Ondas de Alta Energia

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330 Organic Chemistry Solutions Manual This work allowed the synthesis Finally, the weak base bicarbonate is enough to remove a proton from the cation has and trial of morphine analogues as painkillers in the Bristol Laboratories by nitrogen participation in the displacement of bromide. The five-membered heterocycle expands The T of Canada (I. Moncovic et al., J. Am. to six-membered. This alkene is formed because the bond to the tertiary carbon is broker Chem. Soc., 1973, 95, 7910). preferentially. Br NH NH NH NaHCO₃ H The Problem 10 but in Give mechanisms for these reactions, commenting on any regio- and stereoselectivity. What controls the rearrangement? The case OH 0 0 OH 0 0 H OH 1. 0 Ho 0 0 pyridine + H 2. 0 OAc Pro Purpose of the problem Exploring the 'semipinacol' rearrangement with some revision of stereochemistry and acetal formation. Suggested solution The formation of the acetal is by the usual mechanism (p. 342) with one twist. The double bond is in conjugation with the ketone in the starting material but has moved in the acetal. There is no longer any conjugation possible but how and why does it move? How is easy. At several stages in the acetal mechanism, intermediates are formed that can enolize we show one possibility. The Purpo conjugated and the unconjugated intermediate are in equilibrium and either can cyclize to the To sho acetal. So the 'unconjugated' acetal must be more stable. This is partly because one of the way bonds must be axial and it is better for the acetal centre to be on a true chair than on a twist-chair conformation. Sugge H The + 0 0 matter OH OH easy OAc OAc H 0 0 OH 0 The cannot OAc OAc The next stage is a standard dihydroxylation with OsO₄. The two OH groups must be added in a cis fashion but why do they add to the top face of the alkene? This is easier to see in conformational drawing. The top face of the alkene is virtually unhindered whereas the bottom face