Q6(a). Structure of C₇H₉N from UV, IR and NMR: p-toluidine
On the basis of the given spectroscopic data, deduce the structure of C₇H₉N:
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UV &: λ_(max) = 290 nm (in ethanol)
IR (ν) &: 3440 cm⁻¹, 3360 cm⁻¹, 3025 cm⁻¹, 2920 cm⁻¹, 1625 cm⁻¹, 1380 cm⁻¹, 810 cm⁻¹
1H-NMR (ppm) &: δ_(2.16), s, 3H; δ_(3.24), s, 2H; δ_(6.37), d, 2H (J = 8.0 Hz); δ_(6.79), d, 2H (J = 8.0 Hz)
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Answer in brief
C₇H₉N is p-toluidine (4-methylaniline), CH₃-C₆H₄-NH₂: DBE 4 (benzene), primary NH₂ (3440/3360 cm⁻¹, δ 3.24), ring CH₃ (δ 2.16) and a para-disubstituted ring (810 cm⁻¹; two doublets, J = 8 Hz).
Key points
- Introduction. Use the formula first, then let each spectrum fix one piece of the structure.
- Four degrees of unsaturation with only seven carbons point to a benzene ring (one ring + three C=C) with no other multiple bond.
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. Use the formula first, then let each spectrum fix one piece of the structure.
DBE = (2C + 2 + N - H)/(2) = (2(7) + 2 + 1 - 9)/(2) = 4
Four degrees of unsaturation with only seven carbons point to a benzene ring (one ring + three C=C) with no other multiple bond.
| Data | Inference |
| 3440, 3360 cm⁻¹ | two N–H stretches (asymmetric and symmetric): a primary amine, -NH₂ |
| 3025 / 2920 cm⁻¹ | aromatic C–H / aliphatic (CH₃) C–H |
| 1625 cm⁻¹ | N–H bending (scissoring) overlapping aromatic C=C stretch |
| 1380 cm⁻¹ | symmetric C–H bending of a CH₃ group |
| 810 cm⁻¹ | strong C–H out-of-plane bend of two adjacent ring H: 1,4-(para)-disubstituted benzene |
| UV 290 nm | benzene band shifted to longer wavelength by an auxochrome (NH₂ lone pair conjugated with the ring), as in aniline |
| δ 2.16 (s, 3H) | Ar-CH₃ (no neighbouring H, so a singlet) |
| δ 3.24 (s, 2H) | Ar-NH₂: broad singlet, exchanges with D₂O |
| δ 6.37, 6.79 (d, 2H each, J = 8.0 Hz) | four ring H as two doublets with ortho coupling: the AA'BB' pattern of a para-disubstituted ring |
Assembling. C₆H₄ (para) + CH₃ + NH₂ = C₇H₉N. The doublet at δ 6.37 is from the two H ortho to NH₂: the strong +M effect of the amino group shields them. The doublet at δ 6.79 is from the two H ortho to CH₃.
Figure: 4-Methylaniline (p-toluidine).
Isomers ruled out. o- and m-Toluidine would show four different ring H and no 810 cm⁻¹ para band; benzylamine (PhCH₂NH₂) would give five ring H and a CH₂ singlet near δ 3.8; N-methylaniline would show only one N–H stretch, one N–H proton and an N-CH₃ near δ 2.8.
Conclusion. C₇H₉N is p-toluidine (4-methylaniline), CH₃-C₆H₄-NH₂: DBE 4 (benzene), primary NH₂ (3440/3360 cm⁻¹, δ 3.24), ring CH₃ (δ 2.16) and a para-disubstituted ring (810 cm⁻¹; two doublets, J = 8 Hz).
Application. p-Toluidine is an intermediate for azo dyes and for toluidine blue; the same set of bands identifies aromatic amines in drug and dye analysis.
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All parts of Question 6
- 6(a) Structure of C₇H₉N from UV, IR and NMR: p-toluidine15 marks
- 6(b) NBS (radical and ionic), KMnO₄/NaIO₄, OsO₄/H₂O₂15 marks
- 6(c) Nylon-6 from cyclohexanone: Beckmann rearrangement20 marks