Match List I with List II.
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LIST I |
LIST II |
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A. Hoffman bromamide reaction |
I. Detection Test for primary amines |
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B. Carbylamine reaction |
II. Hinsberg reagent |
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C. Gabriel phthalimide synthesis |
III. Amine with lesser number of carbon atoms |
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D. Benzene sulphonyl chloride |
IV. Reaction of phthalimide with KOH and R-X |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-I, C-IV, D-II
The four items are reactions of amines and their preparation. Two of them make primary amines (Hofmann bromamide degradation and Gabriel synthesis), and two are tests that tell amines apart (carbylamine reaction and the Hinsberg test).
A. Hoffman bromamide reaction — This matches III. Amine with lesser number of carbon atoms. An amide with bromine and NaOH gives a primary amine with one carbon atom less; the carbonyl carbon leaves as carbonate.
\( \ce{RCONH2 + Br2 + 4NaOH -> RNH2 + Na2CO3 + 2NaBr + 2H2O} \)
B. Carbylamine reaction — This matches I. Detection Test for primary amines. Only primary amines, aliphatic or aromatic, give foul-smelling isocyanides with chloroform and alcoholic KOH.
\( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{R-NC + 3KCl + 3H2O} \)
C. Gabriel phthalimide synthesis — This matches IV. Reaction of phthalimide with KOH and R-X. Phthalimide with ethanolic KOH gives potassium phthalimide; this is heated with an alkyl halide, and alkaline hydrolysis releases the primary amine.
\( \displaystyle \ce{C6H4(CO)2NH} \xrightarrow{\ce{KOH}\text{ (alc.)}} \ce{C6H4(CO)2N^-K^+} \xrightarrow{\ce{R-X}} \ce{C6H4(CO)2N-R} \xrightarrow{\ce{NaOH}\text{ (aq)}} \ce{R-NH2} \)
D. Benzene sulphonyl chloride — This matches II. Hinsberg reagent. \( \ce{C6H5SO2Cl} \) reacts with primary and secondary amines to form sulphonamides and is used to distinguish primary, secondary and tertiary amines.
\( \ce{C6H5SO2Cl + H-NH-R -> C6H5SO2NHR + HCl} \)
Shortcut: D-II (benzenesulphonyl chloride is Hinsberg’s reagent) appears only in Option (1).