Q.Account for the following:

(i) pKb of aniline is more than that of methylamine.

Concept: Basicity order of amines — aliphatic > aromatic.

In methylamine, the lone pair on nitrogen is freely available for protonation. In aniline, the lone pair is delocalised into the benzene ring via resonance, making it less available. Lower availability → weaker base → higher pKb.

(ii) Ethylamine is soluble in water whereas aniline is not.

Concept: Hydrogen bonding vs hydrophobic bulk.

Ethylamine forms strong H-bonds with water due to its small alkyl group. Aniline has a large hydrophobic benzene ring that dominates over the polar –NH₂ group, making it poorly soluble.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

Concept: Hydrolysis of Fe³⁺ by a base.

Methylamine is a base: $\ce{CH3NH2 + H2O <=> CH3NH3+ + OH-}$. The OH⁻ ions react with Fe³⁺ to form $\ce{Fe(OH)3}$ (hydrated ferric oxide), which precipitates as a reddish-brown solid.

(iv) Although amino group is o- and p- directing, aniline on nitration gives substantial m-nitroaniline.

Concept: In strongly acidic medium, –NH₂ gets protonated to –NH₃⁺, a meta-directing group.

In nitration using conc. $\ce{HNO3/H2SO4}$, aniline is protonated to anilinium ion ($\ce{C6H5NH3+}$). This group is strongly electron-withdrawing and meta-directing, so a significant amount of m-nitroaniline forms.

(v) Aniline does not undergo Friedel-Crafts reaction.

Concept: Aniline forms a complex with Lewis acid catalyst, deactivating the ring.

The lone pair on nitrogen coordinates strongly with AlCl₃ (the Lewis acid), forming a salt. This makes the nitrogen positively charged and the ring highly deactivated, preventing electrophilic substitution.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Concept: Resonance stabilisation in aryl diazonium salts.

In aromatic diazonium salts, the positive charge on the diazonium group is delocalised into the benzene ring via resonance. Aliphatic diazonium salts lack this stabilisation and decompose readily.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Concept: Avoids over-alkylation and gives pure primary amine.

Phthalimide (acidic N–H) is deprotonated, then alkylated, then hydrolysed. The product is exclusively a primary amine because the nitrogen is protected — no secondary or tertiary amine forms.

1. pKb of aniline is more than that of methylamine

   Basicity is inversely related to pKb — a higher pKb means a weaker base.

   In methylamine, the lone pair on nitrogen is fully available for protonation because the methyl group is electron-donating (+I effect).

   In aniline, the lone pair is delocalised into the benzene ring via resonance, making it less available for protonation.

   Resonance in aniline: $\ce{NH2}$ lone pair conjugates with the ring → partial double-bond character → reduced electron density on N.

   Hence, aniline is a weaker base (higher pKb) than methylamine.

2. Ethylamine is soluble in water whereas aniline is not

   Solubility in water depends on hydrogen bonding with water.

   Ethylamine has a small hydrophobic ethyl group and a polar $\ce{-NH2}$ group that forms strong H-bonds with water.

   Aniline has a large hydrophobic benzene ring that dominates the molecule’s behaviour — the nonpolar ring disrupts water structure, and the lone pair is less available for H-bonding due to resonance.

   Watch out

   Don’t confuse solubility with basicity — aniline’s poor solubility is due to the size of the hydrophobic aryl group, not just resonance.

3. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide

   Methylamine is a stronger base than water. In aqueous solution, it accepts a proton from water:

The released $\ce{OH-}$ ions react with $\ce{Fe^{3+}}$ to form a reddish-brown precipitate of hydrated ferric oxide:

Aniline, being a much weaker base, does not produce enough $\ce{OH-}$ to cause precipitation.

4. Although amino group is o- and p- directing, aniline on nitration gives substantial m-nitroaniline

   In strongly acidic conditions (like nitration with $\ce{HNO3/H2SO4}$), the amino group gets protonated to form $\ce{-NH3+}$.

   The $\ce{-NH3+}$ group is strongly electron-withdrawing (inductive effect) and meta-directing.

   Tip

   The directing effect of the free $\ce{-NH2}$ group is o/p, but under nitration conditions, it’s the protonated form that dominates.

   So the product is a mixture, with a significant amount of meta isomer — a classic exam trap.

5. Aniline does not undergo Friedel-Crafts reaction

   Friedel-Crafts reactions require a Lewis acid catalyst (e.g., $\ce{AlCl3}$).

   Aniline’s nitrogen lone pair coordinates strongly with $\ce{AlCl3}$, forming a salt-like complex. This deactivates the catalyst and also makes the nitrogen positively charged, which deactivates the ring.

   Watch out

   It’s not that aniline is “too reactive” — it’s that it poisons the catalyst by forming an unreactive complex.

6. Diazonium salts of aromatic amines are more stable than those of aliphatic amines

   Aromatic diazonium salts (e.g., $\ce{C6H5N2+}$) are stabilised by resonance delocalisation of the positive charge into the benzene ring.

   Aliphatic diazonium salts lack this resonance — they are highly unstable and decompose readily to give carbocations.

   Resonance in benzenediazonium ion: $\ce{+N#N}$ group conjugated with the ring → charge spread over ortho and para positions.

7. Gabriel phthalimide synthesis is preferred for synthesising primary amines

   This method uses phthalimide (which has an acidic N–H) to form a potassium salt, which then undergoes $\ce{S_N2}$ with an alkyl halide, followed by hydrolysis.

   Tip

   The key advantage: it avoids over-alkylation — a common problem in direct alkylation of ammonia (which gives a mixture of primary, secondary, and tertiary amines).

   Gabriel synthesis gives pure primary amines exclusively.

Here is the clear solution method for each part, following the Concept-First Approach.

Method: Structure-Reactivity Analysis (Inductive, Resonance, and Solvation Effects)

This method explains chemical behavior by analyzing how the molecular structure (bonding, lone pairs, aromaticity) influences electron density, stability of intermediates, and interaction with the solvent.

(i) pKb of aniline is more than that of methylamine.

Concept: Basicity depends on the availability of the lone pair on nitrogen for protonation.

Steps:

1. Identify the lone pair environment:
   • In methylamine ($CH_3NH_2$), the lone pair is on an $sp^3$ hybridized N. The methyl group is electron-donating (+I effect), pushing electrons toward N, making the lone pair more available.
   • In aniline ($C_6H_5NH_2$), the lone pair is on an $sp^2$ hybridized N (due to resonance). The lone pair is delocalized into the benzene ring via resonance.
2. Analyze the effect on protonation:
   • Methylamine: High electron density on N → easily accepts $H^+$ → strong base (low pKb).
   • Aniline: Lone pair is "tied up" in resonance, less available for $H^+$ → weaker base (high pKb).
3. Conclusion: Since pKb is inversely proportional to basicity, aniline has a higher pKb than methylamine.

(ii) Ethylamine is soluble in water whereas aniline is not.

Concept: Solubility in water depends on the ability to form hydrogen bonds and the size of the hydrophobic part.

Steps:

1. Analyze the polar group:
   • Both have an $-NH_2$ group capable of forming H-bonds with water.
2. Analyze the hydrophobic part:
   • Ethylamine: Has a small ethyl group ($-C_2H_5$). The hydrophilic $-NH_2$ group dominates, allowing it to dissolve.
   • Aniline: Has a large, non-polar benzene ring ($-C_6H_5$). The hydrophobic ring dominates, preventing effective solvation.
3. Conclusion: The large hydrophobic benzene ring in aniline makes it insoluble in water, while the small ethyl group in ethylamine allows solubility.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

Concept: Amines are bases; they produce $OH^-$ ions in water. Metal ions like $Fe^{3+}$ precipitate as hydroxides in basic conditions.

Steps:

1. Identify the reaction in water:
   • Methylamine ($CH_3NH_2$) acts as a base: $CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-$
2. Identify the interaction with $FeCl_3$:
   • The $OH^-$ ions produced react with $Fe^{3+}$ ions.
3. Write the precipitation reaction:
   • $Fe^{3+} (aq) + 3OH^- (aq) \rightarrow Fe(OH)_3 (s)$ (hydrated ferric oxide, a reddish-brown precipitate).
4. Conclusion: Methylamine provides the $OH^-$ necessary to precipitate $Fe(OH)_3$.

(iv) Although amino group is o- and p- directing, aniline on nitration gives a substantial amount of m-nitroaniline.

Concept: The directing effect of a group can be altered if the group itself gets protonated under the reaction conditions.

Steps:

1. Identify the reaction conditions:
   • Nitration of aniline is done using a strongly acidic mixture (conc. $HNO_3$ + conc. $H_2SO_4$).
2. Analyze the effect of the acid:
   • In strong acid, the $-NH_2$ group gets protonated to form anilinium ion ($-NH_3^+$).
3. Analyze the directing effect of the new group:
   • The $-NH_3^+$ group is a strong deactivating and meta-directing group (due to its positive charge withdrawing electron density from the ring).
4. Conclusion: Under nitration conditions, the active species is the anilinium ion, which directs the incoming nitro group to the meta position, yielding a substantial amount of m-nitroaniline.

(v) Aniline does not undergo Friedel-Crafts reaction.

Concept: Friedel-Crafts reactions require a Lewis acid catalyst ($AlCl_3$), which can be deactivated by basic substrates.

Steps:

1. Identify the catalyst and substrate:
   • Friedel-Crafts uses $AlCl_3$ (a strong Lewis acid). Aniline is a strong Lewis base.
2. Analyze the acid-base interaction:
   • The lone pair on the N of aniline forms a salt/complex with $AlCl_3$: $C_6H_5NH_2 + AlCl_3 \rightarrow C_6H_5NH_2 \cdot AlCl_3$.
3. Analyze the result:
   • The catalyst ($AlCl_3$) is consumed and deactivated.
   • The aniline molecule becomes a strong deactivating group ($-NH_2AlCl_3$), making the ring too deactivated to undergo electrophilic substitution.
4. Conclusion: The basicity of aniline deactivates the Lewis acid catalyst, preventing the Friedel-Crafts reaction.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Concept: Stability of diazonium salts depends on the ability to delocalize the positive charge.

Steps:

1. Identify the structure:
   • Diazonium salt: $R-N^+ \equiv N$.
2. Analyze aliphatic diazonium salts:
   • The positive charge is localized on the terminal N. The alkyl group ($R$) cannot stabilize this charge effectively. They are highly unstable and decompose readily to form carbocations.
3. Analyze aromatic diazonium salts:
   • The positive charge on the diazonium group ($-N^+ \equiv N$) can be delocalized into the $\pi$-electron cloud of the benzene ring via resonance.
4. Conclusion: Resonance stabilization makes aromatic diazonium salts significantly more stable than their aliphatic counterparts.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Concept: The method must avoid over-alkylation (formation of secondary and tertiary amines).

Steps:

1. Identify the problem with direct alkylation:
   • Direct reaction of $NH_3$ with $RX$ gives a mixture of $1^\circ$, $2^\circ$, and $3^\circ$ amines (and quaternary salts) because the product is more nucleophilic than the starting material.
2. Analyze the Gabriel method:
   • It uses phthalimide (which has an acidic N-H). It is first converted to its potassium salt.
   • This salt ($N$-potassiophthalimide) is a single, non-nucleophilic nitrogen source.
3. Analyze the alkylation and hydrolysis:
   • Alkylation: $N$-potassiophthalimide + $R-X$ → $N$-alkylphthalimide. (Only one alkyl group can be added because the N now has no H).
   • Hydrolysis: $N$-alkylphthalimide + $H_2O/H^+$ → Phthalic acid + pure primary amine ($R-NH_2$).
4. Conclusion: The Gabriel synthesis ensures that only one alkyl group is attached to the nitrogen, yielding a pure primary amine without any secondary or tertiary byproducts.

Here is a breakdown of the common mistakes students make for each part of this question, along with the correct conceptual approach to avoid them.

(i) pKb of aniline is more than that of methylamine.

Common Mistake:

Students often confuse pKb with Kb. They think a higher pKb means a stronger base. They also forget that pKb is inversely proportional to base strength ($pK_b = -\log K_b$).

How to Avoid:

• Memorize the relationship: Stronger base = higher $K_b$ = lower $pK_b$.
• Focus on the lone pair: In aniline, the lone pair on nitrogen is delocalized into the benzene ring (resonance), making it less available for donation. In methylamine, the +I effect of the methyl group pushes electron density onto nitrogen, making the lone pair more available.
• Conclusion: Aniline is a weaker base (higher pKb) than methylamine (lower pKb).

(ii) Ethylamine is soluble in water whereas aniline is not.

Common Mistake:

Students think that because aniline has an $-\text{NH}_2$ group (like ethylamine), it should also be soluble. They ignore the size of the hydrophobic part.

How to Avoid:

• Apply the "Like Dissolves Like" rule: Solubility depends on the balance between the hydrophilic ($-\text{NH}_2$) and hydrophobic (alkyl/aryl) parts.
• Compare the hydrophobic groups:
  • Ethylamine: Small ethyl group ($\text{C}_2\text{H}_5$). The $-\text{NH}_2$ group can form strong H-bonds with water, overcoming the small hydrophobic effect. Soluble.
  • Aniline: Large, non-polar benzene ring ($\text{C}_6\text{H}_5$). The hydrophobic ring dominates, preventing effective H-bonding with water. Insoluble.

(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.

Common Mistake:

Students treat this as a simple double displacement reaction (like $\text{NaOH} + \text{FeCl}_3$). They forget that methylamine is a base, not a source of $\text{OH}^-$ ions directly.

How to Avoid:

• Recognize the reaction type: This is a hydrolysis reaction driven by the basicity of methylamine.
• Write the correct mechanism:
  1. Methylamine ($\text{CH}_3\text{NH}_2$) is a base. It accepts a proton from water: $\text{CH}_3\text{NH}_2 + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{NH}_3^+ + \text{OH}^-$.
  2. The $\text{OH}^-$ ions produced then react with $\text{Fe}^{3+}$ ions from ferric chloride: $\text{Fe}^{3+} + 3\text{OH}^- \rightarrow \text{Fe(OH)}_3$ (hydrated ferric oxide precipitate).
• Key takeaway: The base ($\text{RNH}_2$) generates $\text{OH}^-$ in water, which then causes the precipitation.

(iv) Aniline on nitration gives a substantial amount of m-nitroaniline.

Common Mistake:

Students blindly apply the rule that $-\text{NH}_2$ is an activating and o/p-directing group. They forget that the reaction conditions can change the directing group.

How to Avoid:

• Check the reaction conditions: The nitration of aniline is done in strongly acidic medium (conc. $\text{HNO}_3$ + conc. $\text{H}_2\text{SO}_4$).
• Identify the actual species: In strong acid, the $-\text{NH}_2$ group gets protonated to form anilinium ion ($\text{C}_6\text{H}_5\text{NH}_3^+$).
• Analyze the new directing group: The $-\text{NH}_3^+$ group is a strong deactivating and meta-directing group. This is because the positive charge on nitrogen withdraws electron density from the ring by induction.
• Conclusion: The major product is m-nitroaniline because the reaction proceeds via the anilinium ion, not aniline itself.

(v) Aniline does not undergo Friedel-Crafts reaction.

Common Mistake:

Students think aniline should react because it is highly activated. They forget that the catalyst ($\text{AlCl}_3$) is a Lewis acid.

How to Avoid:

• Identify the problem: The Lewis acid catalyst ($\text{AlCl}_3$) is an electron-deficient species.
• Predict the reaction: The lone pair on the nitrogen of aniline is strongly basic. It will form a complex with the Lewis acid $\text{AlCl}_3$ (e.g., $\text{C}_6\text{H}_5\text{NH}_2 \cdot \text{AlCl}_3$).
• Consequences of complex formation:
  1. The nitrogen becomes positively charged ($\text{C}_6\text{H}_5\text{NH}_2^+ \text{AlCl}_3^-$), making the ring strongly deactivated.
  2. The catalyst is consumed and is no longer available to generate the electrophile ($\text{R}^+$ or $\text{RCO}^+$).
• Conclusion: The reaction fails because the catalyst is destroyed by the reactant.

(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines.

Common Mistake:

Students think stability is only about the positive charge on nitrogen. They don't consider the structure of the carbon attached.

How to Avoid:

• Compare the carbon attached to the $-\text{N}_2^+$ group:
  • Aromatic: The $-\text{N}_2^+$ group is attached to an $sp^2$ hybridized carbon of the benzene ring.
  • Aliphatic: The $-\text{N}_2^+$ group is attached to an $sp^3$ hybridized carbon.
• Apply the concept of resonance:
  • Aromatic diazonium salts are stabilized by resonance with the benzene ring. The positive charge can be delocalized onto the ring (e.g., $\text{C}_6\text{H}_5-\text{N}\equiv\text{N}^+ \leftrightarrow \text{C}_6\text{H}_5^+=\text{N}-\text{N}$). This makes them stable at 0-5°C.
  • Aliphatic diazonium salts have no resonance stabilization. The $sp^3$ carbon cannot delocalize the charge. They are extremely unstable and decompose immediately into a carbocation and nitrogen gas.
• Conclusion: Resonance stabilization is the key difference.

(vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.

Common Mistake:

Students think it's preferred simply because it works. They don't compare it to other methods like the reduction of alkyl halides with ammonia.

How to Avoid:

• Identify the problem with other methods: The reaction of $\text{RX}$ with $\text{NH}_3$ gives a mixture of primary, secondary, and tertiary amines (and quaternary salts). This is because the product ($\text{RNH}_2$) is more nucleophilic than $\text{NH}_3$ and reacts further.
• Explain how Gabriel Phthalimide solves this:
  1. It uses a masked ammonia equivalent (phthalimide).
  2. The nitrogen in the phthalimide anion has only one hydrogen to replace (after alkylation).
  3. After alkylation, the product is a single N-alkyl phthalimide.
  4. Hydrolysis releases only the primary amine ($\text{RNH}_2$).
• Conclusion: It is preferred because it gives a pure primary amine without any contamination from secondary or tertiary amines.