Q5(e). IR spectra of two isoxazolone tautomers
How will you distinguish the two using IR-spectroscopy?
[I] [II]
Answer in brief
II shows an N–H band (about 3100–3300 cm⁻¹) and a conjugated C=O near 1700 cm⁻¹; I has no N–H, a high-frequency lactone C=O near 1800 cm⁻¹ and a C=N band near 1630 cm⁻¹.
Key points
- only II shows it. It broadens and shifts in concentrated solution or the solid (intermolecular N–H…O=C hydrogen bonding) and sharpens on dilution in CCl₄.
- ring strain plus the electronegative ring O raise the lactone C=O of I to near 1800 cm⁻¹; in II the push of the N lone pair through C=C into C=O lowers the bond order and the wavenumber by about 100 cm⁻¹.
- The two forms interconvert in solution; the relative intensities of the 1800 and 1700 cm⁻¹ bands show which tautomer dominates in a given solvent.
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 and II are tautomers of 3,4-dimethylisoxazol-5-one. I is the CH form (4H): an sp3 C-4–H, a C=N and an unconjugated lactone C=O in a five-membered ring. II is the NH form (2H): an N–H, a C3=C4 double bond and a C=O conjugated with C=C and with the N lone pair.
| Band | I (CH form) | II (NH form) |
| N–H stretch | absent | present: broad band about 3100–3300 cm⁻¹ (H-bonded) |
| C=O stretch | high, about 1790–1810 cm⁻¹: five-ring lactone, not conjugated, O–N next to O | lower, about 1700 cm⁻¹: conjugated with C=C and with the N lone pair (vinylogous carbamate) |
| C=C / C=N | C=N about 1620–1640 cm⁻¹ (medium) | C=C about 1600–1620 cm⁻¹, stronger (polarised enamine-type C=C) |
| C–H | sp3 C–H only, below 3000 cm⁻¹ (including the ring C-4–H) | sp3 C–H of the two CH₃ only |
- The decisive band is N–H: only II shows it. It broadens and shifts in concentrated solution or the solid (intermolecular N–H…O=C hydrogen bonding) and sharpens on dilution in CCl₄.
- The C=O position: ring strain plus the electronegative ring O raise the lactone C=O of I to near 1800 cm⁻¹; in II the push of the N lone pair through C=C into C=O lowers the bond order and the wavenumber by about 100 cm⁻¹.
- The two forms interconvert in solution; the relative intensities of the 1800 and 1700 cm⁻¹ bands show which tautomer dominates in a given solvent.
Conclusion. II shows an N–H band (about 3100–3300 cm⁻¹) and a conjugated C=O near 1700 cm⁻¹; I has no N–H, a high-frequency lactone C=O near 1800 cm⁻¹ and a C=N band near 1630 cm⁻¹.
Application. IR (with NMR) is the routine way to find the dominant tautomer of heterocycles such as isoxazolones and pyrazolones, the parent rings of several dyes and drugs (e.g. the pyrazolone analgesics).
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All parts of Question 5
- 5(a) DNA bases and Watson–Crick pairing10 marks
- 5(b) Baeyer–Villiger oxidation of an electron-rich aldehyde10 marks
- 5(c) Photochemistry of norbornenone: 1,3-acyl shift and ketene loss10 marks
- 5(d) λ_(max) by Woodward–Fieser and Scott rules10 marks
- 5(e) IR spectra of two isoxazolone tautomers10 marks