Q5(d). IR of esters and amides; Woodward–Fieser calculations
(i) How can the following pairs of compounds be distinguished based on the vibrational frequencies of their C=O and N-H bonds?
[structure] and [structure] [structure] and [structure]
(ii) Using the Woodward–Fieser rules, calculate the λ_(max) values of the following compounds:
[compound 1] [compound 2]
Answer in brief
(i) Phenyl propanoate (about 1765) and the anilide (about 1665) absorb at higher C=O frequency than ethyl benzoate (1720) and the benzamide (1640). (ii) λ_(max): compound 1 = 323 nm, compound 2 = 331 nm.
Key points
- Introduction. The C=O stretching wavenumber falls when conjugation or a lone pair adds single-bond character to C=O, and rises when the lone pair of the neighbouring O or N is drawn away from it. UV λ_(max) of conjugated…
- In each pair the compound with O or N attached to the ring (phenyl ester, anilide) shows the higher C=O frequency. The N–H stretches of the two secondary amides are similar (one sharp band near 3300, about 3450 in dilute…
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. The C=O stretching wavenumber falls when conjugation or a lone pair adds single-bond character to C=O, and rises when the lone pair of the neighbouring O or N is drawn away from it. UV λ_(max) of conjugated dienes and enones is estimated from base values and increments (Woodward–Fieser).
(i) IR distinctions
| Compound | Band (cm⁻¹) | Reason |
| Ethyl benzoate | C=O ≈ 1720 | C=O conjugated with the ring: lowered from the 1740 of saturated esters |
| Phenyl propanoate | C=O ≈ 1765 | O lone pair partly drawn into the phenyl ring, less donation to C=O, so more double-bond character |
| N-Methyl-4-bromobenzamide | C=O ≈ 1640; N–H ≈ 3300 | C=O conjugated with the aryl ring and fully with the N lone pair |
| 4'-Bromoacetanilide | C=O ≈ 1665; N–H ≈ 3290 | N lone pair shared with the aryl ring, less with C=O, so C=O higher |
In each pair the compound with O or N attached to the ring (phenyl ester, anilide) shows the higher C=O frequency. The N–H stretches of the two secondary amides are similar (one sharp band near 3300, about 3450 in dilute solution); the amide II band (N–H bend, near 1540) is present in both.
(ii) Woodward–Fieser calculations
| Compound 1 (triene) | nm | Compound 2 (dienone) | nm |
| Homoannular diene (ring C) | 253 | Six-membered ring enone | 215 |
| Extra conjugated C=C, 1 × 30 | 30 | Extra conjugated C=C, 1 × 30 | 30 |
| Alkyl groups / ring residues, 6 × 5 | 30 | α-ring residue, 1 × 10 | 10 |
| Exocyclic C=C, 2 × 5 | 10 | β-ring residue, 1 × 12 | 12 |
| γ-ring residue, 1 × 18 | 18 | ||
| δ-substituents (CH₃ + ring), 2 × 18 | 36 | ||
| Exocyclic C=C, 2 × 5 | 10 | ||
| Calculated λ_(max) | 323 | Calculated λ_(max) (EtOH) | 331 |
For compound 1 the triene contains a diene with both C=C in one ring (ring C), so the homoannular base value is used. Both of its ring-fusion C=C bonds are exocyclic to a neighbouring ring. Compound 2 is a linear dienone: C=O–Cα=Cβ–Cγ=Cδ, with the δ carbon carrying the methyl group and a ring residue.
Conclusion. (i) Phenyl propanoate (about 1765) and the anilide (about 1665) absorb at higher C=O frequency than ethyl benzoate (1720) and the benzamide (1640). (ii) λ_(max): compound 1 = 323 nm, compound 2 = 331 nm.
Application. Woodward–Fieser estimates were used to assign the structures of steroids and terpenoids (for example, ergosterol and the dienones of cortisone synthesis); IR C=O positions identify ester and amide drugs in quality control.
Practising this answer? Write it in a Mains test and get it checked by VKC. Civil Services Mains test series · Telegram
All parts of Question 5
- 5(a) Proteins as dynamic molecules: haemoglobin10 marks
- 5(b) Two epoxides of opposite face from one cyclopentene10 marks
- 5(c) Photoenolisation of 2-methylbenzophenone10 marks
- 5(d) IR of esters and amides; Woodward–Fieser calculations10 marks
- 5(e) NMR of ethene and ethyne; benzyl to tropylium cation10 marks