Alkene Reactions

1
Alkene Reactions
2
Pi bonds
Reactivity above and below the molecular plane!
Plane of molecule
3
Addition Reactions
Important characteristics of addition
reactions Orientation (Regioselectivity) If the
doubly bonded carbons are not equivalent which
one get the A and which gets the
B. Stereochemistry geometry of the addition. Syn
addition Both A and B come in from the same
side of the alkene. Both from the top or both
from the bottom. Anti Addition A and B come in
from opposite sides (anti addition). No
preference.
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Reaction Mechanisms
Mechanism a detailed, step-by-step description
of how a reaction occurs.
A reaction may consist of many sequential steps.
Each step involves a transformation of the
structure. For the step C A-B ?
C-A B
Transition State
Three areas to be aware of.
Products
Reactants
Energy of Activation. Energy barrier.
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Energy Changes in a Reaction

• Enthalpy changes, DH0, for a reaction arises from
  changes in bonding in the molecule.
• If weaker bonds are broken and stronger ones
  formed then DH0 is negative and exothermic.
• If stronger bonds are broken and weaker ones
  formed then DH0 is positive and endothermic.

6
Gibbs Free Energy

• Gibbs Free Energy controls the position of
  equilibrium for a reaction. It takes into
  account enthalpy, H, and entropy, S, changes.
• An increase in H during a reaction favors
  reactants. A decrease favors products.
• An increase in entropy (eg., more molecules being
  formed) during a reaction favors products. A
  decrease favors reactants.
• DG0 if positive equilibrium favors reactants
  (endergonic), if negative favors products
  (exergonic).
• DG0 DH0 TDS0

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How do enthalpy and entropy changes work out.
Favor Products.
Favor reactants
Low temperatures favors reactants but at higher
temperature favors products.
Favor reactants
Low temperatures favors products but at higher
temperature favors reactants.
Favor Products.
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Multi-Step Reactions
Step 1 is the slow step, the rate determining
step.
Step 1 A B ? Intermediate
Step 2 Intermediate ? C D
Step 2 exergonic, small energy of activation.
Fast Process.
Step 1 endergonic, high energy of activation.
Slow process
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Characteristics of two step Reaction

1. The Intermediate has some stability. It resides
   in a valley.
2. The concentration of an intermediate is usually
   quite low. The Energies of Activation for
   reaction of the Intermediate are low.
3. There is a transition state for each step. A
   transition state is not a stable structure.
4. The reaction coordinate can be traversed in
   either direction AB ?CD or CD ? AB.

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Hammond Postulate
The transition state for a step is close to the
high energy end of the curve. For an endothermic
step the transition state resembles the product
of the step more than the reactants. For an
exothermic step the transition state resembles
the reactants more than the products.
Reaction coordinate.
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Example

• Endothermic
• Transition state resembles the (higher energy)
  products.

Almost formed radical.
Only a small amount of radical character remains.
Almost formed.
Almost broken.
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Electrophilic Additions

• Hydrohalogenation using HCl, HBr, HI
• Hydration using H2O in the presence of H2SO4
• Halogenation using Cl2, Br2
• Halohydrination using HOCl, HOBr
• Oxymercuration using Hg(OAc)2, H2O followed by
  reduction

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Electrophilic Addition
We now address regioselectivity.
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Regioselectivity (Orientation)
The incoming hydrogen attaches to the carbon with
the greater number of hydrogens. This is
regioselectivity. It is called Markovnikov
orientation.
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Mechanism
Step 1
Step 2
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Now examine Step 1 Closely
Electron rich, pi system. Showed this reaction
earlier as an acid/base reaction. Alkene was the
base. New term the alkene is a nucleophile,
wanting to react with a positive species.
Acidic molecule, easily ionized. We had portrayed
the HBr earlier as a Bronsted-Lowry acid. New
term the HBr is an electrophile, wanting to
react with an electron rich molecule
(nucleophile).
Rate Determining Step. The rate at which the
carbocation is formed controls the rate of the
overall reaction. The energy of activation for
this process is critical.
The carbocation intermediate is very reactive. It
does not obey the octet rule (electron deficient)
and is usually present only in low concentration.
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Carbocations
sp2 hybridized. p orbital is empty and can
receive electrons. Flat, planar. Can react on
either side of the plane. Very reactive and
present only in very low concentration.
Electron deficient. Does not obey octet
rule. Lewis acid, can receive electrons. Electroph
ile.
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Step 2 of the Mechanism
Br-
Mirror objects
Br-
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Regioselectivity (Orientation)
Or
Primary carbocation
Secondary carbocation
Secondary carbocation more more stable and more
easily formed.
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Carbocation Stabilities
Order of increasing stability Methyl lt
Primary lt Secondary lt Tertiary
Order of increasing ease of formation Methyl lt
Primary lt Secondary lt Tertiary
Increasing Ease of Formation
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Factors Affecting Carbocation Stability -
Inductive

• Inductive Effect. Electron redistribution due to
  differences in electronegativities of
  substituents.
• Electron releasing, alkyl groups, -CH3, stabilize
  the carbocation making it easier to form.
• Electron withdrawing groups, such as -CF3,
  destabilize the carbocation making it harder to
  form.

d-
d
d-
d-
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Factors Affecting Carbocation Stability -
Hyperconjugation
2. Hyperconjugation. Unlike normal resonance or
conjugation hyperconjugation involves s bonds.
Hyperconjugation spreads the positive charge onto
the adjacent alkyl group
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Hyperconjugation Continued
Another description of the effect.
Drifting of electrons from the filled C-H bond
into the empty p orbital of the carbocation.
Result resembles a pi bond.
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Factors Affecting Carbocation Stability -
Resonance
Note the allylic carbocation can react at
either end!
Utilizing an adjacent pi system.
Positive charge delocalized through resonance.
The benzylic carbocation will react only at the
benzylic position even though delocalization
occurs!
Another very important example.
Positive charge delocalized into the benzene
ring. Increased stability of carbocation.
25
Another Factor Affecting Carbocation Stability
Resonance
Utilizing an adjacent lone pair.
Look carefully. This is the conjugate acid of
formaldehyde, CH2O.
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Production of Chiral Centers. Goal is to see all
the possibilities.
React alkene with HBr. Note that the ends of the
double bond are different.
The H will attach here.
Regioselectivity Analysis the positive charge
will go here and be stabilized by resonance with
the phenyl group.
Enantiomeric carbocations.
What has been made?
Two pairs of enantiomers.
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Production of Chiral Centers - 2
diastereomers
Racemic Mixture 1
Racemic Mixture 2
The product mixture consists of four
stereoisomers, two pairs of enantiomers The
product is optically inactive. Distillation of
the product mixture yields two fractions
(different boiling points). Each fraction is
optically inactive. Rule optically inactive
reactants yield optically inactive products
(either achiral or racemic).
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Acid Catalyzed Hydration of Alkenes
What is the orientation???
Markovnikov
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Mechanism
Step 1
Step 2
Note the electronic structure of the oxonium ion.
Step 3
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Carbocation Rearrangements
Expected product is not the major product
rearrangement of carbon skeleton occurred.
The methyl group moved. Rearranged.
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Also, in the hydration reaction.
The H moved.
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Mechanism including the 1,2 shift
Step 1, formation of carbocation
Step 2, the 1,2 shift of the methyl group with
its pair of electrons.
Reason for Shift Converting a less stable
carbocation (20) to a more stable carbocation
(30).
Step 3, the nucleophile reacts with the
carbocation
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Addition of Br2 and Cl2
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Stereochemistry
Anti Addition (halogens enter on opposite sides)
Stereoselective Syn addition (on same side) does
not occur for this reaction.
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Mechanism, Step 1
Step 1, formation of cyclic bromonium ion.
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Step 2
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Detailed Stereochemistry, addition of Br2
Bromide ion attacked the carbon on the right.
S,S
enantiomers
But can also attack the left-side carbon.
R,R
enantiomers
R,R
Alternatively, the bromine could have come in
from the bottom!
S,S
Only two compounds (R,R and S,S) formed in equal
amounts. Racemic mixture.
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Number of products formed.
S,S
enantiomers
We have formed only two products even though
there are two chiral carbons present. We know
that there is a total of four stereoisomers. Half
of them are eliminated because the addition is
anti. Syn (both on same side) addition does not
occur.
R,R
enantiomers
R,R
S,S
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Attack of the Bromide Ion
In order to preserve a tetrahedral carbon these
two substituents must move upwards. Inversion.
Starts as R
Becomes S
The carbon was originally R with the Br on the
top-side. It became S when the Br was removed
and a Br attached to the bottom.
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Progress of Attack

• Things to watch for
• Approach of the red Br anion from the bottom.
• Breaking of the C-Br bond.
• Inversion of the C on the left Retention of the
  C on the right.

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Using Fischer Projections
Convert to Fischer by doing 180 deg rotation of
top carbon.

Not a valid Fischer projection since top vertical
bond is coming forward.
42
There are many variations on the addition of X2
to an alkene. Each one involves anti addition.
Br -
I -
I -
I -

The iodide can attach to either of the two
carbons.
Instead of iodide ion as nucleophile can use
alcohols to yield ethers, water to yield
alcohols, or amines.
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Regioselectivity
If Br2 is added to propene there is no
regioselectivity issue.
If Br2 is added in the presence of excess
alternative nucleophile, such as CH3OH,
regioselectivity may become important.
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Regioselectivity - 2
Consider, again, the cyclic bromonium ion and the
resonance structures.
Stronger bond
Weaker bond
More positive charge
Expect the nucleophile to attack here. Remember
inversion occurs.
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Regioselectivity, Bromonium Ion

• Bridged bromonium ion from propene.

46
Example
Stereochemistry anti addition Note non-reacting
fragment unchanged
Regioselectivity, addition of Cl and OH
Put in Fisher Projections. Be sure you can do
this!!
Cl, from the electrophile Cl2, goes here
OH, the nucleophile, goes here
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Bromination of a substituted cyclohexene
Consider the following bromination.
Expect to form two bromonium ions, one on top and
the other on bottom.
Expect the rings can be opened by attack on
either carbon atom as before.
But NO, only one stereoisomer is formed. WHY?
48
Addition to substituted cyclohexene
The tert butyl group locks the conformation as
shown.
The cyclic bromonium ion can form on either the
top or bottom of the ring.
How can the bromide ion come in? Review earlier
slide showing that the bromide ion attacks
directly on the side opposite to the ring.
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Progress of Attack
Notice that the two bromines are maintained anti
to each other!!!

• Things to watch for
• Approach of the red Br anion from the bottom.
• Breaking of the C-Br bond.
• Inversion of the C on the left Retention of the
  C on the right.

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Addition to substituted cyclohexene
Attack as shown in red by incoming Br ion will
put both Br into equatorial positions, not anti.
This stereoisomer is not observed. The bromines
have not been kept anti to each other but have
become gauche as displacement proceeds.
Observe Ring is locked as shown. No ring flipping.
Be sure to allow for the inversion motion at the
carbon attacked by the bromide ion.
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Addition to substituted cyclohexene
Attack as shown in green by the incoming Br will
result in both Br being axial and anti to each
other
This is the observed diastereomer. We have kept
the bromines anti to each other.
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Oxymercuration-Reduction
Alkene ? Alcohol
Regioselective Markovnikov Orientation
Occurs without 1,2 rearrangement, contrast the
following
No rearrangement
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Mechanism
1
2
3
4
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Hydroboration-Oxidation
Alkene ? Alcohol
Anti-Markovnikov orientation
Syn addition
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Borane, a digression
Isoelectronic with a carbocation
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Mechanism
Syn stereochemistry, anti-Markovnikov orientation
now established.
Just call the circled group R. Eventually have
BR3.

• Two reasons why anti-Markovnikov
• Less crowded transition state for B to approach
  the terminal carbon.
• A small positive charge is placed on the more
  highly substituted carbon.

Next
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Contd
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Oxidation and Reduction Reactions
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We think in terms of Half Reactions
Will be reduced.
Will be oxidized.

• Write reactants and products of each half
  reaction.

Inorganic half reaction
Cr2O7 2-
2 Cr 3
7 H2O
14 H
6 e -
Balance oxygen by adding water
In acid balance H by adding H
Balance charge by adding electrons
If reaction is in base first balance as above
for acid and then add OH- to both sides to
neutralize H . Cancel extra H2O.
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Contd
Now the organic half reaction
CH3CH2OH
CH3CO2H
H2O
4 H
4 e-
Balance oxygen by adding water
In acid balance H by adding H
Balance charge by adding electrons
Combine half reactions so as to cancel electrons
3 x (
)
CH3CH2OH
CH3CO2H
H2O
4 H
4 e-
2 x (
)
Cr2O7 2-
2 Cr 3
7 H2O
14 H
6 e -
16 H 2 Cr2O7 2- 3 CH3CH2OH
4 Cr 3 3 CH3CO2H 11 H2O
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Formation of glycols with Syn Addition

• Osmium tetroxide

Syn addition
also
KMnO4
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Anti glycols
Using a peracid, RCO3H, to form an epoxide which
is opened by aq. acid.
epoxide
The protonated epoxide is analagous to the cyclic
bromonium ion.
Peracid for example, perbenzoic acid
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An example
Are these unique?
Diastereomers, separable (in theory) by
distillation, each optically active
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Ozonolysis
Reaction can be used to break larger molecule
down into smaller parts for easy identification.
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Ozonolysis Example
For example, suppose an unknown compound had the
formula C8H12 and upon ozonolysis yielded only
3-oxobutanal. What is the structure of the
unknown?
The hydrogen deficiency is 18-12 6. 6/2 3
pi bonds or rings.
The original compound has 8 carbons and the
ozonolysis product has only 4
Conclude Unknown ? two 3-oxobutanal.
Unknown C8H12
ozonolysys
Simply remove the new oxygens and join to make
double bonds.
But there is a second possibility.
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Another Example
d
Hydrogen Deficiency 8. Four pi bonds/rings.
a
c
Unknown has no oxygens. Ozonolysis product has
four. Each double bond produces two carbonyl
groups. Expect unknown to have 2 pi bonds and
two rings.
b
To construct unknown cross out the oxygens and
then connect. But there are many ways the
connections can be made.
Look for a structure that obeys the isoprene rule.
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Mechanism
Consider the resonance structures of ozone.
Electrophile capability.
Nucleophile capability.
These two, charged at each end, are the useful
ones to think about.
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Mechanism - 2
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Mechanism - 3
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Mechanism - 4
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Hydrogenation
No regioselectivity Syn addition
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Heats of Hydrogenation
Consider the cis vs trans heats of hydrogenation
in more detail
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Heats of Hydrogenation - 2
The trans alkene has a lower heat of
hydrogenation.

• Conclusion
• Trans alkenes with lower heats of hydrogenation
  are more stable than cis.
• We saw same kind of reasoning when we talked
  about heats of combustion of isomeric alkanes to
  give CO2 and H2O

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Heats of Hydrogenation
Increasing substitution
Reduced heat of Hydrogenation
By same reasoning higher degree of substitution
provide lower heat of hydrogenation and are,
therefore, more stable.
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Acid Catalyzed Polymerization
Principle Reactive pi electrons (Lewis base)
can react with Lewis acid. Recall
Which now reacts with a Lewis base, such as
halide ion to complete addition of HX yielding
2-halopropane
Variation there are other Lewis bases available.
THE ALKENE.
The new carbocation now reacts with a Lewis base
such as halide ion to yield halide ion to yield
2-halo-4-methyl pentane (dimerization) but could
react with another propene to yield higher
polymers.
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Examples of Synthetic Planning
Give a synthesis of 2-hexanol from any alkene.
Planning
Alkene is a hydrocarbon, thus we have to
introduce the OH group
How is OH group introduced (into an alkene)
hydration

• What are hydration reactions and what are their
  characteristics
• Mercuration/Reduction Markovnikov
• Hydroboration/Oxidation Anti-Markovnikov and syn
  addition

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What alkene to use? Must involve C2 in double
bond.
Which reaction to use with which alkene?
Markovnikov rule can be applied here. CH vs
CH2. Want Markovnikov! Use Mercuration/Reduction!!
!
Markovnkov Rule cannot be used here. Both are
CH. Do not have control over regioselectivity. Do
not use this alkene.
For yourself how would you make 1 hexanol, and
3-hexanol?
78
Another synthetic example
How would you prepare meso 2,3 dibromobutane from
an alkene?
Analysis
Alkene must be 2-butene. But wait that could be
either cis or trans!
We want meso. Have to worry about stereochemistry
Know bromine addition to an alkene is anti
addition (cyclic bromonium ion)
79
This did not work, gave us the wrong
stereochemistry!
This worked! How about starting with the cis?
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Addition Reaction General Rule
Characterize Reactant as cis or trans, C or
T Characterize Reaction as syn or anti, S or
A Characterize Product as meso or racemic
mixture, M or R
Relationship
Characteristics can be changed in pairs and C A R
will remain true. Want meso instead?? Have to use
trans. Two changed!!