Amines practice · Mixed difficulty

Option C (N-Methyl aniline) is correct because it reacts with Hinsberg's reagent to form a sulfonamide that has one acidic hydrogen, which reacts with NaOH to become soluble. N,N-dimethyl aniline does not have any acidic hydrogen.

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.

Option D ($p -$ methyl benzylamine) is correct because the carbylamine test is specific for primary amines. $p-$methyl benzylamine is a primary amine, while the other options are secondary or tertiary amines.

Option B (Colourless liquid) is correct because pure aniline is a colorless liquid that turns brown upon exposure to air and light due to oxidation.

Option A (Aromatic nature) is correct because pyridine follows Huckel's rule and has a cyclic, planar structure with a delocalized pi electron system.

Option A ((${\left( {\mathrm{C}{\mathrm{H}_2}} \right)_6}{\mathrm{N}_4}$)) is correct because the chemical formula for urotropine (hexamethylenetetramine) is (CH2)6N4.

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.

Option D (Dyes) is correct because sulfonic acids are used as intermediates in the synthesis of many dyes due to their ability to impart water solubility and reactivity.

Option B (Acetic acid) is correct because the hydrolysis of methyl cyanide (acetonitrile) yields acetic acid.

Option B (2 – aminobutane) is correct because it correctly identifies the longest carbon chain containing the amine group and the position of the amine substituent.

Option A ($\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}$) is correct because primary amines react with Hinsberg's reagent to form a sulfonamide that has an acidic hydrogen and is therefore soluble in alkali.

Option B (Diazonium salt) is correct because Sandmeyer's reaction involves the replacement of the diazonium group in diazonium salts with other substituents.

Option D (Aniline) is correct because the reaction with chloroform and alcoholic KOH is the carbylamine reaction, which is a test for primary amines like aniline.

Option B (aniline) is correct because the lone pair of electrons on the nitrogen atom in aniline is delocalized into the benzene ring, making it less available for protonation and thus less basic.

Option C (biuret) is correct because heating urea slowly leads to the formation of biuret through a condensation reaction.

Option A (Diazotisation followed by coupling with phenol) is correct because aniline undergoes diazotization to form a diazonium salt, which then couples with phenol to form a colored dye, while methylamine does not form a stable diazonium salt under the same conditions.

Option A (Hydrazine) is correct because hydrazine (N2H4) does not contain carbon, which is required for the formation of cyanide ions in the Lassaigne's test.

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

Option B (Dyes) is correct because diazo-coupling reactions are commonly used in the synthesis of azo dyes, which are widely used as coloring agents.

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.

Option D (Ethylamine and aniline both react with $\mathrm{NaN}{\mathrm{O}_2} + \mathrm{HCl}$ to give hydroxyl compounds in cold) is not correct because while ethylamine reacts to form an alcohol, aniline forms a diazonium salt which requires different conditions to form a hydroxyl compound.

Option C (${\mathrm{C}_n}{\mathrm{H}_{2n + 3}}\mathrm{N}$) is correct because it represents the general formula for aliphatic amines, where n is the number of carbon atoms.

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

Option D (It does not cause acidity in the soil) is correct because the hydrolysis of urea in the soil does not produce acidic products, unlike ammonium sulfate which can lead to soil acidification.

Option B ($R\mathrm{C}{\mathrm{H}_2}\mathrm{N}{\mathrm{H}_2}$) is correct because the reaction sequence involves converting an alkyl halide to an alkyl cyanide, followed by reduction to a primary amine with one additional carbon.

Option C (Alkaline sodium arsenite) is correct because nitrobenzene is reduced to nitrosobenzene using alkaline sodium arsenite.

Option B (o-aminotoluene) is correct because it is not commonly used as an explosive, unlike trinitroglycerin, dynamite, and TNT.

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.

Option B ($\mathrm{AgN}{\mathrm{O}_3}$) is correct because anilium hydrochloride will react with silver nitrate to form a precipitate of silver chloride (AgCl), while p-chloroaniline will not directly produce chloride ions in solution.

Option B (${\left( {{C_2}{H_5}} \right)_2}NH$) is the most basic in aqueous medium because of the combined effect of +I effect of ethyl groups and solvation.