A vacuum ultraviolet photolysis of ethyl bromide at 174.3-174.5 nm was studied over the pressure range of 0.2-38.2 torr at room temperature using a nitrogen resonance lamp.
The pressure effect with and without He as an additive was investigated. A scavenger effect of the reaction was also observed by adding NO as a radical scavenger.
The principal reaction products were $C_2H_4$, $C_2H_6$ and $CH_3CHBr_2$. Increasing pressure without He the product quantum yields, Φ, of $C_2H_4$ and $CH_3CHBr_2$ remain constant, while that of $C_2H_6$ was found to be weak positive pressure dependence. When the pressure of He was varied at a constant pressure of $C_2H_5Br$, however, the quantum yields of all products were found to be pressure independence. Addition of NO completely suppressed the formation of $C_2H_6$ and $CH_3CHBr_2$, and partially reduced that of $C_2H_4$. These results were interpreted in terms of two channel competition between the molecular eliminations and the formation of radicals. Two different decomposition modes were 20% molecular elimination and 80% radical reactions, respectively.
From the aforementioned results, it is suggested that there exist two electronically excited states and that reactant molecule which absorbs a photon proceeds to any one of two different electronically excited states through one of two pathways. One of the electronically formed excited states plays a role as the main source of the molecular elimination products and the other state goes on the radical decomposition modes.