Prévia do material em texto
CHAPTER 23 985 The first possibility is not viable because Heck reactions are stereoselective with respect to the starting alkene, and the E isomer (rather than the Z isomer) would be expected: This is not the desired stereochemical outcome, so we turn our attention to the second possibility. This possibility works, and it has the added advantage that the starting alkene is monosubstituted, with an electron- withdrawing substituent (so we expect the reaction to be very effective). (f) Analyze the structure of the product and determine which bond(s) can be made via a Heck reaction. Recall that a Heck reaction forms a bond between a vinyl position and either an aryl, vinyl or benzyl position. In the desired compound, there is only one bond that fits this criterion: This bond, indicated with a wavy line, is between a benzyl group and a vinyl group. To make this bond via a Heck reaction, we must start with an organohalide and an alkene. Since one of the coupling partners must be an alkene, the other partner (in this case) must be a benzyl halide, as shown: Notice that the alkene is monosubstituted, with an electron-withdrawing substituent, so we expect the reaction to be very effective. 23.25. In a Heck reaction, an alkene is coupled with an organohalide. In this case, the two coupling partners are tethered together in the same compound: Upon treatment with a base and a suitable palladium catalyst, an intramolecular Heck reaction can occur. Notice that the Z configuration of the vinyl iodide is preserved. The new C-C bond is formed regioselectively and stereoselectively, with the E configuration being formed: 23.26. Begin by identifying the carbon atoms that will be joined. The carbon atom connected to the iodide will be joined directly to the proximal carbon of the alkene (a bond will be formed between the two highlighted positions): Next, consider the intermediate that forms when the palladium catalyst interacts with compound 1. www.MyEbookNiche.eCrater.com