A is the proton near the methoxy groups because it keeps relatively the same shift value as before, and now F and G
are both coupling with it, giving it a signal with a dd.
B and C are the protons of the isopropyl groups and should have the same signal as they are identical.
They appear close together in this, but the NMR machine read them as slightly different because they may not rotate
well and spend time in slightly different environments
D shows a ddd because it couples with E, F, and G. Notice that its shift value has
changed significantly from SI-1 because you can no longer draw a resonance structure to withdraw electrons.
E is a doublet because it only has one neighbor. Notice that its JH values match with D.
F and G are the CH2 protons, and they have different signals here because there are now diastereotopic. Before this reaction,
there were no stereocenters in the molecule, so they were enantiotopic and displayed only one signal.
H is the protons of the tert-butyl group, whose shift values have changed very little.