Novel Rearrangement During Alkaline Hydrolysis of 1,1,1-Trichloro-2-alken-4-ones
The action of a five-fold excess of potassium hydroxide in methanol on 1,1,1-trichloro-2-penten-4-one (1), 1,1,1-trichloro-6-methyl-2-hepten-4-one (5), and 1,1,1-trichloro-6,6-dimethyl-2-hepten-4-one (9) gives rise to the potassium salt of 5-chloro-2,4-pentadienoic acid (2), the potassium salt and methyl ester of 5-chloro-6-methyl-2,4-heptadienoic acid (7), and the methyl ester of 5-chloro-6,6-dimethyl-2,4-heptadienoic acid (10), respectively.When the same reaction is performed with an equimolar amount of base, 1 is converted into methyl 5-chloro-2,4-pentadienoate (3) as well as 1,1,1-trichloro-2-methoxy-4-pentanone (4); the latter compound (4) is shown to be a true intermediate in the formation of 3.Neither 1,1,1-trichloro-5-methyl-2-hexen-4-one (11) nor 1,1,1-trichloro-5-ethyl-2-hepten-4-one (14) yields the expected dienoic acid or ester when treated with an equimolar amount of potassium hydroxide. Instead, alkaline methanolysis of 11 leads to the formation of a mixture of 1,1,1-trichloro-2-methoxy-5-methyl-4-hexanone (12) and 1,1-dichloro-2,5-dimethoxy-5-methyl-4-hexanone (13), whereas 14 gives predominantly 1,1,1-trichloro-2-methoxy-5-ethyl-4-heptanone (15) and traces of 1,1-dichloro-2,5-dimethoxy-5-ethyl-4-heptanone (16).On the basis of these observations, a mechanism is proposed for the formation of 2, 7, and 10 which requires the presence of two geminal α-hydrogen atoms in the olefinic starting material.