ChemistryCBSE Class 12 · Amines

Basicity of Amines: Class 12 Chemistry

Updated 15 October 2026 · Fact-checked

Amines are bases because the nitrogen lone pair accepts a proton. Alkyl groups push electron density onto nitrogen, so aliphatic amines are stronger bases than ammonia, while in aniline the lone pair is delocalised into the ring, making it much weaker. In water the order also depends on solvation and steric hindrance: $(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3$.

Understand Basicity of Amines

Why amines are basic. Like ammonia, an amine has a lone pair on nitrogen that can accept a proton. In water: $R - N H_{2} + H_{2} O ⇌ R - N H_{3}^{+} + O H^{-}$ . With acids amines form salts: $R N H_{2} + H C l \to R N H_{3}^{+} C l^{-}$ ; aniline gives anilinium chloride. These salts are soluble in water and give back the amine when treated with NaOH, which is a convenient way to separate amines from neutral compounds.

Measuring basic strength. $K_{b} = \frac{[ R N H _{3}^{+} ] [ O H ^{-} ]}{[ R N H _{2} ]}$ and $p K_{b} = - lo g K_{b}$ . A **larger $K_{b}$ (smaller $p K_{b}$ ) means a stronger base.** NCERT's values in water: $N H_{3}$ 4.75; methanamine 3.38, N-methylmethanamine 3.27, N,N-dimethylmethanamine 4.22; ethanamine 3.29, N-ethylethanamine 3.00, N,N-diethylethanamine 3.25; benzenamine (aniline) 9.38, phenylmethanamine (benzylamine) 4.70, N-methylaniline 9.30, N,N-dimethylaniline 8.92.

Aliphatic amines vs ammonia. Alkyl groups are electron-releasing (+I). They increase the electron density on nitrogen, so the lone pair is more available, and they also stabilise the substituted ammonium ion by spreading its positive charge. So aliphatic amines are stronger bases than ammonia.

Gas phase. With no solvent, only the inductive effect matters, so basicity rises with the number of alkyl groups: $3° > 2° > 1° > N H_{3}$ .

Water. In solution, the ammonium ion formed is stabilised by hydrogen bonding with water (solvation). The more N–H bonds the cation has, the better it is solvated: $R N H_{3}^{+}$ (three H) > $R_{2} N H_{2}^{+}$ (two) > $R_{3} N H^{+}$ (one). This favours primary amines. Steric hindrance from bulky alkyl groups also hinders solvation and proton approach. The observed order is a balance of all three effects:

Methyl series: $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}$ .
Ethyl series: $(C_{2} H_{5})_{2} N H > (C_{2} H_{5})_{3} N > C_{2} H_{5} N H_{2} > N H_{3}$ (ethyl groups give more +I, which helps the tertiary amine).

Aromatic amines. In aniline, the nitrogen lone pair is in conjugation with the benzene ring and is delocalised over it (five resonance structures for aniline). It is less available for protonation. In the anilinium ion there is no lone pair left to delocalise, so it has only two resonance structures (those of the ring). Protonation therefore costs aniline its extra resonance stabilisation, and aniline ( $p K_{b}$ 9.38) is a much weaker base than ammonia. Benzylamine ( $C_{6} H_{5} C H_{2} N H_{2}$ ), in which a $C H_{2}$ group separates N from the ring, behaves like an aliphatic amine ( $p K_{b}$ 4.70).

Substituted anilines. Electron-releasing groups ( $- O C H_{3}$ , $- C H_{3}$ ) on the ring raise basicity; electron-withdrawing groups ( $- N O_{2}$ , $- S O_{3} H$ , $- C O O H$ , $- X$ ) lower it. N-Alkyl groups also raise basicity slightly: N,N-dimethylaniline (8.92) is a bit stronger than aniline.

Basicity of Amines: formulas and key facts

Base dissociation: $R N H_{2} + H_{2} O ⇌ R N H_{3}^{+} + O H^{-}; K_{b} = \frac{[ R N H _{3}^{+} ] [ O H ^{-} ]}{[ R N H _{2} ]}$; Larger $K_{b}$ = stronger base.
pKb: $p K_{b} = - lo g K_{b}$; Smaller $p K_{b}$ = stronger base; a difference of 1 in $p K_{b}$ = factor of 10 in $K_{b}$ .
Gas-phase order: $3° > 2° > 1° > N H_{3}$; Inductive effect only.
Aqueous order (methyl and ethyl): $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}; (C_{2} H_{5})_{2} N H > (C_{2} H_{5})_{3} N > C_{2} H_{5} N H_{2} > N H_{3}$; Inductive effect + solvation + steric hindrance.
Aromatic vs aliphatic: $C_{6} H_{5} N H_{2} (p K_{b} 9.38) < N H_{3} (4.75) < C_{6} H_{5} C H_{2} N H_{2} (4.70) < C H_{3} N H_{2} (3.38)$; Resonance in aniline lowers basicity.
Salt formation: $R N H_{2} + H C l \to R N H_{3}^{+} C l^{-} N a O H R N H_{2} + N a C l + H_{2} O$; Used to separate amines.

How to solve Basicity of Amines questions

Basicity questions are mostly ordering or 'account for' questions. Separate the effects and apply them in order.

1Classify each nitrogen: aliphatic (sp³ N, no conjugation) or aromatic (lone pair conjugated with a ring).
2Aromatic amines are always weaker than ammonia and aliphatic amines; benzylamine counts as aliphatic.
3Among aliphatic amines in the gas phase, more alkyl groups = more basic.
4In water, use the learned orders: methyl series 2° > 1° > 3° > $N H_{3}$ ; ethyl series 2° > 3° > 1° > $N H_{3}$ , explained by +I, solvation and steric effects.
5For substituted anilines, electron-releasing groups increase and electron-withdrawing groups decrease basicity; o/p substituents act through resonance as well as induction.
6If $p K_{b}$ values are given, order by $p K_{b}$ (smallest = strongest) and convert with $K_{b} = 1 0^{- p K_{b}}$ when asked.
7Always write the effect responsible for each position in the order.

Basicity of Amines in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Amines form non-volatile salts with acids

Suppose a cook squeezes lemon juice over a piece of fish that smells strongly. The 'fishy' smell is partly due to volatile amines such as trimethylamine, $(C H_{3})_{3} N$ .

Explain, with an equation, why the lemon juice reduces the smell. Use citric acid as $H^{+}$ donor in a general form.

Try it first, then show the solutionSolution

Trimethylamine is a base because of the lone pair on nitrogen.
Acid from lemon juice protonates it: $(C H_{3})_{3} N + H^{+} \to (C H_{3})_{3} N H^{+}$ .
The trimethylammonium ion forms a salt with the citrate anion; salts are ionic, non-volatile and stay on the fish rather than evaporating.
Fewer amine molecules escape into the air, so the smell is weaker.

Answer: Acid protonates the volatile amine to a non-volatile ammonium salt: $(C H_{3})_{3} N + H^{+} \to (C H_{3})_{3} N H^{+}$ .

Memory hook: Acid ties up the amine's lone pair, so the smell cannot escape.

Tricks to remember Basicity of Amines

Short memory aids. Say each one aloud once, then use it on a question the same day.

2-1-3 for methyl, 2-3-1 for ethyl
In water: methyl amines go secondary, primary, tertiary; ethyl amines go secondary, tertiary, primary; ammonia always last.
Use it when: Aqueous order questions.
Ring steals the lone pair
Attach N directly to a benzene ring and its lone pair goes on holiday into the ring: aniline is a weak base.
Use it when: Aromatic vs aliphatic comparisons.
Small pKb, big base
pKb is like a golf score: the lower, the better (stronger base).
Use it when: Reading pKb tables.

Basicity of Amines: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Strongest base in water · 1 mark · MCQBoard level

Which is the strongest base in aqueous solution?
(a) $N H_{3}$ (b) $C H_{3} N H_{2}$ (c) $(C H_{3})_{2} N H$ (d) $(C H_{3})_{3} N$

Try it first, then show the solutionSolution

Aqueous order for methylamines: $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}$ ( $p K_{b}$ 3.27, 3.38, 4.22, 4.75).

Answer: (c) $(C H_{3})_{2} N H$ .

Example 2: Aniline vs ammonia · 2 marks · Very short answerBoard level

Why is aniline a weaker base than ammonia?

Try it first, then show the solutionSolution

In aniline the lone pair on N is delocalised into the benzene ring by resonance, so it is less available to accept a proton (1 mark).
The anilinium ion formed has fewer resonance structures than aniline, so protonation loses resonance stabilisation; ammonia has no such effect (1 mark).

Answer: Resonance delocalises aniline's lone pair and is lost on protonation.

Example 3: Two orders · 3 marks · Short answerHarder board level

Arrange in increasing order of basic strength in aqueous solution and justify: (a) $C_{2} H_{5} N H_{2}$ , $(C_{2} H_{5})_{2} N H$ , $(C_{2} H_{5})_{3} N$ , $N H_{3}$ (b) aniline, p-toluidine (4-methylaniline), p-nitroaniline, benzylamine.

Try it first, then show the solutionSolution

(a) $N H_{3} < C_{2} H_{5} N H_{2} < (C_{2} H_{5})_{3} N < (C_{2} H_{5})_{2} N H$ (1 mark).
Reason: +I effect raises basicity; solvation of the cation favours fewer alkyl groups; steric crowding hinders the tertiary amine; the balance puts the 2° amine first (½ mark).
(b) p-Nitroaniline < aniline < p-toluidine < benzylamine (1 mark).
Reason: $- N O_{2}$ withdraws electrons; $- C H_{3}$ releases them; in benzylamine the $C H_{2}$ group keeps the lone pair out of conjugation with the ring (½ mark).

Answer: (a) $N H_{3} < C_{2} H_{5} N H_{2} < (C_{2} H_{5})_{3} N < (C_{2} H_{5})_{2} N H$ . (b) p-Nitroaniline < aniline < p-toluidine < benzylamine.

Example 4: Gas phase vs water · 3 marks · Short answerHarder board level

In the gas phase the basicity of methylamines is $(C H_{3})_{3} N > (C H_{3})_{2} N H > C H_{3} N H_{2} > N H_{3}$ , but in water $(C H_{3})_{3} N$ falls to third place. Explain both orders.

Try it first, then show the solutionSolution

Gas phase: only the inductive effect operates; more methyl groups push more electron density onto N and stabilise the cation, so 3° is strongest (1 mark).
Water: the substituted ammonium ion is stabilised by hydrogen bonding with water; $C H_{3} N H_{3}^{+}$ (three N–H) is better solvated than $(C H_{3})_{3} N H^{+}$ (one N–H) (1 mark).
Steric hindrance by three methyl groups also hinders solvation; so the tertiary amine drops below the primary, and the secondary amine, with a good balance of +I and solvation, is strongest (1 mark).

Answer: Gas: inductive effect alone. Water: solvation (more N–H in the cation) and steric hindrance reduce the tertiary amine's basicity.

Example 5: Reading a pKb table · 5 marks · Long answerHOTS / competency

Given $p K_{b}$ values: methanamine 3.38, aniline 9.38, benzylamine 4.70, N,N-dimethylaniline 8.92. (a) Calculate $K_{b}$ for methanamine and aniline. (b) By what factor is methanamine a stronger base than aniline? (c) Why is benzylamine about $5 \times 1 0^{4}$ times stronger than aniline? (d) Why is N,N-dimethylaniline slightly stronger than aniline but still far weaker than trimethylamine?

Try it first, then show the solutionSolution

(a) $K_{b}$ (methanamine) $= 1 0^{- 3.38} = 4.2 \times 1 0^{- 4}$ ; $K_{b}$ (aniline) $= 1 0^{- 9.38} = 4.2 \times 1 0^{- 10}$ (1 mark).
(b) Ratio $= 1 0^{9.38 - 3.38} = 1 0^{6}$ (1 mark).
(c) $1 0^{9.38 - 4.70} \approx 4.8 \times 1 0^{4}$ . In benzylamine the $C H_{2}$ group separates N from the ring, so the lone pair is not delocalised; it behaves like an aliphatic amine (1½ marks).
(d) The two methyl groups on N push electrons (+I), raising basicity a little ( $p K_{b}$ 8.92 vs 9.38), but the lone pair is still conjugated with the ring, so it stays far weaker than trimethylamine ( $p K_{b}$ 4.22) (1½ marks).

Answer: (a) $4.2 \times 1 0^{- 4}$ and $4.2 \times 1 0^{- 10}$ . (b) $1 0^{6}$ . (c) No conjugation in benzylamine. (d) +I helps slightly; resonance still dominates.

Where students lose marks in Basicity of Amines

The slips examiners see most often, and what to write instead.

Mistake: Using the gas-phase order for aqueous questions
Wrong working
In water: $(C H_{3})_{3} N > (C H_{3})_{2} N H > C H_{3} N H_{2}$
Right working
In water: $(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}$
Do this: Read whether the question says 'aqueous' or 'gas phase'.
Usually costs: 1 mark.
Mistake: Saying larger pKb means stronger base
Do this: Smaller $p K_{b}$ (larger $K_{b}$ ) = stronger base.
Usually costs: 1 mark.
Mistake: Treating benzylamine as an aromatic amine
Do this: In benzylamine N is on a $C H_{2}$ , not on the ring; it is as basic as an aliphatic amine.
Usually costs: 1 mark in ordering questions.
Mistake: Explaining aqueous order only by inductive effect
Do this: Mention all three: inductive effect, solvation (H-bonding of the cation) and steric hindrance.
Usually costs: 1 mark.

Exam tips

Learn the two aqueous orders (methyl and ethyl series); they are tested directly.
Use NCERT $p K_{b}$ values in answers when helpful; they show precision.
For aniline, write 'lone pair delocalised into the ring by resonance' explicitly.
In substituted anilines, name the effect (+R, −R, +I, −I) of each substituent.
Give a one-line reason after each position in an order.

Practice questions on Basicity of Amines

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which is the weakest base?
(a) $N H_{3}$ (b) $C H_{3} N H_{2}$ (c) $C_{6} H_{5} N H_{2}$ (d) $(C H_{3})_{2} N H$
Hint
Resonance with the ring.
Answer
(c) Aniline.
Q2 · 2 marks · Very short answerBoard level
How are $K_{b}$ and $p K_{b}$ related? Which has the larger $p K_{b}$ , methanamine or aniline? What does this tell you?
Hint
$p K_{b} = - lo g K_{b}$ .
Answer
$p K_{b} = - lo g K_{b}$ . Aniline has the larger $p K_{b}$ (9.38 vs 3.38), so it has the smaller $K_{b}$ and is the weaker base.
Q3 · 1 mark · Very short answerHarder board level
Arrange in decreasing order of basic strength in aqueous solution: $(C H_{3})_{3} N$ , $(C H_{3})_{2} N H$ , $C H_{3} N H_{2}$ , $N H_{3}$ .
Hint
Use $p K_{b}$ values or the 2-1-3 rule.
Answer
$(C H_{3})_{2} N H > C H_{3} N H_{2} > (C H_{3})_{3} N > N H_{3}$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): Methanamine is a stronger base than ammonia.
Reason (R): The +I effect of the methyl group increases the electron density on nitrogen.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Electron release by alkyl groups.
Answer
(a).
Q5 · 2 marks · Very short answerHarder board level
Why do amines dissolve in dilute HCl? Write the equation for aniline and show how the amine can be recovered.
Hint
Salt formation.
Answer
Amines are bases and form water-soluble ammonium salts: $C_{6} H_{5} N H_{2} + H C l \to C_{6} H_{5} N H_{3}^{+} C l^{-}$ (anilinium chloride). Adding NaOH regenerates aniline: $C_{6} H_{5} N H_{3}^{+} C l^{-} + N a O H \to C_{6} H_{5} N H_{2} + N a C l + H_{2} O$ .
Q6 · 2 marks · Very short answerHarder board level
Using NCERT $p K_{b}$ values (ammonia 4.75, aniline 9.38), calculate $K_{b}$ for each and the ratio $K_{b} (N H_{3}) / K_{b} (C_{6} H_{5} N H_{2})$ .
Hint
$K_{b} = 1 0^{- p K_{b}}$ .
Answer
$K_{b} (N H_{3}) = 1 0^{- 4.75} = 1.8 \times 1 0^{- 5}$ ; $K_{b}$ (aniline) $= 1 0^{- 9.38} = 4.2 \times 1 0^{- 10}$ . Ratio $= 1 0^{4.63} \approx 4.3 \times 1 0^{4}$ .
Q7 · 3 marks · Short answerHOTS / competency
Which is more basic in each pair? Give reasons. (a) 4-methoxyaniline or 4-nitroaniline (b) aniline or cyclohexylamine
Hint
Electron donation vs withdrawal; conjugation.
Answer
(a) 4-Methoxyaniline: $- O C H_{3}$ donates electrons to the ring (+R), increasing the electron density available at N; $- N O_{2}$ withdraws electrons (−R, −I), reducing it. (b) Cyclohexylamine: its N is attached to an $s p^{3}$ carbon with no ring conjugation, so the lone pair is fully available; aniline's lone pair is delocalised.
Q8 · 4 marks · Case-basedHOTS / competency
Read and answer.
Suppose a student prepares a table of $p K_{b}$ values: ethanamine 3.29, N-ethylethanamine 3.00, N,N-diethylethanamine 3.25, ammonia 4.75, aniline 9.38, N-methylaniline 9.30.
(a) Arrange the three ethylamines and ammonia in order of decreasing basic strength. (1 mark)
(b) Why is triethylamine more basic than ethylamine in water, whereas trimethylamine is less basic than methylamine? (1 mark)
(c) Why is N-methylaniline only slightly more basic than aniline? (1 mark)
(d) Which compound in the table has the largest $K_{b}$ ? (1 mark)
Hint
Larger groups, larger +I effect.
Answer
(a) $(C_{2} H_{5})_{2} N H > (C_{2} H_{5})_{3} N > C_{2} H_{5} N H_{2} > N H_{3}$ . (b) Ethyl groups have a larger +I effect, which outweighs the loss of solvation and steric effects for the tertiary amine; with smaller methyl groups the solvation effect wins. (c) The methyl group adds a small +I effect, but the lone pair remains delocalised into the ring, which dominates. (d) N-Ethylethanamine (smallest $p K_{b}$ , 3.00).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. $p K_{b}$ of aniline (NCERT)?Show answer
9.38.
2.Strongest base among ethylamines in water?Show answer
Diethylamine, $(C_{2} H_{5})_{2} N H$ .
3.Basicity order of methylamines in gas phase?Show answer
$(C H_{3})_{3} N > (C H_{3})_{2} N H > C H_{3} N H_{2} > N H_{3}$ .
4.Effect of $- N O_{2}$ on basicity of aniline?Show answer
Decreases it.
5.Product of aniline with HCl?Show answer
Anilinium chloride, $C_{6} H_{5} N H_{3}^{+} C l^{-}$ .
6.Why is benzylamine more basic than aniline?Show answer
Its N lone pair is not conjugated with the ring.

Basicity of Amines: frequently asked questions

Why are amines basic in Class 12 Chemistry?

The nitrogen atom has a lone pair of electrons that can accept a proton, forming a substituted ammonium ion.

Why is aniline less basic than ammonia?

Its lone pair is delocalised into the benzene ring by resonance, and this extra stability is lost when aniline is protonated.

What is the order of basicity of methylamines in water?

(CH_3)_2NH > CH_3NH_2 > (CH_3)_3N > NH_3, due to the combined inductive, solvation and steric effects.

Why is the order of basicity different in the gas phase?

In the gas phase there is no solvation, so only the inductive effect matters and tertiary amines are the strongest bases.

How do substituents affect the basicity of aniline?

Electron-releasing groups such as -CH_3 and -OCH_3 increase it; electron-withdrawing groups such as -NO_2, -COOH and halogens decrease it.