Proton A has the highest signal because it is in the meta position to an electron-rich aromatic methylene group, causing its chemical shift to be more deshielded. Proton E appears at ~3 ppm instead of the expected 1.2-1.6 ppm because it belongs to an aromatic methylene group that is substituted between two phenyl rings, causing it to be downfield shifted by approximately 1.3 ppm. The chemical shifts of protons C and D are shielded compared to A and B due to increased electron density at the para positions of the phenyl and benzylic methylene groups.

1. Hello
Please explain to me why proton A has the highest signal and E is at ~3 when it should be 1.2-
1.6
4 0 HSP-40-226 E) ?B1 (D? (A? ??? ????(ppm) 7·783 7.536 7.367 7.297 3.889
Solution
Generally Aromatic C-H Bond chemical shift region willbe 6.5-8.5
Depending On The Groups attached to the aromatic rings C-H bond chemical shift values varies.
Here for C and D Hydrogens chemical shifts are shielded compared to A and B due to electron
densities are more at para positions compare to ortho position.(due to phenyl ring and benzylic
CH2 groups are at 1and 2 positions)
.D Proton is para to electron rich Aromatic methyl group.(Thats why D shielded more)
C Proton is para to phenyl ring that is why deshielded compare to D.
A proton is meta to electron ric aromatic CH2 electron density is less at meta position Chemical
shift is more at A Proton.
Aromatic CH3 comes at 2.3-2.4 E methylene(CH2) group is Downfielded to 3.88 due to
sustitution on CH2(2 Phenyl rings).Down fields:CH3-0.9ppm CH2:1.3ppm CH:1.7ppm
Downfield=2.4+1.3=3.7(approximately 3.8 region)