Question

(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?

Short answer

In their liquid forms, tert-butyl fluoride exhibits a doublet due to the coupling of the fluorine atom with the methyl group protons, and isopropyl fluoride shows two doublets and two multiplets indicating different chemical environments for protons. When dissolved in liquid \(\mathrm{SbF}_{5}\), new NMR spectra are obtained, suggesting the formation of new compounds or complexes with \(\mathrm{SbF}_{5}\), resulting in the disappearance of some peaks and large downfield shifts in chemical shifts.

Step-by-step solution

(a) Interpreting NMR spectra of tert-butyl fluoride and isopropyl fluoride in their liquid forms

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.

(b) Interpreting NMR spectra of tert-butyl fluoride and isopropyl fluoride dissolved in liquid SbF5

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.

Key concepts

Chemical Shifts

Chemical shifts are crucial in NMR spectroscopy for identifying the type of environment a nucleus is in. They provide a direct indication of how shielded or deshielded a nucleus is by its surrounding electrons. In tert-butyl fluoride and isopropyl fluoride, the chemical shifts tell us how the environment around the fluorine and hydrogen atoms changes, especially when dissolved in different solvents.

When dissolved in \( ext{SbF}_5 \) , a downfield shift occurs. This shift is toward higher \( ext{ppm} \) values and suggests a change in the electron distribution around the nucleus. This change results from interactions with the \( ext{SbF}_5 \) . It highlights the sensitivity of NMR to subtle chemical environment changes.

Coupling Constants

Coupling constants, denoted by \( ext{J} \) , are a measure of the interaction between nuclei, typically through bonds. They appear as the spacing between peaks in a multiplet in an NMR spectrum. The size of these constants gives insight into the structural relationship between the interacting nuclei.

In the spectra of tert-butyl fluoride and isopropyl fluoride, different \( ext{J} \) values indicate varying extents of coupling among the nuclei, informing us about the number of bonds and angles involved. An interesting point is observed when these compounds are dissolved in \( ext{SbF}_5 \), with some coupling constants decreasing. This phenomenon suggests a change in the molecular interactions that affects the spin-spin coupling.

Tert-butyl Fluoride

Tert-butyl fluoride presents an interesting case in NMR analysis. Initially, in its pure form, it displays a doublet, indicating a coupling of fluorine with the protons of the methyl groups. However, when dissolved in \( ext{SbF}_5 \), it transforms into a singlet.

This transformation infers that there is a significant chemical change, possibly including complex formation or reaction with \( ext{SbF}_5 \). This reaction disrupts the coupling originally observed, highlighting the role of its chemical environment in defining its NMR characteristics.

Isopropyl Fluoride

Isopropyl fluoride's NMR spectrum is more complex due to the unique environments of its protons. It shows both doublets and multiplets in its liquid form, giving insights into multiple coupling interactions and environments.

When dissolved in \( ext{SbF}_5 \), some peaks disappear, hinting at a chemical reaction or shift in the environment. The complexity of its spectrum in liquid form and simplification when interacting with other chemicals demonstrate the dynamic nature of molecular interactions detected by NMR.

Fluorine Coupling

Fluorine coupling is key in these analyses because fluorine is highly electronegative, often causing substantial shifts and couplings in its NMR spectrum. This can help diagnose how fluorine interacts within a molecule, as seen in both tert-butyl and isopropyl fluoride.

Both molecules show significant coupling involving fluorine, which reflects its influence on the neighboring protons. Understanding fluorine's role and behavior, particularly in different environments, helps chemists predict the behavior and reactivity of fluorine-containing compounds.

Chemical Environment Changes

NMR spectroscopy is sensitive to changes in the chemical environment of the nucleus. When tert-butyl and isopropyl fluorides dissolve in \( ext{SbF}_5 \), the spectra change markedly. These changes are attributed to modifications in the surrounding electron cloud, possibly due to bonding or interactions with \( ext{SbF}_5 \).

This sensitivity is beneficial for detecting small modifications or transformations in a molecule. The very large shifts occurring for both compounds suggest notable changes in the electronic environment, often signifying new interactions, such as complex formation or even complete reactions.