Benzene ring in phenol is expected to give the characteristic substitution or replacement reactions, i.e., halogenation, nitration, sulphonation, alkylation, etc. In these reactions, the attacking reagent is electrophile. The phenolic group activates the ortho and para positions in the ring due to conjugation. Disubstitution reactions are quite fast as the electron density in the ring has increased considerably. Phenols do not respond to nucleophilic substitution reactions as it is influenced by phenolic group.
Victor Meyer's test gives the following observations. Match List I with List II:
| List I | List II |
| A. Blood red colouration | I. Tertiary alcohol |
| B. Blue colouration | II. Phenols |
| C. Colourless solution | III. Primary alcohols |
| D. Violet colouration with \(FeCl_3\) | IV. Secondary alcohol |
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-IV, C-I, D-II
The Victor Meyer test tells primary, secondary and tertiary alcohols apart. The alcohol is turned into the alkyl iodide (red P and \( \ce{I2} \)), then into the nitroalkane with \( \ce{AgNO2} \). The nitroalkane is treated with nitrous acid and then made alkaline; the colour depends on how many α-hydrogens it has.
\( \ce{R-OH} \xrightarrow{\ce{P/I2}} \ce{R-I} \xrightarrow{\ce{AgNO2}} \ce{R-NO2} \)
(A) Blood red colouration — This matches (III) Primary alcohols. A primary nitroalkane has two α-hydrogens and forms a nitrolic acid, whose sodium salt is blood red.
\( \ce{RCH2NO2 + HNO2 -> RC(=NOH)NO2 + H2O} \)
(B) Blue colouration — This matches (IV) Secondary alcohol. A secondary nitroalkane has one α-hydrogen and forms a blue pseudonitrole.
\( \ce{R2CHNO2 + HNO2 -> R2C(NO)NO2 + H2O} \)
(C) Colourless solution — This matches (I) Tertiary alcohol. A tertiary nitroalkane has no α-hydrogen, so it does not react with nitrous acid and the solution stays colourless.
(D) Violet colouration with FeCl3 — This matches (II) Phenols. Phenols give a violet colour with neutral ferric chloride solution; this is the test for the phenolic –OH group.
Note: The question is from beyond the NCERT textbook.