Match List I with List II
| List I | List II |
(A)  |
(I)  |
(B)  |
(II)  |
(C)  |
(III)  |
(D)  |
(IV)  |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-I, C-III, D-II
The –OH group strongly activates the benzene ring and directs the incoming group to the ortho and para positions. How many groups enter depends on the conditions: mild conditions give mono-substitution, while bromine water or concentrated nitric acid substitute all three o/p positions.
(A) Phenol + \( \ce{Br2} \), \( \ce{H2O} \) — This matches (IV) 2,4,6-tribromophenol. In water (a polar solvent) bromine is strongly polarised and every ortho and para position is brominated; a white precipitate forms.
\( \ce{C6H5OH + 3Br2} \xrightarrow{\ce{H2O}} \ce{C6H2Br3OH v + 3HBr} \)
(B) Phenol + conc. \( \ce{HNO3} \), \( \ce{H2SO4} \) — This matches (I) 2,4,6-trinitrophenol (picric acid). Concentrated nitric acid nitrates all three ortho and para positions.
\( \ce{C6H5OH + 3HNO3} \xrightarrow{\text{conc. }\ce{H2SO4}} \ce{C6H2(NO2)3OH + 3H2O} \)
(C) Phenol + \( \ce{Br2} \), \( \ce{CS2} \) — This matches (III) o- and p-bromophenol. In a solvent of low polarity (\( \ce{CS2} \), \( \ce{CHCl3} \)) at low temperature, only monobromophenols form, mainly the para isomer.
\( \ce{C6H5OH + Br2} \xrightarrow{\ce{CS2}, \ 273\,\mathrm{K}} \ce{BrC6H4OH + HBr} \)
(D) Phenol + dil. \( \ce{HNO3} \) — This matches (II) o- and p-nitrophenol. Dilute nitric acid at low temperature (298 K) gives a mixture of ortho- and para-nitrophenols, separated by steam distillation.
\( \ce{C6H5OH + HNO3} \xrightarrow{\text{dil., }298\,\mathrm{K}} \ce{O2NC6H4OH + H2O} \)
Shortcut: A-IV and B-I appear together in Options (1) and (2); D-II then leaves only Option (2).
NCERT (2026-27 edition):
Electrophilic Aromatic Substitution – “… (i) Nitration: With dilute nitric acid at low temperature (298 K), phenol yields a mixture of ortho and para nitrophenols. … With concentrated nitric acid, phenol is converted to 2,4,6-trinitrophenol. The product is commonly known as picric acid.”
Electrophilic Aromatic Substitution – “… (ii) Halogenation: On treating phenol with bromine, different reaction products are formed under different experimental conditions. (a) When the reaction is carried out in solvents of low polarity such as CHCl3 or CS2 and at low temperature, monobromophenols are formed. … (b) When phenol is treated with bromine water, 2,4,6-tribromophenol is formed as white precipitate.”