Chapter 7
7.1 One problem with elimination reactions is that mixtures of products are often formed. For example,
treatment of 2-bromo-2-methylbutane with KOH in ethanol yields a mixture of two alkene products.
What are their likely structures?
Solution:
major product
Br
7.2 How many alkene products, including E,Z isomers, might be obtained by dehydration of
3-methyl-3-hexanol (3-methyl-3-hydroxyhexane) with aqueous sulfuric acid?
Solution:
dehydration
OH
7.3 What product would you expect to obtain from addition of Cl2 to 1,2-dimethylcyclohexene? Show
the stereochemistry of the product.
Solution:
Cl
CH3
Cl
H3C
H3C
CH3
Cl
CH3
Cl
H3 C
Cl
Cl
Cl
H3C
H3C
Cl
CH3
Cl
Cl
CH3
7.4 Addition of HCl to 1,2-dimethylcyclohexene yields a mixture of two products. Show the
stereochemistry of each, and explain why a mixture is formed.
Solution:
Cl
Cl
CH3
H3C
H3C
CH3
H
H3 C
CH3
Cl
CH3
CH3
H3C
Cl
H3 C
7.5 What product would you expect from the reaction of cyclopentene with NBS and water? Show the
stereochemistry.
Solution:
Cl
H
OH
H
O
H
Br
OH
H
Br
Br
H
H
Br
Br
H
Br
H
Br
OH
H
H
OH
7.6 When an unsymmetrical alkene such as propene is treated with NBS in aqueous DMSO, the major
product has the bromine atom bonded to the less highly substituted carbon atom. Is this
Markovnikov or non-Markovnikov orientation? Explain.
OH
H3C
C
H
CH2
NBS,H2O
DMSO
H3C
C
H
CH2Br
Solution:
Br
H
vs
CH2
H3C
Br
H
CH2
H3C
H 2O
H 2O
The product is Markovnikoff product.
7.7 What products would you expect from oxymercuration of the following alkenes?
Solution:
(a)
H3C
H2
C
H2
C
C
H
CH2
Hg(OAc)2
H3C
H2
C
H2
C
H
C
CH3
H2O
OH
2-pentanol
(b)
OH
H3C
C
CH3
C
H
H2
C
CH3
Hg(OAc)2
H2O
H3C
C
H2
C
H2
C
CH3
2-methyl-2-pentanol
CH3
7.8 What alkanes might the following alcohols have been prepared from?
Solution:
(a)
H3C
C
H2
C
C
H
H2
C
CH3
Hg(OAc)2
OH
CH3
H2O
2-methyl-2-hexene
H3C
Hg(OAc)2
H2
C
H2
C
C
H2C
H2
C
C
H2
C
H2
C
H2
C
CH3
OH
H2O
CH3
CH3
2-methyl-1-hexene
(b)
OH
Hg(OAc)2
H2O
cyclohexylethylene
1.BH3
2.H2O2 OH
ethylidenecyclohexane
7.9 Show the structures of the products you would obtain by hydroboration/oxidation of the following
alkenes:
(a)
CH3
CH3C
(b)
CH3
CHCH2CH3
CH3
Solution: (a)
CH3CH
OH
CHCH2CH3
(b)
OH
CH3
7.10 What alkenes might be used to prepare the following alcohols by hydroboration/oxidation?
(a) (CH3)2CHCH2CH2OH
OH
(b)
(CH3)2CHCHCH3
(c)
CH2OH
Solution: (a)
(CH3)2CHCH CH2
(b)
(CH3 )2C
CHCH3
(c)
CH2
7.11 The following cycloalkene gives a mixture of two alcohols on hydroborotion followed by
oxidation, Draw the structures of both, and explain the result.
H
H3C
CH3
H
Solution:
The structure of the two cycloalcohols:
H
H3C
OH CH3
H
CH3
H
H
1. BH3 / THF
H3C
2. OH-, H2O
OH
H
H3C
H
H
CH3
H
7.12 What products would you expect from the following reactions?
CH2
(a)
Cl
Cl
KOH
?
C Cl
H
Zn(Cu)
(b)
(H3C)2HCH2CHC CHCH3
Solution:
CH2I2
?
Cl
CH2
Cl
(a)
Cl
Cl
C
KOH
HCl
C Cl
H
Zn(Cu)
(b)
CH2I2
(H3C)2HCH2CHC CHCH3
ZnI2
7.13 What product would you obtain from calalytic hydrogenation of the following alkenes?
CH3
(a) H3C
C
C
H
H2
C
CH3
H2
C
CH3
CH3
(b)
CH3
Solution：
CH3
(a) H3C
H2
C
C
H
CH3
(b)
CH3
7.14 What alkene would you start with to prepare each of the following compounds?
H
OH
(a)
OH
(b)
H3C
H2
C
OH
C
H
C
CH3
(c) HO
H2
C
OH
OH
C
H
C
H
H2
C
OH
CH3
CH3
OH
S
olution:
H
(a)
(b)
H3C
H2
C
C
H
C
CH3
CH3
CH3
7.15 What products would you expect?
Solution:
(a) Aqueous acidic KMnO4
Product CH3COCH2CH2CH2CH2COOH
(c)
H2C
C
H
C
H
CH2
(b) O3
Product CH3COCH2CH2CH2CH2CHO
7.16 Propose structure for alkenes.
Solution:
(a): (CH3) 2C=O and H2C=O
Structure
(CH3) 2C =CH2
(b): 2 equiv CH3CH2CH=O
Structure
CH3CH2CH=CHCH2CH3
7.17 Show the monomer units you would use to prepare the following polymers:
OCH3
H2
C
OCH3
H2
C
C
H
C
H
OCH3
H2
C
C
H
(a)
OCH3
Solution: The monomer unit is H2C
CH
.
Cl
Cl
Cl
Cl
Cl
Cl
C
H
C
H
C
H
C
H
C
H
C
H
(b)
Solution: The monomer unit is ClHC
CHCl.
7.18 One of the chain-termination steps that sometimes occurs to interrupt polymerization is the
following reaction between two radicals:
CH2CH2
2
CH2CH3
+
C
H
CH2
Propose a mechanism for this reaction, using fishhook arrows to indicate electron flow.
Solution:
H
C
CH2
C
H
H
H2
C
CH2 +
CH2CH3
CH2
7.19 tert-Butyl vinyl ether is polymerized commercially for use in adhesives by a cationic process.
Draw a segment of poly(tert-butyl vinyl ether), and show the mechanism of the chain-carrying step.
O
tert-butyl vinyl ether
Solution:
Acid catalyst
O
H
O
O
O
O
O
Repeat many times
*
C
H
H2
C
n
*
7.20 Name the following alkenes, and predict the products of their reaction with (i) KMnO4 in aqueous
acid and (ii) O3, following by Zn in acetic acid:
(a)
(b)
Solution: (a)2,5-dimethyl-2-heptene
O
O
OH
and
(i)
O
O
H
and
(ii)
(b)3,3-dimethylcyclopentene
HO
O
OH
(i)
O
H
O
H
(ii)
O
7.21 Draw the structures of alkenes that would yield the following alcohols on hydration (red=O). Tell
in each case whether you would use hydroboration/oxidation.
CH2
Solution: (a)
I would use oxymercuration in this case.
(b)
Both oxymercuration / demercuration and hyboration / oxidation reactions could be used.
7.22 The following alkene undergoes hydroboration/oxidation to yiele a single product rather than a
mixture. Explain the result, and draw the product showing its stereochemistry.
Solution: The reactant is symmetrical, so
OH adding to either of the carbon that is double bonded
will yield the same product. The single product is:
OH
7.23 Predict the products of the following reactions (the aromatic ring is unreactive in all cases).
Indicate regiochemistry when relevant.
(a) H2/Pd
CH
CH2
(b)
Br2
(c)
HBr
1.OsO4
2.NaHSO4
(e) D2/Pd
(d)
CH
H
H C
H
C H
H
H
Br C
H
C Br
H
CH2
H2/Pd
Solution: (a)
CH
CH2
Br2
(b)
CH
H
Br C
CH2
H
C
HBr
H
H
(c)
HO
H
CH
CH2
(d)
C
OH
H
H
1.OsO4
2.NaHSO4
D
H
CH
C
CH2
C
C
D
H
H
D2/Pd
(e)
7.24 Suggest structures for alkenes that give the following reactions products. There may be more than
one answer for some cases.
CH3
?
CH3CHCH2CH2CH2CH3
(a) ?
Br
?
Br 2
(b)
CH3
Cl
CH3CHCHCH2CHCH3
Br
(c)
CH3
CH3
?
HCl
CH3CHCHCH2 CH2CH2CH3
CH3
(d)
OH
1.Hg(OAc)2,H2O
?
CH3CH2CH2CHCH3
(e) 2.NaBH4
CH3CHCH2CH2CH CH2
Solution: (a)
CH3
H3C
H3CHCH2CHC
CH3
H3C
H3CHCHC CHCH2CH3
CH3
H2C
H3CC
CCH2CH2CH2CH3
CH3
CH3
CH3
CH3
(b)
CH3
(c) H3CHC
CHCH3
CHCH2 CHCH3
CHCH2CH2CH3
H2C
CHCHCH2CH2CH2CH3
CH3
(d)
(e) H3CH2CH2CHC
CH2
7.25 Predict the products of the following reactions, indicating both regiochemistry and
stereiochemistry when appropriate:
CH3
1. O3
?
2.Zn, H3O+
H
(a)
O
CH3
CH3
1. O3
2.Zn, H3O+
H
H
O
Solution:
KMnO4
H3 O+
?
(b)
O
OH
KMnO4
H3O+
OH
Solution:
O
CH3
1. BH3
2. H2O2,- OH
?
(c)
CH3
H
CH3
1. BH3
2. H2O2,- OH
CH3
(d)
H
50%
Solution:
2. NaBH4
H
H
OH
+
OH
1. Hg(OAc)2,H2 O
CH3
?
50%
CH3
H
CH3
H
OH
H
OH
H
CH3
1. Hg(OAc)2,H2O
2. NaBH4
50%
Solution:
50%
7.26 How would you carry out the following transformations? Indicate the reagents you would use in
each case.
H
OH
?
OH
(a)
H
H
OH
1.OsO4
2.NaHSO3 ,H2O
OH
Solution:
H
OH
?
(b)
OH
1. Hg(Ac)2
2. NaBH4
Solution:
H
Cl
?
Cl
(c)
H
H
Cl
1. CHCl3
2. KOH
Cl
Solution:
H
CH3
CH3
OH
(d)
?
CH3
CH3
OH 1. H2SO4, H2 O
2. THF, 50 ℃
Solution:
O
CH3
H3CHC
?
CHCHCH3
CH3
+ CH3CHCH
CH3CH
(e)
O
CH3
H3CHC
O
1. O3
2. Zn/H3 O+
CHCHCH3
CH3
CH3CH + CH3 CHCH
Solution:
O
CH3
CH3
(f)
H3CC
?
CH2
CH3CHCH2 OH
CH3
Solution:
H3CC
CH3
CH2
1. BH3 , THF
2. H2O2,OH-
CH3 CHCH2OH
7.27 What product will result from hydroboration/oxidation of 1-methylcyclopentene with deuterated
borane, BD3? Show both the stereochemistry (spatial arrangement) and the regiochemistry
(orientation) of the product.
H
CH3
H
CH3
δ
H
CH3
-
OH
H2O2
B
D
δ
D
D2B
D
D
HO
D
HO
D
BD3
H
CH3
D
D
D
B
H
δ
D2B
δ
D
-
OH
H2O2
CH3
H
CH3
H
CH3
7.28 Draw the structure of an alkene that yield only acetone, (CH3)2C=O, on ozonolysis followed by
treatment with Zn.
O3
Zn/H3O+
O
+ O
7.29 Draw the structure of a hydrocarbon that react with 1 molar equivalent of H2 on catalytic
hydrogenation and gives only pentanal , CH3CH2CH2CH2CHO, on the ozonolysis followed by
treatment with Zn. Write the reaction involved.
Solution:
The structure of the hydrocarbon is showed in the following picture:
The reactions are showed as follow:
Catalyst
H2
O3
H
2
Zn/H3O
O
7.30 Draw the structure of alkenes that give the following products on oxidative cleavage with KMnO4
in the acidic solution:
O
O
O
+
+
CO2
(b).
(a).
O
O
+
(c).
Solution:
(b).
(a).
(c).
7.31 Compound A has the formula C10H16. On catalytic hydrogenation over palladium, it reacts with
only 1 molar equivalent of H2. Compound A also undergoes reaction with ozone, followed by zinc
treatment, to yield a symmetrical diketone, B (C10H16O2).
(a) How many rings does A have?
A have two rings.
(b) What are the structures of A and B?
A
O
B
O
(c) Write the reactions.
+H2 palladium
O
+O3
Zn
O
7.32 An unknown hydrocarbon A, with formula C6H12, reacts with 1molar equivalent of H2 over a
palladium catalyst. Hydrocarbon A also reacts with OsO4 to give a diol, B. When oxidized with
KMnO4 in acidic solution, A gives two fragments. One fragment is propanoic acid, CH3CH2CO2H,
and the other fragment is a ketone. C. What are the structures of A, B, and C? Write all reactions,
and show your reasoning.
A
H2
C
CH3
B H3C
H
HO
O
C
CH3
OH
+
H2
palladium
H2
C
OsO4
CH3
H3C
H
CH3
OH
HO
O
O
KMnO4
+ H3C
C
C
H2
H
O
7.33 Using an oxidative cleavage reaction explain how you would distinguish between the following
two isomeric dienes:
and
Solution: When they react with KMnO4 in acid, the first one will yield two products and the second one
will only yield one product.
7.34 Compound A, C10H18O, undergoes reaction with dilute H2SO4 at 2500C to yield a mixture of two
alkenes, C10H16. The major alkene product, B, gives only cyclopentanone after ozone treatment
followed by reduction with zine in acetic acid. Identify A and B, and write the reaction.
Solution: A:
B:
HO
Reaction:
HO
H2SO4
2500C
major
O3
O
Zn/H3O
7.35 Which reaction would you expect to be faster, addition of HBr to cyclohexene or to
1-methylcyclohexene? Explain.
Solution: The intermediates of the two reactions are not equal in stability. Judge by Hammond postulate,
the more stable intermediate forms faster, so addition of HBr to 1-methylcyclohexene may be
faster.
7.36 Predict the products of the following reactions, and indicate regiochemistry if relevant:
Br
(a)
HBr
(c)
CH2I2 , Zn-Cu
H
(d)
H
CH2I2 , Zn-Cu
H
H
7.37 Iodine azide, IN3, adds to alkenes by an electrophilic mechanism similar to that of bromine. If a
monosubstituted alkene such as 1-butene is used, only one product results:
N
H3CH2CHC
(a)
CH2
+
I
N
N
N
CH3CH2CHCH2 I
N
Add lone-pair electrons to the structure shown for IN3, and draw a second resonance form for the
molecule.
I
(b)
N
N
N
I
N
N
N
N
Calculate formal charges for the atoms in both resonance structures you drew for IN3 in part
For the iodine:
Iodine valence electrons
=7
Iodine bonding electrons
=2
Iodine nonbonding electrons
=6
Formal charge = 7 - 2/2 – 6 = 0
For the left nitrogen:
Nitrogen valence electrons
=5
Nitrogen bonding electrons
=6
Nitrogen nonbonding electrons
=2
Formal charge = 5 – 6/2 – 2 = 0
For the middle nitrogen:
Nitrogen valence electrons
=5
Nitrogen bonding electrons
=8
Nitrogen nonbonding electrons
=0
Formal charge = 5 – 8/2 – 0= 1
For the right nitrogen:
Nitrogen valence electrons
=5
Nitrogen bonding electrons
=4
Nitrogen nonbonding electrons
=4
Formal charge = 5 – 4/2 – 4= - 1
(c)
In light of the result observed when IN3 adds to 1-butebe, what is the polarity of the I-N3bomd?
Propose a mechanism for the reaction using curved arrows to show the electron flow in each step.
H3CH2CHC
CH2
+
I
N
N
N
N
H3CH2CHC
CH2I
N
N
N
N
N
CH3CH2CHCH2I
7.38 Draw the structure of a hydrocarbon that absorbs 2 molar equivalents of H2 on catalytic
hydrogenation and gives only butanedial on ozonolysis.
O
HC
O
H2
C
H2
C
CH
Butanedial
Solution:
7.39 Simmons–Smith reaction of cyclohexene with diiodomethane gives a single cycloproprane
product, but the analogous reaction of cyclohexene with 1,1-diiodoethane gives (in low yield) a
mixture of two isomeric methylcyclopropane products. What are the two products, and how do
they differ?
Solution: The reaction occurs as below:
H
+ I2HC CH3
Zn(Cu)
Ether
H3C
CH3
H
+
7.40 In planning the synthesis of one compound from another, it’s just as important to know what not to
do as to know what to do. The reactions all have serious drawbacks to them. Explain the potential
problems of each.
Solution:
(a)
CH3
H3C C CHCH3
CH3
H3 C C CHCH3
H
I
HI
According to Markovnikov’s rule the reaction should be:
CH3
H3C C CHCH3
CH3
H3C C CH2CH3
I
HI
(b)
H
OH
1.OsO4
2.NaHSO3
H
OH
Actually it should be syn addition as below:
H
OH
1.OsO4
2.NaHSO3
OH
H
(c)
CHO
1.O3
CHO
2.Zn
Actually the reaction should be:
H2
1.O3
2 OHC C CHO
2.Zn
(d)
CH3
H
1.BH3
2.H2O2 , OH
H
CH3
OH
Actually the reaction should be:
H
CH3
1.BH3
2.H2 O2, OH
CH3
H
OH
7.41 Which of the following alcohols could be made selectively by hydroboration/oxidation of an
alkene? Explain.
H
OH
CH3
OH
CH3
OH
OH
(a)
CH3CH2 CH2CHCH3
(b)
(CH3)2CHC(CH3)2
Solution:
(a) No selectivity
(b) Selectively synthesized
(c)Can’t be formed
(c)
H
H
(d)
H
(d) Can’t be formed
7.42 What alkenes might be used to prepare the following cyclopropane?
Cl
CH(CH3) 2
(a)
Cl
(b)
Solution:
+ (CH3)2CHCHI2
(a)
Zn(Cu)
CH(CH3)2
Ether
Cl
+ CHCl3
OH
Cl
(b)
7.43 Predict the products of the following reactions. Don’t worry about the size of the molecule;
concentrate on the functional groups.
Br2
CH3
HBr
CH3
1. OsO4
2. NaHSO3
HO
1. BH3, THF
2. H2O2, OH-
CH2I2, Zn (Cu)
Solution:
CH3
CH3
CH3
CH3
Br2
a)
HO
Br
HO
Br
CH3
CH3
CH3
CH3
HBr
b)
HO
HO
Br
CH3
CH3
CH3
CH3
OsO4
NaHSO3
c)
HO
OH
HO
OH
CH3
CH3
CH3
CH3
BH3,THF
H2O2,OH
d)
HO
HO
OH
CH3
CH3
CH3
CH3
CH2I2
Zn(Cu)
e)
HO
HO
7.44 The sex attractant of the common housefly is a hydrocarbon with the formula C23H46. On
treatment with aqueous acidic KMnO4, two products are obtained, CH3(CH2)12CO2H and
CH3(CH2)7CO2H.Propose a structure.
Solution:
H3C
H2
C
H
C
12
C
H
H2
C
7
CH3
should be the right structure.
7.45 Compound A has the formula C8H8. It reacts rapidly with KMnO4 to give CO2 and carboxylic
acid, B (C7H6O2), but reacts with only 1molar equivalent of H2 on catalytic hydrogenation over a
palladium catalyst. On hydrogenation under conditions that reduce aromatic rings, 4 equivalents of
H2 are taken up and hydrocarbon C (C8H16) is produced. What are the structure of A, B, and C?
Write the reactions.
SOLUTION:
O
H
C
A
C
B
H
OH
C
H
H2
C
C
CH3
The reactions:
O
H
C
C
H
KMnO4
C
OH
CO2
H
H
C
H
C
H2
C
CH3
1mol H2
palladium catalyst
H
H
C
H
C
H2
C
4mol H2
CH3
H
7.46. Plexiglas, a clear plastic used to make many molded articles, is made by polymerization of
methyl methacrylate. Draw a representative segment of Plexiglas.
H2C
SOLUTION:
CH3
O
C
C
OCH3
Methyl methacrylate
The segment of Plexiglas:
CH3
H2
C
C
C
O
OCH3
7.47 Poly (vinyl pyrrolidone), prepared from N-vinylpyrrolidone, is used both in cosmetics and as a
synthetic blood substitute. Draw a representative segment of the polymer.
O
N
N-vinylpyrrolidone
Solution:
*
H
C
H2
C
*
N
O
7.48 Reaction of 2-methylpropene with CH3OH in the presence of H2SO4 catalyst yields methyl
tert-butyl ether, CH3OC(CH3)3, by a mechanism analogous to that of acid-catalyst alkene
hydration. Write the mechanism, using curved arrows for each step.
Solution:
CH3
CH3
H3C
H
+
CH2
C
HSO4
CH3OH
H3C
C
CH3
+
CH3OH
CH3
+
H2SO4
CH3
CH3
H3C
C
CH3
H3C
C
H3C
O
H
H3C
O
HSO4
7.49 When 4-penten-1-ol is treated with aqueous Br2, a cyclic bromo ether is formed, rather than the
expected bromohydrin. Propose a mechanism, using curved arrows to show electron movement.
Solution:
H
HO
O
Br
H
O
O
CH2Br
Br
7.50 How would you distinguish between the following pairs of compounds using simple chemical
tests? Tell what you would do and what you would see.
(a) cyclopentene and cyclopentane
(b) 2-hexene and benzene
Solution: Bromine in carbon tetrachloride solution is reddish-brown color. Using this one to identify
the double bond by electrophilic addition reaction.
7.51
O
C
C
O-Na+
Cl
Cl
Cl
O-
C
C
Cl
2
+
a
N
+
+
O
C
C
l
C
+
Cl
Cl
They have the same intermediate which is CCl3-.
Cl
O
Cl
CCl
+
Cl
Cl
O
C
O
7.52
(a) 3
(b) double bonds:2
(c)
7.53 Evidence that cleavage of 1,2-diols by HIO4 occurs through a five membered cyclic periodate
intermediate is based on kinetic data – the measurement of reaction rates. When diols A and B
were prepared and the rates of their reaction with HIO4 were measured, it was found that diol A
cleaved approximately 1 million times faster than diol B. Make molecular models of A and B and
of potential cyclic periodate intermediates, and then explain the kinetic results.
OH
HO
H
OH
H
H
OH
H
A
cis diol
B
trans diol
Solution:
OH
H
H
A:
O
HIO4
HO
O
I
O
O
OH
O
O
In cis diol, two –OH are in one plane, so it is easier to form a cyclic periodate inetermediate,
while in trans diol, the two –OH aren’t in one plane, there is larger steric strain in the five
membered cyclic periodate intermediate, it is less stable, so the rates of A is more faster.
7.54 Reaction of HBr with 3-methylcyclohexene yields a mixture of four products: cis- and
trans-1-3-methylbromocyclohexane and cis- and trans-1-2-methylbromocyclohexane.
The analogous reaction of HBr with 3-bromocyclohexane trans-1, 2-dibromocyclohexane as the sole product. Draw structures of the possible intermediates, and then
Explain why only a single product is formed in the reaction of HBr with 3-bromocyclohexane.
Solution:
Br
Br
Br
H
Br
Br
Br
H
Br
H
7.55 The following reaction takes place in high yield:
CO2 CH3
CO2 CH3
Hg(OAc)2
AcO
Hg
Use your general knowledge of alkene chemistry to propose a mechanism, even though you’ve
never seen this reaction before.
Solution:
OAc
CO2CH3
CO2CH3
OAc
AcO
Hg
+
Hg
AcO
CO2CH3
CO2CH3
H
AcO
OAc
Hg
AcO
Hg
CO2CH3
AcO
Hg
7.56 Reaction of cyclohexene with mercury(II) acetate in CH3OH rather than H2O, followed by
treatment with NaBH4, yields cyclohexyl methyl ether rather than cyclohexanol. Suggest a
mechanism.
1.Hg(OAc)2, CH3OH
2.NaBH4
Solution:
OCH3
OAc
OAc
Hg
Hg
OAc
HO
CH3
HgOAc
HgOAc
O
CH3
O
CH3
OAc
H
NaBH4
OCH3
7.57 Hydroboration of 2-methyl-2-pentene at 25℃ followed by oxidation with alkaline H2O2 yields
2-methyl-3-pentano, but hydroboration at 160℃ followed by oxidation yields 4-methyl-1-pentanol.
Explain.
CH3
1.BH3,THF,25℃
2.H2O2,OH-
CH3CHCHCH2CH3
OH
CH3
2-Methyl-3-pentanol
CH3C
CHCH2CH3
CH3
1.BH3,THF,160℃
2.H2O2,OH-
CH3CHCH2CH2CH2OH
4-Methyl-1-pentanol
Solution:
Case 1:
CH3
CH3
3 CH3C
CHCH2 CH3 + BH3
THF,25℃
3 CH3CH CHCH2CH3
BH2
Case 2:
H2O2,OH-
CH3
3 CH3CHCHCH2CH3
OH
BH2
BH2
H2O2 / OHOH
BH2
7.58 Alkynes undergo many of the same reactions that alkenes do. What products would you expect
from each of the following reactions?
CH3
H3C
C
H
H2
C
H2
C
C
CH
(1) with 1 equiv Br2;
(2) with 2 equiv H2 ,Pd/C
(3)with 1 equiv HBr
Solutions:
(1)
CH3
H3C
C
H
H2
C
H2
C
1 equiv Br2
C
CH
CH3
H3C
C
H
Br
H2
C
H2
C
C
CH
Br
(2)
CH3
H3C
C
H
H2
C
H2
C
2 equiv H2 ,Pd/C
C
CH
CH3
H3C
C
H
H2
C
H2
C
H2
C
CH3
(3)
CH3
H3C
C
H
H2
C
H2
C
1 equiv HBr
C
CH
CH3
H3C
C
H
Br
H2
C
H2
C
C
CH2
7.59 Explain the observation that hydroxylation of cis-2-butene with OsO4 yields a different product
than hydroxylation of trans -2-butene. First draw the structure and show the stereochemistry of
each product, and then make molecular models.
Solution:
H3C
H3C
C
H
CH3
HC
C
H
CH3
H
C
H3C
C
H
CH
CH3
trans-But-2-ene
cis-But-2-ene
OH
H3C
CH3
C
C
C
C
H
H
OH
H3C
OH
Butane-2,3-diol
meso
compound
CH3
H
H
OH
Butane-2,3-diol
enantiomer