INTEGRATED SPECTRA PROBLEMS

SOLUTION for  PROBLEM #3

• 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

• From the data, we can get a molecular formula of C₅H₁₀O which matches the MW = 86 g / mol
• This has an IHD = 1, which is accounted for by the p bond of the C=O.
• The following pieces can be deduced from the spectra  C=O, 2 different -CH₃,  and 2 different -CH₂-.
• Given these pieces, we can start to fit them together based on the coupling patterns and chemical shifts as:  CH₃-C=O and CH₃-CH₂-CH₂-
• There is only one way they can all fit together:

• 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.