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JEE • CHEMISTRY • AMINES

Amines practice · Easy

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 31-41 of 41 questions
31 Easy
31 of 41
Remembering +4 | -1
Which of the following is an ambidentate ligand?
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Correct option: B
nitrito
Option B (nitrito) is correct because nitrito (NO2−\mathrm{NO_2^-}) can coordinate through either the nitrogen or one of the oxygen atoms, making it an ambidentate ligand.
32 Easy
32 of 41
Understanding +4 | -1
Nitration of aniline is done in:
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Correct option: D
In acidic medium by first converting it into acetanilide before nitration
Option D (In acidic medium by first converting it into acetanilide before nitration) is correct because aniline is first converted to acetanilide to protect the amino group from oxidation and to control the reaction.
33 Easy
33 of 41
Remembering +4 | -1
Which of the following compounds gives positive carbylamine test?
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Correct option: C
Primary amine
Option C (Primary amine) is correct because primary amines react with chloroform and a base to form isocyanides, which have a foul smell and indicate a positive carbylamine test.
34 Easy
34 of 41
Application +4 | -1
Which one of the following compounds will dissolve in an alkali solution after it has undergone reaction with Hinsberg reagent?
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Correct option: A
CH3NH2\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}
Option A (CH3NH2\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}) is correct because primary amines react with Hinsberg reagent to form N-substituted sulfonamides, which are soluble in alkali.
35 Easy
35 of 41
Analyzing +4 | -1
From the following compounds which does not react with C6H5SO2CI{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{S}{\mathrm{O}_2}\mathrm{CI}?
View answer and explanation
Correct option: B
CH3.NH2\mathrm{C}{\mathrm{H}_3}.\mathrm{N}{\mathrm{H}_2}
Option D ((C2H5)3N) is correct because tertiary amines lack a hydrogen atom directly attached to the nitrogen, preventing them from reacting with benzenesulfonyl chloride.
36 Easy
36 of 41
Remembering +4 | -1
Aqueous solution of urea is:
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Correct option: C
Almost neutral
Option C (Almost neutral) is correct because an aqueous solution of urea is nearly neutral, although it can act as a weak monoacid-base.
37 Easy
37 of 41
Application +4 | -1
Major product, P\mathrm{P} of the reaction is

ArN2Cl+HBF4⟶Q⟶ΔP\mathrm{ArN}_2 \mathrm{Cl}+\mathrm{HBF}_4 \longrightarrow \mathrm{Q} \stackrel{\Delta}{\longrightarrow} \mathrm{P}
View answer and explanation
Correct option: A
Ar−F\mathrm{Ar}-\mathrm{F}
The correct answer is A (Ar-F) because the reaction sequence involves the formation of an aryl diazonium tetrafluoroborate salt, followed by thermal decomposition to yield an aryl fluoride.
38 Easy
38 of 41
Understanding +4 | -1
Aniline is weaker base than ethylamine because:
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Correct option: A
Lone pair of electrons of N-atom is not freely available for coordination with a proton due to resonance than in ethylamine
Option A is correct because the lone pair of electrons on the nitrogen atom in aniline is delocalized into the benzene ring through resonance, making it less available for protonation compared to ethylamine.
39 Easy
39 of 41
Remembering +4 | -1
Hinsberg’s reagent is
View answer and explanation
Correct option: C
C6H5SO2Cl{C_6}{H_5}S{O_2}Cl
Option C (C6H5SO2Cl{C_6}{H_5}S{O_2}Cl) is correct because Hinsberg’s reagent is benzene sulphonyl chloride, used to distinguish between primary, secondary, and tertiary amines.
40 Easy
40 of 41
Analyzing +4 | -1
Which one of the following is most basic?
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Correct option: D
C6H5CH2NH2{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{N}{\mathrm{H}_2}
Option D (C6H5CH2NH2{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{N}{\mathrm{H}_2}) is the most basic because the lone pair on the nitrogen is not delocalized into the benzene ring, making it more available for protonation.
41 Easy
41 of 41
Remembering +4 | -1
Schiff’s bases are formed when aniline is condensed with:
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Correct option: B
Aromatic aldehydes
Option B (Aromatic aldehydes) is correct because Schiff's bases are formed by the condensation reaction between primary amines (like aniline) and aldehydes (especially aromatic aldehydes).
Revision notes and key takeaways

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