Q.Arrange the following:

Key idea: Basicity of amines depends on inductive effects, resonance, solvation, and steric hindrance. pKb is inversely related to basic strength (lower pKb = stronger base).

(i) pKb values: stronger base → lower pKb. Alkyl amines are more basic than arylamines due to resonance in arylamines. Among alkyl amines, secondary > primary. Among arylamines, N-methylaniline is more basic than aniline (alkyl group donates electron density).

Decreasing pKb: $C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH$

(ii) Increasing basic strength: weakest base first. Arylamines are weaker than alkylamines. Among arylamines, N,N-dimethylaniline is (slightly) more basic than aniline itself — the two N-methyl groups donate electron density inductively, and even though they add some steric hindrance to solvation of the conjugate acid, the net effect in water still favours N,N-dimethylaniline as the stronger of the two (its conjugate-acid pKa, ~5.1, is higher than aniline's, ~4.6). Among alkylamines, secondary > primary.

Order: $C_6H_5NH_2 < C_6H_5N(CH_3)_2 < CH_3NH_2 < (C_2H_5)_2NH$

(iii)(a) p-Nitroaniline has strong electron-withdrawing –NO₂ group (decreases basicity). p-Toluidine has electron-donating –CH₃ group (increases basicity).

Increasing basic strength: p-nitroaniline < aniline < p-toluidine

(iii)(b) Benzylamine ($C_6H_5CH_2NH_2$) is an alkylamine (no resonance with ring), so strongest. N-Methylaniline is more basic than aniline due to +I of –CH₃.

Increasing basic strength: $C_6H_5NH_2 < C_6H_5NHCH_3 < C_6H_5CH_2NH_2$

(iv) In gas phase, basicity depends only on inductive effect (no solvation). Alkyl groups donate electrons, so tertiary > secondary > primary > ammonia.

Decreasing basic strength: $(C_2H_5)_3N > (C_2H_5)_2NH > C_2H_5NH_2 > NH_3$ …

The Core Idea: What Makes an Amine Basic?

Basicity in amines comes from the lone pair on nitrogen being available to accept a proton. Anything that increases electron density on nitrogen makes it a stronger base (lower pKb). Anything that decreases it (by withdrawing electrons or delocalising the lone pair) makes it weaker (higher pKb).

Three main factors compete:

1. Inductive effect – Alkyl groups push electrons toward nitrogen, making it more basic. More alkyl groups = stronger base, but only in the gas phase.
2. Resonance / Delocalisation – If the lone pair is part of a conjugated system (like in aniline), it's less available for protonation, drastically lowering basicity.
3. Solvation & Steric Hindrance – In water, the protonated ammonium ion is stabilised by hydrogen bonding. Bulky groups around nitrogen hinder solvation, reducing stability of the conjugate acid, and thus lowering basic strength in solution.

(i) Decreasing order of pKb values

pKb is the negative logarithm of the base dissociation constant. Higher pKb = weaker base. So we need to arrange from weakest base (highest pKb) to strongest base (lowest pKb).

Step 1: Identify the compounds

• $C_2H_5NH_2$ – ethylamine (1° aliphatic)
• $C_6H_5NHCH_3$ – N-methylaniline (2° aromatic)
• $(C_2H_5)_2NH$ – diethylamine (2° aliphatic)
• $C_6H_5NH_2$ – aniline (1° aromatic)

Step 2: Compare aromatic vs aliphatic

Aromatic amines are much weaker bases than aliphatic ones because the lone pair on nitrogen is delocalised into the benzene ring. So aniline and N-methylaniline will have higher pKb (weaker) than ethylamine and diethylamine.

Step 3: Within aromatic amines

N-methylaniline has an electron-donating methyl group on nitrogen, which slightly increases electron density compared to aniline. So aniline is weaker (higher pKb) than N-methylaniline.

Step 4: Within aliphatic amines

In water, diethylamine (2°) is a stronger base than ethylamine (1°) due to better inductive effect, but the solvation effect is less severe for 2° than for 3°. So diethylamine has lower pKb than ethylamine.

Step 5: Arrange from highest to lowest pKb …

Method: Electronic Effects + Solvation Analysis

This method uses inductive effect, resonance effect, solvation (hydration) effect, and steric hindrance to compare basicity. For pKb, remember: lower pKb = stronger base.

(i) Decreasing order of pKb:

$C_2H_5NH_2$, $C_6H_5NHCH_3$, $(C_2H_5)_2NH$, $C_6H_5NH_2$

Steps:

1. Identify base strength order first (stronger base → lower pKb).

   • Aliphatic amines are stronger bases than aromatic amines (due to resonance delocalisation of lone pair in aniline).
   • Among aliphatics: $(C_2H_5)_2NH$ (2° amine) > $C_2H_5NH_2$ (1° amine) in aqueous medium (due to +I effect of two alkyl groups + better solvation of 2° ammonium ion).
   • Among aromatics: $C_6H_5NHCH_3$ > $C_6H_5NH_2$ (methyl group donates electron density via +I and hyperconjugation).

2. Order of basic strength (aqueous):

   $(C_2H_5)_2NH > C_2H_5NH_2 > C_6H_5NHCH_3 > C_6H_5NH_2$

3. Convert to pKb order (reverse of basic strength):

   $C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH$

(ii) Increasing order of basic strength:

$C_6H_5NH_2$, $C_6H_5N(CH_3)_2$, $(C_2H_5)_2NH$, $CH_3NH_2$

Steps:

1. Separate aliphatic vs aromatic.

   • $(C_2H_5)_2NH$ and $CH_3NH_2$ are aliphatic → stronger bases.
   • $C_6H_5NH_2$ and $C_6H_5N(CH_3)_2$ are aromatic → weaker bases.

2. Compare within aliphatic:

   • $(C_2H_5)_2NH$ (2°) > $CH_3NH_2$ (1°) in aqueous medium.

3. Compare within aromatic:

   • $C_6H_5N(CH_3)_2$ has two methyl groups donating electrons → stronger than $C_6H_5NH_2$.

4. Final increasing order:

   $C_6H_5NH_2 < C_6H_5N(CH_3)_2 < CH_3NH_2 < (C_2H_5)_2NH$

(iii) Increasing order of basic strength:

(a) Aniline, p-nitroaniline, p-toluidine

Steps:

1. Identify substituent effect:

   • $-\text{NO}_2$ is strong electron-withdrawing (decreases basicity).
   • $-\text{CH}_3$ is electron-donating (increases basicity).

2. Order:

   p-nitroaniline < aniline < p-toluidine

   Answer: p-nitroaniline < aniline < p-toluidine

(b) $C_6H_5NH_2$, $C_6H_5NHCH_3$, $C_6H_5CH_2NH_2$

Steps:

1. Identify type:

   • $C_6H_5CH_2NH_2$ (benzylamine) — aliphatic-like (no direct resonance with ring).
   • $C_6H_5NHCH_3$ (N-methylaniline) — aromatic with +I from methyl.
   • $C_6H_5NH_2$ (aniline) — aromatic.

2. Basicity order:

   Benzylamine > N-methylaniline > Aniline

   Answer: $C_6H_5NH_2 < C_6H_5NHCH_3 < C_6H_5CH_2NH_2$

(iv) Decreasing order of basic strength in gas phase:

$C_2H_5NH_2$, $(C_2H_5)_2NH$, $(C_2H_5)_3N$, $NH_3$

Steps:

1. In gas phase, solvation is absent — only inductive effect matters. …

Here are the most common mistakes students make when solving basicity order problems for amines, along with how to avoid each.

1. Confusing $pK_b$ with Basic Strength

Mistake: Students often treat a higher $pK_b$ as meaning higher basic strength.

• Why it happens: $pK_b = -\log K_b$. A smaller $K_b$ means a weaker base, but a larger $pK_b$.
• How to avoid: Remember the rule:
  • Higher $pK_b$ → Weaker base
  • Lower $pK_b$ → Stronger base

For part (i): You need decreasing $pK_b$ (weakest to strongest base). The correct order is:

2. Ignoring the Difference Between Aqueous and Gas Phase

Mistake: Applying aqueous-phase logic (alkyl groups increase basicity) to gas-phase questions.

• Why it happens: In water, solvation effects dominate. In gas phase, inductive effect (+I) is the only factor.
• How to avoid: For gas phase, more alkyl groups = more electron density on N = stronger base.

For part (iv): Gas phase decreasing basic strength:

3. Forgetting Resonance in Aromatic Amines

Mistake: Treating aniline like an aliphatic amine.

• Why it happens: Students forget that the lone pair on N in aniline is delocalized into the benzene ring, making it less available for protonation.
• How to avoid: Always check if the N lone pair is part of a conjugated system. If yes, basicity drops sharply.

For part (iii)(b): $C_6H_5NH_2$ is weaker than $C_6H_5CH_2NH_2$ (benzylamine) because the lone pair in aniline is resonance-stabilized.

4. Misapplying the +I Effect of Alkyl Groups in Aqueous Medium

Mistake: Assuming that more alkyl groups always mean stronger base in water.

• Why it happens: In water, steric hindrance to solvation reduces basicity for bulky amines like $(C_2H_5)_3N$.
• How to avoid: In aqueous solution, the order is usually:

(due to balance of +I effect and solvation)

For part (ii): Increasing basic strength in water:

5. Ignoring Substituent Effects on Aromatic Rings

Mistake: Not considering whether substituents are electron-donating or electron-withdrawing.

• Why it happens: Students focus only on the amine group and forget the ring substituents.
• How to avoid: Use the rule:
  • Electron-donating groups (e.g., $-\text{CH}_3$) → increase basicity
  • Electron-withdrawing groups (e.g., $-\text{NO}_2$) → decrease basicity

For part (iii)(a): Increasing basic strength:

6. Mixing Up Boiling Point Trends with Basicity

Mistake: Assuming stronger bases have higher boiling points.

• Why it happens: Both depend on intermolecular forces, but differently. …