Based on the given information from the IR and NMR spectra, we can deduce the following:1. The IR spectrum shows a peak at 1720 cm, which is indicative of a carbonyl group C=O present in the compound. This is commonly found in aldehydes, ketones, carboxylic acids, and esters.2. The NMR spectrum shows three distinct peaks: a. A singlet at 1.0 ppm: This is characteristic of a methyl group CH3 that is not adjacent to any other protons. b. A quartet at 2.5 ppm: This suggests a methylene group CH2 that is adjacent to a single proton, as the splitting pattern follows the n+1 rule where n is the number of neighboring protons . c. A doublet at 7.2 ppm: This indicates a proton that is adjacent to two other protons, which is typical of an aromatic proton such as in a benzene ring or an alkenyl proton C=C-H .Considering the information above, the molecular formula of the compound can be deduced as C4H8O. This is because we have one carbonyl group C=O , one methyl group CH3 , one methylene group CH2 , and one proton H from the doublet.A possible structure for the compound is an ethyl acetate CH3COOCH2CH3 . This structure contains a carbonyl group C=O in the ester functional group, a methyl group CH3 adjacent to the carbonyl group, a methylene group CH2 adjacent to the ester oxygen, and the remaining proton H is part of the methylene group.In summary, the unknown organic compound has a molecular formula of C4H8O and a possible structure of ethyl acetate CH3COOCH2CH3 .