Organic Compounds Containing Nitrogen

Amines are organic derivatives of ammonia ($\text{NH}_3$) wherein hydrogen atoms are replaced by alkyl or aryl groups. Unlike alkyl halides or alcohols where classification depends on the nature of the carbon atom attached to the functional group, the classification of amines depends strictly on the number of carbon groups directly bonded to the nitrogen atom:

• Primary ($1^\circ$) Amines: One hydrogen replaced; general formula $R\text{-NH}_2$ or $\text{Ar-NH}_2$ (Functional group: amino group $-\text{NH}_2$). Examples: Methanamine ($\text{CH}_3\text{NH}_2$), Aniline or Benzenamine ($\text{C}_6\text{H}_5\text{NH}_2$).
• Secondary ($2^\circ$) Amines: Two hydrogens replaced; general formula $R\text{-NH-}R'$ or $\text{Ar-NH-}R$ (Functional group: imino group $-\text{NH-}$). Examples: $N$-Methylmethanamine or Dimethylamine ($(\text{CH}_3)_2\text{NH}$), $N$-Methylaniline ($\text{C}_6\text{H}_5\text{NHCH}_3$).
• Tertiary ($3^\circ$) Amines: All three hydrogens replaced; general formula $R_3\text{N}$ or $\text{Ar-N}R_2$ (Functional group: tertiary nitrogen atom). Examples: $N,N$-Dimethylmethanamine or Trimethylamine ($(\text{CH}_3)_3\text{N}$), $N,N$-Dimethylaniline ($\text{C}_6\text{H}_5\text{N(CH}_3)_2$).
• Quaternary Ammonium Salts: Four organic groups covalently bonded to nitrogen, carrying a permanent positive charge balanced by an anion ($[R_4\text{N}]^+X^-$). Example: Tetramethylammonium chloride ($[(\text{CH}_3)_4\text{N}]^+\text{Cl}^-$).

The central nitrogen atom in amines is $sp^3$ hybridized. Three of the $sp^3$ hybrid orbitals overlap with carbon ($C$) or hydrogen ($H$) atomic orbitals to form three $\sigma$-bonds, while the fourth $sp^3$ hybrid orbital accommodates the unshared lone pair of electrons:

• Pyramidal Shape & Bond Angles: Due to greater lone pair-bond pair ($lp-bp$) repulsion compared to bond pair-bond pair ($bp-bp$) repulsion (according to VSEPR theory), the tetrahedral angle compresses from $109.5^\circ$ to approximately $108^\circ$ (e.g., $108^\circ$ in trimethylamine, $(\text{CH}_3)_3\text{N}$).
• Amine Inversion (Umbrella Inversion / Walden Inversion of Amines): In tertiary amines with three distinct substituents ($R_1R_2R_3\text{N}:$), the molecule is theoretically chiral (asymmetric). However, the lone pair rapidly turns inside out through a planar $sp^2$ transition state at room temperature. The energy barrier for this inversion is extremely low ($\approx 25\text{ kJ}\cdot\text{mol}^{-1}$); consequently, enantiomers interconvert too rapidly to be resolved into optically active antipodes. Only chiral quaternary ammonium ions ($[R_1R_2R_3R_4\text{N}]^+X^-$), which lack an unshared pair to invert, can be resolved into stable optical isomers.

Intermolecular Hydrogen Bonding & Boiling Point Hierarchy:
Primary and secondary amines possess polar $\text{N-H}$ bonds ($\Delta \chi \approx 3.0 - 2.1 = 0.9$) capable of forming intermolecular hydrogen bonds ($\text{N-H}\cdots\text{N}$). Because nitrogen is less electronegative than oxygen ($\chi_\text{O} = 3.5$), $\text{N-H}\cdots\text{N}$ hydrogen bonds are substantially weaker than $\text{O-H}\cdots\text{O}$ bonds found in alcohols and carboxylic acids of comparable molecular mass.

Among isomeric amines of formula $\text{C}_3\text{H}_9\text{N}$, the boiling point order is strictly: $$\mathbf{1^\circ\text{ Amine (321.8 K)} > 2^\circ\text{ Amine (309 K)} > 3^\circ\text{ Amine (276 K)}}$$ Reason: Primary amines have two hydrogen atoms bonded to nitrogen and form extensive branched hydrogen-bonded networks; secondary amines have only one $\text{N-H}$ bond, forming linear associations; and tertiary amines have zero hydrogen atoms bonded to nitrogen, precluding any intermolecular hydrogen bonding among their own molecules.

Solubility in Water: Lower aliphatic amines ($\text{CH}_3\text{NH}_2, (\text{CH}_3)_2\text{NH}, \text{C}_2\text{H}_5\text{NH}_2$) are completely miscible with water because they readily form hydrogen bonds with water molecules ($\text{R-N:}\cdots\text{H-O-H}$). Solubility decreases sharply as molar mass increases ($>\text{C}_6$) due to the dominant hydrophobic effect of the non-polar alkyl/aryl chain. Aniline is practically insoluble in water ($\approx 3.4\text{ g}/100\text{ mL}$ at $25^\circ\text{C}$) owing to the bulky hydrophobic phenyl ring.

Nitro compounds are readily reduced to primary amines via catalytic or chemical reduction:

• Catalytic Hydrogenation: Passing $\text{H}_2$ gas over finely divided nickel, platinum, or palladium in ethanol: $$\text{Ar-NO}_2 + 3\text{H}_2 \xrightarrow{\text{Pd / ethanol}} \mathbf{\text{Ar-NH}_2} + 2\text{H}_2\text{O}$$
• Chemical Reduction with Active Metals in Acid: $$\text{Ar-NO}_2 + 6[\text{H}] \xrightarrow{\text{Sn / conc. HCl or Fe / conc. HCl}} \text{Ar-NH}_3^+\text{Cl}^- \xrightarrow{\text{NaOH}} \mathbf{\text{Ar-NH}_2}$$
  Board Favorite Examiner Question: Why is $\text{Fe} + \text{HCl}$ preferred over $\text{Sn} + \text{HCl}$ for industrial preparation of aniline?
  Answer: Reduction with scrap iron ($\text{Fe}$) produces $\text{FeCl}_2$, which undergoes hydrolysis in aqueous medium to regenerate $\text{HCl}$ ($\text{FeCl}_2 + 2\text{H}_2\text{O} \to \text{Fe(OH)}_2 + 2\text{HCl}$). Consequently, only a tiny catalytic amount of hydrochloric acid is required to initiate the reaction, making the process highly economical and environmentally superior.

Heating an alkyl halide with an alcoholic solution of ammonia in a sealed tube at $373\text{ K}$ leads to nucleophilic substitution ($S_N2$) of the halogen atom: $$\text{NH}_3 \xrightarrow{+R\text{X}} R\text{NH}_3^+X^- \xrightarrow{+\text{NH}_3} \mathbf{R\text{NH}_2} \xrightarrow{+R\text{X}} \mathbf{R_2\text{NH}} \xrightarrow{+R\text{X}} \mathbf{R_3\text{N}} \xrightarrow{+R\text{X}} \mathbf{[R_4\text{N}]^+X^-}$$

• Major Drawback: Ammonolysis produces a complex mixture of primary, secondary, and tertiary amines alongside quaternary ammonium salts, which are exceedingly difficult to separate by fractional distillation due to close boiling points.
• Control of Reaction: To obtain the primary amine as the major product, a large excess of ammonia ($\text{NH}_3$) must be employed, so that collision between alkyl halide and $\text{NH}_3$ is overwhelmingly favored over reaction with the newly formed primary amine. Conversely, if alkyl halide is taken in excess, the quaternary ammonium salt ($[R_4\text{N}]^+X^-$) is obtained as the predominant product.
• Reactivity order of halides: $R\text{-I} > R\text{-Br} > R\text{-Cl}$ (governed by decreasing $\text{C-X}$ bond dissociation energy).

• Reduction of Nitriles (Cyanides): Reduction using lithium aluminium hydride ($\text{LiAlH}_4$) or sodium and boiling ethanol ($\text{Na} / \text{C}_2\text{H}_5\text{OH}$, Mendius reaction) yields primary amines containing one more carbon than the starting alkyl halide: $$R\text{-X} \xrightarrow{\text{alc. KCN}} R\text{-C}\equiv\text{N} \xrightarrow{\text{LiAlH}_4 \text{ or } \text{Na/EtOH}} \mathbf{R\text{-CH}_2\text{NH}_2}$$ This sequence is universally used in organic synthesis for ascending the homologous series.
• Reduction of Amides: Amides undergo selective reduction with $\text{LiAlH}_4$ in ether followed by acid workup to yield amines containing the same number of carbon atoms: $$R\text{-CONH}_2 \xrightarrow{\text{(i) }\text{LiAlH}_4\text{ / ether, (ii) }\text{H}_2\text{O}} \mathbf{R\text{-CH}_2\text{NH}_2}$$

Gabriel phthalimide synthesis is the premier industrial and laboratory method for preparing pure primary aliphatic amines without contamination from $2^\circ$ or $3^\circ$ amines:

1. Formation of Potassium Phthalimide: Phthalimide is treated with ethanolic potassium hydroxide ($\text{KOH}$). The acidic imidic hydrogen (sandwiched between two electron-withdrawing carbonyl groups) is abstracted to yield nucleophilic potassium phthalimide: $$\text{C}_6\text{H}_4(\text{CO})_2\text{NH} + \text{KOH} \longrightarrow \mathbf{\text{C}_6\text{H}_4(\text{CO})_2\text{N}^-\text{K}^+} + \text{H}_2\text{O}$$
2. Alkylation ($S_N2$ step): Potassium phthalimide reacts with a primary alkyl halide ($R\text{-X}$) via bimolecular nucleophilic substitution ($S_N2$) to form $N$-alkylphthalimide: $$\text{C}_6\text{H}_4(\text{CO})_2\text{N}^-\text{K}^+ + R\text{-X} \longrightarrow \mathbf{\text{C}_6\text{H}_4(\text{CO})_2\text{N-}R} + \text{K}X$$
3. Alkaline Hydrolysis: Alkaline hydrolysis of $N$-alkylphthalimide with aqueous sodium hydroxide ($\text{NaOH}$) furnishes pure primary aliphatic amine and sodium phthalate: $$\text{C}_6\text{H}_4(\text{CO})_2\text{N-}R + 2\text{NaOH} \longrightarrow \mathbf{R\text{-NH}_2} + \text{C}_6\text{H}_4(\text{COONa})_2$$

When an aliphatic or aromatic primary acid amide is heated with bromine ($\text{Br}_2$) and concentrated aqueous or ethanolic sodium hydroxide ($\text{NaOH}$ or $\text{KOH}$), it degrades to yield a primary amine containing one less carbon atom than the starting amide: $$\mathbf{R\text{-CONH}_2 + \text{Br}_2 + 4\text{NaOH} \xrightarrow{\Delta} R\text{-NH}_2 + \text{Na}_2\text{CO}_3 + 2\text{NaBr} + 2\text{H}_2\text{O}}$$ $$\mathbf{\text{Ar-CONH}_2 + \text{Br}_2 + 4\text{KOH} \xrightarrow{\Delta} \text{Ar-NH}_2 + \text{K}_2\text{CO}_3 + 2\text{KBr} + 2\text{H}_2\text{O}}$$

• Stoichiometry: Exactly 1 mole of $\text{Br}_2$ and 4 moles of $\text{NaOH}$ are consumed per mole of amide.
• Mechanistic Steps:
  1. Base deprotonates amide to form bromoamide intermediate: $R\text{CONH}_2 + \text{Br}_2 + \text{OH}^- \to R\text{CONHBr} + \text{H}_2\text{O} + \text{Br}^-$.
  2. Further base treatment generates acyl nitrene-like species or bromamide anion: $R\text{CON}^-\text{Br}$.
  3. Intramolecular 1,2-rearrangement (r.d.s.): Alkyl or aryl group $R$ migrates with its bonding pair from carbonyl carbon to electron-deficient nitrogen with simultaneous loss of $\text{Br}^-$, yielding an alkyl isocyanate ($R\text{-N=C=O}$): $$R\text{-CO-N}^-\text{-Br} \xrightarrow{\text{migration}} \mathbf{R\text{-N=C=O}} + \text{Br}^-$$
  4. Hydrolysis of isocyanate yields primary amine and carbonate ion: $$R\text{-N=C=O} + 2\text{OH}^- \longrightarrow \mathbf{R\text{-NH}_2} + \text{CO}_3^{2-}$$
• Key Synthetic Role: Used for descending the homologous series (e.g., converting Ethanoic acid $\to$ Methanamine via ethanamide).

Amines act as typical Lewis bases due to the unshared lone pair of electrons on the nitrogen atom ($R\text{-NH}_2 + \text{H}^+ \rightleftharpoons R\text{-NH}_3^+$). In aqueous solution, their basic strength is expressed by the base dissociation constant ($K_b$) or $\text{p}K_b$: $$R\text{-NH}_2 + \text{H}_2\text{O} \rightleftharpoons R\text{-NH}_3^+ + \text{OH}^- \qquad K_b = \frac{[R\text{-NH}_3^+][\text{OH}^-]}{[R\text{-NH}_2]}$$ $$\mathbf{\text{p}K_b = -\log_{10} K_b} \qquad (\text{Smaller }\text{p}K_b \implies \text{Stronger Base})$$

In the gas phase, molecules are isolated and completely free from solvent molecules (no hydration or steric hindrance to solvation exists). Here, the basicity depends solely on the electron-releasing inductive effect ($+I$) of the alkyl groups, which disperses the positive charge on the conjugate ammonium cation ($R_n\text{NH}_{4-n}^+$): $$\mathbf{3^\circ\text{ Amine } (R_3\text{N}) > 2^\circ\text{ Amine } (R_2\text{NH}) > 1^\circ\text{ Amine } (R\text{NH}_2) > \text{NH}_3}$$ Because three alkyl groups release electron density towards nitrogen more effectively than two or one, $(CH_3)_3N$ is the strongest base in the gaseous phase.

In aqueous solution, basicity does NOT follow the simple gas-phase order. It is governed by a delicate balance of three competing factors:

1. Inductive Effect ($+I$): Stabilizes the ammonium cation by electron release: Order: $3^\circ > 2^\circ > 1^\circ$.
2. Solvation (Hydration) Energy: The protonated amine cation ($R_n\text{NH}_{4-n}^+$) is stabilized by hydrogen bonding with water molecules ($\text{N-H}^+\cdots\text{OH}_2$). The greater the number of hydrogens attached to nitrogen, the greater the hydration stabilization: $$\text{Hydration Order: } \mathbf{1^\circ\text{ cation } (R\text{NH}_3^+ \text{ with 3 H's}) > 2^\circ\text{ cation } (R_2\text{NH}_2^+ \text{ with 2 H's}) > 3^\circ\text{ cation } (R_3\text{NH}^+ \text{ with 1 H})}$$
3. Steric Hindrance: Bulky alkyl groups crowd the nitrogen atom, physically hindering incoming protons and obstructing water molecules from approaching the conjugate cation for hydration: Order favoring protonation: $1^\circ > 2^\circ > 3^\circ$.

Why does tertiary trimethylamine drop to 3rd place in methyl? Because in $(CH_3)_3N$, three methyl groups create crowding that severely inhibits hydration of $(CH_3)_3\text{NH}^+$ (which has only 1 H-bonding site), outweighing its $+I$ gain. In the ethyl series, ethyl groups exert a much larger $+I$ inductive effect, pushing $(C_2H_5)_3N$ ahead of the $1^\circ$ ethylamine ($231$). In both series, the secondary ($2^\circ$) amine is the undisputed champion, achieving the perfect compromise between $+I$ induction and hydration stabilization.

Aniline ($\text{C}_6\text{H}_5\text{NH}_2$, $\text{p}K_b = 9.38$) is approximately $10^6$ times weaker as a base than methylamine ($\text{p}K_b = 3.38$) and substantially weaker than ammonia ($\text{p}K_b = 4.75$). This monumental difference is explained by two primary phenomena:

• Resonance Delocalization of Lone Pair in Aniline: The nitrogen lone pair participates in $+R$ (mesomeric) resonance with the aromatic $\pi$-electron cloud of the benzene ring. Five resonance canonical contributors delocalize electron density into the ortho and para positions: $$\text{C}_6\text{H}_5\text{-}\ddot{\text{N}}\text{H}_2 \longleftrightarrow \text{ortho}^- \longleftrightarrow \text{para}^- \longleftrightarrow \text{ortho}^-$$ As a consequence, the lone pair on nitrogen is far less available for donation to a proton.
• Differential Resonance Stabilization of Conjugate Acid: When aniline accepts a proton to form the anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^+$), the nitrogen atom no longer possesses a lone pair. Hence, the anilinium ion has only 2 Kekulé resonance structures. Because neutral aniline has 5 resonance contributors (greater resonance stabilization) while the anilinium cation has only 2, protonation is thermodynamically unfavorable ($\Delta G^\circ > 0$).
• Effect of Substituents on Aniline Basicity:
  • Electron-Donating Groups (EDGs: $-\text{OCH}_3, -\text{CH}_3, -\text{NH}_2$): Release electron density into the ring through $+R / +I$ effects, increasing electron density on nitrogen and stabilizing the anilinium cation $\implies$ Increase Basicity (Lower $\text{p}K_b$). E.g., $p$-Toluidine ($\text{p}K_b = 8.8$) is a stronger base than aniline.
  • Electron-Withdrawing Groups (EWGs: $-\text{NO}_2, -\text{CN}, -\text{COOH}, -\text{X}$): Withdraw electron density through $-R / -I$ effects, destabilizing the anilinium cation $\implies$ Decrease Basicity (Higher $\text{p}K_b$). E.g., $p$-Nitroaniline ($\text{p}K_b = 13.0$) is an exceptionally weak base.
  • Ortho-Effect in Substituted Anilines: Regardless of whether an ortho-substituent is electron-donating or electron-withdrawing (e.g., $o$-toluidine, $o$-nitroaniline), ortho-substituted anilines are almost always weaker bases than aniline itself. This is attributed to steric crowding around the amino group, which inhibits coplanar protonation and disrupts hydration of the resulting conjugate anilinium ion.

When an aliphatic or aromatic primary amine is gently warmed with chloroform ($\text{CHCl}_3$) and alcoholic potassium hydroxide ($\text{KOH}$), it produces an isocyanide or carbylamine ($R\text{-NC}$), characterized by an overwhelmingly foul, repulsive, nauseating odor: $$\mathbf{R\text{-NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \xrightarrow{\Delta} R\text{-N}\equiv\text{C} + 3\text{KCl} + 3\text{H}_2\text{O}}$$ $$\mathbf{\text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{-N}\equiv\text{C (Phenyl isocyanide)} + 3\text{KCl} + 3\text{H}_2\text{O}}$$

Amines are treated with benzenesulfonyl chloride ($\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$, Hinsberg's reagent) in presence of excess aqueous $\text{KOH}$:

• Primary ($1^\circ$) Amines: React to form $N$-alkylbenzenesulfonamide. The remaining hydrogen attached to nitrogen is strongly acidic due to the powerful electron-withdrawing sulfonyl group ($-\text{SO}_2-$). Hence, it dissolves completely in aqueous $\text{KOH}$ to form a clear water-soluble potassium salt. Upon subsequent acidification with dilute $\text{HCl}$, the insoluble sulfonamide precipitates out: $$\text{C}_6\text{H}_5\text{SO}_2\text{Cl} + R\text{NH}_2 \xrightarrow{-\text{HCl}} \text{C}_6\text{H}_5\text{SO}_2\text{NH-}R \xrightarrow{\text{aq. KOH}} [\text{C}_6\text{H}_5\text{SO}_2\text{N-}R]^-\text{K}^+ \text{ (Soluble in alkali)}$$
• Secondary ($2^\circ$) Amines: React to form $N,N$-dialkylbenzenesulfonamide. Because there is no acidic hydrogen remaining on nitrogen, it is completely insoluble in aqueous alkali ($\text{KOH}$) and remains as an undissolved precipitate/solid: $$\text{C}_6\text{H}_5\text{SO}_2\text{Cl} + R_2\text{NH} \xrightarrow{-\text{HCl}} \mathbf{\text{C}_6\text{H}_5\text{SO}_2\text{N}R_2} \xrightarrow{\text{aq. KOH}} \text{Remains Insoluble}$$
• Tertiary ($3^\circ$) Amines: Tertiary amines possess zero replaceable hydrogen atoms on nitrogen and therefore do not react with benzenesulfonyl chloride. They remain insoluble in alkaline solution but dissolve when acidified due to ammonium salt formation.

Nitrous acid ($\text{HNO}_2$) is unstable and prepared in situ by reacting sodium nitrite with cold mineral acid at $273-278\text{ K}$ ($0-5^\circ\text{C}$):

1. Primary Aliphatic Amines: Form highly unstable alkyldiazonium salts ($[R\text{-N}_2^+]\text{Cl}^-$) that instantly decompose even at $0^\circ\text{C}$ to release nitrogen gas ($\text{N}_2\uparrow$) quantitatively and produce a mixture of alcohols, alkenes, and alkyl chlorides: $$R\text{-NH}_2 + \text{HNO}_2 \xrightarrow{\text{NaNO}_2 + \text{HCl}} [R\text{-N}_2^+\text{Cl}^-] \xrightarrow{\text{H}_2\text{O}} \mathbf{R\text{-OH} + \text{N}_2\uparrow + \text{HCl}}$$ Application: Used for quantitative estimation of amino acids and proteins via measuring the volume of evolved $\text{N}_2$ gas (Van Slyke method).
2. Primary Aromatic Amines (Aniline): React at ice-cold temperatures ($273-278\text{ K}$) to yield stable benzenediazonium chloride: $$\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278\text{ K}} \mathbf{\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-} + \text{NaCl} + 2\text{H}_2\text{O}$$
3. Secondary Amines (Aliphatic or Aromatic): React slowly to form $N$-nitrosamines, which separate out as yellow oily insoluble liquids: $$R_2\text{NH} + \text{HONO} \longrightarrow \mathbf{R_2\text{N-N=O (Yellow oily liquid)}} + \text{H}_2\text{O}$$ Liebermann's Nitroso Reaction: Heating the yellow nitrosoamine with phenol and concentrated $\text{H}_2\text{SO}_4$ gives a green or blue solution which turns red on dilution with water and deep blue/green on making alkaline with $\text{NaOH}$.
4. Tertiary Aliphatic Amines: Dissolve in cold nitrous acid to form soluble trialkylammonium nitrite salts ($[R_3\text{NH}]^+\text{NO}_2^-$), which decompose on warming to give nitrosamines and aldehydes.

Primary and secondary amines react with acid chlorides or anhydrides in presence of a base stronger than the amine (e.g., pyridine) to remove $\text{HCl}$ and form $N$-substituted amides: $$\text{CH}_3\text{NH}_2 + \text{CH}_3\text{COCl} \xrightarrow{\text{pyridine}} \mathbf{\text{CH}_3\text{CONHCH}_3 \text{ (}N\text{-Methylacetamide)}} + \text{HCl}$$ Reaction with benzoyl chloride ($\text{C}_6\text{H}_5\text{COCl}$) in presence of aqueous $\text{NaOH}$ is termed the Schotten-Baumann reaction: $$\text{C}_6\text{H}_5\text{NH}_2 + \text{C}_6\text{H}_5\text{COCl} \xrightarrow{\text{aq. NaOH}} \mathbf{\text{C}_6\text{H}_5\text{NHCOC}_6\text{H}_5 \text{ (Benzanilide)}} + \text{NaCl} + \text{H}_2\text{O}$$

The amino group ($-\text{NH}_2$) is an extraordinarily powerful activating group and strong ortho/para-director. The nitrogen lone pair strongly enriches the electron density of the benzene ring through $+R$ resonance. Consequently, aniline reacts instantaneously with electrophiles without needing any Lewis acid catalyst:

• Bromination with Bromine Water: When aniline is shaken with aqueous bromine ($\text{Br}_2/\text{H}_2\text{O}$), simultaneous bromination occurs at both ortho positions and the para position, yielding an immediate white precipitate of 2,4,6-tribromoaniline: $$\text{C}_6\text{H}_5\text{NH}_2 + 3\text{Br}_2 \xrightarrow{\text{H}_2\text{O}} \mathbf{\text{C}_6\text{H}_2\text{Br}_3\text{NH}_2 \text{ (2,4,6-Tribromoaniline, White ppt)}} \downarrow + 3\text{HBr}$$

To prepare mono-substituted derivatives (such as $p$-bromoaniline or $p$-nitroaniline) rather than poly-substituted products, the activating influence of the $-\text{NH}_2$ group must be moderated. This is achieved by protecting (blocking) the amino group through acetylation:

1. Protection (Acetylation): Aniline is reacted with acetic anhydride ($(\text{CH}_3\text{CO})_2\text{O}$) or acetyl chloride in pyridine to form acetanilide: $$\text{C}_6\text{H}_5\text{NH}_2 + (\text{CH}_3\text{CO})_2\text{O} \xrightarrow{\text{pyridine}} \mathbf{\text{C}_6\text{H}_5\text{NHCOCH}_3 \text{ (Acetanilide)}} + \text{CH}_3\text{COOH}$$
   Why is acetanilide less activating than aniline? In acetanilide, the nitrogen lone pair is delocalized into the carbonyl group ($\text{N-C=O} \leftrightarrow \text{N}^+=\text{C-O}^-$) in addition to the benzene ring. Due to this competing cross-conjugation with the electron-withdrawing carbonyl oxygen, the availability of the nitrogen lone pair for donation into the benzene ring is markedly attenuated.
2. Regioselective Electrophilic Substitution: Bromination of acetanilide with $\text{Br}_2$ in glacial acetic acid yields predominantly $p$-bromoacetanilide (the bulky $-\text{NHCOCH}_3$ group sterically blocks the ortho positions): $$\text{C}_6\text{H}_5\text{NHCOCH}_3 + \text{Br}_2 \xrightarrow{\text{CH}_3\text{COOH}} \mathbf{p\text{-Br-C}_6\text{H}_4\text{NHCOCH}_3} + \text{HBr}$$
3. Deprotection (Hydrolysis): Acidic or alkaline hydrolysis of $p$-bromoacetanilide regenerates the free amino group, delivering pure $p$-bromoaniline: $$p\text{-Br-C}_6\text{H}_4\text{NHCOCH}_3 + \text{H}_2\text{O} \xrightarrow{\text{H}^+ \text{ or } \text{OH}^-} \mathbf{p\text{-Br-C}_6\text{H}_4\text{NH}_2 \text{ (}p\text{-Bromoaniline)}} + \text{CH}_3\text{COOH}$$

Direct nitration of aniline with concentrated nitrating mixture ($\text{conc. HNO}_3 + \text{conc. H}_2\text{SO}_4$) at $288\text{ K}$ gives an unexpected distribution of isomeric nitroanilines: $$\text{Aniline} \xrightarrow{\text{conc. HNO}_3 / \text{conc. H}_2\text{SO}_4, 288\text{ K}} \mathbf{p\text{-Nitroaniline (51\%)}} + \mathbf{m\text{-Nitroaniline (47\%)}} + \mathbf{o\text{-Nitroaniline (2\%)}}$$

How to obtain pure $p$-nitroaniline? By protecting aniline first via acetylation ($\text{Ac}_2\text{O}/\text{pyridine} \to$ acetanilide), followed by nitration with $\text{HNO}_3/\text{H}_2\text{SO}_4$ at $288\text{ K}$ (yielding $p$-nitroacetanilide), and final hydrolysis with acid/base to give $p$-nitroaniline as the exclusive major product.

Heating aniline with concentrated sulphuric acid at $453-473\text{ K}$ ($180-200^\circ\text{C}$) yields aniline hydrogen sulfate, which upon prolonged heating undergoes intramolecular rearrangement to produce $p$-aminobenzenesulfonic acid (Sulfanilic acid): $$\text{C}_6\text{H}_5\text{NH}_2 + \text{H}_2\text{SO}_4 \longrightarrow \text{C}_6\text{H}_5\text{NH}_3^+\text{HSO}_4^- \xrightarrow{453-473\text{ K}} \mathbf{p\text{-H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{H}} \text{ (Sulfanilic acid)}$$ Sulfanilic acid contains both an acidic sulfonic group ($-\text{SO}_3\text{H}$) and a basic amino group ($-\text{NH}_2$) in the same molecule. An internal acid-base neutralization occurs, forming a dipolar internal salt known as a Zwitterion: $$\text{H}_2\text{N-C}_6\text{H}_4\text{-SO}_3\text{H} \rightleftharpoons \mathbf{^+\text{H}_3\text{N-C}_6\text{H}_4\text{-SO}_3^-} \text{ (Zwitterion)}$$ Due to this salt-like dipolar ionic structure, sulfanilic acid has an extraordinarily high melting point ($>280^\circ\text{C}$ with decomposition) and is insoluble in water and organic solvents, but dissolves in aqueous alkali.

Aniline does NOT undergo Friedel-Crafts alkylation or acylation in the presence of anhydrous aluminium chloride ($\text{AlCl}_3$).
Reason: Anhydrous $\text{AlCl}_3$ is a powerful Lewis acid (electron-pair acceptor) and reacts with the basic amino group ($-\text{NH}_2$) of aniline to form an insoluble Lewis acid-base coordination complex: $$\text{C}_6\text{H}_5\text{NH}_2 + \text{AlCl}_3 \longrightarrow \mathbf{\text{C}_6\text{H}_5\text{NH}_2^+-\text{AlCl}_3^-}$$ The nitrogen atom acquires a formal positive charge, transforming the strongly activating $-\text{NH}_2$ group into a powerful electron-withdrawing, strongly deactivating group that completely prevents electrophilic aromatic attack by carbocations or acylium ions.

Diazotisation: The conversion of primary aromatic amines into diazonium salts by treatment with nitrous acid ($\text{NaNO}_2 + \text{HCl}$) at ice-cold temperature ($273-278\text{ K}$ or $0-5^\circ\text{C}$): $$\mathbf{\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{273-278\text{ K}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}}$$

• Why is Arenediazonium Salt Stable at $0-5^\circ\text{C}$? In benzenediazonium cation ($\text{C}_6\text{H}_5-\text{N}^+\equiv\text{N}$), the positive charge on nitrogen is delocalized over the $\pi$-system of the benzene ring through resonance: $$\text{C}_6\text{H}_5-\text{N}^+\equiv\text{N} \longleftrightarrow \text{ortho}^+ \longleftrightarrow \text{para}^+ \longleftrightarrow \text{ortho}^+$$ This resonance delocalization provides sufficient stability to allow isolation and synthetic handling in cold aqueous solution.
• Why are Alkyldiazonium Salts Exceptionally Unstable? Aliphatic diazonium salts ($R\text{-N}_2^+X^-$) lack an aromatic $\pi$-system and cannot delocalize the positive charge. They spontaneously lose molecular nitrogen ($\text{N}_2$, one of the best leaving groups in chemistry with high triple-bond thermodynamic stability) even at $-78^\circ\text{C}$, forming reactive carbocations ($R^+$).
• Temperature Control: Diazotisation must be carried out strictly below $5^\circ\text{C}$ ($278\text{ K}$). Above $5^\circ\text{C}$, benzenediazonium chloride undergoes spontaneous hydrolysis with water to form phenol: $$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{H}_2\text{O} \xrightarrow{> 278\text{ K}} \mathbf{\text{C}_6\text{H}_5\text{OH}} + \text{N}_2\uparrow + \text{HCl}$$

Because molecular nitrogen ($\text{N}_2$) is an extraordinary leaving group, diazonium salts serve as the premier synthetic crossroads in organic chemistry:

Arenediazonium cations act as weak electrophiles and undergo electrophilic aromatic substitution with highly activated aromatic substrates (such as phenols and aromatic amines). The diazo group ($-\text{N=N}-$) is retained, linking two aromatic rings into an extended conjugated chromophoric system called an azo dye:

• Coupling with Phenol (Mildly Alkaline pH 9–10):
  In alkaline medium ($\text{pH } 9-10$), phenol is converted into the phenoxide ion ($\text{C}_6\text{H}_5\text{O}^-$), which is far more activating than phenol itself. Coupling occurs exclusively at the para-position to yield $p$-hydroxyazobenzene (brilliant orange dye): $$\mathbf{\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{pH } 9-10, 273-278\text{ K}} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-OH (Orange Dye)} + \text{Cl}^- + \text{H}_2\text{O}}$$
• Coupling with Aniline (Mildly Acidic pH 4–5):
  In mildly acidic medium ($\text{pH } 4-5$), aniline remains predominantly unprotonated with its activating lone pair free to direct electrophilic attack to the para-position, yielding $p$-aminoazobenzene (yellow dye): $$\mathbf{\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{pH } 4-5, 273-278\text{ K}} \text{C}_6\text{H}_5\text{-N=N-C}_6\text{H}_4\text{-NH}_2 \text{ (Yellow Dye)} + \text{HCl}}$$
  Why must pH be carefully controlled? If the medium is too acidic ($\text{pH} < 4$), phenoxide reverts to unionized phenol, or aniline protonates to the deactivated anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^+$), halting coupling. If the medium is too basic ($\text{pH} > 11$), the diazonium cation converts into unreactive diazotate ion ($\text{Ar-N=N-O}^-$).