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આ MCQ મોડ્યુલ આના પર આધારિત છે: Physical Properties Basicity
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Physical Properties Basicity
🎓 Class 12
Chemistry
CBSE
Theory
Ch 9 – Amines
⏱ ~14 min
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આ મૂલ્યાંકન આના પર આધારિત હશે: Physical Properties Basicity
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Amines — Physical Properties and Basic Character
This part answers two questions that carry heavy weight in the board examination. Why do primary amines boil higher than tertiary amines of the same molar mass? And why is the aqueous order of basicity so stubbornly irregular when the inductive effect predicts a clean trend? Both answers turn on hydrogen bonding — in the first case between amine molecules, in the second between water and the cation.
9.5 Physical Properties
Physical state and odour
The lower aliphatic amines are gases with a fishy odour. Primary amines with three or more carbon atoms are liquids, and still higher ones are solids. Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation.
Solubility
Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of the amine, because the hydrophobic alkyl part grows in size. Higher amines are essentially insoluble in water. Amines are soluble in organic solvents such as alcohol, ether and benzene.
Amines versus alcohols. Taking the electronegativity of nitrogen as 3.0 and of oxygen as 3.5, the N–H bond is less polar than the O–H bond. Alcohols are therefore more polar than amines and form stronger intermolecular hydrogen bonds. Out of butan-1-ol and butan-1-amine, butan-1-ol is the more soluble in water — and it also boils considerably higher (390.3 K against 350.8 K).
Boiling points and intermolecular association
Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between the nitrogen of one molecule and the hydrogen of another. This association is greater in primary amines than in secondary amines, because a primary amine has two hydrogen atoms available for hydrogen bond formation. Tertiary amines cannot associate at all, since they have no hydrogen atom attached to nitrogen.
Order of boiling points of isomeric amines:
Primary > Secondary > Tertiary
The reason is the number of N–H hydrogens available for hydrogen bonding: two, then one, then none.
Primary > Secondary > Tertiary
The reason is the number of N–H hydrogens available for hydrogen bonding: two, then one, then none.
Table 9.2 — Boiling points of amines, alcohols and alkanes of similar molar mass
| Sl. No. | Compound | Class | Molar mass | b.p. / K |
|---|---|---|---|---|
| 1. | n-C₄H₉NH₂ | 1° amine | 73 | 350.8 |
| 2. | (C₂H₅)₂NH | 2° amine | 73 | 329.3 |
| 3. | C₂H₅N(CH₃)₂ | 3° amine | 73 | 310.5 |
| 4. | C₂H₅CH(CH₃)₂ | alkane | 72 | 300.8 |
| 5. | n-C₄H₉OH | alcohol | 74 | 390.3 |
Read the table as one story. At essentially constant molar mass the boiling point climbs steadily with hydrogen-bonding capacity: alkane (no H-bonding, 300.8 K) < 3° amine (no N–H, 310.5 K) < 2° amine (one N–H, 329.3 K) < 1° amine (two N–H, 350.8 K) < alcohol (stronger O–H bonding, 390.3 K). The tertiary amine still boils above the alkane because of its dipole moment, even though it cannot self-associate.
9.6 Chemical Reactions — Basic Character of Amines
The difference in electronegativity between nitrogen and hydrogen, together with the presence of the unshared pair of electrons on nitrogen, makes amines reactive. The number of hydrogen atoms attached to nitrogen also decides the course of reaction, which is why primary, secondary and tertiary amines differ in many reactions. Amines behave as nucleophiles because of that unshared electron pair.
1. Basic character of amines
Amines, being basic in nature, react with acids to form salts. Amine salts on treatment with a base like NaOH regenerate the parent amine.
R–NH₂ + HCl → R–NH₃⁺Cl⁻
R–NH₃⁺Cl⁻ + NaOH → R–NH₂ + NaCl + H₂O
R–NH₃⁺Cl⁻ + NaOH → R–NH₂ + NaCl + H₂O
A separation technique, not just a reaction. Amine salts are soluble in water but insoluble in organic solvents like ether. This is the basis for separating amines from non-basic organic compounds that are insoluble in water: shake the mixture with dilute HCl, the amine passes into the aqueous layer as its salt, and adding NaOH to that layer liberates the free amine.
The reaction of amines with mineral acids to form ammonium salts shows that they are basic in nature. Amines have an unshared pair of electrons on nitrogen, due to which they behave as a Lewis base. Basic character is better understood in terms of Kb and pKb:
\[ \mathrm{R-NH_2 + H_2O \rightleftharpoons R-NH_3^+ + OH^-} \]
\[ K_b = \frac{[\mathrm{R-NH_3^+}][\mathrm{OH^-}]}{[\mathrm{R-NH_2}]} \qquad\qquad pK_b = -\log K_b \]
The rule to fix in memory. The larger the value of Kb, or the smaller the value of pKb, the stronger is the base. For reference, pKb of ammonia is 4.75.
Aliphatic amines are stronger bases than ammonia due to the +I effect of alkyl groups, which raises the electron density on nitrogen; their pKb values lie in the range of about 3 to 4.22. Aromatic amines, on the other hand, are weaker bases than ammonia due to the electron-withdrawing nature of the aryl group.
Table 9.3 — pKb values of amines in aqueous phase
| Name of amine | Formula | pKb | Compared with NH₃ (4.75) |
|---|---|---|---|
| Methanamine | CH₃NH₂ | 3.38 | stronger base |
| N-Methylmethanamine | (CH₃)₂NH | 3.27 | stronger base |
| N,N-Dimethylmethanamine | (CH₃)₃N | 4.22 | stronger base |
| Ethanamine | C₂H₅NH₂ | 3.29 | stronger base |
| N-Ethylethanamine | (C₂H₅)₂NH | 3.00 | stronger base |
| N,N-Diethylethanamine | (C₂H₅)₃N | 3.25 | stronger base |
| Benzenamine (aniline) | C₆H₅NH₂ | 9.38 | much weaker base |
| Phenylmethanamine | C₆H₅CH₂NH₂ | 4.70 | slightly stronger |
| N-Methylaniline | C₆H₅NHCH₃ | 9.30 | much weaker base |
| N,N-Dimethylaniline | C₆H₅N(CH₃)₂ | 8.92 | much weaker base |
Note the two extremes in the table. Benzylamine (phenylmethanamine, pKb 4.70) is a fairly normal base because the ring is one carbon away from nitrogen. Aniline (pKb 9.38) is about 40,000 times weaker, purely because the nitrogen is bonded directly to the ring. One CH₂ group makes all the difference.
Structure–basicity relationship of amines
Basicity of amines is related to their structure. The basic character of an amine depends upon the ease of formation of the cation by accepting a proton from the acid. The more stable the cation is relative to the amine, the more basic is the amine. Three factors compete.
(a) Alkanamines versus ammonia
Due to the electron-releasing nature of the alkyl group, it pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing with the proton of the acid. Moreover, the substituted ammonium ion formed from the amine gets stabilised by dispersal of the positive charge through the +I effect of the alkyl group. Hence alkylamines are stronger bases than ammonia.
On this reasoning alone, basicity should increase steadily with the number of alkyl groups. That trend is indeed followed in the gaseous phase:
Gas phase: 3° amine > 2° amine > 1° amine > NH₃ (pure inductive order)
But the trend is not regular in the aqueous state, as the pKb values in Table 9.3 show. In the aqueous phase the substituted ammonium cations are stabilised not only by the electron-releasing effect of the alkyl group but also by solvation with water molecules. The greater the size of the ion, the less the solvation and the less stabilised the ion. The order of stability of the ions by solvation is therefore:
RNH₃⁺ > R₂NH₂⁺ > R₃NH⁺
(decreasing extent of H-bonding with water, hence decreasing stabilisation by solvation)
(decreasing extent of H-bonding with water, hence decreasing stabilisation by solvation)
Greater the stability of the substituted ammonium cation, the stronger the corresponding amine as a base. Thus on the solvation argument alone the order of basicity of aliphatic amines should be primary > secondary > tertiary — which is exactly opposite to the inductive-effect order.
Secondly, when the alkyl group is small like –CH₃ there is no steric hindrance to H-bonding. If the alkyl group is bigger than CH₃ there will be steric hindrance to H-bonding. Therefore changing the nature of the alkyl group, for example from –CH₃ to –C₂H₅, results in a change in the order of basic strength.
The two orders you must be able to reproduce.
Ethyl-substituted: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃
Methyl-substituted: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
In both, the secondary amine wins — it is the best compromise between inductive donation and solvation.
Ethyl-substituted: (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃
Methyl-substituted: (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃
In both, the secondary amine wins — it is the best compromise between inductive donation and solvation.
(b) Arylamines versus ammonia
The pKb value of aniline is quite high (9.38) — it is a much weaker base than ammonia. The reason is that in aniline or other arylamines the –NH₂ group is attached directly to the benzene ring. This puts the unshared electron pair on nitrogen in conjugation with the benzene ring, making it much less available for protonation.
Aniline is a resonance hybrid of five structures. The anilinium ion obtained by accepting a proton can have only two resonating structures (the two Kekulé forms). Since the greater the number of resonating structures the greater the stability, aniline is more stabilised relative to its cation than ammonia is relative to ammonium ion. Hence the proton acceptability — the basic nature — of aniline and other aromatic amines is less than that of ammonia.
Substituent effects on aniline. In substituted anilines, electron-releasing groups such as –OCH₃ and –CH₃ increase basic strength, whereas electron-withdrawing groups such as –NO₂, –SO₃H, –COOH and –X decrease it. So p-toluidine is a stronger base than aniline, while p-nitroaniline is very much weaker.
Worked Example 9.4 — Ordering basic strength
Question. Arrange the following in decreasing order of basic strength: C₆H₅NH₂, C₂H₅NH₂, (C₂H₅)₂NH, NH₃.
Step 1. Separate the aromatic from the aliphatic. C₆H₅NH₂ has its lone pair in conjugation with the ring, so it will be the weakest — weaker even than ammonia.
Step 2. Among the aliphatic amines and ammonia, alkyl groups donate electron density, so both ethylamines beat ammonia. Between them, the secondary amine wins in water (best balance of +I donation and solvation).
Answer. (C₂H₅)₂NH > C₂H₅NH₂ > NH₃ > C₆H₅NH₂
🧪 Activity 9.3 — Reading Table 9.3 against two predictionsL5 Evaluate
Rather than memorising the aqueous orders, derive them. This activity makes the conflict between the inductive and solvation arguments visible in the actual data.
Predict: Using only the +I effect, write the expected order of basic strength for CH₃NH₂, (CH₃)₂NH and (CH₃)₃N. Then look up their pKb values in Table 9.3 and see whether your prediction survives.
- Write the inductive prediction: more methyl groups → more basic, so (CH₃)₃N > (CH₃)₂NH > CH₃NH₂.
- Now read the pKb values from Table 9.3: CH₃NH₂ = 3.38, (CH₃)₂NH = 3.27, (CH₃)₃N = 4.22. Remember lower pKb means stronger base.
- Rank them from the data and compare with your prediction.
- Repeat for the ethyl series: C₂H₅NH₂ = 3.29, (C₂H₅)₂NH = 3.00, (C₂H₅)₃N = 3.25.
- Explain the difference between the two series in one sentence.
The inductive prediction fails for the tertiary amine in both series.
Methyl series from data: (CH₃)₂NH (3.27) > CH₃NH₂ (3.38) > (CH₃)₃N (4.22) > NH₃ (4.75). The trimethylamine, which the inductive argument said should be strongest, is in fact the weakest of the three amines.
Ethyl series from data: (C₂H₅)₂NH (3.00) > (C₂H₅)₃N (3.25) > C₂H₅NH₂ (3.29) > NH₃ (4.75). Here the tertiary amine climbs above the primary — the order is different again.
One-sentence explanation: the tertiary cation R₃NH⁺ has only one N–H hydrogen, so it is the least stabilised by solvation, and the bulky alkyl groups additionally hinder both protonation and hydrogen bonding — the larger ethyl groups shift the balance differently from the compact methyl groups. In the gas phase, where no solvent is present to stabilise anything, the pure inductive order 3° > 2° > 1° > NH₃ is restored exactly as predicted. This is the cleanest proof in the whole chapter that solvation, not just inductive effect, is doing the work.
Intext questions
Intext 9.4 — Increasing order of basic strength
(i) C₆H₅NH₂ < NH₃ < C₆H₅CH₂NH₂ < C₂H₅NH₂ < (C₂H₅)₂NH
(ii) C₆H₅NH₂ < C₂H₅NH₂ < (C₂H₅)₃N < (C₂H₅)₂NH
(iii) C₆H₅NH₂ < C₆H₅CH₂NH₂ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH
Note in (i) and (iii) how benzylamine sits above aniline but below the simple alkylamines — the ring still withdraws a little through the CH₂, but it cannot conjugate with the lone pair.
Intext 9.5 — Complete the acid–base reactions and name the products
(i) CH₃CH₂CH₂NH₂ + HCl → CH₃CH₂CH₂NH₃⁺Cl⁻, propan-1-aminium chloride (propylammonium chloride).
(ii) (C₂H₅)₃N + HCl → (C₂H₅)₃NH⁺Cl⁻, N,N-diethylethanaminium chloride (triethylammonium chloride).
Competency-Based Questions
A separation problem in a teaching laboratory: a student is given a mixture of aniline, phenol and toluene dissolved in ether and asked to recover each component pure. She is provided with dilute HCl, dilute NaOH, distilled water and a separating funnel. In a second experiment she is given three bottles of amines of molar mass 73 — a primary, a secondary and a tertiary amine — and a thermometer.
1. Describe how the aniline can be removed selectively from the ether solution, and how it is then recovered. L3 Apply
Shake the ether solution with dilute HCl. Aniline, being basic, reacts to form anilinium chloride (C₆H₅NH₃⁺Cl⁻), which is soluble in water but insoluble in ether, so it passes into the aqueous layer while phenol and toluene remain in the ether. Separate the aqueous layer and add dilute NaOH to it: C₆H₅NH₃⁺Cl⁻ + NaOH → C₆H₅NH₂ + NaCl + H₂O. The free aniline separates as an oily layer and can be extracted back into fresh ether.
2. Without measuring anything, predict which of the three amines of molar mass 73 will have the highest boiling point, and justify the prediction. L4 Analyse
The primary amine, n-C₄H₉NH₂ (b.p. 350.8 K). A primary amine has two N–H hydrogens available for intermolecular hydrogen bonding, a secondary amine only one, and a tertiary amine none. More extensive intermolecular association means more energy is needed to separate the molecules, so the order is 1° (350.8 K) > 2° (329.3 K) > 3° (310.5 K).
3. Aniline has pKb 9.38 while benzylamine has pKb 4.70. Account for this very large difference. L4 Analyse
In aniline the –NH₂ is bonded directly to the ring, so the nitrogen lone pair is delocalised into the benzene ring by resonance (aniline is a hybrid of five resonating structures). The lone pair is therefore far less available to accept a proton, and protonation also destroys that favourable delocalisation — the anilinium ion has only two resonating structures. In benzylamine the nitrogen is separated from the ring by an sp³ CH₂ group, which blocks conjugation entirely; the lone pair stays localised on nitrogen and is freely available. Hence benzylamine behaves as a normal aliphatic amine and is roughly 40,000 times the stronger base.
4. Arrange in increasing order of basic strength: aniline, p-nitroaniline, p-toluidine. L3 Apply
p-Nitroaniline < aniline < p-toluidine. The –NO₂ group is strongly electron withdrawing, pulling further electron density away from the nitrogen and making p-nitroaniline a much weaker base than aniline. The –CH₃ group is electron releasing (+I), which increases electron density on nitrogen, so p-toluidine (4-methylaniline) is a stronger base than aniline.
5. A student argues: "Since alkyl groups are electron releasing, (CH₃)₃N must be a stronger base than (CH₃)₂NH in every medium." Evaluate this claim using evidence. L5 Evaluate
The claim is only half right, and the word "every" makes it wrong. In the gas phase the inductive argument holds perfectly and the order is 3° > 2° > 1° > NH₃, so (CH₃)₃N is indeed the strongest base there. In aqueous solution the data contradict the claim: pKb of (CH₃)₂NH is 3.27 but that of (CH₃)₃N is 4.22, so trimethylamine is the weaker base. Two additional factors operate in water that are absent in the gas phase: (a) solvation — the cation (CH₃)₃NH⁺ has only one N–H hydrogen and so forms fewer hydrogen bonds with water, leaving it less stabilised than (CH₃)₂NH₂⁺; and (b) steric hindrance — three methyl groups crowd the nitrogen, obstructing both the incoming proton and the solvating water molecules. Basic strength in solution is therefore a compromise between inductive effect, solvation and steric hindrance, not the inductive effect alone.
Assertion–Reason Questions
For each pair choose: (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.
Assertion (A): Ethylamine is soluble in water whereas aniline is not.
Reason (R): Ethylamine forms hydrogen bonds with water, while in aniline the large hydrophobic benzene ring dominates and hydrogen bonding with water is much less effective.
Answer: A. Lower aliphatic amines dissolve because the –NH₂ group hydrogen bonds to water; as the hydrophobic part grows — and a benzene ring is a large hydrophobic group — solubility falls sharply.
Assertion (A): pKb of aniline is greater than that of methylamine.
Reason (R): In aniline the lone pair on nitrogen is in conjugation with the benzene ring and is therefore less available for protonation.
Answer: A. A higher pKb means a weaker base (9.38 for aniline against 3.38 for methylamine), and delocalisation of the lone pair into the ring is exactly why aniline is the weaker base.
Assertion (A): Tertiary amines have the lowest boiling points among isomeric amines.
Reason (R): Tertiary amines have the largest molar mass among isomeric amines.
Answer: C. The assertion is true. The reason is false — isomeric amines have the same molar mass by definition. The correct reason is that a tertiary amine has no hydrogen attached to nitrogen, so it cannot undergo intermolecular hydrogen bonding at all.
Coming next. Part 4 covers the remaining reactions of Section 9.6 — alkylation, acylation and benzoylation, the carbylamine test, reaction with nitrous acid, the Hinsberg test with arylsulphonyl chloride, and electrophilic substitution in aniline including why the –NH₂ group must be protected by acetylation.
Frequently Asked Questions
Why is the boiling point order of isomeric amines primary greater than secondary greater than tertiary?
The order follows the number of N–H hydrogens available for intermolecular hydrogen bonding. A primary amine has two such hydrogens and associates most extensively, a secondary amine has one, and a tertiary amine has none and so cannot self-associate at all. More association means more energy is required to separate the molecules. For molar mass 73 the measured values are 350.8 K for n-C₄H₉NH₂, 329.3 K for (C₂H₅)₂NH and 310.5 K for C₂H₅N(CH₃)₂.
Why is aniline a much weaker base than ammonia?
In aniline the –NH₂ group is attached directly to the benzene ring, so the unshared electron pair on nitrogen is in conjugation with the ring and is delocalised over it. That makes the lone pair far less available for accepting a proton. Aniline is a resonance hybrid of five structures while the anilinium ion has only two, so aniline is stabilised much more than its cation and protonation is disfavoured. Its pKb is 9.38 against 4.75 for ammonia.
Why is the aqueous basicity order of alkylamines irregular?
Three factors compete. The inductive (+I) effect of alkyl groups favours the order tertiary greater than secondary greater than primary. Solvation of the cation favours the opposite order, because RNH₃⁺ has more N–H hydrogens to hydrogen bond with water than R₃NH⁺ does. Steric hindrance from bulky alkyl groups penalises the tertiary amine further. The net result in water is usually that the secondary amine is the strongest base — (CH₃)₂NH greater than CH₃NH₂ greater than (CH₃)₃N greater than NH₃.
What is the gas phase order of basicity of amines and why does it differ from the aqueous order?
In the gas phase the order is tertiary greater than secondary greater than primary greater than ammonia, exactly as the inductive effect predicts. There is no solvent present, so no solvation of the cation is possible and the inductive effect acts alone. In water, solvation and steric hindrance also come into play, which is why the aqueous order is irregular.
How are Kb and pKb used to compare the strength of amines?
Kb is the equilibrium constant for R–NH₂ + H₂O in equilibrium with R–NH₃⁺ + OH⁻, and pKb = −log Kb. The larger the Kb or the smaller the pKb, the stronger the base. Ammonia has pKb 4.75; aliphatic amines fall in the range about 3 to 4.22 and so are stronger bases, while aniline at 9.38 is far weaker.
How can an amine be separated from a mixture of non-basic organic compounds?
Shake the mixture with dilute mineral acid. The amine, being basic, forms an ammonium salt that is soluble in water but insoluble in organic solvents such as ether, so it moves into the aqueous layer while non-basic compounds stay in the organic layer. Separating the aqueous layer and adding a strong base such as NaOH regenerates the free amine, which can then be extracted back into fresh organic solvent.
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