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.
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