Amines practice · Medium

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

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

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

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.

Option B (Gattermann reaction) is correct because the reaction of arenediazonium chloride with Cu/HCl to form aryl chloride is known as the Gattermann reaction.

Option D (All the above) is correct because primary, secondary, and tertiary amines can all react with $NaN{O_2} + HCl$ to form alcohols or related products.

Option A (Liebermann’s nitroso test) is correct because it is specifically used to detect secondary amines, forming a green solution that turns blue upon addition of acid.

Option A (A, B, C) is correct because primary aliphatic amines, urea, and amides react with $HCl + NaN{O_2}$ to liberate $N_2$ gas.

Option C (Hofmann) is correct because the Hofmann degradation reaction converts amides into amines with one less carbon atom.

Option D ($\mathrm{CH_3NH_2}$) is correct because methylamine is soluble in dilute HCl, reacts with NaNO2 to release nitrogen, and the resulting solution smells of wood spirit (methanol).

Option A (In acidic (strong) medium aniline is present as anilinium ion) is correct because in a strongly acidic medium, aniline is protonated to form the anilinium ion, which is meta-directing due to the positive charge on the nitrogen.

Option C (Wöhler’s synthesis of urea from NH4CNO) is correct because it disproved the vital force theory by synthesizing an organic compound (urea) from inorganic starting materials (ammonium cyanate).