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.

The starting compound is:

$$\mathrm{Isopropyl\ Phenyl\ Ether}$$

It reacts with concentrated:

$$\mathrm{HI}$$

under heat.

The ether oxygen first gets protonated.

$$\mathrm{R-O-R' + H^+ \longrightarrow R-O^+H-R'}$$

Cleavage of the ether bond then occurs by attack of:

$$\mathrm{I^-}$$

The:

$$\mathrm{C_{aryl}-O}$$

bond is highly stable due to resonance between oxygen and the benzene ring.

Hence, cleavage occurs only at the alkyl-oxygen bond.

$$\mathrm{C_6H_5OCH(CH_3)_2 + HI \longrightarrow C_6H_5OH + (CH_3)_2CHI}$$

Thus, compound $$\mathrm{A}$$ is:

$$\mathrm{Phenol}$$

Phenol is then heated with zinc dust:

$$\mathrm{Zn/\Delta}$$

Zinc removes oxygen from phenol as:

$$\mathrm{ZnO}$$

$$\mathrm{C_6H_5OH \xrightarrow{Zn/\Delta} C_6H_6}$$

Thus, compound $$\mathrm{B}$$ is:

$$\mathrm{Benzene}$$

Thus, the right option is D.