Hinsberg Test & Carbylamine Reaction (Distinguishing Amines)

Why the N–H count decides everything

Every reaction in this article turns on a single structural fact: how many hydrogen atoms remain on the nitrogen. A primary amine $\ce{R-NH2}$ has two N–H bonds, a secondary amine $\ce{R2NH}$ has one, and a tertiary amine $\ce{R3N}$ has none. NCERT states the principle plainly: the number of hydrogen atoms attached to nitrogen decides the course of reaction, which is why primary, secondary and tertiary amines differ in many reactions.

The lone pair on nitrogen makes all three classes nucleophilic, so all three can begin to react with an electrophile. What separates them is the second step — whether an acidic N–H survives, or whether a hydrogen is even available for the reaction to proceed at all. Keep that lens fixed and the three classic tests stop being a list to memorise and become predictable consequences.

If you are not yet confident labelling an amine as 1°, 2° or 3°, settle that first in amine structure and classification; the rest of this page assumes you can read the N–H count off a structure at a glance.

The Hinsberg test

The Hinsberg test uses benzenesulphonyl chloride, $\ce{C6H5SO2Cl}$, called Hinsberg's reagent. NCERT §9.6 describes how the reagent reacts with primary and secondary amines to form sulphonamides, while tertiary amines do not react at all. The whole test is read off the solubility of the product in alkali.

Primary amine → sulphonamide soluble in alkali

A primary amine attacks the sulphonyl chloride to form an N-substituted benzenesulphonamide. Taking ethanamine as the worked NCERT example:

$$\ce{C2H5NH2 + C6H5SO2Cl -> C6H5SO2-NH-C2H5 + HCl}$$

The product still carries one N–H bond. NCERT notes that this hydrogen is strongly acidic because of the powerfully electron-withdrawing sulphonyl group, so the sulphonamide dissolves in alkali by losing that proton to give a water-soluble salt:

$$\ce{C6H5SO2-NH-C2H5 + KOH -> C6H5SO2-N^{-}(K^{+})-C2H5 + H2O}$$

Secondary amine → sulphonamide insoluble in alkali

A secondary amine such as N-ethylethanamine forms N,N-diethylbenzenesulphonamide:

$$\ce{(C2H5)2NH + C6H5SO2Cl -> C6H5SO2-N(C2H5)2 + HCl}$$

This product has no N–H bond left. With no acidic hydrogen for the alkali to remove, NCERT records that it is not acidic and hence insoluble in alkali — it stays as a precipitate.

Tertiary amine → no reaction

A tertiary amine has no hydrogen on nitrogen to begin with, so it cannot form a sulphonamide. NCERT is explicit that tertiary amines do not react with benzenesulphonyl chloride. NIOS §28.2.7 adds the practical observation: with a water-insoluble tertiary amine, nothing changes on adding the reagent and alkali; only on subsequent acidification does the amine dissolve, because it then forms a water-soluble ammonium salt.

NCERT records one practical refinement worth knowing: these days benzenesulphonyl chloride is often replaced by p-toluenesulphonyl chloride for the same identification. The logic is identical — only the aryl group changes. Because the test gives three visibly distinct outcomes, it also serves to separate a mixture of the three amine classes, not merely identify them.

Why solubility splits 1° from 2°

The single most examined idea on this page is the contrast between the primary and secondary sulphonamides. Both are sulphonamides; both are formed by the same nucleophilic substitution at sulphur. The difference is whether an N–H bond survives the reaction.

In the primary case the nitrogen started with two hydrogens; one is replaced by the bulky $\ce{-SO2C6H5}$ group, leaving one N–H. The adjacent $\ce{-SO2-}$ group is strongly electron-withdrawing and stabilises the conjugate base formed when that proton leaves, so the N–H is acidic enough to be deprotonated by KOH — hence solubility. In the secondary case the nitrogen had only one hydrogen, and acylation by sulphur uses it up entirely; the resulting $\ce{R2N-SO2C6H5}$ has no proton to give, so alkali cannot touch it.

The carbylamine (isocyanide) reaction

The carbylamine reaction, also called the isocyanide test, is the cleanest single test for a primary amine. NCERT §9.6 states that aliphatic and aromatic primary amines, on heating with chloroform and ethanolic potassium hydroxide, form isocyanides (carbylamines), which are foul-smelling substances; secondary and tertiary amines do not show this reaction.

For a generic primary amine the overall equation, in the NIOS form, is:

$$\ce{R-NH2 + CHCl3 + 3KOH ->[\Delta] R-NC + 3KCl + 3H2O}$$

Using aniline, the favourite aromatic example named in NCERT exercise 9.11(i):

$$\ce{C6H5NH2 + CHCl3 + 3KOH ->[\Delta] C6H5NC + 3KCl + 3H2O}$$

The product $\ce{R-NC}$ is an isocyanide (carbylamine). Its extremely offensive odour is the positive signal; because the test is positive only for primary amines, the smell is taken as confirmation of a 1° amine. The reaction is doubly useful — it is also a sensitive test for the chloroform that supplies the carbene, so a foul carbylamine smell betrays the presence of $\ce{CHCl3}$.

Dichlorocarbene mechanism

The reactive species in the carbylamine reaction is dichlorocarbene, $\ce{:CCl2}$ — a neutral, electron-deficient carbon with only six valence electrons. It is generated in two steps from chloroform and the strong base. First, alcoholic KOH removes the relatively acidic chloroform proton to give a trichloromethyl carbanion:

$$\ce{CHCl3 + KOH -> CCl3^{-} + K+ + H2O}$$

The carbanion then ejects a chloride ion (α-elimination) to leave the carbene:

$$\ce{CCl3^{-} -> {:}CCl2 + Cl^{-}}$$

The lone pair on the primary amine nitrogen now attacks the empty orbital of $\ce{:CCl2}$; successive loss of two molecules of HCl (driven by the base) converts the adduct into the isocyanide. Schematically:

$$\ce{R-NH2 + {:}CCl2 -> R-NC + 2HCl}$$

Reaction with nitrous acid

The third distinguishing tool is nitrous acid, $\ce{HNO2}$, generated in situ from a mineral acid and sodium nitrite. NCERT §9.6 states that the three classes of amine react differently, which is exactly what makes it diagnostic. The behaviour also separates the two kinds of primary amine.

Primary aliphatic amines form an aliphatic diazonium salt that is unstable; it decomposes at once, liberating nitrogen quantitatively and giving an alcohol. The brisk, steady effervescence of $\ce{N2}$ is the visible signal:

$$\ce{R-NH2 + HNO2 -> [R-N2^{+}] ->[H2O] R-OH + N2 ^ + H2O}$$

Primary aromatic amines form a diazonium salt that is stable at low temperature (273–278 K) — the basis of the entire diazonium chemistry that follows:

$$\ce{C6H5NH2 + HNO2 ->[273-278\,K] C6H5N2^{+}Cl^{-} + H2O}$$

Secondary and tertiary amines react with nitrous acid in a different manner and do not show the clean nitrogen evolution of primary aliphatic amines, so the absence of brisk effervescence is itself informative. The downstream chemistry of the aromatic diazonium salt — Sandmeyer, hydroxylation, coupling — is developed separately in Diazonium Salts.

The three classes side by side

The table collects every response so the pattern is visible at a glance. Read each column as a fingerprint: any one positive test plus one negative is usually enough to pin a class.

Choosing the right test

For a NEET question, match the test to the discrimination asked for. To single out a primary amine from the other two in one step, the carbylamine test is decisive — it is the only test positive for 1° alone. To separate 1° from 2° specifically, either the carbylamine test (positive vs negative) or the Hinsberg test (soluble vs insoluble) works. To flag a tertiary amine, the Hinsberg test is the natural choice, since 3° amines are the only class that gives no reaction with the reagent.

To distinguish an aliphatic from an aromatic primary amine, lean on nitrous acid: the aliphatic amine fizzes with $\ce{N2}$ at room temperature, while the aromatic amine forms a salt that survives at 273–278 K and can be coupled to a coloured dye. The dye test (azo coupling) is itself a confirmation of an aromatic primary amine.