Chem 351 F10 Final : Spectroscopy

Part 8: SPECTROSCOPY

The following data is available from the question.

Note : Remember to cross reference things to confirm.... e.g. IR may show C=C, use NMR to confirm that etc.

MS: M+ seen at 144 g/mol. MW is even, so N rule implies no N (or an even number of N atoms)

IR: There is a C=O at 1730 cm^-1 which is a little high for a simple ketone. The only other absorbances of any note are 1209 and 1161 cm^-1 which are probably C-O.

13C nmr: The proton decoupled spectrum shows a total of 7 peaks indicating 7 types of C. By analysis of the chemical shifts, we have a C=O at 208ppm, suggesting an aldehyde or ketone and another C=O at 173ppm suggesting a carboxylic acid derivative, peaks at about 62ppm (possibly carbons with an electronegative atoms attached) and then 4 peaks between 0-40 ppm that are most likely from a hydrocarbon portion.

1H nmr: The proton spectrum shows a total of 5 peaks indicating 5 types of H.

d/ppm	multiplicity	integration	Inference
4.12	q	2	CH₂ coupled to 3H, deshielded
2.75	t	2	CH₂ coupled to 2H, slightly deshielded
2.58	t	2	CH₂ coupled to 2H, slightly deshielded
2.2	s	3	CH₃ not coupled
1.25	t	3	CH₃ coupled to 2H

(q = quartet, t = triplet, d = doublet, s = singlet)

The most significant structural information from this are:

there are 3 hydrocarbon units: CH₃-, CH₃CH₂- and -CH₂CH₂-

Summary....
The MS indicated MW = 144, no Cl or Br
The IR showed the presence of C=O, possible C-O
13C NMR shows 2 x C=O, one aldehyde or ketone and one carboxylic acid derivative. A C-O is also possible.
Since there is no H NMR peak between 9-10ppm, it is not an aldehyde, therefore one of the C=O is a ketone.
H NMR gives CH₃-, CH₃CH₂- and -CH₂CH₂-

This information suggests a molecular formula = C₇H₁₂O₃ which matches the MW = 144.

Altogether...

The fragments we have are:

CH₃-, CH₃CH₂- and -CH₂CH₂-, C=O, C=O, and -O-

The chemical shift of the CH₂ in the CH₃CH₂ group requires that it be attached to the O, so we have an ethoxy group, CH₃CH₂O-

The IR and C-NMR data suggest that an ester is present, so given the ethoxy group, it's an ethyl ester: -CO₂CH₂CH₃

The C- and H-NMR data indicate that the other C=O is from a ketone (13C > 190ppm and no H 9-10ppm). This means that the CH₃- and -CH₂CH₂- need to be connected to the other C=O.

The final step should always be to check what you have drawn. The easiest thing to check is usually the coupling patterns you would expect to see, and the chemical shifts of each unit. You should be asking yourself : "Does my answer give me what the H-nmr shows ?"