Spectra Problem #3 Solution

Chapter 13: Spectroscopy

SPECTRA PROBLEM #3 SOLUTION

In the MS, the molecular ion occurs at m/z = 86 indicating the MW = 86 g / mol.
The IR shows an C=O (1715 cm^-1) and but no -OH or -NH (around 3500 cm^-1) or C-O (1250-1000 cm^-1)
13 C-nmr shows 5 types of C, including a C=O (210 ppm), and 4 other hydrocarbon C (55, 34, 22 and 18 ppm)
H-nmr

d/ppm	multiplicity	integration	assignment
2.4	triplet	2	CH₂ coupled to 2H, slightly deshielded
2.1	singlet	3	isolated CH₃
1.7	hextet	2	CH₂ coupled to 5 H
0.9	triplet	3	CH₃ coupled to 2H

The following pieces can be deduced from the H-NMR spectra 2 different -CH₃, and 2 different -CH₂-.
Given these pieces, we can start to fit them together based on the coupling patterns: an uncoupled CH₃- and a CH₃-CH₂-CH₂- group.

Summary....

The MS indicated MW = 86 g/mol.
The IR showed the presence of a C=O bond.
13C peak at 210 ppm suggests that the C=O is an aldehyde or ketone.
H nmr gives an uncoupled -CH₃and a -CH₂-CH₂-CH₃group.
Note we can rule out it being an aldehyde (no peak at 9-10 ppm in the H-NMR).
Use what we have to check the molecular formula : C₅H₁₀O = 5 x 12 + 10 x 1 + 1 x 16 = 86 g/mol
This has an IHD = 1, and is consistent with the p bond of the C=O.

Altogether....

With the pieces we have : C=O, -CH₂-CH₂-CH₃, and -CH₃.
IR and 13C is consistent with a ketone.
Given these pieces, there is only one way they can fit together:

With the H-NMR chemical shifts we can see the CH₃-C=O
Checking the H-NMR, the middle -CH₂- is actually six lines since it has a total of 5 similar neighbours. The C=O slightly deshields the other adjacent groups.

2-pentanone
or
propyl methyl ketone