Q8(a). NMR shift in Hz and ppm; UV bands of 3-buten-2-one
(i) Dichloromethane has a chemical shift of 532 Hz from TMS (tetramethylsilane) on a 100 MHz spectrometer. What will be the chemical shift in ppm? What would the chemical shift be in ppm and Hz on a 600 MHz spectrometer?
(ii) 3-Buten-2-one gives two peaks in its ultraviolet spectrum at 213 nm (ε = 7100) and 320 nm (ε = 27). What type of transition is responsible for each absorbance? Account for the difference in molar absorptivity for the two absorptions.
Answer in brief
(i) δ = 5.32 ppm on both instruments; 3192 Hz from TMS at 600 MHz. (ii) 213 nm: π→π* (allowed, ε 7100); 320 nm: n→π* (forbidden, ε 27); the large difference in ε comes from good orbital overlap in π→π* and orthogonal n and π* orbitals in n→π*.
Key points
- 213 nm, ε = 7100: a π→π* transition of the conjugated C=C-C=O system (K band). Conjugation raises the HOMO and lowers the LUMO, so it falls above 200 nm.
- 320 nm, ε = 27: an n→π* transition (R band): an oxygen lone-pair electron is promoted to π*. The n level lies above π, so this is the lower-energy, longer-wavelength band.
- Why the intensities differ. The molar absorptivity is proportional to the square of the transition dipole moment, which depends on the overlap of the two orbitals involved. π and π* occupy the same region of space (the p…
- The oxygen lone pair (n) lies in the molecular plane, at right angles to the π* orbital. The overlap is almost zero, so n→π* is symmetry-forbidden; it appears only weakly, through vibronic coupling, with ε below 100.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. (i) The chemical shift δ (ppm) is the shift in Hz divided by the spectrometer frequency in MHz; it is independent of the field, while the shift in Hz is proportional to it. (ii) 3-Buten-2-one (methyl vinyl ketone) is an enone: it has a conjugated π system and oxygen lone pairs, so it shows a π→π* and an n→π* band.
(i) Chemical shift of CH₂Cl₂
δ=(Δν (Hz))/(ν₀ (MHz))=(532)/(100)=5.32 ppm
At 600 MHz: δ=5.32 ppm; Δν=5.32×600=3192 Hz
δ = 5.32 ppm (both); 532 Hz (100 MHz), 3192 Hz (600 MHz) The shielding constant is a molecular property, so δ is the same on every instrument. The two electronegative chlorines deshield the protons (inductive withdrawal of electron density), which is why CH₂Cl₂ appears far downfield of CH₄.
(ii) 213 and 320 nm bands of 3-buten-2-one
- 213 nm, ε = 7100: a π→π* transition of the conjugated C=C-C=O system (K band). Conjugation raises the HOMO and lowers the LUMO, so it falls above 200 nm.
- 320 nm, ε = 27: an n→π* transition (R band): an oxygen lone-pair electron is promoted to π*. The n level lies above π, so this is the lower-energy, longer-wavelength band.
Figure: Orbital energies of the enone (sketch).
- Why the intensities differ. The molar absorptivity is proportional to the square of the transition dipole moment, which depends on the overlap of the two orbitals involved. π and π* occupy the same region of space (the p orbitals perpendicular to the molecular plane), so π→π* is symmetry-allowed and intense (ε in the thousands).
- The oxygen lone pair (n) lies in the molecular plane, at right angles to the π* orbital. The overlap is almost zero, so n→π* is symmetry-forbidden; it appears only weakly, through vibronic coupling, with ε below 100.
- Solvent test. On going to a polar, protic solvent the n→π* band shifts to shorter wavelength (blue shift, the lone pair is stabilised by hydrogen bonding), while the π→π* band shifts slightly to longer wavelength.
Conclusion. (i) δ = 5.32 ppm on both instruments; 3192 Hz from TMS at 600 MHz. (ii) 213 nm: π→π* (allowed, ε 7100); 320 nm: n→π* (forbidden, ε 27); the large difference in ε comes from good orbital overlap in π→π* and orthogonal n and π* orbitals in n→π*.
Application. Higher-field instruments spread signals apart in Hz while δ stays fixed, which is why 600 MHz spectra resolve overlapping multiplets; the K and R bands of enones are the basis of the Woodward–Fieser rules for λₘax.
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All parts of Question 8
- 8(a) NMR shift in Hz and ppm; UV bands of 3-buten-2-one15 marks
- 8(b) Epoxidation of a silyl dihydropyran; the Shapiro reaction15 marks
- 8(c) Bullatenone: correcting a structure from IR, NMR and MS20 marks