chap 06 alkene reactions
TRANSCRIPT
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Alkene Reactions
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Pi bonds
Plane of molecule
Reactivity
above and
below the
molecular
plane!
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Addition Reactions
A - B A
B
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
Reactants
Products
Transition State
Energy of
Activation.
Energy
barrier.
Three areas to be aware of.
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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.
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Gibbs Free Energy
Gibbs Free Energy controls the position ofequilibrium for a reaction. It takes into accountenthalpy, 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. Adecrease favors reactants.
DG0: if positive equilibrium favors reactants(endergonic), if negative favors products(exergonic).
DG0 = DH0 TDS0
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Multi-Step ReactionsStep 1: A + B Intermediate
Step 2: Intermediate C + D
Step 1:
endergonic,
high energy of
activation.
Slow process
Step 2:exergonic, small
energy of
activation. Fast
Process.
Step 1 is the slow step, the
rate determining step.
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Characteristics of two step
Reaction 1. The Intermediate hassome stability. Itresides in a valley.
2. The concentration of
an intermediate is
usually quite low. The
Energies of Activationfor 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: A+BC+D
or C+D A+B.
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Hammond Postulate
The transition state for a step isclose 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.
CH3 - H + Br CH3 + H - Br DH = 109 kJ
[H3C H Br]
Almostbroken.
Almost
formed.
Almost
formed
radical. Only a smallamount of radical
character remains.
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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 2
Step 1
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Now examine Step 1 Closely
Rate Determining
Step. The rate at
which the
carbocation is
formed controlsthe rate of the
overall reaction.
The energy of
activation for
this process is
critical.
Electron rich, pi
system.
Showed thisreaction earlier as
an acid/base
reaction. Alkene
was the base.
New term: thealkene is a
nucleophile,
wanting to react
with a positive
species.
Acidic molecule, easily
ionized.
We had portrayed the HBrearlier as a Bronsted-
Lowry acid.
New term: the HBr is an
electrophile, wanting to
react with an electron richmolecule (nucleophile).
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
Electron deficient.
Does not obey octet rule.
Lewis acid, can receive
electrons.
Electrophile.
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.
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Step 2 of the Mechanism
Br
Br
Mirror objects
:Br-
:Br-
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Regioselectivity (Orientation)
H - Br
H
+ Br
H
+ Br
HBr
2-Bromo-propane
HBr
1-Bromo-propane
Secondary
carbocation
Primary
carbocation
Secondary
carbocation more
more stable and
more easily formed.
Or
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Carbocation Stabilities
Order of increasing stability:
Methyl < Primary < Secondary < Tertiary
Order of increasing ease of formation:
Methyl < Primary < Secondary < Tertiary
Increasing Ease of Formation
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Factors Affecting Carbocation Stability -
Inductive
1. 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.
HF
F
FH
d-
d+
d-
d-
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Factors Affecting Carbocation
Stability - Hyperconjugation
2. Hyperconjugation. Unlike normal resonance or
conjugation hyperconjugation involves s bonds.
H
H
HH
HHH
H
HH
ethyl carbocation
Hyperconjugation spreads the positive charge onto the
adjacent alkyl group
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Hyperconjugation Continued
Drifting of electrons from the filled C-H
bond into the empty p orbital of thecarbocation. Result resembles a pi bond.
Another description of the effect.
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Factors Affecting Carbocation Stability -
Resonance
allylic carbocation
Utilizing an adjacent pi system.
H HH H
benzylic carbocation
H H H H
Positive charge delocalized through resonance.
Another
very
important
example.
Positive charge delocalized into the benzene ring. Increased stability of
carbocation.
Note: the allylic
carbocation can react at
either end!
The benzylic
carbocation will
react only at the
benzylic position
even thoughdelocalization
occurs!
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Another Factor Affecting Carbocation
Stability Resonance
Utilizing an adjacent lone pair.
CH2
O
HCH2
O
H
Look carefully.
This is the
conjugate acid
of formaldehyde,
CH2=O.
P d ti f Chi l C t G l i t
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Production of Chiral Centers. Goal is to see
all the possibilities.
The H willattach here.Regioselectivity Analysis:
the positive charge will go here
and be stabilized by resonance
with the phenyl group.
Ph Me
EtMe
Me is methyl group
Et is ethyl group
Ph is phenyl groupPh
Me
Et
Me
H
Ph Me
EtMe
H
mirror plane
Enantiomeric
carbocations.
Br-
PhMe
EtMe
HBr
Br-
Ph
Me
Et
MeH
Br
H+
H+
Br-
Ph MeEt
Me
HBr
Ph
Me
Et
MeH
Br
Br-
What has been made?
Two pairs of enantiomers.
React alkene with HBr.
Note that the ends of the double
bond are different.
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Production of Chiral Centers - 2
PhMe
EtMe
HBr Ph
Me
Et
MeH
Br
Ph MeEt
Me
HBrPh
MeEt
MeH
Br
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 opticallyinactive roducts either achiral or racemic .
diastereomers
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Acid Catalyzed Hydration of
Alkenes
What is the orientation??? Markovnikov
M h i
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Mechanism
Step 1
Step 2
Step 3
Note the electronicstructure of the
oxonium ion.
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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 themethyl group with its pair of
electrons.
Step 3, the nucleophile
reacts with the
carbocation
Reason for Shift: Converting a
less stable carbocation (20) to a
more stable carbocation (30).
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Addition of Br2 and Cl2
St h i t
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Stereochemistry
Ant i Addi t ion (halogens enter on
oppos i te s ides); 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.
St 2
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Step 2
Detailed Stereochemistry addition of Br2
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Detailed Stereochemistry, addition of Br2
H3C CH3
C3H7 C2H5
Br
Br
(S) (R)
H3
C CH3
C3
H7
C2H5
Br
Br
Br
(S)
(S)
H3C
CH3C3H7
C2
H5
Br
Br
(R)
(R)
H3C
CH3
C3
H7
C2H5
Br
Br
(R) (S)
H3CCH3
C3H7 C2H5
Br
Br
Br
Br
(R)
(R)
H3C
CH3
C3H7
C2H5
Br
Br
(S)
(S)
H3C
CH3
C3H7
C2H5
Br
Br
enantiomers
Alternatively, the bromine
could have come in from the
bottom!
enantiomers
S,S
S,S
R,R
R,R
Only two compounds (R,Rand S,S) formed in equal
amounts. Racemic mixture.
Bromide ion
attacked the
carbon on the
right.
But can also
attack theleft-side
carbon.
f f
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Number of products formed.
(S)
(S)
H3C
CH3
C3H7
C2H5
Br
Br
enantiomers
enantiomers
S,S
S,S
R,R
R,R
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)
H3C
CH3
C3H7
C2H5
Br
Br
(R)
(R)
H3C
CH3
C3H7
C2
H5
Br
Br
(S)
(S)
H3C
CH3C3H7
C2H5
Br
Br
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Attack of the Bromide Ion
(S) (R)
H3C CH3
C3H7 C2H5
Br
Br
(S)
(S)
H3C
CH3C3H7
C2H5
Br
Br
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.
In order to
preserve a
tetrahedralcarbon these
two substituents
must move
upwards.
Inversion.
P f A k
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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.
U i Fi h P j ti
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R1R2
R3R4
Br2
anti addition
R1R2
R3R4
Br
Br
+ enantiomer
Using Fischer Projections
Not a validFischer
projection since
top vertical
bond is coming
forward.
Convert to Fischer by
doing 180 deg rotationof top carbon.
+ enantiomer
Br
R1 R2
Br
R4 R3
=
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There are many variations on the addition of X2 to
an alkene. Each one involves anti addition.
Br-
+ enantiomer
Br
R1 R2
Br
R4 R3
R2 R4
R1 R3
Br
I -
+ enantiomer
I
R1 R2
Br
R4 R3
+ enantiomer
Br
R1 R2
I
R4 R3
+
The iodide can attach to either
of the two carbons.
I -I -
Instead of iodide ion as nucleophile can
use alcohols to yield ethers, water to
yield alcohols, or amines.
R1
R2
R3R4
Br2
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Regioselectivity
If Br2 is added to propene there is no regioselectivity issue.
Br2
Br
Br
If Br2 is added in the presence of excess alternative nucleophile, such as
CH3OH, regioselectivity may become important.
Br - Br
OCH3
Br
CH3O-H
Br
OCH3and/or
+ H+
+ Br-
+ H+
+ Br-
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Regioselectivity - 2Consider, again, the cyclic bromonium ion and the resonance structures.
R
BrWeaker
bond
More positive charge
Stronger bond
Expect the nucleophile to attack here. Remember inversion occurs.
R i l ti it B i I
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Regioselectivity, Bromonium Ion
Bridged bromonium ion from propene.
Example
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Et
H Me
Me
Me
Cl2/H2O
H
p
Regioselectivity,
addition of Cl and OH
Cl, from the electrophile Cl2,
goes here
OH, the nucleophile, goes
here
Stereochemistry: anti addition
Note: non-reacting fragment unchangedEt
H Me
Me
Me
Et
H Me
Me
Me
+
Cl
H
H
OH
Cl
OH
Put in Fisher Projections. Be
sure you can do this!!
Et
H Me
Me OH
Me
H Cl
Et
H Me
HO Me
Me
Cl H
+
B i ti f b tit t d l h
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Bromination of a substituted cyclohexene
Consider the following bromination.
C(CH3)3
Br2
Expect to form
two bromonium
ions, one on top
and the other on
bottom.
C(CH3)3
Br
+
C(CH3)3
Br+
+
Expect the rings
can be opened
by attack on
either carbon
atom as before.
But NO, only one
stereoisomer is
formed. WHY?
C(CH3)3
Br-Br
Br
C(CH3)3
Br
Br
+
Addition to substituted cyclohexene
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Addition to substituted cyclohexene
HH
Br2
The tert butyl grouplocks the
conformation as
shown.
Br
Br
H
H
H H
+
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
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.
Notice that the two bromines are
maintained anti to each other!!!
Addition to substituted cyclohexene
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Addition to substituted cyclohexene
Br2
Br
Br
+
Observe
Ring is locked as shown. No ring flipping.
Attack as shown in red by incoming
Br ion will put both Br into equatorial
positions, not anti.
Br
Br
This stereoisomer is not
observed. The bromines
have not been kept anti to
each other but have become
gauche as displacement
proceeds.
Be sure to allow for the
inversion motion at the carbon
attacked by the bromide ion.
Addition to substituted cyclohexene
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Addition to substituted cyclohexene
Br2
Br
Br
+
Attack as shown in green by the
incoming Br will result in both Br being
axial and anti to each other
Br
Br
This is the observed
diastereomer. We have kept
the bromines anti to eachother.
O ti R d ti
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Oxymercuration-Reduction
Regioselective: Markovnikov Orientation
Occurs without 1,2 rearrangement, contrast the following
3,3-dimethylbut-1-ene
H2O
H2SO4
OH
formed via
rearrangement
1 Hg(OAc)2
2. NaBH4OH
No rearrangement
Alkene Alcohol
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Mechanism
1
2
3
4
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Hydroboration-Oxidation
Alkene Alcohol
Anti-Markovnikov orientation
Syn addition
1. BH3
2. H2O2
H
HO
H
HO
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Borane, a digression
Isoelectronic with a carbocation
B B
H
HH
H
HH
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MechanismSyn stereochemistry, anti-
Markovnikov orientationnow established.
Two reasons why anti-Markovnikov:
1. Less crowded transition state for B to approach the
terminal carbon.
2. A small positive charge is placed on the more highly
substituted carbon.
Just call the circled group R.
Eventually have BR3.
Next
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Contd
Oxidation and Reduction Reactions
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Oxidation and Reduction Reactions
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We think in terms of Half Reactions
Write reactants and products of each halfreaction.
Cr2O7 2- + CH3CH2OH Cr3+ + CH3CO2H
Cr2O72- 2 Cr3+
Balance oxygen by adding water
+ 7 H2O
In acid balance H by adding H +
14 H+ +
Balance charge by adding electrons
6 e - +
Inorganic half reaction
If reaction is in base: first balance as above for acid and then add OH- to both sides to
neutralize H +. Cancel extra H2O.
Will be
oxidized.
Will be
reduced.
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Contd
Now the organic half reaction
Balance oxygen by adding water
In acid balance H by adding H +
Balance charge by adding electrons
CH3CH2OH CH3CO2HH2O + + 4 H+ + 4 e-
Combine half reactions so as to cancel electrons
CH3CH2OH CH3CO2HH2O + + 4 H+ + 4 e-
Cr2O72- 2 Cr3+ + 7 H2O14 H
+ +6 e - +
3 x ( )
16 H+ +2 Cr2O72- + 3 CH3CH2OH 4 Cr
3+ + 3 CH3CO2H + 11 H2O
2 x ( )
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Formation of glycols with SynAddition
Osmium tetroxide
Syn addition
KMnO4
cold, dilute, slightly alkaline
also KMnO4
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Anti glycols
PhCO3H, a peracid
O
H+
O
H
H2
O
HO
OH
Using a peracid, RCO3H, to form an epoxide which is opened by aq. acid.
Peracid: for example, perbenzoic acid
O O
OH
The protonated epoxide is analagous
to the cyclic bromonium ion.
epoxide
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An example
chiral, optically active
(S)-3-methylcyclohex-1-ene
PhCO3H
O + O
aq. acid
OH
OH
OH
OHOH
OH
OH
OH
Are these
unique?
Diastereomers, separable (in theory) by
distillation, each optically active
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Ozonolysis
R3
R4
R1
R2
1. O3
2. (CH3)2SR4
R3
O
R1
R2
O+
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
ozonolysysO
O
O
O
Simply remove the new oxygens and join to make double bonds.
But there is a second possibility.
O
O
Another Example
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Another Example
2. An unknown compound (derived from the gall bladder of the gila monster) has the formulaC10H14 . When subjected to ozonolysis the following compound is isolated
O
O O
O
Suggest a reasonable structure for the unknown.
Hydrogen Deficiency = 8. Four pi
bonds/rings.
Unknown has no oxygens. Ozonolysis
product has four. Each double bondproduces two carbonyl groups. Expect
unknown to have 2 pi bonds and two rings.
To construct unknown cross out the oxygens and then connect. But there are many
ways the connections can be made.
a
b
c
d
a-b & c-d
a b
c
da-c & b-d
ac
d
b
a-d & b-c
ad
c
b
Look for a structure
that obeys the
isoprene rule.
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Mechanism
OO
OO
O
OO
O
O O
O
O
Consider the resonance structures of ozone.
These two, charged at
each end, are the useful
ones to think about.
Electrophile
capability.Nucleophile
capability.
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Mechanism - 2
O
O
OO
O
OO
O
O O
O
O
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Mechanism - 3
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Mechanism - 4
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Hydrogenation
No regioselectivity
Syn addition
Heats of Hydrogenation
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Heats of Hydrogenation
Consider the
cis vs trans
heats of
hydrogenationin more
detail
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Heats of Hydrogenation - 2The 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
Heats of Hydrogenation
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Heats of Hydrogenation
Increasingsub
stitution
ReducedheatofHydrogenation
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 additionof HX yielding 2-halopropane
Variation: there are other Lewis bases available. THE ALKENE.
+ HH
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.
the carbocation is an acid!
+
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Examples ofSynthetic Planning
Give a synthesis of 2-hexanol from any alkene.
OH
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?
Another synthetic example
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How would you prepare meso 2,3 dibromobutane from an alkene?
Analysis:
Alkene must be 2-butene. But wait that could be eithercis ortrans!
We want meso. Have to worry about stereochemistry
Know bromine addition to an alkene is anti addition (cyclic bromonium ion)
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trans
Br2
Br
Br
H
Br
Br
rotate lower unit
Br H
Br H
meso
This worked! How about
starting with the cis?
cis
Br2
H Br
Br H
racemic mixture
+ enantiomer
This did not work, gave us
the wrong stereochemistry!
Addition Reaction General Rule
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Characterize Reactant as cis ortrans, C or T
Characterize Reaction as syn oranti, S or A
Characterize Product as meso orracemicmixture, M or R
Relationship
C RA
cis
Br2H Br
Br H
racemic mixture
+ enantiomer
Characteristics can be
changed in pairs and C A R willremain true.
Want meso instead?? Have to
use trans. Two changed!!
A
T M
t
Br H
Br H
Br2