Reaction with HNO2 & Hinsberg Test · Amines - Organic Compounds Containing Nitrogen

CHEReactions Of Amines

Reaction with HNO₂ & Hinsberg Test

Reaction with nitrous acid ( $NaNO_2 + HCl$ , $273\text{-}278\,K$ ) and the Hinsberg test are the two classic ways to tell $1^\circ$ , $2^\circ$ and $3^\circ$ amines apart. The key is that each amine class gives a characteristically different product or solubility behaviour.

Aliphatic vs aromatic

1^\circ

amine with

HNO_2

CH_3NH_2 \xrightarrow{HNO_2} CH_3OH + N_2 \qquad C_6H_5NH_2 \xrightarrow{HNO_2,\ 273\text{-}278\,K} C_6H_5N_2^+Cl^-

alkyl gives alcohol + brisk

N_2

; aryl gives a stable diazonium salt

2^\circ

amine with

HNO_2

(CH_3)_2NH \xrightarrow{HNO_2} (CH_3)_2N\text{-}NO + H_2O

product is a yellow oily N-nitrosamine

Aliphatic $1^\circ$ amine $+ HNO_2 \rightarrow$ alcohol $+ N_2$ (brisk effervescence): $CH_3NH_2 \rightarrow CH_3OH + N_2$ (unstable diazonium decomposes).
Aromatic $1^\circ$ amine $+ HNO_2$ at $273\text{-}278\,K \rightarrow$ stable diazonium salt: $C_6H_5NH_2 \rightarrow C_6H_5N_2^+Cl^-$ (no immediate $N_2$ ).
$2^\circ$ amines $+ HNO_2 \rightarrow$ yellow oily N-nitrosamine, e.g. $(CH_3)_2NH \rightarrow (CH_3)_2N\text{-}NO$ .
$3^\circ$ amines (e.g. trimethylamine) give no N-nitrosamine; they simply form a soluble salt, so the result differs from the $2^\circ$ case.
$HNO_2$ is unstable, so it is always generated in situ from $NaNO_2$ and a mineral acid ( $HCl$ ) at low temperature.
Hinsberg test uses benzenesulphonyl chloride ( $C_6H_5SO_2Cl$ ); the three classes are told apart by what forms and its solubility in alkali.
$1^\circ$ amine + Hinsberg reagent $\rightarrow$ a sulphonamide with an acidic N-H, soluble in $KOH$ (gives a clear solution).
$2^\circ$ amine $\rightarrow$ a sulphonamide with no N-H, insoluble in $KOH$ (remains as a precipitate); $3^\circ$ amine does not react (no N-H to acylate).
Hinsberg summary: $1^\circ$ = precipitate that dissolves in alkali; $2^\circ$ = precipitate insoluble in alkali; $3^\circ$ = no reaction (no precipitate).
The $1^\circ$ sulphonamide is alkali-soluble because the strongly electron-withdrawing $\text{-}SO_2\text{-}$ group makes its remaining N-H acidic, so $KOH$ forms a soluble salt.
Combined check: carbylamine confirms $1^\circ$ only, while Hinsberg separates all three by the soluble / insoluble / no-reaction pattern.

Where the marks go

Treating all $1^\circ$ amines the same with $HNO_2$ — aliphatic ones give alcohol + $N_2$ , but aromatic ones give a stable diazonium salt at low temperature.
Swapping the Hinsberg solubilities: the $1^\circ$ sulphonamide is soluble in alkali, the $2^\circ$ sulphonamide is insoluble — reversing them is a frequent error.
Saying a $3^\circ$ amine gives an N-nitrosamine with $HNO_2$ — it does not (no N-H); only $2^\circ$ amines give the yellow nitrosamine.
Forgetting that $HNO_2$ must be made in situ from $NaNO_2 + HCl$ and used cold; writing it as a stable bottled reagent loses marks.
Claiming the Hinsberg test distinguishes only two classes — modern ISC treatment uses the soluble/insoluble/no-reaction pattern to separate all three.

How the board asks it

Distinguishthe soluble / insoluble / no-reaction pattern of the hinsberg test
How will you distinguish between a primary, a secondary and a tertiary amine using the Hinsberg test? State the observation in each case.
Predict the productamine $+ HNO_2$ generated in situ at $273\text{-}278\,K$
Write the product formed when (i) $CH_3CH_2NH_2$ , (ii) $(CH_3)_2NH$ and (iii) aniline ( $C_6H_5NH_2$ ) each react with nitrous acid ( $NaNO_2/HCl$ ) at $273\text{-}278\,K$ .
Give reasonsacidity of the $1^\circ$ sulphonamide N-H due to the $\text{-}SO_2\text{-}$ group
Account for the fact that the sulphonamide formed from a primary amine in the Hinsberg test dissolves in $KOH$ , whereas that from a secondary amine does not.
Distinguishaliphatic effervescence vs aromatic stable diazonium with $HNO_2$
How will you distinguish between ethylamine and aniline using nitrous acid at $273\text{-}278\,K$ ? Give the reaction in each case.
Conversionaliphatic $1^\circ$ amine $+ HNO_2 \rightarrow$ alcohol $+ N_2$
How will you convert methylamine ( $CH_3NH_2$ ) into methanol ( $CH_3OH$ )? Write the equation involved.

Practise this topic

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Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.