Q.Arrange the following in increasing order of their basic strength:

Concept: Basicity of amines depends on the combined effect of inductive effects (+I of alkyl groups), resonance delocalisation (in aromatic amines), and solvation of the conjugate acid in water. There is no single universal $2^\circ > 1^\circ > 3^\circ$ rule — the observed aqueous order differs between the methyl and ethyl series.

(i) Aniline ($C_6H_5NH_2$) is the weakest because the lone pair is delocalised into the benzene ring. Ammonia comes next (no +I alkyl groups). Benzylamine ($C_6H_5CH_2NH_2$) is stronger than ammonia — the $CH_2$ spacer blocks resonance, leaving only a weak inductive pull from the ring — but weaker than a simple alkylamine. Among the ethylamines, secondary > primary.

Order: $C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2 < (C_2H_5)_2NH$

(ii) For the ethyl series in water, the observed order is $2^\circ > 3^\circ > 1^\circ$: the strong +I effect of two/three ethyl groups outweighs triethylamine's poorer solvation, so triethylamine sits above ethylamine (Table 9.3: $pK_b$ $(C_2H_5)_2NH$ 3.00 < $(C_2H_5)_3N$ 3.25 < $C_2H_5NH_2$ 3.29). Aniline is weakest.

Order: $C_6H_5NH_2 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH$

(iii) For the methyl series in water, the order is $2^\circ > 1^\circ > 3^\circ$ (the smaller +I of methyl cannot compensate trimethylamine's poor solvation). Benzylamine is stronger than aniline but weaker than all the methylamines.

Order: $C_6H_5NH_2 < C_6H_5CH_2NH_2 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$

The Core Idea: What Makes an Amine Basic?

Basicity is about how readily the nitrogen atom donates its lone pair to a proton. In aqueous solution, the equilibrium is:

The stronger the base, the more it shifts right. Three factors compete:

1. Inductive effect — Alkyl groups ($-CH_3$, $-C_2H_5$) are electron-donating. They push electron density toward nitrogen, making the lone pair more available. More alkyl groups = stronger inductive push, and an ethyl group pushes harder than a methyl group.
2. Resonance effect — In aniline ($C_6H_5NH_2$), the lone pair on nitrogen is delocalised into the aromatic ring. This makes it much less available for protonation — aniline is a very weak base.
3. Solvation and steric hindrance — In water, the protonated form is stabilised by hydrogen bonding with water. More hydrogen atoms on the nitrogen (i.e., fewer alkyl groups) means better solvation. Bulky alkyl groups also physically crowd the nitrogen.

(i) $C_2H_5NH_2$, $C_6H_5NH_2$, $NH_3$, $C_6H_5CH_2NH_2$, $(C_2H_5)_2NH$

1. Identify the weakest — $C_6H_5NH_2$ (aniline) has its lone pair delocalised into the benzene ring. This is a massive drop in basicity. It is by far the weakest here.

2. Next weakest — $NH_3$ has no alkyl groups to donate electron density. It is a weaker base than any alkylamine.

3. Benzylamine — $C_6H_5CH_2NH_2$ has the amino group separated from the ring by a $-CH_2-$ spacer. The ring cannot delocalise the lone pair (too far away), but it does exert a weak electron-withdrawing inductive effect through the chain. So benzylamine is a weaker base than a simple alkylamine like ethylamine, but stronger than ammonia and much stronger than aniline.

4. Ethylamine vs diethylamine — $(C_2H_5)_2NH$ is secondary, $C_2H_5NH_2$ is primary. In water, secondary > primary. So diethylamine is the strongest here.

Order: $C_6H_5NH_2 < NH_3 < C_6H_5CH_2NH_2 < C_2H_5NH_2 < (C_2H_5)_2NH$

(ii) $C_2H_5NH_2$, $(C_2H_5)_2NH$, $(C_2H_5)_3N$, $C_6H_5NH_2$

1. Aniline is weakest — same reason as before. Lone pair delocalised into the ring.

2. Among the ethylamines — this is the ethyl series, so the aqueous order is $2^\circ > 3^\circ > 1^\circ$. The two (or three) ethyl groups exert a strong enough +I push that triethylamine, despite its poorly solvated conjugate acid, stays above ethylamine. Diethylamine, which enjoys both a strong inductive push and reasonable solvation, tops the list.

Order: $C_6H_5NH_2 < C_2H_5NH_2 < (C_2H_5)_3N < (C_2H_5)_2NH$

(iii) $CH_3NH_2$, $(CH_3)_2NH$, $(CH_3)_3N$, $C_6H_5NH_2$, $C_6H_5CH_2NH_2$

1. Aniline is weakest — resonance delocalisation, as before.

2. Benzylamine — the $-CH_2-$ spacer prevents resonance but the ring still pulls electron density inductively. So it's weaker than any of the simple methylamines here, though far stronger than aniline.

3. Methylamines — the classic aqueous order for the methyl series: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$. Dimethylamine (secondary) is strongest, then methylamine (primary), then trimethylamine (tertiary, demoted by poor solvation).

Order: $C_6H_5NH_2 < C_6H_5CH_2NH_2 < (CH_3)_3N < CH_3NH_2 < (CH_3)_2NH$

Method: Inductive Effect + Solvation Effect Analysis (for aliphatic amines) and Resonance Effect (for aromatic amines)

This is the standard approach for comparing basic strength of amines in aqueous medium. The one refinement that matters: the inductive and solvation effects pull in opposite directions, and their balance comes out differently for the methyl and ethyl series — so identify the series before applying an order.

The two aqueous orders to know (NCERT Table 9.3):

• Methyl series: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$ — i.e. $2^\circ > 1^\circ > 3^\circ$
• Ethyl series: $(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3$ — i.e. $2^\circ > 3^\circ > 1^\circ$ (the stronger +I of ethyl keeps the tertiary amine above the primary)

(i) $C_2H_5NH_2$, $C_6H_5NH_2$, $NH_3$, $C_6H_5CH_2NH_2$, $(C_2H_5)_2NH$

Step 1 — Identify the type of each amine

• $C_2H_5NH_2$ — 1° aliphatic (ethylamine)
• $(C_2H_5)_2NH$ — 2° aliphatic (diethylamine)
• $NH_3$ — ammonia
• $C_6H_5NH_2$ — aromatic (aniline)
• $C_6H_5CH_2NH_2$ — aralkyl (benzylamine)

Step 2 — Apply the rules

• Aromatic amines ($C_6H_5NH_2$) are weakest due to resonance delocalisation of the lone pair into the benzene ring.
• $NH_3$ has no +I alkyl group, so it is weaker than every alkyl/aralkyl amine here — but far stronger than aniline.
• Aralkyl amines ($C_6H_5CH_2NH_2$) sit between ammonia and the simple alkylamines: the $CH_2$ spacer blocks resonance, but the ring's weak inductive pull keeps benzylamine below ethylamine.
• Ethylamines: secondary > primary, so $(C_2H_5)_2NH > C_2H_5NH_2$.

Step 3 — Arrange in increasing order

$C_6H_5NH_2$ < $NH_3$ < $C_6H_5CH_2NH_2$ < $C_2H_5NH_2$ < $(C_2H_5)_2NH$

(ii) $C_2H_5NH_2$, $(C_2H_5)_2NH$, $(C_2H_5)_3N$, $C_6H_5NH_2$

Step 1 — Identify types

• $C_6H_5NH_2$ — aromatic (weakest)
• $C_2H_5NH_2$ — 1° aliphatic
• $(C_2H_5)_2NH$ — 2° aliphatic
• $(C_2H_5)_3N$ — 3° aliphatic

Step 2 — Apply the ETHYL-series aqueous order

This is the ethyl series, so in water: $2^\circ > 3^\circ > 1^\circ$

So: $(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2$

(Table 9.3 confirms it: $pK_b$ 3.00 < 3.25 < 3.29.)

Step 3 — Arrange

$C_6H_5NH_2$ < $C_2H_5NH_2$ < $(C_2H_5)_3N$ < $(C_2H_5)_2NH$

(iii) $CH_3NH_2$, $(CH_3)_2NH$, $(CH_3)_3N$, $C_6H_5NH_2$, $C_6H_5CH_2NH_2$

Step 1 — Identify types

• $C_6H_5NH_2$ — aromatic (weakest)
• $C_6H_5CH_2NH_2$ — aralkyl (stronger than aromatic)
• $CH_3NH_2$ — 1° aliphatic
• $(CH_3)_2NH$ — 2° aliphatic
• $(CH_3)_3N$ — 3° aliphatic

Step 2 — Apply the METHYL-series aqueous order

This is the methyl series, so in water: $2^\circ > 1^\circ > 3^\circ$

So: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N$ — and benzylamine ($pK_b$ 4.70) slots in just below trimethylamine ($pK_b$ 4.22).

Step 3 — Arrange

$C_6H_5NH_2$ < $C_6H_5CH_2NH_2$ < $(CH_3)_3N$ < $CH_3NH_2$ < $(CH_3)_2NH$

Key Concept Summary

Exam tip: First check the medium (aqueous vs gas phase), then check which alkyl series you are ordering. In water the secondary amine tops both series, but the $1^\circ$/$3^\circ$ positions swap between the methyl and ethyl series — quoting the book's $pK_b$ values (Table 9.3) is the safest justification.

🧠 Core Concept First

Basicity of amines depends on electron density on the nitrogen atom. More electron density → more available to donate → stronger base.

Key factors (in order of importance for these problems):

1. Inductive effect — alkyl groups are electron-donating (+I), aryl groups are electron-withdrawing (–I and resonance). An ethyl group donates more strongly than a methyl group.
2. Resonance effect — in aniline, the lone pair is delocalised into the ring, drastically reducing basicity.
3. Solvation & steric hindrance — in aqueous solution, more H atoms on N allow better solvation of the conjugate acid, increasing basicity.

Because factors 1 and 3 oppose each other, the aqueous order is a compromise that differs by series: methylamines follow $2^\circ > 1^\circ > 3^\circ$, ethylamines follow $2^\circ > 3^\circ > 1^\circ$.

✗ Common Mistake #1: Forgetting that Aliphatic > Aromatic always

Example from (i):

$C_6H_5NH_2$ (aniline) is much weaker than $NH_3$, $C_2H_5NH_2$, etc.

Why students go wrong:

They compare only inductive effects and forget that in aniline, the lone pair is delocalised into the benzene ring via resonance — making it far less available.

✓ How to avoid:

Always check: is the nitrogen directly attached to an aromatic ring? If yes → resonance delocalisation → very weak base. Place it last (or first in increasing order).

Correct order for (i):

✗ Common Mistake #2: Applying one fixed $2^\circ > 1^\circ > 3^\circ$ rule to every alkyl series

Example from (ii):

Students memorise "in water: $2^\circ > 1^\circ > 3^\circ$" from the methylamines and mechanically write $C_2H_5NH_2$ above $(C_2H_5)_3N$.

Why that's wrong here:

The aqueous order is a tug-of-war between the +I push (favours $3^\circ$) and solvation of the conjugate acid (favours $1^\circ$). For methyl groups the +I push is weak, so solvation wins and $3^\circ$ drops below $1^\circ$. For ethyl groups the +I push is stronger, so triethylamine stays above ethylamine. NCERT Table 9.3 confirms it: $pK_b$ $(C_2H_5)_2NH$ 3.00 < $(C_2H_5)_3N$ 3.25 < $C_2H_5NH_2$ 3.29.

✓ How to avoid:

Learn both series explicitly:

• Methyl (aq.): $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$
• Ethyl (aq.): $(C_2H_5)_2NH > (C_2H_5)_3N > C_2H_5NH_2 > NH_3$ (In the gas phase, with no solvation, both series follow $3^\circ > 2^\circ > 1^\circ > NH_3$.)

Correct order for (ii) in aqueous medium:

✗ Common Mistake #3: Forgetting that benzylamine is aliphatic in behaviour

Example from (iii):

$C_6H_5CH_2NH_2$ (benzylamine) is not like aniline — the nitrogen is not directly attached to the ring.

Why students go wrong:

They see a benzene ring and immediately assume "weak base".

✓ How to avoid:

Check the attachment:

• $C_6H_5$—$NH_2$ → aniline (weak, resonance)
• $C_6H_5$—$CH_2$—$NH_2$ → benzylamine (no resonance; behaves like an alkylamine slightly weakened by the ring's inductive pull)

Benzylamine ($pK_b$ 4.70) is more basic than aniline by far, and slightly weaker than trimethylamine ($pK_b$ 4.22) — so in (iii) it sits just below the three methylamines.

Correct order for (iii):

✗ Common Mistake #4: Placing benzylamine below ammonia in (i)

Why students go wrong:

They over-count the ring's electron-withdrawing pull and drop $C_6H_5CH_2NH_2$ below $NH_3$.

✓ How to avoid:

The $CH_2$ spacer insulates the nitrogen from the ring's resonance; only a weak inductive pull remains. Benzylamine ($pK_b$ 4.70) is slightly more basic than ammonia ($pK_b$ 4.75) — above $NH_3$, below ethylamine, exactly as the printed NCERT answer for (i) has it.

📋 Quick Summary Table

✓ Final Exam Tip

When asked "increasing order of basic strength":

1. Separate aromatic from aliphatic — aromatic goes last (weakest).
2. Identify the alkyl series: methyl → $2^\circ > 1^\circ > 3^\circ$; ethyl → $2^\circ > 3^\circ > 1^\circ$ (in water).
3. Benzylamine = aliphatic in behaviour, just above $NH_3$.
4. Always check the medium — if not specified, assume aqueous; in the gas phase both series revert to $3^\circ > 2^\circ > 1^\circ$.