Compound I is heated with Conc. HI to give a hydroxy compound A which is further heated with Zn dust to give compound B. Identify A and B.

$\underset{∆}{\overset{\mathrm{Conc}.\mathrm{HI}}{\to }}\mathrm{A}\stackrel{\mathrm{Zn}, ∆}{\to }\mathrm{B}$

Reaction Analysis

1. Reaction with Concentrated HI (Hydroiodic Acid):

   The compound is a phenyl ether derivative of isopropyl alcohol. Upon treatment with concentrated HI and heating, the ether undergoes cleavage. This forms two products:

   • Phenol (C₆H₅OH) from the phenyl group.
   • Isopropyl iodide (C₃H₇I) from the isopropyl group.

2. Reduction of Phenol:

   The phenol (C₆H₅OH) formed in the first step undergoes reduction with zinc (Zn) in the presence of heat. This process removes the hydroxyl (-OH) group, converting phenol into benzene (C₆H₆).

Final Answer:

The overall reaction leads to the formation of benzene (C₆H₆) as the final product.