Abstract

The photodissociation of p-methoxytoluene and p-methoxybenzyl alcohol at 266 nm in n-heptane solution is studied by nanosecond fluorescence and absorption spectroscopy. The formation of a p-methoxybenzyl radical is identified by its fluorescence which is induced by excitation at 308 nm. The yields of the radical are of the order of ∼10−3 for dissociation of p-methoxytoluene and p-methoxybenzyl alcohol. The growth rate of 1.5×108 s−1 for the radical is equal to the decay rate of (1.5±0.3)×108 s−1 for the precursor fluorescence in dissociation of p-methoxytoluene, whereas the growth rate of >1.0×109 s−1 for the radical is much faster than the decay rate of (1.8±0.3)×108 s−1 for the precursor fluorescence in dissociation of p-methoxybenzyl alcohol. The formation of the radical depends linearly on the photolysis pulse fluence for dissociation of p-methoxytoluene and p-methoxybenzyl alcohol. The data show existence of two distinct dissociation channels. p-Methoxytoluene dissociates from thermally equilibrated levels of the S1 state after vibrational relaxation, whereas p-methoxybenzyl alcohol dissociates from vibrationally excited levels of the S1 state in competition with vibrational relaxation. The difference of these channels is explained on a model of electronic coupling between the precursor and product states in the geometry where the C–H and C–O bonds are stretched in a plane perpendicular to the benzene rings. For p-methoxytoluene, the S1 state does not correlate adiabatically to the ground state of the C–H bond fission products, so intersystem crossing or internal conversion precedes dissociation. For p-methoxybenzyl alcohol, avoided crossing between the ππ* (benzene) configuration and the np(O)σ*(C–O) repulsive configuration results in the adiabatic potential-energy surface which evolves to the ground state of the C–O bond fission products allowing rapid dissociation.

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