Abstract

The pathway in the formation of cyanogen chloride from the reaction of 1-naphthol (1) and 4-phenylimidazole with chloramine was investigated. The intermediates isolated in the reaction of 1-naphthol (1) with chloramine were N-chloro-1, 2-naphthoquinone 2-imine (2) and o-carboxycinnamonitrile (6), the latter of which, liberating cyanogen chloride, was finally converted tophthalide-3-carboxylic acid (8). The products obtained in the reaction of 4-phenylimidazole (10) with chloramine were benzonitrile (14), benzoylformic acid (22) and benzoic acid (23) besides cyanogen chloride. Cyanogen chloride formed by the reaction of 4-methylimidazole with [15N]- chloramine was C14NCl. From these results the pathway of cyanogen chloride formation was elucidated.

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