✍ Scribed by Bruno M. Vittimberga; Donald F. Sears
No coin nor oath required. For personal study only.
When 6‐phenanthridinecarbonitrile (3) is irradiated at 2537 Å in neutral 9:1 2‐propanol/water, three major products are formed. These are dimethyl‐(6‐phenanthridinyl)methanol (4), phenanthridine (5) and 6,6′‐biphenanthridine (6). When benzophenone is present in the reaction mixture, diphenyl‐(6‐phenanthridinyl)‐methanol is also formed. 6‐Phenanthridinyl radical which is common to the formation of all these products, is formed by a monophotonic process involving hydrogen atom abstraction from an alcohol molecule by an excited state of 3. Unlike what is generally found with other nitrogen‐heterocycles, the photochemistry of 3 appears to involve only a π,π* singlet state. The fluorescence of 3 is quenched with the triplet quencher cis/trans‐piperylene as a function of the concentration of the diene without the accompaniment of an exci‐plex emission.
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## Abstract The physical and photophysical properties of 6‐phenanthridinecarbonitrile (**1**) have been examined. We previously reported that when **1** is irradiated in aqueous 2‐propanol, three products are formed [3]. These include dimethyl‐6‐phenanthridinylcarbinol (2), phenanthridine (3) and 6
The C6H 6 and I z molecular product fragments of the C6H6-I2 charge-transfer state have been investigated. Wavelength resolved studies show considerable rotational excitation of the benzene fragment and corresponding velocity spectra indicate a degree of translational excitation as well, consistent