INTEGRATED SPECTRA PROBLEMS

SOLUTION for  PROBLEM #4

• 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 6 types of C, including a C=O (aldehyde or ketone, 213 ppm), and 3 ArC (178, 143 and 126 ppm) plus 2 other hydrocarbon C (63 and 34 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 suggests a para-disubstituted benzene
• This gives the following pieces C₆H₄, C=O, and 2 different -CH₃
• Leads to a suggested molecular formula of C₉H₁₀O which gives a MW = 134 g / mol which is 16 short, suggesting an extra O, probably as part of -OCH₃ (peak at 3.9 ppm)\
• So, molecular formula of C₉H₁₀O₂ which gives a MW = 150 g / mol and IHD = 6
• The ¹³C C=O at 213 ppm is a little high for an ester, so the methoxy group is the second aromatic substituent.
• This gives the structure as:

• 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.