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Solutions to Problems 265 Thus, for humulene we cleave at the three double bonds, giving three "dicarbonyl" compounds that we may draw by copying the structural portions from the starting molecule, as follows: cleave cleave 0 1. 2. O cleave 74. = 30 + 2 = 32; degree of unsaturation = (32 - 24)/2 = 4 π bonds or rings. Reaction 1 tells you two π bonds are present (hydrogenation only adds two H₂), so there must be two rings. Reaction 2 forms two pieces: formaldehyde (CH₂O) and the triketone shown, which is C₁₄H₂₂O₃. These account for all the carbons and hydrogens in caryophyllene. The oxygens came from the ozonolysis. All that's left to decide is how the four carbonyl carbons originally connected up into two alkene double bonds. Reaction 3 gives the answer to this last question. Hydroboration converts one of the caryophyllene double bonds into an alcohol. Then, ozonolysis cleaves the other to give the diketo-alcohol shown. Working backward from this you can write the following: O CH₃ C CH₃ CH₃ CH₃ CH₃ CH₃ H HC O CH₂OH CH₂OH Before ozonolysis CH₃ CH₃ CH₃ H = C₁₅H₂₄! CH₂ Before hydroboration The only question left is whether the double bond in the nine-membered ring is cis or trans (more properly, Z or E)-it could be either. In fact, this is the difference between caryophyllene (the E isomer) and the isocaryophyllene (Z): CH₃ CH₃ CH₃ CH₃ H H H H CH₃ CH₃ CH₂ CH₂ Caryophyllene Isocaryophyllene