Abstract

The photolysis of 1-chloropropane was studied at 184.9, 163.3 and 147 nm in the presence or the absence of a radical scavenger. The main fragmentation process observed at 184.9 nm involves the rupture of the CCl bond (Φ ⩾ 0.5). The n-propyl fragment has a low internal energy content compared with the excess energy of the incident photon so it does not itself fragment (or does so with a low probability). The effect of the irradiation time (or conversion effect) showed the presence of hydrogen chloride although this product was not analysed. The addition of 5% ammonia eliminated this effect. At 147 nm, the main product is ethylene. However, this product does not seem to be the result of the secondary decomposition of the excited n-propyl radical. It could be formed in a primary fragmentation process of the photoexcited molecule leading also to the formation of the methylene radical.

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