Arrange the following compounds as asked : (a) in decreasing order of pKb values : C₂H₅NH₂, (C₂H₅)₂NH, C₆H₅NHC

Question

Q. Arrange the following compounds as asked :

(a) in decreasing order of pKb values :

C_2H_5NH_2

(C_2H_5)_2NH

C_6H_5NHCH_3

C_6H_5NH_2

(b) increasing order of boiling point :

C_2H_5OH

C_2H_5NH_2

(CH_3)_2NH

C_6H_5NH_2

(C_2H_5)_2NH

C_2H_5NH_2

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Amine Basicity Order

Amine Basicity Order – From Intuition to Precision

You already know that a base is something that accepts a proton ( $H^+$ ). An amine does this through the lone pair on its nitrogen. The more readily that lone pair grabs a proton, the stronger the base. So the question becomes: what makes that lone pair more or less willing to accept a proton?

The answer depends on three competing effects: inductive effect (electron pushing/pulling from nearby groups), steric hindrance (bulky groups blocking the proton), and solvation (how water molecules stabilise the protonated amine). And critically, the order flips depending on whether you are in the gas phase or in water.

The Intuition – What Would You Expect?

Imagine a nitrogen with three hydrogens: ammonia ( $NH_3$ ). Now replace one hydrogen with a methyl group ( $CH_3$ ). Methyl is an electron-donating group — it pushes electron density toward the nitrogen. That makes the lone pair richer, so it should grab a proton more easily. So methylamine ( $CH_3NH_2$ ) should be a stronger base than ammonia.

Replace another hydrogen with a second methyl group. Now you have dimethylamine ( $(CH_3)_2NH$ ). Even more electron density on nitrogen — stronger base still. Replace the third hydrogen: trimethylamine ( $(CH_3)_3N$ ). Maximum electron donation — so you would expect it to be the strongest base of all.

That is the inductive effect prediction: more alkyl groups → stronger base. In the gas phase, this is exactly what happens. The order is:

Gas phase: $NH_3 < 1^\circ < 2^\circ < 3^\circ$

But in water, the observed order is different. Why?

The Complication – Solvation and Steric Hindrance

When an amine accepts a proton, it becomes a positively charged ammonium ion ( $RNH_3^+$ ). In water, that positive charge is stabilised by hydrogen bonding with water molecules. The more hydrogens on the nitrogen, the more hydrogen bonds it can form — and the more stable the protonated form becomes.

A primary amine ( $1^\circ$ ) has three N–H hydrogens after protonation. A secondary ( $2^\circ$ ) has two. A tertiary ( $3^\circ$ ) has only one. So solvation stabilises the protonated form in the order: $1^\circ > 2^\circ > 3^\circ$ .

At the same time, bulky alkyl groups physically block the approach of a proton to the nitrogen — this is steric hindrance. A tertiary amine has three bulky groups crowding the nitrogen, making it harder for $H^+$ to reach the lone pair.

So in water, two effects oppose the inductive effect: solvation (which favours more N–H bonds) and steric hindrance (which favours less crowding). The result is a compromise.

The Precise Statement – Aqueous Phase Order

In water, the typical order of basic strength for aliphatic amines is:

Aqueous phase: $NH_3 < 3^\circ < 1^\circ < 2^\circ$

That is, secondary amines are the strongest bases in water, followed by primary, then tertiary, then ammonia.

Important

The order in water is not simply "more alkyl groups = stronger base". The secondary amine wins because it has a good balance: enough electron donation from two alkyl groups, but still two N–H hydrogens for solvation, and not too much steric hindrance.

The Full Picture – A Table

Amine	Gas-phase order (inductive only)	Aqueous order (all effects)	Reason
$NH_3$	Weakest	Weakest	No alkyl donation; only 3 H-bonds
$1^\circ$ ( $RNH_2$ )	Second	Second	One alkyl group donates; 3 H-bonds after protonation
$2^\circ$ ( $R_2NH$ )	Third	Strongest	Two alkyl groups donate; 2 H-bonds; steric hindrance still low
$3^\circ$ ( $R_3N$ )	Strongest	Third	Three alkyl groups donate most, but only 1 H-bond and high steric hindrance

Start your 14-day free trial to unlock the full solution →

pKb is inversely related to base strength; stronger bases have lower pKb. For (a), the decreasing pKb order is $C_6H_5NH_2 > C_6H_5NHCH_3 > C_2H_5NH_2 > (C_2H_5)_2NH$ . For (b), increasing boiling point: $(CH_3)_2NH < C_2H_5NH_2 < C_2H_5OH$ . For (c), increasing solubility in water: $C_6H_5NH_2 < (C_2H_5)_2NH < C_2H_5NH_2$ .

The Concept: What pKb, Boiling Point, and Solubility Really Mean

Before we touch a single compound, let’s get the logic straight. pKb is the negative logarithm of the base dissociation constant Kb. A smaller pKb means a stronger base — the compound accepts protons more readily. So “decreasing pKb” means we list from the weakest base (highest pKb) to the strongest base (lowest pKb).

Boiling point depends on intermolecular forces: hydrogen bonding dominates here, but the number and strength of H-bonds, plus molecular size and shape, matter. Solubility in water is about how well a molecule can form hydrogen bonds with water — and how much of its nonpolar part resists dissolving.

Now let’s tackle each part.

(a) Decreasing order of pKb: $C_2H_5NH_2$ , $(C_2H_5)_2NH$ , $C_6H_5NHCH_3$ , $C_6H_5NH_2$

1. Identify the families. Two are aliphatic amines (ethylamine and diethylamine), two are aromatic amines (aniline and N-methylaniline). Aliphatic amines are generally much stronger bases than aromatic ones because the lone pair on nitrogen in aromatic amines is delocalised into the benzene ring — it’s less available to accept a proton.

Watch out

A common mistake is to think that more alkyl groups always mean stronger base. That’s true in the gas phase, but in water, solvation effects matter. For aliphatic amines in water, the order of base strength is usually secondary > primary > tertiary (due to a balance of inductive effect and solvation of the ammonium ion). So $(C_2H_5)_2NH$ is a stronger base than $C_2H_5NH_2$ .

2. Compare the aromatic amines. $C_6H_5NH_2$ (aniline) has its lone pair fully conjugated with the ring. $C_6H_5NHCH_3$ has an electron-donating methyl group on nitrogen, which pushes electron density toward nitrogen, making the lone pair more available — so it’s a stronger base than aniline. Hence pKb of aniline > pKb of N-methylaniline.

3. Compare the aliphatic amines. Diethylamine is a secondary amine; ethylamine is primary. In water, secondary aliphatic amines are slightly stronger bases than primary ones (the +I effect of two ethyl groups outweighs the slight steric hindrance to solvation). So $(C_2H_5)_2NH$ has a lower pKb than $C_2H_5NH_2$ .

4. Combine the two groups. Aromatic amines are much weaker bases than aliphatic amines. So the weakest base (highest pKb) is $C_6H_5NH_2$ , then $C_6H_5NHCH_3$ , then $C_2H_5NH_2$ , and the strongest base (lowest pKb) is $(C_2H_5)_2NH$ .

Tip

Approximate pKb values (in water) for reference: aniline ~9.4, N-methylaniline ~9.2, ethylamine ~3.4, diethylamine ~3.0. This confirms the order.

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

(b) Increasing order of boiling point: $C_2H_5OH$ , $C_2H_5NH_2$ , $(CH_3)_2NH$

1. What determines boiling point here? All three can form hydrogen bonds, but the strength and number differ. Ethanol has an O–H group — oxygen is more electronegative than nitrogen, so O–H···O hydrogen bonds are stronger than N–H···N bonds. Also, ethanol can form two H-bonds per molecule (one donor, one acceptor), but so can primary and secondary amines.

2. Compare the amines. $C_2H_5NH_2$ (ethylamine) is a primary amine — it has two N–H hydrogens, so it can form more extensive H-bonding networks than $(CH_3)_2NH$ (dimethylamine), which is secondary and has only one N–H. Also, ethylamine is slightly larger and has a more linear shape, which allows better packing. So ethylamine boils higher than dimethylamine. …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 19 on this concept.

CBSE 2026Set V11 markMCQ
$Q. Which of the following groups when present at para position increases the basic strength of aniline?$
(a) $-\text{NO}_2$
(b) $-\text{Br}$
(c) $-\text{NH}_2$
(d) $-\text{COOH}$
›Reveal solutionSolution
An electron-donating para substituent raises aniline's basicity; among the options only –NH $_2$ is electron-donating.

The basic strength of aniline depends on the availability of the lone pair on the nitrogen of the –NH $_2$ group. Groups that push electron density into the ring (electron-donating, +R/+I) increase the electron density on nitrogen and therefore increase basicity; electron-withdrawing groups do the opposite.

Assessing the para substituents:

$-\text{NO}_2$ : strongly electron-withdrawing ( $-R,\,-I$ ) → decreases basicity.

$-\text{Br}$ : electron-withdrawing by induction → slightly decreases basicity. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: Amines behaves as ______ due to presence of an unshared pair of electron on Nitrogen atom.$
›Reveal solutionSolution

Because of the nitrogen lone pair, amines behave as bases (and nucleophiles).

In an amine, the nitrogen atom carries an unshared (lone) pair of electrons. It can donate this lone pair to a proton (H+) or to an electron-deficient centre. Donating the lone pair to H+ makes the amine act as a Bronsted/Lewis base; donating it t …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Arrange the following in increasing order of their basic strength- C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2$
›Reveal solutionSolution

Aniline is the weakest (its lone pair is delocalised into the ring); among the ethylamines in water the accepted order is secondary > tertiary > primary, so diethylamine is the strongest and the increasing order is C6H5NH2 < C2H5NH2 < (C2H5)3N < (C2H5)2NH.

Basicity in aqueous solution is decided by the net effect of three factors: the electron-donating alkyl groups (+I effect, which raises the electron density on nitrogen and increases basicity), the steric hindrance to protonation together with the solvation (hydrogen bonding) of the resulting ammonium cation, and - for aniline only - resonance delocalisation of the nitrogen lone pair into the benzene ring, which sharply lowers its availability and hence its basicity.

$C_6H_5NH_2$ (aniline) is the weakest base of the four, because its lone pair is delocalised into the aromatic ring and is much less available for protonation.

Among the three ethylamines, the order actually observed in aqueous solution (the balance of +I effect, steric crowding and cation solvation) is:

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

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Identify the most basic among the given amines:$
(a) C6H3NH2
(b) (C2H5)2NH
(c) (C2H5)3N
(d) C2H5NH2
›Reveal solutionSolution
In water, basicity of these amines follows (C2H5)2NH > (C2H5)3N > C2H5NH2, so the most basic is diethylamine, (C2H5)2NH.

Basicity of an amine in aqueous solution is governed by three competing effects: (i) the +I (electron-releasing) effect of alkyl groups, which raises electron density on N and favours protonation; (ii) solvation of the resulting ammonium cation by water through H-bonding, which stabilises the cation and favours basicity; and (iii) steric hindrance of bulky alkyl groups around N, which hampers solvation.

C2H5NH2 (1°): one ethyl group gives a modest +I effect; good solvation (two N-H bonds available). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Arrange the following compounds in increasing order of their basic strength in aqueous solution: NH3, C2H5NH2, (C2H5)2NH, (C2H5)3N$
›Reveal solutionSolution
In AQUEOUS solution, basicity of amines depends not just on the electron-donating (+I) effect of alkyl groups but also on steric hindrance and how well the resulting ammonium ion is stabilised by hydrogen-bonding with water (solvation); this combination gives the order NH3 < 3 degree < 1 degree < 2 degree amine for the ethyl series.

Factors affecting basicity in water:

Inductive (+I) effect of alkyl groups: increases electron density on N, favouring basicity; increases with number of alkyl groups (3 degree > 2 degree > 1 degree > NH3 on this factor alone).

Steric hindrance: bulky alkyl groups around N hinder the approach of a proton and hinder solvation of the resulting ammonium cation.

Solvation (H-bonding) of the ammonium ion R-NH3+ (or its substituted forms) by water: more N-H bonds allow more hydrogen bonding with water, better stabilising the cation and boosting the observed basicity.

NH3 has no alkyl groups at all (weakest +I effect) - it is the weakest base of the four.

…
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$Q. Why do amines react as nucleophiles?$
›Reveal solutionSolution

An available, reasonably accessible lone pair on N is exactly what defines a good nucleophile/Lewis base.

In an amine, R–NH₂ (or R₂NH, R₃N), the nitrogen is sp³ hybridized (approximately) with one lone pair of electrons not involved in any bond. This lone pair is available to attack an electron-deficient (electrophilic) carbon or other centre, forming a new bond — the defining behaviour of a nucleophile (and, correspondingly, amines also behave as Lewis/Brønsted bases, accepting a proton with this same lone pair). This is why amines readi …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Which of the following is least basic ?$
(i) (CH₃)₂NH
(ii) NH₃
(iii) C₆H₅NH₂
(iv) (CH₃)₃N
›Reveal solutionSolution
Aniline is least basic because the lone pair on its N atom is delocalised into the benzene ring, making it far less available to accept a proton than in the aliphatic amines.

Basicity of an amine depends on how freely available the nitrogen lone pair is to accept a proton (H⁺):

(CH₃)₂NH (dimethylamine, 2° aliphatic amine): two +I (electron-donating) methyl groups increase electron density on N → quite basic.

NH₃: no alkyl groups, moderate basicity, used as the reference.

(CH₃)₃N (trimethylamine, 3° aliphatic amine): three +I methyl groups would suggest highest basicity electronically, though steric hindrance and poorer solvation of the resulting cation slightly reduce it — but it remains far more basic than aniline. …
Start your 14-day free trial to see the full solution →
CBSE 2024Set 56/1/11 markMCQ
$Q. Assertion (A): Aniline is a stronger base than ammonia. Reason (R): The unshared electron pair on nitrogen atom in aniline becomes less available for protonation due to resonance. [Codes (A)-(D) as in the Assertion-Reason instruction.]$
›Reveal solutionSolution
Aniline is a weaker base than ammonia because the lone pair on nitrogen in aniline is delocalized into the benzene ring via resonance, making it less available for protonation. Therefore, Assertion (A) is false, and Reason (R) is true.

To understand the basicity of amines, we need to consider the availability of the lone pair of electrons on the nitrogen atom. A base, according to the Brønsted-Lowry definition, is a proton acceptor, and according to the Lewis definition, it is an electron pair donor. Therefore, the more readily available the lone pair on nitrogen is to accept a proton (or donate to an acid), the stronger the base.

Factors that increase the electron density on the nitrogen atom or localize the lone pair make it a stronger base. Conversely, factors that decrease the electron density on nitrogen or delocalize the lone pair make it a weaker base.

Let's analyze ammonia and aniline based on these principles.

Basicity of Ammonia ( $\text{NH}_3$ )

In ammonia, the nitrogen atom has one lone pair of electrons.

This lone pair is localized on the nitrogen atom and is readily available for donation to a proton ( $\text{H}^+$ ).

There are no significant electron-withdrawing or electron-donating groups directly attached to the nitrogen that would extensively delocalize or concentrate this lone pair. Thus, ammonia serves as a reference point for basicity.

Basicity of Aniline ( $\text{C}_6\text{H}_5\text{NH}_2$ )

Aniline consists of an amino ( $\text{-NH}_2$ ) group attached to a benzene ring.

The nitrogen atom in the amino group also possesses a lone pair of electrons.

However, this lone pair is in conjugation with the $\pi$ -electron system of the benzene ring. This leads to resonance.

$\text{C}_6\text{H}_5\text{NH}_2 \longleftrightarrow \text{C}_6\text{H}_5\text{NH}_2^+ \text{ (with negative charge on ortho/para positions)}$

Due to this resonance, the lone pair of electrons on the nitrogen atom is not entirely localized on nitrogen; it is delocalized over the nitrogen and the ortho and para positions of the benzene ring.

This delocalization makes the lone pair less available for donation to a proton.

Comparison of Basicity …
Start your 14-day free trial to see the full solution →
CBSE 2024Set FZ1 markMCQ
$Q. The compound having most basic strength is:$
(a) $(CH_3)_2NH$
(b) $CH_3NH_2$
(c) $(CH_3)_3N$
(d) $NH_3$
›Reveal solutionSolution

In water, dimethylamine is the strongest base of the four → option (a).

Concept. Basicity of an amine in water is decided by three factors: the electron-releasing (+I) effect of alkyl groups (raises basicity), steric hindrance around N (lowers it), and stabilisation of the protonated cation by H-bonding to water (more H on N = better solvation).

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. The compound showing highest basic strength in aqueous solution is -$
(a) (C2H5)3N
(b) (C2H5)2NH
(c) C2H5NH2
(d) NH3
›Reveal solutionSolution

In water, basicity of alkylamines is governed by a balance of +I (electron-donating) effect, steric hindrance, and how well the protonated ammonium ion is stabilised by hydrogen bonding with water; for ethylamines this gives the order secondary > primary > tertiary > ammonia.

Electron-donating alkyl groups increase electron density on N (favouring basicity), which would suggest (C2H5)3N should be most basic. …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$Q. Which of the following compounds is the most basic?$
(a) benzylamine (C6H5-CH2NH2)
(b) aniline (C6H5-NH2)
(c) p-nitroaniline (4-NH2-C6H4-NO2)
(d) m-nitroaniline (3-NH2-C6H4-NO2)
›Reveal solutionSolution

Benzylamine is most basic because its nitrogen lone pair is not conjugated with the benzene ring, while all the anilines lose basicity because their -NH $_2$ lone pair delocalises into the ring (and nitro groups withdraw electron density further).

In aniline, the nitrogen lone pair is conjugated (delocalised via resonance) into the aromatic ring. This delocalisation makes the lone pair much less available to accept a proton, so aniline is far less basic than an ordinary alkyl amine.

In benzylamine, the -CH $_2$ - group between the ring and -NH $_2$ breaks this conjugation — the nitrogen behaves essentially like an aliphatic amine, with its lone pair fully available for protonation. This makes benzylamine considerably more basic than aniline.

The two nitroanilines are even less basic than aniline itself, because the electron-withdrawing -NO $_2$ group further reduces electron density on nitrogen: …

Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$Q. The amine having lowest pKb value is:$
(a) CH3NH2
(b) (CH3)2NH
(c) (CH3)3N
(d) C6H5NH2
›Reveal solutionSolution
In aqueous solution, basicity of methylamines follows (CH3)2NH > CH3NH2 > (CH3)3N > C6H5NH2, because a balance of the electron-donating (+I) effect of methyl groups and steric hindrance to solvation (hydration) of the resulting ammonium ion determines the base strength; the secondary amine gives the best balance.

Basicity of an amine in water depends on two opposing factors:

+I (electron-donating) effect of alkyl groups - increases electron density on N, increasing basicity as more methyl groups are added (favours 3-degree > 2-degree > 1-degree).

Steric hindrance and reduced solvation (hydrogen bonding with water) of the ammonium cation - a more crowded ammonium ion is less stabilised by water, decreasing basicity (favours 1-degree > 2-degree > 3-degree). …
Start your 14-day free trial to see the full solution →

Q.Arrange the following compounds as asked :

Concept understanding — Amine Basicity Order

Amine Basicity Order – From Intuition to Precision

The Intuition – What Would You Expect?

The Complication – Solvation and Steric Hindrance

The Precise Statement – Aqueous Phase Order

The Full Picture – A Table

(a) Decreasing order of pKb values

The Concept: What pKb, Boiling Point, and Solubility Really Mean

(a) Decreasing order of pKb: $C_2H_5NH_2$ , $(C_2H_5)_2NH$ , $C_6H_5NHCH_3$ , $C_6H_5NH_2$

(b) Increasing order of boiling point: $C_2H_5OH$ , $C_2H_5NH_2$ , $(CH_3)_2NH$

Q.Arrange the following compounds as asked :

The Concept: What pKb, Boiling Point, and Solubility Really Mean

(a) Decreasing order of pKb: C2H5NH2C_2H_5NH_2C2​H5​NH2​, (C2H5)2NH(C_2H_5)_2NH(C2​H5​)2​NH, C6H5NHCH3C_6H_5NHCH_3C6​H5​NHCH3​, C6H5NH2C_6H_5NH_2C6​H5​NH2​

(b) Increasing order of boiling point: C2H5OHC_2H_5OHC2​H5​OH, C2H5NH2C_2H_5NH_2C2​H5​NH2​, (CH3)2NH(CH_3)_2NH(CH3​)2​NH

🎓Unlock everything free for 14 days

(a) Decreasing order of pKb: $C_2H_5NH_2$ , $(C_2H_5)_2NH$ , $C_6H_5NHCH_3$ , $C_6H_5NH_2$

(b) Increasing order of boiling point: $C_2H_5OH$ , $C_2H_5NH_2$ , $(CH_3)_2NH$

Unlock everything free for 14 days