Organic Compounds Containing Nitrogen MCQ

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    mcq on Organic Compounds Containing Nitrogen for NEET Students

    Which of the following compounds does not participate in azo coupling with benzene diazonium chloride?

    a) Aniline

    b) Phenol

    c) Anisole

    d) Nitrobenzene

    Explanation:
    Azo coupling is a reaction where a diazonium ion reacts with an activated aromatic ring to form azo compounds. This reaction requires the aromatic system to be rich in electron density so that it can easily undergo electrophilic substitution. The presence of electron-donating groups enhances the reactivity of the ring by increasing electron density through resonance or inductive effects, especially at ortho and para positions.

    When substituents strongly withdraw electrons from the aromatic ring, the electron density decreases significantly, making the ring less reactive toward electrophilic attack. Such deactivation reduces or prevents the formation of azo coupling products under normal conditions. The reaction is highly dependent on how substituents influence the stability of the intermediate sigma complex formed during substitution.

    In general, aromatic compounds with strong activating groups readily participate in coupling reactions, while those with strong deactivating groups fail to react efficiently. The key is understanding how substitution affects resonance stability and nucleophilicity of the benzene ring. The compound that strongly reduces ring activation will not undergo azo coupling effectively with diazonium Salts.

    Option d – Nitrobenzene

    When t-butyl bromide is treated with alcoholic ammonia, the reaction it undergoes is:

    a) Nucleophilic substitution

    b) Elimination

    c) Addition

    d) Electrophilic substitution

    Explanation:
    Tertiary alkyl halides behave differently from primary and secondary halides due to steric hindrance and carbocation stability. In a polar protic or alcoholic medium, the leaving group can detach more easily, leading to the formation of a relatively stable tertiary carbocation intermediate. This intermediate plays a key role in determining the reaction pathway.

    Once the carbocation is formed, nucleophiles or Bases present in the medium can interact with it. However, due to steric hindrance around the tertiary carbon, direct backside attack is highly unfavorable, making direct substitution difficult. Instead, reactions tend to proceed through pathways involving carbocation intermediates, where elimination can compete strongly with substitution depending on reaction conditions.

    Alcoholic ammonia acts both as a weak nucleophile and a weak Base, which means it can participate in more than one possible pathway. In such systems, elimination often becomes significant because removal of a proton from a neighboring carbon leads to formation of a more stable unsaturated product. The overall outcome depends on the balance between carbocation stability and steric effects around the reaction center.

    Option b – Elimination

    Which primary amine can be synthesized by reducing both nitriles and amides?

    a) Methylamine

    b) Benzylamine

    c) Aniline

    d) Isopropylamine

    Explanation:
    Primary amines are commonly prepared through reduction reactions of nitrogen-containing functional groups such as nitriles and amides. In Organic Chemistry, nitriles (–C≡N) undergo catalytic hydrogenation or metal hydride reduction to yield primary amines after complete reduction of the triple bond. Similarly, amides can also be reduced using strong reducing agents to form amines by converting the carbonyl group into a methylene group.

    The key idea is identifying a compound that can be obtained from both types of precursors through reduction while maintaining the same carbon skeleton. Not all amines are formed this way because some require different synthetic routes like aromatic substitution or rearrangements. Therefore, the correct structure must be one that can arise from both a nitrile precursor and an amide precursor without structural modification of the carbon framework.

    Reduction reactions typically preserve the number of carbon atoms present in the starting compound, meaning the product must be compatible with both linear and substituted precursor routes. The amine formed must also be stable under reduction conditions and represent a common endpoint for both functional group transformations.

    Option b – Benzylamine

    N,N-Dimethylmethanamine is a structural isomer of:

    a) Ethylamine

    b) Ethyldimethylamine

    c) Ethylmethylamine

    d) n-Butylamine

    Explanation:
    Structural isomerism occurs when compounds have the same Molecular formula but differ in the connectivity of atoms. In amines, this often involves rearrangement of alkyl groups around the nitrogen Atom while maintaining the same total number of carbon, hydrogen, and nitrogen atoms. Tertiary amines are particularly useful in illustrating such isomerism because multiple alkyl group distributions can lead to the same Molecular formula.

    To analyze isomerism, one must compare Molecular compositions rather than names alone. A compound with a single carbon attached to nitrogen and two methyl groups attached elsewhere can be compared with other amines having the same total carbon count but different Bonding arrangements. The key is ensuring identical Molecular formula while varying structural connectivity.

    This concept highlights how nitrogen substitution patterns influence isomer formation. The presence of different alkyl group distributions around nitrogen can generate multiple structural possibilities even when the total atoms remain unchanged. Understanding this requires careful accounting of carbon distribution across all substituents attached to nitrogen.

    Option c – Ethylmethylamine

    Which of the following pairs can be differentiated using the isocyanide test?

    a) Ethylamine and aniline

    b) Benzylamine and dimethylamine

    c) Diethylamine and trimethylamine

    d) Cyclohexylamine and aniline

    Explanation:
    The isocyanide (carbylamine) test is a qualitative test used to distinguish primary amines from secondary and tertiary amines. Primary amines react with chloroform and alcoholic potassium hydroxide to form isocyanides, which have a characteristic foul smell. Secondary and tertiary amines do not undergo this reaction because they lack the required structural feature necessary for isocyanide formation.

    This test relies on the presence of a free –NH₂ group attached to either an alkyl or aryl group. Aromatic primary amines may react more slowly compared to aliphatic ones, but they still respond under suitable conditions. Secondary and tertiary amines fail to form the intermediate required for the reaction pathway.

    Therefore, when comparing two amines, one must identify whether at least one contains a primary amino group capable of undergoing this transformation. The difference in reactivity allows clear identification between similar-looking amines based on functional group classification and reaction behavior.

    Option b – Benzylamine and dimethylamine

    Acetaldoxime reacts with sodium in ethanol, forming compound P. P, when treated with nitrous Acid at low temperature, produces compound D. Heating D with concentrated H₂SO₄ at 443 K gives compound S. Identify D and S.

    a) Ethanol, Ethene

    b) Ethylamine, Ethene

    c) Ethanol, Ethane

    d) Propan-2-ol, Propene

    Explanation:
    This sequence involves multiple functional group transformations starting from an oxime derivative. Oximes can undergo reduction in the presence of suitable reagents to yield amines. Sodium in ethanol acts as a reducing system that converts the oxime functional group into a corresponding amine by reducing the nitrogen–oxygen bond.

    Once a primary amine is formed, treatment with nitrous Acid at low temperature typically leads to the formation of diazonium intermediates or related nitrogen-containing species depending on whether the amine is aliphatic or aromatic. Aliphatic amines often form unstable diazonium compounds that rapidly decompose into Alcohols with Evolution of nitrogen gas.

    Further heating of the resulting Alcohol with strong Acid leads to dehydration reactions, producing alkenes. Concentrated sulfuric Acid at elevated temperature promotes elimination by removing water from Alcohols, yielding unsaturated Hydrocarbons. The sequence therefore connects reduction, diazotization, and elimination processes in a stepwise transformation pathway.

    Option a – Ethanol, Ethene

    How many of the following amines can undergo diazotization? tert-Butylamine, ethanamine, aniline, N-methylaniline, p-toluidine, m-chloroaniline, 2-phenylethanamine, 2,4,6-tribromoaniline.

    a) 4

    b) 5

    c) 6

    d) 8

    Explanation:
    Diazotization is the process in which primary aromatic amines react with nitrous Acid at low temperature to form diazonium Salts. This reaction is characteristic mainly of aromatic primary amines because the resulting diazonium ions are stabilized by resonance within the aromatic ring. Aliphatic amines generally do not form stable diazonium Salts and instead decompose rapidly.

    Secondary and tertiary amines do not undergo diazotization because they lack the required free primary amino group necessary for forming diazonium intermediates. Substituted aromatic amines may still undergo diazotization if the amino group is primary in nature, even if other substituents are present on the ring.

    The key is identifying which compounds contain a free –NH₂ group attached directly to an aromatic system. Steric and electronic effects may influence reaction rate but do not prevent the fundamental ability to form diazonium Salts. The count depends on selecting only those structures that meet this structural requirement.

    Option a – 4

    What is the major product formed when nitrobenzene is treated with tin and concentrated hydrochloric Acid?

    a) Benzene

    b) Toluene

    c) Chlorobenzene

    d) Aniline

    Explanation:
    Reduction of aromatic nitro compounds is a fundamental transformation in Organic Chemistry. Nitrobenzene undergoes reduction in the presence of metal and Acid systems such as tin with concentrated hydrochloric Acid. This reagent combination provides a strongly reducing Environment that converts the nitro group stepwise through intermediate stages such as nitroso and hydroxylamine derivatives.

    The final product of complete reduction is an aromatic amine, where the nitro group is fully converted into an amino group. This transformation significantly increases the electron density on the aromatic ring due to the strong electron-donating nature of the amino group. The reaction is widely used in laboratory and industrial synthesis of aromatic amines.

    The mechanism involves multiple electron transfer steps under acidic conditions, ensuring full reduction rather than partial intermediates. The aromatic ring remains intact throughout the process, with only functional group conversion occurring at the substitution site.

    Option d – Aniline

    Which of the following is more basic compared to aniline?

    a) N,N-Diphenylaniline

    b) N-Phenylaniline

    c) p-Nitroaniline

    d) Phenylmethanamine

    Explanation:
    Basicity in amines depends on the availability of the lone pair of electrons on nitrogen for protonation. In aniline, the lone pair is partially delocalized into the benzene ring through resonance, which reduces its availability for Bonding with protons and therefore decreases basicity compared to aliphatic amines.

    When comparing related structures, electron-withdrawing or resonance-stabilizing effects significantly influence basic strength. Substituents that pull electron density away from nitrogen reduce basicity, while those that increase electron density enhance it. Aromatic amines generally show lower basicity than aliphatic amines due to resonance delocalization.

    Thus, structures where nitrogen is attached to non-aromatic carbon centers tend to have higher basicity because the lone pair remains localized and more available for protonation. The overall comparison depends on how effectively the nitrogen lone pair is retained versus delocalized within the Molecular framework.

    Option d – Phenylmethanamine

    Given that Butan-1-amine (I), N-ethylethanamine (II), and N,N-dimethylethanamine (III) have the same Molecular weight, their boiling points increase in the order:

    a) III < II < I

    b) I < II < III

    c) II < III < I

    d) II < I < III

    Explanation:
    Boiling point trends in amines are strongly influenced by hydrogen Bonding and intermolecular forces. Primary amines can form more extensive hydrogen Bonding networks because they have two hydrogen atoms attached to nitrogen, allowing stronger intermolecular association. Secondary amines form fewer hydrogen bonds, while tertiary amines cannot form intermolecular hydrogen Bonding through nitrogen due to the absence of N–H bonds.

    As branching increases, surface area decreases and van der Waals forces weaken, further reducing boiling point. The balance between hydrogen Bonding ability and Molecular structure determines the overall trend. Primary amines generally have the highest boiling points, followed by secondary and then tertiary amines.

    Molecular weight being the same ensures that differences arise mainly from structural effects rather than size. Therefore, the variation in boiling points is primarily governed by hydrogen Bonding capability and steric hindrance around the nitrogen Atom, which influences intermolecular interactions.

    Option a – III < II < I

    Among the given amines, which one exhibits the highest basicity in aqueous medium?

    a) N,N-diethylethanamine

    b) N-ethylethanamine

    c) N-methylmethanamine

    d) Ethanamine

    Explanation:
    Basicity in aqueous solution depends on how readily a nitrogen Atom donates its lone pair to accept a proton and how stable the resulting conjugate Acid becomes in water. Aliphatic amines are generally more basic than aromatic amines because the lone pair on nitrogen in aliphatic amines remains localized and is not delocalized into any resonance system. This makes it more available for protonation.

    In aqueous medium, solvation also plays a major role. Smaller alkyl groups allow better hydration of the protonated amine, stabilizing the conjugate Acid through hydrogen Bonding with water molecules. However, excessive steric hindrance around nitrogen can reduce solvation and slightly decrease basicity despite strong electron donation.

    Tertiary amines have strong electron-releasing effects due to alkyl groups, but their bulky structure reduces solvation of the protonated form. Primary amines, on the other hand, balance electron donation and good solvation. Secondary amines often show slightly higher basicity due to an optimal balance between inductive effects and solvation stability.

    Thus, the most basic amine in water is typically the one that combines strong electron donation with effective solvation of its conjugate acid.

    Option b – N-ethylethanamine

    Which compound gives a positive carbylamine test?

    a) N,N-dimethylbenzenamine

    b) N,N-diethylethanamine

    c) N-methylbenzenamine

    d) Phenylmethanamine

    Explanation:
    The carbylamine test is used to identify primary amines. In this reaction, a primary amine reacts with chloroform in the presence of alcoholic potassium hydroxide to form an isocyanide (carbylamine), which has a very unpleasant odor. This reaction is highly specific because only primary amines possess the required structural feature to form the intermediate necessary for isocyanide formation.

    Secondary and tertiary amines do not respond to this test because they lack the required –NH₂ group. Aromatic primary amines can also give this test, though sometimes under more vigorous conditions due to resonance stabilization of the lone pair. The key requirement is the presence of at least one hydrogen attached directly to nitrogen.

    The reaction proceeds through the formation of a dichlorocarbene intermediate from chloroform, which then reacts with the amine. Only primary amines can successfully complete the sequence to form the isocyanide product, making this a diagnostic test for identifying such compounds.

    Option d – Phenylmethanamine

    The reaction of ethylamine with acetyl chloride in the presence of pyridine yields:

    a) N-ethylethanamide

    b) N,N-diethylethanamide

    c) N-methylethanamide

    d) N-ethylpropanamide

    Explanation:
    Acylation of amines involves the reaction of an amine with an acid chloride to form an amide. In this process, the lone pair on nitrogen attacks the carbonyl carbon of the acid chloride, leading to substitution of the chlorine Atom. Pyridine acts as a Base to neutralize the hydrogen chloride formed during the reaction, driving the reaction forward.

    Primary amines undergo acylation to form secondary amides, where one hydrogen on nitrogen is replaced by an acyl group. This transformation reduces the basicity of nitrogen because the lone pair becomes involved in resonance with the carbonyl group. The reaction is widely used to protect amine groups in multistep synthesis.

    The product formed retains the carbon skeleton of both reactants but creates a new amide linkage. The mechanism involves nucleophilic substitution at the carbonyl carbon, followed by elimination of chloride ion. The final compound is stabilized by resonance between nitrogen and oxygen in the amide functional group.

    Option a – N-ethylethanamide

    Which substance will dissolve in an alkaline medium after reacting with Hinsberg’s reagent?

    a) CH₃NH₂

    b) (CH₃)₃N

    c) (C₂H₅)₂NH

    d) C₆H₅NHC₆H₅

    Explanation:
    Hinsberg’s reagent (benzenesulfonyl chloride) is used to distinguish between primary, secondary, and tertiary amines based on their behavior in alkaline medium. Primary amines react with the reagent to form sulfonamides that contain an acidic hydrogen, allowing them to dissolve in alkali due to Salt formation. Secondary amines form sulfonamides that lack acidic hydrogen and therefore remain insoluble.

    Tertiary amines do not form sulfonamides at all but may dissolve in acid due to protonation. The key factor is whether the derivative formed has an N–H bond capable of ionization in basic medium. Only compounds that can form stable anionic Salts in alkali show solubility after reaction.

    Thus, identifying the correct substance depends on recognizing which amine leads to a sulfonamide with acidic hydrogen. The solubility behavior after reaction is directly tied to structural features of the amine and its ability to form ionizable derivatives under alkaline conditions.

    Option a – CH₃NH₂

    The correct sequence of chemical transformations required to convert p-nitrophenol into quinol is:

    a) Reduction, diazotization, acidic hydrolysis

    b) Acidic hydrolysis, diazotization, reduction

    c) Acidic hydrolysis, reduction, diazotization

    d) Diazotization, reduction, acidic hydrolysis

    Explanation:
    This transformation involves multiple functional group changes including reduction, substitution, and rearrangement steps. Nitro groups are typically reduced to amino groups under suitable reducing conditions, which significantly alters the electronic nature of the aromatic ring. Once an amino group is present, it can undergo diazotization in the presence of nitrous acid at low temperature to form diazonium Salts.

    Diazonium Salts are highly versatile intermediates that can be replaced by various nucleophiles or undergo hydrolysis depending on reaction conditions. Acidic hydrolysis often replaces the diazonium group with a hydroxyl group, leading to phenolic compounds. Further transformations can modify the substitution pattern to achieve the final quinol structure.

    The sequence of reactions must be carefully controlled because diazonium intermediates are unstable and temperature-sensitive. Each step builds upon the previous functional group modification, gradually converting nitro-substituted phenolic systems into dihydroxybenzene derivatives through controlled substitution and reduction pathways.

    Option a – Reduction, diazotization, acidic hydrolysis

    Aniline reacts with bromine water, followed by sodium nitrite and dilute HCl to form a diazonium Salt. This Salt is then converted into diazonium tetrafluoroborate and heated. What is the final product?

    a) 1-Bromo-4-fluorobenzene

    b) 4-Bromoaniline

    c) 1,3,5-Tribromo-2-fluorobenzene

    d) 1,3,5-Tribromobenzene

    Explanation:
    This sequence involves electrophilic substitution followed by diazotization and thermal decomposition of a diazonium Salt. Initially, aniline undergoes rapid bromination due to the strong activating effect of the amino group, typically leading to poly-substitution on the aromatic ring. The amino group strongly directs substitution to ortho and para positions.

    After bromination, the amino group is converted into a diazonium Salt using nitrous acid at low temperature. This diazonium Salt can then be transformed into a more stable tetrafluoroborate derivative, which allows controlled thermal decomposition. Heating such diazonium Salts leads to replacement of the diazonium group with other substituents depending on reaction conditions.

    This pathway is commonly used in synthetic Organic Chemistry to introduce halogen atoms into aromatic rings via diazonium intermediates. The final product depends on substitution patterns already present on the aromatic system and the nature of the decomposition step, which replaces the diazonium group with a substituent.

    Option c – 1,3,5-Tribromo-2-fluorobenzene

    When benzene diazonium chloride reacts with Phenol in the presence of an alkali, the resulting dye has what color?

    a) Red

    b) Yellow

    c) Orange

    d) Green

    Explanation:
    Azo coupling between diazonium Salts and Phenols in alkaline medium produces azo dyes, which are highly conjugated systems responsible for vivid coloration. The phenoxide ion formed under basic conditions is strongly activated toward electrophilic substitution, making it an excellent coupling partner for diazonium ions.

    The extended conjugation formed between the aromatic rings through the azo linkage (–N=N–) allows absorption of visible Light. The wavelength of absorbed Light depends on the extent of conjugation and the nature of substituents on the aromatic rings. Electron-donating groups enhance conjugation and shift absorption toward longer wavelengths, producing characteristic colors.

    The resulting dye exhibits a distinct color due to π-electron delocalization across both aromatic systems. The exact shade depends on the substituents and degree of conjugation but generally falls within the visible Spectrum of azo compounds commonly used in dye Chemistry.

    Option c – Orange

    A test tube labeled ‘X’ contains N-methylaniline. Upon treatment with Hinsberg’s reagent and excess potassium hydroxide, the correct observation is:

    a) A clear solution forms which yields an insoluble product upon acidification

    b) An insoluble material forms that dissolves upon acidification

    c) An insoluble substance forms that does not dissolve upon acidification

    d) The clear solution remains unchanged even after acidification

    Explanation:
    The Hinsberg test differentiates amines based on their ability to form sulfonamides with benzenesulfonyl chloride. Secondary amines like N-methylaniline react to form sulfonamide derivatives, but these products lack acidic hydrogen and therefore remain insoluble in alkaline medium. Primary amines, in contrast, form soluble Salts under basic conditions due to the presence of an N–H bond.

    The behavior in alkali is a key diagnostic feature. If the sulfonamide formed is not acidic enough to ionize, it remains as an insoluble precipitate. Acidification does not change this behavior significantly for secondary amine derivatives, as they do not form soluble anionic species in the same way primary amines do.

    Thus, the observation depends on whether the product can undergo deprotonation in basic medium. Structural features of the amine determine whether solubility changes occur during the test sequence.

    Option c – An insoluble substance forms that does not dissolve upon acidification

    Which amide forms propanamine when it undergoes Hofmann bromamide reaction?

    a) Propanamide

    b) Pentanamide

    c) Ethanamide

    d) Butanamide

    Explanation:
    The Hofmann bromamide reaction converts amides into amines with one carbon Atom less than the original amide. This reaction involves treatment with bromine and strong Base, leading to rearrangement and formation of an isocyanate intermediate, which then hydrolyzes to form a primary amine.

    The key feature is carbon loss during the rearrangement step, meaning the amine product always has one fewer carbon Atom than the starting amide. Therefore, to obtain a specific amine such as propanamine, the starting amide must contain one additional carbon Atom in its structure.

    This transformation is widely used in synthetic Chemistry for preparing primary amines from carboxylic acid derivatives. The mechanism involves migration of the alkyl group from carbonyl carbon to nitrogen, followed by rearrangement and hydrolysis. The structural relationship between amide and amine is essential for determining the correct precursor.

    Option d – Butanamide

    Which of the following accurately describes the Hofmann bromamide reaction?

    a) Produces unsymmetrical amines

    b) Yields tertiary amines

    c) Produces primary amines

    d) Does not form aromatic amines

    Explanation:
    The Hofmann bromamide reaction is a rearrangement reaction in which an amide is converted into a primary amine using bromine and a strong Base. During this process, the carbonyl carbon is lost as carbon dioxide, resulting in an amine with one fewer carbon Atom than the starting amide.

    The mechanism involves formation of an N-bromoamide intermediate, followed by rearrangement to an isocyanate. This intermediate then undergoes hydrolysis to yield the final amine product. The reaction is particularly useful for synthesizing primary amines that are otherwise difficult to prepare directly.

    A key characteristic of this reaction is that it always produces primary amines regardless of whether the starting amide is primary, secondary, or tertiary. The rearrangement ensures loss of one carbon Atom while preserving the nitrogen-containing framework in the final product.

    Option c – Produces primary amines

    Which compound cannot establish intermolecular hydrogen Bonding?

    a) Propan-2-amine

    b) Triethylamine

    c) Diphenylamine

    d) o-Toluidine

    Explanation:
    Intermolecular hydrogen Bonding requires the presence of a hydrogen Atom directly bonded to a highly electronegative atom such as nitrogen, oxygen, or fluorine. This allows molecules to attract each other through strong dipole interactions, significantly affecting boiling point, solubility, and physical state.

    In amines, primary and secondary amines can form intermolecular hydrogen bonding because they possess N–H bonds. This allows one Molecule to donate hydrogen while another accepts via the lone pair on nitrogen. Aromatic amines and substituted amines may still participate depending on whether N–H bonds are present and accessible.

    However, tertiary amines lack N–H bonds entirely. Even though they contain a nitrogen atom with a lone pair, they cannot donate hydrogen for intermolecular bonding. As a result, their intermolecular forces are weaker compared to primary and secondary amines, leading to lower boiling points and different physical behavior.

    Thus, the key factor is whether the structure contains a hydrogen attached directly to nitrogen. Without it, intermolecular hydrogen bonding cannot occur.

    Option b – Triethylamine

    Which of the following is true regarding the acylation of amines?

    a) Involves substitution of the nitrogen hydrogen atom by an alkyl group

    b) It’s an electrophilic substitution reaction

    c) Requires a strong Base like pyridine

    d) Secondary amines do not participate in acylation

    Explanation:
    Acylation of amines is a nucleophilic substitution reaction where an amine reacts with an acylating agent such as an acid chloride or anhydride. The nitrogen atom donates its lone pair to the carbonyl carbon, forming a tetrahedral intermediate, followed by elimination of a leaving group to form an amide.

    This reaction is not electrophilic substitution; instead, it proceeds through nucleophilic attack on the carbonyl carbon. During the process, a hydrogen atom attached to nitrogen is replaced by an acyl group, reducing the basicity of the nitrogen due to resonance delocalization of its lone pair with the carbonyl group.

    A Base like pyridine is often used to neutralize the acid formed during the reaction, preventing protonation of the amine and driving the reaction forward. Secondary amines can also undergo acylation, forming N-substituted amides, while tertiary amines do not typically form amides due to lack of N–H bond.

    Thus, the reaction involves substitution at the acyl carbon rather than aromatic electrophilic substitution.

    Option c – Requires a strong Base like pyridine

    Gabriel phthalimide synthesis is mainly used to prepare:

    a) Aromatic primary amines

    b) Aliphatic primary amines

    c) Aliphatic secondary amines

    d) Aromatic secondary amines

    Explanation:
    Gabriel phthalimide synthesis is a method used to prepare primary amines in a controlled manner. It involves the use of phthalimide, which upon reaction with a strong Base forms a stable phthalimide anion. This anion then undergoes alkylation with alkyl halides, followed by hydrolysis or hydrazinolysis to release the primary amine.

    The method is particularly useful for synthesizing aliphatic primary amines because it avoids over-alkylation, which is a common problem in direct alkylation of ammonia. Aromatic amines are generally not prepared using this method due to the difficulty in forming appropriate alkylated intermediates on aromatic systems.

    The process ensures that only one alkyl group is introduced to nitrogen, giving clean formation of primary amines. It is widely used in Organic synthesis where selective preparation of primary amines is required without formation of secondary or tertiary by-products.

    Thus, the method is best suited for producing primary amines from alkyl halides under controlled conditions.

    Option b – Aliphatic primary amines

    Among the following, which is the least basic according to Brønsted theory?

    a) Aniline

    b) Cyclohexylamine

    c) Phenol

    d) Cyclohexanol

    Explanation:
    Brønsted basicity depends on the ability of a species to accept a proton. The strength of a Base is influenced by the availability of lone pair electrons and their tendency to bind with a proton. Factors such as resonance, electronegativity, and hybridization significantly affect this availability.

    Aromatic compounds with electron-withdrawing effects or resonance delocalization of lone pairs show reduced basicity. In particular, when the lone pair is involved in resonance with an aromatic ring or adjacent functional group, it becomes less available for protonation. This greatly reduces basic strength compared to aliphatic compounds.

    In contrast, Alcohols and aliphatic amines may show relatively higher basicity due to localized electron density, though oxygen-based compounds still differ from nitrogen-based ones in proton affinity. The least basic species is typically the one where electron density is most strongly delocalized or stabilized, making proton acceptance least favorable.

    Thus, comparison depends on evaluating how strongly each structure stabilizes or withdraws electron density from the lone pair responsible for basicity.

    Option c – Phenol

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