Following are \({ }^{1} \mathrm{H}-\mathrm{NMR}\) and \({ }^{13} \mathrm{C}-\mathrm{NMR}\) spectral data for compound \(\mathrm{F}\left(\mathrm{C}_{12} \mathrm{H}_{16} \mathrm{O}\right)\). From this information, deduce the structure of compound \(F\). $$ \begin{array}{lll} \hline{ }^{1} \text { H-NMR } & \multicolumn{2}{c}{{ }^{13} \text { C-NMR }} \\\ \hline 0.83(\mathrm{~d}, 6 \mathrm{H}) & 207.82 & 50.88 \\ 2.11(\mathrm{~m}, 1 \mathrm{H}) & 134.24 & 50.57 \\ 2.30(\mathrm{~d}, 2 \mathrm{H}) & 129.36 & 24.43 \\ 3.64(\mathrm{~s}, 2 \mathrm{H}) & 128.60 & 22.48 \\ 7.2-7.4(\mathrm{~m}, 5 \mathrm{H}) & 126.86 & \\ \hline \end{array} $$

First, we will analyze the 1H-NMR chemical shifts and their splitting: 0.83 (d, 6H): A doublet with a chemical shift of 0.83 ppm indicates a group of protons which are likely attached to a carbon next to CH3. In this case, there are 6 protons, meaning we have two methyl groups sharing a chemical environment. These protons are most likely on two equivalent secondary or tertiary carbons. 2.11 (m, 1H): A multiplet at 2.11 ppm indicates a proton that is likely near a carbon with sp3 hybridization, possibly adjacent to a double bond or an electronegative atom. 2.30 (d, 2H): A doublet at 2.30 ppm represents two equivalent protons that might be on the carbon attached to the oxygen atom (which is probably an ether group). 3.64 (s, 2H): A singlet at 3.64 ppm represents two equivalent protons that are likely part of an O-CH2 group. 7.2-7.4 (m, 5H): The multiplet between 7.2-7.4 ppm corresponds to five protons which are likely to be attached to aromatic carbons.

We will now analyze the 13C-NMR chemical shifts: 207.82 ppm: A carbon with a chemical shift of 207.82 ppm is likely a carbonyl (C=O) group. The five chemical shifts between approximately 126-134 ppm correspond to the aromatic carbons. 50.88 ppm and 50.57 ppm: These two chemical shifts are likely due to the carbons attached to sp3 hybridized carbons. 24.43 ppm and 22.48 ppm: These chemical shifts likely represent carbons bonded to the oxygen atom or neighboring the oxygen atom.

Now we will combine the information from the 1H-NMR and 13C-NMR to propose the structure of compound F (C12H16O). From the 1H-NMR, we know that there are likely five aromatic protons and an O-CH2 group. We also have a carbonyl group due to the chemical shift at 207.82 ppm in the 13C-NMR. From the 1H-NMR data at 0.83 ppm (d, 6H), we know that there are two methyl groups attached to secondary or tertiary carbons. On 13C-NMR, we can match these protons with the carbons' chemical shifts at 24.43 ppm and 22.48 ppm. Thus, the proposed structure of compound F is: [C6H5-C-(O-CH2)-(CH(CH3)2)].