(i) $(\text{CH}_3)_2\text{CHNHCH}_3$:

The longest continuous carbon chain attached to nitrogen has 3 carbons (propane). The nitrogen atom holds a methyl substituent and is attached at position 2 of the propane chain.

IUPAC Name: N-Methylpropan-2-amine (Secondary amine)

(ii) $\text{C}_6\text{H}_5\text{N(CH}_3)_2$:

The parent aromatic structure is benzenamine (aniline). The nitrogen atom holds two methyl groups.

IUPAC Name: N,N-Dimethylbenzenamine (or N,N-Dimethylaniline, Tertiary amine)

Hinsberg's Reagent: Benzenesulfonyl chloride ($\text{C}_6\text{H}_5\text{SO}_2\text{Cl}$).

1. Reaction with Primary ($1^\circ$) Amines:

Primary amines react with Hinsberg's reagent to form an $N$-alkylbenzenesulfonamide:

The hydrogen attached to nitrogen in this compound is **highly acidic** due to the presence of the strong electron-withdrawing sulfonyl group ($-\text{SO}_2-$). Thus, it dissolves readily in sodium hydroxide ($\text{NaOH}$) solution, forming a clear, soluble sodium salt.

2. Reaction with Secondary ($2^\circ$) Amines:

Secondary amines react with Hinsberg's reagent to form $N,N$-dialkylbenzenesulfonamide:

Since this molecule has **no acidic hydrogen** attached to the nitrogen atom, it is completely **insoluble in aqueous alkali** and precipitates out of solution as a solid suspension.

Note: Tertiary ($3^\circ$) amines do not react with Hinsberg's reagent.

Chemical Explanation:

Aniline ($\text{C}_6\text{H}_5\text{NH}_2$) acts as a strong Lewis base due to the presence of a lone pair of electrons on the nitrogen atom. The catalyst anhydrous aluminium chloride ($\text{AlCl}_3$) used in Friedel-Crafts reactions is a powerful, electron-deficient Lewis acid.

When mixed, aniline immediately reacts with the anhydrous $\text{AlCl}_3$ catalyst to form an insoluble, highly polar **salt complex**:

In this complex, the nitrogen atom acquires a positive charge. This highly electron-withdrawing positive nitrogen strongly deactivates the aromatic ring towards electrophilic attack, preventing any subsequent alkylation or acylation from taking place.

(i) Gaseous Phase Order:

Reason: In the gas phase, solvent hydration is absent. Basicity is dictated strictly by the $+I$ inductive effect of the electron-releasing methyl groups, which increases the electron density on nitrogen as the number of alkyl groups increases.

(ii) Aqueous Solution Order:

Reason: In water, basicity is governed by a combination of three competing factors:

1. Inductive Effect ($+I$): Favors $3^\circ \gt 2^\circ \gt 1^\circ$.
2. Solvation (Hydration) Effect: Water molecules stabilize the conjugate cation via hydrogen bonding. The smaller the cation, the more hydrogen bonds it forms, releasing greater hydration energy, favoring $1^\circ \gt 2^\circ \gt 3^\circ$.
3. Steric Hindrance: Bulky methyl groups block the proton approach to nitrogen, favoring $1^\circ \gt 2^\circ \gt 3^\circ$.

The combination of these factors makes the secondary amine $(\text{CH}_3)_2\text{NH}$ the most basic, while the tertiary amine $(\text{CH}_3)_3\text{N}$ drops in basicity due to steric hindrance and poor solvation.

Increasing Order:

Explanation:

Similar to methyl-substituted amines, basicity is controlled by inductive, solvation, and steric effects in water.

Because the ethyl group ($-\text{C}_2\text{H}_5$) is larger and more electron-releasing than the methyl group, its stronger $+I$ inductive effect in the tertiary amine $(\text{C}_2\text{H}_5)_3\text{N}$ is able to overcome the steric hindrance and lower solvation to a greater extent. This makes $(\text{C}_2\text{H}_5)_3\text{N}$ more basic than the primary amine $\text{C}_2\text{H}_5\text{NH}_2$. However, the secondary amine $(\text{C}_2\text{H}_5)_2\text{NH}$ remains the strongest base because it represents the optimal combination of induction and steric accessibility.

1. Resonance Effect: In aniline ($\text{C}_6\text{H}_5\text{NH}_2$), the lone pair of electrons on the nitrogen atom is in conjugation with the $\pi$-system of the benzene ring. This lone pair is delocalized over the ortho and para positions of the ring via resonance, making it less available for donation to a proton ($\text{H}^{+}$). In methylamine ($\text{CH}_3\text{NH}_2$), the methyl group increases electron density on nitrogen through its $+I$ inductive effect, making the lone pair highly available.
2. Relative Stability of Conjugate Cations: Aniline is resonance-stabilized by 5 distinct structures, whereas its protonated conjugate acid, the anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^{+}$), is stabilized by only 2 resonance structures. Protonation of aniline therefore causes a significant loss of resonance stabilization energy, making the reaction thermodynamically unfavorable compared to aliphatic amines.

Increasing Order:

Explanation based on Ring Substituents:

• $p$-Nitroaniline (Weakest Base): The nitro group ($-\text{NO}_2$) is a powerful electron-withdrawing group via resonance ($-M$) and induction ($-I$). It strongly pulls electron density away from the $-\text{NH}_2$ group, making nitrogen's lone pair least available for protonation.
• Aniline: Baseline reference.
• $p$-Toluidine: The methyl group ($-\text{CH}_3$) is electron-donating via induction ($+I$) and hyperconjugation. This increases electron density on nitrogen, making it a stronger base than aniline.
• $p$-Methoxyaniline (Strongest Base): The methoxy group ($-\text{OCH}_3$) is strongly electron-donating through resonance ($+M$) in the para position, maximizing the electron density on nitrogen and making it the most basic compound of the group.

Hoffmann Bromamide Degradation: This reaction is a laboratory method used to convert a primary amide to a primary amine containing **one less carbon atom** than the starting amide, by treating it with bromine ($\text{Br}_2$) in an aqueous or ethanolic solution of sodium hydroxide ($\text{NaOH}$).

Balanced Equation for Aniline:

Why "Degradation":

During the reaction mechanism, an alkyl/aryl group migration occurs where the carbonyl carbon ($-\text{C}=\text{O}$) of the amide is eliminated as a carbonate ion ($\text{CO}_3^{2-}$). The resulting amine has one less carbon atom than the starting material, hence the name "degradation".

The Gabriel Phthalimide Synthesis is used to prepare pure **primary aliphatic amines** via three main chemical steps:

1. Salt Formation: Phthalimide reacts with ethanolic $\text{KOH}$ to form potassium phthalimide.
2. Alkylation ($S_N2$): Potassium phthalimide is heated with an alkyl halide ($\text{R-X}$) to yield $N$-alkylphthalimide.
3. Hydrolysis: $N$-alkylphthalimide is heated with aqueous sodium hydroxide ($\text{NaOH}$) to yield the primary amine ($\text{R-NH}_2$) and sodium phthalate.

Why Aromatic Amines Cannot Be Prepared:

The second step of this synthesis relies on a nucleophilic substitution ($S_N2$) reaction. To synthesize aniline, potassium phthalimide would need to react with an aryl halide ($\text{C}_6\text{H}_5\text{-X}$). However, aryl halides do **not** undergo nucleophilic substitution under normal conditions due to resonance stabilization imparting partial double bond character to the $\text{C-X}$ bond, making it too strong to cleave. Therefore, aniline cannot be prepared by this method.

Carbylamine Reaction: Aliphatic and aromatic primary ($1^\circ$) amines, when warmed with chloroform ($\text{CHCl}_3$) in the presence of ethanolic potassium hydroxide ($\text{KOH}$), undergo a reaction to produce highly foul-smelling compounds called **isocyanides** (or carbylamines).

Chemical Equation for Ethylamine:

Qualitative Diagnostic Use:

Because only primary ($1^\circ$) amines undergo this reaction to produce a characteristic offensive, sharp odor, the Carbylamine reaction is used as a highly selective diagnostic test to distinguish primary amines from secondary and tertiary amines in qualitative analysis.

• Arenediazonium Salts Stability: In benzenediazonium chloride ($\text{C}_6\text{H}_5\text{N}_2^{+}\text{Cl}^{-}$), the positive diazo group is attached to an $sp^2$-hybridized carbon of the benzene ring. The positive charge on the nitrogen atom is stabilized by **resonance delocalization** into the aromatic ring system, allowing it to exist at low temperatures ($0\text{-}5^\circ\text{C}$).
• Alkyldiazonium Salts Instability: Alkyl groups cannot undergo resonance stabilization. Alkyldiazonium salts ($\text{R-N}_2^{+}\text{Cl}^{-}$) are highly unstable and decompose immediately (even below $0^\circ\text{C}$) to release nitrogen gas ($\text{N}_2$) and form highly reactive carbocations, which react with water to yield alcohols:
  $$ \text{R-N}_2^{+}\text{Cl}^{-} + \text{H}_2\text{O} \rightarrow \text{R-OH} + \text{N}_2(g)\uparrow + \text{HCl} $$

Azo coupling is an electrophilic aromatic substitution reaction where the diazonium cation acts as an electrophile, attacking the para position of activated rings:

(i) Coupling with Phenol (Basic Medium, pH 9-10):

The product **$p$-hydroxyazobenzene** is an **Orange Dye**.

(ii) Coupling with Aniline (Acidic Medium, pH 4-5):

The product **$p$-aminoazobenzene** is a **Yellow Dye**.

(i) Conversion to Fluorobenzene (Balz-Schiemann Reaction):

React benzenediazonium chloride with fluoroboric acid ($\text{HBF}_4$) to form a precipitate of benzenediazonium fluoroborate, which is then dried and heated to decompose into fluorobenzene:

(ii) Conversion to Benzonitrile (Sandmeyer Reaction):

Treat benzenediazonium chloride with cuprous cyanide ($\text{CuCN}$) dissolved in aqueous potassium cyanide ($\text{KCN}$):

(i) Methylamine ($1^\circ$) and Dimethylamine ($2^\circ$):

Use the **Carbylamine Test**. Methylamine, being a $1^\circ$ amine, reacts on warming with chloroform and alcoholic $\text{KOH}$ to form a foul-smelling isocyanide. Dimethylamine, being a $2^\circ$ amine, does not react.

(ii) Aniline and Ethylamine:

Use the **Azo-Dye Test**. Dissolve the compound in dilute $\text{HCl}$ at $0\text{-}5^\circ\text{C}$ and add sodium nitrite ($\text{NaNO}_2$). Pour this diazonium mixture into an alkaline solution of $\beta$-naphthol.

• Aniline forms a stable arenediazonium salt that couples with $\beta$-naphthol to produce a brilliant **red/orange azo dye**.
• Ethylamine forms an unstable alkyldiazonium salt that decomposes to form alcohol and nitrogen gas without forming any dye.

Step 1: Nitration of Benzene.

Treat benzene with a nitrating mixture containing concentrated nitric acid ($\text{HNO}_3$) and concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) at $323\text{-}333\text{ K}$ to yield nitrobenzene:

Step 2: Reduction of Nitrobenzene.

Reduce nitrobenzene to aniline using iron filings ($\text{Fe}$) and concentrated hydrochloric acid ($\text{HCl}$):

Note: Fe/HCl is preferred over Sn/HCl because the hydrated $\text{FeCl}_2$ formed releases hydrochloric acid on hydrolysis, requiring only a tiny starting initiator amount of acid.

Step 1: Amide Formation.

React benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$) with ammonia ($\text{NH}_3$) and heat the resulting ammonium salt strongly to obtain benzamide:

Step 2: Hoffmann Bromamide Degradation.

Treat benzamide with bromine and sodium hydroxide to degrade it to aniline:

Explanation:

In an amide, the nitrogen atom is bonded directly to a highly electronegative, electron-withdrawing carbonyl group ($-\text{C}=\text{O}$). This carbonyl group strongly withdraws electron density via both resonance and induction, polarizing the $\text{N-H}$ bond and making it easier to lose a proton ($\text{H}^{+}$).

Furthermore, the conjugate anion formed after losing a proton is highly stabilized by resonance as the negative charge is delocalized onto the highly electronegative carbonyl oxygen atom:

In contrast, in an amine, there is no electron-withdrawing carbonyl group. The alkyl group is actually electron-donating ($+I$), which increases electron density on nitrogen, making the $\text{N-H}$ bond less polar and its conjugate anion highly unstable. Thus, amides are much more acidic than amines.

(i) Reduction of Propanamide:

Lithium aluminium hydride ($\text{LiAlH}_4$) reduces the amide carbonyl group directly to a methylene group, preserving the carbon count to yield a primary aliphatic amine:

(ii) Bromination of Aniline:

The amino group ($-\text{NH}_2$) is an exceptionally powerful activating and ortho/para-directing group due to resonance. Aniline reacts instantly with bromine water at room temperature to undergo electrophilic trisubstitution, yielding a white precipitate:

Why Meta Product Forms:

Nitration is carried out using a mixture of concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$. Under these strongly acidic conditions, aniline (a base) gets heavily protonated to form the **anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^{+}$)**:

Unlike the amino group ($-\text{NH}_2$), which is activating and ortho/para-directing, the positive charge on the anilinium ion makes it a highly deactivating and **meta-directing** group. Consequently, the reaction yields a significant amount of the meta product ($\approx 47\% m$-nitroaniline, $51\% p$-nitroaniline, and $2\% o$-nitroaniline).

How to Bypass the Problem:

The activating power of the $-\text{NH}_2$ group is controlled ("protected") by **acetylation** using acetic anhydride in pyridine to form acetanilide ($\text{C}_6\text{H}_5\text{-NHCOCH}_3$). The nitrogen lone pair is shared by resonance with the acetyl carbonyl oxygen, reducing its activating power.

Acid nitration of acetanilide selectively yields the para product due to steric hindrance at the ortho position. Subsequent acid/base hydrolysis removes the protecting group to yield pure $p$-nitroaniline:

1. Step 1 ($\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Fe / HCl}} \text{A}$): Reducing the nitro group yields aniline:
   $$ \text{A} = \text{C}_6\text{H}_5\text{NH}_2 \quad \text{(Aniline)} $$
2. Step 2 ($\text{A} \xrightarrow{\text{NaNO}_2 + \text{HCl}, 273\text{-}278\text{ K}} \text{B}$): Diazotization of aniline under cold conditions yields benzene diazonium chloride:
   $$ \text{B} = \text{C}_6\text{H}_5\text{N}_2^{+}\text{Cl}^{-} \quad \text{(Benzenediazonium Chloride)} $$
3. Step 3 ($\text{B} \xrightarrow{\text{H}_2\text{O} / \Delta} \text{C}$): Warming the diazonium salt in water hydrolyzes the diazo group, releasing nitrogen gas and forming phenol:
   $$ \text{C} = \text{C}_6\text{H}_5\text{OH} \quad \text{(Phenol)} $$

Answer: Intermediate A is aniline, B is benzenediazonium chloride, and C is phenol.