= Solution
The two <methyl groups> give one six-<proton> <NMR doublet> centred at $(1.319+1.296)/2=1.3075$ ppm. The methine gives a one-<proton> <NMR septet> centred near $2.976$ ppm. The latter is split by six equivalent methyl <protons>, producing the first-order $1:6:15:20:15:6:1$ intensity pattern.
$$
\boxed{{}^3J_{\mathrm{CH,CH_3}}\simeq6.9\,\mathrm{Hz}.}
$$
This follows exactly from the methyl doublet spacing, $300(0.023)$, and approximately from the successive septet spacings, which alternate between $0.022$ and $0.024$ ppm within the precision of the printed labels. There is no methylene triplet or quartet: the six-<proton> integral and septet establish the branched skeleton.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-acylchloride-assignment.png]
{title=Independent original carbon and proton assignments for B; both equivalent methyl groups couple to the methine}
Back to article page
