Chapter 13: Spectroscopy

SPECTRA PROBLEM #4 SOLUTION

• In the MS, the molecular ion occurs at m/z = 88 indicating the MW = 88 g/mol (even, no isotope pattern for Cl or Br)
• The IR shows a carbonyl C=O (1740 cm^-1), and possible C-O (between 1250-1000 cm^-1). No -OH or -NH (above 3200 cm^-1).
• 13 C-nmr shows 4 peaks indicating 6 types of C, a C=O (171 ppm, probably an acid derivative), a C-O at 60 ppm and 2 hydrocarbon C (21 and 14 ppm)
• H-nmr

d/ppm	multiplicity	integration	assignment
4.1	quartet	2	CH2 coupled to 3H, deshielded by O ?
2.0	singlet	3	CH3 with no adjacent H, slightly deshielded
1.3	triplet	3	CH3 coupled to 2H

• The most significant structural information in the H nmr data is that we have an ethyl group : -CH₂-CH₃ most likely as an ethoxy group: -O-CH₂-CH₃ and an isolated methyl group : -CH₃

Summary....

• The MS indicated MW =  88 g/mol.
• The IR showed the presence of  C=O and C-O bonds, possibly an ester
• 13C peak at 171 ppm suggests that the C=O is a carboxylic acid derivative, an ester ?
• H nmr also gives a -O-CH₂-CH₃ and an uncoupled  -CH₃.
• Use this to check the molecular formula : C₄H₈O₂ = 4 x 12 + 8 x 1 + 2 x 16 = 88 g/mol
• So with all this information we have the following pieces: C=O, -O-CH₂-CH₃, and -CH₃.

Altogether....

With the pieces we have :  C=O, -O-CH2-CH3, and -CH3.  IR and 13C suggest an ester so we have : CO2CH2-CH3 This only leaves the -CH3 group.  Therefore, it must be CH3-CO2CH2-CH3

ethyl ethanoate or ethyl acetate

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