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

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

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

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

In the chemical reaction, ${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{N}}{{\rm{H}}_2} + {\rm{CHC}}{{\rm{I}}_3} + 3{\rm{KOH}} \to \left( A \right) + \left( B \right) + 3{{\rm{H}}_2}{\rm{O}}$ The compounds $\left( A \right)$ and $\left( B \right)$ are respectively

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{NC\;}}$ and $3{\rm{KCI}}$

${\rm{C}}{{\rm{H}}_3}{\rm{C}}{{\rm{H}}_2}{\rm{CON}}{{\rm{H}}_2}$ and $3{\rm{KCl}}$

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{NC}}$ and ${{\rm{K}}_2}{\rm{C}}{{\rm{O}}_3}$

${{\rm{C}}_2}{{\rm{H}}_5}{\rm{CN}}$ and $3{\rm{KCI}}$

$C{H_3}C{H_2}Cl\xrightarrow{{NaCN}}A\xrightarrow{{Ni/{H_2}}}B\xrightarrow[{{\rm{anhydrlde}}}]{{{\rm{acetic}} - }}C,\,C\,{\rm{is}}$

$C{H_3}\,C{H_2}\,C{H_2}\,N{H_2}$

$C{H_3}C{H_2}C{H_2}CO\,NHC{H_3}$

$C{H_3}C{H_2}C{H_2}CO\,NH\,COC{H_3}$

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JEE • CHEMISTRY • AMINES

Amines practice · Mixed difficulty

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Showing 1-30 of 132 questions

1 Medium

1 of 132

Application +4 | -1

Identify '

\mathrm{Z}

' in the sequence

{{\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

\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}

Option C (

\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 132

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 Easy

3 of 132

Analyzing +4 | -1

Assertion: The main product of reaction of alcoholic silver nitrite and ethyl bromide is nitroethane.

Reason: Silver nitrite is predominantly covalent compound.

View answer and explanation

Correct option: A

If both assertion and reason are true and reason is the correct explanation of assertion

Option A (

\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}

- Diamagnetic) is the wrong match because

\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}

has two unpaired electrons and is therefore paramagnetic, not diamagnetic.

4 Easy

4 of 132

Remembering +4 | -1

Hofmann’s bromamide reaction is to convert

View answer and explanation

Correct option: C

Amide to amine

Option C (Amide to amine) is correct because Hofmann's bromamide reaction converts an amide to an amine with one less carbon atom.

5 Easy

5 of 132

Remembering +4 | -1

Gabriel phthalamide synthesis is used for preparation of :-

View answer and explanation

Correct option: A

{RNH}_2

Option A is correct because the Gabriel phthalamide synthesis is specifically used for the preparation of primary amines (RNH2).

6 Easy

6 of 132

Remembering +4 | -1

Elements found in explosive are:

View answer and explanation

Correct option: C

Both S and N

Option C (Both S and N) is correct because both sulfur and nitrogen are commonly found in explosive compounds, contributing to their explosive properties.

7 Medium

7 of 132

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.

8 Hard

8 of 132

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 it's a primary amine that reacts with

HNO_2

to form an alcohol, which then dehydrates to form an acid.

9 Easy

9 of 132

Understanding +4 | -1

In which reaction, nitrene is not the intermediate?

View answer and explanation

Correct option: D

Gabriel’s phthalimide

Option D (Gabriel’s phthalimide) is correct because the Gabriel phthalimide synthesis proceeds through a carbanion intermediate, not a nitrene.

10 Easy

10 of 132

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

{\mathrm{H}_3}\mathrm{P}{\mathrm{O}_2} + {\mathrm{H}_2}\mathrm{O}

\mathrm{N}{\mathrm{a}_2}\mathrm{Sn}{\mathrm{O}_2} + \mathrm{NaOH}

, and

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{OH}

can all reduce benzene diazonium chloride to benzene.

11 Medium

11 of 132

Application +4 | -1

The IUPAC name of,

C{H_3} - C{H_2} - \mathop {\mathop C\limits_| H}\limits_{C{H_3}} - N{H_2}

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.

12 Medium

12 of 132

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.

13 Medium

13 of 132

Understanding +4 | -1

IUPAC name of the compound

\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.

14 Medium

14 of 132

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.

15 Easy

15 of 132

Understanding +4 | -1

Gas evolved during the reaction of sodium metal on ethyl amine is:

View answer and explanation

Correct option: C

{\mathrm{H}_2}

Option D (Both (A) and (R) are true but (R) is the correct explanation of (A).) is correct because Lithium salts are hydrated due to the high charge density and polarizing power of the small Li+ ion.

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16 Easy

16 of 132

Analyzing +4 | -1

The basic strength order of the following in

\mathrm{H}_2 \mathrm{O}

\mathop {{\text{MeN}}{{\text{H}}_{\text{2}}}{\text{,}}}\limits_{{\text{(I)}}} \,\,\,\,\,\mathop {{\text{M}}{{\text{e}}_{\text{2}}}{\text{NH}}}\limits_{{\text{(II)}}} {\text{,}}\,\,\,\,\,\mathop {{\text{M}}{{\text{e}}_{\text{3}}}{\text{N}}}\limits_{{\text{(III)}}}

View answer and explanation

Correct option: A

>

>

III

Option B (II > I > III) is correct because the basic strength order in water is Me2NH > MeNH2 > Me3N due to the combined effects of solvation and steric hindrance.

17 Easy

17 of 132

Analyzing +4 | -1

When

\mathrm{NaN}{\mathrm{O}_2}

and dilute

\mathrm{HCl}

were added to an amine at

\mathrm{0}^\circ \mathrm{C}

, a colourless gas was evolved and an ionic compound is formed. The amine is

View answer and explanation

Correct option: B

An aromatic primary amine

Option B is correct because an aromatic primary amine reacts with NaNO2 and HCl at 0°C to form a diazonium salt, which is an ionic compound, and evolves nitrogen gas upon warming.

18 Medium

18 of 132

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^{\circ}

amine is diazotized with

\mathrm{HNO}_2

and

\mathrm{NaCl}

0^{\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.

19 Medium

19 of 132

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.

20 Medium

20 of 132

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.

21 Easy

21 of 132

Understanding +4 | -1

In aqueous solutions, the basic strength of amines decreases in the order

View answer and explanation

Correct option: D

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

Option D (

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

) is correct because in aqueous solutions, the basic strength of amines depends on the electron-donating effect and steric hindrance.

22 Hard

22 of 132

Analyzing +4 | -1

Assertion:-

{\text{H}} - \mathop {\text{C}}\limits_{\begin{array}{cccccccccccccccccccc} {||} \\ {\text{O}} \end{array}} - \mathop {\text{N}}\limits_{\begin{array}{cccccccccccccccccccc} | \\ {{\text{C}}{{\text{H}}_{\text{3}}}} \end{array}} - {\text{C}}{{\text{H}}_{\text{3}}}

and

{\text{C}}{{\text{H}}_{\text{3}}} - \mathop {\text{C}}\limits_{\begin{array}{cccccccccccccccccccc} {||} \\ {\text{O}} \end{array}} - {\text{NH}} - {\text{C}}{{\text{H}}_{\text{3}}}

are metameres

Reason:-There is adjustment of

-\mathrm{CH}_2-

group in the molecule.

View answer and explanation

Correct option: D

If both Assertion & Reason are false

Option D is correct because both the assertion and the reason are false. The given compounds are not metamers, and the reason provided is not a valid explanation of metamerism.

23 Easy

23 of 132

Analyzing +4 | -1

The basicity of compounds I, II, III and IV

\mathop {\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}}\limits_\mathrm{I} ,\,\,\,\mathop {{{\left( {\mathrm{C}{\mathrm{H}_3}} \right)}_2}\mathrm{NH}}\limits_{\mathrm{II}} ,\mathrm{\;}\,\,\mathop {{{\left( {\mathrm{C}{\mathrm{H}_3}} \right)}_3}\mathrm{N}}\limits_{\mathrm{III}} ,\mathrm{\;}\mathop {{\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{N}{\mathrm{H}_2}}\limits_{\mathrm{IV}}

varies in the order

View answer and explanation

Correct option: B

II > I > III > IV

Option B is correct because the basicity order is II > I > III > IV due to the combined effects of +I effect, steric hindrance, and the electron-withdrawing nature of the phenyl group.

24 Medium

24 of 132

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.

25 Easy

25 of 132

Application +4 | -1

In the chemical reaction,

\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{N}{\mathrm{H}_2} + \mathrm{CHC}{\mathrm{I}_3} + 3\mathrm{KOH} \to \left( A \right) + \left( B \right) + 3{\mathrm{H}_2}\mathrm{O}

The compounds

\left( A \right)

and

\left( B \right)

are respectively

View answer and explanation

Correct option: C

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{NC\;}

and

3\mathrm{KCI}

Option C (

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{NC\;}

and

3\mathrm{KCI}

) is correct because this reaction is the carbylamine reaction, which produces an isocyanide and potassium chloride.

26 Easy

26 of 132

Analyzing +4 | -1

Which of the following statements is correct?

View answer and explanation

Correct option: B

Methylamine is a stronger base than aniline and ammonia

Option B (Methylamine is a stronger base than aniline and ammonia) is correct because alkylamines are generally stronger bases than ammonia due to the +I effect of the alkyl group, and aniline is a weaker base due to resonance.

27 Hard

27 of 132

Application +4 | -1

C{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

C{H_3}C{H_2}C{H_2}\,NH\,COC{H_3}

Option C (

C{H_3}C{H_2}C{H_2}\,NH\,COC{H_3}

) is correct because it represents the product of the given reaction sequence involving nucleophilic substitution, reduction, and acylation.

28 Medium

28 of 132

Analyzing +4 | -1

Assertion: Basicity of

{CH}_3 {CH}_2 {NH}_2(\text{I}), {NH}_3 \text{(II)}

and

{C}_6 {H}_5 {NH}_2 \text{(III)}

is in order I

>

>

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.

29 Medium

29 of 132

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.

30 Medium

30 of 132

Analyzing +4 | -1

Which one will have lowest value of

\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.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 132 single-correct-answer MCQs on AMINES.
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