Match List - I with List - II.
| List - I | List - II |
| (A) Ethanol and Diethyl ether | (I) Lucas reagent |
| (B) Methanol and Ethanol | (II) Sodium metal |
| (C) n-Butyl alcohol and tert. Butyl alcohol | (III) Iodoform test |
| (D) Phenol and Cyclohexanol | (IV) Ferric Chloride |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Each pair is told apart by a test that only one member gives: an O–H group reacts with sodium, a \( \ce{CH3CH(OH)} \)– group gives the iodoform test, tertiary alcohols react at once with Lucas reagent, and phenols give a violet colour with neutral ferric chloride.
(A) Ethanol and Diethyl ether — This matches (II) Sodium metal. Ethanol has an O–H group and gives hydrogen with sodium; the ether has no O–H and does not react.
\( \ce{2C2H5OH + 2Na -> 2C2H5ONa + H2 ^} \)
(B) Methanol and Ethanol — This matches (III) Iodoform test. Ethanol has the \( \ce{CH3CH(OH)} \)– group and gives a yellow precipitate of iodoform; methanol does not.
\( \ce{C2H5OH + 4I2 + 6NaOH -> CHI3 v + HCOONa + 5NaI + 5H2O} \)
(C) n-Butyl alcohol and tert. Butyl alcohol — This matches (I) Lucas reagent. The tertiary alcohol gives turbidity at once; the primary alcohol gives none at room temperature.
\( \ce{(CH3)3COH + HCl} \xrightarrow{\ce{ZnCl2}} \ce{(CH3)3CCl + H2O} \)
(D) Phenol and Cyclohexanol — This matches (IV) Ferric Chloride. Phenol gives a violet colour with neutral ferric chloride solution; cyclohexanol, an alcohol, does not.