¹H NMR: δ 7.61 (5H, m), 2.89 (2H, d, J = 7 Hz), 2.29 (1H, nonet, J = 7 Hz), and 1.34 (6H, d, J = 7 Hz)

¹³C NMR: 141.8, 129.4, 128.4, 126.0, 45.8, 30.5, and 22.7 ppm.

The degree of unsaturation is 4. The ¹³C spectrum shows only 7 of 10 carbons. There are 3 pairs of identical carbons. The doublet in the ¹H spectrum at δ 1.34 is two equivalent methyl groups coupled by 7 Hz to a single hydrogen, the one at δ 2.29. This gives us (CH₃)₂CH-. Since the signal at δ 2.29 is a nonet with J = 7 Hz, it must be coupled to eight other protons. We have accounted for six, so the remaining two are at δ 2.89. Now we have (CH₃)₂CHCH₂-, (C₄H₉). This isobutyl group has two equivalent methyl groups. The residual formula is C₆H₅, which must contain the four degrees of unsaturation. The chemical shift of these five hydrogens suggests a phenyl group as does the appearance of only four different carbon singlets in the aromatic region of the the ¹³C spectrum. Return to Menu.