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.

Solution

Chemistry ma’am · 4 steps

HI cuts the ether on the alkyl side, never the ring side, so phenol stays and Zn dust strips its OH.

1. Protonate the oxygen. Conc. HI first protonates the ether oxygen of isopropyl phenyl ether. This turns the oxygen into a good leaving group.
2. Iodide picks a side. I⁻ can only attack the isopropyl carbon, which is sp³. The ring carbon is sp², and its C–O bond has partial double bond character from resonance, so it does not break. The products are phenol and isopropyl iodide, and phenol is the hydroxy compound A.

   C₆H₅-O-CH(CH₃)₂ + HI →(Δ) C₆H₅OH + (CH₃)₂CHI

3. Zinc dust. Heating phenol with Zn dust removes the oxygen as ZnO: C₆H₅OH + Zn → C₆H₆ + ZnO. So B is benzene.
4. So. A is phenol and B is benzene, which is option (d).

The idea to keep. In an alkyl aryl ether, HI always gives phenol plus alkyl iodide, because the aryl C–O bond will not break. Zn dust with heat reduces a phenol to the parent arene.

Answer from the NTA final answer key. Solution by PrepWiser.