In order to differentiate between N-methylmethanamine and N,N-Dimethylmethanamine, the reaction to be used is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Hinsberg’s test
Hinsberg’s reagent, benzenesulphonyl chloride (\( \ce{C6H5SO2Cl} \)), reacts in a different way with primary, secondary and tertiary amines, so it is used to tell them apart.
N-Methylmethanamine, \( \ce{(CH3)2NH} \), is a secondary amine and forms a sulphonamide that does not dissolve in alkali:
\( \displaystyle \ce{C6H5SO2Cl + HN(CH3)2 -> C6H5SO2N(CH3)2 + HCl} \)
N,N-Dimethylmethanamine, \( \ce{(CH3)3N} \), is a tertiary amine and gives no reaction. So Hinsberg’s test distinguishes them.
Why the other options are incorrect
Option (1) Carbylamine reaction — Incorrect. Only primary amines give the foul-smelling isocyanide with \( \ce{CHCl3} \) and ethanolic KOH; both given amines (2° and 3°) give a negative test, so it cannot tell them apart.
\( \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{R-NC + 3KCl + 3H2O} \)
Option (2) Gabriel Phthalimide reaction — Incorrect. This is a method of preparing primary amines, not a test. Phthalimide is converted into its potassium salt, which is alkylated and then hydrolysed:
\( \ce{C6H4(CO)2NH} \xrightarrow{\text{KOH (alc.)}} \ce{C6H4(CO)2N^-K^+} \xrightarrow{\ce{R-X}} \ce{C6H4(CO)2N-R} \xrightarrow{\text{NaOH (aq)}} \ce{R-NH2 + C6H4(COO^-Na^+)2} \)
Option (3) Hofmann Bromamide reaction — Incorrect. This prepares a primary amine from an amide, with one carbon atom fewer; it is not a test that tells amines apart.
\( \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
NCERT (2026-27 edition):
Reaction with Arylsulphonyl Chloride – “Benzenesulphonyl chloride (C6H5SO2Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides. …
(b) In the reaction with secondary amine, N,N-diethyl-benzenesulphonamide is formed. … Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.
(c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines.”