The question involves using the reagent \(\text{MnO}_2\)to oxidize a compound, resulting in compound A. Let's solve and understand the reaction step-by-step:
In the given compound, we have a benzene ring with a hydroxyl group \((\text{-OH})\)directly attached to it as a primary alcohol.
\(\text{MnO}_2\)is a selective oxidizing agent that primarily oxidizes allylic and benzylic alcohols to the corresponding carbonyl compounds.
The primary alcohol on the benzene ring, being a benzylic alcohol, will be oxidized by \(\text{MnO}_2\)to form the corresponding aldehyde.
Thus, the product formed, compound A, will have a carbonyl group \((\text{-CHO})\)instead of the hydroxyl group \((\text{-OH})\)on the benzene ring, leading to the formation of benzaldehyde.
Upon examining the options, the correct structure of compound A corresponds to option (b), which is benzaldehyde:
(b) Benzaldehyde, which has a formyl group \((\text{-CHO})\)attached to the benzene ring.
