Richard F. Daley and Sally J. Daley
www.ochem4free.com
Organic
Chemistry
Chapter 14
Electrophilic Addition to
Unsaturated Carbons
14.1 Addition Reaction Mechanisms
713
14.2 Direction and Stereochemistry of Addition
Reactions
715
14.3 Addition of Hydrogen Halides
718
14.4 Addition of Water and Alcohols
722
14.5 Hydroboration-Oxidation
726
Synthesis of (-)-Isopinocampheol 730
14.6 Electrophilic Addition of Halogens
732
14.7 Addition of Hydrogen
736
14.8 Dihydroxylation Reactions
742
14.9 Addition of Carbenes
745
Synthesis of 7,7-Dichlorobicyclo[4.1.0]heptane
14.10 Oxidation of Alkenes
750
Key Ideas from Chapter 14
754
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Organic Chemistry - Ch 14
711
Daley & Daley
Copyright 1996-2005 by Richard F. Daley & Sally J. Daley
All Rights Reserved.
No part of this publication may be reproduced, stored in a retrieval system, or
transmitted in any form or by any means, electronic, mechanical, photocopying,
recording, or otherwise, without the prior written permission of the copyright
holder.
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Chapter 14
Electrophilic Addition to
Unsaturated Carbons
Chapter Outline
14.1
Addition Reaction Mechanisms
An introduction to the AdE2 and AdE3 reaction
mechanisms
14.2
Direction and Stereochemistry of Addition
Reactions
The regiochemistry and stereochemistry of addition
reactions
14.3
Addition of Hydrogen Halides
The scope of the addition of HCl and HBr to alkenes
14.4
Addition of Water and Alcohols
The reaction of water and alcohols with alkenes
14.5
Hydroboration-Oxidation
An introduction to the hydroboration reaction
14.6
Electrophilic Addition of Halogens
The addition of bromine and chlorine to an alkene
14.7
Addition of Hydrogen
The mechanism of hydrogenation of alkenes
14.8
Dihydroxylation Reactions
Formation of vicinal diols from alkenes
14.9
Addition of Carbenes
The formation of carbenes and their reaction with alkenes
to form cyclopropane derivatives
14.10
Oxidation of Alkenes
Oxidative cleavage of alkenes
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Objectives
✔ Understand the AdE2 and AdE3 reaction mechanisms
✔ Know the regiochemistry and stereochemistry of addition reaction
mechanisms
✔ Know the scope and reactivity of addition reactions of hydrogen
halides, water, alcohols, and halogens
✔ Understand the hydroboration-oxidation reaction
✔ Know the catalytic hydrogenation reaction
✔ Understand the formation of vicinal diols
✔ Know how carbenes form and how they react with alkenes
✔ Recognize the utility of the oxidation of alkenes
The grand aim of all science is to cover
the greatest number of empirical facts by
logical deduction from the smallest
number of hypotheses or axioms.
- Albert Einstein
T
he addition, substitution, and elimination reactions you
studied in earlier chapters all involved nucleophilic
reagents that reacted with electrophilic substrates. In those reactions,
the nucleophilic reagent formed a new bond with an atom in the
substrate that was, or readily became, electron deficient. Nucleophilic
reagents are all Lewis bases.
This chapter considers the reaction of an electrophilic reagent
with an electron-rich atom of the substrate. Such a reaction is called
an electrophilic addition reaction. Electrophilic reagents are all Lewis
acids. The most important electrophilic addition reactions involve
carbon—carbon double or triple bonds in alkenes and alkynes. These
functional groups provide the required regions of high electron
density. Usually, this electron density is evenly distributed over the
entire functional group.
The mechanism of an addition reaction of an electrophile with
a π bond is the reverse of the elimination reactions covered in Chapter
13. Because a σ bond is lower in energy than a π bond, addition
reactions commonly follow a much lower energy pathway than do
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elimination reactions. When chemists use an addition reaction, they
use milder reaction conditions than with an elimination reaction. The
final state of an addition/elimination reaction pair depends on the
reaction conditions.
This chapter covers only the 1,2-addition reactions. 1,2Addition reactions are the most common type of addition reactions,
although 1,4-addition reactions do frequently occur with conjugated
systems. Chapter 16 discusses the 1,4-addition reactions. 1,2-Addition
reactions follow one of two mechanistic pathways with one accounting
for the majority of the 1,2-addition reactions.
14.1 Addition Reaction Mechanisms
An electrophilic
addition reaction
occurs when an
electron deficient
reagent adds to an
electron-rich substrate
in the rate determining
step of the reaction.
Electrophilic addition reactions to unsaturated carbon
atoms follow two general mechanisms. Both mechanistic pathways
involve the addition of an electrophile, E, to a π bond, followed by a
reaction with a nucleophile, Nu, resulting in the formation of two new
σ bonds. Although there are a variety of possible nucleophiles, the
most common electrophile is a hydrogen ion. The following is the
generalized reaction for an electrophilic addition reaction.
C
The Ad stands for
addition, the E for
electrophilic, and the
number for the kinetics
of the reaction.
C
E
Nu
C
C
E
Nu
The two electrophilic addition mechanisms are designated as the
AdE2 and AdE3 mechanisms. The AdE2 is bimolecular, or the reaction
has two reagents involved in the rate-determining step. The AdE3 is
trimolecular, or a reaction with three reagents involved in the ratedetermining step. The AdE2 mechanism corresponds closely to the E1
mechanism, and the AdE3 mechanism corresponds closely to the E2
mechanism.
The AdE2 mechanism, shown in Figure 14.1, is a two-step
mechanism that proceeds through a carbocation intermediate. The
energy profile of an AdE2 mechanism is the mirror of that of an E1
mechanism. In the first step of the mechanism, the electrophilic
portion of the reagent adds to the substrate to form a carbocation. The
first step is a 1,3-electron pair abstraction operation. In the second
step, which is a fast step, the nucleophile adds to the carbocation
intermediate to complete the addition reaction. The second step is a
neutralization operation.
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C
C
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Daley & Daley
C
+ H
Slow
C
H
Nu
C
+ Nu
C
C
•
•
Fast
C
H
H
Figure 14.1. The AdE2 reaction mechanism. The rate of formation of the carbocation
intermediate determines the rate of the reaction.
To review the
mechanism of an
addition to a carbonyl
group see Section 7.1,
page 000.
The AdE2 mechanism is similar to the pathway proposed for an
addition reaction across the carbon—oxygen double bond of a carbonyl
group. Chemists call that addition reaction a nucleophilic addition;
they call the addition to a carbon—carbon double bond an electrophilic
addition. Why this difference in terminology when each of these two
reaction types involves the addition of both a nucleophile and an
electrophile to a π bond?
The difference in terminology is due to the difference in the
mechanisms of the two processes. In an addition reaction to the
carbonyl double bond, the nucleophile adding to the partial positive
charge on the carbon of the carbonyl group determines the rate of the
reaction. The presence of the electrophile normally only slightly
accelerates the rate of reaction. Thus, the term used to describe this
reaction type is nucleophilic addition. In an addition reaction to a
carbon—carbon double bond, the electrophile adding to a carbon of the
double bond determines the rate of reaction. The presence of the
nucleophile has only a small effect on the rate. Thus, the term used to
describe this reaction type is electrophilic addition.
The AdE3 mechanism, shown in Figure 14.2, requires a
simultaneous collision of three reactive species. These species include
the substrate molecule containing the double bond, the electrophile,
and the nucleophile. In an AdE3 mechanism the electrophile and the
nucleophile come from two separate molecules. This is a 1,5-electron
pair displacement operation. Because of the low probability of three
separate reactive species approaching each other with the correct
orientation and energy to react, addition reactions seldom follow this
mechanism. However, an addition reaction may occur with a
nonionized reagent.
Nu
H
Nu
C
C
C
Nu
C
H
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C
C
H
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Figure 14.2. The AdE3 mechanism.
14.2 Direction and Stereochemistry of Addition
Reactions
The addition of an unsymmetrical reagent to an unsymmetrical
alkene or alkyne can, in principle, take place in either of two
directions. The electrophilic portion of the reagent could add to either
of the carbons involved in the double bond. Then the remaining
nucleophilic portion could add to the other carbon.
E
H E
R
C
H
Nu
C
R'
H
C
R
Nu
C
R'
H
Nu
H
C
R
R'
H
C
E
It turns out, however, that the AdE2 reaction is normally regiospecific.
Thus, the electrophilic reagent adds to a specific area of the substrate.
For example, the following reactions produce the products shown as
the major product.
CH3CH
HI
CH2
CH3CHCH3
I
2-Iodopropane
(87%)
I
CH3C
CH3
CH2
HI
CH3CCH3
CH3
2-Iodo-2-methylpropane
(89%)
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Markovnikov’s rule
states that the
electrophile adds to the
less substituted carbon
atom of an alkene.
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Daley & Daley
In 1870, Russian chemist Vladimir V. Markovnikov formalized
this regioselectivity with a rule that bears his name, Markovnikov's
rule. Markovnikov's rule states that the addition of HX to a carbon—
carbon double bond proceeds in such a way that the hydrogen adds to
the carbon atom already bearing the greater number of hydrogen
atoms.
Markovnikov's rule readily translates to the modern
understanding of chemical mechanisms. An AdE2 reaction mechanism
shows that the incoming electrophile controls the regioselectivity of
the reaction. The electrophile adds to the double bond in such a way
that the most stable carbocation intermediate forms. Recall from
Chapters 12 and 13 that the most stable carbocation is the carbon
with the greatest number of alkyl group substituents. Thus, in modern
terms, according to Markovnikov's rule the electrophile adds to the
double bond in such a way to form the most stable carbocation.
Because the electron density in a double bond is above and
below the plane of the molecule, the most favorable approach of an
electrophile is to those electron-rich regions. This approach is
perpendicular to the plane of the π bond.
E
C
C
E
Figure 14.3. The approach of an electrophile to the π orbitals of the double bond is
perpendicular to the plane of the molecule.
Anti addition means
the electrophile and
nucleophile add 180o
apart. With a syn
addition they add 0o
apart.
In an electrophilic addition reaction, two groups (the
electrophile and the nucleophile) actually add to the double bond. The
electrophilic portion of the reagent first adds to the double bond,
followed by the addition of the nucleophilic portion of the reagent. The
electrophilic and nucleophilic parts of a reagent add either to the
opposite sides of the substrate, an anti addition, or to the same side of
the substrate, a syn addition.
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E
Daley & Daley
E
C
C
C
C
Nu
Nu••
Anti addition
E
Nu••
C
Nu
E
C
C
C
Syn addition
Many addition reactions are stereoselective with a number of
these being stereospecific. Most electrophilic addition reactions favor
anti stereochemistry indicating that the reaction mechanism is more
complex than that shown in Figure 14.1. The mechanism in Figure
14.1 shows little or no preference between the syn and anti additions.
To show the preference, the mechanism must show more reagent—
substrate interaction. For example, rather than producing a free
carbocation, the first step in the bromination of a double bond is
probably the formation of an intermediate bromonium ion.
••
C
•
Br•
C
A bromonium ion
This bromonium ion blocks the incoming nucleophile from reacting
with one side of the ion, forcing it to react with the other side. This
reaction is an anti addition. Other examples of reagent—substrate
interactions that force the reagent to add either anti or syn are
presented later in this chapter.
Solved Exercise 14.1
Chemists use the following reagents in electrophilic addition reactions.
Identify the electrophilic species available from each reagent.
a) HBr
Solution
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Daley & Daley
- ions. Thus, the electrophile is the
HBr readily ionizes forming H⊕ and Brc
⊕
H ion.
b) HOCl
Solution
HOCl has two possible bonds that could break: the H—O and the O—Cl
bonds. Of these two, the O—Cl bond is the least stable. Thus, although Cl⊕
- OH ions
does not really form, HOCl reacts as if it ionizes to form Cl⊕ and c
⊕
with the Cl ion being the electrophile.
c) Cl2
Solution
Polarization of Cl2 allows the substrate to react with the Cl—Cl bond
- Cl ion. The reaction occurs as if the Cl⊕ ion is the
displacing the c
electrophile.
14.3 Addition of Hydrogen Halides
A hydrohalogenation
reaction is the addition
of HX to a double bond.
X is a halogen.
Section 14.4, page 000,
discusses the addition
of water to alkenes and
alkynes.
Hydrogen halides add to the π bonds of alkenes to form alkyl
halides and to alkynes to form vinyl halides or gem-dihalides
depending on the ratio of the reactants. These reactions are called
hydrohalogenation
reactions.
Chemists
perform
hydrohalogenation reactions by bubbling the gaseous hydrogen halide
through the reaction mixture in a non-nucleophilic solvent. With the
addition of aqueous solutions of the hydrogen halide, water also adds
to the double bond. This reaction works best with HBr or HI. The
reaction of HCl is generally not acidic enough to react, unless a
particularly stable carbocation intermediate can form.
CH2
CH
A protonium ion is a
complex involving a
proton and an alkene.
CH
HBr
CH2
CH2
HI
CHBr
CH3CH2I
HBr
CH3CHBr2
The hydrohalogenation reaction follows Markovnikov's rule
and predominantly gives the anti addition product. Chemists do not
clearly understand the reasons for the stereoselectivity of the
hydrohalogenation reaction. What chemists have proposed is that with
a polar-protic solvent, which ionizes the hydrogen halide, the reaction
proceeds through a complex between the proton and the alkene. They
call this complex a protonium ion.
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H
H
C
C
or
C
C
A protonium ion
The protonium ion forms rapidly in the presence of a proton source.
The protonium ion then reacts with the nucleophile. Because the
proton blocks the nucleophile from attacking the ion on the side to
which the proton is attached, the nucleophile forms a bond anti to the
proton.
H
C
H
C
C
C
Br
••
•
•
Br
••
•
•
Not all reactions that ionize the hydrogen halide form a
protonium ion. In some cases, the protonium ion is unstable and the
substrate rearranges. This rearrangement suggests that the
protonium ion has significant carbocation character, thus, permitting
hydride or alkyl shifts. These shifts form more stable intermediates.
The following reaction shows a hydride shift.
Cl
CH3CHCH
CH3
CH2
HCl
CH3CCH2CH3
CH3
Cl
+
CH3CHCHCH3
CH3
2-Chloro-2-methylbutane 2-Chloro-3-methylbutane
(40%)
(60%)
A polar-aprotic solvent allows the hydrogen halide to remain
nonionized. Thus, when a polar-aprotic solvent is used, the ratecontrolling step involves two molecules of hydrogen halide and one
molecule of alkene. The anti stereochemistry for the following example
of an AdE3 mechanism arises because each molecule of hydrogen
bromide approaches the alkene from opposite sides of the double bond.
In this case, the rate of reaction is much slower than in a polar-protic
solvent because the reaction requires three molecules to collide with
the proper energy and orientation.
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•
•
C
C
C
•
•
Daley & Daley
••
Br H
••
•
•
C
C
••
Br••
C
H
••
Br
H
••
The following are examples of hydrohalogenation reactions:
CH3
CH3
H
CH3
HBr
pentane, 0oC
CH3
Br
1-Bromo-trans-1,2-dimethylcyclohexane
(100%)
CH3CH
CHBr
HBr
CH3CH2CHBr2
CH3NO2
1,1-Dibromopropane
(93%)
In the second example above, the bromine adds to the carbon already
bearing a bromine. The nonbonding electrons on the bromine are
donated to the carbocation intermediate to stabilize it. Thus the more
stable carbocation is the one bearing the bromine atom.
H
CH3
•
•
C
H
••
H
Br ••
C
CH3
H
C
•
•
••
Br••
C
H
H
H
CH3
C
•
•
••
Br
C
H
H
Reactions of hydrogen halides with alkynes follow the same
rules of regiochemistry and stereochemistry that alkenes follow. In a
reaction of one equivalent of hydrogen halide with an alkyne, the
resulting vinyl halide has the hydrogen and the halogen trans to one
another. For example, the addition of one equivalent of hydrogen
bromide to 2-butyne produces Z-2-bromo-2-butene, a vinyl halide:
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CCH3
CH3C
HBr
Daley & Daley
H3C
Br
C
C
H
CH3
Z-2-Bromo-2-butene
(98%)
If the vinyl halide then reacts with another molecule of hydrogen
halide, the result is a gem dihalide. For example, reacting Z-2-bromo2-butene with another mole of HBr produces 2,2-dibromobutane.
CH3
Br
C
H
C
Br
HBr
CH3
CH3CH2CCH3
Br
2,2-Dibromobutane
(100%)
The second bromine adds to the carbon bearing the bromine in a
Markovnikov addition reaction. The reaction is a Markovnikov
addition because the positively charged intermediate is resonancestabilized:
H
CH3
•
•
C
H
••
H
Br••
C
CH3
••
H
Br••
•
•
C
C
CH3
•
•
H
CH3
C
CH3
••
•
•
Br
••
•
•
C
H
••
Br
H
H
CH3
CH3 C
•
•
••
Br••
C
•
•
CH3
Br••
••
Exercise 14.1
Predict the major product(s) for each of the following reactions.
a)
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Organic Chemistry - Ch 14
CH3
723
Daley & Daley
HBr
b)
(CH3)3CCH
HI
CH2
c)
HCl
d)
CH3
H
C
H
C
HCl
NO2
Sample Solution
b)
CH3
(CH3)3CCH
CH2
HI
(CH3)3CCHCH3
+
(CH3)2CCHCH3
I
I
Minor
Major
14.4 Addition of Water and Alcohols
An acid-catalyzed
hydration reaction
results in a net
addition of water to a
double bond.
Under acidic conditions, water adds to alkenes to form alcohols
in an acid-catalyzed hydration reaction. The product of an acidcatalyzed hydration reaction is regiospecific, following Markovnikov's
rule, but it is not stereoselective. The process, as expected with a
carbocation intermediate, usually gives relatively equal amounts of
both the anti and syn addition products. Because a carbocation is
planar, the adding group can come in from either side. Acid-catalyzed
hydration reactions are widely used in industry for producing large
quantities of commercially important alcohols, but chemists seldom
use them in the laboratory.
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CH3C
724
CH2
Daley & Daley
OH
H3O
CH3CCH3
CH3
CH3
2-Methyl-2-propanol
(tert-Butyl alcohol)
(71%)
OH
H3O
Cyclohexanol
(77%)
An oxymercuration
reaction is the net
addition of water to a
double bond via an
organomercury
intermediate. Other
methods are replacing
oxymercuration
because mercury
presents a hazardous
waste disposal
problem.
A more important laboratory method for the Markovnikov
addition of water to an alkene is the oxymercuration reaction. The
oxymercuration reaction is a two-step process. In the first step, an
aqueous solution of mercury (II) acetate is added to a solution of the
alkene in THF. During a period of reaction time, the mercury (II)
acetate adds to the double bond. In the second step, a solution of
NaBH4 in aqueous base is added to the reaction mixture.
OH
Hg(OAc)2
H2O, THF
HgOAc
Not normally isolated
OH
NaBH4
H2O,
OH
(99%)
The above reaction, which produces cyclohexanol, gives a reaction
yield of 99%. For most reactions of simple alkenes with water, the
alcohol yields are in excess of 90%. Compounds that tend to rearrange
when using the strongly acid-catalyzed hydration reaction do not
rearrange in an oxymercuration reaction.
Oxymercuration is both regiospecific and stereospecific. The
stereospecificity results from the formation of an intermediate
mercurinium ion.
HgOAc
C
C
A mercurinium ion
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Daley & Daley
Attack on this mercurinium ion by water produces an organomercurial
alcohol. The water reacts with the more substituted carbon atom
because is has a larger partial positive charge.
HgOAc
C
HgOAc
C
C
HgOAc
C
C
••
HO
••
HO
H
H2O••
C
•
•
••
The reaction then gets rid of the mercury in a demercuration reaction.
In the demercuration reaction, sodium borohydride reduces the
mercury to metallic mercury and replaces the mercury with hydrogen.
HgOAc
C
C
H2O,
HO
H
NaBH4
C
C
OH HO
Chemists prefer the oxymercuration reaction for the
Markovnikov synthesis of an alcohol because it gives better yields
than an acid-catalyzed hydration and does not involve rearrangements
of the substrate.
H3C
CH3
C
H
C
H
1) Hg(OAc)2, H2O
2) NaBH4, OH
H3C
H
OH
C
H
C
H
CH3
HO
+
C
H
H3C
C
CH3
H
H
Racemic 2-Butanol
(72%)
CH3
CH3
1) Hg(OAc)2, H2O
OH
2) NaBH4, OH
1-Methylcyclohexanol
(86%)
Both acid-catalyzed hydration reactions and oxymercuration
reactions take place with alkyne substrates. The addition follows
Markovnikov's rule. The acid-catalyzed hydration reaction is slow, but
speeds up when mercury (II) salts are used as a catalyst. The initial
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Daley & Daley
product of a hydration reaction is a vinyl alcohol, but the vinyl alcohol
rapidly tautomerizes to form a carbonyl compound.
Tautomerism is
introduced in Section
8.7, page 000, and 19.1,
page 000.
OH
CH3CH2CH2C
CH3CHC
H3O
CH
CH
CH3
HgSO4
CH3CH2CH2C
O
CH2
Unstable vinyl alcohol
1) Hg(OAc)2, H2O
2) NaBH4, OH
CH3CH2CH2CCH3
2-Pentanone
(86%)
OH
CH3CHC
O
CH2
CH3
CH3CHCCH3
CH3
3-Methyl-2-butanone
(83%)
Exercise 14.2
Predict the major products for each of the following reactions.
a)
H3O
CH2
b)
C
CH
1) Hg(OAc)2, H2O
2) NaBH4,
OH
c)
CH3
CH3
1) Hg(OAc)2, H2O
2) NaBH4,
OH
d)
CH3
H3O
CH3
e)
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Organic Chemistry - Ch 14
CH
727
CH2
Daley & Daley
H3O
Sample Solution
a)
CH2
Alcoholysis is much
like hydrolysis except
that the reaction is
with an alcohol instead
of water.
CH3
H3O
OH
Using an alcohol nucleophile in both the acid-catalyzed and the
oxymercuration reaction yields an ether. Chemists seldom use an acidcatalyzed alcoholysis because of the number of side products that it
forms. However, the alkoxymercuration reaction is a good synthetic
method that chemists use with simple alcohols.
CH3
CH3
1) Hg(OAc)2, CH3OH
2) NaBH4,
OH
OCH3
1-Methoxy-1-methylcyclohexane
(88%)
14.5 Hydroboration-Oxidation
In an antiMarkovnikov reaction,
the product appears to
have the opposite
regiochemistry to the
‘normal’ mode of
addition.
The hydroborationoxidation reaction is a
method for
synthesizing alcohols
via an organoboron
intermediate. The net
reaction is an antiMarkovnikov addition
of water to the double
bond.
Section 14.4 covered two methods for adding water to a double
bond to form an alcohol following Markovnikov's rule: acid-catalyzed
hydration reactions and oxymercuration reactions. In both methods
the proton added to the carbon that would leave the most stable
carbocation. In most reactions, this is the carbon that contained the
greater number of hydrogens. Sometimes, however, chemists want to
form an alcohol with the —OH group bonded to that carbon. Neither of
these methods forms such an alcohol. In the early 1950s, H. C. Brown
of Purdue University discovered that diborane (B2H6) adds to the
double bond of an alkene to form a product called an organoborane.
The boron of the B—H bond is the electrophilic portion of the reagent.
Thus, the boron adds to the carbon that would be the less stable
carbocation and the hydrogen adds to the other carbon of the alkene.
If the organoborane is oxidized, it forms an alcohol that is the
anti-Markovnikov addition product. In this reaction the hydrogen
adds to the carbon containing the fewer hydrogens and the —OH
group ends up on the carbon with the greater number of hydrogens.
This reaction is a hydroboration-oxidation reaction. For the
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Organic Chemistry - Ch 14
A monomer is a
building-block
molecule which make
up a dimer or larger
molecule.
728
Daley & Daley
discovery and exploration of the synthetic utility of organoboranes,
Brown received the Nobel Prize in chemistry in 1979.
Diborane is a dimer—a molecule made up of two identical
simpler molecules, or monomers. Diborane consists of two monomer
units of borane (BH3). These two monomers bond together in an
unusual type of bond. Instead of a single bond between the two boron
atoms, both boron atoms are involved in two three-centered bonds.
Each bond includes three atoms: the two boron atoms and the
hydrogen atom between them.
H
H
H
B
H
B
H
H
Three-centered bonding occurs with diborane because each boron atom
in BH3 has only six electrons and an empty orbital in its valence shell.
Boron has an orbital that readily accepts an electron pair, so borane is
a strong Lewis acid. Sharing the electrons in a B—H bond with the
empty orbital on a second boron atom fills the octet of the second
boron atom. The second boron shares the electrons in a B—H bond
with the first boron atom to satisfy its octet as well.
As a reagent, diborane is a difficult to handle flammable gas,
that is not very reactive with alkenes. However, Brown found that
dissolving diborane in THF (tetrahydrofuran) or diglyme
(CH3OCH2CH2OCH2CH2OCH2CH2OCH3) produces a convenient,
easy to handle solution that is very reactive with alkenes. In this
solution, the THF and the diborane form a complex in which the
oxygen of the ether shares a pair of electrons with a boron atom.
H
O
B
H
H
Borane-THF complex
Chemists think that the hydroboration reaction is a concerted
reaction with both the boron and the hydrogen adding simultaneously
in a syn addition.
C
H
C
C
C
H
B
B
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The B—H bond is a polar bond but, because the boron has a lower
electronegativity than hydrogen, the hydrogen is the negative end of
the dipole. When adding to the double bond, the boron adds to the
least substituted carbon atom because it is the electrophile. Thus,
although the reaction seems to add contrary to Markovnikov's rule, it
really does not. In a hydroboration reaction, all three of the B—H
bonds are reactive.
CH3CH2CH
CH2
BH3
(CH3CH2CH2CH2)3B
Tributylborane
THF
Hydroboration is very sensitive to steric factors. The hydroboration of
butene places about 95% of the boron at C1 of the butyl group and
about 5% at C2. For 2-methyl-1-propene, the amount of boron at C1 is
in excess of 99%.
Alkylboranes are generally pyrophoric. That is, they ignite
spontaneously in air. Therefore, chemists seldom isolate them but use
them directly as reactive intermediates. Although chemists use
organoboranes as the intermediates for many functional groups, the
one they use it for the most is the oxidation reaction that forms an
alcohol. The oxidation reaction proceeds by adding an alkaline
solution of hydrogen peroxide to the organoborane solution. Note that
the stereochemistry of the borane reaction is retained in the oxidation
step.
CH3
CH3
1) BH3/THF
2) H2O2, H2O, NaOH
OH
trans-2-Methylcyclopentanol
(95%)
CH3
C
CH3
CH2
CHCH2OH
1) BH3/THF
2) H2O2, H2O, NaOH
2-Phenyl-1-propanol
(97%)
The net reaction obtained with a hydroboration-oxidation
reaction is an anti-Markovnikov addition of water to a double bond.
The stereochemical configuration of the carbon originally bearing the
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For more on the
equilibrium between
geminal diol and
carbonyl group see
Section 7.5, page 000.
730
Daley & Daley
boron is retained during the reaction. The result is a stereospecific syn
addition of water to a double bond with anti-Markovnikov
regioselectivity.
Alkynes also undergo hydroboration reactions. The resulting
product from a terminal alkyne is a geminal diol that rapidly loses
water to form an aldehyde.
H
R
H
BH3
THF
H
B
R
H
H
H
H2O2
NaOH, H2O
OH
R
H
B
OH
H
H
O
R
H
For more on tautomers,
see Section 19.1, page
000.
An internal alkyne forms a vinyl alcohol that rapidly tautomerizes to a
ketone.
H
R
R'
BH3
THF
H
B
R
H2O2
NaOH, H2O R
OH
R'
R'
H
H
O
R
R'
Following are some examples of hydroboration of alkynes:
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O
1) BH3/THF
H
2) H2O2, NaOH, H2O
Pentanal
(74%)
O
1) BH3/THF
2) H2O2, NaOH, H2O
3-Hexanone
(67%)
Synthesis of (–)-Isopinocampheol
H3C
CH3
H3C
CH3
CH3
CH3
1) BH3, THF
OH
2) H2O2, NaOH
(+)-a-Pinene
(-)-Isopinocampheol
(91%)
Preparation of the BH3/THF Solution
Fit a dry 100 mL round bottom flask with a Claisen adapter. Fit a rubber septum on
the arm directly over the opening of the flask. Connect a drying tube filled with
calcium chloride to the other arm of the Claisen adapter. Gently heat the glass drying
tube with a flame to dry the calcium chloride. Place 520 mg (0.014 mol) of sodium
borohydride in the flask and add 30 mL of freshly distilled dry THF. Cool the flask in
an ice-water bath. Slowly add a solution of 1.7 g (0.014 mol) of iodine in 20 mL of dry
THF. The addition should require at least 45 minutes. Maintain the solution at 0oC
during the addition.
(–)-Isopinocampheol
To the solution of BH3 in THF prepared above, add 2.8 g (0.02 mol) of (+)-α-pinene
over the course of 10 minutes. Allow the mixture to warm to room temperature and
stir. At the end of 1.5 hr., slowly add 2 mL of water to destroy any excess BH3. Mix 20
mL of 30% hydrogen peroxide with 20 mL of 3M sodium hydroxide. Add this solution
cautiously to the reaction mixture. Stir for an additional 5-10 minutes. Transfer the
reaction mixture to a separatory funnel and separate the layers. Wash the aqueous
layer with three 15 mL portions of ethyl ether. Combine the ether and the organic
layers. Wash these combined layers with 10 mL portions of water and saturated
sodium chloride solutions. Dry with anhydrous sodium sulfate. Remove the drying
agent and evaporate the solvent. Recrystallize the solid product by dissolving it in a
minimum of boiling ethanol and adding water until the solution just becomes cloudy.
Cool this mixture and collect the crystals. Yield of product is 2.9 g (93%), m.p. 5053oC.
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Discussion Questions
1. Why must the solvent and glassware be completely dry before beginning the
synthesis of BH3?
2. This hydroboration reaction is stereoselective with borane reacting with only one
side of the π system in (+)-α-pinene. Explain this stereoselectivity.
Solved Exercise 14.2
Show how you would prepare 1- and 2-methylcyclohexanol from 1methylcyclohexene.
Solution
1-Methylcyclohexanol can be prepared two different ways: acid-catalyzed
hydrolysis or the oxymercuration-demercuration reaction. In this case, acidcatalyzed hydrolysis is simpler because the desired product comes from the
most stable carbocation.
H2O
OH
H2SO4
2-Methylcyclohexanol is synthesized using the hydroboration-oxidation
reaction.
1) BH3/THF
2) H2O2, NaOH
OH
(Both enantiomers formed)
Exercise 14.3
Predict the major products for each of the following reactions.
a)
CH3CH2CH2C
CH3
CH2
1) BH3, THF
2) H2O2, H2O, NaOH
b)
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1) BH3, THF
2) H2O2, H2O, NaOH
c)
1) BH3, THF
2) H2O2, H2O, NaOH
d)
1) BH3, THF
2) H2O2, H2O, NaOH
e)
1) BH3, THF
2) H2O2, H2O, NaOH
Sample Solution
d)
1) BH3, THF
2) H2O2, H2O, NaOH
OH
14.6 Electrophilic Addition of Halogens
Halogens readily add to alkenes to form vicinal dihalides. Of
the various halogens, chemists commonly use bromine and chlorine.
They do not use fluorine because it reacts explosively and produces
many side reactions. They do not use iodine because vicinal diiodides
decompose readily.
Formation of vicinal dihalides is a stereospecific anti-addition
reaction. The first step in the mechanism of an electrophilic addition
of a halogen to an alkene is the formation of a halonium ion. To form a
halonium ion, the π electrons of the double bond react with one
halogen atom from the halogen molecule. The following carbocation is
not the minimum energy intermediate—the halonium ion is lower in
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energy. Although shown here as a carbocation, many chemists
consider this reaction to be a concerted reaction.
C
••
•
•
••
X
C
••
•
•
X
••
•
•
C
••
X••
•• ••
C
C
X
C
Chlorine and bromine are both electrically neutral and nonpolar.
However, both are polarizible. The interaction of a chlorine or bromine
molecule with an alkene leads to polarization of the molecule and loss
of a chloride or bromide ion, a good leaving group. The formation of
the chloronium or bromonium ion depends on the polarizibility of the
halogens and the fact that chloride or bromide ions are good leaving
groups. The net result is the equivalent to a reaction with Cl⊕ or Br⊕.
Because chlorine is smaller and more electronegative than bromine,
the chloronium ion is not as stable as the bromonium ion.
The halonium ion is a three-membered ring with considerable
ring strain. This ring strain, combined with the positive charge on the
electronegative halogen atom, makes the halonium ion intermediate
very electrophilic. In the second step of the mechanism, reaction with
a nucleophile, such as the halide ion released when the halonium ion
was formed, opens the ring to give a stable vicinal dihalide.
•
•
C
••
X
••
•
• •
•
X
C
C
C
•
•
•
•
•
•
X
••
••
X••
••
The incoming nucleophile reacts only from the side opposite the
positive halogen of the halonium ion to form the anti vicinal dihalide.
Br2
Br
Br
trans-1,2-Dibromocyclopentane
(83%)
Because the double bond region of most molecules is symmetrical and
planar, the addition of the halogen produces a pair of enantiomers or a
meso product.
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CH3
CH3
C
Br2
C
H
735
H
Daley & Daley
Br
H C
CH3
H
H
CH
3
C
C CH3 +
Br
Br
Br
C
H
CH3
Racemic 2,3-Dibromobutane
(88%)
H
H3C
C
C
H
Br
Br2
H C
CH3
CH3
CH3
C H
Br
meso-2,3-Dibromobutane
(87%)
The Cl⊕ ion is more reactive than the Br⊕ ion, so the addition of
chlorine is less stereoselective than the addition of bromine. In most
cases, there is still some preference for the anti addition. However,
significant amounts of syn addition also occur.
Exercise 14.4
The addition reaction involving cis-2-butene forms two products, but
trans-2-butene forms only one. Why?
Chlorine and bromine react with alkynes in much the same
way as they do with alkenes. One mole of halogen added to an alkyne
forms a vicinal dihaloalkene product. The stereochemistry of this
reaction is either syn or anti, and the products of the reaction are
often a mixture of Z and E isomers.
CH3CH2CH2C CCH3
Br2
Br
CH3CH2CH2
C
Br
+
C
CH3
E-2,3-Dibromo-2-hexene
77%
Br
Br
C
CH3CH2CH2
C
CH3
Z-2,3-Dibromo-2-hexene
23%
Because the bromonium ion intermediate formed from an alkyne is a
three-membered ring that contains a double bond, it is less stable
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than the ion formed from an alkene. The reaction is also less
stereospecific. Thus, the reaction proceeds partially via the halonium
ion and partially via a vinyl carbocation.
•
•
R
C
C
R
••
••
Br
••
Br
••
••
•
Br •
•
•
C
R
•
•
+
C
R
••
Br
••
C
C
R
R
Reactions of compounds containing triple bonds with two moles
of the halogen per mole of substrate produce a tetrahaloalkane. Yields
for electrophilic addition reactions with triple bonds are often nearly
100%.
Br Br
CH3CH2CH2C CCH3
Br2
CH3CH2CH2
C
C
CH3
Br Br
2,2,3,3-Tetrabromohexane
(94%)
Exercise 14.5
When treating 2-hexyne with one equivalent of bromine to produce a
dibromoalkene, should you add the bromine to the 2-hexyne or the 2hexyne to the bromine? Explain.
In a halohydrin
reaction, a halogen and
water add to a double
bond forming an αhalo alcohol.
When other nucleophiles are present in the reaction mixture,
they compete with the halide ion in reacting with the halonium ion.
For example, when chemists dissolve the halogen in water and react
this solution with the alkene, the water acts as a nucleophile and
reacts with the halonium ion to form a molecule with a halogen on one
carbon atom and a hydroxyl group on the adjacent atom. These
products are called halohydrin compounds. The reaction is a
stereospecific anti-addition to the alkene.
Cl
Cl2
H2O
OH
trans-2-Chlorocyclohexanol
(a chlorohydrin)
(89%)
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In reactions involving an unsymmetrical alkene, the reaction is
also regiospecific. With an unsymmetrical alkene, a Markovnikov
addition reaction occurs with the electrophile adding to the carbon
with the most hydrogens. The water reacts with the most electrophilic,
usually the most substituted, carbon of the halonium ion.
Br2
CH3
OH
H2O
Br
trans-2-Bromo-1-methylcyclohexanol
(a bromohydrin)
(89%)
Exercise 14.6
Predict the major products for each of the following reactions.
a)
Br2
H2O
b)
Cl2
c)
Br2
d)
Br2
Sample Solution
b)
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Cl
Cl2
Cl
14.7 Addition of Hydrogen
In a catalytic
hydrogenation
reaction, hydrogen
adds to a double bond
using a catalyst.
To adsorb is to
accumulate hydrogen
atoms on the surface of
the catalyst.
With heterogeneous
catalysis, the catalyst
is in a different phase
from the reactants.
Adding hydrogen to a double or triple bond is called a reduction
reaction. As the hydrogen adds to the π bond, the reaction follows a
stereospecific syn addition to the bond. A direct addition of hydrogen
requires the presence of a metal catalyst such as platinum, palladium,
rhodium, or nickel. Without the catalyst, the energy of activation is
too high to make the reaction of practical value as a synthetic method.
Adding hydrogen with the use of a catalyst is called a catalytic
hydrogenation.
C
C
H2
catalyst
H
H
C
C
When running a catalytic hydrogenation, chemists usually use
a finely powdered catalyst suspended in an alkane, alcohol, or acetic
acid solvent. They place the reaction mixture in a closed system, as
this allows them to measure the amount of hydrogen reacted. They
then stir or shake the whole reaction mixture, usually at room
temperature.
As the reaction begins, the surface of the catalyst adsorbs a
number of hydrogen molecules from the hydrogen gas. This adsorption
is a reversible process. Experimental evidence for reversible
adsorption was obtained by mixing two isotopes of hydrogen in the
presence of a catalyst. When a mixture of hydrogen (H2) and
deuterium (D2) was placed in contact with a platinum catalyst, the gas
quickly reached an equilibrium with H2, D2, and HD present. This
scrambling did not take place in the absence of the catalyst. Catalytic
hydrogenation is an example of heterogeneous catalysis. With
heterogeneous catalysis, the catalyst is in a different phase than the
reactants. The catalyst is a solid, but the reactants are liquids or
gasses.
After adsorbing the hydrogen onto its surface, the catalyst
binds to one face of the π bond of the double bond in the substrate. The
catalyst then transfers one hydrogen to the double bond, followed by a
rapid transfer of a second hydrogen that frees the product from the
catalyst. When this transfer is complete, the catalyst adsorbs more
hydrogen and repeats the process. The result is that the hydrogens
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add stereospecifically in a syn fashion to the π bond. Figure 14.4 shows
this reaction.
C
C
H
C
H
C
C
H
H
Pt
C
H
Pt
H
Pt
Figure 14.4. The three steps in catalytic hydrogenation.
Following are some examples of catalytic hydrogenation.
CH3CH2CH
CH3
CH2
H2 (50 PSI)
o
Pt, 25 C
CH3CH2CH2CH3
Butane
(97%)
CH3
D
D2 (50 PSI)
o
Pt, 25 C
D
CH3
CH3
cis-1,2-Dideutero-1,2-Dimethylcyclohexane
(95%)
Isolated double or triple carbon—carbon bonds hydrogenate
more readily than carbon—oxygen double bonds. Hydrogenations
involving aldehydes and ketones require extended reaction times, and
hydrogenations involving carboxylic acids require high temperature as
well as extended reaction times. Benzene is very difficult to
hydrogenate and requires very high pressures and temperatures.
O
O
H2 (50 PSI)
Pt, 25oC
Cyclohexanone
(97%)
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CH2CH
CH2
Daley & Daley
CH2CH2CH3
H2 (50 PSI)
o
Pt, 30 C
Propylbenzene
(98%)
Alkynes
hydrogenate
more
rapidly
than
alkenes.
Hydrogenation with a platinum or palladium catalyst reduces the
alkyne all the way to the alkane.
CH2C
CH
H2 (50 PSI)
o
Pt, 25 C
CH2CH2CH3
Propylbenzene
(97%)
Lindlar’s catalyst is a
deactivated
hydrogenation catalyst
suitable for
hydrogenating triple,
but not double,
carbon—carbon bonds.
Sometimes chemists want to stop the reaction of an alkyne at the
alkene. To do so, they use a deactivated, or “poisoned,” catalyst. A
deactivated, or “poisoned,” catalyst is a catalyst of palladium coated
with particles of barium sulfate treated with quinoline. This catalyst,
called Lindlar's catalyst, reduces an alkyne to a cis alkene.
CH3CH2C
CCH2CH3
H2 (55 PSI)
Pd/BaSO4
CH3CH2
o
Quinoline, 30 C
CH2CH3
C
C
H
H
Z-3-Hexene
(81%)
C
CCH3
H2 (50 PSI)
Pd/BaSO4
H
H
C
C
CH3
o
Quinoline, 30 C
Z-1-Phenyl-1-propene
(87%)
To form a trans alkene from an alkyne, the addition of
hydrogen must take place with anti stereochemistry. Chemists
accomplish the anti stereochemistry by reacting the alkyne with
sodium in liquid ammonia.
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741
C
C
Daley & Daley
H
Na
NH3
R
R
C
C
R
H
A solution of sodium in liquid ammonia has a deep blue color due to
the solvation of electrons from the sodium with the ammonia.
NH3
(NH3)e
Blue
Na
+
+
Na
The mechanism for this reaction initially involves the addition of an
electron to the triple bond to form a radical anion. The radical anion
forms with the nonbonding pair and unpaired electrons trans to one
another because the trans configuration places the electrons farther
apart than they would be if they were in the cis configuration.
R
C
C
R
e
R'
•
C
••
C
R'
Next in the mechanism, the radical anion intermediate removes a
proton from the ammonia solvent to become a neutral radical.
R
•C
H
••
C
NH2 R
R'
H
•C
C
R'
The neutral radical then picks up another electron from the solution to
form a vinyl anion.
R'
C
C
H
R'
e
•
C
••
C
H
R
R
Finally, the ammonia solvent protonates this vinyl anion to form the
trans double bond.
R'
C
H
H
••
NH2
R'
H
C
C
R
H
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5 July 2005
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Here are some examples of this reaction.
CH3CH2CH2C CCH3
CH3CH2CH2
Na
NH3
H
C
C
H
CH3
E-2-Hexene
(84%)
C
H
Na
C
NH3
C
C
H
E-1,2-Dicyclohexylethene
(93%)
Exercise 14.7
Predict the major products for each of the following reactions.
a)
C
D2
CH
Pt
b)
CH3CH2
H
C
H2
C
H
Pt
CH3
c)
C
C
CH3
Na, NH3
d)
CH3CH2
H
C
H
D2
C
CH3
Pt
Sample Solution
c)
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H
C
C
C
Na, NH3
CH3
CH3
C
H
14.8 Dihydroxylation Reactions
A dihydroxylation
reaction forms a diol
from, in this case, a
double bond.
A dihydroxylation reaction adds two —OH groups to a
double bond to form a diol. The reaction of an alkene with cold
alkaline potassium permanganate (KMnO4) or with osmium tetroxide
(OsO4) in an aqueous hydrogen peroxide solution readily yields a 1,2dihydroxyalkane. 1,2-Dihydroxyalkanes are often called glycols. With
KMnO4 and OsO4, the dihydroxylation reaction yields a syn diol.
C
C
OsO4
H2O, H2O2
C
C
OH OH
OsO4 and KMnO4 react in similar ways to form an
intermediate cyclic ester from an alkene. In a concerted reaction, two
of the oxygens from the reagent add to the double bond of an alkene
forming the cyclic ester.
C
C
O
O
O
Os
O
Os
O
O
O
O
The hydrogen peroxide in the reaction mixture then readily oxidizes
the cyclic ester to produce the diol and regenerate the osmium
tetroxide.
H2O2
O
O
Os
O
+
HO
OsO4
OH
O
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Osmium tetroxide is an expensive, highly toxic reagent that
gives high yields and is very selective for carbon—carbon multiple
bonds. Potassium permanganate is much less expensive, less toxic,
gives lower yields and oxidizes a number of other functional groups.
However, in most cases, chemists prefer to use potassium
permanganate.
OH
CH3
OsO4
CH3
CH3
OH
H2O, H2O2
CH3
cis-1,2-Dimethylcyclopentane-1,2-diol
(81%)
HO
OH
KMnO4
KOH, H2O
H
H
erythro-Octane-4,5-diol
(73%)
Exercise 14.8
Chemists often use permanganate ion to detect the presence of an
alkene. They do so because the solution is purple and, as it reacts with
an alkene, the solution becomes colorless and a brown precipitate of
MnO2 forms. Reasoning by analogy to the OsO4 reaction, write a
mechanism for the oxidation of an alkene by cold alkaline potassium
permanganate solution.
Three-membered cyclic ethers, known as epoxides, react to
form trans diols. The cyclic ether is very strained and readily reacts
with aqueous acid to produce a diol. The most common method of
forming an epoxide is the reaction of an alkene with a peroxy acid.
O
H
CH3COOH
H
H
CHCl3
H
O
cis-3,4-Epoxyhexane
(67%)
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An epoxidation
reaction forms a threemembered cyclic ether
from an alkene.
745
Daley & Daley
An epoxidation reaction with a peracid accomplishes a syn
addition of oxygen to a double bond. This reaction is stereospecific,
indicating that the addition of the oxygen to the double bond is a
concerted reaction. The generally accepted mechanism for an
epoxidation reaction involves the transfer of the oxygen farthest from
the carbonyl carbon to the double bond.
C
C
••
••
O
C
+
•• ••
O
C
C
••O••
••
•
•
O
H
C
O
••
•
•
H
••
O••
H3C
H3C
Because the reaction is concerted, the substrate has no opportunity to
rotate and change its geometry. Thus, a cis alkene becomes a cis
epoxide, and a trans alkene becomes a trans epoxide.
CH3
O
CH3COOH
CHCl3
CH3
CH3
O
CH3
1,2-Dimethyl-1,2-epoxycyclohexane
(52%)
O
O
CH3COOH
CHCl3
H
H
(62%)
Exercise 14.9
Chemists commonly use meta-chloroperoxybenzoic acid (MCPBA) to
form epoxides because of its desirable solubility characteristics. What
products form from the reaction of MCPBA with Z-2-methyl-3-hexene?
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O
Cl
COOH
m-Chloroperoxybenzoic acid
Any moderately strong aqueous acid can protonate the epoxide
oxygen. Three-membered rings have so much ring strain that they
react more readily than larger rings or acyclic compounds. Because of
this ring strain, a protonated epoxide reacts with a nucleophile. The
reaction is essentially the reaction of a bromonium ion with a
nucleophile, in this case water, adding to the backside of the epoxide.
The reaction product is a vicinal diol with a net anti hydroxylation.
••
••
•
•
O
H
OH
OH2
••
••
O
••
H
••
OH2
••
H2O••
-H
••
OH
••
••
OH
••
Solved Exercise 14.3
Any strong acid can open the ring of an epoxide when forming the vicinal
diol. Why is hydrochloric acid never used to form the diol? What product
forms in the reaction of epoxycyclohexane with hydrochloric acid?
Solution
- ion—a
Although hydrochloric acid is a strong acid, it also produces the Clc
- ion competes with water to form significant
good nucleophile. The Clc
amounts of 2-chlorocyclohexanol.
OH
OH
O
HCl
H2O
+
Cl
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Exercise 14.10
What reagents would you use to form meso-3,4-hexanediol and dl-3,4hexanediol from Z-3-hexene?
14.9 Addition of Carbenes
A carbene is a neutral
disubstituted carbon
atom with two
nonbonding electrons.
It needs two more
electrons to complete its
octet.
This chapter repeatedly discusses occasions when a threemembered ring is either an intermediate or, in the case of an epoxide,
the final product of an addition reaction. The addition of a carbene to
a double bond is another of these occasions. When a carbene adds to a
double bond, the product is a cyclopropane ring.
Carbenes are unlike almost all other reagents presented in this
book in that the carbon of a carbene possesses only six electrons in its
valence shell and is electrically neutral. There are two forms of
carbenes. Both are sp2 hybridized and are relatively close in energy to
each other. The first, called a singlet carbene, has a pair of electrons
in an sp2 hybrid orbital, although the p orbital is empty. The second,
called a triplet carbene, has the one of electron each in the sp2 and p
orbitals. With some exceptions (halogens bonded to the deficient
carbon) the triplet state is somewhat more stable than the singlet
state because the two nonbonding electrons are further apart.
H C
H
Singlet
H C
H
Triplet
The two forms of carbene
Although carbenes are electrically neutral, they are very
reactive electrophiles because the carbon is electron deficient. In an
addition reaction, the carbene adds to the electron-rich π bond of an
alkene to form a cyclopropane ring.
C
C
•
•
CH2
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A photolysis reaction is
run by shining light of
the proper wavelength
onto the reaction.
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Methylene (:CH2) is the simplest example of a carbene. The methods
used to produce methylene involve heating diazomethane or exposing
it to ultraviolet light in a process called photolysis. The products of
both methods are methylene and nitrogen.
•
•
CH2
N
N••
N
CH2
••
•
•
N
or
UV light
•
•
CH2 +
N2
Diazomethane
Diazomethane, however, is difficult to work with because it is a very
toxic, unstable compound. It is so unstable that it sometimes
detonates without warning. Methylene generated from diazomethane
is also extremely reactive and not only forms the cyclopropane ring by
adding to the double bond but inserts itself into any C—H bonds in the
molecule.
CH3CH
CH2
CH2N2
H
CH3
+
CH3
C
CH3
H
E-2-Butene
26%
Methylcyclopropane
39%
CH3
CH3
C
H
C
+
CH3CH2CH
CH2
H
Z-2-Butene
20%
The Simmons-Smith
reagent is the product
of a reaction between
zinc and methylene
iodide.
A carbenoid reagent
forms products as
though it were a
carbene.
+
C
1-Butene
15%
Diazomethane's instability and lack of selectivity kept it from
being very useful as a routine laboratory reagent. Then two chemists
at DuPont discovered a synthetically useful alternate to
diazomethane. This reagent, the Simmons-Smith reagent, involves
the addition of methylene iodide (CH2I2) to copper or silver activated
zinc. In an addition reaction, the Simmons-Smith reagent forms an
organometallic reagent that reacts with alkenes similarly to a
carbene. Thus, chemists call a Simmons-Smith reagent a carbenoid
because it reacts similarly to a carbene.
CH2I2
Zn(Cu)
ICH2ZnI
Simmons-Smith
reagent
The Simmons-Smith reagent is both regiospecific and
stereospecific. Thus, the product of a Simmons-Smith reaction
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contains only cyclopropane rings that retain the stereochemistry of the
double bond in the substrate.
CH2I2
Zn(Cu)
H
CH3CH2
H
CH3
1-Ethyl-2-methylcyclopropane
(74%)
In an addition reaction, a Simmons-Smith reagent gives moderate to
good yields of the cyclopropane from an alkene. It also gives few, if
any, side products.
CH2I2
Zn(Cu)
Bicyclo[4.1.0]heptane
(63%)
A halogenated carbene forms by a reaction of a base with a
halogenated alkane. The reaction requires one hydrogen and at least
two halogens. The most commonly used halogens are CHCl3, CHBr3,
and RCHBr2. The presence of the halogens makes the hydrogen
slightly acidic and reactive with a 50% aqueous solution of sodium or
potassium hydroxide.
Br3C H
An α elimination
reaction eliminates
atoms or groups from
the same atom.
••
•
•
••
OH
Br3C ••
–Br
•
•
CBr2
In this dehydrohalogenation reaction, called an α elimination
reaction, the hydrogen and the halogen are eliminated from the same
carbon atom. All the eliminations covered in Chapter 13 are β
eliminations because the groups leaving are on adjacent carbon atoms.
The halocarbene that results from the α elimination adds to a double
bond to form a halogenated cyclopropane.
CHBr3
Br
KOH
Br
7,7-Dibromobicyclo[4.1.0]heptane
(87%)
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Phase Transfer Catalysis
The formation of a halogenated carbene usually involves a two-phase reaction.
Because the base is insoluble in the organic solvent, a phase transfer catalyst is used
to move the hydroxide ion from the aqueous layer into the organic layer. A common
type of phase transfer catalyst is a quaternary ammonium salt. An example is
benzyltriethylammonium chloride. As the chloride, benzyltriethylammonium chloride
is insoluble in chloroform but quite soluble in water. However, as the quaternary
ammonium hydroxide, it is somewhat soluble in the organic solvent. Thus, it can
transport the hydroxide ion from the aqueous layer to the organic where the
hydroxide ion reacts. After reacting, the halide ion is transported back to the organic
layer.
CH2CH3 Cl
CH2N
CH2CH3
OH
CH2CH3
CH2N
Cl
CH2CH3
More soluble in water
than in chloroform.
OH
CH2CH3
CH2CH3
More soluble in chloroform
than the quaternary ammonium
chloride.
Synthesis of 7,7-Dichlorobicyclo[4.1.0]heptane
Cl
CHCl3
NaOH
Cl
7,7-Dichlorobicyclo[4.1.0]heptane
(83%)
Place 0.5 mL (0.005 mol) of cyclohexene in a flask. Add 1.25 mL of 50% aqueous
sodium hydroxide and 1.25 mL (0.012 mol) of chloroform. Place a magnetic stir bar
into the flask and add 50 mg of benzyltriethylammonium chloride. Cap the flask to
prevent loss of vapors and stir vigorously for 12 hours at room temperature. An
emulsion should form during the reaction time. Add 4 mL of water and 1.5 mL of
methylene chloride to the reaction mixture. Stir gently to break the emulsion.
Separate the layers. Add another 1.5 mL of methylene chloride to the reaction
mixture. Stir for 5 minutes and separate the layers. Repeat with a third portion of
methylene chloride. Combine the organic layers and wash with 2 mL of saturated
sodium chloride solution. Separate the layers and dry the organic layer over
anhydrous sodium sulfate. Remove the drying agent and evaporate the solvent using
a rotary evaporator. Yield of product is 0.68 g (83%), b.p. 195-200oC.
Discussion Questions
1. Why is it necessary to stir the reaction vigorously? What would happen to the
product yield if the reaction mixture was not stirred vigorously?
2. Near the end of the isolation of the product, the reaction mixture is washed with
saturated sodium chloride solution. Why?
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Exercise 14.11
Predict the major products for each of the following reactions.
a)
CH3
CH3
C
H
CH3CHBr2
C
CH2CH3
NaOH
b)
CHCl3
NaOH
c)
CH2I2
Zn(Cu)
d)
CHBr2
KOH
Sample Solution
b) This reaction produces two isomers in which the threemembered rings are either cis or trans to each other.
Cl
Cl
NaOH Cl
Cl
CHCl3
14.10 Oxidation of Alkenes
The oxidation of an alkene generally involves breaking the
carbon—carbon double bond to form two carbonyl-containing
compounds.
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R3
R1
C
R2
For more about the
reaction of cold
alkaline potassium
permanganate, see
Section 14.8, page 000.
C
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R3
R1
Oxidation
C
O
+ O
C
H (or OH)
R2
H
Whether the carbonyl compounds produced are aldehydes, ketones, or
carboxylic acids depends on the oxidizing reagent and on the alkene
used.
Cold alkaline potassium permanganate forms a vicinal diol. If
the solution is too concentrated, too warm, or too acidic, oxidative
cleavage occurs, and mixtures of ketones and carboxylic acids form.
H3O
KMnO4, OH
O +
H2O, warm
O
OH
Ozonolysis is the
reaction of ozone (O3)
with an alkene.
A more widely used method for the oxidative cleavage of double
bonds is ozonolysis. Ozone is readily prepared by passing an electric
discharge through a stream of oxygen gas. This mixture of ozone and
oxygen is then bubbled into a solution of the alkene forming a
compound called a molozonide. The molozonide is quite unstable and
spontaneously rearranges to form an ozonide.
C
•
•
O
••
C
••
••
O
O••
••
•
•
C
C
•
•
•
•
O
••
O••
O
C
•
•
C
O
•
•
••
O••
O
••
••
••
Molozonide
•• ••
O
C
C
••
O
••
••
O••
Ozonide
Ozonides are not much more stable than the molozonides, so
they are rarely isolated. Low molecular weight ozonides are frequently
explosive. Ozonides are treated with a reducing agent to convert them
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to carbonyl compounds. One of the most common reducing agents is
zinc metal in acid.
•• ••
O
C
•• ••
O
••
Zn
C
O•
H3O
••
O•• +
C
••
O
•
•
C
•
Following are some examples of ozonolysis.
1) O3
2) Zn, H3O
O
+ O
H
2-Pentanone
Propanal
(63%)
1) O3
CHO
2) Zn, H3O
CHO
Hexanedial
(72%)
Exercise 14.12
Predict the major products for each of the following reactions.
a)
1) O3
2) Zn, H3O
b)
KMnO4, OH H3O
H2O, warm
c)
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O
1) O3
OCH3 2) Zn, H O
3
Sample Solution
a)
1) O3
CHO
CHO
2) Zn, H3O
Before the development of spectroscopy, ozonolysis was widely
used to determine the position of double bonds in alkenes. After an
ozonolysis reaction, the structure of the alkene can be reconstructed
by the position of the two oxygen atoms from the carbonyl groups.
Simply connect the carbon atoms to which the oxygen atoms are
bonded to form a double bond. The only uncertainty is whether the
double bond has an E or Z configuration.
Ozonolysis
or
O
+
O
H
Exercise 14.13
Propose structures for the alkenes that would yield the following
carbonyl compounds on reaction with ozone followed by zinc metal in
acid.
a)
O
+
O
b)
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O
+
c)
O
O
Sample solution
b)
Or
Ozonolysis
O
O
+
Key Ideas from Chapter 14
❏
A carbon—carbon double or triple bond undergoes an addition
reaction bonding a new atom or group of atoms to each of the
carbons of the original multiple bond.
❏
One of the groups that adds to the multiple (π) bond is an
electrophile, the other is a nucleophile.
❏
An addition reaction to a multiple bond follows one of two
possible reaction mechanisms, AdE2 and AdE3. In the AdE2
mechanism, the electrophile adds to the π bond forming a
cation. In the AdE3 mechanism, both the nucleophile and the
electrophile add to the π bond in a concerted addition.
❏
The rate of reaction for an AdE2 mechanism depends on the
concentration of both the electrophile and the substrate. Thus,
the mechanism follows second order kinetics.
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❏
Many AdE2 reactions proceed through a carbocation. The ones
that do lose the stereochemistry contained in the reaction site
of the substrate.
❏
Other AdE2 reactions, especially those where the electrophilic
reagent has nonbonding electron pairs, form a three-membered
ring intermediate. The net reaction is usually an anti addition
to the multiple bond.
❏
The rate of reaction for the AdE3 mechanism depends on the
concentrations of the electrophile, nucleophile, and substrate.
Thus, the mechanism follows third-order kinetics and is a
concerted reaction. The net reaction is normally an anti
addition to the multiple bond.
❏
As an addition to a carbon—carbon multiple bond proceeds, the
electrophile adds to the less highly substituted carbon, and the
nucleophile adds to the more highly substituted carbon. This
direction of adding is called Markovnikov's rule.
❏
Hydrogen halides add to the π bonds of alkenes and alkynes to
form organohalogen compounds. In a polar-protic solvent this
reaction proceeds through a carbocation. In a polar-aprotic
solvent the reaction proceeds through an AdE3 mechanism.
❏
An acid-catalyzed addition of water is an AdE2 reaction and
proceeds via a carbocation.
❏
The oxymercuration reaction proceeds through an AdE2
reaction via a cyclic mercurinium ion intermediate.
Nucleophilic substitution by water followed by reduction with
NaBH4 produces an alcohol with net anti addition. The reaction
is both stereospecific and regiospecific.
❏
Hydroboration is a stereospecific and regiospecific syn addition
to an alkene forming an organoborane. The organoborane can
be oxidized to form an alcohol. The alcohol is the antiMarkovnikov product.
❏
Halogens add to alkenes to form vicinal dihalides. For bromine,
the reaction proceeds stereospecifically via a three-membered
ring bromonium ion. With chlorine, the reaction is less
stereospecific. The reaction does not work well with iodine.
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❏
In a water solvent, the water competes with the bromine
nucleophile and reacts with the bromonium ion to form a
halohydrin product.
❏
Catalytic hydrogenation reduces a multiple bond by adding
hydrogen to the π bond. Catalysts such as platinum, palladium,
or nickel are the most common ones in use. Hydrogen adds to
the multiple bond to form an alkane.
❏
Lindlar's catalyst catalyzes the reduction of a triple bond to a
double bond in a syn addition to form a cis alkene. The reaction
stops there; the double bond is not reduced.
❏
Reaction of a triple bond with sodium in liquid ammonia
reduces a triple bond to a double bond in an anti addition to
form a trans alkene.
❏
Reaction of an alkene with osmium tetroxide or potassium
permanganate forms a vicinal diol. The reaction is syn
stereospecific.
❏
An epoxide forms from an alkene by a syn addition of oxygen
using a peroxy carboxylic acid as the reagent.
❏
Hydrolysis of an epoxide forms a vicinal diol. The diol is a net
anti addition of the hydroxyl groups.
❏
The addition of a carbene to a double bond forms a
cyclopropane ring. Carbene generated from diazomethane also
inserts a methylene group into any carbon—hydrogen bond in
the molecule.
❏
The Simmons-Smith reagent is called a carbenoid reagent
because it adds to a double bond as a carbene and forms a
cyclopropane ring. Chemists usually choose the SimmonsSmith reagent in preference to diazomethane because the
Simmons-Smith reagent is regiospecific for a double bond.
❏
Halogenated carbenes can form from an α elimination of a
halogenated hydrocarbon. A carbon bearing at least one
hydrogen and two halogens eliminates the hydrogen and one of
the halogens to form a carbene.
❏
Alkenes readily oxidize with warm potassium permanganate to
form ketones or carboxylic acids. Ozonolysis, followed by
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reaction with zinc and acid, forms ketones and aldehydes from
alkenes.
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