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NEET • CHEMISTRY • AMINES

Amines practice · Medium

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Showing 1-30 of 48 questions
1 Medium
1 of 48
Understanding +4 | -1
The gas evolved when methyl amine reacts with nitrous acid is
View answer and explanation
Correct option: B
N2{N_2}
Option B (N2{N_2}) is correct because methyl amine reacts with nitrous acid to produce nitrogen gas.
2 Medium
2 of 48
Application +4 | -1
Aniline was diazotised and subsequently reduced with stannous chloride and hydrochloric acid to yield:
View answer and explanation
Correct option: B
Phenyl hydrazine
Option B (Phenyl hydrazine) is correct because the reduction of diazotized aniline with stannous chloride and hydrochloric acid yields phenyl hydrazine.
3 Medium
3 of 48
Remembering +4 | -1
The reaction is
Ar N2+ Cl−→Cu/HCl ArCl+N2+ CuCl  Ar\,\mathop {{N_2}}\limits^ + \,C{l^ - }\xrightarrow{{Cu/HCl}}\,ArCl + {N_2} + \,CuCl\,\,
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.
4 Medium
4 of 48
Application +4 | -1
Identify AA and BB in the reaction given below.
Ethanenitrile    →  Hydrolysisaq.H2SO4+2H2O−NH3  A  →DecarboxylationSodalimeΔ−CO2  B\mathrm{Ethane nitrile}\,\,\mathop {\mathrm{\;}\mathop \to \limits^{\mathrm{\;Hydrolysis}} }\limits_{\begin{array}{cccccccccccccccccccc} {aq.{\mathrm{H}_2}S{\mathrm{O}_4}}\\ { + 2{\mathrm{H}_2}O}\\ { - N{\mathrm{H}_3}} \end{array}} \;A\;\mathop {\mathop \to \limits^{\mathrm{Decarboxylation}} }\limits_{\begin{array}{cccccccccccccccccccc} {Sodalime}\\ \Delta \\ { - C{\mathrm{O}_2}} \end{array}} \mathrm{\;}B
View answer and explanation
Correct option: D
Ethanoic acid, methane
Option D (Ethanoic acid, methane) is correct because hydrolysis of ethane nitrile yields ethanoic acid, and decarboxylation of ethanoic acid with sodalime yields methane.
5 Medium
5 of 48
Application +4 | -1
Consider the following reaction sequence CH3CH2ClABCDC{H_3}C{H_2}ClABCD Major product D is
View answer and explanation
Correct option: A
CH3CH2NH2C{H_3}C{H_2}N{H_2}
Option A (CH3CH2NH2C{H_3}C{H_2}N{H_2}) is correct. The reaction sequence involves converting ethyl chloride to ethyl cyanide, then to propylamine, and finally back to ethylamine via Hofmann degradation.
6 Medium
6 of 48
Analyzing +4 | -1
Decreasing order of basic nature in aqueous solutions
View answer and explanation
Correct option: C
(CH3)2NH>CH3NH2>NH3>C6H5NH2{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{NH} > C{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2} > N{\mathrm{H}_3} > {\mathrm{C}_6}{\mathrm{H}_5}\mathrm{N}{\mathrm{H}_2}
Option C ((CH3)2NH>CH3NH2>NH3>C6H5NH2( {\mathrm{C}{\mathrm{H}_3}} )_2\mathrm{NH} > C{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2} > N{\mathrm{H}_3} > {\mathrm{C}_6}{\mathrm{H}_5}\mathrm{N}{\mathrm{H}_2}) is correct because in aqueous solution, the basicity order is influenced by factors like solvation, steric hindrance, and electronic effects, leading to the observed trend.
7 Medium
7 of 48
Understanding +4 | -1
Which of the following reagents is useful for separating aniline from a mixture of aniline and nitrobenzene?
View answer and explanation
Correct option: D
HCl(aq.)\mathrm{HCl}\left( {aq.} \right)
Option D (HCl(aq.)\mathrm{HCl}\left( {aq.} \right)) is correct because aniline, being a base, reacts with HCl to form a water-soluble salt, while nitrobenzene remains unaffected and can be separated.
8 Medium
8 of 48
Analyzing +4 | -1
Which one of the following is least basic?
View answer and explanation
Correct option: D
Aniline
Option D (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 the least basic.
9 Medium
9 of 48
Analyzing +4 | -1
CH3NH2  C{H_3}N{H_2}\; is heated with sodium and extracted with water and then AgNO3AgN{O_3} is added. The white ppt. obtained is of:
View answer and explanation
Correct option: A
AgCN\mathrm{AgCN}
Option A (AgCN\mathrm{AgCN}) is correct because when methylamine is heated with sodium, it forms sodium cyanide, which then reacts with silver nitrate to form silver cyanide precipitate.
10 Medium
10 of 48
Remembering +4 | -1
The diazonium ion of Benzenediazonium chloride can be replaced by Cl−,Br−{Cl}^{-},{Br}^{-} or CN−{CN}^{-} in presence of
View answer and explanation
Correct option: B
Cu{Cu} (I) ions
Option B (Cu{Cu} (I) ions) is correct because the diazonium ion is replaced by Cl−,Br−{Cl}^{-},{Br}^{-} or CN−{CN}^{-} in the presence of Cu{Cu} (I) ions in Sandmeyer's reaction.
11 Medium
11 of 48
Understanding +4 | -1
Which of the following reaction is appropriate for converting acetamide to methanamine?
View answer and explanation
Correct option: C
Hoffmann hypobromamide reaction
Option C (Hoffmann hypobromamide reaction) is correct because the Hoffmann hypobromamide reaction converts amides to amines with one less carbon atom.
12 Medium
12 of 48
Analyzing +4 | -1
R−N=→C+HgO→  A+Hg2OR - N\mathop = \limits_ \to C + HgO\xrightarrow{\,}\,A + H{g_2}O ; What is AA ?
View answer and explanation
Correct option: C
R−NCOR - {\text{NCO}}
The correct answer is C (R−NCOR - \text{NCO}) because the reaction sequence leads to the formation of an isocyanate.
13 Medium
13 of 48
Understanding +4 | -1
p−p - amine and s−s -amine are distinguished by:
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Correct option: C
HNO2\mathrm{HN}{\mathrm{O}_2}
Option C (HNO2\mathrm{HN}{\mathrm{O}_2}) is correct because primary, secondary, and tertiary amines react differently with nitrous acid (HNO2\mathrm{HN}{\mathrm{O}_2}), allowing for their distinction.
14 Medium
14 of 48
Application +4 | -1
Acetanilide on nitration followed by alkaline hydrolysis mainly gives
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Correct option: B
P-Nitroaniline
Option B (P-Nitroaniline) is correct because nitration of acetanilide occurs primarily at the para position due to the directing effect of the acetamido group, and subsequent hydrolysis removes the acetyl group to yield p-nitroaniline.
15 Medium
15 of 48
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.

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16 Medium
16 of 48
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.
17 Medium
17 of 48
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.
18 Medium
18 of 48
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.
19 Medium
19 of 48
Analyzing +4 | -1
Among the following, the strongest base is:
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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 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.
20 Medium
20 of 48
Application +4 | -1
Which of the following reagents will be fruitful for separating a mixture of nitrobenzene and aniline?
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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.
21 Medium
21 of 48
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.
22 Medium
22 of 48
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.
23 Medium
23 of 48
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.
24 Medium
24 of 48
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.
25 Medium
25 of 48
Analyzing +4 | -1
C3H9N{C_3}{H_9}N cannot represent:
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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.
26 Medium
26 of 48
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.
27 Medium
27 of 48
Application +4 | -1
Which reacts with Heinsberg’s reagent and gives a compound insoluble in NaOH due to lack of acidic hydrogen is
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Correct option: C
N-Methyl aniline
Option C (N-Methyl aniline) is correct because it reacts with Hinsberg's reagent to form a sulfonamide that has one acidic hydrogen, which reacts with NaOH to become soluble. N,N-dimethyl aniline does not have any acidic hydrogen.
28 Medium
28 of 48
Understanding +4 | -1
A positive carbylamine test is given by:
View answer and explanation
Correct option: D
p−p - methyl benzylamine
Option D (p−p - methyl benzylamine) is correct because the carbylamine test is specific for primary amines. p−p-methyl benzylamine is a primary amine, while the other options are secondary or tertiary amines.
29 Medium
29 of 48
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 (hexamethylenetetramine) is (CH2)6N4.
30 Medium
30 of 48
Understanding +4 | -1
Methyl cyanide gives on hydrolysis
View answer and explanation
Correct option: B
Acetic acid
Option B (Acetic acid) is correct because the hydrolysis of methyl cyanide (acetonitrile) yields acetic acid.
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