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342 Organic Chemistry Solutions Manual K. Kondo et al., Tetrahedron The second reaction is base-catalysed and starts with the hydrolysis of the ester by Lett., 1978, 907. This fragmentation also needs 'push', though only a three-membered ring is being brok because the leaving group is an enolate, nowhere near as electron-withdrawing as the molecule or even the carbocation of the first example. Are they fragmentations? In both C-C bond is broken but we would understand if you felt the first was not strictly fragmentation, especially if it goes stepwise. Neither breaks the molecule into three pieces it's all a matter of opinion. 0 0 0 NaOH product 0 Problem 11 CO₂Me What steps would be necessary to carry out an Eschenmoser fragmentation on this ketone and what products would be formed? 0 Purpose of the problem Revision of an important and complex reaction involving a fragmentation. Suggested solution The Eschenmoser fragmentation uses a tosyl hydrazone of an α,β-epoxy-ketone (p. 1008). The epoxide can be made with alkaline hydrogen peroxide and the tosylhydrazone needs just tosylhydrazine to form what is simply an imine. Then the fun can begin. The doesn't matter for once. CO₂Me CO₂Me TsNH.NH₂ NaOH 0 0 D N-NH Ts The fragmentation is initiated with base that removes a proton from the NHTs group. This fragments the molecule one way with the epoxide as leaving group and then the oxyanion the molecule the other way with nitrogen gas and Ts⁻ as leaving groups. It comes as no surprise the same Eschenmoser invented the other double fragmentation in Problem 6 of this chapter. The product is an acetylenic ketone or, as here, aldehyde. CO₂Me first base fragmentation 0 0 N-NH N-N Ts Ts CO₂Me second 0 fragmentation 0 H N=N Ts