Question

(i) Explain Hinsberg test for the distinction between primary, secondary and tertiary amines. (ii) Write a short note on Carbylamine reaction. (2½+2½=5)
OR
What happens when (write chemical equations only): (i) Aqueous solution of benzene diazonium chloride is heated (ii) Acetamide is reacted with aqueous KOH in the presence of bromine (iii) Aniline reacts with sodium nitrite and dil. HCl at 0°C (iv) Aniline reacts with acetic anhydride in the presence of pyridine (v) Aniline reacts with Bromine water. (1+1+1+1+1=5)

Hint:

Option 1: Hinsberg's reagent is benzenesulphonyl chloride; the number of N-H bonds left decides solubility in KOH (1° soluble, 2° insoluble, 3° no reaction). Carbylamine (isocyanide) test with CHCl\(_3\) + alc. KOH is specific for primary amines. Option 2: think phenol formation, Hofmann degradation, diazotisation, acetylation, and 2,4,6-tribromoaniline.

Answer 1

Option 1

(i) Hinsberg test:
Step 1: The reagent is benzenesulphonyl chloride, C\(_6\)H\(_5\)SO\(_2\)Cl (Hinsberg's reagent). The amine is shaken with this reagent in presence of aqueous KOH; the behaviour of the product towards alkali distinguishes the three classes of amine.
Step 2 (primary amine): It reacts to form an N-alkylbenzenesulphonamide. The H atom still attached to nitrogen is acidic (activated by the \(-SO_2-\) group), so the product dissolves in KOH to give a clear solution.
\[ C_6H_5SO_2Cl + H_2N-R \rightarrow C_6H_5SO_2NHR \xrightarrow{KOH} C_6H_5SO_2N^-R\;(\text{soluble}) \]
Step 3 (secondary amine): It forms an N,N-dialkylbenzenesulphonamide. There is no H left on nitrogen, so it cannot form a salt and remains insoluble in KOH (a precipitate/oily layer).
\[ C_6H_5SO_2Cl + HNR_2 \rightarrow C_6H_5SO_2NR_2\;(\text{insoluble in KOH}) \]
Step 4 (tertiary amine): It has no replaceable H on nitrogen and does not react with the reagent at all; it remains as an insoluble layer that dissolves in acid.
Summary: primary \(\rightarrow\) product soluble in alkali; secondary \(\rightarrow\) product insoluble in alkali; tertiary \(\rightarrow\) no reaction.

(ii) Carbylamine reaction:
Step 1: Aliphatic and aromatic primary amines, when heated with chloroform (CHCl\(_3\)) and alcoholic KOH, produce isocyanides (carbylamines) which have an extremely offensive (foul) smell.
Step 2: The reaction:
\[ R-NH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} R-NC + 3KCl + 3H_2O \]
(e.g. \(C_6H_5NH_2 \rightarrow C_6H_5NC\), phenyl isocyanide).
Step 3: Secondary and tertiary amines do NOT give this reaction, so the carbylamine (isocyanide) test is a specific test used to detect primary amines.

Option 2

(i) Warming the aqueous diazonium salt replaces \(-N_2^+\) by \(-OH\), giving phenol:
\[ C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2 + HCl \]
(ii) Hofmann bromamide degradation; the amide loses one carbon to give an amine with one carbon less:
\[ CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O \]
(iii) Diazotisation at 273–278 K gives benzene diazonium chloride:
\[ C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\,K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O \]
(iv) Acetylation of the amino group gives acetanilide:
\[ C_6H_5NH_2 + (CH_3CO)_2O \xrightarrow{\text{pyridine}} C_6H_5NHCOCH_3 + CH_3COOH \]
(v) Bromine water gives the white precipitate of 2,4,6-tribromoaniline:
\[ C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3(NH_2) + 3HBr \]