Abstract
In the reduction of 1-methyl-3- and 4-methoxycarbonylpyridinium iodides in methanol by NaBH4, it was found that the 4-substitute gave only 1, 2, 5, 6-tetrahydropyridine, and that the 3-substitute gave 1, 2, 5, 6-tetrahydropyridine and 1, 6-dihydropyri dine compounds. When this reaction was carried out in sodium hydroxide methanol, no tetrahydro compound was obtained, but 1, 2- and 1, 6-dihydro compounds were obtained. When the molar ratio of pyridinium salt to NaBH4 was varied, the product was examined quantitatively and the dihydro derivative isolated was found not to undergo NaBH4 reduction in methanol.
The result showed that the reductive reaction of pyridinium salt by NaBH4 was taken place at the first stage, between pyridinium cation and BH4 anion, followed by the reaction between dihydro compound produced and BH3. When the methoxy carbonyl group was present at the terminal of the double bond in enamine part of the dihydro compound, and when the reaction was taken place in sodium hydroxide alkali medium, any reduction into tetrahydro compound by BH3 has not been found to take place. This fact may be explained by the following reactivity with BH3
OH->>N-CH=CH->MeOH>N-CH=C-COOMe
