Addition of Br2 and Cl2

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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.
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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.

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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?
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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?
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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)
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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!!