Chemical reactions of amines

1. Alkylation. Amines react with alkyl halides via an $S_N2$ mechanism, climbing the substitution ladder from primary to quaternary:

where: $R\text{-}NH_2$ is the starting primary amine, $R'X$ is the alkyl halide supplying successive alkyl groups, and each arrow is one nitrogen-lone-pair attack on the electrophilic carbon of $R'X$, displacing halide $X^-$ in a backside ($S_N2$) attack. Because the product of each step is itself a better nucleophile than the starting amine (more alkyl groups push electron density onto nitrogen), the reaction rarely stops cleanly at one stage — a large excess of amine relative to alkyl halide is needed to favour mono-alkylation.

2. Acylation. First and second amines react with acid chlorides or anhydrides, replacing an N–H with an N-acyl group to form an amide; third amines cannot be acylated because no N–H remains to replace:

where: $R\text{-}NH_2$ is the amine, $CH_3COCl$ is acetyl chloride, and $R\text{-}NHCOCH_3$ is the resulting amide (an N-substituted acetamide). Mechanistically, the nitrogen lone pair attacks the electrophilic carbonyl carbon, forming a tetrahedral intermediate that collapses by expelling chloride (or a carboxylate, if an anhydride is used instead).

3. Reaction with nitrous acid. $HNO_2$ is generated in situ from $NaNO_2 + HCl$ at $0$–$5\ ^\circ C$, and each amine class gives a diagnostically distinct outcome. A first amine gives an unstable diazonium salt that decomposes at once with brisk effervescence, releasing nitrogen gas and forming an alcohol: $R\text{-}NH_2 + HNO_2 \rightarrow R\text{-}OH + N_2\uparrow + H_2O$. A second amine gives a yellow, oily N-nitrosamine: $R_2NH + HNO_2 \rightarrow R_2N\text{-}N{=}O + H_2O$. A third amine simply forms an unstable salt, with no substitution possible at a nitrogen that already carries three alkyl groups, and is largely recovered unchanged.

4. Carbylamine (isocyanide) reaction. This is a diagnostic test specific to primary amines only. Heated with chloroform and alcoholic $KOH$, a primary amine gives a foul-smelling isocyanide:

Second and third amines give no reaction at all, since forming the isocyanide requires an N–H bond to be lost, and neither class has the right combination available.

5. Hinsberg's test. The amine is shaken with benzenesulphonyl chloride, $C_6H_5SO_2Cl$, and aqueous $KOH$. A first amine forms N-alkylbenzenesulphonamide, which still carries one acidic N–H and so dissolves in excess KOH as a soluble potassium salt. A second amine forms N,N-dialkylbenzenesulphonamide with no N–H left, which stays insoluble in KOH as a solid. A third amine does not react at all — there is no N–H for the sulphonyl chloride to substitute — and it simply remains as an unreacted, insoluble oily layer.

All five reactions trace back to two structural facts about an amine nitrogen: it carries a lone pair, and it carries a specific number of N–H bonds depending on its class. The lone pair makes nitrogen nucleophilic, so alkylation, acylation, and attack on nitrous acid are all variations of the same event — nitrogen donates its lone pair to an electrophilic centre (an alkyl halide carbon, a carbonyl carbon, or the nitrogen of $HNO_2$) and a leaving group departs. Basicity and nucleophilicity run together here: anything that makes the lone pair more available (more alkyl groups feeding electron density onto nitrogen, up to steric limits) tends to make the amine react faster in these steps too. The N–H count is why the carbylamine test and Hinsberg's test are class-selective rather than universal. Both hinge on whether an N–H bond is available to be lost or substituted: a first amine has two N–H bonds and can form the isocyanide and, after Hinsberg substitution, still retain one acidic N–H to dissolve in excess base; a second amine has one N–H, enough to react with benzenesulphonyl chloride but not enough left over to stay soluble; a third amine has none, so it fails the carbylamine test and never reacts with benzenesulphonyl chloride at all. Reading any of these five reactions is really just asking two questions — is the lone pair available, and how many N–H bonds are left to give up?

Problem: Ethylamine is treated with $NaNO_2/HCl$ at $0$–$5\ ^\circ C$. Write the equation, name the gas evolved, and state the analytical significance of this reaction. Then explain how the same sample would behave differently if it were diethylamine or triethylamine instead.

Step 1. Write the equation for the primary amine. $CH_3CH_2NH_2 + HNO_2 \rightarrow CH_3CH_2OH + N_2\uparrow + H_2O$. Mechanistically, the diazonium salt $CH_3CH_2N_2^+Cl^-$ forms transiently but, unlike the aromatic analogue, is too unstable at this temperature and decomposes immediately.

Step 2. Identify the gas and its significance. The gas evolved is $N_2$ — colourless, odourless, and released with brisk effervescence, visually confirming a primary amine is present. Together with the carbylamine test, this reaction is a standard analytical marker for a primary amino group: the vigorous, immediate gas evolution is what distinguishes a 1° amine from 2° and 3° amines under the same conditions.

Step 3. Contrast with diethylamine (2°). Diethylamine would instead give a yellow oily N-nitrosamine, $(C_2H_5)_2N\text{-}N{=}O$, with no gas evolved at all — a visibly different, non-effervescent outcome that immediately rules out a primary amine.

Step 4. Contrast with triethylamine (3°). Triethylamine has no N–H bond to be substituted, so it merely forms an unstable ammonium-type salt with nitrous acid and is largely recovered unchanged on working up — again, no gas and no coloured oily product. Answer: ethylamine gives $CH_3CH_2OH + N_2\uparrow + H_2O$ (brisk effervescence, confirms 1° amine); diethylamine gives a yellow oily nitrosamine with no gas; triethylamine gives only an unreactive salt.

Competitive stretch

The same three amines can also be told apart by Hinsberg's test: ethylamine gives a product that dissolves in excess KOH (soluble salt of the mono-substituted sulphonamide), diethylamine gives an insoluble solid (no N–H left to deprotonate), and triethylamine gives no reaction at all, remaining as an unreacted oily layer that separates from the aqueous phase and, on acidification, regenerates the free amine unchanged. Running nitrous acid and Hinsberg's test side by side on an unknown mixture is exactly how a lab distinguishes all three amine classes with no ambiguity.

• Trying to acylate a tertiary amine — impossible, since acylation requires an N–H bond to be replaced, and a third amine has none.
• Forgetting that only primary amines give a positive carbylamine (isocyanide) test; a negative result rules out a 1° amine but says nothing else.
• Mixing up the nitrous acid products across the three classes: 1° gives an alcohol plus $N_2$ gas; 2° gives an N-nitrosamine with no gas; 3° (aliphatic) gives only a simple, largely unreactive salt.
• In Hinsberg's test, judging solubility before adding excess KOH — the initial sulphonamide product from a 1° amine can look insoluble at first and only dissolves once the base deprotonates its remaining N–H.

Ask the tutor: want to work through why the aromatic diazonium salt from aniline is stable at $0$–$5\ ^\circ C$ while the aliphatic one from ethylamine is not — ask the AI tutor to connect it to the diazonium salts article in this series.