= Solution
\b[P is H2], the acetal methine between two oxygens; its only resolved coupling is to the attached methyl, giving the quartet. \b[S is that C2 methyl], whose doublet has the same 6.9 Hz coupling. \b[T is tert-butyl], integrating to nine equivalent <hydrogens> with no adjacent <carbon>-bound methine <proton> across its quaternary attachment.
\b[U is equatorial H5], coupled to the four <hydrogens> at C4 and C6. The two equivalent equatorial methylene <hydrogens> are \b[Q], each a doublet of doublets with a 12.3 Hz <geminal coupling> to its axial partner and a 1.3 Hz coupling to H5. The two equivalent axial methylene <hydrogens> are \b[R], each with the same 12.3 Hz <geminal coupling> and a 3.7 Hz <vicinal coupling> to H5. Equivalence between C4 and C6 makes two pairs rather than four separately shifted methylene <hydrogens>.
$$
\boxed{\begin{aligned}
J_{P,S}&=6.9\ \mathrm{Hz},\\
|J_{Q_4,R_4}|=|J_{Q_6,R_6}|&=12.3\ \mathrm{Hz},\\
J_{U,R_4}=J_{U,R_6}&=3.7\ \mathrm{Hz},\\
J_{U,Q_4}=J_{U,Q_6}&=1.3\ \mathrm{Hz}.
\end{aligned}}
$$
Coupling U to two equivalent R <hydrogens> and two equivalent Q <hydrogens> produces its triplet of triplets. The different small <vicinal couplings> are consistent with oblique H–C–C–H torsions for equatorial H5; the 12.3 Hz value is geminal and must not be mistaken for a large vicinal trans-diaxial coupling.
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