In the \(^1\mathrm{H}\) NMR spectrum of compound \(\mathbf{X}\), the coupling constant (\(J_{ab}\)) between \(H_a\) and \(H_b\) is 3 Hz. The amino alcohol(s) that give(s) \(\mathbf{X}\) on reaction with methyl iodide followed by heating is(are)
Show Hint
MeI quaternizes $NMe_2$; heating closes the ring by intramolecular $S_N2$ at the non-stereogenic $CH_2$ carbon, so the relative configuration of the two Ph-bearing centers is unchanged. Use the Karplus relation ($J$ small near $90^\circ$ dihedral) to pick the diastereomer, then include both its enantiomers.
Approach this from the reaction type first, then bring in the NMR data only at the end, rather than jumping straight to the Karplus equation.
Recognize the reaction as an intramolecular Williamson ether synthesis. Methyl iodide is a strong, small alkylating agent; a tertiary amine like the $NMe_2$ group reacts with it fast and irreversibly to give a quaternary ammonium iodide. On warming, this quaternary center becomes an excellent intramolecular leaving group ($NMe_3$), and the nearby alcohol oxygen, positioned exactly right for a six-membered ring transition state, displaces it in a single $S_N2$ step. This is the standard way an ortho-(aminomethyl)benzylic alcohol like this is converted into a cyclic ether without needing any external base or catalyst.
Track the stereochemistry through the mechanism. An $S_N2$ step inverts configuration only at the carbon being attacked, here the benzylic $CH_2$ that is not a stereocenter to begin with (two identical $H$'s on it). The two carbons that do carry stereochemistry, the ones bearing the phenyl groups, are spectators in this step: their configurations, and therefore their relative configuration to each other, pass unchanged from starting amino alcohol into the isochroman product $\mathbf{X}$.
Bring in the NMR clue last. Now that $\mathbf{X}$ is a rigid bicyclic ring, $J_{ab}=3$ Hz is a small vicinal coupling, which by the Karplus curve means the $H_a-C-C-H_b$ dihedral in the ring is close to $90^\circ$. Only one of the two possible relative configurations of the phenyl-bearing stereocenters places the ring hydrogens near that dihedral once the ring is locked; the other diastereomer would push $H_a$ and $H_b$ closer to antiperiplanar and give a coupling several times larger.
Bring back the starting materials. Because the relative configuration is preserved (step 2) and only relative configuration controls $J$ (step 3), whichever amino alcohol diastereomer has the correct relative disposition of the two phenyls is the one that answers the question, and its mirror image (enantiomer) answers it equally well, since enantiomers give identical, superimposable NMR spectra.
Reading the four drawn structures as two enantiomeric pairs, the pair carrying the relative configuration consistent with $J_{ab}=3$ Hz is B and C; A and D belong to the other, non-matching diastereomeric pair.
Let's summarize:
The cyclization is a simple intramolecular Williamson ether synthesis at a non-stereogenic carbon.
Relative stereochemistry survives the reaction untouched, so the NMR coupling constant of the product directly reports on the starting diastereomer.
Both enantiomers of the correct diastereomer, B and C, are valid answers.
