Physical properties of amines

Hydrogen-bonding capacity sets the whole story. A primary amine $R\text{-}NH_2$ has two N-H bonds, so it can both donate and accept hydrogen bonds — to other amine molecules and to water. A secondary amine $R_2NH$ has only one N-H bond, so it still hydrogen-bonds but less extensively. A tertiary amine $R_3N$ has no N-H bond at all: it cannot donate a hydrogen bond to another amine molecule, so tertiary amines rely only on weaker dipole-dipole and van der Waals forces between themselves, though the nitrogen lone pair still lets a tertiary amine accept a hydrogen bond from water.

Boiling point trend among isomers. For amines of the same molar mass (same carbon skeleton, different substitution pattern), boiling points fall in the order

where: $1^\circ$, $2^\circ$, $3^\circ$ label primary, secondary, and tertiary amines of comparable molecular formula. All three isomers have essentially the same van der Waals contribution, so the ranking is decided purely by how many N-H bonds are left to hydrogen-bond with.

Amines versus alcohols of the same carbon count. An alcohol boils consistently higher than an amine of comparable molar mass, because the O-H bond is more polar than the N-H bond (oxygen is more electronegative than nitrogen), so O-H⋯O hydrogen bonds are stronger than N-H⋯N hydrogen bonds. Ethanol (b.p. $\approx 78\,^\circ C$) boils far above ethylamine (b.p. $\approx 17\,^\circ C$) even though both have two carbons and both can hydrogen-bond.

Water solubility. Lower amines, up to roughly three or four carbons, are freely miscible with water because the N-H (donor) and lone pair (acceptor) let them hydrogen-bond extensively with water molecules. Solubility drops off as the alkyl chain lengthens and the hydrophobic hydrocarbon part starts to dominate — the same shape of trend seen with alcohols, though at any given chain length an amine is somewhat less water-soluble than the matching alcohol.

Physical state and odour. The smallest amines — methylamine, dimethylamine, trimethylamine, and ethylamine — are gases at room temperature, all sharing a fishy, ammonia-like smell. Slightly larger amines are liquids, and the largest are solids, following the same size-driven state progression seen across many homologous series.

Aniline, the aromatic exception. Aniline ($C_6H_5NH_2$) is a colourless oily liquid with a notably high boiling point ($\approx 184\,^\circ C$) for its size, because the large, flat, polarisable benzene ring adds substantial van der Waals attraction on top of ordinary N-H hydrogen bonding. Despite having an $-NH_2$ group, aniline is only slightly soluble in water: the bulky hydrophobic phenyl ring dominates over the modest hydrogen-bonding pull of a single amine group. Aniline also darkens on standing in air or light, slowly oxidising.

Why does the count of N-H bonds, not total mass, decide the ranking within a set of isomers? Boiling means supplying enough energy to pull molecules apart against every attractive force holding them together. Within a family of isomeric $1^\circ/2^\circ/3^\circ$ amines the molecular formula, and therefore the van der Waals contribution, is identical — so the only variable left is hydrogen bonding. A primary amine's two N-H bonds let each molecule act as both donor and acceptor, building an extended hydrogen-bonded network that takes real energy to break. A secondary amine has one such donor site, so it forms fewer links. A tertiary amine has none, so its molecules cling together only through weaker dipole-dipole and dispersion forces — hence the lowest boiling point of the three. The same reasoning, run across elements instead of across substitution, explains why alcohols beat amines. Electronegativity governs how polar a bond is and how strong a hydrogen bond built from it can be. Oxygen ($\approx 3.5$) pulls electron density from its bonded hydrogen more strongly than nitrogen ($\approx 3.0$) does, so the O-H bond is more polar and its hydrogen bond correspondingly stronger. That single electronegativity gap is why an alcohol always boils higher than an amine of the same carbon count, even though both can hydrogen-bond. Water solubility follows the same logic in reverse: a molecule dissolves well when it can hydrogen-bond with water about as effectively as water hydrogen-bonds with itself. Small amines manage this because the polar $N-H_2$ or $N-H$ group is a large fraction of a small molecule. As the alkyl chain grows, the hydrophobic carbon skeleton grows with it, and past three or four carbons it outweighs the single hydrogen-bonding site — solubility falls.

Arrange $CH_3CH_2CH_2NH_2$ (1°, propylamine), $(CH_3CH_2)_2NH$ (2°, diethylamine), $(CH_3)_3N$ (3°, trimethylamine), and $CH_3CH_2CH_2OH$ (propanol) in increasing order of boiling point, and justify each step using hydrogen-bonding capacity and electronegativity — all four compounds have essentially the same molar mass.

Step 1. Note that all four compounds have similar molar mass (each built from three carbons plus one heteroatom-bearing group), so van der Waals attraction is roughly equal across all four; the differentiator must be hydrogen-bonding capacity.

Step 2. Trimethylamine, a tertiary amine, has no N-H bond, so it can only attract its neighbours through weak dipole-dipole and van der Waals forces — this places it lowest among the amines.

Step 3. Diethylamine, a secondary amine, has exactly one N-H bond, giving it some intermolecular hydrogen bonding — enough to boil higher than trimethylamine but not as high as a primary amine with two such bonds.

Step 4. Propylamine, a primary amine, has two N-H bonds and therefore the most extensive hydrogen-bonded network among the three amines, giving it the highest boiling point of the amine trio.

Step 5. Propanol has an O-H bond rather than an N-H bond; because oxygen is more electronegative than nitrogen, its hydrogen bonds are stronger than any N-H⋯N bond among the amines, so it boils higher than even the primary amine.

Final order (increasing boiling point): $(CH_3)_3N < (CH_3CH_2)_2NH < CH_3CH_2CH_2NH_2 < CH_3CH_2CH_2OH$.

Competitive stretch

Now explain why aniline ($C_6H_5NH_2$, b.p. $\approx 184\,^\circ C$) boils far higher than propylamine (b.p. $\approx 48\,^\circ C$) even though both are primary amines with two N-H bonds each. Since hydrogen-bonding capacity is identical, it cannot explain the roughly $136\,^\circ C$ gap. The decisive factor is size and polarisability: aniline's benzene ring is far larger and more polarisable than propylamine's short chain, adding substantial van der Waals attraction, and aniline's higher molar mass ($93\ \text{g/mol}$ versus $59\ \text{g/mol}$) means more thermal energy is needed to separate its molecules. Aniline's high boiling point comes from its aromatic ring's extra van der Waals contribution stacked on ordinary primary-amine hydrogen bonding, not from any difference in hydrogen-bonding ability.

• Treating tertiary amines as always lowest-boiling overall. The $1^\circ > 2^\circ > 3^\circ$ rule holds only within a set of amines of comparable molar mass; a small tertiary amine can still boil lower than a much bigger primary amine, and a large aromatic tertiary amine can outboil a tiny primary one.
• Forgetting amines lose to alcohols of the same size. The gap comes from oxygen's higher electronegativity strengthening O-H hydrogen bonds relative to N-H hydrogen bonds — missing this electronegativity link is the most common mark lost on this topic.
• Assuming tertiary amines cannot hydrogen-bond at all. They cannot donate a hydrogen bond (no N-H), but the nitrogen lone pair can still accept one from water's O-H, which is why tertiary amines remain somewhat water-soluble.
• Reading aniline's low water solubility as proof it cannot hydrogen-bond. It can, through its N-H bonds and lone pair; the large hydrophobic phenyl ring simply dominates the overall solubility outcome.

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