Skip to main content
Back to AMINES

Your Progress

0 / 104 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • 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

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 31-60 of 104 questions

Preparing practice and saved progress…

31 Medium
31 of 104
Application +4 | -1
CH3CH2Cl→NaCNA→Ni/H2B→anhydrldeacetic−C, C isC{H_3}C{H_2}Cl\xrightarrow{{NaCN}}A\xrightarrow{{Ni/{H_2}}}B\xrightarrow[{\mathrm{anhydrlde}}]{{\mathrm{acetic} - }}C,\,C\,\mathrm{is}
View answer and explanation
Correct option: C
CH3CH2CH2 NH COCH3C{H_3}C{H_2}C{H_2}\,NH\,COC{H_3}
Option C (CH3CH2CH2 NH COCH3C{H_3}C{H_2}C{H_2}\,NH\,COC{H_3}) is correct because the reaction sequence involves cyanide substitution, reduction to an amine, and then acylation.
32 Hard
32 of 104
Application +4 | -1
1,4 – dichloro butane →C2H5OHNaCNA→H2/NiB,  B  is\xrightarrow[{{C_2}{H_5}OH}]{{NaCN}}A\xrightarrow[{}]{{{H_2}/Ni}}B,\,\,{\text{B}}\,\,{\text{is}}
View answer and explanation
Correct option: A
Hexan -1,6 – diamine
Option A (Hexan -1,6 – diamine) is correct because the reaction sequence involves converting 1,4-dichlorobutane to hexanedinitrile, then reducing it to hexan-1,6-diamine.
33 Hard
33 of 104
Remembering +4 | -1
Gabriel pthalimide reaction is used in the synthesis of
View answer and explanation
Correct option: C
Primary aliphatic amines
Option C (Primary aliphatic amines) is correct because the Gabriel phthalimide reaction is specifically designed for the synthesis of primary aliphatic amines.
34 Hard
34 of 104
Analyzing +4 | -1
An aromatic amine (A)\left( A \right) was treated with alcoholic potash and another compound (Y)\left( Y \right) when a foul smeling gas was formed with formula C6H5NC{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{NC}. (Y)\left( Y \right) was formed by reacting a compound (Z)\left( Z \right) with Cl2\mathrm{C}{\mathrm{l}_2} in the presence of slaked lime. Compound (Z)\left( Z \right) is:
View answer and explanation
Correct option: C
CH3COCH3\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}
Option C (CH3COCH3\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}) is correct because the reaction sequence describes the haloform reaction, where acetone reacts with chlorine in the presence of a base to form chloroform.
35 Easy
35 of 104
Understanding +4 | -1
Aniline, chloroform and alc. KOH\mathrm{KOH} on heating give:
View answer and explanation
Correct option: A
Phenyl isocyanide
Option A (Phenyl isocyanide) is correct because the reaction of aniline with chloroform and alcoholic KOH is the carbylamine reaction, which produces phenyl isocyanide.
36 Easy
36 of 104
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}
Option A (Ar-F) is correct because the reaction sequence represents the Balz-Schiemann reaction, which converts an aryl diazonium salt to an aryl fluoride.
37 Medium
37 of 104
Analyzing +4 | -1
Which one does not liberate NH3\mathrm{N}{\mathrm{H}_3} when undergoes hydrolysis?
View answer and explanation
Correct option: D
Phenyl isocyanide
Option D (Phenyl isocyanide) is correct because isocyanides do not liberate ammonia upon hydrolysis; they form amines and formic acid.
38 Easy
38 of 104
Analyzing +4 | -1
The value of Kb{K_b} is highest in case of:
View answer and explanation
Correct option: A
p−p -methoxy aniline
Option A (p−p -methoxy aniline) is correct because the methoxy group is an electron-donating group, which increases the electron density on the nitrogen atom, making it more basic and thus increasing the KbK_b value.
39 Medium
39 of 104
Understanding +4 | -1
The compound which on reaction with cold HNO2\mathrm{HN}{\mathrm{O}_2} gives only nitrosoamine is:
View answer and explanation
Correct option: B
(CH3)2NH{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{NH}
Option B ((CH3)2NH{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{NH}) is correct because secondary amines react with cold nitrous acid to form nitrosoamines.
40 Medium
40 of 104
Application +4 | -1
Consider the following reaction, C6H5NO2→Sn/HCIX→C6H5COCIY+HCI{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{N}{\mathrm{O}_2}\mathop \to \limits^{\mathrm{Sn}/\mathrm{HCI}} X\mathop \to \limits^{{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{COCI}} Y + \mathrm{HCI} What is YY?
View answer and explanation
Correct option: B
Benzanilide
Option B (Benzanilide) is correct because the reaction sequence involves reducing nitrobenzene to aniline, which then reacts with benzoyl chloride to form benzanilide.
41 Easy
41 of 104
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 (NaNO2+HCl\mathrm{NaN}{\mathrm{O}_2} + \mathrm{HCl}) to produce ethanol and nitrogen gas, which is colorless, odorless, and non-combustible.
42 Hard
42 of 104
Understanding +4 | -1
Choose the incorrect option from the following.
View answer and explanation
Correct option: C
Aryl diazonium salts are less stable than alkyl diazonium salts
Option D (Aryl diazonium salts are less stable than alkyl diazonium salts) is the incorrect statement because aryl diazonium salts are more stable than alkyl diazonium salts due to resonance.
43 Medium
43 of 104
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 correct because the benzyl group reduces the electron-withdrawing effect of the phenyl group on the nitrogen, making it more basic than aniline and its nitro-substituted derivatives.
44 Medium
44 of 104
Application +4 | -1
Which of the following reagents will be fruitful for separating a mixture of nitrobenzene and aniline?
View answer and explanation
Correct option: C
Aq. HCl
Option C (Aq. HCl) is correct because aniline, being a base, will react with HCl to form a water-soluble salt, while nitrobenzene will remain insoluble and can be separated.
45 Easy
45 of 104
Remembering +4 | -1
Nitrobenzene is generally used for:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because nitrobenzene can be used as a solvent in shoe polish, floor polish, and is a precursor to aniline production.
46 Medium
46 of 104
Analyzing +4 | -1
An organic compound ‘A’ having molecular formula C2H3N{\mathrm{C}_2}{\mathrm{H}_3}\mathrm{N} on reduction gave another compound B, upon treatment with nitrous acid ‘B’ gave ethyl alcohol. On warming with chloroform and alcoholic KOH\mathrm{KOH}, it formed an offensive smelling compound ‘C’. The compound ‘C’ is
View answer and explanation
Correct option: B
CH3CH2N=→C\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{N}\mathop = \limits^ \to \mathrm{C}
Option B (CH3CH2N=→C\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{N}\mathop = \limits^ \to \mathrm{C}) is correct because the reaction sequence describes the carbylamine reaction, which produces an isocyanide.
47 Medium
47 of 104
Remembering +4 | -1
The reaction, C6H5N2Cl →Cu2Cl2/HCl C6H5Cl+N2{C_6}{H_5}{N_2}Cl\,\xrightarrow{{C{u_2}C{l_2}/HCl}}\,{C_6}{H_5}Cl + {N_2} is called:
View answer and explanation
Correct option: B
Sandmeyer’s reaction
Option B (Sandmeyer’s reaction) is correct because the reaction of benzenediazonium chloride with Cu2Cl2/HClCu_2Cl_2/HCl to form chlorobenzene is the Sandmeyer reaction.
48 Medium
48 of 104
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 NaNO2NaN{O_2} and HClHCl forms an unstable diazonium salt, which then reacts with water to yield ethanol.
49 Medium
49 of 104
Remembering +4 | -1
The reduction of CH3CN    to   CH3CH2NH2  \mathrm{C}{\mathrm{H}_3}\mathrm{CN\;}\,\,\mathrm{to}\,\mathrm{\;C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{N}{\mathrm{H}_2}\mathrm{\;} is called:
View answer and explanation
Correct option: C
Mendius reduction
Option C (Mendius reduction) is correct because it specifically refers to the reduction of nitriles to primary amines using reducing agents.
50 Easy
50 of 104
Remembering +4 | -1
Schiff’s bases are formed when aniline is condensed with:
View answer and explanation
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.
51 Easy
51 of 104
Understanding +4 | -1
The most basic compound among the following is:
View answer and explanation
Correct option: A
Benzylamine
Option A (Benzylamine) is the most basic compound because the nitrogen's lone pair is not delocalized into a ring, unlike aniline, acetanilide, and p-nitroaniline.
52 Medium
52 of 104
Analyzing +4 | -1
C3H9N{C_3}{H_9}N cannot represent:
View answer and explanation
Correct option: D
Quaternary salt
The correct answer is D (Quaternary salt) because C3H9N{C_3}{H_9}N does not have enough carbons to form a quaternary ammonium salt.
53 Hard
53 of 104
Application +4 | -1
CH3CH2Cl→NaCNA→Ni/H2B→anhydrldeacetic−C, C isC{H_3}C{H_2}Cl\xrightarrow{{NaCN}}A\xrightarrow{{Ni/{H_2}}}B\xrightarrow[{\mathrm{anhydrlde}}]{{\mathrm{acetic} - }}C,\,C\,\mathrm{is}
View answer and explanation
Correct option: C
CH3CH2CH2 NH COCH3C{H_3}C{H_2}C{H_2}\,NH\,COC{H_3}
Option C (CH3CH2CH2 NH COCH3C{H_3}C{H_2}C{H_2}\,NH\,COC{H_3}) is correct because the reaction sequence involves converting ethyl chloride to butanenitrile, then to butan-1-amine, and finally to N-ethylacetamide.
54 Easy
54 of 104
Remembering +4 | -1
The reagent used for conversion of benzene diazonium chloride to benzene is:
View answer and explanation
Correct option: D
All of these
Option D (All of these) is correct because H3PO2+H2O{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}, Na2SnO2+NaOH\mathrm{N}{\mathrm{a}_2}\mathrm{Sn}{\mathrm{O}_2} + \mathrm{NaOH}, and C2H5OH{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH} can all reduce benzene diazonium chloride to benzene.
55 Hard
55 of 104
Analyzing +4 | -1
The compound will react most readily with NaOH\mathrm{NaOH} to form methanol is
View answer and explanation
Correct option: A
(CH3)4N+I−{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_4}{\mathrm{N}^ + }{\mathrm{I}^ - }
Option A ((CH3)4N+I−{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_4}{\mathrm{N}^ + }{\mathrm{I}^ - }) is correct because it's a quaternary ammonium salt that undergoes Hoffmann elimination upon reaction with NaOH to form methanol and trimethylamine.
56 Easy
56 of 104
Application +4 | -1
The product AA and BB in the reaction are:
C2H5NH2+CHCl3+3KOH  →A+B+3H2O{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{N}{\mathrm{H}_2} + \mathrm{CHC}{\mathrm{l}_3} + 3\mathrm{KOH\;} \to A + B + 3{\mathrm{H}_2}\mathrm{O}
View answer and explanation
Correct option: A
C2H5NC+3KCl{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{NC} + 3\mathrm{KCl}
Option A (C2H5NC+3KCl{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{NC} + 3\mathrm{KCl}) is correct because this reaction is the carbylamine reaction (or isocyanide test), where a primary amine reacts with chloroform and a base to form an isocyanide, which has a foul odor.
57 Medium
57 of 104
Application +4 | -1
In the following sequence of reaction CH3−Br→KCNA→H3O+B→etherLiAlH4CC{H_3} - Br\xrightarrow{{KCN}}A\xrightarrow{{{H_3}{O^ + }}}B\xrightarrow[{\mathrm{ether}}]{{LiAl{H_4}}}C The end product C is
View answer and explanation
Correct option: D
C2H5OH{C_2}{H_5}OH
Option D (C2H5OH{C_2}{H_5}OH) is correct because the reaction sequence leads to the reduction of a nitrile to a primary amine, which is then protonated to form ethanol.
58 Easy
58 of 104
Understanding +4 | -1
The reaction of aniline with acetyl chloride in presence of NaOH\mathrm{NaOH} gives:
View answer and explanation
Correct option: A
Acetanilide
Option A (Acetanilide) is correct because the reaction of aniline with acetyl chloride in the presence of NaOH is an acylation reaction, forming acetanilide.
59 Easy
59 of 104
Understanding +4 | -1
Reduction of alkyl nitriles, produces
View answer and explanation
Correct option: B
Primary amine
Option B (Primary amine) is correct because the reduction of alkyl nitriles results in the formation of primary amines.
60 Hard
60 of 104
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 N-nitrosoamines, which are known carcinogens.
Revision notes and key takeaways

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

Showing 31-60 of 104 questions

Your practice

0 / 104 answered · 104 remaining · 0% accuracy

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

✓ Correct · ✗ Incorrect · ○ Unanswered

Answer feedback and difficulty

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.