Amines practice · Mixed difficulty

Option A ($\mathrm{C}{\mathrm{H}_3}\mathrm{N}{\mathrm{H}_2}$) is correct because primary amines react with Hinsberg reagent to form N-substituted sulfonamides, which are soluble in alkali.

Option C (glycine and aminocaproic acid) is correct because nylon-2-nylon-6, a biodegradable polyamide, is made from these two monomers.

Option B is correct because anilinium hydrochloride will react with $\mathrm{AgN}{\mathrm{O}_3}$ to produce a white precipitate of AgCl, while p-chloroaniline will not.

The correct answer is A because amines are more basic than esters and ethers due to nitrogen being less electronegative than oxygen, making it better at accommodating the positive charge on the proton.

Option D ((C2H5)3N) is correct because tertiary amines lack a hydrogen atom directly attached to the nitrogen, preventing them from reacting with benzenesulfonyl chloride.

Option C (Mendius reduction) is correct because it specifically refers to the reduction of nitriles (like CH3CN) to primary amines (like CH3CH2NH2).

Option A (Both statement I and II are correct) is correct because primary and secondary amines can donate and accept hydrogen bonds, and low molecular weight amines are soluble in water due to their polarity.

Option B (Decreases) is correct because conductivity is the conductance of a unit volume of solution. Upon dilution, the number of ions per unit volume decreases, hence conductivity decreases.

Option C (Almost neutral) is correct because an aqueous solution of urea is nearly neutral, although it can act as a weak monoacid-base.

Option A (p-Phenylenediamine) is correct because the presence of two electron-donating amino groups in the para position increases the electron density on the nitrogen atoms, making it the most basic.

Option D (If both Assertion & Reason are False) is correct because a 2px-2py transition does not produce a spectral line, and not only Balmer lines are observed in the visible region.

Option B (${N_2}$) is correct because methyl amine reacts with nitrous acid to form methyl diazonium ion, which decomposes to give methanol and nitrogen gas.

Option B (${R}-\stackrel{+}{{N}} \equiv {C}^{-}$) is correct because N-alkyl formamides, when dehydrated with ${POCl}_3$ in the presence of pyridine, give isocyanides.

Option C (Both (1) and (2)) is correct because the basic character of an amine depends on the ease of cation formation and the stability of the resulting cation.

Option C is correct. The assertion that Gabriel phthalimide synthesis can be used to convert R-Cl to pure R-NH2 is true, but the reason that it can be used to prepare 1°, 2°, or 3° amines is false. It is only suitable for preparing primary amines.

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.

The correct answer is A (Ar-F) because the reaction sequence involves the formation of an aryl diazonium tetrafluoroborate salt, followed by thermal decomposition to yield an aryl fluoride.

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.