Chapter 13: Spectroscopy

SPECTRA PROBLEM #6 SOLUTION

• In the MS, the molecular ion occurs at m/z = 108 indicating the MW = 108 g/mol (even, no isotope pattern for Cl or Br)
• The IR shows an O-H (broad 3330 cm^-1) and a C-O at 1020 cm^-1. There is no carbonyl C=O (around 1715 cm^-1)
• 13 C-nmr shows 5 types of C, including 4 ArC between 125-145 ppm and a deshielded C (65 ppm)
• H-nmr

d/ppm	multiplicity	integration	assignment
7.2-7.4	"singlet"	5	5 ArH = monosubstituted
4.6	singlet	2	deshielded CH2 no coupling
2.3	singlet	1	H  no coupling (-OH ?)

• The H-NMR gives us a monosubstituted benzene ring, C₆H₅-, a -CH₂- and a peak that could match the -OH seen in the IR.

Summary....

• The MS indicated MW =  108 g/mol.
• The IR gave us an -OH but no C=O, so looks like some type of alcohol.
• The 13C and H NMR show a benzene system C₆H₅-.
• The H NMR gives another -CH₂-
• Use this to check the molecular formula : C₇H₈O = 7 x 12 + 8 x 1 + 1 x 16 = 108 g/mol
• This confirms that we have the molecular formula

Altogether....

With the pieces we have :  C6H5-, -CH2- and -OH These can actually only be put together in one way....

benzyl alcohol or phenylmethanol

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