Chapter 13: Spectroscopy

SPECTRA PROBLEM #10 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 (1720 cm^-1), C=C (1640 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 (167 ppm, probably an acid derivative), 2 types in the region of 125-140ppm for C=C, a C-O at 61ppm and 2 hydrocarbon C (18 and 14 ppm)
• H-nmr

d/ppm	multiplicity	integration	assignment
6.1	"multiplet"	1	C=C-H
5.5	"multiplet"	1	C=C-H
4.2	quartet	2	CH2 coupled to 3H, deshielded by O ?
1.9	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 disubstituted alkene (not a benzene system), 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 =  114 g/mol.
• The IR showed the presence of  C=O, C=C and C-O bonds.
• 13C peak at 167 ppm suggests that the C=O is a carboxylic acid derivative
• H nmr also gives a disubstituted alkene, a -O-CH₂-CH₃ and an uncoupled  -CH₃.
• Use this 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: C=O, -O-CH₂-CH₃, -CH₃ and H-C=C-H or H₂C=C.

Altogether....

With the pieces we have :  C=O, -O-CH2-CH3, -CH3 and H-C=C-H or H2C=C.  IR and 13C suggest an ester so we have : CO2CH2-CH3 This means the two substituents on the alkene C=C are the ester group and the -CH3 There are three possible arrangements: 1,1- , cis-1,2- or trans-1,2-  If it were 1,2- then the methyl group would have to have an H neighbour and therefore appear as a doublet (i.e. in a CH3-CH= system).  Therefore, it must be the 1,1-isomer where is has no adjacent H.....

ethyl methacrylate or ethyl 2-methylpropenoate

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