The degree of unsaturation is 4. The 2H doublets at δ 6.79 and 7.08 are coupled to each other (confirm the coupling constants in the spectrum). They are aromatic hydrogens. We have accounted for C₆H₄-, leaving C₂H₆O (CH₃- and CH₃O-). Only the p-isomer of methoxytoluene fits the data. Why are the blue hydrogens assigned as ortho to the methyl group and not ortho to the methoxy group? Check the solution to the previous exercise.

The ¹³C spectrum of 4-methylanisole (3) has six singlets and eight carbons. Because of the plane of symmetry in the molecule, two pairs of aromatic carbons are identical. Two signals, 157.6 ppm and 129.8 ppm, are weak signals. These signals are assigned to the substituted carbons of the aromatic ring (no hydrogens attached to aid in relaxation). Which one is which? Examining the shifts in anisole (2), we see that 159.7 ppm is a good match for carbon 157.6 ppm in 3. Thus, 129.8 ppm must be the carbon (ipso) attached to the methyl in 3. The signals at 113.8, 130.0 and 55.1 ppm in 3 match with their respective signals in 2: 114.0, 129.1, and 55.1 ppm. Notice that the ortho, meta, and para carbons of toluene (1) have about the same chemical shift whereas anisole (2), bearing the stronger electron-donating methoxy group, has the ortho and para carbons at higher field (shielded) while the meta carbon of both 1 and 2 are at nearly the same value (128.3 and 129.5 ppm, respectively). The signal, 129.8 ppm, in 3 does not match the 137.8 ppm signal in toluene 1 because the electron donation of the para-methoxy group in 3 shields (upfield shift) this position. The 137.8 ppm carbon (ipso) in 1 and the 120.7 ppm carbon (para) in 2 average to a value of 129.2 ppm. This value is in good agreement with the observed value for the equivalent carbon (129.8 ppm) in 3. You may wish to compare the ¹³C chemical shifts here with those in p-xylene. Return to Menu.