The compound shown is brominated with $Br X_{2}$ and $FeBr X_{3}$ (the methoxy group is the stronger director), and the product is then heated with excess concentrated $HI$ . Which structure is the final aromatic product?

Answer: A

The compound is 4-methylanisole. Bromination goes ortho to the methoxy group (the stronger activator), giving 2-bromo-4-methylanisole. Hot

HI

then cleaves the aryl alkyl ether at the methyl carbon, giving 2-bromo-4-methylphenol and

CH X_{3} I

. The aryl-oxygen bond and the aryl bromide are untouched.