Abstract
The gas‐phase photodissociation of (CH3)2Zn, (C2H5)2Zn, and (n‐C3H7)2Zn has been examined at 248 nm using laser‐induced fluorescence to detect the monoalkyl zinc radical and zinc atom photoproducts. For each compound, the monoalkyl zinc radical is the primary photoproduct and is formed sufficiently hot that it spontaneously dissociates to an alkyl radical and a Zn atom without absorption of a second photon. Photodissociation was examined in the presence of He buffer gas to measure the probability of quenching the secondary spontaneous dissociation of the monoalkyl zinc species. For all three dialkyl zinc compounds, the probability of quenching the secondary dissociation step increases substantially over the He pressure range of 0–400 Torr. The quenching probability vs He pressure was fit using RRKM theory in conjunction with a time‐dependent master equation, treating the nascent vibrational energy distribution of the monoalkyl zinc product as an adjustable function. The quenching data for C2H5Zn and n‐C3H7Zn can be fit only if it is assumed that these species are formed with a hot, narrow vibrational energy distribution, much narrower than that predicted by phase‐space theory. A dissociation mechanism involving crossover from an optically prepared singlet state to a repulsive triplet state is proposed to explain this observation. Spontaneous dissociation of CH3Zn is quenched much more strongly by He than is calculated using any reasonable vibrational energy distribution function for CH3Zn. This is attributed to the inapplicability of RRKM theory to reactions involving very low‐state‐density molecules like CH3Zn.