Amines – JEE Main ChemistryPractice Questions & PYQs

The intermediate labeled $A$ is $p$-methoxy benzene diazonium chloride, which functions as the central reactive species throughout the chemical transformation. Within the progression of the reaction, the reagent $X$ corresponds to acetaldehyde, denoted as $CH_3CHO$, while the byproduct $Y$ is identified as hydrochloric acid, $HCl$. Recognizing these specific components is essential for correctly determining the reaction pathway and the chemical nature of each species involved.

The conversion of aniline into phenol relies on the process of diazotization, where primary aromatic amines react with nitrous acid-generated in situ by $\mathrm{NaNO}_{2}$ and $\mathrm{HCl}$-at temperatures between 273-278 K to produce the reactive benzene diazonium chloride intermediate, $\mathrm{C_6H_5N_2^+ Cl^-}$. This intermediate is subsequently transformed into phenol through hydrolysis using $\mathrm{H_2O}$ under heating, completing the chemical pathway.

This reaction sequence relies on transforming a primary aromatic amine into a nitrile, followed by its controlled reduction to an aldehyde. During the first phase, treating the starting amine with $\mathrm{NaNO}_2$ and $\mathrm{HCl}$ at a low temperature of 0-5 °C creates a diazonium intermediate, which is then displaced by the cyanide ion from $\mathrm{KCN}$ to synthesize cyanobenzene. In the second phase, the reduction of this nitrile is achieved using the Stephen's reduction method, where $\mathrm{SnCl}_2$ and $\mathrm{HCl}$ effectively convert the cyano group into the aldehyde functional group, ultimately producing benzaldehyde.

The reaction of benzonitrile with $CH_3MgBr$ followed by hydrolysis yields acetophenone ($C_6H_5COCH_3$), as the Grignard reagent attacks the electrophilic carbon of the nitrile group to form an intermediate imine that subsequently hydrolyzes to the ketone. Acetophenone contains the characteristic acetyl group ($CH_3CO-$) required to produce a yellow precipitate of iodoform ($CHI_3$) upon treatment with iodine and an alkali in the iodoform test. This compound does not react with Schiff's reagent or Tollens' reagent, which are specific to aldehydes, nor does it yield a positive ninhydrin test, which is specific to amino acids.

Rhumann's purple is the resonance-stabilized chromophore resulting from the condensation of ammonia, generated by the oxidative deamination and decarboxylation of amino acids by ninhydrin, with two additional ninhydrin molecules. The resulting structure consists of two indane-1,3-dione rings connected through a central imine linkage, an arrangement which creates an extended conjugated system responsible for the compound's characteristic intense purple color. This specific molecular geometry, involving the nitrogen-bridged tricyclic system, confirms the accuracy of structure D.

The Sandmeyer reaction facilitates the replacement of the diazonium group in an arenediazonium salt with a halogen atom through a radical mechanism initiated by copper(I) halides. In the presence of $CuCl$ and $HCl$, the diazonium group is displaced by a chloride ion with the evolution of nitrogen gas to form chlorobenzene. This process yields the specific aryl halide structure identified in option D.

The formation of a cyanohydrin from a carbonyl compound necessitates an alkaline medium where X represents $\mathrm{NaOH}$, as it generates the reactive $\mathrm{CN}^{-}$ nucleophile via the equilibrium $\mathrm{HCN} + \mathrm{OH}^{-} \rightleftharpoons \mathrm{CN}^{-} + \mathrm{H}_{2}\mathrm{O}$. To complete the sequence, the resulting nitrile group is reduced to a primary amine using Y, which corresponds to the strong reducing agent $\mathrm{LiAlH}_{4}$ that effectively transforms $-\mathrm{C}\equiv\mathrm{N}$ into $-\mathrm{CH}_{2}-\mathrm{NH}_{2}$.

In a strongly acidic medium, aniline undergoes protonation to yield the anilinium ion, represented as $C_6H_5-NH_3^+$. This transformation significantly alters the electronic nature of the ring because the $-NH_3^+$ group functions as an electron-withdrawing substituent with a distinct $-I$ effect. Consequently, the substituted ring becomes meta-directing, which ensures that the electrophilic nitration reaction occurs effectively to produce the meta-nitroaniline isomer as the major product.

Basic strength in nitrogenous compounds is fundamentally dictated by the availability of the nitrogen lone pair, as this electron density is what allows the species to act as a proton acceptor. When this lone pair engages in resonance, or delocalization, with neighboring electron-withdrawing groups, its availability drops significantly, which directly reduces the basicity of the molecule. While alkylamines such as $(\mathrm{C}_2\mathrm{H}_5)_2\mathrm{NH}$ and $(\mathrm{C}_2\mathrm{H}_5)_3\mathrm{N}$ are relatively strong bases due to the inductive electron-donating effect of their alkyl groups, the structure $(\mathrm{CH}_3\mathrm{CO})_2 \mathrm{NH}$ presents a stark contrast. In this molecule, the nitrogen lone pair is simultaneously delocalized across two carbonyl groups, resulting in the most severe depletion of electron density on the nitrogen and making it the least basic species among those listed.

In the strongly acidic conditions required for the nitration of aniline, the amino group becomes protonated to form the $\mathrm{NH}_3^+$ ion, which acts as a meta-directing group. Despite the influence of this group, steric and electronic factors lead to the formation of 51% para-nitroaniline, 47% meta-nitroaniline, and only 2% ortho-nitroaniline. Consequently, the relative percentage yields follow the order of 51% > 47% > 2%, corresponding to the sequence C > B > A.

The ammonolysis of alkyl halides is a nucleophilic substitution process that releases hydrogen halide ($HX$) as a byproduct during each successive alkylation step. Because this byproduct is acidic, it can react with the resulting primary, secondary, or tertiary amines to form stable ammonium salts, which interferes with the intended reaction pathway. Introducing $\mathrm{NaOH}$ into the mixture serves the crucial role of neutralizing these acidic impurities, thereby facilitating the continuous formation of the desired free amines.

The relative basicity of nitrogen-containing compounds depends on the availability of the nitrogen lone pair to accept a proton, which is primarily determined by the extent of electron delocalization. While Statement II accurately describes how the lone pair in aniline participates in resonance with the benzene ring, Statement I is incorrect because aniline is actually more basic than acetamide. This difference occurs because in acetamide, the nitrogen lone pair is strongly delocalized toward the highly electronegative oxygen atom, which suppresses its basicity much more effectively than the resonance stabilization observed in aniline.

Aniline reacts with acetic anhydride through a nucleophilic substitution process that replaces a hydrogen on the amino group with an acetyl group, yielding acetanilide as the intermediate product. During the subsequent bromination with $Br_2$ in acetic acid, the $-\mathrm{NHCOCH}_3$ substituent acts as an ortho/para directing group because the nitrogen lone pair activates the aromatic ring. Steric hindrance at the ortho position makes the para-substituted isomer, p-bromoacetanilide, the major product of the reaction.

The synthesis of $m$-dibromobenzene from nitrobenzene depends on exploiting the meta-directing influence of the nitro group followed by functional group interconversion via a diazonium salt. Since the $-\mathrm{NO}_2$ group is a meta-directing substituent that decreases electron density at the ortho and para positions, the first step is an electrophilic aromatic substitution using $\mathrm{Br}_2$ and $\mathrm{Fe}$ to introduce the first bromine atom at the meta position. To install the second bromine atom, the nitro group is reduced to an amino group using $\mathrm{Sn}$ and $\mathrm{HCl}$, which yields $m$-bromoaniline. This amino group is subsequently converted into a diazonium salt through reaction with $\mathrm{NaNO}_2$ and $\mathrm{HCl}$. Finally, the diazonium group is replaced by a bromine atom using $\mathrm{CuBr}$ and $\mathrm{HBr}$, completing the transformation into $m$-dibromobenzene.

The acidic hydrolysis of an ester involves breaking the ester linkage through reaction with water, facilitated by an acid catalyst such as $\mathrm{H}_2\mathrm{SO}_4$. This process effectively converts the ester $\mathrm{CH}_3\mathrm{COOC}_2\mathrm{H}_5$ into its corresponding alcohol, $\mathrm{C}_2\mathrm{H}_5\mathrm{OH}$. In the case of reducing an ester to an aldehyde, $\mathrm{DIBAL-H} / \mathrm{H}_2\mathrm{O}$ is the appropriate reagent because it selectively halts the reduction process at the aldehyde stage rather than reducing the molecule further to an alcohol.

The Stephen reduction is a specific method for converting nitriles into aldehydes. In this reaction, $\mathrm{SnCl}_2$ and $\mathrm{HCl}$ are employed to reduce the nitrile $\mathrm{CH}_3\mathrm{C}\equiv\mathrm{N}$ to an iminium salt, which is then quenched with $\mathrm{H}_2\mathrm{O}$ to release the aldehyde $\mathrm{CH}_3\mathrm{CHO}$. Finally, Grignard reagents such as $\mathrm{CH}_3\mathrm{MgBr}$ act as strong nucleophiles toward the electrophilic carbon of a nitrile, forming an intermediate that, when hydrolyzed with $\mathrm{H}_3\mathrm{O}^{+}$, results in a ketone.

The Hinsberg test utilizes benzene sulphonyl chloride to classify amines based on the solubility of the resulting sulphonamide in an alkaline medium. Primary amines react with this reagent to form an N-substituted sulphonamide that possesses an acidic hydrogen atom attached to the nitrogen, allowing the product to dissolve in dilute $\mathrm{NaOH}$. In contrast, secondary amines form sulphonamides that lack this acidic hydrogen and are therefore insoluble in alkali, while tertiary amines do not undergo this reaction. Identifying the structure $\mathrm{C}_6\mathrm{H}_5-\mathrm{CH}(\mathrm{CH}_3)-\mathrm{NH}_2$ as the only primary amine among the given options explains why it is the specific compound that yields a product soluble in dilute $\mathrm{NaOH}$.

The reaction commences with the Grignard reagent acting upon the keto and cyanide groups, where subsequent hydrolysis generates an alcohol from the former and yields a keto group from the latter. When this intermediate is heated in the presence of $\mathrm{H}_{2}\mathrm{SO}_{4}$, the hydroxyl group undergoes acid-catalyzed dehydration to eliminate water, resulting in the formation of an alkene as the major product.

The reaction sequence initiates with the interaction between ammonia and ethanoyl chloride ($CH_3COCl$) to form ethanamide ($CH_3CONH_2$) and hydrogen chloride ($HCl$). This amide subsequently undergoes Hoffmann's bromamide degradation to produce a primary amine, which is further processed via diazotization to create an unstable diazonium salt intermediate. This intermediate quickly decomposes to yield an alcohol as the major product, alongside smaller quantities of alkene and alkyl halide byproducts.

The basicity of amines is primarily determined by the availability of the nitrogen atom's lone pair, which is far more accessible in aliphatic compounds where the lone pair is not involved in resonance. Phenyl methanamine is effectively an aliphatic amine because the nitrogen is attached to an $sp^3$ hybridized carbon, preventing its lone pair from delocalizing into the benzene ring and making it the most basic in the set. Among the remaining aromatic amines, the basic strength increases with the addition of electron-donating alkyl groups that partially offset the electron-withdrawing resonance effect of the ring; thus, N,N-dimethylaniline with two methyl groups is more basic than N-methyl aniline with one, while benzenamine remains the least basic due to the absence of any stabilizing inductive groups.

The nitration of aniline occurs in an acidic medium composed of concentrated $HNO_3$ and $H_2SO_4$, where the basic amino group readily gains a proton to form the anilinium ion. This resulting $-NH_3^+$ group is strongly electron-withdrawing and acts as a meta-directing substituent on the benzene ring. Since this protonation occurs extensively in the presence of strong acids, the reaction mixture produces a significant amount of the meta-nitro product along with the expected ortho and para isomers.