As this is clearly a homework type question, this answer will just give you some helpful pointers to head you in the right direction, and leave you to piece together the answer.
My usual approach for these types of problems is to have a look at the NMR data first, and ignore all the IR data. We might use that later for double checking.
From both spectra, it is possible to identify the following spin systems:
- -C6H5 (monosubstituted benzene).
We also know we must have a -OC(O)- group in there also. That satisfies our molecular formula.
Looking at Isomer C first:
- CH2 group at 2.7, indicative of it being adjacent to the carbonyl of an ester
- Ortho aromatics at 7.0 and upfield of the meta, indicative of substitution with an electron donating (or aromatic activating) group, Ar-O-
I'll let you piece together the structure of C.
Looking at Isomer D:
- CH2 group at 4.4, indicative of being adjacent to a heteroatom, so -O-CH2
- Ortho protons at 8.0 and downfield of the meta , indicative of substitution with an electron withdrawing (or aromatic deactivating) group; Ar-C(O)-
I'll let you put this structure together also.
And what about the IR data? We know the following, which is consistent with what we propose from NMR:
- Al-C(O)-O-Ar 1740-1770cm-1
- Ar-C(O)-O-Al 1715-1730cm-1
The general rule with IR absorptions is that conjugation lowers the frequency - I'll leave you to rationalise this with some fancy arrow-pushing from the aromatic ring to the carbonyl group.