Match List-I with List-II: Match the test used for distinguishing the organic compounds
| List-I | List-II |
| (A) \((CH_3)_3N, CH_3CH_2CH_2NH_2\) | (I) Tollen's test |
| (B) \(HCOOH, CH_3COOH\) | (II) Lucas test |
 |
(III) Carbylamine test |
 |
(IV) Iodoform test |
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(A-III, B-I, C-IV, D-II\)
A distinguishing test works when one compound of the pair has the group the test needs and the other does not: the carbylamine test needs a primary amine, Tollens’ reagent needs an oxidisable H–C=O group, the iodoform test needs a CH3CO– group, and the Lucas test separates alcohols by how fast they form a carbocation (3° fastest).
(A) \( \ce{(CH3)3N} \), \( \ce{CH3CH2CH2NH2} \) — This matches (III) Carbylamine test. Propan-1-amine (1°) gives a foul-smelling isocyanide with chloroform and alcoholic KOH; trimethylamine (3°) does not.
\( \ce{CH3CH2CH2NH2 + CHCl3 + 3KOH} \xrightarrow{\Delta} \ce{CH3CH2CH2NC + 3KCl + 3H2O} \)
(B) HCOOH, \( \ce{CH3COOH} \) — This matches (I) Tollen’s test. Methanoic acid has an H on the carbonyl carbon, so it reduces Tollens’ reagent to a silver mirror; ethanoic acid does not.
\( \ce{HCOOH + 2[Ag(NH3)2]^+ + 2OH^- -> 2Ag v + CO2 + 2H2O + 4NH3} \)
(C) \( \ce{CH3COCH2CH2CH3} \), \( \ce{CH3CH2COCH2CH3} \) — This matches (IV) Iodoform test. Pentan-2-one is a methyl ketone and gives yellow \( \ce{CHI3} \) with NaOI; pentan-3-one has no \( \ce{CH3CO} \)– group.
\( \ce{CH3COC3H7 + 3NaOI -> CHI3 v + C3H7COONa + 2NaOH} \)
(D) \( \ce{CH3CH2CH2CH2OH} \), \( \ce{(CH3)3COH} \) — This matches (II) Lucas test. With conc. HCl and \( \ce{ZnCl2} \), 2-methylpropan-2-ol (3°) gives turbidity at once; butan-1-ol (1°) gives none at room temperature.
\( \ce{(CH3)3COH + HCl} \xrightarrow{\ce{ZnCl2}} \ce{(CH3)3CCl + H2O} \)
Shortcut: the amine pair can only be told apart by the carbylamine test (A-III) and the 1°/3° alcohols by the Lucas test (D-II); only Option (3) has both.