Abia Polytechnic University Reactions of Alkene Questions

Chapter 7 (review)Presentation Lecture
Structure and Synthesis
of Alkenes
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Slide – 1
Introduction
• Alkenes are hydrocarbons with carbon-carbon double
bonds.
• Alkenes are also called olefins, meaning
“oil-forming gas.”
• The functional group of alkenes is the
carbon-carbon double bond, which gives this group its
reactivity.
• Alkenes are characterized by the reactions of their double
bonds
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Slide – 2
Ethylene is the largest-volume industrial organic compound, used to make
polyethylene and a variety of other industrial and consumer chemicals.
Pinene is a major component of turpentine, the paint solvent distilled from extracts of
evergreen trees.
Muscalure (cis-tricos-9-ene) is the sex attractant of the common housefly.
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Orbital Description of the Alkene
Double bond.
• Sigma bonds around the double-bonded carbon are sp2
hybridized.
• Angles are approximately 120° and the molecular
geometry is trigonal planar.
• Unhybridized p orbitals with one electron will overlap,
forming the double bond (pi bond).
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Slide – 4
Sigma Bonds of Ethylene
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Slide – 5
Bond Lengths and Angles
• sp2 hybrid orbitals have more s character than the
sp3 hybrid orbitals.
• Pi overlap brings carbon atoms closer, shortening the C—C
bond from 1.54 Å in alkanes down to 1.33 Å in alkenes.
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Slide – 6
Pi Bonding in Ethylene
• The pi bond in ethylene is formed by overlap of the unhybridized p orbitals of the sp2 hybrid carbon atoms.
• For pi overlap to occur, these p orbitals must be parallel, which requires that the two carbon atoms be oriented with
all their C−H bonds in a single plane. Half of the pi-bonding orbital is above the C−C sigma bond, and the other half
is below the sigma bond
• Each carbon has one unpaired electron in the p orbital.
• This overlap requires the two ends of the molecule to be coplanar.
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Slide – 7
Cis-Trans Interconversion
• Cis and trans isomers cannot be interconverted.
• No rotation around the carbon-carbon bond is possible without breaking the pi bond (264 kJ/mole).
• Unlike single bonds, a carbon–carbon double bond does not permit rotation. Six atoms, including
the double-bonded carbon atoms and the four atoms bonded to them, must remain in the same
plane. This is the origin of cis-trans isomerism. If two groups are on the same side of a double bond
(cis), they cannot rotate to opposite sides (trans) without breaking the pi bond.
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Slide – 8
IUPAC Nomenclature
• Find the longest continuous carbon chain that includes the double-bonded carbons.
• Ending -ane changes to -ene.
• Number the chain so that the double bond has the lowest possible number.
• In a ring, the double bond is assumed to be between carbon 1 and carbon 2.
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Slide – 9
IUPAC and New IUPAC
When the chain contains more than three carbon atoms, a number is used to give the location of the
double bond. The chain is numbered starting from the end closest to the double bond, and the double
bond is given the lower number of its two double-bonded carbon atoms. Cycloalkenes are assumed to
have the double bond in the number 1 position.
In 1993, the IUPAC recommended a logical change in the positions of the numbers used in names. Instead
of placing the numbers before the root name (1-butene), it recommended placing them immediately before
the part of the name they locate (but-1-ene). The new placement is helpful for clarifying the names of
compounds containing multiple functional groups.
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Slide – 10
Ring Nomenclature
In a ring, the double bond is assumed to be
between carbon 1 and carbon 2.
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Slide – 11
Multiple Double Bonds
• A compound with two double bonds is a diene. A triene has three double bonds, and a tetraene
has four. Numbers are used to specify the locations of the double bonds.
• Give the double bonds the lowest numbers possible.
• Use di-, tri-, or tetra- before the ending -ene to specify how many double bonds are present.
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Slide – 12
• Each alkyl group attached to the main chain is listed with a number to give its location.
Note that the double bond is still given preference in numbering, however.
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Slide – 13
Alkenes As Substitutents
When a double bond is not part of the main chain, we name the group containing the double bond as a
substituent called an alkenyl group. Alkenyl groups can be named systematically (ethenyl, propenyl,
etc.) or by common names. Common alkenyl substituents are the vinyl, allyl, methylene, and phenyl
groups. The phenyl group (Ph) is different from the others because it is aromatic (see ​Chapter 16​) and
does not undergo the typical reactions of alkenes.
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Slide – 14
Cis-Trans Isomers
•
A double bond that gives rise to cis-trans stereoisomerism is called stereogenic, and it has two
possible configurations. If two of the same groups (often hydrogen) are bonded to the same side
of the carbons of the double bond, the alkene is the cis isomer.
•
Also called geometric isomerism
•
If there are similar groups on same side of the double bond, alkene is cis.
•
If there are similar groups on opposite sides of the double bond, alkene is trans.
•
Not all alkenes show cis-trans isomerism.
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Slide – 15
E-Z Nomenclature
• Use the Cahn-Ingold-Prelog rules to assign priorities to
groups attached to each carbon in the double bond.
• If high-priority groups are on the same side, the name is
Z (for zusammen).
• If high-priority groups are on opposite sides, the name is
E (for entgegen).
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Slide – 16
Example
• Assign priority to the
substituents according
to their atomic number
(1 is highest priority).
• If the highest priority groups
are on opposite sides, the
isomer is E.
• If the highest priority groups
are on the same side, the
isomer is Z.
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Slide – 17
Cyclic Stereoisomers
• Double bonds outside the ring can show stereoisomerism.
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Slide – 18
Stereochemistry in Dienes
• If there is more than one double bond in the molecule,
the stereochemistry of all the double bonds should be
specified.
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Slide – 19
Organic Chemistry
Lecture Presentation
Chapter 8
Reactions of Alkenes
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Slide – 1
Reactivity of the Carbon–Carbon Double Bond
• C=C contains two pair of electrons
• Is shorter than C-C single bond: C=C double bonds are electron rich.
• Alkenes react with electrophiles (Lewis acids) (opposite of nucleophiles)
• Net result is typically addition
• The most common reactions of double bonds transform the pi bond into a
sigma bond
•
Typical mechanism (there are several):
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Slide – 2
Thermodynamics of Addition (sec 8-1)
A
C
C
+ A-Z
C
C
Z
• Bonds made: s (C-A) and s (C-Z).
• Bonds broken: s (A-Z) and p (C=C).
MOST ADDITIONS ARE EXOTHERMIC.
Thermodynamics of Addition (sec 8-1)
• catalytic hydrogenation converts the C=C pi bond and the H−H sigma bond into two C−H sigma bonds (
Section 7-8). The reaction is exothermic (ΔH°= about −80 to −120 kJ/mol or about −20 to −30 kcal/mol),
showing that the product is more stable than the reactants.
For example, the reaction of ethylene with hydrogen (to give ethane) is strongly exothermic, but the rate
is very slow. A mixture of ethylene and hydrogen can remain for years without appreciable reaction.
Adding a catalyst such as platinum, palladium, or nickel allows the reaction to take place at a rapid rate.
Thermodynamics of Addition (sec 8-1)
So a reaction is exothermic: does it go?
• Not necessarily: activation energy may be low, or very high.
• Left (low activation energy): fast reaction.
• Right (high activation energy): slow reaction
.
• In principle, many different reagents could add to a double bond to form more stable
products; that is, the reactions are energetically favorable. Not all of these reactions have
convenient rates, however
8-2 Electrophilic Addition to Alkenes
• Some reagents react with carbon–carbon double bonds without the aid of a catalyst. HOW ?
the pi bond is delocalized above and below the sigma bond. The pibonding electrons are spread farther from the carbon nuclei, and they
are more loosely held.
A strong electrophile has an affinity for these loosely held electrons. It
can pull them away to form a new bond, leaving one of the carbon
atoms with only three bonds and a positive charge: a carbocation. In
effect, the double bond has reacted as a nucleophile, donating a pair of
electrons to the electrophile.
Most addition reactions involve a second step in which a nucleophile
attacks the carbocation (as in the second step of the SN1 reaction),
forming a stable addition product.
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Slide – 6
8-2 Electrophilic Addition to Alkenes
KEY MECHANISM: Electrophilic Addition to Alkenes
• Step 1: Attack of the pi bond on the electrophile forms a carbocation..
First, a strong electrophile attracts the loosely held electrons from the
pi bond of an alkene. The electrophile forms a sigma bond to one of
the carbons of the (former) double bond, while the other carbon
becomes a carbocation.
• Step 2: Nucleophile attacks the carbocation.
The carbocation (a strong electrophile) reacts with a nucleophile (often
a weak nucleophile) to form another sigma bond
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Slide – 7
8-2 Electrophilic Addition to Alkenes
EXAMPLE: Ionic addition of HBr to but-2-ene
• Step 1: Protonation of the double bond forms a carbocation.
The proton in HBr is electrophilic; it reacts with the alkene
to form a carbocation
• Step 2: Bromide ion attacks the carbocation.
Bromide ion reacts rapidly with the carbocation
to give a stable product in which the elements of
HBr have added to the ends of the double bond.
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Slide – 8
Types of Additions to Alkenes
Table 8-1 summarizes the classes of additions we will cover.
Note that the table shows what elements have added across the
double bond in the final product, but it says nothing about reagents or
mechanisms.
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Slide – 9
8-3 Addition of Hydrogen Halides to Alkenes
• Step 1 is the protonation of the double bond.
• The protonation step forms the most stable carbocation possible.
• In step 2, the nucleophile attacks the carbocation, forming an alkyl halide.
• HBr, HCl, and HI can be added through this reaction.
Step 1: Protonation of the pi bond forms a carbocation.
Step 2: Attack by the halide ion gives the addition
product.
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Slide – 10
8-3 Addition of Hydrogen Halides to Alkenes
8-3A Orientation of Addition: Markovnikov’s Rule
• Lets consider the addition of HBr to 2-methylbut-2-ene
the addition of HBr to 2-methylbut-2-ene could lead to either of two products,
yet only one is observed. WHY??
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Slide – 11
8-3 Addition of Hydrogen Halides to Alkenes
8-3A Orientation of Addition: Markovnikov’s Rule
addition of HBr to 2-methylbut-2-ene
The first step is protonation of the double
bond. If the proton adds to the secondary
carbon, the product will be different from
the one formed if the proton adds to the
tertiary carbon.
The tertiary carbocation is more stable so the
first reaction is favored
Remember the Stability order of carbocations ?
3°>2°>1°>CH3+
The proton adds to the end of the double bond
that is less substituted to give the more
substituted carbocation (the more stable
carbocation).
An electrophile adds to a double bond to give the
most stable carbocation in the intermediate.
The second half of the mechanism
produces the final product of the
addition of HBr to 2-methylbut-2ene.
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Slide – 12
8-3 Addition of Hydrogen Halides to Alkenes
8-3A Orientation of Addition: Markovnikov’s Rule
• Lets again look at the addition of HBr to 2-methylbut-2-ene
What can we conclude ?
The addition of HBr (and other hydrogen halides) is said to be regioselective because in each case, one of the two
possible orientations of addition results preferentially over the other.
The regiochemistry a reaction, also called the orientation of addition, meaning which part of the reagent adds to
which end of the double bond.
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Slide – 13
8-3 Addition of Hydrogen Halides to Alkenes
8-3A Orientation of Addition: Markovnikov’s Rule
• EXAMPLE: The ionic addition
EXAMPLE: The ionic addition of HBr to propene shows protonation of
the less substituted carbon to give the more substituted carbocation.
Reaction with bromide ion completes the addition.
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Slide – 14
8-3 Addition of Hydrogen Halides to Alkenes
8-3A Orientation of Addition: Markovnikov’s Rule
• Markovnikov’s rule: The addition of a proton acid to the double bond of an alkene results in a
product with the acid proton bonded to the carbon atom that already holds the greater number of
hydrogens.
• We are often interested in adding electrophiles other than proton acids to the double bonds of
alkenes. Markovnikov’s rule can be extended to include a wide variety of other additions, based on
the addition of the electrophile in such a way as to produce the most stable carbocation.
Markovnikov’s rule (extended): In an electrophilic addition to an alkene, the electrophile adds in
such a way that it generates the most stable intermediate.
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Slide – 15
8-3 Addition of Hydrogen Halides to Alkenes
8-3A Orientation of Addition: Markovnikov’s Rule
• HBr adds to
HBr adds to 1-methylcyclohexene to give the product with an additional
hydrogen bonded to the carbon that already had the most bonds to hydrogen
(one) in the alkene. Note that this orientation results from addition of the proton
in the way that generates the more stable carbocation.
An electrophile adds to the less substituted end of the double bond to
give the more substituted (and therefore more stable) carbocation.
Other examples:
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Slide – 16
Rearrangements in H-Cl Additions
•The alkene shown adds the elements of H-Cl (notice
that the starting material has one degree of
unsaturation but the product has none), but chlorine
comes out one carbon away from the more
substituted end of the original C=C bond
• See the next slide.
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Slide – 17
Rearrangement, ctd.
First, the alkene is protonated in the normal way.
This generates a secondary cation next to a tertiary center.
In Organic I, you saw E1 and SN1 reactions in which cations of this type
rearranged. A single shift of H gives a tertiary cation, which is more
stable. (The two H’s shown are the two involved in the chemistry: as an
exercise, locate the other H’s in the molecule.)
Finally, the tertiary cation is captured by chloride.
8-3B Free-Radical Addition of HBr: Anti-Markovnikov Addition
• In 1933, M. S. Kharasch and F. W. Mayo found that some additions of HBr (but not HCl or HI) to alkenes gave products that
were opposite those expected from Markovnikov’s rule.
• These anti-Markovnikov reactions were most likely when the reagents or solvents came from old supplies that had
accumulated peroxides from exposure to the air.
• Peroxides give rise to free radicals that initiate the addition, causing it to occur by a radical mechanism.
H-Br Addition: a Poorly Behaved System
Why is this observed ?
Peroxides: the Key Ingredient
that the key ingredients are compounds called peroxides. These contain O-O single bonds. If they are present, H-Br adds in antiMarkovnikov fashion; if they are absent, H-Br obeys Markovnikov’s rule.
To get clean Markovnikov addition, you must exclude peroxides explicitly, by treating your solvent with a reducing agent. The
problem was difficult because hydrocarbons normally contain traces of peroxides
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Slide – 19
8-3B Free-Radical Addition of HBr: Anti-Markovnikov Addition
The Peroxide Effect
MECHANISM 8-3: Free-Radical Addition of HBr to Alkenes
Initiation: Formation of radicals.
Peroxides give rise to free radicals that initiate the addition, causing it to
occur by a radical mechanism. The oxygen–oxygen bond in peroxides is
rather weak, (Like all single bonds between two electronegative atoms, OO bonds are weak.) so it can break to give two alkoxyl radicals.
Alkoxyl radicals (R−O ⋅) initiate the anti-Markovnikov addition of HBr.
Propagation: A radical reacts to generate another radical.
Step 1: A bromine radical adds to the double bond to generate an alkyl radical on the
more substituted carbon atom.
Radicals are electrophilic because they do not have octets. In most
cases, a radical needs one more electron for its octet.
Remember the Stability of radicals:
3°>2°>1°>⋅CH3
A radical adds to a double bond to give the most
stable radical in the intermediate.
Step 2: The alkyl radical abstracts a hydrogen atom from HBr to generate the product
and a bromine radical.
The bromine radical generated in Step 2 goes on to react with another molecule of
alkene in Step 1, continuing the chain.
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Slide – 20
8-3B Free-Radical Addition of HBr: Anti-Markovnikov Addition
The Peroxide Effect
• EXAMPLE: Free-radical addition of HBr to propene.
Initiation: Radicals are formed.
Propagation: A radical reacts to generate another radical.
Step 1: A bromine radical adds to the double bond to generate an alkyl radical on the secondary carbon atom.
Step 2: The alkyl radical abstracts a hydrogen atom from HBr to generate the product and a bromine radical.
The bromine radical generated in Step 2 goes on to react with another molecule of the alkene in another Step 1, continuing the chain.
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Slide – 21
8-3B Free-Radical Addition of HBr: Anti-Markovnikov Addition
The Peroxide Effect
• The reversal of orientation in the presence of peroxides is called the peroxide effect. It occurs only with the addition of HBr to
alkenes. The peroxide effect is not seen with HCl or with HI, why ?
• Compare the second step :
The peroxide effect is not seen with HCl because the second step, the reaction of an alkyl radical with HCl and HI, is strongly endothermic.
Only HBr has just the right reactivity for each step of the free-radical chain reaction to take place.
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Slide – 22
8-3B Free-Radical Addition of HBr: Anti-Markovnikov Addition
The Peroxide Effect
Radical Addition of HBr to Unsymmetrical Alkenes
Adding the bromine radical to the secondary end of the double bond forms a tertiary radical.
The electrophile (in this case, Br⋅) adds to the less substituted end of the double bond, and the unpaired
electron appears on the more substituted carbon to give the more stable free radical.
This intermediate tertiary radical reacts with HBr to give the anti-Markovnikov product, in which H has added to
the more substituted end of the double bond: the end that started with fewer hydrogens.
Note that both mechanisms for the addition of HBr to an alkene (with and without peroxides) follow our extended statement of Markovnikov’s rule: In both cases, the
electrophile adds to the less substituted end of the double bond to give the more stable intermediate, either a carbocation or a free radical. In the ionic reaction, the
electrophile is H+. In the peroxide-catalyzed free-radical reaction, Br⋅ is the electrophile.
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Slide – 23
Solved Problem
Show how you would accomplish the following synthetic conversions.
(a) Convert 1-methylcyclohexene to 1-bromo-1-methylcyclohexane.
SOLUTION
This synthesis requires the addition of HBr to an alkene with Markovnikov orientation. Ionic addition of HBr gives
the correct product.
(b) Convert 1-methylcyclohexanol to 1-bromo-2-methylcyclohexane.
SOLUTION
This synthesis requires the conversion of an alcohol to an alkyl bromide with the bromine atom at the neighboring
carbon atom. This is the anti-Markovnikov product, which could be formed by the radical-catalyzed addition of HBr
to 1-methylcyclohexene.
Organic Chemistry, 9th Edition
L.G. Wade, Jr.
© 2017 Pearson Education, Inc.
Problem
• Predict the major products of the following reactions, and propose mechanisms to
support your predictions.
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Slide – 25
8-4 Addition of Water: Hydration of Alkenes
An alkene may react with water in the presence of a strongly acidic catalyst to form an alcohol. Formally, this reaction is a hydration (the
addition of water), with a hydrogen atom adding to one carbon and a hydroxy group adding to the other.
• Addition of water to the double bond forms an alcohol.
• The addition follows Markovnikov’s rule.
• This is the reverse of the dehydration of alcohol.
• It uses dilute solutions of H2SO4 or H3PO4 to drive equilibrium toward hydration.
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Slide – 26
8-4 Addition of Water: Hydration of Alkenes
• Hydration of an alkene is the reverse of the dehydration of alcohols
• For dehydrating alcohols, a concentrated dehydrating acid (such as H2SO4 or H3PO4) is used to drive the equilibrium to favor
the alkene.
• Hydration of an alkene, on the other hand, is accomplished by adding excess water to drive the equilibrium toward the alcohol
and It uses dilute solutions of H2SO4 or H3PO4
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Slide – 27
8-4 Addition of Water: Hydration of Alkenes
Mechanism
Step 1: Protonation of the double bond forms a carbocation.
Step 2: Nucleophilic attack of water gives a protonated alcohol
Step 3: Deprotonation gives the alcohol
H+ is consumed in the first step but released in the last, so it is a true catalyst
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Slide – 28
8-4 Addition of Water: Hydration of Alkenes
Orientation of Hydration.
The protonation follows Markovnikov’s rule: The proton adds to the less substituted end of the double bond, so the positive
charge appears at the more substituted end (most stable carbocation).
Step 1 of the hydration mechanism is similar to the first step in the addition of HBr. The proton adds to the less substituted end of
the double bond to form the more substituted carbocation.
Water attacks the carbocation to give (after loss of a proton) the alcohol with the −OH group on the more substituted carbon.
Like the addition of hydrogen halides, hydration is regioselective: It follows Markovnikov’s rule, giving a product in which the new
hydrogen has added to the less substituted end of the double bond
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Slide – 29
8-4 Addition of Water: Hydration of Alkenes
Example
• Acid-catalyzed hydration of propene.
Step 1: Protonation of the double bond forms a secondary carbocation.
Step 2: Nucleophilic attack by water gives a protonated alcohol.
Step 3: Deprotonation gives the alcohol.
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Slide – 30
8-4 Addition of Water: Hydration of Alkenes
Rearrangements Are Possible
Like other reactions that involve carbocation intermediates, hydration may take place with rearrangement. For
example, when 3,3-dimethylbut-1-ene undergoes acid-catalyzed hydration, the major product results from
rearrangement of the carbocation intermediate.
WHY??
• A methyl shift after protonation will produce the more stable tertiary carbocation.
Another rearrangement
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Slide – 31
8-4 Addition of Water: Hydration of Alkenes
Possibility of Ploymerization
• The cation that’s generated first can react with many different nucleophiles, one of which is another alkene. This
generates a cation, which can capture another alkene; which generates a cation, which can capture another alkene;
and so forth. Cationic polymerization creates many commercial plastics
• To control the chemistry of cations requires close attention to specific conditions: concentrations, temperatures,
stirring rates, catalysts. Chemical engineers make good money by doing this.
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Slide – 32
8-4 Addition of Water: Hydration of Alkenes
• Direct hydration is used in industry, but it’s a poor laboratory (small-scale) synthesis,
because as we have seen its cationic intermediate is prone to two kinds of side
reactions, rearrangement and ploymerization.
Also many alkenes do not easily undergo hydration in aqueous acid. Some alkenes are
nearly insoluble in aqueous acid
• Is there a solution ???
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Slide – 33
8-5 Hydration by Oxymercuration–Demercuration
Laboratory Hydration: Markovnikov
A well-behaved, laboratory-scale Markovnikov hydration of alkenes has been developed by Purdue chemist H.C. Brown, ). It
is called oxymercuration/demercuration. First, here it is in recipe format:
An alkene reacts with a mercury(II) salt (doesn’t matter which one) in aqueous solution. The crude product then reacts
with sodium borohydride (NaBH4). The elements of water add across the original C=C bond in Markovnikov fashion,
without rearrangement or polymerization.
• OK, what’s going on? And what does “oxymercuration/demercuration” mean?
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Slide – 34
8-5 Hydration by Oxymercuration–Demercuration
• .Oxymercuration–demercuration works with many alkenes that do not easily undergo direct hydration, and it takes
place under milder conditions. No free carbocation is formed, so there is no opportunity for rearrangements or
polymerization.
• The reagent for mercuration is mercuric acetate, Hg(OCOCH3)2, abbreviated Hg(OAc)2.
• Mercuric acetate dissociates slightly to form a positively charged mercury species, +Hg(OAc).
• Oxymercuration involves an electrophilic attack on the double bond by the positively charged mercury species. The
product is a mercurinium ion, an organometallic cation containing a three-membered ring.In the second step, water
from the solvent attacks the mercurinium ion to give (after deprotonation) an organomercurial alcohol.
• Demercuration, the subsequent reaction is to remove the mercury (demercuration) , Sodium borohydride (NaBH4, a
reducing agent) replaces the mercuric acetate fragment with a hydrogen atom.
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Slide – 35
8-5 Hydration by Oxymercuration–Demercuration
Mechanism
Step 1: Electrophilic attack forms a mercurinium ion.
Step 2: Water opens the ring to give an organomercurial alcohol.
Demercuration replaces the mercuric fragment with hydrogen to give the alcohol (“Demercuration” must mean getting rid of mercury
somehow. And indeed ):
In the demercuration reaction, a hydride furnished by the sodium borohydride (NaBH4) replaces the mercuric acetate.
To make the demercuration reaction easier , think of NaBH4 as a source of “nucleophilic hydride” (H-) It reacts with the organomercury
compound from the previous step (or its solution, if you didn’t isolate it) to release mercury. Counting electrons in the standard way, the leaving
group is mercury(0): elemental mercury.
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Slide – 36
Stereochemistry of the Mercurinium Ion Opening
• Water adds to the more substituted carbon to form the Markovnikov product.
• Water approaches the mercurinium ion from the side opposite the ring (anti addition).
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Slide – 37
Oxymercuration–Demercuration of 3,3,-Dimethylbut-1-ene
oxymercuration–demercuration of 3,3-dimethylbut-1-ene gives the Markovnikov product, 3,3-dimethylbutan-2-ol, in excellent yield.
Contrast this unrearranged product with the rearranged product formed in the acid-catalyzed hydration of the same alkene.
Oxymercuration–demercuration reliably adds water across the double bond of an alkene with Markovnikov orientation and without
rearrangement.
The reaction does not suffer from rearrangements because there is no carbocation intermediate.
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Slide – 38
Solved Problem
Show the intermediates and products that result from alkoxymercuration–demercuration of 1-methylcyclopentene,
using methanol as the solvent.
SOLUTION
Mercuric acetate adds to 1-methylcyclopentene to give the cyclic mercurinium ion. This ion has a considerable amount
of positive charge on the more substituted tertiary carbon atom. Methanol attacks this carbon from the opposite side,
leading to anti addition: The reagents (HgOAc and OCH3) have added to opposite faces of the double bond.
Reduction of the intermediate gives the Markovnikov product, 1-methoxy-1-methylcyclopentane.
Organic Chemistry, 9th Edition
L.G. Wade, Jr.
© 2017 Pearson Education, Inc.
8-6 Alkoxymercuration–Demercuration
• When mercuration takes place in an alcohol solvent, the alcohol serves as a nucleophile to attack the mercurinium ion. The
resulting product contains an alkoxy (−O−R)-O-R group.
•
alkoxymercuration–demercuration converts alkenes to ethers by adding an alcohol across the double bond of the alkene.
an alkene reacts to form a mercurinium ion that is attacked by the nucleophilic solvent. Attack by an alcohol solvent gives an organomercurial
ether that can be reduced to the ether.
The solvent attacks the mercurinium ion at the more substituted end of the double bond (where there is more δ+charge), giving Markovnikov
orientation of addition. The Hg(OAc) group appears at the less substituted end of the double bond. Reduction gives the Markovnikov product, with
hydrogen at the less substituted end of the double bond.
Example:
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Slide – 40
More alkoxymercuration/demercuration
• Alkoxymercuration/demercuration works just like
oxymercuration/demercuration: as an exercise, work through it.
• The result is that RO-H is added in Markovnikov fashion. The
notation RO-H means the elements of alcohol are added, splitting
between the O and the H.
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Slide – 41
Organic Chemistry
Lecture Presentation
Chapter 8
Reactions of Alkenes
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Slide – 1
8-7 Hydroboration of Alkenes
What if we need to convert an alkene to the anti-Markovnikov alcohol?
For example, how can we accomplish the following reaction:
•
Alkenes can be hydrated in anti-Markovnikov fashion also, by a process called hydroboration/oxidation, developed by H.C.
Brown of Purdue University.
•
Borane adds to alkenes with anti-Markovnikov orientation to form alkylboranes, which can be oxidized to give antiMarkovnikov alcohols.
In general:
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Slide – 2
8-7A Mechanism of Hydroboration
• Borane is an electron-deficient compound. It has only six valence electrons, so the boron atom in BH3 cannot have an octet.
• As an electron-deficient compound, BH3 is a strong electrophile, capable of adding to a double bond. This hydroboration of the
double bond is thought to occur in one step, with the boron atom adding to the less substituted end of the double bond.
• In the transition state, the electrophilic boron atom withdraws electrons from the pi bond, and the carbon at the other end of the
double bond acquires a partial positive charge.
• This partial charge is more stable on the more substituted carbon atom. The product shows boron bonded to the less
substituted end of the double bond and hydrogen bonded to the more substituted end.
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Slide – 3
8-7 Hydroboration of Alkenes
Example:
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Slide – 4
8-7C Stereochemistry of Hydroboration
The simultaneous addition of boron and hydrogen to the double bond leads to a syn addition: Boron and hydrogen
add across the double bond on the same side of the molecule.
Oxidation of the trialkylborane replaces boron with a hydroxy group in the same stereochemical position.
In general:
•
The alkene reacts with borane, which adds H-BR2 across the C=C bond anti-Markovnikov and stereospecifically syn; and then the resulting
organoborane is oxidized with cold basic aqueous H2O2, which replaces the C-B bond with a C-OH bond, with retention of
stereochemistry.
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Slide – 5
8-8 Addition of Halogens to Alkenes
Halogens add to alkenes to form vicinal dihalides.
• The reaction is run in chloroform (CHCl3), or methylene chloride (CH2Cl2) or carbon tetrachloride CCl4 as solvent
For example : cyclooctene +Br2 in CCl4
Br2
Br
CCl4
cold
dark
Br
• Cl2, Br2, and sometimes I2 add to a double bond to form a vicinal dihalide.
• This is an anti addition of halides
.
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Slide – 6
8-8A Mechanism of Halogen Addition
• A halogen molecule (Br2,Cl2,or I2) is electrophilic; a nucleophile can react with a halogen, displacing a halide ion.
the pi electrons of an alkene can attack the bromine molecule, expelling bromide ion. A bromonium ion results, containing a threemembered ring with a positive charge on the bromine atom.
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Slide – 7
Mechanism of Halogen Addition to Alkenes (Continued)
Step 1: Electrophilic attack forms a halonium ion.
Step 2: The halide ion opens the halonium ion.
EXAMPLE: Addition of Br2 to propene.
• Step 1: Electrophilic attack forms a bromonium ion.
Step 2: Bromide ion opens the bromonium ion.
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Slide – 8
8-8B Stereochemistry of Halogen Addition
The addition of bromine to cyclopentene is a stereospecific anti addition.
• Anti stereochemistry results from the back-side attack of the nucleophile on the bromonium ion.
• This back-side attack assures anti stereochemistry of addition.
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Slide – 9
8-9 Formation of Halohydrins
Addition of Halogens in Water: Addition of X-OH
• A halohydrin is an alcohol with a halogen on the adjacent carbon atom.
• In the presence of water, halogens add to alkenes to form halohydrins.
•
You may have encountered “bromine water” or “chlorine water”. These agents add to alkenes in a way that’s straightforward – up to a
point. A halonium ion forms, and is opened by water. Water successfully competes with halide anion as nucleophile, because (being the
solvent) there’s so much more of it.
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Slide – 10
MECHANISM 8-8: Formation of Halohydrins
Step 1: Electrophilic attack forms a halonium ion.
Step 2: Water opens the halonium ion; deprotonation gives the halohydrin.
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Slide – 11
MECHANISM 8-8: Formation of Halohydrins
EXAMPLE: Addition of Cl2 to propene in water.
Step 1: Electrophilic attack forms a chloronium ion.
Step 2: Back-side attack by water opens the chloronium ion (why H2O
attacks on this carbon and not the other neighboring carbon ? )
Answer to the question in step 2:
• The more substituted carbon of the chloronium ion bears
more positive charge than the less substituted carbon
since the positive charge is more stable on the more
substituted carbon.
• Attack by water occurs on the more substituted carbon to
give the Markovnikov product.
Step 3: Water removes a proton to give the chlorohydrin.
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Slide – 12
Solved Problem.
Propose a mechanism for the reaction of 1-methylcyclopentene with bromine water.
Solution
1-Methylcyclopentene reacts with bromine to give a bromonium ion. Attack by water could occur at
either the secondary carbon or the tertiary carbon of the bromonium ion. Attack actually occurs at the
more substituted carbon, which bears more of the positive charge. The product is formed as a racemic
mixture.
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Slide – 13
8-10 Catalytic Hydrogenation of Alkenes
• Hydrogen (H2) can be added across the double bond in a process known as catalytic hydrogenation.
• The reaction only takes place if a catalyst is used, the catalyst used could be platinum (Pt), palladium (Pd) or Nickel
(Ni)
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Slide – 14
8-10 Catalytic Hydrogenation of Alkenes
Mechanism of Catalytic Hydrogenation
Syn Stereochemistry in Catalytic Hydrogenation
• Lets consider the following reaction:
H
H
Pt
+H-H
+
(1 atm)
H
H
(same compound)
H2
Pt-Pt-Pt
H H
Pt-Pt-Pt
First H2 enters the solution, and binds to the surface of the platinum, forming some Pt-H bonds. This adsorption of hydrogen gas on the
surface of the of the metal catalyst weakens the H-H bond.
Next the alkene reacts as follows :
H H
H H
H H
Pt-Pt-Pt
Pt-Pt-Pt
Pt-Pt-Pt
One face of the alkene pi bond binds to the catalyst, which has hydrogen adsorbed on its surface. Hydrogen inserts into the pi bond, and
the product is freed from the catalyst.
Because the two hydrogen atoms add from a solid surface, they add with syn stereochemistry.
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Slide – 15
8-11 Addition of Carbenes to Alkenes
• Methylene (:CH2) is the simplest of the carbenes: uncharged, reactive intermediates that have a carbon atom with
two bonds and two nonbonding electrons.
• Methylene is a potent electrophile because it has an unfilled octet. It adds to the electron-rich pi bond of an alkene
to form a cyclopropane.
• Heating or photolysis of diazomethane gives nitrogen gas and methylene:
Problem:
diazomethane is very toxic and explosive,
Is there a safer way o make cyclopropane? Yes, by using the Simmons–Smith reagent.
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Slide – 16
8-11 Addition of Carbenes to Alkenes
8-11A The Simmons–Smith Reaction
• Two chemists at DuPont found a reaction which is named for them: the Simmons-Smith reaction. The active
species is ICH2ZnI; it gives the product you’d expect from carbene without actually containing divalent
carbon, so it’s called a carbenoid.
The Simmons-Smith reagent adds to double bonds much like methylene (:CH2) would, except that the reagent is
easier to use, is much more stable, and gives better yields without as many side reactions.
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Slide – 17
8-11 Addition of Carbenes to Alkenes
8-11B Formation of Carbenes by Alpha Elimination
• In the presence of a base, chloroform or bromoform can be dehydrohalogenated to form a carbene.
• Notice that the hydrogen in chloroform and bromoform is slightly acidic.
This dehydrohalogenation is called an alpha elimination (α elimination) because the hydrogen and the halogen are lost from the same
carbon atom.
Example : Dibromocarbene formed from CHBr3 can add to a double bond to form a dibromocyclopropane.
Notice the reagent in this example
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Slide – 18
8-11 Addition of Carbenes to Alkenes
8-11B Formation of Carbenes by Alpha Elimination
• Another example :
CHCl3
NaOH
(aq)
H
H
Cl
Cl
H
HO
H CCl3
Cl CCl2
:CCl2
H
Cl
Cl
The anion loses chloride. (Loss of H and Cl from the same carbon is called aelimination.) The product is :CCl2, dichlorocarbene. Its carbon is neutral but
octet-deficient – has six electrons in its valence shell – so it’s highly reactive
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Slide – 19
Carbene Examples
Simmons-Smith Reaction
Alpha Elimination Reaction
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Slide – 20
8-12 Epoxidation of Alkenes
• An alkene reacts with a peroxyacid to form an epoxide (also called oxirane). Epoxidation of alkenes is an oxidation,
because an oxygen atom is added.
• An alkene is converted to an epoxide by a, peroxyacid a carboxylic acid that has an extra oxygen atom in a −O−O−
(peroxy) linkage.
What are some examples of peroxyacids?
m-chloroperoxybenzoic acid (mCPBA)
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Slide – 21
8-12 Epoxidation of Alkenes
Examples
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Slide – 22
8-13 Acid-Catalyzed Opening of Epoxides
Epoxides undergo ring opening under acid catalysis.
Any moderately strong acid protonates the epoxide, however. Water attacks the protonated epoxide, opening the ring and
forming a 1,2-diol, commonly called a glycol.
EXAMPLE: Acid-catalyzed hydrolysis of propylene oxide (epoxypropane).
Step 1: Protonation of the epoxide.
Steps 2 and 3: Back-side attack by water, then deprotonation of the product
Another example:
Notice in both examples that because glycol formation involves a back-side attack on a protonated
epoxide, the result is anti orientation of the hydroxy groups on the double bond.
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Slide – 23
8-13 Acid-Catalyzed Opening of Epoxides
• In general , epoxidation of an alkene followed by acidic hydrolysis gives us anti
dihydroxylation of the double bond, the product is a glycol with the two hydroxyl
groups anti to each other.
For example:
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Slide – 24
8-14 Syn Dihydroxylation of Alkenes
By epoxidation followed by acid-catalyzed ring opening, alkenes can be converted in two steps to anti-1,2-dialcohols as we have seen in the
previous slides.
But, is there a way we can get syn-1,2 alcohols instead of anti-1,2-dialcohols ?
YES we can, how ??
This can be done in one step, by reacting the alkene either with a combination of osmium tetroxide and hydrogen peroxide, or with cold
basic aqueous potassium permanganate (the potassium permanganate is a dilute solution in this case)
• Alkene is converted to a syn-1,2-diol
• Two reagents can be used to accomplish this:
– Osmium tetroxide, OsO4, followed by hydrogen peroxide or
– Cold, dilute solution of KMnO4 in base
Example :
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Slide – 25
8-15 Oxidative Cleavage of Alkenes
• By using stronger oxidants, C=C double bonds can be cleaved completely. Ozone, followed by a reducing agent (zinc metal,
or dimethyl sulfide), will cleave a C=C double bond to a pair of C=O double bonds (to different oxygen atoms, of course). If
a double bond is within a ring, you get one product with two C=O double bonds. The other groups attached to the doublebonded carbons remain unchanged.
Ozonolysis: ozone cleaves double bonds to give ketones and aldehydes.
Examples:
O
(1) O3
(2) (CH3)2S
(1) O3
+ CH2=O
O
O
(2) (CH3)2S
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Slide – 26