Chapter 9
Addition Reactions of Alkenes
Review of Concepts
Fill in the blanks below. To verify that your answers are correct, look in your textbook at
the end of Chapter 9. Each of the sentences below appears verbatim in the section
entitled Review of Concepts and Vocabulary.
• Addition reactions are thermodynamically favorable at ____ temperature and
disfavored at _____ temperature.
• Hydrohalogenation reactions are regioselective, because the halogen is generally
placed at the ______ substituted position, called _______________ addition.
• In the presence of _____________, addition of HBr proceeds via an antiMarkovnikov addition.
• The regioselectivity of an ionic addition reaction is determined by the preference
for the reaction to proceed through ____________________________________.
• Acid-catalyzed hydration is inefficient when ____________________________
are possible. Dilute acid favors formation of the ___________ and while
concentrated acid favors the ___________.
• Oxymercuration-demercuration achieves hydration of an alkene without
_______________________________________.
• _____________-_______________ can be used to achieve an anti-Markovnikov
addition of water across an alkene. The reaction is stereospecific and proceeds
via a _____ addition.
• Asymmetric hydrogenation can be achieved with a ________ catalyst.
• Bromination proceeds through a bridged intermediate, called a
________________ ______, which is opened by an SN2 process that produces an
_____ addition.
• A two-step procedure for anti dihydroxylation involves conversion of an alkene to
an _________, followed by acid-catalyzed ring opening.
• Ozonolysis can be used to cleave a double bond and produce two ______ groups.
• The position of a leaving group can be changed via ______________ followed by
________________.
• The position of a π bond can be changed via ______________ followed by
________________.
Review of Skills
Fill in the blanks and empty boxes below. To verify that your answers are correct, look
in your textbook at the end of Chapter 9. The answers appear in the section entitled
SkillBuilder Review.
9.1 Drawing a Mechanism for Hydrohalogenation
STEP 1 - DRAW TWO CURVED ARROWS SHOWING
PROTONATION OF THE ALKENE, AND DRAW THE
CARBOCATION THAT IS FORMED.
STEP 2 - DRAW ONE CURVED ARROW THAT SHOWS THE
HALIDE ION ATTACKING THE CARBOCATION, AND DRAW
THE PRODUCT.
H X
+ X
CHAPTER 9
9.2 Drawing a Mechanism for Hydrohalogenation with a Carbocation Rearrangement
STEP 1 - DRAW TWO CURVED ARROWS
SHOWING PROTONATION OF THE
ALKENE AND DRAW THE CARBOCATION
THAT IS INITIALLY FORMED.
H
STEP 2 - DRAW ONE CURVED ARROW
SHOWING A CARBOCATION
REARRANGEMENT AND DRAW THE
RESULTING, MORE STABLE CARBOCATION.
STEP 3 - DRAW ONE CURVED ARROW
SHOWING THE HALIDE ION
ATTACKING THE CARBOCATION, AND
DRAW THE PRODUCT.
Cl
Cl
+
Cl
9.3 Drawing a Mechanism for an Acid-Catalyzed Hydration
STEP 1 - DRAW TWO CURVED ARROWS
SHOWING PROTONATION OF THE
ALKENE, AND DRAW THE RESULTING
CARBOCATION.
H
H
STEP 2 - DRAW ONE CURVED ARROW
SHOWING WATER ATTACKING THE
CARBOCATION, AND DRAW THE
RESULTING OXONIUM ION
O
H
H
O
STEP 3 - DRAW TWO CURVED ARROWS
SHOWING DEPROTONATION OF THE
OXONIUM ION, AND DRAW THE
RESULTING PRODUCT.
H
H
O
H
9.4 Predicting the Products of Hydroboration-Oxidation
DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP
1) BH3 THF
+
2) H2O2, NaOH
RELATIONSHIP = _______________________
9.5
Predicting the Products of Catalytic Hydrogenation
DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP
H2
+
Pt
RELATIONSHIP = _______________________
9.6 Predicting the Products of Halohydrin Formation
DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP
Br2
H2O
+
RELATIONSHIP = _______________________
173
174
CHAPTER 9
9.7 Drawing the Products of Anti Dihydroxylation
DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP
H
1) MCPBA
+
2) H3O+
RELATIONSHIP = _______________________
9.8 Predicting the Products of Ozonolysis
DRAW THE EXPECTED PRODUCTS OF THE
FOLLOWING REACTION.
1) O3
+
2) DMS
9.9 Predicting the Products of an Addition Reaction
DRAW THE EXPECTED PRODUCTS OF THE
FOLLOWING REACTION.
1) BH3 THF
+
2) H2O2, NaOH
9.10 Proposing a One-Step Synthesis
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
OH
9.11 Changing the Position of a Leaving Group
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
Br
1)
Br
2)
9.12 Changing the Position of a π Bond
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
1)
2)
175
CHAPTER 9
Review of Reactions
Identify the reagents necessary to achieve each of the following transformations. To
verify that your answers are correct, look in your textbook at the end of Chapter 9. The
answers appear in the section entitled Review of Reactions.
O
X
H
O
OH
+ En
Br
OH
OH
OH
+ En
OH
OH
+ En
Br
+ En
OH
Br
+ En
Br
Solutions
9.1.
HBr
HBr
Br
a)
b)
HBr
HCl
Br
d)
c)
I
HI
e)
f)
Br
HBr
ROOR
Br
ROOR
Cl
176
CHAPTER 9
9.2.
Br
Br
HBr
HBr
a)
ROOR
b)
9.3.
H Br
Br
Br
+
a)
Cl
H Cl
Cl
+
b)
H Cl
c)
Cl
Cl
+
9.4.
a)
b)
c)
d)
9.5. In this case, the less-substituted carbocation is more stable because it is resonancestabilized:
Cl
H
O
O
O
O
+
Cl
Cl
resonance-stabilized
9.6.
HBr
+
a)
Br
Br
Cl
HCl
b)
HBr
c)
Br
+
Br
177
CHAPTER 9
HI
+
I
d)
I
Cl
HCl
e)
Cl
HCl
Cl
+
f)
9.7.
H
Br
Br
Hydride
Shift
H
a)
H
Br
Br
Br
Hydride
Shift
H
b)
Br
Cl
H
c)
Cl
Methyl
Shift
Cl
9.8.
Br
H
Ring
Expansion
Br
Br
178
CHAPTER 9
9.9.
H
H Br
Methyl
Shift
This rearrangement converts a
secondary carbocation into a more
stable tertiary carbocation.
Hydride Shift
This rearrangement converts
a tertiary carbocation into a
more stable, resonancestabilized, tertiary
carbocation.
Br
Br
9.10.
a)
, because the reaction proceeds via a tertiary carbocation, rather than a
secondary carbocation.
b) 2-methyl-2-butene, because the reaction proceeds via a tertiary carbocation, rather
than a secondary carbocation.
9.11.
a) To favor the alcohol, dilute sulfuric acid (mostly water) is used. Having a high
concentration of water favors the alcohol according to Le Chatelier’s principle.
b) To favor the alkene, concentrated sulfuric acid (which has very little water) is used.
Having a low concentration of water favors the alkene according to Le Chatelier’s
principle.
9.12.
H
H O
H
H
O
H
O H
H
H
O
OH
H
a)
H
H O
H
H
O
H
O H
H
H
O
OH
H
b)
H
H
H O
H
c)
H
O
H
O
H
H
O
OH
H
CHAPTER 9
9.13.
O
H O S O H
O
H
O
Me
O H
Me
9.14.
O
H O S O H
O
OH
OH
H
O
O
H
H
O
(even concentrated H2SO4
has some water present)
9.15.
a)
OH
1) Hg(OAc)2, H2O
2) NaBH4
HO
+
H3O
b)
OH
1) Hg(OAc)2, H2O
2) NaBH4
H 3O +
OH
H
O
Me
OMe
179
180
CHAPTER 9
c)
1) Hg(OAc)2, H2O
OH
2) NaBH4
H 3O +
9.16.
1) Hg(OAc)2 , EtOH
2) NaBH4
OEt
a)
1) Hg(OAc) 2 , EtNH2
2) NaBH4
H
b)
N
Et
9.17.
1) BH3 THF
a)
2) H2O2, NaOH
OH
1) BH3 THF
b)
OH
2) H2O2, NaOH
1) BH3 THF
c)
OH
2) H2O2, NaOH
9.18.
1) BH3 THF
OH
2) H2O2, NaOH
9.19.
1) BH3 THF
a)
2) H2O2, NaOH
+ En
OH
CHAPTER 9
181
OH
1) BH3 THF
b)
+ En
2) H2O2, NaOH
OH
1) BH3 THF
c)
2) H2O2, NaOH
1) BH3 THF
d)
+ En
2) H2O2, NaOH
1) BH3 THF
e)
HO
HO
2) H2O2, NaOH
H
1) BH3 THF
f)
+ En
2) H2O2, NaOH
OH
9.20.
1) BH3 THF
2) H2O2, NaOH
+ En
OH
9.21. Only one chirality center is formed, so both possible stereoisomers (enantiomers)
are obtained, regardless of the configuration of the starting alkene:
1) BH3 THF
2) H2O2, NaOH
+
1) BH3 THF
2) H2O2, NaOH
9.22.
OH
OH
182
CHAPTER 9
9.23.
H2
Ni
a)
H2
Pd
b)
H2
+ En
Pt
c)
H2
+ En
Ni
d)
H2
Pt
e)
H2
Pd
(meso)
f)
9.24.
D
D2
+ En
Pt
D
9.25.
a)
1) BH3 THF
b)
2) H2O2, NaOH
OH
CHAPTER 9
9.26.
Br
Br2
+ En
Br
a)
Br
Br2
+ En
Br
b)
Br
Br
Br2
+ En
c)
Br
Br2
+ En
Br
d)
9.27.
Br
+ En
a)
OH
OH
HO
Br
Br
c)
Br
+ En
d)
OH
9.28.
Br2
OEt
+ En
OH
a)
Br
Et
Br2
N H
+ En
EtNH 2
b)
+ En
+ En
b)
Br
183
184
CHAPTER 9
9.29. The bromonium ion can open (before a bromide ion attacks), forming a resonance
stabilized carbocation. This carbocation is trigonal planar and can be attacked from
either side:
Br
Br2
Br
resonance-stabilized
9.30.
OH
HO
OH
HO
+ En
OH
+ En
a)
b)
OH
c)
OH
OH
OH
OH
OH
d)
e)
+ En
(meso)
OH
f)
9.31.
OEt
1) MCPBA
+ En
2) [H2SO4] ,
OH
a)
OH
O
OH
O
b)
[H2SO4]
OH
9.32.
a)
MCPBA
H3O+
O
HO
OH
no chirality centers
MCPBA
O
H3O+
OH
OH
no chirality centers
+ En
CHAPTER 9
b)
H
O
MCPBA
H
Et
H
Et
+ En
H3O+
HO
H
Et
H
Et
H
OH
meso
9.33.
OH
OsO4 (catalytic)
+ En
NMO
OH
a)
OH
1) OsO4
b)
OH
+ En
2) NaHSO3 / H2O
OH
KMnO4, NaOH
cold
OH
(meso)
c)
OH
KMnO4, NaOH
cold
OH
d)
OH
OsO4 (catalytic)
OH
+ En
OOH , NaOH
e)
OH
OsO4 (catalytic)
f)
NMO
+ En
OH
185
186
CHAPTER 9
9.34.
a)
H
1) O3
O
O
O
2) DMS
O
b)
1) O3
O
O
2) DMS
O
O
c)
H
1) O3
O
O
2) DMS
H
d)
H
1) O3
O
O
2) DMS
H
e)
H
O
H
1) O3
2) DMS
H
meso
O
O
H
O
f)
1) O3
O
2) DMS
O
9.35.
O
1) O3
a)
2) DMS
O
O
1) O3
b)
2) DMS
CHAPTER 9
O
1) O3
c)
O
2) DMS
9.36.
1) BH3 THF
a)
OH
2) H2O2, NaOH
H2
b)
+ En
Pt
HO
1) CH3CO3H
c)
OH
2) H3O+
OH
1) OsO4
d)
+ En
+ En
2) NaHSO3 / H2O
H3O+
OH
OH
+ En
e)
Br
HBr
f)
1) MCPBA
OH
+ En
2) H3O+
OH
g)
OH
1) BH3 THF
h)
2) H2O2, NaOH
OsO4 (catalytic)
+ En
OH
+ En
NMO
i)
HO
187
188
CHAPTER 9
9.37.
OH
KMnO4 , NaOH
OH
+
cold
OH
OH
Diastereomers
9.38. The products are the same:
HO
KMnO4 , NaOH
+ En
cold
OH
(2R,3R)
HO
1) MCPBA
OH
+ En
2) H3O+
(2R,3R)
9.39.
O
+
O
H
1) O3
1) BH3 THF
2) DMS
2) H2O2, NaOH
Compounds E + F
Compound A
Compounds B + C
HBr
Br
Compound D
9.40.
OH
1) BH3 THF
+ En
2) H2O2, NaOH
a)
Br
t-BuOK
b)
HBr
c)
ROOR
Br
+
OH
OH
CHAPTER 9
H2
Pt
d)
Cl
HCl
e)
1) BH3 THF
OH
+ En
2) H2O2, NaOH
f)
Br
NaOEt
g)
Br
HBr
h)
9.41.
Br
HBr
ROOR
a)
HBr
Br
b)
HO
KMnO4 , NaOH
OH
cold
c)
1) MCPBA
HO
2) H3O+
d)
OH
+ En
9.42.
Cl
1) NaOMe
2) HCl
a)
Cl
189
190
CHAPTER 9
1) TsCl, py
OH
2) t-BuOK
HO
3) BH3 THF
4) H2O2, NaOH
b)
Br
1) t-BuOK
HO
2) BH3 THF
3) H2O2, NaOH
c)
OH
1) conc. H2SO4
+ En
2) BH3 THF
OH
3) H2O2, NaOH
d)
9.43.
Br
1) NaOMe
2) HBr, ROOR
a)
Br
1) NaOMe
2) HBr, ROOR
b)
Br
Br
9.44.
1) NaOMe
2) HBr
Br
3) t-BuOK
4) HBr, ROOR
9.45.
a)
1) HBr
2) NaOMe
b)
1) HBr
2) t-BuOK
1) HBr
2) NaOMe
Br
CHAPTER 9
191
9.46.
1) HBr, ROOR
2) t-BuOK
a)
1) HBr
2) NaOMe
b)
9.47.
a)
1) HBr
2) NaOMe
1) HBr
2) NaOMe
3) HBr, ROOR
4) t-BuOK
3) HBr, ROOR
4) t-BuOK
b)
9.48. A reaction is only favorable if ∆G is negative. Recall that ∆G has two
components: (∆H) and (-T∆S). The first term (∆H) is positive for this reaction (two
sigma bonds are converted into one sigma bond and one pi bond). The second term (T∆S) is negative because ∆S is positive (one molecule is converted into two molecules).
Therefore, the reaction is only favorable if the second term is greater in magnitude than
the first term. This only occurs at high temperature.
9.49.
HO
OH
+ En
KMnO4
1) Hg(OAc)2 , H2O
NaOH, cold
HCl
2) NaBH4
H2
Pt
Cl
HO
Br2
H2O
HO
Br
+ En
192
CHAPTER 9
9.50.
HO
1) BH3 THF
1) MCPBA
OH
2) H3O
+ En
+
OH
+ En
2) H2O2, NaOH
HBr
Br2
H2
Pt
Br
Br
Br
+ En
9.51.
a)
H
H O
H
H
O
H
O H
H
H
O
OH
H
b)
H
H O
Hydride
H
H
H
O
H
O H
H
Shift
H
O
OH
c)
H
Br
Br
Br
d)
H
Br
Br
Methyl
Shift
Br
H
CHAPTER 9
9.52.
Br
NaOMe
H2
Pt
Compound A
9.53.
1) HBr
2) NaOMe
a)
1) HBr, ROOR
2) t-BuOK
b)
9.54.
OH
1) Conc. H2SO4
2) HBr, ROOR
3) t-BuOK
9.55. Two different alkenes will produce 2,4-dimethylpentane upon hydrogenation:
9.56.
OH
1) MCPBA
2) H3O+
OH
Compound A
9.57.
a)
1) Conc. H2SO4
2) dilute H2SO4
OH
1) Conc. H2SO4
2) BH3 THF
3) H2O2, NaOH
OH
193
194
CHAPTER 9
b)
1) NaOMe
2) HBr, ROOR
Br
Br
1) NaOMe
2) HBr
c)
Cl
1) NaOMe
2) H2, Pt
d)
OH
OH
OH
1) Conc. H2SO4
2) OsO4, NMO
9.58.
1) HBr, ROOR
2) t-BuOK
a)
OH
1) Conc. H2SO4
OH
2) dilute H2SO4
b)
9.59.
Br
Compound A
dilute H2SO4
NaOMe
Compound B
OH
Compound C
CHAPTER 9
195
9.60.
HBr, ROOR
+ En
Br
HBr
Br
1) BH3 THF
2) H2O2, NaOH
H3O+
OH
+ En
HO
9.61.
excess H2
+
Pt
Diastereomers
9.62. Markovnikov addition of water without carbocation rearrangements can be
achieved via oxymercuration-demercuration:
OH
1) Hg(OAc)2, H2O
2) NaBH4
racemic
9.63.
O
O
H O S O H
O
H
O
H
O
Me
OH
Me O
H
O
Me
OH
MeO
H
196
CHAPTER 9
9.64.
a)
H
H O
H
Methyl
Shift
H
H
OH
H
O
H
b)
Br
H Br
+
Br
9.65.
H2
(PPh3)3RhCl
(meso)
a)
H3O+
OH
b)
1) BH3 THF
c)
OH
2) H2O2, NaOH
HO
1) MCPBA
d)
2) H3O+
OH
+ En
O
H
O
H
197
CHAPTER 9
9.66.
a) Hydroboration-oxidation gives an anti-Markovnikov addition. If 1-propene is the
starting material, the OH group will not be installed in the correct location. Acidcatalyzed hydration of 1-propene would give the desired product.
b) Hydroboration-oxidation gives a syn addition of H and OH across a double bond. This
compound does not have a proton that is cis to the OH group, and therefore,
hydroboration-oxidation cannot be used to make this compound.
c) Hydroboration-oxidation gives an anti-Markovnikov addition. There is no starting
alkene that would yield the desired product via an anti-Markovnikov addition.
9.67.
Br2
Br
Br
Br
Br
(meso)
9.68.
H
H
H
a)
b)
c)
d)
9.69. The reaction proceeds via a resonance-stabilized carbocation, which is even lower
in energy than a tertiary carbocation:
O
H Br
O
resonance-stabilized
O
Br
Br
O
198
CHAPTER 9
9.70.
OH
Br
HBr
ROOR
1) BH3 THF
2) H2O2, NaOH
OH
OsO4, NMO
Cl2
Br2
H2O
HBr
OH
Cl
OH
Cl
Br
Br
OH
H2
Pt
NaOMe
+ En
Br
Br2
H2O
1) O3
O
2) DMS
H
1) BH3 THF
2) H2O2, NaOH
O
+ En
OH
CHAPTER 9
199
9.71.
Br
O
HBr
ROOR
HO
1) BH3 THF
2) H2O2, NaOH
1) O3
2) DMS
HO
OH
HO
Br
OsO4, NMO
Br2
H3O+
HBr
OH
Br
NaOMe
H2 O
H2
Pt
OH
OH
OsO4, NMO
+ En
1) MCPBA
2) H3O+
OH
OH
+ En
9.72. Addition of HBr to 2-methyl-2-pentene should be more rapid because the reaction
can proceed via a tertiary carbocation. In contrast, addition of HBr to 4-methyl-1pentene proceeds via a less stable, secondary carbocation.
9.73.
HS
Br2
+ En
Br
H2S
9.74.
Br
1) NaOMe
2) HBr
3) NaOMe
4) BH3 THF
5) H2O2, NaOH
OH
+ En
200
CHAPTER 9
9.75.
H2
1) BH3 THF
2) H2 O2 , NaOH
Pt
Compound X
OH
2,4-d imethy lpentan-1-ol
H3O+
OH
9.76.
Cl
t-BuOK
HBr
Br
Br
+
+
Br
9.77.
HBr, ROOR
1) O 3
2) DMS
Br
Compound Y
C7H12
H2
Pt
9.78.
H
H
H
H
H
O
+
O
H
CHAPTER 9
9.79.
a)
O
H O S O H
O
OH
OH
O
O
H
O
H O S O
O
O
HO
O
H
HO
b)
O
H O S O H
O
O
O
H
H
O
H
H
HO
HO
(even concentrated H2SO4
has some water present)
9.80.
Cl
O
1) NaOMe
2) O3
3) DMS
H
O
H
201
202
CHAPTER 9
9.81.
a)
Br
Br
Br
Br
H
OH
Br
Br
O
O
OH
b)
O
H O S O H
O
OH
OH
O
H O S O
O
O
H
O
9.82.
I
I
I
I
O
OH
OH
I
H
O
O
I
O
O
O
9.83.
Br
Br
Br
H Br
+ En
+ En
+ En
Br
Br
Br
+ En