Question

Identify the product A obtained in the following reaction.
Phenol + Conc. Nitric acid $\xrightarrow{\text{Conc. } H_2SO_4}$ A

• A. o-Nitrophenol
• B. p-Nitrophenol
• C. 2,4,6-Trinitrophenol
• D. Mixture of ortho and para-nitrophenol

Hint:

Logic Tip: The common name for 2,4,6-trinitrophenol is \textbf{Picric Acid}. It is a strong acid despite not having a carboxylic acid group, due to the intense electron-withdrawing effect of the three nitro groups. Remember: Dilute $HNO_3$ yields a mixture of ortho/para mononitrophenols; Concentrated $HNO_3$ yields picric acid.

Answer 1

The Correct Option is C

Concept:
The hydroxyl group ($-OH$) on a phenol ring is a highly activating, ortho/para-directing group. When treated with a strongly acidic nitrating mixture (concentrated $HNO_3$ and concentrated $H_2SO_4$), the benzene ring undergoes exhaustive electrophilic aromatic substitution at all available ortho and para positions simultaneously.
Step 1: Identify the reagents and reaction type.
The reaction of phenol with a mixture of concentrated nitric acid ($HNO_3$) and concentrated sulfuric acid ($H_2SO_4$) is a nitration reaction. The sulfuric acid acts as a catalyst to generate the highly reactive nitronium ion electrophile ($NO_2^+$).
Step 2: Determine the directing effect of the -OH group.
The $-OH$ group donates electron density into the benzene ring via resonance (+R effect), making the ring highly reactive towards electrophiles. It specifically increases electron density at the \textit{ortho} (positions 2 and 6) and \textit{para} (position 4) positions.
Step 3: Predict the final product.
Because the conditions are severe (concentrated acids) and the ring is highly activated, nitration does not stop at mono-substitution. It continues until all ortho and para positions are substituted with nitro groups. $$C_6H_5OH + 3HNO_3 \xrightarrow{H_2SO_4} C_6H_2(NO_2)_3OH + 3H_2O$$ The final product has nitro groups at carbon positions 2, 4, and 6. Its chemical name is 2,4,6-trinitrophenol.