MHTCET 11 May 2024 Shift 1 Paper

Concept:Phenol undergoes oxidation with $\mathrm{CrO_3}$ to form $p$-benzoquinone.Explanation:Phenol is an aromatic alcohol in which $\mathrm{-OH}$ is attached directly to a benzene ring.$\mathrm{CrO_3}$ is a strong oxidising agent, especially in acidic medium.It attacks the electron-rich ring of phenol rather than the $\mathrm{-OH}$ group alone.The $\mathrm{-OH}$ group activates the ring, favouring oxidation at the para position.This leads to the formation of two carbonyl groups at positions $1$ and $4$ of the six-membered ring.The product is therefore $p$-benzoquinone, also called $1,4$-benzoquinone, with molecular formula $\mathrm{C_6H_4O_2}$.The reaction can be written as:$$\mathrm{C_6H_5OH \xrightarrow{\mathrm{CrO_3}} p\text{-benzoquinone}}$$Among the given structures, select the one that shows a quinone ring with two $=O$ groups at opposite para positions.Answer:The product is $p$-benzoquinone ($1,4$-benzoquinone).