Correct answer Carries: 4.
Wrong Answer Carries: -1.
The product of \( \ce{CH3CH2Cl} \) reacting with excess \( \ce{NH3} \) is:
Ethyl chloride (\( \ce{CH3CH2Cl} \)) with excess \( \ce{NH3} \) undergoes ammonolysis to form ethylamine (\( \ce{CH3CH2NH2} \)), as per the PDF’s preparation methods.
The product of aniline reacting with \( \ce{Br2} \) in acetic acid is:
Aniline (\( \ce{C6H5NH2} \)) with \( \ce{Br2} \) in acetic acid (less reactive medium) forms p-bromoaniline predominantly due to the ortho-para directing effect of \( \ce{-NH2} \), as per the PDF.
Which amine produces \( \ce{CH3CH2CH2NC} \) with \( \ce{CHCl3} \) and alcoholic KOH?
Propan-1-amine (\( \ce{CH3CH2CH2NH2} \)) forms propyl isocyanide (\( \ce{CH3CH2CH2NC} \)) in the carbylamine reaction, as per the PDF.
Which amine has no hydrogen atoms directly attached to nitrogen?
Tertiary amines like \( \ce{(CH3CH2)3N} \) have no N-H bonds, unlike primary and secondary amines, as per the PDF’s classification.
The product of the reaction \( \ce{C6H5NH2 + CH3COCl ->} \) is:
Aniline reacts with acetyl chloride to form acetanilide (\( \ce{C6H5NHCOCH3} \)) via acetylation, releasing HCl.
Which amine cannot form a salt with \( \ce{C6H5SO2Cl} \) in the Hinsberg test?
Tertiary amines like \( \ce{(CH3CH2)3N} \) lack an N-H bond and do not react with \( \ce{C6H5SO2Cl} \) to form a sulphonamide salt, as per the PDF.
The product of \( \ce{C6H5NH2} \) reacting with \( \ce{CH3CH2Cl} \) in the presence of a base is:
Aniline (\( \ce{C6H5NH2} \)) can be alkylated with \( \ce{CH3CH2Cl} \) and a base to form N-ethylaniline (\( \ce{C6H5NHCH2CH3} \)), as per the PDF’s alkylation.
The IUPAC name of \( \ce{CH3CH2NHCH2CH3} \) is:
For secondary amines, the larger alkyl group is the parent. Here, it’s ethanamine with an N-ethyl group, named N-ethylethanamine, as per the PDF.
Which reagent protects the \( \ce{-NH2} \) group of aniline during electrophilic substitution?
Acetic anhydride (\( \ce{(CH3CO)2O} \)) acetylates aniline to acetanilide, reducing reactivity of the \( \ce{-NH2} \) group, as per the PDF’s chemical properties.
The reaction of \( \ce{CH3CH2NH2} \) with \( \ce{CH3CH2COCl} \) yields:
Ethylamine (\( \ce{CH3CH2NH2} \)) reacts with propanoyl chloride (\( \ce{CH3CH2COCl} \)) to form N-ethylpropanamide (\( \ce{CH3CH2CONHCH2CH3} \)), as per the PDF’s acylation.
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