Hinsberg Test: Distinguishing & Separating Amines
The Hinsberg Test is a chemical test used to distinguish between Primary ($1^\circ$), Secondary ($2^\circ$), and Tertiary ($3^\circ$) amines. It utilizes Benzenesulfonyl Chloride ($C_6H_5SO_2Cl$) as the reagent, also known as the Hinsberg Reagent.
1. The Reagent
Hinsberg Reagent: Benzenesulfonyl Chloride ($C_6H_5SO_2Cl$).
Sometimes p-Toluenesulfonyl chloride (Tosyl chloride) is used for better crystallization of products.
2. Reactions and Observations
The test is based on the formation of sulfonamides and their solubility in alkali (aqueous NaOH or KOH).
A. Primary Amines ($R-NH_2$)
Primary amines react with Benzenesulfonyl chloride to form N-Alkylbenzenesulfonamide.
Observation
The hydrogen atom attached to the nitrogen in the sulfonamide is strongly acidic (due to the electron-withdrawing sulfonyl group). Therefore, the product dissolves in aqueous NaOH to form a clear solution.
$$ C_6H_5SO_2-NH-R + NaOH \rightarrow [C_6H_5SO_2-N-R]^-Na^+ \text{ (Soluble)} $$B. Secondary Amines ($R_2NH$)
Secondary amines react to form N,N-Dialkylbenzenesulfonamide.
Observation
The product has no acidic hydrogen on the nitrogen atom. Therefore, it is insoluble in aqueous NaOH and remains as a precipitate/oil even after adding alkali.
C. Tertiary Amines ($R_3N$)
Tertiary amines do not have any hydrogen attached to the nitrogen atom.
Observation
The amine remains unreacted and insoluble in water/alkali. However, it dissolves in dilute acid ($HCl$) because it forms a soluble ammonium salt.
3. Summary Table
| Amine Type | Reaction with Reagent | Solubility in NaOH | Solubility in Acid |
|---|---|---|---|
| Primary ($1^\circ$) | Forms Sulfonamide | Soluble | Re-precipitates on acidification |
| Secondary ($2^\circ$) | Forms Sulfonamide | Insoluble | Insoluble |
| Tertiary ($3^\circ$) | No Reaction | Insoluble | Soluble (forms salt) |
4. Separation of Amine Mixtures
The Hinsberg test can also be used to separate a mixture of $1^\circ, 2^\circ, 3^\circ$ amines.
- Treat mixture with $PhSO_2Cl$ and $NaOH$.
- Tertiary amine remains unreacted (insoluble oil). Extract with ether or distill off.
- Secondary amine forms insoluble sulfonamide. Filter it out. Hydrolysis regenerates $2^\circ$ amine.
- Primary amine forms soluble salt. Acidify the filtrate to precipitate the sulfonamide. Hydrolysis regenerates $1^\circ$ amine.
5. Reaction Mechanism
The reaction is a Nucleophilic Substitution at the Sulfur atom.
- The amine nitrogen (Nucleophile) attacks the electrophilic sulfur atom.
- Chloride ion acts as the leaving group.
- Similar to acyl substitution but happens at the sulfonyl group ($S_{N}-S$).
Hinsberg Quiz
Test your concepts on Amine differentiation. 10 MCQs with explanations.
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