Match List-I with List-II
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List-I Test |
List-II Identification |
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(A) Hinsberg's Test |
(I) Primary, Secondary and tertiary Amines |
|
(B) Isocyanide test |
(II) Aliphatic and Aromatic primary amines |
|
(C) Azo dye test |
(III) Primary Aromatic amines |
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(D) Tollen's Test |
(IV) Reducing sugars |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Each test depends on a reaction that only one class of compound gives, with a visible result: a precipitate, a smell, a coloured dye or a silver mirror.
(A) Hinsberg's Test — This matches (I) Primary, Secondary and tertiary Amines. Benzenesulphonyl chloride (Hinsberg’s reagent) reacts with a primary amine to give an N-alkylbenzenesulphonamide that dissolves in alkali (it still has an acidic H on N), with a secondary amine to give a sulphonamide that does not dissolve in alkali, and does not react with a tertiary amine. So it tells all three classes apart.
\( \displaystyle \ce{C6H5SO2Cl + H2N-R -> C6H5SO2NH-R + HCl} \)
(B) Isocyanide test — This matches (II) Aliphatic and Aromatic primary amines. On warming with chloroform and ethanolic KOH, any primary amine, aliphatic or aromatic, gives an isocyanide (carbylamine) with a very unpleasant smell; secondary and tertiary amines do not.
\( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\text{heat}} \ce{R-NC + 3KCl + 3H2O} \)
(C) Azo dye test — This matches (III) Primary Aromatic amines. A primary aromatic amine is diazotised with \( \ce{NaNO2} \) and HCl at 273–278 K, and the diazonium salt couples with alkaline β-naphthol to give an orange-red azo dye. Aliphatic diazonium salts are not stable and do not give the dye.
\( \displaystyle \ce{C6H5NH2 + NaNO2 + 2HCl} \xrightarrow{\text{273--278 K}} \ce{C6H5N2^+Cl^- + NaCl + 2H2O} \)
(D) Tollen's Test — This matches (IV) Reducing sugars. Tollens’ reagent (ammoniacal silver nitrate) is reduced to a silver mirror by aldehydes and by reducing sugars such as glucose, which are oxidised.
\( \displaystyle \ce{RCHO + 2[Ag(NH3)2]^+ + 3OH^- -> RCOO^- + 2Ag v + 2H2O + 4NH3} \)
Shortcut: (A)-(I) and (B)-(II) appear in Options (1) and (3); (D)-(IV) then leaves only Option (1).