 Solution

The bromines ortho and para to OH are removed, leaving the meta bromine at C3. A suitable polar mechanism begins with electrophilic ipso protonation of a brominated carbon. protonation at C4 or C6 gives a Wheland intermediate stabilized by donation of the phenolic oxygen lone pair; an important contributor has an O-protonated cyclohexadienone-like framework. Corresponding protonation at C3 lacks this direct stabilization by resonance in chemistry from OH.

iodide can capture/reduce the departing electrophilic bromine as the C–Br bond electrons restore the aromatic pi system. Thus the incoming hydrogen remains on carbon while bromine is removed, unlike ordinary electrophilic substitution in which the incoming electrophile remains and a proton leaves. bromine trapping can be represented through IBr formation followed by reaction with further iodide to give iodine and bromide. For each replacement the balanced net reduction isApplying it at C4 and C6 gives 3-bromo-2,5-dimethylphenol, with formula ${\mathrm{C}}_8{\mathrm{H}}_9\mathrm{BrO}$. This explains why the position least activated by OH is the one at which bromine survives. It is a proposed acid-assisted protodebromination mechanism consistent with the selectivity; the preparative sequence alone does not prove every proton-transfer or halogen-transfer intermediate. It should not be drawn as iodide simply displacing bromide from an unactivated aryl carbon.