= Solution
\b[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 (chemistry)> is
$$
\mathrm{ArBr+2HI\longrightarrow ArH+HBr+I_2}.
$$
Applying it at C4 and C6 gives \b[3-bromo-2,5-dimethylphenol], with formula $\mathrm{C_8H_9BrO}$. This explains why the position least activated by OH is the one at which <bromine> survives. It is a proposed <acid (chemistry)>-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>.
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