Q7(a). Structure of C₈H₄O₃ from IR, NMR and mass spectrum: phthalic anhydride
An organic compound with molecular formula C₈H₄O₃ exhibited the following spectroscopic data:
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IR (cm⁻¹) &: 1830, 1760, 1120
1H-NMR &: δ_(8.049), 2H, s; δ_(7.635), 2H, s
Mass &: m/z = 148, 104, 76, 50
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Deduce the structure of the compound and justify the formation of the above fragment ions.
Answer in brief
C₈H₄O₃ is phthalic anhydride: DBE 7; two coupled anhydride C=O bands (1830, 1760 cm⁻¹) and C–O–C (1120 cm⁻¹); four aromatic H in two equivalent pairs (ortho-disubstituted, symmetrical). The mass spectrum shows M⁺˙ 148 and the losses CO₂ (104), CO (76) and C₂H₂ (50).
Key points
- m/z 148 → 104 (loss of 44, CO₂). The anhydride ring opens at an acyl–O bond and expels carbon dioxide, a very stable neutral, leaving the radical cation C₇H₄O⁺˙ (C₆H₄CO⁺˙, a benzoyl-type / cyclic ketene ion).
- m/z 104 → 76 (loss of 28, CO). The remaining carbonyl is lost as carbon monoxide: C₆H₄⁺˙, the benzyne radical cation. This ion is typical of ortho-disubstituted benzenes.
- m/z 76 → 50 (loss of 26, C₂H₂). The C₆H₄ ring fragments with loss of acetylene: C₄H₂⁺˙, a common low-mass ion of benzene derivatives.
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. Start from the formula, then match each band, signal and peak.
DBE = (2C + 2 - H)/(2) = (2(8) + 2 - 4)/(2) = 7
Seven degrees of unsaturation: a benzene ring (4) + two C=O (2) + one more ring (1).
| Data | Inference |
| 1830 and 1760 cm⁻¹ | two C=O bands about 70 cm⁻¹ apart, at high wavenumber: the symmetric and asymmetric stretches of the coupled C=O pair of a cyclic anhydride (five-membered ring) |
| 1120 cm⁻¹ | C–O–C stretch of the anhydride |
| No band above 3000 cm⁻¹ except weak aromatic C–H | no O–H: not a carboxylic acid |
| δ 8.05 (2H), 7.64 (2H) | only four H, all aromatic, in two sets of two equivalent H: an ortho-disubstituted ring with identical substituents (H-3/H-6 and H-4/H-5) |
| m/z 148 | molecular ion C₈H₄O₃⁺˙ (M = 148) |
C₆H₄ (ortho) + -CO-O-CO⁻ bridging the two ring carbons = C₈H₄O₃: phthalic anhydride (isobenzofuran-1,3-dione). The molecule has a mirror plane, so H-3 and H-6 (next to C=O, deshielded, δ 8.05) are equivalent, as are H-4 and H-5 (δ 7.64). The AA'BB' multiplets appear as two narrow signals in the data.
Figure: Phthalic anhydride.
Fragmentation
Scheme: Successive losses of neutral, stable molecules from the molecular ion.
- m/z 148 → 104 (loss of 44, CO₂). The anhydride ring opens at an acyl–O bond and expels carbon dioxide, a very stable neutral, leaving the radical cation C₇H₄O⁺˙ (C₆H₄CO⁺˙, a benzoyl-type / cyclic ketene ion).
- m/z 104 → 76 (loss of 28, CO). The remaining carbonyl is lost as carbon monoxide: C₆H₄⁺˙, the benzyne radical cation. This ion is typical of ortho-disubstituted benzenes.
- m/z 76 → 50 (loss of 26, C₂H₂). The C₆H₄ ring fragments with loss of acetylene: C₄H₂⁺˙, a common low-mass ion of benzene derivatives.
Conclusion. C₈H₄O₃ is phthalic anhydride: DBE 7; two coupled anhydride C=O bands (1830, 1760 cm⁻¹) and C–O–C (1120 cm⁻¹); four aromatic H in two equivalent pairs (ortho-disubstituted, symmetrical). The mass spectrum shows M⁺˙ 148 and the losses CO₂ (104), CO (76) and C₂H₂ (50).
Application. Phthalic anhydride is made by air oxidation of o-xylene or naphthalene over V₂O₅ and is used for phthalate plasticisers, alkyd resins and phenolphthalein; the 148/104/76 pattern identifies it in the mass spectra of phthalate esters.
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All parts of Question 7
- 7(a) Structure of C₈H₄O₃ from IR, NMR and mass spectrum: phthalic anhydride15 marks
- 7(b) Borane: acid reduced, ester kept; δ-lactone15 marks
- 7(c) SeO₂/H₂O₂, Pb(OAc)₄, OsO₄, periodate20 marks