CHEM 313 Chapter : Chem_313_PS5_2014_answers.pdf
Document Summary
Chem 313, problem set #5, answers 2014: mechanism of this rearrangement, it is due to epimerization of c-2 going through enediol. Ch2oh: retro-aldol where here b represent oh-: In step (a), it is ok if you use a general base (b) instead of the lysine (lys420) and using proton transfer instead of describing the water capturing the proton. In step (b), it is acceptable to again use a general base to abstract the proton instead of describing the histidine (his306"). Than, you can use the conjugated acid (bh+) or proton transfer to put back the proton in step d instead of describing the histidine (his306") and glutamic (glu285). The sugar cannot ring open because the anomeric carbon is linked to bn, thus no reaction with nabh4 as well as with br2. There is a typo mistake in product of phcocl/pyridine, on c-2 hydroxyl, it should be. The most stable conformation of the two fused ring is described below.
