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

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

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

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

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

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

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

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

An aromatic amine $\left( A \right)$ was treated with alcoholic potash and another compound $\left( Y \right)$ when a foul smeling gas was formed with formula ${{\rm{C}}_6}{{\rm{H}}_5}{\rm{NC}}$. $\left( Y \right)$ was formed by reacting a compound $\left( Z \right)$ with ${\rm{C}}{{\rm{l}}_2}$ in the presence of slaked lime. Compound $\left( Z \right)$ is:

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

${{\rm{C}}_6}{{\rm{H}}_5}{\rm{N}}{{\rm{H}}_2}{\rm{\;}}$

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

${\rm{CHC}}{{\rm{l}}_3}{\rm{\;}}$

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

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

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

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

The compound will react most readily with ${\rm{NaOH}}$ to form methanol is

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

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

${\left( {{\rm{C}}{{\rm{H}}_3}} \right)_3}{{\rm{S}}^ + }{{\rm{I}}^ - }$

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

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NEET • CHEMISTRY • AMINES

Amines practice · Hard

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1 Hard

1 of 23

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.

2 Hard

2 of 23

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.

3 Hard

3 of 23

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.

4 Hard

4 of 23

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.

5 Hard

5 of 23

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.

6 Hard

6 of 23

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.

7 Hard

7 of 23

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.

8 Hard

8 of 23

Application +4 | -1

1,4 – dichloro butane

\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.

9 Hard

9 of 23

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.

10 Hard

10 of 23

Analyzing +4 | -1

An aromatic amine

\left( A \right)

was treated with alcoholic potash and another compound

\left( Y \right)

when a foul smeling gas was formed with formula

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

\left( Y \right)

was formed by reacting a compound

\left( Z \right)

with

\mathrm{C}{\mathrm{l}_2}

in the presence of slaked lime. Compound

\left( Z \right)

is:

View answer and explanation

Correct option: C

\mathrm{C}{\mathrm{H}_3}\mathrm{COC}{\mathrm{H}_3}

Option C (

\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.

11 Hard

11 of 23

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.

12 Hard

12 of 23

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 the reaction sequence involves converting ethyl chloride to butanenitrile, then to butan-1-amine, and finally to N-ethylacetamide.

13 Hard

13 of 23

Analyzing +4 | -1

The compound will react most readily with

\mathrm{NaOH}

to form methanol is

View answer and explanation

Correct option: A

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_4}{\mathrm{N}^ + }{\mathrm{I}^ - }

Option A (

{\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.

14 Hard

14 of 23

Remembering +4 | -1

Carcinogens are the products of the reaction between:

View answer and explanation

Correct option: A

{R_2}\mathrm{NH} + \mathrm{HN}{\mathrm{O}_2}

Option A (

{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.

15 Hard

15 of 23

Analyzing +4 | -1

A\mathop \to \limits^{{\mathrm{H}_2}\mathrm{NOH\;}} B\mathop \to \limits^{\mathrm{Reduction\;}} C\mathop \to \limits^{\mathrm{NOCl\;}} \mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{Cl}

In the above sequence

A

and

C

are

View answer and explanation

Correct option: D

Acetaldehyde, ethyl amine

Option D (Acetaldehyde, ethyl amine) is correct because acetaldehyde reacts with hydroxylamine to form an oxime, which upon reduction gives ethylamine, and ethylamine reacts with NOCl to give ethyl chloride.

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16 Hard

16 of 23

Application +4 | -1

C{H_3}C{H_2}CON{H_2}\xrightarrow[{B{r_2}}]{{NaoH}}A

Aqueos solution of

A

View answer and explanation

Correct option: B

Turns red litmus to blue

Option B (Turns red litmus to blue) is correct because the reaction produces ethylamine, which is a basic amine that turns red litmus paper blue.

17 Hard

17 of 23

Understanding +4 | -1

Which of the following compound reacts with chloroform and a base to form phenyl isocyanide?

View answer and explanation

Correct option: B

Aniline

Option B (Aniline) is correct because aniline reacts with chloroform and a base (KOH) in the carbylamine reaction to form phenyl isocyanide.

18 Hard

18 of 23

Analyzing +4 | -1

Which of the following orders is wrong with respect to property indicated?

View answer and explanation

Correct option: D

Aniline > Cyclohexylamine > Benzamide (Base strength)

Option D (Aniline > Cyclohexylamine > Benzamide (Base strength)) is the wrong order because cyclohexylamine is a stronger base than aniline, and benzamide is a very weak base due to resonance stabilization.

19 Hard

19 of 23

Application +4 | -1

The product formed in the reaction

n - hexanamide + \mathrm{B}{\mathrm{r}_2} + \mathrm{KOH,}

View answer and explanation

Correct option: D

pentanamine

Option D (pentanamine) is correct because the reaction is the Hoffmann bromamide degradation, which results in an amine with one less carbon atom.

20 Hard

20 of 23

Application +4 | -1

A compound with molecular mass 180 is acylated with

C{H_3}COCl

to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is

View answer and explanation

Correct option: B

Option B (5) is correct because each acylation adds a CH3CO group (42 mass units), and the total mass increase (390-180=210) divided by 42 gives the number of amino groups.

21 Hard

21 of 23

Analyzing +4 | -1

(A)

{\mathrm{C}_2}{\mathrm{H}_5}\mathrm{N}{\mathrm{H}_2}\mathop {\mathop \to \limits^{\left( \mathrm{i} \right)\mathrm{NOCl}} }\limits_{\left( {\mathrm{ii}} \right)\mathrm{AgN}{\mathrm{O}_2}} \left[ W \right]

(B)

\,{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHN}{\mathrm{H}_2}\mathop {\mathop \to \limits^{\left( \mathrm{i} \right)\mathrm{NOCl}} }\limits_{\left( {\mathrm{ii}} \right)\mathrm{AgN}{\mathrm{O}_2}} \left[ X \right]

(C)

{\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathrm{\;CN}{\mathrm{H}_2}\mathop {\mathop \to \limits^{\left( \mathrm{i} \right)\mathrm{NOCl}} }\limits_{\left( {\mathrm{ii}} \right)\mathrm{AgN}{\mathrm{O}_2}} \left[ Y \right]

(D)

\mathrm{C}{\mathrm{H}_3}\mathrm{CH}\left( {\mathrm{N}{\mathrm{H}_2}} \right){\mathrm{C}_2}{\mathrm{H}_5}\mathop {\mathop \to \limits^{\left( \mathrm{i} \right)\mathrm{NOCl}} }\limits_{\left( {\mathrm{ii}} \right)\mathrm{AgN}{\mathrm{O}_2}} \left[ Z \right]

Which product will not show tautomerism?

View answer and explanation

Correct option: C

Y

Option C (

Y

) is correct because it is a tertiary nitroalkane, which does not have an alpha-hydrogen and therefore cannot exhibit tautomerism.

22 Hard

22 of 23

Application +4 | -1

How many of the following compounds can be methylated by diazomethane ?

\mathrm{C}_2 \mathrm{H}_5 \mathrm{COOH}, \mathrm{C}_2 \mathrm{H}_5 \mathrm{NH}_2, \mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}

and

\mathrm{CH}_3 \mathrm{COCH}_2 \mathrm{COOC}_2 \mathrm{H}_5

View answer and explanation

Correct option: A

Option A (3) is correct because carboxylic acids, phenols, and compounds with active methylene groups can be methylated by diazomethane.

23 Hard

23 of 23

Understanding +4 | -1

Potassium cyanate is heated with ammonium sulphate. The product formed is

View answer and explanation

Correct option: A

Urea

Option A (Urea) is correct because potassium cyanate reacts with ammonium sulphate upon heating to produce urea.

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Key Takeaways

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