Amines practice · Medium

The correct answer is B because the electronegativity of nitrogen is 3.0 and oxygen is 3.5.

Option D (All of these) is correct because the basic nature of amines in aqueous medium is influenced by inductive effect, solvation effect, and steric hindrance.

Option B is correct because aniline reacts with acetyl chloride in the presence of NaOH to produce acetanilide.

Option B (Sandmeyer’s reaction) is correct because the given reaction, where ${C_6}{H_5}{N_2}Cl$ is converted to ${C_6}{H_5}Cl$ using $C{u_2}C{l_2}/HCl$, is the Sandmeyer's reaction.

Option B ($\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{NH}_2$) is correct because the cyanide group is reduced to a primary amine by the reducing agent.

Option D (Phenyl isocyanide) is correct because isocyanides on hydrolysis form primary amines, not ammonia.

Option A (${CH}_3 {CN}$) is correct because reduction of ${CH}_3 {CN}$ gives ethylamine, which upon treatment with $HNO_2$ gives ethanol.

Option C (${Ph}-{NH}_2+{CHCl}_3+3 {KOHO} \text{(alc.)} \stackrel{\Delta}{\longrightarrow} {Ph}-{NC}+3 {KCl}+3 {H}_2 {O}$) is correct because it represents the carbylamine reaction where a primary amine reacts with chloroform and a base to form an isocyanide.

Option A is correct because aniline is stabilized by resonance due to the delocalization of the lone pair of electrons on the nitrogen atom into the benzene ring, resulting in more resonating structures compared to the anilinium ion.

Option C (${\text{C}}{{\text{H}}_3}{\text{N}}\mathop {\text{ = }}\limits^ \to {\text{C}},:{\text{CC}}{{\text{l}}_2}$) is correct because the reaction is the carbylamine reaction, and dichlorocarbene is the intermediate.

Option C is correct because the Biuret test detects the presence of peptide bonds, which are characteristic of amides and proteins.

The correct answer is Ethanamine. The reaction sequence involves dehydration of acetamide to ethanenitrile, followed by reduction to ethanamine.

Option C is correct because the final product 'D' is propane, which has two types of chemically distinct hydrogens, leading to two monochloro derivatives.

Option B (Urea nitrate) is correct because urea reacts with nitric acid to form urea nitrate, which is a salt.

Option B (Acetanilide) is correct because aniline reacts with glacial acetic acid in the presence of anhydrous $ZnCl_2$ to form acetanilide, which is an acetylation reaction.

Option D (In purification of water) is correct because urea is primarily used as a fertilizer, in plastic manufacturing, and in medicines, but not for water purification.

Option C (2,4,6-tribromoaniline) is correct because aniline is highly reactive and undergoes tribromination at the ortho and para positions when treated with bromine water.

Option B is correct. Aniline does form meta derivatives during nitration, and the amino group is activating, but the reason isn't the direct explanation for meta substitution.

Option C is correct because diethylamine $(\mathrm{C}_2 \mathrm{H}_5)_2 \mathrm{NH}$ is the most basic among the given options in aqueous medium due to a combination of inductive effect and solvation effects.

Option C is correct because the assertion is correct (tertiary amines have lower boiling points), but the reason is wrong (tertiary amines cannot form intermolecular hydrogen bonds).

Option A (${(C{H_3})_2}NH$) is correct because secondary aliphatic amines generally have the smallest $pK_b$ value (highest basicity) in aqueous solutions due to a combination of inductive effects and solvation.

The correct answer is B (Mullikan barker test) because the Mullikan-Barker test is specifically used to detect the presence of a nitro group in organic compounds like nitrobenzene.