The view is prevalent that oxidation in organic chemistry is very often the equivalent of dehydrogenation (1), and this has led to the belief that the intimate mechanistic representation of such reactions should portray the removal of a pair of neutral hydrogen atoms per se from the substance oxidized and their transfer to the substance reduced (2-6). Theories of organic oxidation founded on the free-radical thesis (2-6), which require hemolytic cleavage of valence bonds with the accompanying unpairing of electron spins fail to take into account the fact that the great majority of oxidations brought about by the common strong inorganic oxidants exhibit little of the character of radical reactions,2 and are most effective in strongly polar media where the intermediate formation of ions and their incipient solvolysis is a process requiring a considerably smaller expenditure of energy ( 9) than formation of a radical pair.3A cursory consideration of the simple fact that paraffin hydrocarbons are the most oxidatively resistant organic compounds known suffices to reveal immediately that the abstraction of hydrogen atoms (of which the paraffins have the most) has little if, indeed, anything to do with the process of oxidation.The organic compounds most easily oxidized, such as olefins, acetylenes, alcohols, phenols, aldehydes, amines, alkyl halides, thiols, sulfides, disulfides, and sulfoxides, all have one thing in common, namely-an excess of valence electrons over and above those needed for bonding. This, of course, suggests at once that the true mechanism of organic oxidation may well be identical with the well-recognized fundamental definition of oxidation, i.e., a loss of electrons.But if this be the case, how, then, can there be a plausible explanation for the fact that the partially oxidized terminal carbon atom of RCH2OH or RCH2NH2 is more susceptible to oxidation than the corresponding C atom in the paraffin RCH3; for, based on a consideration of the relative electronegativities of the atoms involved, one would necessarily reach the conclusion that, by reason of «+ a-j+ athe inductive effect, the resulting polarities are RCH2OH and RCH2NH2, (both having C's with a lower electron density than in RCH3) and that consequently, the alcohol and amine should be less susceptible to oxidative attack at