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Discussion

A compound, C15H16O2, has the following spectral data;

1H NMR (ppm): 9.16 (s), 6.89 (d, J = 8 Hz), 6.64 (d, J = 8 Hz), 1.53 (s)
13C NMR (ppm): 154.7, 140.9, 127.1, 114.4, 40.7, 30.7

The structure of the compound is

  • A.2,2-Bis(4-hydroxyphenyl)propane, two para-hydroxyphenyl rings joined by a CMe2 carbon
  • B.Bis(4-methoxyphenyl)methane, two para-methoxyphenyl rings joined by a CH2 group
  • C.Diphenyl ketal, Ph2C(OMe)2, a carbon bearing two phenyl and two methoxy groups
  • D.Bis(4-hydroxy-3-methylphenyl)methane, two rings each with OH and an ortho methyl, joined by CH2

Answer: A

The compound is (A), bisphenol A, 2,2-bis(4-hydroxyphenyl)propane.

  • 1H: 9.16 (s) is the two phenolic OH protons (typical in DMSO); 6.89 and 6.64 (d, J = 8 Hz) are the two pairs of ortho-coupled ring protons of a para-disubstituted benzene, the 6.64 signal being ortho to OH; 1.53 (s) is the six protons of two equivalent methyl groups on a quaternary carbon.
  • 13C: six signals for a symmetric molecule: 154.7 (C-OH), 140.9 (ring carbon bearing the CMe2), 127.1 and 114.4 (ring CH), 40.7 (quaternary CMe2) and 30.7 (CH3).

All four structures fit C15H16O2, so the spectra decide. (B) would show OCH3 and ArCH2Ar singlets near 3.7 to 3.8 ppm and no OH. (C) has monosubstituted phenyl rings, not two doublets, and methoxy singlets near 3 ppm. (D) would show three aromatic signals per ring, an aryl methyl near 2.1 ppm and a CH2 near 3.7 ppm, not a six-proton singlet at 1.53.

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