Compound \(\mathrm{K}\), molecular formula \(\mathrm{C}_{8} \mathrm{H}_{14} \mathrm{O}\), readily undergoes acid-catalyzed dehydration when warmed with phosphoric acid to give compound L, molecular formula \(\mathrm{C}_{6} \mathrm{H}_{12}\), as the major organic product. The \({ }^{1} \mathrm{H}\)-NMR spectrum of compound \(\mathrm{K}\) shows signals at \(\delta 0.90(\mathrm{t}, 6 \mathrm{H}), 1.12(\mathrm{~s}, 3 \mathrm{H}), 1.38(\mathrm{~s}, 1 \mathrm{H})\), and \(1.48(\mathrm{q}, 4 \mathrm{H})\). The \({ }^{19} \mathrm{C}-\mathrm{NMR}\) spectrum of compound \(\mathrm{K}\) shows signals at \(\delta 72.98,33.72,25.85\), and 8.16. Deduce the structural formulas of compounds \(\mathrm{K}\) and \(\mathrm{L}\).

Calculate the degrees of unsaturation for both compound K and L using their molecular formulas. With degrees of unsaturation, we can have an idea of whether double bonds or rings are present. Degree of unsaturation (\(DU\)) for compound K: \(DU = \frac{2 \times (\text{number of carbons})+2 - \text{number of hydrogens}}{2}\) Compound K: \(C_{8}H_{14}O\) \(DU_K = \frac{2 \times (8) + 2 - 14}{2} = \frac{16}{2} = 4\) Degree of unsaturation (\(DU\)) for compound L: \(DU = \frac{2 \times (\text{number of carbons})+2 - \text{number of hydrogens}}{2}\) Compound L: \(C_{6}H_{12}\) \(DU_L = \frac{2 \times (6) + 2 - 12}{2} = \frac{10}{2} = 2\) Compound K has 4 degrees of unsaturation and compound L has 2 degrees of unsaturation.

Analyze the 1H-NMR spectrum of compound K, including the chemical shifts and splitting patterns: - Signal at \(\delta\) 0.90 (t, 6 H): Indicates protons on a methyl group (\(-CH_3\)) with coupling to nearby protons due to the triplet splitting pattern. - Signal at \(\delta\) 1.12 (s, 3 H): Indicates protons on a methyl group (\(-CH_3\)) that does not couple with any other protons due to the singlet splitting pattern. - Signal at \(\delta\) 1.38 (s, 1 H): Indicates a single proton that is isolated from other protons (could be on a quaternary carbon) due to the singlet splitting pattern. - Signal at \(\delta\) 1.48 (q, 4 H): Indicates protons on two methylene groups (\(-CH_2-CH_2-\)) that are coupled to nearby protons due to the quartet splitting pattern.

Analyze the 19C-NMR spectrum of compound K, focusing on the chemical shifts: - Signal at δ 72.98: Indicates carbons in the alcohol (OH) or ether (\(O-CH_2-\)) region. - Signal at δ 33.72: Indicates carbons in the aliphatic region (C-C and C-H bonds). - Signal at δ 25.85: Indicates carbons in the aliphatic region (C-C and C-H bonds). - Signal at δ 8.16: Indicates carbons in the alkyl region (C-H bonds).

Combine the information obtained from both the 1H-NMR and 19C-NMR spectra to propose a structure for compound K. Considering 4 degrees of unsaturation, we can infer the presence of a ring and a double bond. Taking into account the chemical shifts and splitting patterns, the proposed structure for compound K could be a tert-butyl group (\(-C(CH_3)_3\)) connected to a cyclohexene ring with an alcohol group on the double bond.

Compound K undergoes acid-catalyzed dehydration to form compound L. This means the alcohol group in compound K is removed in a dehydration reaction, and a carbon-carbon double bond is formed due to the loss of water. The resulting structure for compound L is a 3,3-dimethylcyclohexene.