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CHAPTER 8 245 
 
water, H2O, is used as the reagent). If ethanol (EtOH) is 
used as the reagent instead of water, then Z = OEt, so we 
expect Markovnikov addition of ethanol (H and OEt) 
across the alkene, with the ethoxy (OEt) group being 
installed at the more substituted (secondary) position, 
rather than the less substituted (primary) position. 
 
 
 
(b) Oxymercuration-demercuration involves the 
addition of H-Z across the double bond (where Z = OH 
when water, H2O, is used as the reagent). If ethylamine 
(EtNH2) is used as the reagent instead of water, then Z = 
NHEt, so we expect Markvnikov addition of H and 
NHEt across the alkene, with the ethylamino group 
(NHEt) being installed at the more substituted 
(secondary) position, rather than the less substituted 
(primary) position. 
 
 
 
8.14. 
(a) Hydroboration-oxidation results in the anti-
Markovnikov addition of water (H and OH) across the  
bond. That is, the OH group is installed at the less-
substituted (primary) position, rather than the more 
substituted (tertiary) position: 
 
 
 
(b) Hydroboration-oxidation results in the anti-
Markovnikov addition of water (H and OH) across the  
bond. That is, the OH group is installed at the less-
substituted (primary) position, rather than the more 
substituted (tertiary) position: 
 
 
 
(c) Hydroboration-oxidation results in the anti-
Markovnikov addition of water (H and OH) across the  
bond. That is, the OH group is installed at the less-
substituted (primary) position, rather than the more 
substituted (secondary) position: 
 
 
 
8.15. There is only one alkene that will afford the 
desired product upon hydroboration-oxidation: 
 
 
 
8.16. 
(a) The reagents indicate a hydroboration-oxidation. 
The net result of this two-step process is the anti-
Markovnikov addition of H and OH across the  bond. 
That is, the OH group is installed at the less-substituted 
position, while the H is installed at the more substituted 
position. In this case, two chiral centers are created. 
Therefore, the stereochemical requirement for syn 
addition determines that the H and OH are added on the 
same face of the alkene, giving the following products: 
 
 
 
In this case, it might seem as if there was an anti 
addition, rather than a syn addition, because we see that 
the product has one wedge and one dash. But this is an 
optical illusion. Recall, that most hydrogen atoms are 
not drawn in bond-line drawings, so the H that was 
added during the process has not been drawn. However, 
if we draw that hydrogen atom, we will see that the H 
and OH were indeed added in a syn fashion: 
 
 
 
(b) The reagents indicate a hydroboration-oxidation. 
The net result of this two-step process is the anti-
Markovnikov addition of H and OH across the  bond. 
That is, the OH group is installed at the less-substituted 
position, while the H is installed at the more substituted 
position. In this case, only one chiral center is created. 
Since syn addition can take place from either face of the 
alkene with equal likelihood, we expect a pair of 
enantiomers, as shown: 
 
 
(c) The reagents indicate a hydroboration-oxidation. 
The net result of this two-step process is the anti-
Markovnikov addition of H and OH across the  bond. 
That is, the OH group is installed at the less-substituted 
position, while the H is installed at the more substituted 
position. In this case, no chiral centers are created, so 
the requirement for syn addition is irrelevant. 
 
 
 
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