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JEE • CHEMISTRY • AMINES

Amines practice · Medium

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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 1-30 of 82 questions
1 Medium
1 of 82
Application +4 | -1
Identify 'Z\mathrm{Z}' in the sequence

C6H5NH2→NaNO2+HClX→CuCNY→H+/H2OZ{{\text{C}}_6}{{\text{H}}_5}{\text{N}}{{\text{H}}_2}\xrightarrow{{{\text{NaN}}{{\text{O}}_2} + {\text{HCl}}}}{\text{X}}\xrightarrow{{{\text{CuCN}}}}{\text{Y}}\xrightarrow{{{{\text{H}}^ + }/{{\text{H}}_2}{\text{O}}}}{\text{Z}}
View answer and explanation
Correct option: C
C6H5COOH\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}
Option C (C6H5COOH\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}) is correct because the sequence of reactions converts aniline to benzoic acid via diazotization, cyanation, and hydrolysis.
2 Medium
2 of 82
Understanding +4 | -1
The primary, secondary and tertiary amines can be distinguished by:
View answer and explanation
Correct option: A
Hinsberg's reagent
Option A (Hinsberg's reagent) is correct because primary, secondary, and tertiary amines react differently with Hinsberg's reagent, allowing for their distinction based on the solubility of the products.
3 Medium
3 of 82
Remembering +4 | -1
Which of the following name reaction used to form Amines through amides?
View answer and explanation
Correct option: C
Hoffmannbromamide
Option C (Hoffmannbromamide) is correct because the Hoffmann bromamide reaction converts amides to amines with one less carbon atom.
4 Medium
4 of 82
Application +4 | -1
The IUPAC name of, CH3−CH2−C∣HCH3−NH2C{H_3} - C{H_2} - \mathop {\mathop C\limits_| H}\limits_{C{H_3}} - N{H_2} is
View answer and explanation
Correct option: B
2 – aminobutane
Option B (2 – aminobutane) is correct because it follows the IUPAC naming rules for organic compounds, correctly identifying the longest carbon chain and the position of the amine group.
5 Medium
5 of 82
Remembering +4 | -1
What is the name of reactive intermediate formed in Hoffmann bromamide degradation?
View answer and explanation
Correct option: C
Isocyanate
Option C (Isocyanate) is correct because the reactive intermediate formed in the Hoffmann bromamide degradation is an isocyanate.
6 Medium
6 of 82
Understanding +4 | -1
IUPAC name of the compound CHO⋅NHCH3\mathrm{CHO} \cdot \mathrm{NHCH}_3 is :
View answer and explanation
Correct option: C
N-methyl methanamide
Option C (N-methyl methanamide) is correct because it accurately represents the IUPAC nomenclature for the given compound, considering the amide functional group and N-methyl substituent.
7 Medium
7 of 82
Remembering +4 | -1
Isocyanide is prepared by
View answer and explanation
Correct option: B
Carbylamine reaction
Option B (Carbylamine reaction) is correct because isocyanides are prepared by the carbylamine reaction, which involves the reaction of a primary amine with chloroform in the presence of a base.
8 Medium
8 of 82
Analyzing +4 | -1
Statement-I: The presence of amino group can be detected by the azo dye test.

Statement-II: Amino group of a 2∘2^{\circ} amine is diazotized with HNO2\mathrm{HNO}_2 and NaCl\mathrm{NaCl} at 0∘C0^{\circ} \mathrm{C} which gives a red coloured dye with p-naphthol.
View answer and explanation
Correct option: C
Statement I is correct but statement II is incorrect
Option C is correct. Statement I is correct because the azo dye test detects amino groups. Statement II is incorrect because the azo dye test is primarily for 1° aromatic amines, not 2° amines.
9 Medium
9 of 82
Analyzing +4 | -1
Assertion: Alkyl isocyanides in acidified water give alkyl formamides.

Reason: In isocyanides, carbon first acts as a nucleophile and then as an electrophile.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A is correct because the assertion that alkyl isocyanides give alkyl formamides in acidified water is true, and the reason that carbon in isocyanides acts as a nucleophile and then as an electrophile correctly explains this reaction mechanism.
10 Medium
10 of 82
Analyzing +4 | -1
Arrange the following compounds in decreasing order of basicity
I. ethylamine \quad II. 2-amino ethanol \quad III. 3-amino-1-propanol
View answer and explanation
Correct option: C
I > III > II
Option C (I > III > II) is correct because the basicity of amines decreases with the presence of electron-withdrawing groups (-OH) and with increasing proximity to the amino group.
11 Medium
11 of 82
Analyzing +4 | -1
Assertion: Anilinium chloride is more acidic than ammonium chloride.

Reason: Anilinium ion is resonance-stabilised.
View answer and explanation
Correct option: C
If Assertion is true but reason is false
Option C is correct because the assertion is true, anilinium chloride is more acidic than ammonium chloride, but the reason is false; the anilinium ion is not resonance-stabilized in a way that increases acidity.
12 Medium
12 of 82
Analyzing +4 | -1
Assertion: Basicity of CH3CH2NH2(I),NH3(II){CH}_3 {CH}_2 {NH}_2(\text{I}), {NH}_3 \text{(II)} and C6H5NH2(III){C}_6 {H}_5 {NH}_2 \text{(III)} is in order I >> II >> III.

Reason: Electron-donating groups (such as alkyl group) increase the basicity of amines, and electron-withdrawing groups (such as aryl group) decrease the basicity of amines.
View answer and explanation
Correct option: A
If both assertion and reason are true and reason is the correct explanation of assertion
Option A (If both assertion and reason are true and reason is the correct explanation of assertion) is correct because the basicity of amines is indeed affected by electron-donating and electron-withdrawing groups as described.
13 Medium
13 of 82
Remembering +4 | -1
Diazo-coupling is useful to prepare:
View answer and explanation
Correct option: B
Dyes
Option B (Dyes) is correct because diazo-coupling reactions are widely used in the synthesis of azo dyes.
14 Medium
14 of 82
Analyzing +4 | -1
Which one will have lowest value of pKb\mathrm{pK}_{\mathrm{b}} in aqueous phase?
View answer and explanation
Correct option: B
N-Methylmethanamine
Option B (N-Methylmethanamine) is correct because secondary amines are generally more basic than primary or tertiary amines in the aqueous phase due to solvation effects.
15 Medium
15 of 82
Analyzing +4 | -1
On hydrolysis of a "compound", two compounds are obtained. One of which on treatment with sodium nitrite and hydrochloric acid gives a product which does not respond to iodoform test. The second one reduces Tollen's reagent and Fehling's solution. The "compound" is
View answer and explanation
Correct option: B
CH3CH2CH2NC{CH}_3 {CH}_2 {CH}_2 {NC}
Option B (CH3CH2CH2NC) is correct because the hydrolysis of propyl isocyanide yields propylamine and formic acid. Propylamine, upon treatment with NaNO2 and HCl, gives propanol, which does not respond to the iodoform test. Formic acid reduces Tollen's reagent and Fehling's solution.

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16 Medium
16 of 82
Understanding +4 | -1
Statement-1 : Nitrating mixture used for carrying out nitration of benzene consists of conc. HNO3+{HNO}_3+ conc. H2SO4{H}_2 {SO}_4.

Statement-2 : In presence of H2SO4,HNO3{H}_2 {SO}_4, {HNO}_3 acts as a base and produces NO2+{NO}_2^{+}ions.
View answer and explanation
Correct option: A
Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement -1
Option A (Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for Statement -1) is correct because the nitrating mixture consists of concentrated nitric acid and concentrated sulfuric acid, where sulfuric acid acts as a catalyst and helps generate the nitronium ion (NO2+{NO}_2^{+}).
17 Medium
17 of 82
Application +4 | -1
The major product PP in the following reaction is
CH3−CH2NH2→0−5oCNaNo2,  HClA→H2OP(Major)C{H_3} - C{H_2}N{H_2}\xrightarrow[{0 - {5^\mathrm{o}}C}]{{NaN{o_2},\,\,HCl}}A\xrightarrow{{{H_2}O}}P(\mathrm{Major})
View answer and explanation
Correct option: A
CH3−CH2−OHC{H_3} - C{H_2} - OH
Option A (CH3−CH2−OHC{H_3} - C{H_2} - OH) is correct because the reaction of ethylamine with NaNO2/HCl followed by hydrolysis yields ethanol.
18 Medium
18 of 82
Understanding +4 | -1
Benzene diazonium chloride reacts with hypophosphor ous acid to produce
View answer and explanation
Correct option: B
Benzene
Option B (Benzene) is correct because benzene diazonium chloride reacts with hypophosphorous acid (H3PO2H_3PO_2) to yield benzene, nitrogen gas, hydrochloric acid, and phosphorous acid.
19 Medium
19 of 82
Analyzing +4 | -1
Select corerct statement(s).
View answer and explanation
Correct option: D
All of the above
Option D (All of the above) is correct because all three statements regarding cyanamide ion, magnesium carbide, and calcium carbide are accurate.
20 Medium
20 of 82
Analyzing +4 | -1
p−p -chloro aniline and anilinium hydrogen chloride can be distinguished by
View answer and explanation
Correct option: D
AgNO3\mathrm{AgN}{\mathrm{O}_3}
Option D (AgNO3\mathrm{AgN}{\mathrm{O}_3}) is correct because anilinium hydrogen chloride reacts with AgNO3\mathrm{AgN}{\mathrm{O}_3} to give a white precipitate of AgCl, while p-chloroaniline does not.
21 Medium
21 of 82
Remembering +4 | -1
The reaction, ArN2+Cl−→Cu/HClArCl+N2+CuCl\mathrm{ArN}_2^{+} \mathrm{Cl}^{-} \xrightarrow{\mathrm{Cu} / \mathrm{HCl}} \mathrm{ArCl}+\mathrm{N}_2+\mathrm{CuCl} is named as
View answer and explanation
Correct option: B
Gattermann reaction
Option B (Gattermann reaction) is correct because the reaction of arenediazonium chloride with Cu/HCl to form aryl chloride is known as the Gattermann reaction.
22 Medium
22 of 82
Analyzing +4 | -1
Given below are two statements:

Statement I: Aniline reacts with con. H2SO4\mathrm{H}_2 \mathrm{SO}_4, followed by heating at 453−473 K453-473 \mathrm{~K} gives p−\mathrm{p}- aminobenzene sulphonic acid, which gives blood red colour in the 'Lassaigne's test'.
Statement II: In Friedel-Craft's alkylation and acylation reactions, aniline forms salt with the AlCl3\mathrm{AlCl}_3 catalyst. Due to this, nitrogen of aniline acquires a positive charge and acts as deactivating group.

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: C
Both statement I and statement II are true
The correct answer is C (Both statement I and statement II are true) because aniline does form p-aminobenzene sulfonic acid and gives a positive Lassaigne's test, and it also forms a salt with AlCl3, deactivating the ring.
23 Medium
23 of 82
Understanding +4 | -1
Statement I: In Sandmeyer reaction nucleophiles like Cl−,Br−{Cl}^{-}, {Br}^{-}and CN−{CN}^{-}are introduced in benzene ring in the presence of Cu+{Cu}^{+} ion.

Statement-II : In Gattermann rection nucleophiles are introduced in benzene ring in the presence of copper powder and HCl.HCl.
View answer and explanation
Correct option: A
Both statement I and II are correct
Both statement I and II are correct. Sandmeyer reaction uses Cu+ ions to introduce nucleophiles, and Gattermann reaction uses copper powder and HCl to introduce nucleophiles into the benzene ring.
24 Medium
24 of 82
Analyzing +4 | -1
C2H5NH2→HNO2A→PCI3B→NH3C{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{N}{\mathrm{H}_2}\mathop \to \limits^{\mathrm{HN}{\mathrm{O}_2}} A\mathop \to \limits^{\mathrm{PC}{\mathrm{I}_3}} B\mathop \to \limits^{\mathrm{N}{\mathrm{H}_3}} C
Recognize the compound C from the following
View answer and explanation
Correct option: C
Ethylamine
Option C (Ethylamine) is correct because the reaction sequence involves conversion of ethylamine to ethanol, then to ethyl chloride, and finally back to ethylamine.
25 Medium
25 of 82
Analyzing +4 | -1
Arrange the following in increasing order of their basic strength?

p-nitroaniline (1); m-nitroaniline (2); 2,6-trimethylaniline(3); 3-methylanline(4).
View answer and explanation
Correct option: D
1,2,4,31,2,4,3
Option D (1,2,4,31,2,4,3) is correct because electron-donating groups increase basicity and electron-withdrawing groups decrease it.
26 Medium
26 of 82
Understanding +4 | -1
When aniline is heated with benzaldehyde, the product is:
View answer and explanation
Correct option: B
Schiff’s base
Option B (Schiff’s base) is correct because aniline reacts with benzaldehyde to form a Schiff's base.
27 Medium
27 of 82
Remembering +4 | -1
Urotropine has the composition:
View answer and explanation
Correct option: C
(CH2)6N4{\left( {\mathrm{C}{\mathrm{H}_2}} \right)_6}{\mathrm{N}_4}
Option A ((CH2)6N4{\left( {\mathrm{C}{\mathrm{H}_2}} \right)_6}{\mathrm{N}_4}) is correct because the chemical formula for urotropine (hexamethylene tetramine) is (CH2)6N4{\left( {\mathrm{C}{\mathrm{H}_2}} \right)_6}{\mathrm{N}_4}.
28 Medium
28 of 82
Analyzing +4 | -1
Assertion: Tert Butyl amine can be formed by Gabriel phthalimide synthesis

Reason: It follow SN1S_N1 mechanism
View answer and explanation
Correct option: D
If both the Assertion and Reason are incorrect.
Option D is correct. The assertion is incorrect because tert-butyl amine cannot be synthesized by the Gabriel phthalimide synthesis due to steric hindrance. The reason is also incorrect because the reaction does not follow an SN1 mechanism.
29 Medium
29 of 82
Understanding +4 | -1
Which one of the following does not react with Heinsberg reagent.
View answer and explanation
Correct option: A
R3N\text{R}_3\text{N}
Option A (R3N\text{R}_3\text{N}) is correct because tertiary amines do not have a replaceable hydrogen atom on the nitrogen and thus cannot react with Hinsberg's reagent.
30 Medium
30 of 82
Remembering +4 | -1
The discovery that shocked the vital force theory was:
View answer and explanation
Correct option: C
Wӧhler’s synthesis of urea from NH4CNO\mathrm{N}{\mathrm{H}_4}\mathrm{CNO}
Option C (Wöhler’s synthesis of urea from NH4CNO\mathrm{N}{\mathrm{H}_4}\mathrm{CNO}) is correct because it disproved the vital force theory by synthesizing an organic compound from inorganic materials.
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