(a) In the liquid form, tert-buty1 fluoride and isopropyl fluoride gave the following nmr spectra; tert-butyl fluoride: doublet, \(\delta 1.30, \mathrm{~J}=20 \mathrm{~Hz}\) isopropyl fluoride: two doublets, \(\delta 1,23,6 \mathrm{H}\), \(\mathrm{J}=23 \mathrm{~Hz}\) and \(4 \mathrm{~Hz}\) two multiplets, \(\delta 4.64,1 \mathrm{H}\) \(\mathrm{J}=48 \mathrm{~Hz}\) and \(4 \mathrm{~Hz}\) How do you account for each of these spectra? (b) When the alkyl fluorides were dissolved in liquid \(\mathrm{SBF}_{5}\), the following nmr spectra were obtained* tert-butyl fluoride: singlet, \(\delta 4.35\) isopropyl fluoride: doublet, \(\delta 5.06,6 \mathrm{H}, \mathrm{J}=4 \mathrm{~Hz}\) multiplet, \(\delta 13.5,1 \mathrm{H}, \mathrm{J}=4 \mathrm{~Hz}\) To what molecule is each of these spectra due? (Hint: What does the disappearance of just half the peaks observed in part (a) suggest?) Is the very large downfield shift what you might have expected for molecules like these?

In order to explain the NMR spectra of tert-butyl fluoride and isopropyl fluoride in their liquid forms, we can analyze the chemical shifts and coupling constants in each case. For tert-butyl fluoride, the NMR spectrum shows a doublet, \(\delta 1.30, \mathrm{J}=20 \mathrm{~Hz}\). This could be due to the coupling of the fluorine atom with the methyl group protons in the molecule. For isopropyl fluoride, the NMR spectrum shows two doublets, \(\delta 1.23,6 \mathrm{H}\), \(\mathrm{J}=23 \mathrm{~Hz}\) and \(4 \mathrm{~Hz}\). This suggests that the isopropyl group protons couple with the fluorine atom at two different chemical shifts, indicating the presence of two chemical environments for these protons. Additionally, there are two multiplets: \(\delta 4.64,1 \mathrm{H}\), with \(\mathrm{J}=48 \mathrm{~Hz}\) and \(4 \mathrm{~Hz}\). These multiplets could correspond to the methine proton in isopropyl fluoride that is directly connected to the carbon bearing the fluorine atom. The methine proton experiences different coupling constants due to the varying chemical environments in the molecule.

When the alkyl fluorides are dissolved in liquid \(\mathrm{SbF}_{5}\), new NMR spectra are obtained, and we need to identify the molecules responsible for these spectra. For tert-butyl fluoride, the NMR spectrum shows a singlet at \(\delta 4.35\). The singlet suggests that the molecule has lost its coupling to the neighboring protons, possibly due to the formation of a new compound with \(\mathrm{SbF}_{5}\). For isopropyl fluoride, the NMR spectrum shows a doublet at \(\delta 5.06,6 \mathrm{H}, \mathrm{J}=4 \mathrm{~Hz}\) and a multiplet at \(\delta 13.5,1 \mathrm{H}, \mathrm{J}=4 \mathrm{~Hz}\). The disappearance of half of the peaks observed in part (a) suggests that the molecule has reacted with \(\mathrm{SbF}_{5}\), resulting in a new compound or complex. The very large downfield shift observed in both cases may not be expected for molecules such as alkyl fluorides, since fluorine is a highly electronegative atom. However, the interaction with \(\mathrm{SbF}_{5}\) could cause this shift due to changes in the electronic environment of the protons in these molecules.