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266 CHAPTER 8 
 
8.73. 
(a) Compound X reacts with H2 in the presence of a 
catalyst, so compound X is an alkene. The product of 
hydrogenation is 2-methylbutane, so compound X must 
have the same carbon skeleton as 2-methylbutane: 
 
 
 
We just have to decide where to place the double bond in 
compound X. Keep in mind that the following two 
positions are identical: 
 
 
 
So, there are only three possible locations where we can 
place the double bond: 
 
 
 
 
 
 
(b) Upon hydroboration-oxidation, only one of the three 
proposed alkenes will be converted to an alcohol without 
any chiral centers, shown below. Each of the other two 
compounds will be converted into an alcohol with a 
chiral center. 
 
 
 
 
 
 
 
 
 
 
 
8.74. The following is one possible suggested route: 
 
 
 
Other acceptable solutions are certainly possible. For 
example, after the first step (elimination with tert-
butoxide), the next two steps (addition of HBr, followed 
by elimination) could be replaced with acid-catalyzed 
hydration, followed by elimination with conc. H2SO4. 
 
 
8.75. There is only one alkene (compound X, shown 
below) that can be converted to 2,4-dimethyl-1-pentanol 
via hydroboration-oxidation. Treatment of that alkene 
with aqueous acid affords an alcohol (via Markvonikov 
addition): 
 
Compound X
1) BH3 THF
2) H2O2,
NaOH
H2
Pt
OH
2,4-dimethyl-1-pentanol
H3O+
OH
 
 
8.76. The substrate is a secondary alkyl halide, and treatment with tert-butoxide gives the less substituted alkene. 
When that alkene is treated with HBr, the  bond is protonated to give a secondary carbocation (rather than a primary 
carbocation). This carbocation can either be captured by a bromide ion, giving products A and B below, or the 
carbocation can undergo a rearrangement (hydride shift) to give a tertiary carbocation, which is then captured by a 
bromide ion, affording product C. 
 
 
 
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