Reaction of Phenol with Na2Cr2O7 and H2SO4

Phenol, a benzene ring with a hydroxyl group, reacts with sodium dichromate (Na2Cr2O7) and sulfuric acid (H2SO4) to form a red-brown precipitate of chromium(III) oxide (Cr2O3) and water.

Reaction Equation:
C6H5OH + Na2Cr2O7 + H2SO4 → Cr2O3 + C6H5OH + H2O + Na2SO4

Explanation:

Step 1: Acid-Catalyzed Oxidation
Sulfuric acid acts as a catalyst in the reaction, protonating the hydroxyl group of phenol, making it more reactive. The chromate ion (Cr2O7^2-) is reduced to chromium(III) oxide (Cr2O3) while phenol is oxidized to benzoquinone.

Step 2: Formation of Chromium (III) Oxide
Chromium(III) oxide is produced as a red-brown precipitate in the reaction. This compound is insoluble in water and can be easily separated from the reaction mixture.

Step 3: Formation of Water and Sodium Sulfate
Water is formed as a byproduct of the reaction between phenol and the oxidizing agent. Additionally, sodium sulfate is produced as a soluble salt in the reaction.

Conclusion:
The reaction of phenol with sodium dichromate and sulfuric acid results in the oxidation of phenol to benzoquinone and the formation of chromium(III) oxide as a red-brown precipitate. This reaction is a common method used in organic chemistry to oxidize phenolic compounds.