ABSTRACT
The EI-MS of this compound is unusually simple. Since the molecular ion shows up at m/z 95, the compound has an odd number of nitrogen atoms. In addition, the high relative abundance of the M-1 peak (59%) indicates that there is a hydrogen that is easily lost. From the rule of thirteen, C6H9N might be proposed, but the 5 non-equivalent protons in the1H NMR indicate that the formula must be revised. The narrow doublet at 9.5 ppm in1H NMR and the absorptions at 1650 (νC=O), 2833, and 2983 (C-H comb.) cm–1 in the infrared spectrum suggest that there is an aldehyde and, therefore, an oxygen in the molecule. So, after introducing an oxygen in the previous formula we find out that the molecular formula of the unknown compound 7 is C5H5NO, and therefore it has 4 unsaturations. The compound is likely an aldehyde and, thus it has to be a five-membered nitrogen hetero-aromatic compound in order to fit the molecular formula and unsaturations.
