of the following compounds will not undergo electrophilic substitution?
a) Nitrobenzene
b) Aniline
c) Toluene
d) Benzene
Answer: a) Nitrobenzene
Explanation: Nitrobenzene is deactivated toward electrophilic substitution due to the strong electron-withdrawing effect of the nitro group, making the ring less reactive compared to benzene, toluene, or aniline.
47. The major product when phenylhydroxylamine is reduced is:
a) Aniline
b) Hydrazobenzene
c) Nitrobenzene
d) Azobenzene
Answer: a) Aniline
Explanation: Phenylhydroxylamine (C₆H₅NHOH) is reduced to aniline (C₆H₅NH₂) through catalytic hydrogenation or by treatment with reducing agents.
48. The reaction of aniline with benzoyl chloride in the presence of a base gives:
a) Benzoic acid
b) Benzamide
c) N-Phenylbenzamide
d) Phenylhydrazine
Answer: c) N-Phenylbenzamide
Explanation: Aniline reacts with benzoyl chloride (C₆H₅COCl) in the presence of a base (Schotten-Baumann reaction) to form N-phenylbenzamide (C₆H₅NHCOC₆H₅) by acylation of the amino group.
49. The reaction of a primary amine with carbon disulfide (CS₂) followed by hydrolysis produces:
a) Thiourea
b) Isocyanide
c) Ammonia
d) Dithiocarbamic acid
Answer: d) Dithiocarbamic acid
Explanation: Primary amines react with carbon disulfide to form dithiocarbamates, which on hydrolysis yield dithiocarbamic acid (RNH-CS₂H). This reaction is utilized in the synthesis of organosulfur compounds.
50. In the Gabriel phthalimide synthesis, the phthalimide is treated with:
a) HCl
b) NaOH
c) NaOH and alkyl halide
d) Ammonium hydroxide
Answer: c) NaOH and alkyl halide
Explanation: In the Gabriel phthalimide synthesis, phthalimide is deprotonated by NaOH, forming a phthalimide anion, which then reacts with an alkyl halide (R-X) to form a primary amine after hydrolysis.
51. What happens when nitrobenzene is reduced using Zn/NaOH?
a) Aniline is formed
b) Phenylhydroxylamine is formed
c) Benzylamine is formed
d) Nitrobenzene is converted into azobenzene
Answer: b) Phenylhydroxylamine is formed
Explanation: When nitrobenzene (C₆H₅NO₂) is reduced using zinc and sodium hydroxide (Zn/NaOH), phenylhydroxylamine (C₆H₅NHOH) is formed as the intermediate product.
52. Which of the following reagents can convert a nitro group into an amino group?
a) NaBH₄
b) H₂/Pd
c) KMnO₄
d) Na₂SO₄
Answer: b) H₂/Pd
Explanation: Hydrogen gas (H₂) in the presence of a palladium catalyst (Pd) can reduce a nitro group (-NO₂) to an amino group (-NH₂), converting nitrobenzene into aniline.
53. Which of the following is the correct order of basicity in the gas phase?
a) NH₃ > CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
b) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
c) (CH₃)₂NH > (CH₃)₃N > NH₃ > CH₃NH₂
d) NH₃ > (CH₃)₂NH > (CH₃)₃N > CH₃NH₂
Answer: b) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
Explanation: In the gas phase, the basicity of amines increases with the number of alkyl groups because alkyl groups donate electron density to the nitrogen atom, increasing its ability to accept a proton.
54. Aniline reacts with acetyl chloride to form:
a) Acetaldehyde
b) Acetamide
c) Acetanilide
d) Benzanilide
Answer: c) Acetanilide
Explanation: When aniline reacts with acetyl chloride (CH₃COCl), it undergoes acetylation to form acetanilide (C₆H₅NHCOCH₃). This reaction is commonly used to protect the amino group.
55. Which of the following is used in the nitration of benzene?
a) HCl and NaOH
b) HNO₃ and H₂SO₄
c) Br₂ and Fe
d) NaNO₂ and HCl
Answer: b) HNO₃ and H₂SO₄
Explanation: Nitration of benzene is carried out by treating it with a mixture of concentrated nitric acid (HNO₃) and sulfuric acid (H₂SO₄), which generates the nitronium ion (NO₂⁺) that acts as the electrophile.
56. Which of the following reactions is a nucleophilic substitution reaction?
a) Bromination of aniline
b) Diazotization of aniline
c) Conversion of benzyl chloride to benzylamine
d) Nitration of toluene
Answer: c) Conversion of benzyl chloride to benzylamine
Explanation: The conversion of benzyl chloride to benzylamine is a nucleophilic substitution reaction where the chloride ion (Cl⁻) is replaced by the amine group (-NH₂) through a nucleophilic attack.
57. Which of the following reactions does not involve an amine?
a) Gabriel phthalimide reaction
b) Sandmeyer reaction
c) Hoffmann rearrangement
d) Nitration of benzene
Answer: d) Nitration of benzene
Explanation: Nitration of benzene involves the introduction of a nitro group (-NO₂) onto a benzene ring using nitric acid and sulfuric acid, and it does not involve any amine compound.
58. Which reagent is used to convert aniline into iodobenzene?
a) I₂ and HNO₃
b) NaI and NaNO₂
c) KI and H₂SO₄
d) NaNO₂, HCl, and KI
Answer: d) NaNO₂, HCl, and KI
Explanation: Aniline is first diazotized with NaNO₂ and HCl to form the diazonium salt, which then reacts with potassium iodide (KI) to give iodobenzene (C₆H₅I).
59. Which of the following is most basic?
a) Aniline
b) p-Nitroaniline
c) m-Toluidine
d) Benzylamine
Answer: d) Benzylamine
Explanation: Benzylamine is more basic than aniline and its derivatives because the benzyl group (CH₂C₆H₅) does not significantly delocalize the nitrogen’s lone pair, making it more available for protonation compared to the aromatic amines.
60. What is the major product of the reaction of aniline with H₂SO₄ and HNO₃?
a) Nitrobenzene
b) p-Nitroaniline
c) o-Nitroaniline
d) m-Nitroaniline
Answer: b) p-Nitroaniline
Explanation: The nitration of aniline is controlled by the strong activating effect of the amino group (-NH₂), leading to substitution at the para position, resulting in p-nitroaniline as the major product.
61. When nitrobenzene is reduced with Sn/HCl, it gives:
a) Nitroethane
b) Benzene
c) Aniline
d) Benzylamine
Answer: c) Aniline
Explanation: Reduction of nitrobenzene (C₆H₅NO₂) with tin and hydrochloric acid (Sn/HCl) converts the nitro group (-NO₂) into an amino group (-NH₂), forming aniline.
62. Which of the following reagents is used in the Hofmann rearrangement?
a) Br₂ and NaOH
b) H₂SO₄ and NaNO₂
c) KMnO₄ and NaOH
d) NaOH and H₂O₂
Answer: a) Br₂ and NaOH
Explanation: The Hofmann rearrangement uses bromine (Br₂) and sodium hydroxide (NaOH) to convert amides into primary amines with the loss of one carbon atom.
63. Aniline undergoes diazotization with:
a) NaNO₂/HCl
b) NaOH/H₂O₂
c) Br₂/Fe
d) HNO₃/H₂SO₄
Answer: a) NaNO₂/HCl
Explanation: Aniline reacts with sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at low temperatures to form a diazonium salt, in a process known as diazotization.
64. Which product is obtained when aniline reacts with acetic anhydride?
a) Acetanilide
b) Aniline hydrochloride
c) p-Acetaniline
d) Aniline oxide
Answer: a) Acetanilide
Explanation: Aniline reacts with acetic anhydride to form acetanilide, a product of acetylation that introduces an acyl group (-COCH₃) onto the amino group.
65. Which of the following reagents is used for the reduction of nitrobenzene to aniline?
a) Br₂ in CCl₄
b) HNO₃ in H₂SO₄
c) Zn/NaOH
d) Sn/HCl
Answer: d) Sn/HCl
Explanation: Nitrobenzene is reduced to aniline using tin (Sn) and hydrochloric acid (HCl), which effectively reduce the nitro group (-NO₂) to an amino group (-NH₂).
66. Which of the following is the most basic compound?
a) Aniline
b) p-Nitroaniline
c) p-Methoxyaniline
d) Benzylamine
Answer: d) Benzylamine
Explanation: Benzylamine is more basic than aromatic amines like aniline or p-nitroaniline due to the lesser electron-withdrawing effect of the benzyl group compared to the nitro group in p-nitroaniline.
67. Which reagent is used to convert aniline to diazonium salt?
a) NaNO₂/HCl at 0-5°C
b) Br₂/Fe
c) KMnO₄/NaOH
d) H₂SO₄ and NaNO₃
Answer: a) NaNO₂/HCl at 0-5°C
Explanation: Aniline is converted into a diazonium salt using sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at low temperatures (0-5°C), generating a diazonium ion (N₂⁺).
68. Which of the following amines can undergo Hoffmann bromamide degradation?
a) Aniline
b) Benzamide
c) Diethylamine
d) Trimethylamine
Answer: b) Benzamide
Explanation: In Hoffmann bromamide degradation, primary amides such as benzamide are converted to primary amines with one fewer carbon atom using Br₂ and NaOH.
69. What is the product when aniline is reacted with chloroform and alcoholic KOH (Carbylamine reaction)?
a) Acetanilide
b) Phenyl isocyanide
c) Benzylamine
d) Methylamine
Answer: b) Phenyl isocyanide
Explanation: The Carbylamine reaction involves the reaction of an amine with chloroform (CHCl₃) and alcoholic KOH to produce an isocyanide (in this case, phenyl isocyanide, C₆H₅NC) with a characteristic foul odor.
70. When nitrobenzene is subjected to nitration, the major product is:
a) o-Dinitrobenzene
b) m-Dinitrobenzene
c) p-Dinitrobenzene
d) Nitroaniline
Answer: b) m-Dinitrobenzene
Explanation: Nitration of nitrobenzene leads to the formation of meta-dinitrobenzene because the nitro group is a strong electron-withdrawing group, deactivating the ortho and para positions for electrophilic substitution.
71. Which of the following is formed when aniline reacts with bromine water?
a) Mono-bromoaniline
b) Di-bromoaniline
c) 2,4,6-Tribromoaniline
d) 2,6-Dibromoaniline
Answer: c) 2,4,6-Tribromoaniline
Explanation: Aniline reacts with bromine water to form 2,4,6-tribromoaniline because the amino group (-NH₂) is strongly activating, directing bromine substitution to the ortho and para positions.
72. The reduction of nitrobenzene with Zn and NH₄Cl gives:
a) Azobenzene
b) Aniline
c) Hydrazobenzene
d) Phenylhydroxylamine
Answer: d) Phenylhydroxylamine
Explanation: Reduction of nitrobenzene using zinc and ammonium chloride (NH₄Cl) yields phenylhydroxylamine (C₆H₅NHOH) as the intermediate product.
73. Which of the following is used to detect primary amines?
a) Hinsberg’s reagent
b) Lucas reagent
c) Tollen’s reagent
d) Benedict’s reagent
Answer: a) Hinsberg’s reagent
Explanation: Hinsberg’s reagent (benzenesulfonyl chloride) is used to differentiate between primary, secondary, and tertiary amines. Primary amines form sulfonamides, which are soluble in alkali.
74. Which of the following compounds gives a foul smell of isocyanide when heated with chloroform and alcoholic KOH?
a) Ethylamine
b) Acetamide
c) Aniline
d) Benzylamine
Answer: c) Aniline
Explanation: Aniline (C₆H₅NH₂) reacts with chloroform and alcoholic KOH in the Carbylamine reaction to produce phenyl isocyanide (C₆H₅NC), which has a characteristic foul odor.
75. Which of the following reactions will produce a primary amine with one less carbon atom?
a) Sandmeyer reaction
b) Hoffmann bromamide reaction
c) Gabriel phthalimide synthesis
d) Acetylation of aniline
Answer: b) Hoffmann bromamide reaction
Explanation: The Hoffmann bromamide reaction converts an amide to a primary amine with one less carbon atom by using bromine (Br₂) and sodium hydroxide (NaOH).
76. Which of the following amines reacts with nitrous acid to form a yellow oily liquid?
a) Aniline
b) Methylamine
c) Diethylamine
d) Benzylamine
Answer: b) Methylamine
Explanation: Methylamine (a primary aliphatic amine) reacts with nitrous acid to form methyl diazonium salt, which decomposes to form nitrogen gas and a yellow oily liquid (methanol).
77. Which reagent is used to carry out the reduction of nitro compounds to amines?
a) Fe/HCl
b) NaOH
c) H₂O₂
d) KMnO₄
Answer: a) Fe/HCl
Explanation: Iron (Fe) and hydrochloric acid (HCl) are commonly used to reduce nitro compounds to amines, as in the reduction of nitrobenzene to aniline.
78. What is the major product when aniline reacts with acetic anhydride?
a) p-Nitroaniline
b) Acetanilide
c) Aniline hydrochloride
d) Benzanilide
Answer: b) Acetanilide
Explanation: When aniline reacts with acetic anhydride, it forms acetanilide as the product through acetylation of the amino group (-NH₂).
79. Which of the following is least reactive towards nucleophilic substitution?
a) Nitrobenzene
b) Benzyl chloride
c) 2,4-Dinitrochlorobenzene
d) Benzylamine
Answer: a) Nitrobenzene
Explanation: Nitrobenzene is least reactive towards nucleophilic substitution due to the strong electron-withdrawing effect of the nitro group (-NO₂), which deactivates the benzene ring for such reactions.
80. Which of the following reactions involves the formation of a diazonium salt?
a) Nitration of benzene
b) Acetylation of aniline
c) Diazotization of aniline
d) Bromination of aniline
Answer: c) Diazotization of aniline
Explanation: Diazotization of aniline involves the reaction of aniline with NaNO₂ and HCl at low temperatures (0-5°C) to form a diazonium salt (C₆H₅N₂⁺Cl⁻).
81. Which of the following amines is least basic?
a) Ethylamine
b) Aniline
c) Dimethylamine
d) Benzylamine
Answer: b) Aniline
Explanation: Aniline is less basic than aliphatic amines like ethylamine and dimethylamine because the lone pair of electrons on the nitrogen is delocalized into the aromatic ring, reducing its availability for protonation.
82. Which of the following compounds does not give a positive Hinsberg test?
a) Ethylamine
b) Diethylamine
c) Triethylamine
d) Aniline
Answer: c) Triethylamine
Explanation: Triethylamine, a tertiary amine, does not react with Hinsberg’s reagent and, therefore, does not give a positive Hinsberg test. The test is used to differentiate primary and secondary amines.
83. Which of the following products is obtained when nitrobenzene is reduced by lithium aluminium hydride (LiAlH₄)?
a) Azobenzene
b) Aniline
c) Benzylamine
d) Phenylhydroxylamine
Answer: b) Aniline
Explanation: Lithium aluminium hydride (LiAlH₄) is a strong reducing agent that reduces nitrobenzene to aniline by converting the nitro group (-NO₂) to an amino group (-NH₂).
84. Which of the following amines does not undergo diazotization?
a) Methylamine
b) Aniline
c) p-Toluidine
d) o-Nitroaniline
Answer: a) Methylamine
Explanation: Methylamine, a primary aliphatic amine, does not form a stable diazonium salt. Instead, it reacts with nitrous acid to release nitrogen gas, forming methanol as a byproduct.
85. The basicity of amines in aqueous solution follows which order?
a) NH₃ > CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
b) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
c) (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
d) CH₃NH₂ > NH₃ > (CH₃)₃N > (CH₃)₂NH
Answer: c) (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
Explanation: In aqueous solution, secondary amines like dimethylamine ((CH₃)₂NH) are the most basic due to better solvation and availability of the lone pair on nitrogen, followed by primary amines, tertiary amines, and ammonia.
86. Which of the following reagents is used to carry out the diazotization of aniline?
a) NaNO₂/HCl at 273 K
b) H₂SO₄/NaNO₃ at 300 K
c) NaOH/H₂O₂
d) Br₂ and FeCl₃
Answer: a) NaNO₂/HCl at 273 K
Explanation: The diazotization of aniline is carried out by treating it with sodium nitrite (NaNO₂) and hydrochloric acid (HCl) at low temperatures (273 K or 0-5°C) to form a diazonium salt.
87. Which of the following amines can be prepared by Gabriel phthalimide synthesis?
a) Methylamine
b) Ethylamine
c) Diethylamine
d) Trimethylamine
Answer: b) Ethylamine
Explanation: The Gabriel phthalimide synthesis is used to prepare primary amines such as ethylamine by treating phthalimide with an alkyl halide, followed by hydrolysis.
88. Which of the following products is obtained when nitrobenzene is reduced with hydrogen and nickel catalyst?
a) Azobenzene
b) Aniline
c) Phenylhydroxylamine
d) Benzylamine
Answer: b) Aniline
Explanation: Nitrobenzene is reduced to aniline using hydrogen (H₂) in the presence of a nickel catalyst (Ni). The nitro group (-NO₂) is fully reduced to an amino group (-NH₂).
89. Which of the following compounds will not react with nitrous acid at low temperatures to form a diazonium salt?
a) Aniline
b) Benzylamine
c) p-Nitroaniline
d) m-Toluidine
Answer: b) Benzylamine
Explanation: Benzylamine, a primary aliphatic amine, does not form a stable diazonium salt with nitrous acid. It decomposes to release nitrogen gas.
90. Which of the following amines undergoes the Sandmeyer reaction?
a) Diethylamine
b) Aniline
c) Benzylamine
d) Ethylamine
Answer: b) Aniline
Explanation: Aniline undergoes diazotization to form a diazonium salt, which can participate in the Sandmeyer reaction, where it reacts with CuCl or CuBr to form chlorobenzene or bromobenzene, respectively.
91. Which product is formed when acetamide is treated with bromine and sodium hydroxide?
a) Methylamine
b) Aniline
c) Ethylamine
d) Acetanilide
Answer: a) Methylamine
Explanation: The Hoffmann bromamide degradation reaction converts acetamide into methylamine when treated with bromine (Br₂) and sodium hydroxide (NaOH).
92. Which of the following is a correct product for the reaction of aniline with CH₃COCl?
a) Acetaldehyde
b) Acetanilide
c) Aniline acetate
d) Aniline hydrochloride
Answer: b) Acetanilide
Explanation: Aniline reacts with acetyl chloride (CH₃COCl) to form acetanilide, where the amino group (-NH₂) is acetylated to form -NHCOCH₃.
93. Which of the following statements is correct about aromatic amines?
a) They are less basic than aliphatic amines.
b) They are more basic than aliphatic amines.
c) They are strong acids.
d) They are neutral compounds.
Answer: a) They are less basic than aliphatic amines.
Explanation: Aromatic amines, such as aniline, are less basic than aliphatic amines because the lone pair on the nitrogen is delocalized into the aromatic ring, reducing its availability for protonation.
94. The diazonium salt of aniline is stable at:
a) Room temperature
b) Below 5°C
c) Above 50°C
d) 100°C
Answer: b) Below 5°C
Explanation: Diazonium salts are stable at low temperatures, typically below 5°C, because they tend to decompose at higher temperatures, releasing nitrogen gas.
95. Which of the following does not participate in nucleophilic substitution reactions?
a) Nitrobenzene
b) Benzyl chloride
c) Chlorobenzene
d) 2,4-Dinitrochlorobenzene
Answer: a) Nitrobenzene
Explanation: Nitrobenzene is highly deactivated towards nucleophilic substitution reactions due to the electron-withdrawing nature of the nitro group (-NO₂), which reduces the electron density on the aromatic ring.
96. Which of the following reactions converts an amide into an amine with the loss of one carbon atom?
a) Sandmeyer reaction
b) Gabriel synthesis
c) Hoffmann bromamide reaction
d) Diazotization
Answer: c) Hoffmann bromamide reaction
Explanation: The Hoffmann bromamide reaction converts an amide into a primary amine with one less carbon atom by using bromine (Br₂) and sodium hydroxide (NaOH).
97. What is the major product when nitrobenzene is reduced using Sn/HCl?
a) Phenylhydroxylamine
b) Benzylamine
c) Aniline
d) Phenylamine
Answer: c) Aniline
Explanation: When nitrobenzene is reduced using tin (Sn) and hydrochloric acid (HCl), it is converted into aniline (C₆H₅NH₂), where the nitro group (-NO₂) is reduced to an amino group (-NH₂).
98. Which of the following compounds gives a positive Carbylamine test?
a) Methylamine
b) Acetamide
c) Diethylamine
d) Trimethylamine
Answer: a) Methylamine
Explanation: Methylamine, a primary amine, gives a positive Carbylamine test by reacting with chloroform and alcoholic KOH to form methyl isocyanide (CH₃NC), which has a foul odor.
99. Which of the following is the correct decreasing order of basicity for the following amines in the gas phase?
a) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
b) NH₃ > CH₃NH₂ > (CH₃)₂NH > (CH₃)₃N
c) (CH₃)₂NH > (CH₃)₃N > CH₃NH₂ > NH₃
d) CH₃NH₂ > (CH₃)₃N > (CH₃)₂NH > NH₃
Answer: a) (CH₃)₃N > (CH₃)₂NH > CH₃NH₂ > NH₃
Explanation: In the gas phase, the basicity of amines increases with the number of alkyl groups due to increased electron-donating effects.
100. Which of the following compounds can be used to distinguish between primary, secondary, and tertiary amines?
a) Lucas reagent
b) Benedict’s reagent
c) Hinsberg’s reagent
d) Fehling’s solution
Answer: c) Hinsberg’s reagent
Explanation: Hinsberg’s reagent (benzenesulfonyl chloride) is used to distinguish between primary, secondary, and tertiary amines based on their reactivity and the solubility of the products formed in the reaction
Leave a comment