3. Determine the structure of a compound with a molecular formula of C6H12O2 and exhibitsthe following 'H NMR spectrum. No peaks exchange in D20. The °C NMR has a peakat 170 ppm. Each red box indicates a single integration value (i.e., the peak at 4.05 ppmhas an integration = 2). The signal at 2.05 ppm is a singlet.111510976.543.6 (ppm)frequency

The NMR spectrum provides the following details about the structure of the molecules: The number of signals: The number of signals gives information about the different types of hydrogen atoms. Position of signal: The position of the signal tells us about the electronic environment of hydrogen. splitting of signal: Splitting of the signal tells us about the number of neighboring hydrogen atom present. If the number of the neighboring hydrogen atom is n then the signal split in (n+1) peak. Area of signal: The area of the signal provides information about the number of hydrogen atoms responsible for the signal.The compound has no hydrogen that exchanged with D2o, this confirms the absence of an alcohol functional group. The singlet at 2.05 ppm,3H, is a characteristic signal of the CH3 group attached with a carbonyl functional group. Triplet at 4.05 ppm, 2H, is a characteristic signal of a CH2 group attached with electronegative atom oxygen and CH2 group. Two multiple at 1.5 and 1.6 ppm due to two CH2 groups attached with sp3 hybrid carbon atom. Triplet at 0.9 ppm,3H, is a characteristic signal of a CH3 group attached with an sp3 carbon atom.The structure of the unknown compound is as follows:The NMR signal of this compound explained as follows: