How do you differentiate between primary, secondary and tertiary amines?
A. Reaction with Br2 water.
B. Reaction with sulfonamide.
C. Reaction with Hinsberg's reagent followed by alkali.
D. Reaction with acetic anhydride.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → C only
Hinsberg’s reagent, benzenesulphonyl chloride (\( \ce{C6H5SO2Cl} \)), reacts with an N–H bond of an amine, so the three classes of amines give three different results:
\( \displaystyle \ce{C6H5SO2Cl + H2N-R -> C6H5SO2NH-R + HCl}\quad(1^{\circ}) \)
\( \displaystyle \ce{C6H5SO2Cl + HNR2 -> C6H5SO2NR2 + HCl}\quad(2^{\circ}) \)
A. Reaction with \( \ce{Br2} \) water. — Incorrect. Aniline gives a white precipitate of 2,4,6-tribromoaniline. This tests for an activated aromatic ring, not for the class of an amine.
\( \displaystyle \ce{C6H5NH2 + 3Br2} \xrightarrow{\ce{H2O}} \ce{C6H2Br3NH2 v + 3HBr} \)
B. Reaction with sulfonamide. — Incorrect. A sulphonamide is the product of Hinsberg’s test, not a reagent that reacts with amines.
C. Reaction with Hinsberg's reagent followed by alkali. — Correct. The 1° product dissolves in alkali, the 2° product does not, and the 3° amine does not react, so all three classes are told apart.
D. Reaction with acetic anhydride. — Incorrect. Both 1° and 2° amines are acylated, because both have an N–H bond, and only 3° amines fail to react, so 1° and 2° amines cannot be told apart.
\( \displaystyle \ce{R-NH2 + (CH3CO)2O -> R-NHCOCH3 + CH3COOH} \)
NCERT (2026-27 edition):
Reaction with Arylsulphonyl Chloride
