Match List-I with List-II
| Organic Compounds | Positive Tests |
| (A) Formaldehyde | (I) Bromine water test |
| (B) Phenol | (II) Tollens’ test |
| (C) 3° Alcohol | (III) Iodoform test |
| (D) Acetophenone | (IV) Lucas test |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A) - (II), (B) - (I), (C) - (IV), (D) - (III)
Each of these tests picks out one functional group or structural feature, so the list is matched by first identifying what kind of compound each item is:
(A) Formaldehyde — This matches (II) Tollens’ test. Formaldehyde (HCHO) is an aldehyde, and aldehydes are easily oxidised. Tollens’ reagent (ammoniacal silver nitrate) oxidises it, and metallic silver is deposited as a bright silver mirror.
\( \displaystyle \ce{HCHO + 2[Ag(NH3)2]^+ + 3OH^- -> HCOO^- + 2Ag + 2H2O + 4NH3} \)
(B) Phenol — This matches (I) Bromine water test. The –OH group strongly activates the benzene ring, so phenol decolourises bromine water at once and gives a white precipitate of 2,4,6-tribromophenol.
\( \displaystyle \ce{C6H5OH + 3Br2 -> C6H2Br3OH v + 3HBr} \)
(C) 3° Alcohol — This matches (IV) Lucas test. Lucas reagent (conc. HCl + anhydrous \( \ce{ZnCl2} \)) converts an alcohol into an alkyl chloride, which is insoluble and makes the solution turbid. A tertiary alcohol forms the most stable carbocation, so turbidity appears immediately; secondary alcohols take about five minutes and primary alcohols do not react at room temperature.
\( \displaystyle \ce{(CH3)3C-OH + HCl} \xrightarrow{\text{anhyd. }\ce{ZnCl2}} \ce{(CH3)3C-Cl + H2O} \)
(D) Acetophenone — This matches (III) Iodoform test. Acetophenone (\( \ce{C6H5COCH3} \)) is a methyl ketone. The \( \ce{CH3CO} \)– group reacts with iodine and sodium hydroxide to give a yellow precipitate of iodoform.
\( \displaystyle \ce{C6H5COCH3 + 3I2 + 4NaOH -> C6H5COONa + CHI3 v + 3NaI + 3H2O} \)
Shortcut: The pair (A)–(II) appears only in Option (2).