1 (261)

Prévia do material em texto

CHAPTER 8 253 
 
8.33. syn-Dihydroxylation of trans-2-butene results in 
the same products as anti-dihydroxylation of cis-2-
butene, as shown below. The configuration of each 
chiral center has been assigned to demonstrate that the 
products are indeed the same for these two reaction 
sequences: 
 
KMnO4 , NaOH
cold
HO
OH
+ En
1) RCO3H
2) H3O+
+ En
HO OH
(2R,3R)
(2R,3R)
 
 
8.34. Compound A must be an alkene (because it 
undergoes reactions that are typically observed for 
alkenes, such as hydroboration-oxidation, 
hydrobromination and ozonolysis). So, we begin by 
drawing all possible alkenes with the molecular formula 
C5H10 (using a methodical approach similar to the one 
described in the solution to Problem 4.3): 
 
 
Among these isomers, only the last two will afford a 
tertiary alkyl halide upon treatment with HBr. And 
among these two alkenes, only the latter will undergo 
ozonolysis to produce a compound with three carbon 
atoms and another compound with two carbon atoms. 
Now that we have identified the starting alkene, we can 
draw the products B-F, as shown here: 
 
 
8.35. 
(a) The two groups being added across the alkene are H 
and OH. The OH group is installed at the less-
substituted carbon atom, so we must use conditions that 
give an anti-Markovnikov addition of H and OH. This 
can be accomplished via hydroboration-oxidation. The 
reaction proceeds via a syn addition, which can occur on 
either face of the alkene, giving a pair of enantiomers: 
 
 
 
(b) This reaction involves elimination of H and Br to 
give the less-substituted alkene, so a sterically hindered 
base (such as tert-butoxide) is required: 
 
 
 
(c) The two groups being added across the alkene are H 
and Br. The Br group is installed at the less-substituted 
carbon atom, so we must use conditions that give an 
anti-Markovnikov addition of H and Br. This can be 
accomplished by treating the alkene with HBr in the 
presence of peroxides. 
 
 
 
(d) The two groups being added across the alkene are H 
and H, which can be accomplished by treating the alkene 
with molecular hydrogen (H2) in the presence of a 
suitable catalyst. 
 
 
 
(e) The two groups being added across the alkene are H 
and Cl. The latter is installed at the more-substituted 
carbon atom, so we must use conditions that give a 
Markovnikov addition of H and Cl. This can be 
accomplished by treating the alkene with HCl. 
 
 
 
(f) The two groups being added across the alkene are H 
and OH. The OH group is installed at the less-
substituted position, so we must use conditions that give 
an anti-Markovnikov addition of H and OH. Also, the H 
and OH are added in a syn fashion (this can be seen more 
clearly if you draw the H that was installed, as shown 
www.MyEbookNiche.eCrater.com