Amines – JEE Main ChemistryPractice Questions & PYQs

In a molecule of $N(CH_2CH_3)_3$, the nitrogen atom adopts a pyramidal arrangement dictated by the lone pair of electrons and the three surrounding ethyl groups. Although standard pyramidal amines typically exhibit bond angles near 107 degrees, the increased steric hindrance from the bulkier ethyl substituents forces these groups further apart to minimize repulsion. Consequently, the observed C-N-C bond angle in gaseous triethylamine widens slightly to 108 degree.

Gabriel phthalimide synthesis is a specialized method used exclusively for synthesizing $1^\circ$ aliphatic or alicyclic amines, as it relies on a nucleophilic substitution reaction between potassium phthalimide and an alkyl halide. Successful preparation requires the presence of an $\alpha$-carbon attached to the halide group, which inherently rules out the synthesis of aromatic amines like aniline $(\mathrm{C}_{6}\mathrm{H}_{5}\mathrm{NH}_{2})$ due to their lack of a suitable aliphatic site. Because three of the four candidates provided contain the necessary $\alpha$-carbon structural feature, they can be synthesized using this method, whereas the aromatic compound cannot.

The Hinsberg test distinguishes primary, secondary, and tertiary amines by treating them with para-toluene sulphonyl chloride to observe the solubility of the resulting sulfonamide products in $NaOH$. Primary amines form a precipitate that dissolves in $NaOH$ because they retain an acidic hydrogen, whereas secondary amines create a precipitate that remains insoluble in the basic solution. Tertiary amines do not react with the reagent, allowing for the clear separation of all three types based on these distinct chemical characteristics.

The Gabriel phthalimide synthesis relies on a nucleophilic substitution reaction where the phthalimide anion acts as a nucleophile to displace a halide group, typically from an alkyl halide, $R-X$. This method is ineffective for synthesizing aromatic primary amines because aryl halides do not readily undergo nucleophilic substitution, largely due to the partial double bond character of the carbon-halogen bond caused by resonance and the electronic repulsion between the nucleophile and the electron-rich aromatic ring. Because the required nucleophilic displacement cannot occur when using an aryl halide, the synthesis cannot produce the intermediate product necessary for forming aromatic primary amines.

The oxidation of aniline using potassium dichromate ($\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7$) is driven by the release of nascent oxygen, which acts as a potent oxidizing agent in the reaction mixture. Through this mechanism, the aniline molecule undergoes a structural transformation to yield 1,4-benzoquinone, while simultaneously removing the nitrogen group as an ammonium ion ($\mathrm{NH}_4^+$). This specific reaction highlights the susceptibility of aromatic amines to strong oxidizing conditions, where the end result is consistently the formation of a quinone derivative.

The reaction involves the selective reduction of a nitrile, $\mathrm{R}-\mathrm{CN}$, using diisobutylaluminium hydride (DIBAL-H), which is specifically chosen for its ability to stop the reduction process before a primary amine is formed. During the first stage, the reagent attacks the electrophilic carbon atom to create an imine intermediate. When water is added in the subsequent hydrolysis step, this intermediate is converted into an aldehyde, represented as $\mathrm{R}-\mathrm{CHO}$.

Aniline reacts with sodium nitrite and ${HCl}$ in cold condition to form benzene diazonium chloride $(X)$ and on further reaction, it form azo dye $(Y)$. Chemical reactions are as follows

The formation of a semicarbazone results from the condensation reaction between acetone and semicarbazide, which possesses the molecular structure $NH_2NHCONH_2$. When these reactants combine, the product formed is $(CH_3)_2C=NNHCONH_2$, where the original carbonyl oxygen is replaced. By examining this final structure, one can identify one nitrogen atom involved in the double bond at the central carbon and two additional nitrogen atoms within the amide-like side chain, yielding a total of 3 nitrogen atoms per molecule.

The Hoffmann bromamide degradation is a specific chemical transformation used to convert primary amides into amines with one fewer carbon atom by reacting them with bromine and an aqueous base. While pathways originating from benzamide, nitriles, or acid chlorides all facilitate this degradation by generating an amide intermediate, the reaction starting from a methyl ketone follows an entirely different mechanism. In the case of a methyl ketone, a haloform reaction produces a carboxylic acid salt, which forms an amide upon heating with ${NH}_{3}$ and is subsequently reduced to an amine using ${LiAlH}_{4}$ rather than undergoing Hoffmann bromamide degradation.

Hinsberg reagent, benzene sulfonyl chloride ($C_6H_5SO_2Cl$), reacts exclusively with primary and secondary amines to form sulfonamides because these molecules possess at least one hydrogen atom attached to the nitrogen atom, allowing for the elimination of hydrogen chloride. Tertiary amines lack any hydrogen atom bonded to the nitrogen, which makes them incapable of undergoing the condensation reaction required to form a sulfonamide. Since the structure in option B is a tertiary amine, it does not react with the Hinsberg reagent.

The Hofmann bromamide degradation serves as a method to convert primary amides into primary amines by eliminating the carbonyl carbon atom. During this reaction, the carbonyl moiety undergoes a sequence of steps including rearrangement and hydrolysis, which results in the expulsion of the carbon as a carbonate ion, $\mathrm{CO}_3^{2-}$. This transformation is represented by the ionic equation $\mathrm{RCONH}_2 + 4\mathrm{BrO}^{-} + 2\mathrm{H}^{+} \longrightarrow \mathrm{RNH}_2 + \mathrm{CO}_3^{2-} + 4\mathrm{Br}^{-} + \mathrm{OH}^{-}$.

Electrophilic addition reactions to asymmetric alkenes are governed by Markovnikov's rule, which dictates that the nucleophile attaches to the more substituted carbon atom to form a more stable carbocation intermediate. This mechanism proceeds through a rate-determining step involving the formation of the most stable carbocation, stabilized by hyperconjugation and inductive effects from adjacent alkyl groups, represented generally as $R-C^{+} + Nu^{-} \rightarrow R-C-Nu$. Consequently, the product "B" arises from the selective attack of the nucleophile on this highly stable cationic center, minimizing the activation energy and ensuring the structural stability of the resulting compound.

The stability of diazonium salts relies heavily on the resonance stabilization provided by an attached aromatic ring, which makes aryl diazonium species uniquely persistent at low temperatures. Conversely, primary aliphatic amines reacting with $\mathrm{NaNO}_2$ and $\mathrm{HCl}$ fail to yield stable diazonium salts, instead decomposing to form products like N-alkylnitroso amines. Consequently, the aromatic structure found in option B allows for the formation of the most stable diazonium salt, as it avoids the inherent instability characteristic of aliphatic diazonium intermediates.

The reaction of a primary amine with chloroform ($CHCl_3$) and alcoholic potassium hydroxide (KOH) is the Carbylamine reaction, which yields an alkyl isocyanide (isonitrile) as the final product. These isocyanide compounds are highly toxic and possess a characteristic foul odor, but they can be decomposed via hydrolysis in the presence of a strong mineral acid such as concentrated hydrochloric acid ($HCl$). Consequently, compound A is a primary amine, compound B is an isonitrile, and compound C is concentrated hydrochloric acid.

The selectivity of this reaction is determined primarily by the relative basicity of the amine groups within the molecule. Since the $-\mathrm{CH}_2 - \mathrm{NH}_2$ group is inherently more basic than other nucleophilic sites, the reagent preferentially targets this position, leading exclusively to imide formation at the primary amine. This transformation halts at the amide stage because the resulting product possesses significantly reduced basicity, preventing any further reaction from occurring.

Zaitsev's rule dictates that in elimination reactions, the most substituted alkene is the major product because it possesses the lowest activation energy and represents the most thermodynamically stable configuration due to hyperconjugation. During the reaction, the elimination of the leaving group followed by the removal of a proton from the most hindered $\beta$-carbon yields the most highly substituted double bond, $R_2C=CR_2$, which corresponds to the structure of product P.

Hypophosphorous acid ({H}{3}{PO}{2}$) serves as a specific reducing agent used \in the transformation of aromatic diazonium salts into their corresponding arenes, facilitating the removal of the diazonium group via a radical mechanism that releases nitrogen gas. In the context of alkyl synthesis, chloroethane ({CH}_{3}{CH}_{2}{Cl}$) is commonly produced through the nucleophilic substitution of ethanol ({CH}{3}{CH}{2}{OH}$) using reagents such as phosphorus pentachloride or thionyl chloride, which replaces the hydroxyl moiety with a chlorine atom. These reagents represent standard chemical components utilized in functional group interconversion and reduction steps frequently observed in organic reaction pathways.

The reduction of a standard nitrile with $\mathrm{LiAlH}_4$ occurs via the nucleophilic attack of a hydride ion on the electrophilic carbon to form an imine anion, which subsequently undergoes hydrolysis to produce a primary amine as product A. In contrast, the specific acidic hydrolysis of 2-hydroxy-2-methylpropionitrile converts the cyano group into a carboxylic acid, resulting in the intermediate 2-hydroxy-2-methylpropionic acid. Treating this intermediate with $\mathrm{H}_2\mathrm{SO}_4$ promotes dehydration, whereby the elimination of a water molecule generates 2-methylacrylic acid as the final product (B).

The synthesis of $p$-aminoazobenzene relies on the coupling reaction between a diazonium salt and an aromatic amine, a process that strictly requires an acidic medium to proceed effectively. While reaction pathways $A$ and $C$ provide the necessary acidic environment for this coupling to occur, the combination of ${NaBH}_{4}$, ${NaOH}$, and aniline creates a reducing and basic environment that inhibits this transformation. Therefore, this specific set of reagents is unable to facilitate the formation of the desired product.

Friedel-Crafts alkylation of aniline fails because the lone pair on the nitrogen atom makes aniline a potent Lewis base, which immediately undergoes an acid-base reaction with the ${AlCl}_{3}$ catalyst to form the anilinium ion. Because the nitrogen atom coordinates with the Lewis acid, it essentially creates a strongly electron-withdrawing group on the benzene ring, causing severe deactivation of the aromatic system. This significant reduction in electron density renders the ring unreactive toward the electrophilic conditions required for Friedel-Crafts alkylation, meaning the reaction does not proceed.