The degree of unsaturation is 4 and the integration is only half the number of hydrogens in the formula. There is symmetry in the structure. The signal at δ 7.03 indicates 4 equivalent hydrogens on an aromatic ring (deg. of unsat. = 4). This accounts for C₆H₄-, leaving C₂H₆ (2 x CH₃ due to symmetry). There is only one way to place the two methyl groups ( δ 2.28) on an aromatic ring and have the four aromatic hydrogens equivalent. p -Xylene, and not o- or m-xylene is the answer.

The ¹³C NMR spectrum displays only three signals: Two in the aromatic region (134.7 and 129.1 ppm) and one at high field (21.1 ppm). One of the aromatic signals is of weak intensity (no hydrogens attached). The high symmetry excludes m-xylene (5 unique carbons) and o-xylene (4 unique carbons). Return to Menu.