= Solution
\b[The product is 1,4-dimethoxy-2,5-bis(2-methylbutan-2-yl)<benzene>.] Each new substituent is $\mathrm{-C(CH_3)_2CH_2CH_3}$, a <tert-amyl group>, rather than the unrearranged secondary alkyl group of the <alcohol>. The two substituents are related by the <symmetry> of the ring.
The <proton> assignments are:
|| <chemical shift> / ppm
|| Assignment
|| Reason
| $6.75$
| \b[Two aromatic H, at positions 3 and 6]
| <symmetry> makes them equivalent; neither has an adjacent aromatic H to give a resolved ordinary ortho splitting.
| $3.79$
| \b[Two <methoxy groups>, six H]
| Equivalent $\mathrm{OCH_3}$ groups give a <NMR singlet>.
| $1.81$
| \b[Two ethyl $\mathrm{CH_2}$ groups, four H]
| Each couples to a terminal three-proton <methyl group>, giving a <NMR quartet>.
| $1.33$
| \b[Four <methyl groups> attached to the two quaternary benzylic <carbons>, twelve H]
| These <methyl groups> have no neighboring <methine group> <proton> and give a <NMR singlet>.
| $0.65$
| \b[Two terminal ethyl <methyl groups>, six H]
| Each couples to two <methylene group> <protons>, giving a <NMR triplet> with the same coupling constant as the <NMR quartet>.
The integrals add to 30 H. The unusually shielded terminal ethyl methyl signal is compatible with an aryl-containing environment; its <NMR quartet>–<NMR triplet> partner and integral are the decisive assignment evidence. The <molecular formula> and <degree of unsaturation>, $\mathrm{DBE}=4$, fit a <benzene> ring with saturated side chains.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2021/paper-1-b1-products.png]
{title=Original structures of the rearranged aromatic alkylation product and dialkylated beta-ketoester}
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