Question

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

• A. o-Nitrophenol
• B. p-Nitrophenol
• C. mixture of o-Nitrophenol and p-Nitrophenol
• D. 2, 4, 6-Trinitrophenol

Hint:

Remember this crucial reagent distinction for phenol nitration: 1. $\text{Dilute } \mathrm{HNO_3} \rightarrow \text{Mono-substitution (ortho + para mixture)}.$ 2. $\text{Concentrated } \mathrm{HNO_3} + \text{conc. } \mathrm{H_2SO_4} \rightarrow \text{Tri-substitution (Picric acid)}.$

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The reaction describes the nitration of phenol using a nitrating mixture consisting of concentrated nitric acid ($\mathrm{HNO_3}$) and concentrated sulfuric acid ($\mathrm{H_2SO_4}$). We need to identify the major organic product (A) formed under these highly concentrated, rigorous conditions.

Step 2: Key Formula or Approach:
The hydroxyl group ($\mathrm{-OH}$) attached to the benzene ring in phenol is a powerful activating group and an ortho/para-director due to the strong resonance donation of electron density ($(+)\mathrm{R}$ effect). The choice of nitrating reagents determines the degree of substitution:

• Dilute $\mathrm{HNO_3}$ at low temperature yields a mixture of ortho and para-mononitrophenols.

• Concentrated $\mathrm{HNO_3}$ in the presence of concentrated $\mathrm{H_2SO_4}$ causes multi-substitution, introducing nitro groups at all available ortho and para positions simultaneously.

Step 3: Detailed Explanation:
When phenol is treated with a potent nitrating mixture of concentrated $\mathrm{HNO_3}$ and concentrated $\mathrm{H_2SO_4}$, electrophilic aromatic substitution occurs rapidly at multiple positions.
Because the ring is highly activated by the $\mathrm{-OH}$ group, electrophilic nitronium ions ($\mathrm{NO_2^+}$) attack both ortho positions (positions 2 and 6) and the para position (position 4) of the ring.
This complete substitution pathway replaces three hydrogen atoms, yielding 2,4,6-trinitrophenol, commonly known as picric acid. $$\mathrm{C_6H_5OH + 3HNO_3 \xrightarrow{conc. H_2SO_4} C_6H_2(NO_2)_3OH + 3H_2O}$$

Step 4: Final Answer:
The product obtained is 2, 4, 6-Trinitrophenol, which corresponds to option (D).