Chapter 13: Spectroscopy

SPECTRA PROBLEM #8 SOLUTION

• In the MS, the molecular ion occurs at m/z = 114 indicating the MW = 114 g / mol (even, no isotope pattern for Cl or Br)
• The IR shows a carbonyl C=O (1770 cm^-1), C=C (1650 cm^-1) and possible C-O (between 1200-1000 cm^-1). No -OH or -NH (above 3200 cm^-1).
• 13 C-nmr shows 6 peaks indicating 6 types of C, a C=O (171 ppm, probably an acid derivative), 2 types in the region of  for C=C, one at 141 ppm the other at 97 ppm plus 3 other hydrocarbon C (61, 18 and 14 ppm)
• H-nmr

d/ppm	multiplicity	integration	assignment
7.4	doublet of doublets	1	=CH coupled to 2 different H
4.9	doublet	1	=CH coupled to H
4.6	doublet	1	=CH coupled to H
2.4	triplet	2	CH2 coupled to 2H, slightly deshielded
1.7	sextet	2	CH2 coupled to 5H
1.0	triplet	3	CH3 coupled to 2H

• The most significant structural information in the H nmr data is that we have an monosubstituted alkene HC=CH₂ (not a benzene system), a propyl group : -CH₂-CH₂-CH₃

Summary....

• The MS indicated MW =  114 g/mol.
• The IR showed the presence of  C=O, C=C and C-O bonds.
• 13C peak at 171 ppm suggests that the C=O is a carboxylic acid derivative, probably an ester : RCO₂R'
• H nmr gives a monosubstituted alkene HC=CH₂,  and  -CH₂-CH₂-CH₃.
• Use these pieces to check the molecular formula : C₆H₁₀O₂ = 6 x 12 + 10 x 1 + 2 x 16 = 114 g/mol
• So with all this information we have the following pieces: -CO₂-, -CH₂-CH₂-CH₃, and HC=CH₂.

Altogether....

We have :  -CO2-, -CH2-CH2-CH3, and HC=CH2.  IR and 13C suggest an ester so we have : RCO2R'  The key is to sort out which group is connected to the carbonyl C=O and which is connected to the -O- of the ester....  The -CH2- at 2.4 ppm is not deshielded enough to be -OCH2- (nearer to 4 ppm), so we must have CH3-CH2-CH2-C=O and O-HC=CH2 these connect to give the answer.

vinyl butanoate or ethenyl butanoate

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