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Determine the structure of an unknown organic compound using nuclear magnetic resonance (NMR) spectroscopy, given the following information: - The compound has a molecular formula of C6H12O2 - The NMR spectrum shows signals at 1.2 ppm (triplet), 2.3 ppm (quartet), and 3.7 ppm (doublet).

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To determine the structure of the unknown organic compound, we will analyze the given information and use the NMR spectroscopy data.1. Molecular formula: C6H12O2This formula suggests that the compound may contain an ester, ether, or alcohol functional group, as well as some alkyl groups.2. NMR spectrum signals:- 1.2 ppm  triplet : This signal is characteristic of a methyl group  CH3  adjacent to a methylene group  CH2 . The triplet splitting pattern indicates that the CH3 group is next to a carbon with two hydrogens  n+1 rule .- 2.3 ppm  quartet : This signal is characteristic of a methylene group  CH2  adjacent to a carbonyl group  C=O  or an oxygen atom. The quartet splitting pattern indicates that the CH2 group is next to a carbon with three hydrogens  n+1 rule .- 3.7 ppm  doublet : This signal is characteristic of a methylene group  CH2  adjacent to an oxygen atom, such as in ethers or esters. The doublet splitting pattern indicates that the CH2 group is next to a carbon with one hydrogen  n+1 rule .Based on the molecular formula and the NMR data, we can propose the following structure for the unknown compound:CH3CH2COOCH2CH2CH3This structure is consistent with the molecular formula C6H12O2 and the NMR signals. The compound is an ester, specifically ethyl propionate.

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