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Anion Formation and Ring Opening of 9-Substituted Purines in Liquid Ammonia Containing Potassium Amide
Kos, Nico J.,Plas, Henk C. van der,Blees, Wouter J. F.
, p. 850 - 855 (2007/10/02)
Reaction of 9-methylpurine, 6-chloro-9-methylpurine, and 2',3'-O-isopropylidenenebularine with potassium amide in liquid ammonia leads to opening of the imidazole ring, yielding, after hydrolysis during the workup, 4-(substituted amino)-5-formamidopyrimidines. 6-Chloro-9-methylpurine gives, besides 6-chloro-4-(methylamino)-5-formamidopyrimidine as main product, small amounts of 9-methyladenine and 6-chloro-7,8-dihydro-8-oxo-9-methylpurine.The ring opening will involve adduct formation at position 8.Nebularine, adenosine, and 2',3'-O-isopropylideneadenosine do not react .With a greater excess of potassium amide 2',3'-O-isopropylideneadenosine loses the sugar moiety.The existence of an anion at position 8 can be proved in 9-methylpurine via scavenging with bromobenzene in liquid ammonia containing potassium amide, yielding the 8-phenyl derivative.With 6-chloro-9-(2-tetrahydropyranyl)purine this reaction gives 6-anilino-9-(2-tetrahydropyranyl)purine.Scavenging of 9-methyladenine with bromobenzene gives 6-anilino-9-methylpurine. 1H and 13C NMR spectroscopy confirm that in this strongly basic medium 7- and 9-methyladenine and 6-(methylamino)-9-methylpurine deprotonate at C-8 and lose a proton from the amino group.Both 8-(methylthio)- and 8-amino-9-methylpurine give with potassium amide in liquid ammonia opening of the imidazole ring, yielding 5-(cyanoamino)-4-(methylamino)pyrimidine, which can react further to give either 8-amino-9-methylpurine or 7,8-dihydro-8-oxo-9-methylpurine.
