Correct answer Carries: 4.
Wrong Answer Carries: -1.
An amine \( X \) forms \( \ce{CH3CH2CH2OH} \) with \( \ce{HNO2} \) and \( \ce{CH3CH2CH2NC} \) with \( \ce{CHCl3} \) + KOH. \( X \) is:
Propan-1-amine (\( \ce{CH3CH2CH2NH2} \)) reacts with \( \ce{HNO2} \) to form propan-1-ol and with \( \ce{CHCl3} \) + KOH to form propyl isocyanide, as per the PDF.
The product of \( \ce{CH3CH2CH2NH2} \) with \( \ce{HNO2} \) at 273-278 K is:
Propan-1-amine (\( \ce{CH3CH2CH2NH2} \)) reacts with \( \ce{HNO2} \) to form propan-1-ol (\( \ce{CH3CH2CH2OH} \)) and \( \ce{N2} \), as per the PDF’s chemical properties.
The number of hydrogen bonds \( \ce{(CH3)2NH} \) can form with water is:
Dimethylamine (\( \ce{(CH3)2NH} \)), a secondary amine, has one N-H bond to donate and one lone pair to accept, forming two hydrogen bonds, as per the PDF’s solubility.
Which amine forms a sulphonamide soluble in alkali with benzenesulphonyl chloride?
Primary amines like \( \ce{CH3CH2CH2NH2} \) form sulphonamides with an acidic hydrogen, soluble in alkali, as per the PDF’s Hinsberg test.
How many moles of \( \ce{Br2} \) react with one mole of \( \ce{C6H5NH2} \) in excess bromine water?
Aniline reacts with excess \( \ce{Br2} \) in water to form 2,4,6-tribromoaniline, requiring 3 moles of \( \ce{Br2} \) for three substitutions, as per the PDF.
The product of \( \ce{C6H5CH2Cl} \) reacting with excess \( \ce{NH3} \) is:
Benzyl chloride (\( \ce{C6H5CH2Cl} \)) with excess \( \ce{NH3} \) forms benzylamine (\( \ce{C6H5CH2NH2} \)) via ammonolysis, as per the PDF’s preparation methods.
Which property of \( \ce{-NH2} \) in aniline directs electrophilic substitution to ortho and para positions?
The \( \ce{-NH2} \) group’s resonance effect donates electrons to the ring, activating ortho and para positions, as per the PDF’s chemical properties.
Which amine forms a precipitate with benzenesulphonyl chloride that is insoluble in excess NaOH?
Secondary amines like \( \ce{CH3NHCH2CH3} \) form sulphonamides insoluble in alkali, unlike primary amines, as per the PDF’s Hinsberg test.
Which amine is produced from \( \ce{CH3CH2CH2CH2CONH2} \) via Hoffmann bromamide reaction?
Pentanamide (\( \ce{CH3CH2CH2CH2CONH2} \)) loses one carbon in the Hoffmann reaction to form butan-1-amine (\( \ce{CH3CH2CH2CH2NH2} \)), as per the PDF.
Which amine does not form a nitrosamine with \( \ce{HNO2} \)?
Tertiary amines like \( \ce{(CH3CH2)3N} \) do not form nitrosamines with \( \ce{HNO2} \), unlike secondary amines, as per the PDF’s chemical properties.
Are you sure you want to submit your answers?
