Physical and Chemical Properties of Amines

Why amines are basic. The lone pair on nitrogen is available to accept a proton, converting the neutral amine into a positively charged ammonium ion:

where: $K_b$ is the base dissociation constant (larger $K_b$, smaller $pK_b = -\log_{10}K_b$, means a stronger base); the equilibrium lies further right when the lone pair is more available and when the resulting ammonium ion is better stabilised by the surrounding water.

Three factors, one competition. Three effects decide how available that lone pair is in aqueous solution. The inductive (+I) effect of alkyl groups pushes electron density onto nitrogen, so on electronics alone more substitution should mean a stronger base — the gas-phase order is $3° > 2° > 1° > \text{NH}_3$. Steric hindrance works the other way: bulky alkyl groups crowd the ammonium ion once formed, blocking water from hydrogen-bonding to it and stabilising the positive charge. Solvation favours fewer alkyl groups too — more N–H bonds mean more hydrogen bonds to water, so $1°$ ammonium ions solvate best. In water the last two effects dominate for the bulkiest amine, giving the observed order for simple alkylamines:

The secondary amine sits on top because it gets both the $+I$ boost of two alkyl groups and keeps one N–H for reasonable solvation — a compromise the bulkier, poorly solvated tertiary amine cannot match.

Aromatic amines break the pattern entirely. In aniline, the nitrogen lone pair is not just sitting on an isolated atom — it is delocalised into the benzene ring by resonance, spreading itself across the ring's ortho and para carbons. A delocalised lone pair is a less available lone pair, so aniline is a much weaker base than ammonia, let alone an alkylamine: $\text{alkylamines} > \text{NH}_3 > \text{aniline}$.

Physical properties follow the same lone pair and N–H bonds. Lower amines are gases or low-boiling liquids with a fishy smell. Primary and secondary amines hydrogen-bond (N–H···N) with each other and water, so their boiling points sit above similar-mass alkanes but below alcohols of similar size, since N–H···N bonds are weaker than O–H···O. Tertiary amines have no N–H bond, so they cannot hydrogen-bond to each other and boil lower than isomeric 1°/2° amines, though all lower amines stay water-soluble by accepting hydrogen bonds from water via the lone pair.

Key chemical reactions, all driven by the nitrogen lone pair or, for primary amines, the N–H bond:

• Acylation: amines react with acid chlorides or anhydrides to form amides, a nucleophilic substitution at the carbonyl carbon.
• Alkylation: reaction with alkyl halides; the product can react further, over-alkylating to a quaternary ammonium salt.
• Carbylamine test: primary amines heated with $CHCl_3$ and alcoholic $KOH$ give a foul-smelling isocyanide, unique to primary amines.

• Reaction with nitrous acid ($HNO_2$, from $NaNO_2/HCl$): primary aliphatic amines form unstable diazonium salts that decompose at once with brisk $N_2$ effervescence; primary aromatic amines form a diazonium salt stable only at $0$–$5°C$; secondary amines give an oily N-nitrosamine; tertiary amines show no clean reaction.
• Hinsberg test: the amine reacts with benzenesulfonyl chloride ($C_6H_5SO_2Cl$) and aqueous $KOH$; the outcome depends on how many N–H bonds are left to react.

There is no separate formula to derive for the Hinsberg test — its power comes from ordinary acid–base solubility reasoning applied after one predictable substitution, so it is worth walking through rather than memorising the outcome. This is the reasoning an examiner wants reproduced, and it is the second most-tested idea in this chapter after the basicity order.

Step 1 — the same substitution happens for every amine class. Benzenesulfonyl chloride is an electrophile at sulfur, much like an acid chloride is at carbon. Whatever N–H bonds the amine has, its lone pair attacks the sulfonyl chloride and displaces chloride ion, forming a sulfonamide. A primary amine, with two N–H bonds, loses one to give a sulfonamide that still has one N–H left; a secondary amine, with only one N–H, loses it and is left with none; a tertiary amine has no N–H to begin with, so it cannot react this way at all.

Step 2 — solubility in excess KOH follows from what N–H remains. The primary amine's sulfonamide still carries one acidic N–H, made acidic by the neighbouring electron-withdrawing sulfonyl group, so excess $KOH$ removes it, giving a soluble anion. The secondary amine's sulfonamide has no N–H left to deprotonate, so it stays insoluble regardless of how much $KOH$ is added. The unreacted tertiary amine is insoluble in the alkaline layer but dissolves on acidification, since protonating its lone pair regenerates a soluble ammonium salt.

Step 3 — the three outcomes are mutually exclusive. Only the primary amine's product dissolves in excess alkali; only the secondary amine reacts yet stays insoluble; only the tertiary amine fails to react at all. No two classes share a combination of "does it react" and "does the product dissolve," so one set of observations identifies all three.

Question: Rank aniline, methylamine and ammonia in order of increasing basicity, and support the ranking with the following aqueous $pK_b$ values: ammonia $4.75$, methylamine $3.38$, aniline $9.4$. Explain why the aniline value is so much larger.

Step 1 — Recall what $pK_b$ measures. A smaller $pK_b$ means a larger $K_b$, a stronger base. Ranking basicity from weakest to strongest is the same as ranking $pK_b$ from largest to smallest.

Step 2 — Order the given values. Aniline's $pK_b = 9.4$ is largest, ammonia's $4.75$ is next, methylamine's $3.38$ is smallest — largest to smallest $pK_b$ gives weakest to strongest base directly.

Step 3 — Explain methylamine over ammonia. The methyl group's $+I$ effect pushes electron density onto nitrogen, making methylamine's lone pair more available to bond with $H^+$ than ammonia's — consistent with its smaller $pK_b$.

Step 4 — Explain aniline's large $pK_b$. The nitrogen lone pair is delocalised into the benzene ring by resonance (draw the structures placing negative charge at ortho and para carbons); a spread-out lone pair is far less available to accept a proton. This resonance effect outweighs any inductive contribution, pushing aniline's $pK_b$ almost two orders of magnitude above ammonia's. Answer: aniline (weakest) < ammonia < methylamine (strongest), matching $9.4 > 4.75 > 3.38$.

Competitive stretch

Now rank methylamine, dimethylamine and trimethylamine in water using $pK_b$ values of $3.38$, $3.27$ and $4.19$ respectively. Smallest $pK_b$ is strongest, so dimethylamine (2°, $pK_b = 3.27$) is the strongest base, methylamine (1°, $pK_b = 3.38$) is next, and trimethylamine (3°, $pK_b = 4.19$) is weakest — exactly the $2° > 1° > 3°$ aqueous pattern, with the tertiary amine's steric hindrance and poorer solvation outweighing its extra $+I$ substitution.

• Assuming basicity in water always follows $3° > 2° > 1°$ — that is the gas-phase order; in water, steric hindrance and solvation push it to $2° > 1° > 3°$ for simple alkylamines.
• Underestimating how weak aniline is — it is weaker than ammonia itself, not just weaker than methylamine, since resonance outweighs any inductive donation.
• Getting the carbylamine reagent wrong — it is $CHCl_3$ with alcoholic $KOH$, positive only for primary amines.
• Swapping which Hinsberg product dissolves — the primary amine's sulfonamide dissolves in excess KOH (it still has an N–H); the secondary amine's product does not.

Ask the tutor: if the resonance argument for aniline's weak basicity or the Hinsberg solubility logic still feels shaky, ask the AI tutor to walk through the resonance structures or the step-by-step reasoning with you.