Chapter 13: Spectroscopy

SPECTRA PROBLEM #1 SOLUTION

• In the MS, the molecular ion occurs at m/z = 72 indicating the MW = 72 g / mol.
• The IR shows a carbonyl C=O (1720 cm^-1), no -OH or -NH (above 3200 cm^-1), no C=C (1600 cm^-1)
• 13 C-nmr shows 4 types of C, including a C=O (210 ppm), and 3 hydrocarbon C (37, 29 and 8 ppm)
• H-nmr

d/ppm	multiplicity	integration	assignment
2.4	quartet	2	CH2 coupled to 3H
2.1	singlet	3	isolated CH3
1.1	triplet	3	CH3 coupled to 2H

• The following pieces can be deduced from the H-NMR spectra : an isolated -CH₃, and a -CH₂CH₃.

Summary....

• The MS indicated MW =  72 g/mol.
• The IR showed the presence of a C=O bond.
• 13C peak at 210 ppm suggests that the C=O is an aldehyde or ketone.
• H nmr gives an uncoupled  -CH₃ and an -CH₂-CH₃
• Note we can rule out it being an aldehyde (no peak at 9-10 ppm in the H-NMR)
• Use what we have to check the molecular formula : C₄H₈O = 4 x 12 + 8 x 1 + 1 x 16 = 72 g/mol
• This has an IHD = 1, and is consistent with the p bond of the C=O.

Altogether....

With the pieces we have :  C=O, -CH2-CH3, and -CH3.  IR and 13C is consistent with a ketone.  Given these pieces, there is only one way they can fit together:

2-butanone or ethyl methyl ketone

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