Question

Give the structure that corresponds to the following molecular formula and NMR spectra: C6H14O 1H NMR: ? 0.91 (6H, d, J = 7 Hz); ? 1.17 (6H, s); ? 1.48 (1H, s; disappears after D2O shake); ? 1.65 (1H, septet, J = 7 Hz). 13C NMR: ? 17.6, ? 26.5, ? 38.7, ? 73.2.

Answer 1

To assign a structure to the molecular formula C6H14O based on the given NMR spectra.

Step-by-step explanation:

The molecular formula C6H14O corresponds to a compound with six carbon atoms, fourteen hydrogen atoms, and one oxygen atom. The 1H NMR spectrum for this compound shows the following signals: 0.91 (6H, d, J = 7 Hz), 1.17 (6H, s), 1.48 (1H, s; disappears after D2O shake), and 1.65 (1H, septet, J = 7 Hz). The 13C NMR spectrum shows signals at 17.6, 26.5, 38.7, and 73.2. To determine the structure of the compound, we analyze the NMR spectra and consider the possible functional groups and chemical shifts.

Answer 2

The molecular formula C6H14O, combined with NMR spectral data, corresponds to the structure of 2-methyl-2-pentanol, an alcohol with characteristic NMR peaks.

Step-by-step explanation:

The student is asking to identify the structure of a compound with the molecular formula C6H14O based on given 1H NMR and 13C NMR spectra data. Given the NMR spectral data, we can deduce that the compound is an alcohol due to the singlet peak at δ 1.48, which disappears after shaking with D2O, indicating the presence of an -OH proton. Moreover, the coupling patterns and the carbon signals suggest that the structure is 2-methyl-2-pentanol. This structure correlates with the NMR signals presented: a doublet at δ 0.91 (6H, suggesting -CH3 groups adjacent to each other), a singlet at δ 1.17 (6H, indicative of two -CH3 groups on the same carbon), a septet at δ 1.65 (1H, characteristic of a CH group next to two equivalent -CH3 groups), and the carbon signals at δ 17.6, 26.5, 38.7, and 73.2 corresponding to -CH3, -CH3, -CH, and the alcohol-bearing carbon, respectively.