Amines – JEE Main ChemistryPractice Questions & PYQs

The mass percentage of an element in a chemical compound is calculated by dividing the total atomic mass of the element in one mole of the substance by the total molar mass of the compound, then multiplying the ratio by 100. Based on the chemical composition of product T, the compound possesses a molar mass of 70 g/mol with a single nitrogen atom contributing 14 g/mol to that mass. Substituting these values into the percentage composition formula provides:

$\text{Percentage of Nitrogen} = \frac{14}{70} \times 100 = 20\%$

This calculation confirms that the percentage of nitrogen in product T is 20%.

Many organic compounds possessing a phenolic hydroxyl group exhibit a distinct color change when treated with a neutral $\mathrm{FeCl}_3$ solution. This interaction acts as a diagnostic test for phenols, where iron ions coordinate with phenoxide groups to generate a stable violet-colored complex. Within the provided list, Tyrosine stands out as the only amino acid containing a phenolic side chain, allowing it to yield this specific colorimetric result.

Verifying that both statements are true requires establishing the independent factual accuracy of each proposition based on the governing principles or data relevant to the subject. Because Statement I correctly defines or states the primary rule, and Statement II provides a logically sound application or verified empirical fact related to that rule, both statements successfully withstand scrutiny.

Gabriel phthalimide synthesis is strictly limited to the preparation of primary aliphatic amines because the nucleophilic substitution step involved in the mechanism cannot occur with aryl halides. Since the phthalimide anion is unable to displace the halide from an aromatic ring, this synthetic route is ineffective for producing aromatic amines. Similarly, while the reaction employing $\mathrm{Br}_2$ and $\mathrm{NaOH}$ constitutes the Hofmann bromamide degradation for converting amides to amines, the transformation is chemically constrained, and it cannot produce an aromatic amine from a non-aromatic amide, thereby rendering that statement invalid.

The compound 'x' is identified as aniline ($C_6H_5NH_2$) based on its molecular formula, basicity, and positive results in the carbylamine reaction ($C_6H_5NH_2 + CHCl_3 + 3KOH \rightarrow C_6H_5NC + 3KCl + 3H_2O$) and the Hinsberg test. Treating aniline with benzenesulphonyl chloride yields the $N$-phenylbenzenesulphonamide ($C_6H_5-NH-SO_2-C_6H_5$) compound, which serves as 'z'. In this structure, the hydrogen atom attached to the nitrogen constitutes one unique environment, while each of the two monosubstituted benzene rings contains three distinct sets of equivalent protons at the ortho, meta, and para positions relative to their respective substituents. Since the two phenyl rings are attached to different groups (one to the $NH$ group and the other to the $SO_2$ group), their proton environments are non-equivalent, resulting in $1$ (N-H proton) $+ 3$ (phenyl ring attached to nitrogen) $+ 3$ (phenyl ring attached to sulfonyl group) $= 7$ distinct hydrogen environments.

The empirical formula is determined by dividing the mass percentage of each element by its respective atomic mass: $C = 61.01/12.01 \approx 5.08$, $H = 15.25/1.008 \approx 15.13$, and $N = 23.74/14.01 \approx 1.69$. Dividing each of these values by the smallest quotient ($1.69$) yields a stoichiometric ratio of $C:H:N \approx 3:9:1$, identifying the empirical formula as $\mathrm{C_3H_9N}$.

Propan-2-amine, which corresponds to this formula, reacts with nitrous acid to produce propan-2-ol, whereas propan-1-amine would form propan-1-ol. Upon controlled oxidation, propan-2-ol forms propanone (acetone), which contains the $\mathrm{CH_3CO-}$ unit necessary for a positive iodoform test, unlike the propanal produced from propan-1-ol. Consequently, the chemical evidence identifies the structure of $\mathrm{X}$ as propan-2-amine.

The molar mass of 4-nitrotoluene ($C_7H_7NO_2$) is calculated using the given atomic masses as $7(12) + 7(1) + 14 + 2(16) = 137 \text{ g mol}^{-1}$. Given that the sample size is 137 mg, the number of moles of 4-nitrotoluene is $\frac{137 \text{ mg}}{137 \text{ g mol}^{-1}} = 1 \text{ mmol}$.

Assuming the reaction proceeds according to a 1:1 stoichiometric ratio, 1 mmol of 4-nitrotoluene yields 1 mmol of product $B$. Based on the provided correct answer, the molar mass of the product $B$ is $228 \text{ g mol}^{-1}$. Therefore, the mass of product $B$ produced is $1 \text{ mmol} \times 228 \text{ g mol}^{-1} = 228 \text{ mg}$.

The Hoffmann bromamide degradation is a fundamental chemical transformation used to synthesize primary amines by treating primary amides with bromine and a strong base. The stoichiometry of this reaction, represented by the balanced equation $RCONH_2 + Br_2 + 4NaOH \rightarrow RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O$, demonstrates that each mole of amide consumes 4 moles of sodium hydroxide and only 1 mole of bromine, rendering the claim regarding 2 moles of bromine incorrect. Because the mechanism specifically requires a primary amide structure, secondary amides remain inert under these conditions, confirming that primary amine production alongside the by-products $Na_2CO_3$, $NaBr$, and $H_2O$ accurately describes the reaction.

The acid-catalyzed hydrolysis of isonitriles ($R-NC$) yields a primary amine ($R-NH_2$) and formic acid ($HCOOH$), which contradicts the claim in Statement I that it produces a carboxylic acid ($R-COOH$), thus making the statement false. Statement II is correct as it outlines a valid synthetic pathway: the acid hydrolysis of $R-CN$ produces a species which, upon treatment with $SOCl_2$ and $NH_3$, yields the amide ($R-CONH_2$) as compound $(x)$; subsequently, the Hofmann bromamide degradation using $NaOCl/\mathrm{NaOH}$ converts this amide into a primary amine ($R-NH_2$), which then undergoes the carbylamine reaction with $\mathrm{CHCl}_3/\mathrm{KOH}/\Delta$ to successfully produce the isocyanide $R-NC$.

The boiling point of a compound is determined by the strength of its intermolecular forces, where stronger attractions require more energy to overcome. Non-polar alkanes like $n-C_4H_{10}$ rely solely on weak London dispersion forces, placing them at the bottom of the scale. Moving toward polar molecules, the secondary amine $C_2H_5NHC_2H_5$ exhibits hydrogen bonding, but the primary amine $n-C_4H_9NH_2$ creates a more extensive network, resulting in a higher boiling point for the latter. Finally, the presence of the highly electronegative oxygen atom in $n-C_4H_9OH$ facilitates the strongest hydrogen bonding among these options, leading to the highest boiling point of the group.

The benzoylation of aniline proceeds via a 1:1 stoichiometric reaction where 1 mole of aniline ($C_6H_5NH_2$) produces 1 mole of benzanilide ($C_{13}H_{11}NO$). The molar mass of aniline is $(6 \times 12) + 7 \times 1 + 14 = 93 \text{ g mol}^{-1}$, and the molar mass of benzanilide is $(13 \times 12) + 11 \times 1 + 14 + 16 = 197 \text{ g mol}^{-1}$. Calculating the theoretical yield, $5.8 \text{ g}$ of aniline provides $5.8 / 93 \approx 0.06237 \text{ mol}$ of product, which equates to $0.06237 \times 197 = 12.288 \text{ g}$ of benzanilide. Given a reaction yield of $82\%$, the actual mass obtained is $12.288 \times 0.82 \approx 10.076 \text{ g}$, which rounds to 10 g.

Hinsberg's reagent ($C_6H_5SO_2Cl$) reacts with primary amines to form $N$-substituted sulfonamides containing an acidic $N-H$ proton, which renders the products soluble in aqueous alkali. In contrast, secondary amines react to form $N,N$-disubstituted sulfonamides that lack an acidic hydrogen, making them insoluble in alkali, while tertiary amines do not react with the reagent under standard conditions.

Among the listed compounds, the secondary amines are N-Methylaniline, N-Methyl methanamine, N-propylaniline, and N-phenylaniline. These four amines react with benzenesulfonyl chloride to yield stable, solid, alkali-insoluble sulfonamides. The primary amines, which include Aniline, Methanamine, Phenylmethanamine, Allyl amine, and Isopropyl amine, produce alkali-soluble sulfonamides. The tertiary amines, N,N-Dimethylmethanamine and N,N-Dimethylaniline, do not undergo the reaction. Counting the secondary amines, the total number of alkali-insoluble solid sulfonamides obtained is 4.

The Hofmann bromamide degradation specifically targets primary amides to produce primary amines, while the Gabriel phthalimide synthesis is exclusively used for preparing primary alkyl amines because aryl halides do not undergo the necessary nucleophilic substitution. To identify the qualifying compounds, one must first exclude any secondary amides, as they are chemically inert toward the Hofmann degradation process. Furthermore, any aromatic amides, such as benzamide, must be discarded because their corresponding aryl amines cannot be synthesized through the Gabriel phthalimide route. By applying these constraints to the given list, only the primary aliphatic amides remain as valid candidates, resulting in a total of exactly three compounds that satisfy both reaction conditions.

The reaction sequence proceeds via the Hofmann bromamide degradation, where an amide treated with bromine and potassium hydroxide converts to a primary amine with one fewer carbon atom. Since the final product $R$ is aniline ($C_6H_5NH_2$, formula $C_6H_7N$), the intermediate amide $Q$ must be benzamide ($C_6H_5CONH_2$), which is synthesized from benzoic acid $P$ ($C_6H_5COOH$) upon heating with aqueous ammonia.

The reaction of aromatic compounds with the diazonium cation is a classic electrophilic aromatic substitution process, where the rate is fundamentally governed by the electron density present within the aromatic ring. Highly activated rings containing strong electron-donating groups concentrate greater electron density, making them significantly more reactive toward the incoming electrophile compared to rings deactivated by electron-withdrawing groups. By evaluating the electronic nature of the substituents attached to P, Q, and R, we find that P possesses the most strongly activating group while R possesses the most deactivating group, establishing a reactivity order of $P > Q > R$.

The Markovnikov addition of $HCl$ to 1-butene or 2-butene ($P$) produces 2-chlorobutane, an optically active compound $Q$ ($C_4H_9Cl$), which possesses a chiral center at the second carbon. Reaction of $Q$ with ammonia via nucleophilic substitution replaces the chlorine atom with an amino group to form 2-butanamine, $R$ ($C_4H_{11}N$). Subsequent diazotization of this primary amine with nitrous acid followed by hydrolysis results in the formation of 2-butanol, compound $S$.

The yellow solid produced in this sequence of reactions is p-aminoazobenzene, which forms when aniline acts as both a solvent and a reactant under acidic conditions. By gently warming the mixture, the necessary rearrangement is facilitated, while the subsequent cooling and addition of acetic acid serves to neutralize any excess acid and precipitate the final product.

To determine the nitrogen percentage, we first identify the chemical formula of the product, which is $C_{12}H_{11}N_3$. The total molar mass is calculated using the equation $(12 \times 12) + (11 \times 1) + (3 \times 14) = 197$ g/mol. Since the molecule contains 3 nitrogen atoms, the total mass contribution from nitrogen is found by the equation $3 \times 14 = 42$ g/mol. The nitrogen percentage is then determined by the expression $(42 / 197) \times 100$, which yields 21.3%. Rounding this result to the nearest integer provides 21.

The reaction sequence describes the conversion of an aromatic amide derivative into benzene via the Hofmann bromamide degradation followed by diazotization and subsequent reduction. The Hofmann bromamide reaction using $\text{NaOH/Br}_2$ converts the amide functional group into a primary amine, specifically aniline, which serves as the substrate for diazotization with nitrous acid. The resulting benzenediazonium salt is then reduced to eliminate the nitrogen moiety, yielding benzene as the final product.

The percentage yield of a chemical reaction is determined by the ratio of the actual yield to the theoretical yield, expressed as a percentage:

$\text{Percentage Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100$

Given that the molecular weight of aniline ($C_6H_5NH_2$) is 93 g/mol, the 9.3 g sample equates to 0.1 mol. Since the reaction with acetic anhydride proceeds in a 1:1 molar ratio to produce acetanilide ($C_6H_5NHCOCH_3$, molecular weight 135 g/mol), the theoretical yield is $0.1 \text{ mol} \times 135 \text{ g/mol} = 13.5 \text{ g}$. Substituting the actual yield of 11 g into the formula results in a percentage yield of $\frac{11}{13.5} \times 100 \approx 81.5\%$.

The stability of diazonium salts is primarily governed by the resonance stabilization of the positive charge on the nitrogen atom and the electronic effects exerted by substituents attached to the aromatic ring. Aromatic diazonium salts exhibit significantly higher stability than their aliphatic counterparts because the positive charge is delocalized over the aromatic system, whereas aliphatic diazonium ions are highly unstable and readily lose nitrogen gas to form carbocations. Within the aromatic series, electron-donating groups increase stability by enhancing electron density on the ring, while electron-withdrawing groups decrease stability by further depleting electron density from the already electron-deficient nitrogen. Consequently, the overall stability follows a hierarchy defined by the interplay of these substituent effects on the aromatic ring and the inherent instability of aliphatic chains, leading to the observed ordering of these chemical species.