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  4. Amines Chapter-Wise Test 3

Amines Chapter-Wise Test 3

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which method is used to prepare \( \ce{CH3CH2CH2NH2} \) from \( \ce{CH3CH2CH2Cl} \)?

Ammonolysis of \( \ce{CH3CH2CH2Cl} \) with excess \( \ce{NH3} \) yields \( \ce{CH3CH2CH2NH2} \), as per the PDF’s preparation methods.

Hoffmann bromamide reaction
Reduction of nitro compounds
Gabriel phthalimide synthesis
Ammonolysis
4

The product of aniline with \( \ce{H2SO4} \) at 453-473 K is:

Aniline (\( \ce{C6H5NH2} \)) with \( \ce{H2SO4} \) at high temperature forms p-aminobenzene sulphonic acid (sulphanilic acid), as per the PDF’s chemical properties.

o-Aminobenzene sulphonic acid
m-Aminobenzene sulphonic acid
Anilinium sulphate
p-Aminobenzene sulphonic acid
4

Which of the following is not a method to prepare primary amines as per the PDF?

The PDF lists reduction of nitro compounds, Hoffmann bromamide, and Gabriel synthesis for primary amines. Friedel-Crafts alkylation does not produce amines.

Reduction of nitro compounds
Hoffmann bromamide reaction
Gabriel phthalimide synthesis
Friedel-Crafts alkylation
4

Which amine does not react with \( \ce{HNO2} \) to form an alcohol?

Tertiary amines like \( \ce{(CH3)3N} \) do not react with \( \ce{HNO2} \) to form alcohols, unlike primary aliphatic amines, as per the PDF.

\( \ce{CH3NH2} \)
\( \ce{CH3CH2NH2} \)
\( \ce{(CH3)3N} \)
\( \ce{CH3CH2CH2NH2} \)
3

The nitrogen atom in \( \ce{(CH3)2NH} \) has how many sigma bonds?

In \( \ce{(CH3)2NH} \), nitrogen forms two sigma bonds with carbon atoms and one with hydrogen, totaling three sigma bonds, as per the PDF’s structure discussion.

3
2
4
1
1

The product of \( \ce{CH3CONH2} \) undergoing Hoffmann bromamide reaction is:

Hoffmann bromamide reaction converts \( \ce{CH3CONH2} \) (ethanamide) to \( \ce{CH3NH2} \) (methylamine), reducing the carbon chain by one, as detailed in the PDF.

\( \ce{CH3CH2NH2} \)
\( \ce{CH3NH2} \)
\( \ce{(CH3)2NH} \)
\( \ce{CH3OH} \)
2

An alkyl halide \( X \) with \( \ce{NH3} \) gives \( \ce{CH3CH2CH2CH2NH2} \). \( X \) is:

1-Chlorobutane (\( \ce{CH3CH2CH2CH2Cl} \)) reacts with \( \ce{NH3} \) via ammonolysis to form butan-1-amine (\( \ce{CH3CH2CH2CH2NH2} \)), as per the PDF.

\( \ce{CH3CH2Cl} \)
\( \ce{CH3CH2CH2Cl} \)
\( \ce{CH3Cl} \)
\( \ce{CH3CH2CH2CH2Cl} \)
4

Which amine does not react with benzenesulphonyl chloride?

Tertiary amines like \( \ce{(CH3CH2CH2)3N} \) lack an N-H bond to react with benzenesulphonyl chloride, as per the PDF’s Hinsberg test.

\( \ce{CH3CH2NH2} \)
\( \ce{(CH3CH2)2NH} \)
\( \ce{(CH3CH2CH2)3N} \)
\( \ce{C6H5NH2} \)
3

Why does aniline not undergo Friedel-Crafts reaction?

Aniline’s nitrogen forms a salt with \( \ce{AlCl3} \) (Lewis acid catalyst), acquiring a positive charge and deactivating the ring for electrophilic substitution.

Due to steric hindrance
Due to low reactivity
Due to resonance
Due to salt formation with \( \ce{AlCl3} \)
4

A compound \( X \) on reduction with \( \ce{H2/Ni} \) gives \( \ce{CH3CH2CH2NH2} \). \( X \) is:

Reduction of 1-nitropropane (\( \ce{CH3CH2CH2NO2} \)) with \( \ce{H2/Ni} \) yields propan-1-amine (\( \ce{CH3CH2CH2NH2} \)), as per the PDF’s preparation methods.

\( \ce{CH3CH2CH2NO2} \)
\( \ce{CH3CH2CH2OH} \)
\( \ce{CH3CH2CHO} \)
\( \ce{CH3CH2CH2Cl} \)
1

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