An aromatic compound is isolated from a plant extract. The data are:
- MS: molecular ion at m/z 152.
- IR: broad absorption near 3200 cm⁻¹; strong band at 1665 cm⁻¹; bands at 1595 and 1510 cm⁻¹.
- ¹H NMR: δ 9.82 (1H, s); δ 7.42 (1H, d); δ 7.40 (1H, dd); δ 7.04 (1H, d); δ 6.2 (1H, broad s, exchangeable); δ 3.95 (3H, s).
- ¹³C NMR: δ 191 (one carbon); six aromatic carbons between δ 109 and δ 152; δ 56 (one carbon).
Step 1: molecular formula. A molecular ion at 152 is consistent with C₈H₈O₃, since 8 × 12 + 8 × 1 + 3 × 16 = 96 + 8 + 48 = 152.
Step 2: degrees of unsaturation. DBE = C − H/2 + 1 = 8 − 4 + 1 = 5. Four account for a benzene ring, leaving one more double bond.
Step 3: assign the functional groups.
| Evidence | Conclusion |
|---|
| δ 9.82 (1H, s) and ¹³C δ 191 | Aldehyde proton and carbonyl carbon |
| IR 1665 cm⁻¹, lower than a typical aldehyde | Carbonyl conjugated to the ring |
| IR 3200 cm⁻¹ broad, δ 6.2 exchangeable | Hydroxyl, and its breadth indicates phenol |
| δ 3.95 (3H, s) with ¹³C δ 56 | Methoxy group |
| IR 1595 and 1510 cm⁻¹ | Aromatic ring |
Step 4: substitution pattern. Three aromatic protons as a doublet, a doublet of doublets and a doublet give the classic 1,2,4-trisubstituted benzene pattern: one proton with only a meta coupling, one with both ortho and meta, one with only an ortho coupling.
Step 5: assemble. A benzene ring carrying an aldehyde, a hydroxyl and a methoxy group, in a 1,2,4-arrangement, accounts for C₈H₈O₃ and all five degrees of unsaturation. The compound is 4-hydroxy-3-methoxybenzaldehyde, better known as vanillin.
The method matters more than the answer. Work from formula, to degrees of unsaturation, to functional groups, to substitution pattern, and only then propose a structure. Guessing a structure first and defending it afterwards is how marks are lost.