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 anti- Markovnikov 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 S N 2 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 X H 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. + X
Transcript
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 anti-
Markovnikov 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
XH
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.
+ X
CHAPTER 9 173
9.2 Drawing a Mechanism for Hydrohalogenation with a Carbocation Rearrangement
H Cl
Cl+
Cl
STEP 1 - DRAW TWO CURVED ARROWS
SHOWING PROTONATION OF THE
ALKENE AND DRAW THE CARBOCATION
THAT IS INITIALLY FORMED.
STEP 3 - DRAW ONE CURVED ARROW
SHOWING THE HALIDE ION
ATTACKING THE CARBOCATION, AND
DRAW THE PRODUCT.
STEP 2 - DRAW ONE CURVED ARROW
SHOWING A CARBOCATION
REARRANGEMENT AND DRAW THE
RESULTING, MORE STABLE CARBOCATION.
9.3 Drawing a Mechanism for an Acid-Catalyzed Hydration
H O
H
H
OHH
OHH
STEP 1 - DRAW TWO CURVED ARROWS
SHOWING PROTONATION OF THE
ALKENE, AND DRAW THE RESULTING
CARBOCATION.
STEP 3 - DRAW TWO CURVED ARROWS
SHOWING DEPROTONATION OF THE
OXONIUM ION, AND DRAW THE
RESULTING PRODUCT.
STEP 2 - DRAW ONE CURVED ARROW
SHOWING WATER ATTACKING THE
CARBOCATION, AND DRAW THE
RESULTING OXONIUM ION
9.4 Predicting the Products of Hydroboration-Oxidation
DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP
+
RELATIONSHIP = _______________________
1) BH3 THF
2) H2O2, NaOH
9.5 Predicting the Products of Catalytic Hydrogenation
H2
Pt
DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP
+
RELATIONSHIP = _______________________
9.6 Predicting the Products of Halohydrin Formation
Br2
H2O
RELATIONSHIP = _______________________
DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR RELATIONSHIP
+
174 CHAPTER 9
9.7 Drawing the Products of Anti Dihydroxylation
H+
RELATIONSHIP = _______________________
1) MCPBA
2) H3O+
DRAW THE EXPECTED PRODUCTS OF THE FOLLOWING REACTION, AND DETERMINE THEIR 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
OH
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
9.11 Changing the Position of a Leaving Group
Br
Br
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
1)
2)
9.12 Changing the Position of a π Bond
IDENTIFY REAGENTS THAT WILL ACHIEVE THE FOLLOWING TRANSFORMATION:
1)
2)
CHAPTER 9 175
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.
OH
OH
OH
OH
O
O
H
OH
Br
X
Br
OH
Br
Br
OH
+ En
+ En
+ En
+ En
+ En
Solutions
9.1.
a)
HBr Br
b)
HBr
ROORBr
c)
HBr Br
d)
HCl Cl
e) HI
I
f)
HBr
ROOR
Br
176 CHAPTER 9
9.2.
a)
BrHBr
b)
BrHBr
ROOR
9.3.
a)
H Br BrBr+
b)
ClH Cl
Cl+
c) Cl
H Cl
Cl+
9.4.
a) b) c) d)
9.5. In this case, the less-substituted carbocation is more stable because it is resonance-
stabilized:
O
HCl
ClO
+O
resonance-stabilized
O
Cl
9.6.
a)
HBr
Br Br
+
b) HCl Cl
c)
HBr Br
Br+
CHAPTER 9 177
d)
HI
I
I
+
e) HCl
Cl
f) HCl
Cl Cl+
9.7.
a)
H Br BrBr
H
Hydride
Shift
b)
H Br BrHydride
ShiftH
Br
c)
H Cl ClMethyl
Shift
Cl
9.8.
BrHBr
BrRing
Expansion
178 CHAPTER 9
9.9.
H
Br
Br
H BrMethyl
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, resonance-stabilized, tertiary
carbocation.
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.
a)
O OHH O
H
H HO
H
H
H
HO
H
b)
H O
H
H HO
HHO
H O
H
H OH
c)
OHH O
H
H HO
H HO
HO
H H
CHAPTER 9 179
9.13.
H O S O
O
O
H
HO
Me O
Me
H OMeHO
Me
9.14.
O
OH
H O S O
O
O
H
OH
OHH
OH
(even concentrated H2SO4
has some water present)
9.15.
a)
2) NaBH4
1) Hg(OAc)2, H2O
H3O+
HO
OH
b)
2) NaBH4
1) Hg(OAc)2, H2O
H3O+
OH
OH
180 CHAPTER 9
c)
2) NaBH4
1) Hg(OAc)2, H2O
H3O+
OH
9.16.
a)
2) NaBH4
1) Hg(OAc)2 , EtOH
OEt
b)
2) NaBH4
1) Hg(OAc)2 , EtNH2
NEtH
9.17.
a)
1) BH3 THF
2) H2O2, NaOHOH
b)
1) BH3 THF
2) H2O2, NaOH
OH
c)
1) BH3 THF
2) H2O2, NaOH
OH
9.18.
OH1) BH3 THF
2) H2O2, NaOH
9.19.
a)
1) BH3 THF
2) H2O2, NaOHOH
+ En
CHAPTER 9 181
b)
1) BH3 THF
2) H2O2, NaOH
OH
+ En
c)
1) BH3 THF
2) H2O2, NaOH
OH
d)
1) BH3 THF
2) H2O2, NaOH
HO
+ En
e)
1) BH3 THF
2) H2O2, NaOH
HO
f)
1) BH3 THF
2) H2O2, NaOH
H
OH+ En
9.20.
1) BH3 THF
2) H2O2, NaOHOH
+ En
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
OH1) BH3 THF
2) H2O2, NaOH
OH
+
9.22.
182 CHAPTER 9
9.23.
a)
H2
Ni
b)
H2
Pd
c)
H2
Pt+ En
d)
H2
Ni+ En
e)
H2
Pt
f)
H2
Pd
(meso)
9.24.
D2
Pt+ En
D
D
9.25.
a)
b)
1) BH3 THF
2) H2O2, NaOH OH
CHAPTER 9 183
9.26.
a)
Br2
Br
Br+ En
b)
Br2
Br
Br
+ En
c)
Br2
Br
Br
+ En
d)
Br2Br
Br
+ En
9.27.
a) HO
Br
+ En
b)
OH
Br+ En
c)
OH
Br
+ En
d)
Br
OH
+ En
9.28.
a)
Br2
OHBr
OEt
+ En
b)
Br2
EtNH2
Br
N
+ En
Et
H
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:
Br2
Br
Br
resonance-stabilized
9.30.
a)
HOOH
+ En
b)
OHHO
+ En c)
OH
OH+ En
d)
OH
OH
e)
OH
OH
(meso) f)
OH
OH
+ En
9.31.
a)
1) MCPBA
OH
OEt
OH2) [H2SO4] ,
+ En
b)
O
OH
[H2SO4] OH
O
9.32.
a)
MCPBA O H3O+
H3O+
MCPBA O
OHHO
OH
OH
no chirality centers
no chirality centers
CHAPTER 9 185
b)
MCPBA O
HEt H
Et
H3O+
meso
H
H
+ En
HO
HEt
HEt
OH
9.33.
a)
OsO4 (catalytic)
NMO
OH
OH
+ En
b)
1) OsO4
2) NaHSO3 / H2O
OH
OH
+ En
c)
KMnO4, NaOH
cold
OH
OH
(meso)
d)
KMnO4, NaOH
cold
OH
OH
e)
OsO4 (catalytic)
OOH , NaOH
OH
OH
+ En
f)
OsO4 (catalytic)
NMO
OH
OH
+ En
186 CHAPTER 9
9.34.
a)
O
O
OO
H
1) O3
2) DMS
b)
1) O3
2) DMS
O O
O O
c)
1) O3
2) DMS
O
H
O
H
d)
1) O3
2) DMS
O
O
H
H
e)
1) O3
2) DMSOO
H
H
HO
HO
meso
f) 1) O3
2) DMSO O
9.35.
a)
O O1) O3
2) DMS b)
1) O3
2) DMS
O
CHAPTER 9 187
c)
O O1) O3
2) DMS
9.36.
a)
1) BH3 THF
2) H2O2, NaOH
OH
b)
H2
Pt+ En
c)
1) CH3CO3H
2) H3O+
HO
OH + En
d)
1) OsO4
2) NaHSO3 / H2O
OH
OH
+ En
e)
H3O+OH
+ En
f)
HBrBr
g)
1) MCPBA
2) H3O+
OH
OH
+ En
h)
1) BH3 THF
2) H2O2, NaOH
OH
+ En
i)
OsO4 (catalytic)
NMO
OH
HO
+ En
188 CHAPTER 9
9.37.
KMnO4 , NaOH
cold+
OH
OH
OH
OH
Diastereomers
9.38. The products are the same:
KMnO4 , NaOH
cold
HO
OH
+ En
1) MCPBA
2) H3O+ + En
HO OH
(2R,3R)
(2R,3R)
9.39.
Compound A
1) BH3 THF
2) H2O2, NaOHOH
+OH
HBr
1) O3
2) DMSH
OO
Compounds E + F Compounds B + C
+
Br
Compound D
9.40.
a)
OH
+ En1) BH3 THF
2) H2O2, NaOH
b)
Br t-BuOK
c)
HBr
ROOR
Br
CHAPTER 9 189
d)
H2
Pt
e)
Cl
HCl
f)
OH+ En
1) BH3 THF
2) H2O2, NaOH
g)
Br NaOEt
h)
Br
HBr
9.41.
a)
HBr
ROOR
Br
b)
HBr Br
c)
KMnO4 , NaOH
cold
HO OH
d)
1) MCPBA
2) H3O+ + En
HO OH
9.42.
a) Cl
Cl1) NaOMe
2) HCl
190 CHAPTER 9
b)
OH
HO
1) TsCl, py
2) t-BuOK
3) BH3 THF
4) H2O2, NaOH
c)
Br
HO1) t-BuOK
2) BH3 THF
3) H2O2, NaOH
d)
OH
OH
+ En1) conc. H2SO4
2) BH3 THF
3) H2O2, NaOH
9.43.
a)
Br 1) NaOMe
2) HBr, ROORBr
b)
1) NaOMe
2) HBr, ROOR
Br
Br
9.44. 1) NaOMe
2) HBr
3) t-BuOK
4) HBr, ROOR
Br
Br
9.45.
a)
2) NaOMe
1) HBr
b)
2) NaOMe
1) HBr
2) t-BuOK
1) HBr
CHAPTER 9 191
9.46.
a)
1) HBr, ROOR
2) t-BuOK
b)
1) HBr
2) NaOMe
9.47.
a)
1) HBr
2) NaOMe
3) HBr, ROOR
4) t-BuOK b)
1) HBr
2) NaOMe
3) HBr, ROOR
4) t-BuOK
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.
KMnO4
H2
Pt
HCl
2) NaBH4
Br2
H2O
1) Hg(OAc)2 , H2O
NaOH, cold
HO OH
+ En
ClBr
HO
+ En
HO
192 CHAPTER 9
9.50.
H2
Pt
HBr Br2
1) MCPBA
2) H3O+
1) BH3 THF
2) H2O2, NaOH
OH
HO
+ En
Br Br
Br
+ En
OH+ En
9.51.
a)
H O
H
H HO
H HO
HO
H
H OH
b)
H O
H
H
H
HO
H O
H
H
OH
HO
H
Hydride
Shift
c)
BrH Br Br
d)
H Br BrBr
Methyl
Shift
CHAPTER 9 193
9.52.
NaOMeBr H2
Pt
Compound A
9.53.
a) 2) NaOMe
1) HBr
b) 2) t-BuOK
1) HBr, ROOR
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.
1) MCPBA
2) H3O+
Compound A
OH
OH
9.57.
a)
OH
OH
1) Conc. H2SO4
2) dilute H2SO4
1) Conc. H2SO4
2) BH3 THF
3) H2O2, NaOH
194 CHAPTER 9
b)
Br
Br
1) NaOMe
2) HBr, ROOR
1) NaOMe
2) HBr
c)
Cl
1) NaOMe
2) H2, Pt
d)
OH
OHOH
1) Conc. H2SO4
2) OsO4, NMO
9.58.
a)
1) HBr, ROOR
2) t-BuOK
b)
OH OH1) Conc. H2SO4
2) dilute H2SO4
9.59.
NaOMe
Compound A
dilute H2SO4
Compound B
Br
Compound C
OH
CHAPTER 9 195
9.60.
Br + En
Br
OHHO + En
1) BH3 THF
2) H2O2, NaOHH3O+
HBr
HBr, ROOR
9.61.
excess H2
Pt+
Diastereomers
9.62. Markovnikov addition of water without carbocation rearrangements can be
achieved via oxymercuration-demercuration:
OH
racemic
2) NaBH4
1) Hg(OAc)2, H2O
9.63.
O
H O S O
O
O
H
O
HH
OMe OH
O
H
Me MeO
OHHO
Me
196 CHAPTER 9
9.64.
a)
OH
H O
H
H
HO
H
HO
H
OH H
Methyl
Shift
b) Br
H BrBr+
9.65.
a)
H2
(PPh3)3RhCl
(meso)
b)
H3O+ OH
c)
1) BH3 THF
2) H2O2, NaOHOH
d) 2) H3O+
1) MCPBA
OHHO
+ En
CHAPTER 9 197
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. Acid-
catalyzed 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.
a)
H H
b)
H
c) d)
9.69. The reaction proceeds via a resonance-stabilized carbocation, which is even lower
in energy than a tertiary carbocation:
O OBr
O OH Br Br
resonance-stabilized
198 CHAPTER 9
9.70.
OHBr
HBr
ROOR
OH
OH
Cl2HBr Br2
H2O
Cl
Cl
OH
Br
Br
OH
BrH2
PtBr2
NaOMe H2O
O
O
H
OH
+ En
+ En
1) BH3 THF
2) H2O2, NaOH
OsO4, NMO
1) O3
2) DMS
1) BH3 THF
2) H2O2, NaOH
CHAPTER 9 199
9.71.
O
BrOH
HO Br
HO OH
HO
Br
OHOH
+ En
OHOH
+ En
HBr
ROOR
HBr
H2
Pt
Br2
NaOMe
H2O
1) BH3 THF
2) H2O2, NaOH
OsO4, NMO
1) MCPBA
2) H3O+
1) O3
2) DMS
OsO4, NMO
H3O+
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-1-
pentene proceeds via a less stable, secondary carbocation.
