Physical State and the Fishy Smell
The lower aliphatic amines are gases at room temperature and carry a characteristic fishy odour; methylamine and ethylamine in particular smell strongly of ammonia, the parent molecule from which amines are derived. As the carbon count grows, the trend follows the familiar pattern of organic homologous series. Primary amines with three or more carbon atoms are liquids, and the still higher members are solids.
Aromatic amines behave a little differently in appearance. Aniline ($\ce{C6H5NH2}$) and other arylamines are usually colourless when pure, but they acquire a brown-to-red colour on storage. This darkening is not decomposition of the amine itself but the result of slow atmospheric oxidation of the sensitive amino group, a point NCERT raises explicitly. It is also why bottles of aniline used in the laboratory are kept tightly closed and often dark-coloured. The trend in state across the families is summarised below.
The reason the odour and state follow this pattern is the interplay between molecular mass and intermolecular forces. The lightest members have low molar masses and weak overall attractions, so they escape into the vapour phase readily and reach the nose as gases — hence both their volatility and their pronounced smell. As the chain lengthens, dispersion forces accumulate and the members condense to liquids and then solids. The fishy character itself is a hallmark of small amines; many of the volatile amines responsible for the smell of decaying fish, such as trimethylamine, belong to exactly this low-mass group.
| Family / member | Physical state (room temp.) | Sensory / appearance note |
|---|---|---|
| Lower aliphatic amines (≤2–3 C), e.g. $\ce{CH3NH2}$, $\ce{C2H5NH2}$ | Gases | Fishy / ammoniacal smell |
| Primary amines with ≥3 carbon atoms | Liquids | Fishy odour persists |
| Higher amines | Solids | Weaker odour with size |
| Aniline and arylamines | Liquids / low-melting solids | Colourless, darken on storage (air oxidation) |
Why Amines Hydrogen-Bond
Every physical trend that follows depends on whether an amine molecule can form a hydrogen bond. A hydrogen bond requires two things: a hydrogen atom attached to a small, electronegative atom (here N), and a lone pair on a neighbouring electronegative atom to accept it. The nitrogen of a primary or secondary amine supplies both — an N–H bond to act as donor and a lone pair to act as acceptor.
Primary and secondary amines therefore engage in intermolecular association: the N–H of one molecule hydrogen-bonds to the nitrogen lone pair of another, written $\ce{N-H\bond{...}N}$. This network of weak attractions must be partly disrupted before the liquid can boil, raising the boiling point above what dispersion forces alone would give. Tertiary amines, by contrast, have no N–H bond — all three positions on nitrogen are occupied by alkyl or aryl groups — and so cannot self-associate by hydrogen bonding at all. They still possess a lone pair and can accept a hydrogen bond from a partner that has an N–H or O–H, but in the pure liquid there is no donor, so no self-association occurs. The schematic below shows the donor–acceptor arrangement in a chain of primary amine molecules.
It is worth being precise about what "extent of association" means. A primary amine, with two N–H bonds, can act as a hydrogen-bond donor twice over and as an acceptor through its lone pair, so each molecule can be tied into several neighbours at once, building a relatively extended hydrogen-bonded structure in the liquid. A secondary amine, with a single N–H, can donate only once per molecule, halving the donor capacity and weakening the overall network. This graded capacity — two, one, then zero N–H donors — is the master variable behind both the boiling-point order of isomers and, in part, the solubility behaviour discussed later.
Each primary amine donates an N–H to the lone pair of its neighbour. With two N–H bonds per molecule, primary amines build the most extensive network — the structural root of their high boiling points.
Boiling Points: Alkane < Amine < Alcohol
The single most examined fact about amines is where their boiling points sit relative to other families of similar molar mass. The answer is a three-rung ladder: for comparable molecular masses, alkanes boil lowest, amines in the middle, and alcohols (and carboxylic acids) highest.
The lower rung is straightforward. Alkanes are non-polar and held together only by weak London dispersion forces; amines add the stronger $\ce{N-H\bond{...}N}$ hydrogen bonding on top of dispersion, so an amine always boils higher than an alkane of similar mass.
The upper rung is the conceptual heart of the topic. Amines hydrogen-bond more weakly than alcohols because nitrogen is less electronegative than oxygen (about 3.0 versus 3.5). A less electronegative donor atom makes the N–H bond less polar, the partial charges smaller, and the resulting hydrogen bond weaker than the O–H···O bond in alcohols. NCERT states this directly: alcohols are more polar than amines and form stronger intermolecular hydrogen bonds. Hence an amine boils below an alcohol of the same molar mass. The energy ladder is shown in Figure 2.
All five compounds have nearly the same molar mass (≈ 73–74 g mol⁻¹). The branched alkane (only dispersion forces) boils lowest; butan-1-ol (strong O–H···O bonds) boils highest; the three amines lie in between, ordered by how many N–H bonds each has.
"More electronegative N should mean stronger H-bond." No.
A common slip is to reason that because nitrogen pulls the lone pair tightly it must hydrogen-bond strongly. The opposite is true for boiling point. Oxygen (3.5) is more electronegative than nitrogen (3.0), so the O–H bond is more polar and forms a stronger hydrogen bond than N–H. That is exactly why alcohols out-boil amines of equal mass.
Order of intermolecular H-bond strength: O–H···O (alcohol) > N–H···N (amine) > none (alkane).
The 1° > 2° > 3° Isomer Order
Within a set of isomeric amines — same molecular formula, same molar mass — the boiling point depends only on the extent of hydrogen-bonded association, and that in turn depends on how many N–H bonds the molecule offers. The decisive count is:
| Type | General form | N–H bonds available | Intermolecular association |
|---|---|---|---|
| Primary (1°) | $\ce{R-NH2}$ | 2 | Most extensive |
| Secondary (2°) | $\ce{R2NH}$ | 1 | Moderate |
| Tertiary (3°) | $\ce{R3N}$ | 0 | None (no N–H) |
A primary amine has two N–H bonds and so participates in the most hydrogen bonding; a secondary amine has only one; a tertiary amine has none and therefore cannot self-associate at all. NCERT states the resulting order of boiling points of isomeric amines explicitly:
Boiling point of isomeric amines: Primary > Secondary > Tertiary.
The 1°/2°/3° classification used throughout this page is set up in Structure and Classification of Amines.
Boiling-point order vs basicity order — do not confuse them.
The boiling-point order of isomeric amines is fixed by the count of N–H bonds: 1° > 2° > 3°. This is a clean structural rule with no exceptions. The basicity order is a different question entirely — it depends on inductive effect, solvation and steric hindrance, and in water gives the irregular sequence $\ce{(CH3)2NH > CH3NH2 > (CH3)3N}$. Examiners often place both in the same question to see whether you reach for the wrong trend.
Boiling point of isomers → count N–H bonds (1°>2°>3°). Basicity → see the basicity page.
Arrange in increasing order of boiling point: $\ce{C2H5OH}$, $\ce{(CH3)2NH}$, $\ce{C2H5NH2}$ (NCERT Exercise 9.4 v).
All three carry a hydrogen-bonding group. $\ce{(CH3)2NH}$ is a secondary amine (one N–H); $\ce{C2H5NH2}$ is a primary amine (two N–H), so it associates more and boils higher than the secondary amine. $\ce{C2H5OH}$ has O–H, the strongest H-bond, so it boils highest. Order: $\ce{(CH3)2NH < C2H5NH2 < C2H5OH}$.
Comparison Table of Boiling Points
NCERT Table 9.2 places four compounds of nearly the same molar mass side by side so the trend can be read off directly. The first three are isomeric/near-isomeric amines (a primary, a secondary and a tertiary), the fourth a branched alkane, and the fifth an alcohol — together they demonstrate both the alkane < amine < alcohol ladder and the 1° > 2° > 3° isomer order in one dataset.
| # | Compound | Type | b.p. / K | Why |
|---|---|---|---|---|
| 1 | $\ce{n-C4H9NH2}$ (butan-1-amine) | 1° amine | 350.8 | Two N–H bonds — most association |
| 2 | $\ce{(C2H5)2NH}$ (N-ethylethanamine) | 2° amine | 329.3 | One N–H bond — moderate association |
| 3 | $\ce{C2H5N(CH3)2}$ (N,N-dimethylethanamine) | 3° amine | 310.5 | No N–H — no self H-bonding |
| 4 | $\ce{C2H5CH(CH3)2}$ (a branched alkane) | Alkane | 300.8 | Only dispersion forces |
| 5 | $\ce{n-C4H9OH}$ (butan-1-ol) | Alcohol | 390.3 | Strong O–H···O bonding |
Reading down the table, every step is explained by the same logic. The branched alkane (300.8 K) sits at the bottom with dispersion forces only. The tertiary amine (310.5 K) barely beats it — having no N–H, it gains little from its nitrogen, but its polarity still adds a small lift over the non-polar alkane. The secondary amine (329.3 K) and primary amine (350.8 K) climb steadily as N–H bonds are added. The alcohol (390.3 K) tops the list because O–H···O is the strongest hydrogen bond of the set.
Two numerical features of this dataset are worth committing to memory because they recur in NEET options. First, the gap between the tertiary amine and the alkane is small (about 10 K) — a reminder that a tertiary amine, lacking N–H, behaves almost like a polar non-associating liquid. Second, the gap between the primary amine and the alcohol is large (nearly 40 K) — the visible consequence of O–H···O being substantially stronger than N–H···N. When an exam pits an amine against an alcohol of equal mass, the alcohol always wins on boiling point; when it pits isomeric amines against one another, only the number of N–H bonds matters.
Solubility in Water
Lower amines are soluble in water for the same reason they hydrogen-bond among themselves: the N–H bond and the nitrogen lone pair both form hydrogen bonds with water molecules. A small amine such as methanamine or ethanamine is fully miscible with water. NCERT Exercise 9.3(ii) draws the contrast sharply — ethylamine dissolves in water while aniline does not.
Solubility, however, decreases as the molar mass rises. Each additional $\ce{-CH2-}$ unit enlarges the hydrophobic alkyl portion of the molecule, which water cannot accommodate. Once this non-polar tail dominates the hydrogen-bonding head, solubility collapses: higher amines are essentially insoluble in water. The same competition between a hydrogen-bonding head and a growing hydrophobic tail governs the solubility of alcohols and is a recurring theme across organic functional groups.
A subtle point separates solubility in water from the self-association that fixes boiling point. When an amine dissolves, the partner supplying the second half of the hydrogen bond is water, which is itself an excellent donor and acceptor through its O–H bonds. A tertiary amine, although it cannot self-associate, still carries a lone pair and can therefore accept a hydrogen bond from water; small tertiary amines are accordingly water-soluble despite their low boiling points. This is why the 1° > 2° > 3° rule is a boiling-point rule and must not be carried over wholesale to aqueous solubility — the donor that matters for dissolution is water, not the amine.
Out of butan-1-ol and butan-1-amine, which is more soluble in water and why? (NCERT §9.5 question.)
Both form hydrogen bonds with water and both have the same four-carbon chain, so hydrophobic effects are comparable. The deciding factor is the head group: the O–H of butan-1-ol forms stronger hydrogen bonds with water than the N–H of butan-1-amine, because oxygen (3.5) is more electronegative than nitrogen (3.0). Hence butan-1-ol is more soluble.
Aromatic Amines and Organic Solvents
Aromatic amines such as aniline are far less soluble in water than the lower aliphatic amines. The benzene ring is a large hydrophobic group, and although the $\ce{-NH2}$ group can hydrogen-bond to water, this contribution is swamped by the bulky non-polar ring. This is precisely the contrast NCERT highlights: ethylamine dissolves but aniline does not.
All amines — aliphatic and aromatic alike — are nonetheless readily soluble in organic solvents such as alcohol, ether and benzene, where the dominant interactions match the largely non-polar molecular framework. This solubility behaviour, paired with the basic character of amines, underlies their separation chemistry: amine salts formed with mineral acids are water-soluble but ether-insoluble, allowing amines to be separated from non-basic organic compounds.
| System | Solubility | Reason |
|---|---|---|
| Lower aliphatic amines in water | Soluble | H-bonding with water; small hydrophobic part |
| Higher aliphatic amines in water | Essentially insoluble | Large hydrophobic alkyl chain dominates |
| Aniline / arylamines in water | Sparingly soluble | Bulky hydrophobic benzene ring |
| Any amine in alcohol / ether / benzene | Soluble | Compatible non-polar / weakly polar environment |
Physical Properties of Amines — at a glance
- Lower aliphatic amines are gases with a fishy/ammoniacal smell; ≥3 C primary amines are liquids, higher ones solids. Aniline is colourless but darkens on air oxidation.
- Amines hydrogen-bond ($\ce{N-H\bond{...}N}$) only if they have an N–H bond, so they boil higher than alkanes of similar mass.
- They boil lower than alcohols/acids because N (3.0) is less electronegative than O (3.5), making N–H···N weaker than O–H···O.
- Among isomers, boiling point follows 1° > 2° > 3° (two, one, then zero N–H bonds). Tertiary amines cannot self-associate.
- Lower amines are water-soluble via H-bonding; solubility falls with size; aromatic amines are sparingly soluble. All amines dissolve in alcohol, ether and benzene.
- Do not confuse the clean 1°>2°>3° boiling-point rule with the irregular aqueous basicity order.