Chapter 13: Spectroscopy

SPECTRA PROBLEM #7 SOLUTION

• In the MS, the molecular ion occurs at 164 indicating the MW = 164 g/mol (even, no isotope pattern for Cl or Br)
• In the IR there are significant absorptions at 1720cm^-1 due to a C=O and at 1600cm^-1 due to a C=C, C-O possible due to bands at 1275 and 1100cm^-1. There is no OH (about 3500cm^-1).
• 13 C-nmr shows 8 types of C. By analysis of the chemical shifts, we have a C=O (probably an acid derivative) at 167 ppm, 4 types in the region 125-145 ppm is typical of aromatic C, 61 ppm is typical of a C-O and those at 22 and 14 ppm most likely from hydrocarbon.
• H-nmr

 

d/ppm	multiplicity	integration	assignment
7.8	"doublet"	2	Ar-H, must be disubstituted, most likely para
7.3	"doublet"	2
4.3	quartet	2	CH2 coupled to 3H, deshielded by O ?
2.4	singlet	3	CH3 with no adjacent H, slightly deshielded
1.4	triplet	3	CH3 coupled to 2H

• The most significant structural information in the H nmr data is that we have a disubstituted phenyl group, 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 =  164 g/mol.
• The IR gave us the C=O and the C-Owhich the C nmr suggests is an acid derviative, such as an ester rather than a ketone (typically > 200ppm).
• 13C peak at 167 ppm suggests that the C=O is a carboxylic acid derivative, an ester ?
• Together the IR and 13C NMR suggests it is an acid derviative, such as an ester rather than a ketone (typically > 200ppm).
• The H nmr gives us -O-CH₂-CH₃ , -CH₃ and C₆H₄-
• Use all the pieces to check the molecular formula = C₁₀H₁₂O₂ = 10 x 12 + 12 x 1 + 1 x 16 = 164 g/mol
• The molecular formula of C₁₀H₁₂O₂ implies an IHD = 5.

Altogether....

So with this information we have the following pieces: C=O, -O-CH2-CH3, -CH3 and C6H4-.  IR and 13C suggest an ester so we have : CO2CH2-CH3  The -CH2- in the ethyl group is deshielded by O and coupled to the -CH3, while the other -CH3 is isolated and only slightly deshielded by the phenyl ring. The number of types of ArC (4) and the coupling in the Ar region of the H nmr (7-8ppm) suggests the para substitution pattern.

ethyl p-toluate or ethyl 4-methylbenzoate

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