Read the passage carefully and answer the questions.
Alcohols and phenols are formed when a hydrogen atom in a hydrocarbon, aliphatic and aromatic respectively is replaced by - OH group. These are further classified on the basis of number of hydroxy groups and according to the hybridization of the carbon atom \( sp^2 \) or \( sp^3 \) to which the -OH group is attached. They find applications in industry as well as in day-to-day life. One of the most commercially obtained alcohol also known as wood spirit is obtained by destructive distillation of wood. Both alcohols and phenols are acidic in nature. The acidity of alcohols and phenols vary according to the nature of substituent(s) attached to them. Alcohols and phenols react with carboxylic acids, acid chlorides and acid anhydrides to form esters. Introduction of acetyl group in alcohols and phenols is called acetylation. Phenol undergo electrophilic aromatic substitution reaction because the -OH activates the benzene ring. It undergoes nitration, halogenation and sulphonation reactions.
When phenol is treated with concentrated sulphuric acid, then it gets converted into phenol-2,4-disulphonic acid (X). On further treatment of X with conc. nitric acid, product Y is formed. Identify Y.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Picric acid
The –OH group strongly activates the benzene ring towards electrophilic substitution and directs the incoming group to the ortho and para positions, that is, positions 2, 4 and 6. Concentrated nitric acid can nitrate all three, but it also oxidises the very reactive phenol, so direct nitration gives a poor yield. Phenol is therefore sulphonated first.
Concentrated sulphuric acid introduces two –\( \ce{SO3H} \) groups at positions 2 and 4, giving phenol-2,4-disulphonic acid (X):
\( \displaystyle \ce{C6H5OH + 2H2SO4} \xrightarrow{\text{conc.}} \ce{C6H3(OH)(SO3H)2 + 2H2O} \)
The two electron withdrawing –\( \ce{SO3H} \) groups protect the ring from oxidation. Concentrated nitric acid then replaces both –SO3H groups by –NO2 groups and nitrates the free position 6, giving 2,4,6-trinitrophenol (Y):
\( \displaystyle \ce{C6H3(OH)(SO3H)2 + 3HNO3} \xrightarrow{\text{conc.}} \ce{C6H2(OH)(NO2)3 + 2H2SO4 + H2O} \)
So Y is picric acid. With three electron withdrawing nitro groups at the ortho and para positions, it is a strong acid.
Why the other options are incorrect
Option (1) Salicylic acid — Incorrect. It is made from phenol by Kolbe’s reaction, in which sodium phenoxide takes up carbon dioxide; no nitric acid is involved.
\( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2}} \ce{HOC6H4COOH} \)
Option (3) o-Nitrophenol — Incorrect. It is formed, together with p-nitrophenol, when phenol is nitrated with dilute nitric acid at 298 K; concentrated acid nitrates all three activated positions.
\( \displaystyle \ce{C6H5OH + HNO3} \xrightarrow[\text{298 K}]{\text{dil.}} o\text{-}\ce{O2NC6H4OH} + p\text{-}\ce{O2NC6H4OH} + \ce{H2O} \)
Option (4) 2,4-dinitrophenol — Incorrect. Position 6 is also activated by the –OH group, so with concentrated nitric acid the nitration does not stop at two nitro groups.
NCERT (2026-27 edition):
Electrophilic Aromatic Substitution – “… With concentrated nitric acid, phenol is converted to 2,4,6-trinitrophenol. The product is commonly known as picric acid. The yield of the reaction product is poor. Nowadays picric acid is prepared by treating phenol first with concentrated sulphuric acid which converts it to phenol-2,4-disulphonic acid, and then with concentrated nitric acid to get 2,4,6-trinitrophenol.”