Abstract
Photodissociation of the i-C3H7O radical is investigated using fast beam photofragment translational spectroscopy. Neutral i-C3H7O radicals are produced through the photodetachment of a fast beam of i-C3H7O- anions and are subsequently dissociated using 248 nm (5.0 eV). The dominant product channels are CH3 + CH3CHO and OH + C3H6 with some contribution from H + C3H6O. CH3 and H loss are attributed to dissociation on the ground electronic state of i-C3H7O, but in a nonstatistical manner because RRKM dissociation rates exceed the rate of energy randomization. Translational energy and angular distributions for OH loss are consistent with ground state dissociation, but the branching ratio for this channel is considerably higher than predicted from RRKM rate calculations. These results corroborate what has been observed previously in C2H5O dissociation at 5.2 eV that yields CH3, H, and OH loss. Additionally, i-C3H7O undergoes three-body fragmentation to CH3 + CH3 + HCO and CH3 + CH4 + CO. These three-body channels are attributed to dissociation of i-C3H7O to CH3 + CH3CHO, followed by secondary dissociation of CH3CHO on its ground electronic state.