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: The whole test hinges on one idea: how many N–H bonds remain after the amine attacks benzenesulphonyl chloride, C\(_6\)H\(_5\)SO\(_2\)Cl. A primary amine (R–NH\(_2\)) keeps one N–H in its sulphonamide product; the strongly electron-withdrawing sulphonyl group makes that N–H acidic, so it is deprotonated by KOH and the solid dissolves — a clear solution signals a primary amine.
\[ RNH_2 + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NHR \xrightarrow{KOH} \text{soluble salt} \]
A secondary amine (R\(_2\)NH) leaves no N–H after reaction, so its sulphonamide cannot form a salt and stays as an alkali-insoluble solid:
\[ R_2NH + C_6H_5SO_2Cl \rightarrow C_6H_5SO_2NR_2\;(\text{insoluble}) \]
A tertiary amine (R\(_3\)N) has no N–H to begin with and does not react, remaining as a separate layer that only dissolves on adding acid. Thus soluble-in-alkali = 1°, insoluble product = 2°, no reaction = 3°.

(ii) Carbylamine reaction: Only primary amines respond. Chloroform under alcoholic KOH generates dichlorocarbene (:CCl\(_2\)), the reactive species that inserts into the primary amine to build an isocyanide (carbylamine) after loss of HCl:
\[ RNH_2 + CHCl_3 + 3KOH \xrightarrow{\Delta} RNC + 3KCl + 3H_2O \]
The isocyanides produced are intolerably foul-smelling, which is why the appearance of that odour is used as a confirmatory test for a primary amine. Secondary and tertiary amines cannot form the required intermediate and give no isocyanide.

Option 2

(i) Hydrolysis of the diazonium ion on warming → phenol:
\[ C_6H_5N_2^+Cl^- + H_2O \xrightarrow{\Delta} C_6H_5OH + N_2\uparrow + HCl \]
(ii) Amide shortened by one carbon (Hofmann degradation) → methylamine:
\[ CH_3CONH_2 + Br_2 + 4KOH \rightarrow CH_3NH_2 + K_2CO_3 + 2KBr + 2H_2O \]
(iii) Low-temperature diazotisation → benzene diazonium chloride:
\[ C_6H_5NH_2 + NaNO_2 + 2HCl \xrightarrow{273-278\,K} C_6H_5N_2^+Cl^- + NaCl + 2H_2O \]
(iv) N-acylation → acetanilide:
\[ C_6H_5NH_2 + (CH_3CO)_2O \xrightarrow{\text{pyridine}} C_6H_5NHCOCH_3 + CH_3COOH \]
(v) Ring bromination by bromine water → 2,4,6-tribromoaniline (white ppt):
\[ C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3(NH_2) + 3HBr \]