Answer the question on the basis of passage given below:
Strong oxidizing agents, oxidize toluene and its derivatives to benzoic acid. However, it is possible to stop the oxidation at the aldehyde stage with suitable reagents that convert methyl group to an intermediate that is difficult to oxidize further.
Match List I with List II
| List I | List II |
| (A) Benzyl alcohol and phenol | (I) Neutral \(FeCl_3\) |
| (B) Phenol and alcohol | (II) Lucas reagent |
| (C) Secondary butyl alcohol and isobutyl alcohol | (III) Blue litmus |
| (D) Propan-1-ol and propan-2-ol | (IV) Iodoform test |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-III, B-I, C-IV, D-II
(A) Benzyl alcohol and phenol — This matches (III) Blue litmus. Phenol is acidic and turns blue litmus red; benzyl alcohol is neutral and does not.
(B) Phenol and alcohol — This matches (I) Neutral FeCl3. Phenol gives a violet colour with neutral ferric chloride; an alcohol gives none.
(C) Secondary butyl alcohol and isobutyl alcohol — This matches (IV) Iodoform test. Butan-2-ol has the CH3CH(OH)– group and gives yellow iodoform; isobutyl alcohol, \( \ce{(CH3)2CHCH2OH} \), does not:
\( \ce{CH3CH(OH)CH2CH3 + 4I2 + 6NaOH -> CHI3 + CH3CH2COONa + 5NaI + 5H2O} \)
(D) Propan-1-ol and propan-2-ol — This matches (II) Lucas reagent. The secondary alcohol turns turbid within about five minutes; the primary propan-1-ol shows no turbidity at room temperature:
\( \ce{(CH3)2CHOH + HCl} \xrightarrow{\ce{ZnCl2}} \ce{(CH3)2CHCl + H2O} \)