Chapter 13: Spectroscopy

SPECTRA PROBLEM #9 SOLUTION

• In the MS, the molecular ion occurs at m/z = 150 indicating the MW = 150 g / mol.
• The IR shows an C=O (1680 cm^-1) and possible C-O (1250-1000 cm^-1) but no -OH or -NH  (around 3500 cm^-1)
• 13 C-nmr shows 7 types of C, including a C=O (196 ppm), and 3 ArC (163, 131, 130 and 114 ppm) plus 2 other hydrocarbon C (55 and 26 ppm)
• H-nmr

d/ppm	multiplicity	integration	assignment
8.0	doublet	2	2 x ArH
7.0	doublet	2	2 x ArH
3.9	singlet	3	isolated CH3 deshielded, possibly by O
2.6	singlet	3	isolated CH3 slightly deshielded

• The lack of a peak at about 9 ppm rules out an aldehyde and therefore suggests a ketone.
• The peaks at 7-8 ppm in the H-nmr and the 4 peaks in the aromatic region of the ¹³C-nmr suggests a para-disubstituted benzene
• This gives the following pieces C₆H₄, C=O, and 2 different -CH₃

Summary....

• The MS indicated MW =  150 g/mol.
• The IR showed the presence of  C=O and maybe C-O bonds.
• The ¹³C C=O at 196 ppm is a little high for an ester, and is more indicative of an aldehyde or a ketone.
• H nmr gave the following pieces C₆H₄, C=O, and 2 different -CH₃
• Leads to a suggested molecular formula of C₉H₁₀O = 9 x 12 + 10 x 1 + 1 x 16 = 134 g/mol.
• This is 16 short, suggesting an extra O, which if we look at the H-NMR is probably as part of -OCH₃ (peak at 3.9 ppm).
• So with all this information we have the following pieces: C₆H₄, C=O, -CH₃ and -OCH₃

Altogether....

With the pieces we have :  C6H4, C=O, -CH3 and -OCH3 IR and 13C suggest a ketone rather than an ester. H-nmr and 13C-nmr suggest a para-disubstituted benzene Checking the H-nmr, we can see the two uncoupled methyl groups, one deshielded by O, the other slightly deshielded by the C=O and the para-disubstituted benzene.

p-methoxyacetophenone

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