Identify $X$ in the sequence, ${\rm{X}}\,\xrightarrow{{{\rm{HN}}{{\rm{O}}_{\rm{2}}}}}{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{8}}}{\rm{O}}\,\xrightarrow[{{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}}]{}\,{{\rm{C}}_{\rm{3}}}{{\rm{H}}_{\rm{6}}}{{\rm{O}}_{\rm{2}}}{\rm{:}}$

${\rm{C}}{{\rm{H}}_3} - {\rm{C}}{{\rm{H}}_2} - {\rm{C}}{{\rm{H}}_2} - {\rm{N}}{{\rm{H}}_2}$

${\rm{C}}{{\rm{H}}_3} - {\rm{NH}} - {\rm{C}}{{\rm{H}}_2} - {\rm{C}}{{\rm{H}}_3}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{\rm{N}}$

Which compound will liberate ${\rm{C}}{{\rm{O}}_2}$ from ${\rm{NaHC}}{{\rm{O}}_3}$ solution?

${\rm{C}}{{\rm{H}}_3}{{\rm{N}}^ + }{{\rm{H}}_3}{\rm{C}}{{\rm{L}}^ - }$

${\rm{C}}{{\rm{H}}_3}{\rm{CON}}{{\rm{H}}_2}$

${\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{H}}_2}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_4}{{\rm{N}}^ + }{\rm{O}}{{\rm{H}}^ - }$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{Br}}\mathop {\mathop \to \limits^{aq.{\rm{KOH}}} }\limits_\Delta A{\rm{\;}}\mathop {\mathop \to \limits^{{\rm{KMn}}{{\rm{O}}_4}/{{\rm{H}}^ + }} }\limits_\Delta B\mathop {\mathop \to \limits^{{\rm{N}}{{\rm{H}}_3}} }\limits_\Delta C\mathop {\mathop \to \limits^{{\rm{B}}{{\rm{r}}_2}} }\limits_{{\rm{alkali}}} D,\,\,\,'D'$ is

${\rm{C}}{{\rm{H}}_3}{\rm{N}}{{\rm{H}}_2}$

${\rm{C}}{{\rm{H}}_3}{\rm{Br}}$

${\rm{C}}{{\rm{H}}_3}{\rm{CON}}{{\rm{H}}_2}$

From the following compounds which does not react with ${{\rm{C}}_6}{{\rm{H}}_5}{\rm{S}}{{\rm{O}}_2}{\rm{CI}}$?

${\rm{C}}{{\rm{H}}_3}.{\rm{N}}{{\rm{H}}_2}$

${{\rm{C}}_2}{{\rm{H}}_5}.{\rm{N}}{{\rm{H}}_2}$

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_2}{\rm{NH}}$

${\left( {{{\rm{C}}_2}{{\rm{H}}_5}} \right)_3}{\rm{N}}$

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NEET • CHEMISTRY • AMINES

Amines practice · Mixed difficulty

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Showing 1-30 of 104 questions

1 Easy

1 of 104

Understanding +4 | -1

Which statement is not correct regarding aniline?

View answer and explanation

Correct option: D

It is soluble in water

Option D (It is soluble in water) is the incorrect statement because aniline is only slightly soluble in water due to the large hydrophobic phenyl group.

2 Medium

2 of 104

Understanding +4 | -1

The gas evolved when methyl amine reacts with nitrous acid is

View answer and explanation

Correct option: B

{N_2}

Option B (

{N_2}

) is correct because methyl amine reacts with nitrous acid to produce nitrogen gas.

3 Hard

3 of 104

Understanding +4 | -1

Diethylamine on oxidation with

\mathrm{KMn}{\mathrm{O}_4}

gives:

View answer and explanation

Correct option: C

Tetraethyl hydrazine

Option C (Tetraethyl hydrazine) is correct because diethylamine on oxidation with

\mathrm{KMn}{\mathrm{O}_4}

gives tetraethyl hydrazine.

4 Hard

4 of 104

Understanding +4 | -1

An organic amino compound reacts with aqueous nitrous acid at low temperature to produce an oily nitroso amine. The compound is

View answer and explanation

Correct option: D

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2} - \mathrm{NHC}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}

Option D (

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2} - \mathrm{NHC}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}

) is correct because secondary amines react with nitrous acid to form oily nitroso amines.

5 Medium

5 of 104

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.

6 Medium

6 of 104

Remembering +4 | -1

The reaction is

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.

7 Medium

7 of 104

Application +4 | -1

Identify

A

and

B

in the reaction given below.

\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.

8 Hard

8 of 104

Understanding +4 | -1

Which of the following would undergo Hofmann reaction to give a primary amine?

View answer and explanation

Correct option: C

R\mathrm{CON}{\mathrm{H}_2}

Option C (

R\mathrm{CON}{\mathrm{H}_2}

) is correct because the Hofmann reaction involves the reaction of an amide with

Br_2

and a base to yield a primary amine with one less carbon atom.

9 Medium

9 of 104

Application +4 | -1

Consider the following reaction sequence

C{H_3}C{H_2}ClABCD

Major product D is

View answer and explanation

Correct option: A

C{H_3}C{H_2}N{H_2}

Option A (

C{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.

10 Hard

10 of 104

Analyzing +4 | -1

Identify

X

in the sequence,

\mathrm{X}\,\xrightarrow{{\mathrm{HN}{\mathrm{O}_\mathrm{2}}}}{\mathrm{C}_\mathrm{3}}{\mathrm{H}_\mathrm{8}}\mathrm{O}\,\xrightarrow[{{\mathrm{H}_\mathrm{2}}\mathrm{S}{\mathrm{O}_\mathrm{4}}}]{}\,{\mathrm{C}_\mathrm{3}}{\mathrm{H}_\mathrm{6}}{\mathrm{O}_\mathrm{2}}\mathrm{:}

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{H}_3} - \mathrm{C}{\mathrm{H}_2} - \mathrm{C}{\mathrm{H}_2} - \mathrm{N}{\mathrm{H}_2}

Option B (

\mathrm{C}{\mathrm{H}_3} - \mathrm{C}{\mathrm{H}_2} - \mathrm{C}{\mathrm{H}_2} - \mathrm{N}{\mathrm{H}_2}

) is correct because propan-1-ol is formed from the reaction of an amine with

HNO_2

, and

C_3H_6O_2

is propanoic acid, formed by oxidation.

11 Medium

11 of 104

Analyzing +4 | -1

Decreasing order of basic nature in aqueous solutions

View answer and explanation

Correct option: C

{\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 (

( {\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.

12 Easy

12 of 104

Understanding +4 | -1

The amine which is most basic in aqueous solution is

View answer and explanation

Correct option: A

{\left( {C{H_3}} \right)_2}NH

Option A (

{\left( {C{H_3}} \right)_2}NH

) is correct because secondary amines are generally more basic than primary or tertiary amines in aqueous solution due to a combination of inductive effect and solvation effects.

13 Medium

13 of 104

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

\mathrm{HCl}\left( {aq.} \right)

Option D (

\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.

14 Medium

14 of 104

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.

15 Easy

15 of 104

Understanding +4 | -1

The reduction of which of the following compound would yield secondary amine?

View answer and explanation

Correct option: B

Carbylamine

Option B (Carbylamine) is correct because the reduction of carbylamines (isocyanides) yields secondary amines.

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16 Easy

16 of 104

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, and is not used for separating amines.

17 Easy

17 of 104

Understanding +4 | -1

The correct order of the basic strength of methyl substituted amines in aqueous solution is:

View answer and explanation

Correct option: A

{(C{H_3})_2}NH\, > \,C{H_3}N{H_2}\, > \,{(C{H_3})_3}N

Option A (

{(C{H_3})_2}NH\, > \,C{H_3}N{H_2}\, > \,{(C{H_3})_3}N

) is correct because in aqueous solution, the basic strength is influenced by both the +I effect of methyl groups and solvation effects. Secondary amines are generally more basic than primary amines due to better solvation, while tertiary amines are less basic due to steric hindrance and reduced solvation.

18 Hard

18 of 104

Understanding +4 | -1

Diazomethane reacts with carboxylic acids to produce:

View answer and explanation

Correct option: A

Ester

Option A (Ester) is correct because diazomethane reacts with carboxylic acids to form esters, with nitrogen gas as a byproduct.

19 Easy

19 of 104

Remembering +4 | -1

R - N{H_2} + CHC{l_3} + 3KOH\xrightarrow{{{\text{Heat}}}}R - NC + 3KCl + 3{H_2}O

Above reaction is known as:

View answer and explanation

Correct option: B

Carbylamine reaction

Option B (Carbylamine reaction) is correct because the reaction of a primary amine with chloroform and a base (KOH) to form an isocyanide (R-NC) is the defining characteristic of the Carbylamine reaction.

20 Medium

20 of 104

Analyzing +4 | -1

C{H_3}N{H_2}\;

is heated with sodium and extracted with water and then

AgN{O_3}

is added. The white ppt. obtained is of:

View answer and explanation

Correct option: A

\mathrm{AgCN}

Option A (

\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.

21 Medium

21 of 104

Remembering +4 | -1

The diazonium ion of Benzenediazonium chloride can be replaced by

{Cl}^{-},{Br}^{-}

{CN}^{-}

in presence of

View answer and explanation

Correct option: B

{Cu}

(I) ions

Option B (

{Cu}

(I) ions) is correct because the diazonium ion is replaced by

{Cl}^{-},{Br}^{-}

{CN}^{-}

in the presence of

{Cu}

(I) ions in Sandmeyer's reaction.

22 Medium

22 of 104

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.

23 Easy

23 of 104

Analyzing +4 | -1

Which of the following is more basic than aniline?

View answer and explanation

Correct option: B

Benzylamine

Option B (Benzylamine) is correct because the lone pair on the nitrogen in benzylamine is not delocalized into the aromatic ring, making it more available for protonation than aniline and its derivatives.

24 Hard

24 of 104

Analyzing +4 | -1

I. Diazonium salts are very good intermediates for the introduction of

\mathrm{F}, \mathrm{Cl}, \mathrm{Br}, \mathrm{I}, \mathrm{CN}, \mathrm{OH}, \mathrm{NO}_2

groups into the aromatic ring.

II. Aryl fluorides and iodides can be prepared by direct halogenation.

III. Cyano group cannot be introduced by nucleophilic substitution of chlorine in chlorobenzene.

IV. The replacement of diazo group by other groups is helpful in preparing those substituted aromatic compounds which cannot be prepared by direct substitution in benzene or substituted benzene.

Select the incorrect statements,

View answer and explanation

Correct option: B

II and III

The correct answer is B (II and III) because aryl fluorides can be prepared by Schiemann reaction and cyano group can be introduced using CuCN.

25 Medium

25 of 104

Analyzing +4 | -1

R - N\mathop = \limits_ \to C + HgO\xrightarrow{\,}\,A + H{g_2}O

; What is

A

View answer and explanation

Correct option: C

R - {\text{NCO}}

The correct answer is C (

R - \text{NCO}

) because the reaction sequence leads to the formation of an isocyanate.

26 Easy

26 of 104

Application +4 | -1

Which compound will liberate

\mathrm{C}{\mathrm{O}_2}

from

\mathrm{NaHC}{\mathrm{O}_3}

solution?

View answer and explanation

Correct option: D

\mathrm{C}{\mathrm{H}_3}{\mathrm{N}^ + }{\mathrm{H}_3}\mathrm{C}{\mathrm{L}^ - }

Option D (

\mathrm{C}{\mathrm{H}_3}{\mathrm{N}^ + }{\mathrm{H}_3}\mathrm{C}{\mathrm{L}^ - }

) is correct because it is a salt of a weak base (methylammonium chloride) and will react with

\mathrm{NaHC}{\mathrm{O}_3}

to release

\mathrm{C}{\mathrm{O}_2}

27 Medium

27 of 104

Understanding +4 | -1

p -

amine and

s -

amine are distinguished by:

View answer and explanation

Correct option: C

\mathrm{HN}{\mathrm{O}_2}

Option C (

\mathrm{HN}{\mathrm{O}_2}

) is correct because primary, secondary, and tertiary amines react differently with nitrous acid (

\mathrm{HN}{\mathrm{O}_2}

), allowing for their distinction.

28 Hard

28 of 104

Application +4 | -1

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{Br}\mathop {\mathop \to \limits^{aq.\mathrm{KOH}} }\limits_\Delta A\mathrm{\;}\mathop {\mathop \to \limits^{\mathrm{KMn}{\mathrm{O}_4}/{\mathrm{H}^ + }} }\limits_\Delta B\mathop {\mathop \to \limits^{\mathrm{N}{\mathrm{H}_3}} }\limits_\Delta C\mathop {\mathop \to \limits^{\mathrm{B}{\mathrm{r}_2}} }\limits_{\mathrm{alkali}} D,\,\,\,'D'

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}

Option C (

\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}

) is correct because the series of reactions converts ethyl bromide to ethyl alcohol, then to acetic acid, then to acetamide, and finally to methylamine via Hoffmann degradation.

29 Easy

29 of 104

Analyzing +4 | -1

From the following compounds which does not react with

{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{S}{\mathrm{O}_2}\mathrm{CI}

View answer and explanation

Correct option: B

\mathrm{C}{\mathrm{H}_3}.\mathrm{N}{\mathrm{H}_2}

Option D (

{{\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)}_3}\mathrm{N}

) is correct because it is a tertiary amine and does not have any replaceable hydrogen atoms on the nitrogen, thus it cannot react with Hinsberg's reagent.

30 Medium

30 of 104

Application +4 | -1

Acetanilide on nitration followed by alkaline hydrolysis mainly gives

View answer and explanation

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.

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