An oxidation of phenol with chromic acid produces

The correct answer is Option (4) → Benzoquinone

Phenol is easily oxidised because its ring is made electron-rich by the –OH group. With a strong oxidising agent such as chromic acid (sodium dichromate in sulphuric acid), the –OH carbon and the para carbon are both oxidised to C=O groups, and the product is a conjugated diketone, p-benzoquinone (cyclohexa-2,5-diene-1,4-dione):

    \( \displaystyle \ce{C6H5OH} \xrightarrow[\ce{H2SO4}]{\ce{Na2Cr2O7}} \ce{O=C6H4=O}\ (p\text{-benzoquinone}) \)

Even air slowly oxidises phenols to dark coloured mixtures containing quinones, which is why old samples of phenol turn pink or brown.

Why the other options are incorrect

• Option (1) Benzoic acid — Incorrect. It needs a carbon side chain on the ring; it is obtained by oxidising toluene, not phenol:

      \( \displaystyle \ce{C6H5CH3} \xrightarrow[\text{(ii) }\ce{H3O+}]{\text{(i) alk. }\ce{KMnO4}} \ce{C6H5COOH} \)

• Option (2) Salicylic acid — Incorrect. It is formed from phenol by Kolbe’s reaction (sodium phenoxide and carbon dioxide), which is a carboxylation, not an oxidation:

      \( \displaystyle \ce{C6H5ONa} \xrightarrow[\text{(ii) }\ce{H+}]{\text{(i) }\ce{CO2}} o\text{-}\ce{HOC6H4COOH} \)

• Option (3) 2,4,6-Tribromophenol — Incorrect. It is the white precipitate formed when phenol is treated with bromine water (substitution, not oxidation):

      \( \displaystyle \ce{C6H5OH + 3Br2 -> C6H2Br3OH v + 3HBr} \)

NCERT (2026-27 edition):
Oxidation

Note: The same/similar question has been asked in different years by NTA as per the details given below: