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.
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.
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.
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.
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.
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.
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.
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.
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.
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.
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