Q8(a). Structure of C₇H₅Cl₃O from its NMR spectrum
An organic compound having molecular formula C₇H₅Cl₃O exhibited the following 1H-NMR data (ppm):
δ_(3.9), s; δ_(6.8), d (J = 8.0 Hz); δ_(7.3), d (J = 8.0 Hz)
Deduce the structure of the compound.
Answer in brief
The compound is 2,3,4-trichloroanisole (1,2,3-trichloro-4-methoxybenzene): OCH₃ singlet at δ 3.9 (3H); two mutually ortho-coupled ring H (J = 8 Hz), H-6 at δ 6.8 (ortho to OMe) and H-5 at δ 7.3 (ortho to Cl).
Key points
- δ 3.9, singlet. The only non-aromatic signal. A singlet at 3.9 with no neighbours is an aryl methoxy group, Ar-OCH₃ (3H). The extra C and the O are thus used up as OCH₃.
- the ring carries two H and three Cl. The two doublets are these two H, one each.
- J = 8.0 Hz. Both H are doublets with the same 8 Hz coupling: an ortho coupling (3J, 7–9 Hz). The two H are on adjacent carbons and each has no other H neighbour (meta coupling would be about 2–3 Hz, para about 0).
- Placing OMe. One H is at δ 6.8, shielded by the +M effect of OMe: it is ortho to OMe. The other, at δ 7.3, is ortho to Cl. So the H atoms are at C-6 and C-5, and Cl occupies C-2, C-3 and C-4.
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. Work from the formula: unsaturation, then the kind of each signal, then the substitution pattern from the coupling.
DBE = (2C + 2 - H - X)/(2) = (2(7) + 2 - 5 - 3)/(2) = 4
Four degrees of unsaturation with seven carbons: a benzene ring and no other multiple bond. Six C are in the ring, so one C, the O and five H remain outside or on the ring.
- δ 3.9, singlet. The only non-aromatic signal. A singlet at 3.9 with no neighbours is an aryl methoxy group, Ar-OCH₃ (3H). The extra C and the O are thus used up as OCH₃.
- Ring H count. C₇H₅ - OCH₃ leaves C₆H₂: the ring carries two H and three Cl. The two doublets are these two H, one each.
- J = 8.0 Hz. Both H are doublets with the same 8 Hz coupling: an ortho coupling (3J, 7–9 Hz). The two H are on adjacent carbons and each has no other H neighbour (meta coupling would be about 2–3 Hz, para about 0).
- Placing OMe. One H is at δ 6.8, shielded by the +M effect of OMe: it is ortho to OMe. The other, at δ 7.3, is ortho to Cl. So the H atoms are at C-6 and C-5, and Cl occupies C-2, C-3 and C-4.
Isomers ruled out. Of the six trichloroanisoles, only 2,3,4- and 2,3,6- have two ortho ring H. In the 2,3,6-isomer neither H is ortho to OMe (both near δ 7.1–7.2). 2,4,6- and 3,4,5- give one 2H singlet; 2,4,5- gives two singlets; 2,3,5- gives two meta doublets (J≈2.5 Hz).
Figure: 2,3,4-Trichloroanisole: H-6 at δ 6.8, H-5 at δ 7.3.
Another formula fitting C₇H₅Cl₃O. 4-(Trichloromethyl)phenol (HO-C₆H₄-CCl₃) also has a para pattern (two 2H doublets) and an OH singlet. It is rejected because the H ortho to CCl₃ would lie well above δ 7.5 and the OH signal is broad, concentration-dependent and lost on shaking with D₂O, unlike a sharp methoxy singlet at 3.9. Supporting test. The mass spectrum would show M⁺˙ at m/z 210 with a Cl₃ cluster M: M+2: M+4: M+6 ≈ 100: 96: 31: 3.
Conclusion. The compound is 2,3,4-trichloroanisole (1,2,3-trichloro-4-methoxybenzene): OCH₃ singlet at δ 3.9 (3H); two mutually ortho-coupled ring H (J = 8 Hz), H-6 at δ 6.8 (ortho to OMe) and H-5 at δ 7.3 (ortho to Cl).
Application. Chloroanisoles (formed by fungal methylation of chlorophenol wood preservatives) are the musty “cork taint” compounds of wine; NMR coupling patterns of this kind identify the exact substitution pattern of such pollutants.
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All parts of Question 8
- 8(a) Structure of C₇H₅Cl₃O from its NMR spectrum15 marks
- 8(b) 1H NMR of the three bromochloroethylenes15 marks
- 8(c)(i) Monomers of PS, LDPE, PP, PMMA and Buna-S10 marks
- 8(c)(ii) Photodecarbonylation; trapping the o-xylylene10 marks