Skip to main content
Back to AMINES

Your Progress

0 / 132 Answered
0
Correct
0
Incorrect
0%
Accuracy
JEE • CHEMISTRY • AMINES

Amines practice · Mixed difficulty

Choose an answer to check it, or open “View answer” to study the explanation. Sign in to save progress to your account.

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.

Take a chapter quiz →
Showing 61-90 of 132 questions

Preparing practice and saved progress…

61 Easy
61 of 132
Understanding +4 | -1
Which one of the following methods is neither meant for the synthesis nor for separation of amines?
View answer and explanation
Correct option: B
Wurtz reaction
Option B (Wurtz reaction) is correct because the Wurtz reaction is used for the synthesis of alkanes, not amines, nor for their separation.
62 Medium
62 of 132
Remembering +4 | -1
In sodium fusion test of organic compounds, the nitrogen of the organic compound is converted into
View answer and explanation
Correct option: B
Sodium cyanide
Option B (Sodium cyanide) is correct because in the sodium fusion test, nitrogen from the organic compound reacts with sodium to form sodium cyanide.
63 Medium
63 of 132
Understanding +4 | -1
Ethylamine is heated with CS2{CS}_2 in the presence of HgCl2{HgCl}_2. The product formed is
View answer and explanation
Correct option: D
Ethyl isothiocyanate
Option D (Ethyl isothiocyanate) is correct because the reaction of ethylamine with CS2CS_2 in the presence of HgCl2HgCl_2 results in the formation of ethyl isothiocyanate, known as the Hoffmann mustard oil reaction.
64 Medium
64 of 132
Understanding +4 | -1
Which of the following groups does not decrease the basic strength of aniline?
View answer and explanation
Correct option: A
OCH3\mathrm{OCH}_3
Option A (OCH3\mathrm{OCH}_3) is correct because electron-releasing groups like methoxy (OCH3\mathrm{OCH}_3) increase the electron density on the nitrogen atom of aniline, thus increasing its basic strength.
65 Medium
65 of 132
Analyzing +4 | -1
Which of the following will produce isopropyl amine-

(I)(CH3)2CO→NH2OHX→LiAlH4({{I) (C}}{{{H}}_3}{)_2}CO\xrightarrow{{N{H_2}OH}}X\xrightarrow{{LiAl{H_4}}}

(II)CH3−CH2−CHO→heatNH3X→LiAlH4({{II) C}}{{{H}}_3} - C{H_2} - CHO\xrightarrow[{heat}]{{N{H_3}}}X\xrightarrow{{LiAl{H_4}}}

(III)(CH3)2CH−OH+PCl5→X→NH3({{III) (C}}{{{H}}_3}{)_2}CH - OH + PC{l_5} \to X\xrightarrow{{N{H_3}}}

(IV)CH3−CH2−CH2−NH2→heat({{IV) C}}{{{H}}_3} - C{H_2} - C{H_2} - N{H_2}\xrightarrow{{heat}}
View answer and explanation
Correct option: C
I, III
Option C (I, III) is correct because both reactions produce isopropyl amine ((CH3)2CHNH2(CH_3)_2CHNH_2). Reaction I involves the reduction of an oxime derived from acetone, and Reaction III involves the reaction of isopropyl chloride with ammonia.
66 Easy
66 of 132
Analyzing +4 | -1
In the reaction of (S) 2-phenylpropamide with NaBr/H2O\mathrm{NaBr}/{\mathrm{H}_2}\mathrm{O} to give 1-phenylethylamine
View answer and explanation
Correct option: A
There is retention of configuration
Option B is correct because benzoic acid is more acidic than acetic acid due to resonance stabilization of the benzoate ion, and both are more acidic than phenol.
67 Easy
67 of 132
Analyzing +4 | -1
The basicity of the compounds in aqueous medium increases in the order
CH3NH2I,  (CH3)2NHII,  C6H5NH2III,  (CH3)3NIV\mathop {C{H_3}N{H_2}}\limits_{\text{I}} ,\,\,\mathop {{{\left( {C{H_3}} \right)}_2}NH}\limits_{{\text{II}}} ,\,\,\mathop {{C_6}{H_5}N{H_2}}\limits_{{\text{III}}} ,\,\,\mathop {{{\left( {C{H_3}} \right)}_3}N}\limits_{{\text{IV}}}
View answer and explanation
Correct option: B
III << IV << I << II
The correct answer is B (III < IV < I < II) because basicity in aqueous medium depends on the inductive effect and solvation effects of the alkyl groups, with steric hindrance also playing a role.
68 Hard
68 of 132
Remembering +4 | -1
Carcinogens are the products of the reaction between:
View answer and explanation
Correct option: A
R2NH+HNO2{R_2}\mathrm{NH} + \mathrm{HN}{\mathrm{O}_2}
Option A (R2NH+HNO2{R_2}\mathrm{NH} + \mathrm{HN}{\mathrm{O}_2}) is correct because secondary amines react with nitrous acid to form nitrosoamines, which are known carcinogens.
69 Medium
69 of 132
Understanding +4 | -1
Name the product(s) formed during the reaction of primary aliphatic amines with nitrous acid at room temperature?
View answer and explanation
Correct option: C
Both (1) and (2)
Option C (Both (1) and (2)) is correct because primary aliphatic amines react with nitrous acid to form diazonium salts, which are unstable and decompose to form alcohols and nitrogen gas.
70 Hard
70 of 132
Analyzing +4 | -1
Assertion:- Acetamide on reaction with KOH\mathrm{KOH} and bromine gives acetic acid.
Reason:- Bromine catalyses hydrolysis of acetamide,
View answer and explanation
Correct option: D
If both Assertion &amp; reason are False
Option D (If both Assertion & reason are False) is correct because acetamide reacts with KOH and bromine to give methylamine (Hoffmann bromamide degradation), not acetic acid, and bromine does not catalyze the hydrolysis of acetamide.
71 Medium
71 of 132
Remembering +4 | -1
Ceric ammonium nitrate and CHCl3/alc. KOHCHC{l_3}/alc.\,KOH are used for the identification of functional groups present in ….. and …… respectively.
View answer and explanation
Correct option: C
alcohol, amine
Option C (alcohol, amine) is correct because ceric ammonium nitrate is used to identify alcohols, while CHCl3/alc. KOHCHC{l_3}/alc.\,KOH is used to identify amines.
72 Easy
72 of 132
Analyzing +4 | -1
The weakest base among the following is
View answer and explanation
Correct option: B
aniline
Option B (aniline) is correct because the lone pair of electrons on the nitrogen atom in aniline is delocalized into the benzene ring, making it less available for protonation and thus a weaker base.
73 Medium
73 of 132
Understanding +4 | -1
Hinsber’s reagent does not react with
View answer and explanation
Correct option: C
3∘−Amine3^\circ - {\text{Amine}}
Option C is correct because Hinsberg's reagent reacts with primary and secondary amines, but not with tertiary amines due to the absence of an acidic hydrogen atom on the nitrogen.
74 Medium
74 of 132
Remembering +4 | -1
The test that is done for the differentiation of primary amines from secondary and tertiary amine is:
View answer and explanation
Correct option: D
Carbylamine test
Option D (Carbylamine test) is correct because the carbylamine test is specifically used to detect the presence of primary amines, differentiating them from secondary and tertiary amines.
75 Medium
75 of 132
Remembering +4 | -1
An aliphatic amine on treatment with alcoholic carbon disulphide and mercuric chloride forms isothiocyanate, the reaction is known as
View answer and explanation
Correct option: C
Hofmann&#39;s mustard oil reaction
Option C (Hofmann's mustard oil reaction) is correct because this reaction involves the conversion of an aliphatic amine to an isothiocyanate using alcoholic carbon disulfide and mercuric chloride.
76 Medium
76 of 132
Remembering +4 | -1
When aniline reacts with oil of bitter almonds (C6H5CHO)\left({C}_6 {H}_5 {CHO}\right) condensation takes place and benzal derivative is formed. This is known as
View answer and explanation
Correct option: C
Schiff's base
Option C (Schiff's base) is correct because the condensation product of an aldehyde (like benzaldehyde) and a primary amine (like aniline) is called a Schiff's base.
77 Easy
77 of 132
Application +4 | -1
A colourless, odourless and non-combustible gas is liberated when ethylamine reacts with:
View answer and explanation
Correct option: C
NaNO2+HCl\mathrm{NaN}{\mathrm{O}_2} + \mathrm{HCl}
Option C (NaNO2+HCl\mathrm{NaN}{\mathrm{O}_2} + \mathrm{HCl}) is correct because ethylamine reacts with nitrous acid (generated from NaNO2+HClNaNO_2 + HCl) to produce ethanol and nitrogen gas, which is colorless, odorless, and non-combustible.
78 Medium
78 of 132
Understanding +4 | -1
Why are amines less acidic than alcohols of comparable molecular masses?
View answer and explanation
Correct option: C
Because N−H\mathrm{N}-\mathrm{H} bond is less polar than O−H\mathrm{O}-\mathrm{H} bond
Option C is correct because the O-H bond is more polar than the N-H bond due to the higher electronegativity of oxygen compared to nitrogen, making alcohols more acidic.
79 Medium
79 of 132
Analyzing +4 | -1
C3H9N{C_3}{H_9}N cannot represent:
View answer and explanation
Correct option: D
Quaternary salt
Option D (Quaternary salt) is correct because a quaternary ammonium salt requires four alkyl or aryl groups attached to the nitrogen atom, which cannot be achieved with the given formula C3H9N.
80 Easy
80 of 132
Remembering +4 | -1
For carbylamine reaction. We need hot alcoholic KOHKOH and
View answer and explanation
Correct option: A
Any primary amine & chloroform
Option A (Any primary amine & chloroform) is correct because the carbylamine reaction involves the reaction of a primary amine, chloroform, and a base like hot alcoholic KOH.
81 Easy
81 of 132
Remembering +4 | -1
Which of the following is an ambidentate ligand?
View answer and explanation
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.
82 Medium
82 of 132
Analyzing +4 | -1
Acetamide and ethylamine can be distinguished by reacting with
View answer and explanation
Correct option: B
Aqueous NaOH{NaOH} and heat
Option B (Aqueous NaOH{NaOH} and heat) is correct because acetamide will produce ammonia gas upon heating with NaOH, while ethylamine will not react.
83 Medium
83 of 132
Understanding +4 | -1
What would be the side products formed with primary amine in the Hofmann bromamide degradation reaction?
View answer and explanation
Correct option: C
NaBr+H2O+Na2CO3\mathrm{NaBr}+\mathrm{H}_2 \mathrm{O}+\mathrm{Na}_2 \mathrm{CO}_3
Option C (NaBr+H2O+Na2CO3\mathrm{NaBr}+\mathrm{H}_2 \mathrm{O}+\mathrm{Na}_2 \mathrm{CO}_3) is correct because the Hofmann bromamide degradation reaction produces a primary amine along with sodium bromide, water, and sodium carbonate as side products.
84 Medium
84 of 132
Analyzing +4 | -1
A gaseous carbon compound is soluable in dilute HCI\mathrm{HCI}. The solution on treating with NaNO2\mathrm{NaN}{\mathrm{O}_2} gives off nitrogen leaving behind a solution which smells of wood spirit. The carbon compound is
View answer and explanation
Correct option: D
CH3NH2\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}
Option D (CH3NH2\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}) is correct because methylamine dissolves in dilute HCl, and upon treatment with NaNO2, it releases nitrogen and forms methyl alcohol, which has a wood spirit smell.
85 Easy
85 of 132
Understanding +4 | -1
Nitration of aniline is done in:
View answer and explanation
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.
86 Medium
86 of 132
Analyzing +4 | -1
Among the following, the strongest base is:
View answer and explanation
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 strongest base because aliphatic amines are more basic than aromatic amines due to resonance stabilization in the latter.
87 Medium
87 of 132
Remembering +4 | -1
Nitrobenzene on reduction with Al−HgAl - Hg and water gives:
View answer and explanation
Correct option: D
Phenylhydroxylamine
The correct answer is Phenylhydroxylamine because nitrobenzene undergoes reduction with Al−HgAl-Hg and water to form phenylhydroxylamine.
88 Hard
88 of 132
Understanding +4 | -1
Assertion: Hoffmann's ammonolysis can be used to prepare pure primary amines.

Reason: Ammonolysis of haloalkanes lead to multiple substitution of alkyl groups on nitrogen.
View answer and explanation
Correct option: D
Assertion is wrong, but reason is correct statement.
The correct answer is D (Assertion is wrong, but reason is correct statement) because Hoffmann's ammonolysis does not yield pure primary amines due to multiple alkylation.
89 Easy
89 of 132
Remembering +4 | -1
Which of the following compounds gives positive carbylamine test?
View answer and explanation
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.
90 Medium
90 of 132
Understanding +4 | -1
Given below are two statements:

Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction.

Statement II: Aniline cannot be prepared through Gabriel synthesis.

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: D
Both statement I and Statement II are true
Option D (Both statement I and Statement II are true) is correct because aniline does not undergo Friedel-Crafts alkylation and cannot be prepared via Gabriel synthesis.
Revision notes and key takeaways

Review AMINES in your textbook or class notes, then use this set to check your understanding.

Showing 61-90 of 132 questions

Your practice

0 / 132 answered · 132 remaining · 0% accuracy

Progress is stored on this device. Sign in to save to your account.

✓ Correct · ✗ Incorrect · ○ Unanswered

Practice settings

Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.