[1]Annulene is an aromatic compound containing 18 carbon atoms. The annulene molecule has an almost — Organic Chemistry Chemistry Question
Nuclear magnetic resonance (NMR) spectrum of [1] annulene
[1]Annulene is an aromatic compound containing 18 carbon atoms. The annulene molecule has an almost planar structure with 6 inner hydrogens (Hin) and 12 outer hydrogens (Hout). The 1H NMR spectra of [1]annulene at 213 K and 383 K are shown in Fig. 31.1.
Hin
Hout
[1]annulene
Fig. 31.1 [VISUAL]
In the spectrum obtained at 213 K, estimate the area ratio of the peaks at 9.3 and -3.0 ppm.
Model Answer
[1]Annulene has 6 Hin and 12 Hout. Therefore, the larger resonance at 9.3 ppm can be assigned to Hout, while the smaller resonance at –3.0 ppm can be assigned to Hin.
Thus, A(9.3 ppm) / A(–3.0 ppm) = 12 / 6 = 2
Note:
The ring current from the aromatic 18π system of [1]annulene enhances the magnetic field outside the ring and diminishes the field inside the ring. This phenomenon is also responsible for the clear peak assignments. The downfield peaks (9.3 ppm) and the upfield peaks (-3.0 ppm) are assigned to Hout and Hin, respectively.
Explain why the spectrum obtained at 383 K has only one singlet peak while that obtained at 213 K has two multiplet peaks.
Model Answer
The conformarional mobility of [1]annulene allows the exchange of Hin and Hout by ring inversion. At 213 K, the slow exchange of Hin and Hout does not show any effect on the NMR spectra. Therefore, the Hin and Hout resonances are observed at -3.0 ppm and 9.3 ppm, respectively. In contrast, at 383 K, this exchange occurs rapidly and repeatedly; therefore, the resonance is observed at the weighted average of the chemical shifts of Hin and Hout.
Estimate the position of the singlet peak in the spectrum obtained at 383 K.
Model Answer
[9.3 ppm × 12 + (–3.0 ppm) × 6 ] / [12 + 6 ] = 5.2 ppm
Notice the difference in the numbers of Hin and Hout.