The reactions of polycyclic aromatic hydrocarbons with OH

The OH radical adds to naphthalene and naphthalene cation without a barrier. For the neutrals, the most favorable path for this intermediate is the loss of the OH, and the next most favorable option is the loss of an H atom to form the alcohol. For the cation, the most favorable path appears to be a hydrogen migration followed by the loss of a hydrogen to form the alcohol. The OH at carbon atom 1 is energetically most favorable for both the inital complex and final product. This is true for both the neutrals and cations. (C) 2000 Elsevier Science B.V. All rights reserved.