Based on 13C-NMR:
one carbon (at about 180) is present as C=O.
And one peak between 50-100 ppm implies, one saturated carbon is next to Oxygen, like CH2O or CH3O.
Rest all carbons are at less than 50 ppm meaning they are saturated carbons like CH3, CH2 or CH.
Based on 1H-NMR
The maximum deshielding (about 5 ppm) will be observed by the H-atom attached on C, which is directly attached to red colored O atom of the ester group. (CH-O-C=O). Note that this is a single H. so option D) is eliminated, as 2 such H are there.
Again this is a heptet, meaning 6 neighboring and equivalent Hydrogens should be there for this H-atom, so option C) is eliminated, as only 4 neighboring hydrogens are there.
For options a) and b) we will analyze by drawing all the hydrogens as below:
Option a) is the correct answer.
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