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CHAPTER 5 157 
 
 
 
 
 
5.80. 
(a) Glucuronolactone 1L is the enantiomer of 1D, which is shown in the problem statement. To draw the enantiomer of 
1D, we simply redraw it, except that we replace all dashes with wedges, and all wedges with dashes, as shown: 
 
 
 
 
(b) There are 5 chiral centers, so there are 32 (or 25) possible stereoisomers. 
 
(c) 
 
 
 
(d) The four products that are accessible from either of the reactants are the four products shown on the right in the 
solution to part c, as indicated above. Recall that the synthetic protocol allows for control of configurations at C2, C3 
and C5, but not at C4. Therefore, in order for a specific stereoisomer to be accessible from either 1D or from 1L, that 
stereoisomer must display a specific feature. To understand this feature, we must draw one of the ten stereoisomers 
and then redraw it again after rotating it 180 degrees about a vertical axis. For example, let’s do this for one of the 
meso compounds: 
 
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