= Solution
Number the phenolic <carbon> C1, with methyl groups at C2 and C5. \b[All three remaining aromatic <hydrogens> are replaced: $H$ is 3,4,6-tribromo-2,5-dimethylphenol.] The starting <phenol> is $\mathrm{C_8H_{10}O}$; replacing three <hydrogens> gives $\mathrm{C_8H_7Br_3O}$, agreeing with the intermediate's composition. Bromination of methyl side chains would not follow the stated <Lewis acid>-promoted aromatic pathway.
<aluminium tribromide> polarizes Br–Br. The <aromatic ring> attacks the <electrophilic> <bromine> end, forming a <Wheland intermediate>; removal of the <proton> on the attacked <carbon> restores <aromaticity>. Repetition fills C4 and C6, the available para and ortho positions activated by OH, and C3, which is activated by the methyl <substituents> when the other positions are occupied. The formula fixes the extent of substitution without requiring an experimentally unreported order for the individual brominations. The preparative perbromination/selective-debromination sequence is documented in the primary synthetic work.<Total Synthesis of (Nor)illudalane Sesquiterpenes Based on a C(sp3)-H Activation Strategy by Romain Melot and others>[2019, Scheme 6, upper route from 2,5-xylenol]{ref}
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