Lone Pair acting as Base. Note the change in formal charges. As reactant oxygen had complete...
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Lone Pair acting as Base.
Note the change in formal charges. As reactant oxygen had complete ownership of lone pair. In product it is shared. Oxygen more positive by 1.
Similarly, B has gained half of a bonding pair; more negative by 1.
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An example: pi electrons as bases
Bronsted Lowry Acid
Bronsted Lowry Base
The carbocations are conjugate acids of the alkenes.
For the moment, just note that there are two possible carbocations formed.
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Sigma bonding electrons as bases. Much more unusual!!
Super acid
A very, very electronegative F!!
A very positive S!! The OH becomes very acidic because that would put a negative charge adjacent to the S.
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Trends for Relative Acid Strengths
Totally ionized in aqueous solution.
Aqueous Solution
Totally unionized in aqueous solution
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Example
Ethanol, EtOH, is a weaker acid than phenol, PhOH.
It follows that ethoxide, EtO-, is a stronger base than phenolate, PhO-.
For reaction PhOH + EtO- PhO- + EtOH where does equilibrium lie?
pKa = 9.95
Stronger acid
H2O + PhOH H3O+ + PhO-
Ka = [H3O+][PhO-]/[PhOH] = 10-9.95
OH
phenol, PhOH
CH3CH2OH
ethanol, EtOH
Recall
H2O + EtOH H3O+ + EtO-
Ka = [H3O+][EtO-]/[EtOH] = 10-15.9
pKa = 15.9
Weaker acid
Stronger base
Weaker base.
Query: What makes for strong (or weak) acids?
K = 10-9.95 / 10-15.9 = 106.0
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What affects acidity?1. Electronegativity of the atom holding the negative charge.
CH3OH CH3O - + H+
CH3NH2 CH3NH - + H+
CH3CH3 CH3CH2- + H+
Increasing electronegativity of atom bearing negative charge. Increasing stability of anion.
Increasing acidity.
Increasing basicity of anion.
2. Size of the atom bearing the negative charge in the anion.
CH3OH CH3O - + H+; pKa = 16
CH3SH CH3S - + H+; pKa = 7.0
Increasing size of atom holding negative charge. Increasing stability of anion.
Increasing acidity.
Increasing basicity of anion.
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OO
What affects acidity? - 23. Resonance stabilization, usually of the anion.
OH
phenol, PhOH
OO
ethanol, EtOHCH3CH2OH CH3CH2O
- + H+
Increasing resonance stabilization. Increased anion stability.
Aci
dit
y
Increasing basicity of the anion.
No resonance structures!!
OH OH
etc.
Note that phenol itself enjoys resonance but charges are generated, costing energy, making the resonance less important. The more important resonance in the anion shifts the equilibrium to the right making phenol more acidic.
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An example: competitive Bases & Resonance
• Two different bases or two sites in the same molecule may compete to be protonated (be the base).
O
O H
acetic acid
H+O
O H
HH+
O
O H
H
Acetic acid can be protonated at two sites.
Which conjugate acid is favored?
The more stable one! Which is that?
Recall resonance provides additional stability by moving pi or non-bonding electrons.
Pi bonding electrons converted to non-bonding.
O
O H
H
O
O H
H
Non-bonding electrons converted to pi bonding.
No valid resonance structures for this cation.
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An example: competitive Bases & Resonance
H+O
O H
H
O
O H
H
O
O H
H
O
O H
acetic acid
All atoms obey octet rule!
All atoms obey octet rule!
The carbon is electron deficient – 6 electrons, not 8.
Lesser importance
Comments on the importance of the resonance structures.
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What affects acidity? - 34. Inductive and Electrostatic Stabilization.
F3CCH2O - + H+
H3CCH2O - + H+H3CCH2OH
F3CCH2OH
Due to electronegativity of F small positive charges build up on C resulting in stabilization of the anion.
Increasing anion stability.Acidity.Increasing anion basicity.
Effect drops off with distance. EtOH pKa = 15.9
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What affects acidity? - 45. Hybridization of the atom bearing the charge. H-A H+ + A:-.
sp3 sp2 sp
More s character, more stability, more “electronegative”, H-A more acidic, A:- less basic.
Incr
easi
ng
Aci
dit
y o
f H
A
Incr
easi
ng
B
asic
ity
of
A-
Note. The NH2-
is more basic than the RCC-
ion.
Know this order.
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Example of hybridization Effect.
RCCH + AgNO3 AgCCR (ppt)
acid base
terminal alkyne
non-terminal alkyne
RCCR + LiCH2CH2CH2CH3 No Reaction
RCCH + LiCH2CH2CH2CH3 HCH2CH2CH2CH3 + RCCLi
RCCR + AgNO3 NR
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What affects acidity? - 5
6. Stabilization of ions by solvents (solvation).
H
O RO R + H
H
O
H
H
O
HH
OH
Solvation provides stabilization.
OH
ethanol
OH
propan-2-ol
OH
2-methylpropan-2-olCrowding inhibiting solvation
Solvation, stability of anion, acidity
pKa = 15.9 17 18
(CH3)3CO -, crowded
Comparison of alcohol acidities.
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Example
Para nitrophenol is more acidic than phenol. Offer an explanation
OH
OH
N
O O
O
O
N
O O
+ H
+ H The lower lies further to the right.
Why? Could be due to destabilization of the unionized form, A, or stabilization of the ionized form, B.
A B
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OH
N
O O
Examine the equilibrium for p-nitrophenol. How does the nitro group increase the acidity?
O
N
O O
+ H
Resonance structures A, B and C are comparable to those in the phenol itself and thus would not be expected to affect acidity. But note the + to – attraction here
OH
N
O O
OH
N
O O
OH
N
O O
OH
N
O O
A B C D
Structure D occurs only due to the nitro group. The stability it provides will slightly decrease acidity.
Examine both sides of equilibrium. What does the nitro group do?
First the unionized acid.
Note carefully that in these resonance structures charge is created: + on the O and – in the ring or on an oxygen. This decreases the importance of the resonance.
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OH
N
O O
O
N
O O
+ H
Resonance structures A, B and C are comparable to those in the phenolate anion itself and thus would not be expected to affect acidity. But note the + to – attraction here
Structure D occurs only due to the nitro group. It increases acidity. The greater amount of significant resonance in the anion accounts for the nitro increasing the acidity.
Now look at the anion. What does the nitro group do? Remember we are interested to compare with the phenol phenolate equilibrium.
In these resonance structures charge is not created. Thus these structures are important and increase acidity. They account for the acidity of all phenols.
O
N
O O
O
N
O O
O
N
O O
O
N
O O
A B C D
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3. (3 pts) Which is the stronger base and why?
HNvs
HN O
Sample Problem
H2N H2N O H2N O
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Alkenes 1
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ShapeAlkenes, 6 coplanar atoms.
All atoms in same plane except for
these hydrogens on sp3 carbon.
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Arene shapesPlanar ring structure. 12 atoms coplanar.
2-phenyl propane
Phenyl group, C6H5,, Ph
Ph
Ph = C6H5
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Pi bonds
pi orbitalsum
pi* orbitaldifference
Ene
rgy
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Nomenclature
but-1-ene 3,4-dimethylhexa-1,5-diene
cis / trans
cis trans
Z / E generalization of cis / trans
Use R, S priorities to compare substituents on same carbon.
High priority on same side, Z. Opposite, E.
H Br
FCl
H F
BrCl
(E)-1-bromo-2-chloro-1-fluoroethene(Z)-1-bromo-2-chloro-1-fluoroethene
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Cis / Trans in CycloalkenesFor small rings normally have cis double bonds.
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trans cyclooctene
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Terpenes and the isoprene Rule
• A terpene is composed of isoprene units joined head to tail (the isoprene rule).
This moleculehas additional cross links.
Note that location of functional groups such as OH or double bonds is not addressed.
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Vitamin A
Four isprene units joined head to tail
One cross link (non-head to tail) linkage.
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Fatty Acids
• Animal fats and vegetable oils are both triesters of glycerol, hence the name triglyceridetriglyceride.– Hydrolysis of a triglyceride in aqueous base followed by
acidification gives glycerol and three fatty acids.
– Fatty acids with no C=C double bonds are called saturated fatty acid.
– Those with one or more C=C double bonds are called unsaturated fatty acids.
CH2OCR
CH2OCR''
R'COCHO
O
1. NaOH, H2O
2. HCl, H2O
CH2OH
CH2OH
HOCH
RCOOH
R'COOH
R''COOH
+
Fatty acids
O
1,2,3-Propanetriol(glycerol)
A triglyceride(a triester of glycerol
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Fatty Acids– The most common fatty acids have an even number of
carbons, and between 12 and 20 carbons in an unbranched chain.
– The C=C double bonds in almost all naturally occurring fatty acids have a cis configuration.
– The greater degree of unsaturation, the lower the melting point.
– Triglycerides rich in unsaturated fatty acids are generally liquid at room temperature and are called oilsoils.
– Triglycerides rich in saturated fatty acids are generally semisolids or solids at room temperature and are called fatsfats.
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Fatty Acids
– the four most abundant fatty acidsCOOH
COOH
COOH
COOH
Stearic acid (18:0)(mp 70°C)
Oleic acid (18;1)(mp 16°C)
Linoleic acid (18:2)(mp-5°C)
Linolenic acid (18:3)(mp -11°C)
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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 ReactionsA - 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
ReactantsProducts
Transition State
Energy of Activation. Energy barrier.
Three areas to be aware of.
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Energy Changes in a Reaction
• Enthalpy changes, H0, for a reaction arises from changes in bonding in the molecule.– If weaker bonds are broken and stronger ones
formed then H0 is negative and exothermic.– If stronger bonds are broken and weaker ones
formed then H0 is positive and endothermic.
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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.
G0: if positive equilibrium favors reactants (endergonic), if negative favors products (exergonic).
G0 = H0 – TS0
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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 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: A+B C+D or C+D A+B.
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Hammond PostulateThe 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.
CH3 - H + Br CH3 + H - Br H = 109 kJ
[H3C H Br]
Almost broken.
Almost formed.
Almost formed radical. Only a small
amount 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 controls the rate of the overall reaction. The energy of activation for this process is critical.
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).
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
F H
-
+
-
-
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Factors Affecting Carbocation Stability - Hyperconjugation
2. Hyperconjugation. Unlike normal resonance or conjugation hyperconjugation involves bonds.
H
H
H H
HHH
H
H H
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 the carbocation. 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 though delocalization 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.
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Production of Chiral Centers. Goal is to see all the possibilities.
The H will attach here.Regioselectivity Analysis:
the positive charge will go here and be stabilized by resonance
with the phenyl group.
Ph Me
EtMe
Me is methyl groupEt is ethyl groupPh is phenyl group
Ph
MeEt
Me
H
Ph MeEt
MeH
mirror plane
Enantiomeric carbocations.
Br-
Ph MeEtMe
HBr
Br-
Ph
MeEt
Me H
Br
H+
H+
Br-
Ph MeEtMe
HBr
Ph
MeEt
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
Ph MeEtMe
HBr Ph
MeEt
Me H
Br
Ph MeEtMe
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 optically inactive products (either achiral or racemic).
diastereomers
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Acid Catalyzed Hydration of Alkenes
What is the orientation??? Markovnikov
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Mechanism
Step 1
Step 2
Step 3
Note the electronic structure 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 the methyl 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
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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 1Step 1, formation of cyclic bromonium ion.
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Step 2
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Detailed Stereochemistry, addition of Br2
H3C CH3
C3H7 C2H5
Br
Br
(S) (R)H3C CH3
C3H7 C2H5
Br
Br
Br
(S)
(S)
H3C
CH3C3H7
C2H5
Br
Br
(R)
(R)
H3C
CH3
C3H7
C2H5
Br
Br
(R) (S)
H3C CH3
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,R and S,S) formed in equal amounts. Racemic mixture.
Bromide ion attacked the carbon on the right.
But can also attack the left-side carbon.
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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
C2H5
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 tetrahedral carbon these two substituents must move upwards. Inversion.
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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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R1R2
R3R4
Br2
anti addition
R1R2
R3R4
Br
Br
+ enantiomer
Using Fischer Projections
Not a valid Fischer projection since top vertical bond is coming forward.
Convert to Fischer by doing 180 deg rotation of 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.
R1R2
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 - BrOCH3
BrCH3O-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.
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Regioselectivity, Bromonium Ion
– Bridged bromonium ion from propene.
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Et
H Me
Me
Me
Cl2/H2O
H
Example
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
HH
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
+
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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
+
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Addition to substituted cyclohexene
HH
Br2
The tert butyl group locks 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!!!
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Addition to substituted cyclohexene
Br2
Br
Br
+
ObserveRing 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.
Br-
Br-
Be sure to allow for the inversion motion at the carbon attacked by the bromide ion.
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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 each other.
Br-
Br-
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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 viarearrangement
1 Hg(OAc)2
2. NaBH4OH
No rearrangement
Alkene Alcohol
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Mechanism1
2
3
4
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Hydroboration-Oxidation
Alkene Alcohol
Anti-Markovnikov orientation
Syn addition
1. BH3
2. H2O2
HHO
HHO
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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 orientation now 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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Cont’d
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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 half reaction.
Cr2O7 2- + CH3CH2OH Cr 3+ + CH3CO2H
Cr2O7 2- 2 Cr 3+
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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Cont’d
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-
Cr2O7 2- 2 Cr 3+ + 7 H2O14 H+ +6 e - +
3 x ( )
16 H+ + 2 Cr2O7 2- + 3 CH3CH2OH 4 Cr 3+ + 3 CH3CO2H + 11 H2O
2 x ( )
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Formation of glycols with Syn Addition
Osmium tetroxide
Syn addition
KMnO4cold, dilute, slightly alkaline
also KMnO4
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Anti glycols
PhCO3H, a peracid
O
H+
O
H
H2O
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
PhCO3HO + 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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OzonolysisR3
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
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Another Example
2. An unknown compound (derived from the gall bladder of the gila monster) has the formula C10H14 . 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 bond produces 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
bc
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
O OO
O OO
O OO
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
OO
O OO
O OO
O OO
O
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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 - 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
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Heats of Hydrogenation
Incr
easi
ng s
ubst
itutio
n
Red
uced
hea
t of
Hyd
roge
natio
n
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.
+ 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 of Synthetic 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?
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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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trans
Br2
BrBr
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!
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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
C RA
cis
Br2H Br
Br H
racemic mixture
+ enantiomer
Characteristics can be changed in pairs and C A R will remain true.
Want meso instead?? Have to use trans. Two changed!!
AT M
trans
Br H
Br H
meso
Br2