Transcript Alkanes - City University of New York

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.
An example: pi electrons as bases
Bronsted
Lowry Base
Bronsted
Lowry Acid
For the moment, just note that
there are two possible
carbocations formed.
The carbocations are
conjugate acids of the
alkenes.
Sigma bonding electrons as bases.
Much more unusual!!
A very, very
electronegative F!!
Super acid
A very positive S!! The
OH becomes very
acidic because that
would put a negative
charge adjacent to the
S.
Trends for Relative Acid Strengths
Totally unionized
in aqueous
solution
Aqueous Solution
Totally ionized in
aqueous solution.
Example
OH
Recall
CH3CH2OH
pKa = 15.9
pKa = 9.95
Stronger acid
H2O + PhOH
ethanol, EtOH
phenol, PhOH
H3O+ + PhO-
H2O + EtOH
Weaker acid
H3O+ + EtO-
Ka = [H3O+][EtO-]/[EtOH] = 10-15.9
Ka = [H3O+][PhO-]/[PhOH] = 10-9.95
Ethanol, EtOH, is a weaker acid than phenol, PhOH.
It follows that ethoxide, EtO-, is a stronger base than phenolate, PhO-.
For reaction PhOH + EtOStronger
base
PhO- + EtOH where does equilibrium lie?
Weaker base.
K=
10-9.95 / 10-15.9
=
106.0
Query: What makes for strong (or weak) acids?
What affects acidity?
1. Electronegativity of the atom holding the negative charge.
Increasing
acidity.
CH3O - + H+
CH3OH
CH3NH2
CH3CH3
CH3NH - + H+
CH3CH2- + H+
Increasing
Increasing basicity
electronegativity
of
of anion.
atom bearing negative
charge. Increasing
stability of anion.
2. Size of the atom bearing the negative charge in the anion.
Increasing
acidity.
CH3OH
CH3O - + H+; pKa = 16
CH3SH
CH3S - + H+; pKa = 7.0
Increasing
Increasingsize
basicity
of
atom
of anion.
holding negative
charge. Increasing
stability of anion.
What affects acidity? - 2
3. Resonance stabilization, usually of the anion.
Acidity
OH
O
O
O
O
Increasing
resonance
basicity
of
stabilization.
the
anion.
Increased
anion
stability.
phenol, PhOH
CH3CH2OH
CH3CH2O - + H+
ethanol, EtOH
No resonance structures!!
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.
OH
OH
etc.
An example: competitive Bases &
Resonance
• Two different bases or two sites in the same molecule may
compete to be protonated (be the base).
Acetic acid can be protonated at two sites.
O
H
O
Pi bonding
electrons
converted to
non-bonding.
O
H
H+
O
H
acetic acid
Non-bonding
electrons
converted to
pi bonding.
Which conjugate acid is
favored?
H
O
O
H
O
O
H
+
H
H
The more stable one!
Which is that?
Recall resonance
provides additional
stability by moving pi or
non-bonding electrons.
O
O
H
H
No valid resonance
structures for this
cation.
An example: competitive Bases &
Resonance
Comments on the importance of the resonance structures.
All atoms obey octet
rule!
H
O
H+
O
H
O
O
H
acetic acid
The carbon is electron
deficient – 6 electrons, not 8.
Lesser importance
H
O
O
H
H
O
All atoms obey octet
rule!
O
H
What affects acidity? - 3
4. Inductive and Electrostatic Stabilization.
Acidity.
H3CCH2OH
d+
H3CCH2O - + H+
d+
F3CCH2OH
Increasing anion
Increasing anion
stability.
basicity.
F3CCH2O - + H+
Due to electronegativity
of F small positive
charges build up on C
resulting in stabilization
of the anion.
Effect drops off with distance. EtOH pKa = 15.9
What affects acidity? - 4
Increasing Acidity of HA
Increasing Basicity of A-
Note. The NH2-.
5. Hybridization of the atom bearing the charge. H-A  H+ + A:is
more basic
than the RCCsp3
sp2
sp
ion.
More s character, more stability, more “electronegative”, H-A
more acidic, A:- less basic.
Know this order.
Example of hybridization Effect.
terminal alkyne
RCCH + LiCH2CH2CH2CH3
base
acid
RCCH + AgNO3
HCH2CH2CH2CH3 + RCCLi
AgCCR (ppt)
non-terminal alkyne
RCCR + LiCH2CH2CH2CH3
RCCR + AgNO3
NR
No Reaction
What affects acidity? - 5
O
H
H
H
6. Stabilization of ions by solvents (solvation).
O
H
H
O
O
R
+
Solvation provides
stabilization.
H
R
Comparison of alcohol acidities.
OH
OH
OH
H
O
pKa = 15.9
ethanol
17
propan-2-ol
18
H
2-methylpropan-2-ol
Crowding inhibiting solvation
Solvation, stability of anion, acidity
(CH3)3CO -,
crowded
Example
Para nitrophenol is more acidic than phenol. Offer an explanation
OH
O
+ H
Why? Could be due to destabilization
of the unionized form, A, or
stabilization of the ionized form, B.
O
OH
+ H
N
N
O
O
A
O
O
B
The lower lies further to the right.
Examine the equilibrium for p-nitrophenol. How does
the nitro group increase the acidity?
O
OH
+ H
N
N
O
O
OH
O
A
O
O
B
N
N
N
O
O
C
O
O
D
O
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.
OH
OH
OH
N
O
O
Examine both sides
of equilibrium. What
does the nitro group
do?
Structure D occurs only due to the
nitro group. The stability it
provides will slightly decrease
acidity.
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
Now look at the anion. What does the nitro group do? Remember we
are interested to compare with the phenol phenolate equilibrium.
O
OH
+ H
N
N
O
O
A
O
O
B
N
N
N
O
O
O
O
C
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
O
O
N
O
O
O
O
O
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.
D
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
Sample Problem
3. (3 pts) Which is the stronger base and why?
vs
O
HN
HN
H2N
H2N
O
H2N
O
Alkenes 1
Shape
Alkenes, 6 coplanar atoms.
All atoms in same
plane except for
these hydrogens on
sp3 carbon.
Arene shapes
Planar ring structure. 12 atoms coplanar.
Phenyl group,
C6H5,, Ph
Ph
Ph = C6H5
2-phenyl propane
Pi bonds
pi* orbital
Energy
difference
pi orbital
sum
Nomenclature
but-1-ene
3,4-dimethylhexa-1,5-diene
Z / E generalization of cis / trans
Use R, S priorities to compare
substituents on same carbon.
cis / trans
High priority on same side, Z.
Opposite, E.
H
cis
Br
H
F
trans
Cl
F
Cl
Br
(E)-1-bromo-2-chloro-1-fluoroethene
(Z)-1-bromo-2-chloro-1-fluoroethene
Cis / Trans in Cycloalkenes
For small rings normally have cis double bonds.
trans cyclooctene
Terpenes and the isoprene Rule
• A terpene is composed of isoprene units
joined head to tail (the isoprene rule).
Note that location of
functional groups such as
OH or double bonds is not
addressed.
This
molecule
has
additional
cross
links.
Vitamin A
Four isprene units joined head to tail
One cross link (non-head to tail) linkage.
Fatty Acids
• Animal fats and vegetable oils are both
triesters of glycerol, hence the name
triglyceride.
– Hydrolysis of a triglyceride in aqueous base
followed by acidification gives glycerol and
three fatty acids.
O
CH 2 OH
O CH 2 OCR
1 . Na OH, H 2 O
R'COCH
HOCH
+
O
2 . HCl, H 2 O
CH 2 OH
CH 2 OCR' '
1,2,3-Propan etriol
A trigl yce ri de
(gl yce rol)
(a triest er of glycerol
RCOOH
R' COOH
R' ' COOH
Fatty acids
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 oils.
– Triglycerides rich in saturated fatty acids are
generally semisolids or solids at room
temperature and are called fats.
Fatty Acids
– the four most abundant fatty acids
COOH Stearic acid (18:0)
(mp 70°C)
COOH Oleic acid (18;1)
(mp 16°C)
COOH Linoleic acid (18:2)
(mp-5°C)
COOH Lin olen ic acid (18:3)
(mp -11°C)
Alkene Reactions
Pi bonds
Reactivity
above and
below the
molecular
plane!
Plane of molecule
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.
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.
Transition State
For the step
C + A-B  C-A + B
Three areas to be aware of.
Products
Reactants
Energy of
Activation.
Energy
barrier.
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.
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
Multi-Step Reactions
 Intermediate
Step 1:
A + B
Step 2:
Intermediate  C + D
Step 1:
endergonic,
high energy of
activation.
Slow process
Step 1 is the “slow step”, the
rate determining step.
Step 2:
exergonic, small
energy of
activation. Fast
Process.
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.
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.
Example
CH3 - H + Br
CH3 + H - Br
DH = 109 kJ
•Endothermic
•Transition state resembles the (higher
energy) products.
Only a small
[H3C
H Br]
amount of radical
character remains.
Almost
formed
radical.
Almost
broken.
Almost
formed.
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
Electrophilic Addition
We now address regioselectivity….
Regioselectivity (Orientation)
The incoming hydrogen attaches to the carbon with the
greater number of hydrogens. This is regioselectivity.
It is called Markovnikov orientation.
Mechanism
Step 1
Step 2
Now examine Step 1 Closely
Electron rich, pi
system.
Acidic molecule, easily
ionized.
Showed this
reaction earlier as
an acid/base
reaction. Alkene
was the base.
We had portrayed the HBr
earlier as a BronstedLowry acid.
New term: the
alkene is a
nucleophile,
wanting to react
with a positive
species.
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.
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.
Carbocations
Electron deficient.
sp2 hybridized.
Does not obey octet rule.
p orbital is empty and can receive
electrons.
Lewis acid, can receive
electrons.
Electrophile.
Flat, planar. Can react on either
side of the plane.
Very reactive and present only in
very low concentration.
Step 2 of the Mechanism
:Br-
Br
Mirror objects
:Br-
Br
Regioselectivity (Orientation)
H - Br
Or
H
H
+ Br
Br
H
2-Bromo-propane
Secondary
carbocation
Secondary
carbocation more
more stable and
more easily formed.
Primary
carbocation
+ Br
H
Br
1-Bromo-propane
Carbocation Stabilities
Order of increasing stability:
Methyl < Primary < Secondary < Tertiary
Order of increasing ease of formation:
Methyl < Primary < Secondary < Tertiary
Increasing Ease of Formation
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.
d-
F
H
d+
d-
F
F
d-
H
Factors Affecting Carbocation
Stability - Hyperconjugation
2. Hyperconjugation. Unlike normal resonance or
conjugation hyperconjugation involves s bonds.
H
H
H
H
H
H
H
H
H
H
ethyl carbocation
Hyperconjugation spreads the positive charge onto the
adjacent alkyl group
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.
Factors Affecting Carbocation Stability Resonance
Note: the allylic
carbocation can react at
either end!
Utilizing an adjacent pi system.
allylic carbocation
Positive charge delocalized through resonance.
H
H
Another
very
important
example.
benzylic carbocation
H
H
H
H
H
H
The benzylic
carbocation will
react only at the
benzylic position
even though
delocalization
occurs!
Positive charge delocalized into the benzene ring. Increased stability of
carbocation.
Another Factor Affecting Carbocation
Stability – Resonance
Utilizing an adjacent lone pair.
O
H
O
CH2
H
CH2
Look carefully.
This is the
conjugate acid
of formaldehyde,
CH2=O.
Production of Chiral Centers. Goal is to see
all the possibilities.
React alkene with HBr.
Note that the ends of the double
H+
bond are different.
Ph
H
Me
Ph
H
Me
Ph
Me
Br
Br-
Et
Me
Me
Ph
Et
Br-Br
mirror plane
Me
Me is methyl group
Et is ethyl group
Ph is phenyl group
Me
+
H
Br
Et
Me
Regioselectivity Analysis:
the positivePhcharge willH go here
and be stabilized
by resonance
Enantiomeric
Br
with thecarbocations.
phenyl group.
Br
The H will
attach here.
Me
Et
Me
H
Et
Me
Et
Me
Ph
H
Me
Ph
Et
Me
Br
H
What has been made?
Two pairs of enantiomers.
Production of Chiral Centers - 2
Br
Me
Ph
H
Me
Ph
Et
Me
Me
Ph
Et
Me
Br
diastereomers
H
Racemic Mixture 1
H
Br
Et
Me
Br
Et
Me
Me
Ph
H
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).
Acid Catalyzed Hydration of
Alkenes
What is the orientation??? Markovnikov
Mechanism
Step 1
Step 2
Note the electronic
structure of the
oxonium ion.
Step 3
Carbocation Rearrangements
Expected product is not the major product; rearrangement of carbon
skeleton occurred.
The methyl group moved. Rearranged.
Also, in the hydration reaction.
The H moved.
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
Addition of Br2 and Cl2
Stereochemistry
Anti Addition (halogens enter on
opposite sides); Stereoselective
Syn addition (on same side) does not
occur for this reaction.
Mechanism, Step 1
Step 1, formation of cyclic bromonium ion.
Step 2
Detailed Stereochemistry, addition of Br2
Bromide ion
attacked the
carbon on the
right.
Br
C2H5
(S)
Br
C3H7
H3C
CH3
(S)
Br
Br
S,S
Br
Br
C3H7
H3C
(S)
C2H5
CH3
(R)
C3H7
H3C
enantiomers
Br
(R)
C2H5
CH3
(R)
C3H7
C2H5
H3C
CH3
Br
Br
R,R
C3H7
Br
Br
H3C
C3H7
C2H5
CH3
H3C
Br
Alternatively, the bromine
could have come in from the
bottom!
(R)
Br
(R)
C2H5
(R)
CH3
Br
(S)
R,R
enantiomers
Br
C2H5
Br
CH3
(S)
C3H7
(S)
H3C
Only two compounds (R,R and S,S) formed in equal
amounts. Racemic mixture.
But can also
attack the
left-side
carbon.
S,S
Br
Number of products formed.
Br
C2H5
(S)
C3H7
H3C
CH3
(S)
Br
S,S
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.
C3H7
H3C
enantiomers
Br
(R)
C2H5
(R)
CH3
Br
R,R
C3H7
Br
H3C
(R)
C2H5
(R)
CH3
Br
R,R
enantiomers
C2H5
CH3
Br
(S)
C3H7
(S)
H3C
S,S
Br
Attack of the Bromide Ion
Br
(S)
C3H7
H3C
C2H5
CH3
(S)
Br
Br
C3H7
H3C
(S)
(R)
C2H5
CH3
Br
Starts as R
Becomes S
In order to
preserve a
tetrahedral
carbon these
two substituents
must move
upwards.
Inversion.
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.
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.
Using Fischer Projections
Convert to Fischer by
doing 180 deg rotation
of top carbon.
Br
Br
R2
R1
R2
Br2
R1
R1
R2
R4
R3
=
anti addition
R4
R3
R4
R3
Br
Br
+ enantiomer
+ enantiomer
Not a valid
Fischer
projection since
top vertical
bond is coming
forward.
There are many variations on the addition of X2 to
an alkene. Each one involves anti addition.
R2
Br
R1
R1
R2
R4
R3
Br2
R4
Br
R3
Br
+ enantiomer
Br
R1
R2
R3
R4
I-
Br
I
R1
R2
R4
R3
R1
R2
R4
R3
+
Instead of iodide ion as nucleophile can
use alcohols to yield ethers, water to
yield alcohols, or amines.
Br
I
+ enantiomer
+ enantiomer
The iodide can attach to either
of the two carbons.
Regioselectivity
If Br2 is added to propene there is no regioselectivity issue.
Br
Br2
Br
If Br2 is added in the presence of excess alternative nucleophile, such as
CH3OH, regioselectivity may become important.
OCH3
Br
Br - Br
Br
and/or
OCH3
CH3O-H
+ H + + Br -
+ H + + Br -
Regioselectivity - 2
Consider, again, the cyclic bromonium ion and the resonance structures.
Weaker
bond
More positive charge
Br
Stronger bond
R
Expect the nucleophile to attack here. Remember inversion occurs.
Regioselectivity, Bromonium Ion
– Bridged bromonium ion from propene.
Example
Stereochemistry: anti addition
Note: non-reacting fragment unchanged
Et
H
Cl2/H2O
Et
Me
H
Et
Me
OH
Me
H
H
H
Me
OH
+
Me
Me
H
Cl
Me
Cl
Me
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
Me
Et
Et
H
Me
Me
OH
H
Cl
Me
+
H
Me
HO
Me
Cl
H
Me
Bromination of a substituted cyclohexene
Consider the following bromination.
Br+
Br2
Br
Br
C(CH3)3
C(CH3)3
Br
+
+
Br
Expect to form
two bromonium
ions, one on top
and the other on
bottom.
Br
Br
-
+
C(CH3)3
C(CH3)3
C(CH3)3
Expect the rings
can be opened
by attack on
either carbon
atom as before.
But NO, only one
stereoisomer is
formed. WHY?
Addition to substituted cyclohexene
H
H
Br2
Br
H
+
H
Br
The tert butyl group
locks the
conformation as
shown.
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.
Progress of Attack
Things to watch for:
Notice that the two bromines are
maintained anti to each other!!!
•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.
Addition to substituted cyclohexene
Attack as shown in red by incoming
Br ion will put both Br into equatorial
positions, not anti.
Br2
Br
Br
+
Br
Br
Observe
Ring is locked as shown. No ring flipping.
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
Attack as shown in green by the
incoming Br will result in both Br being
axial and anti to each other
Br2
Br
Br
+
Br
Br
This is the observed
diastereomer. We have kept
the bromines anti to each
other.
Oxymercuration-Reduction
Alkene  Alcohol
Regioselective: Markovnikov Orientation
Occurs without 1,2 rearrangement, contrast the following
OH
1 Hg(OAc)2
2. NaBH4
OH
No rearrangement
H2 O
3,3-dimethylbut-1-ene
H2SO4
formed via
rearrangement
Mechanism
1
2
3
4
Hydroboration-Oxidation
Alkene  Alcohol
Anti-Markovnikov orientation
Syn addition
1. BH3
2. H2O2
H
HO
H
HO
Borane, a digression
H
H
H
H
B
B
H
Isoelectronic with a carbocation
H
Mechanism
Syn stereochemistry, antiMarkovnikov orientation
now established.
Next…
Just call the circled group R.
Two reasons why anti-Markovnikov:
Eventually have BR3.
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.
Cont’d
Oxidation and Reduction Reactions
We think in terms of Half Reactions
Cr2O7 2- + CH3CH2OH
Will be
reduced.
Cr 3+ + CH3CO2H
Will be
oxidized.
• Write reactants and products of each half
reaction.
Inorganic half reaction…
6 e - + 14 H+ +
Cr2O7 2-
2 Cr 3+
+ 7 H2O
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.
Cont’d
Now the organic half reaction…
H2O + CH3CH2OH
CH3CO2H + 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 ( H2O + CH3CH2OH
2 x ( 6 e - + 14 H+ + Cr2O7 216 H+ + 2 Cr2O7 2- + 3 CH3CH2OH
CH3CO2H + 4 H+ + 4 e2 Cr 3+
)
+ 7 H2O )
4 Cr 3+ + 3 CH3CO2H + 11 H2O
Formation of glycols with Syn Addition
Osmium tetroxide
Syn addition
also
KMnO4
KMnO4
cold, dilute, slightly alkaline
Anti glycols
Using a peracid, RCO3H, to form an epoxide which is opened by aq. acid.
H+
PhCO3H, a peracid
H2O
O
O
H
HO
epoxide
The protonated epoxide is analagous
to the cyclic bromonium ion.
O
O
OH
Peracid: for example, perbenzoic acid
OH
An example
PhCO3H
O
+
O
(S)-3-methylcyclohex-1-ene
aq. acid
chiral, optically active
OH
OH
OH
OH
OH
OH
Diastereomers, separable (in theory) by
distillation, each optically active
OH
OH
Are these
unique?
Ozonolysis
R3
R1
1. O3
R1
R3
O
R4
R2
2. (CH3)2S
R4
+
O
R2
Reaction can be used to break larger molecule down into smaller parts
for easy identification.
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 ozonolysys
C8H12
O
O
O
O
O
Simply remove the new oxygens and join to make double bonds.
But there is a second possibility.
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
Suggest a reasonable structure for the unknown.
d
Hydrogen Deficiency = 8. Four pi
bonds/rings.
O
a
O
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
O
To construct unknown cross out the oxygens and then connect. But there are many
ways the connections can be made.
a-b & c-d
d
a-d & b-c
a-c & b-d
c
a
c
Look for a structure
that obeys the
isoprene rule.
d
d
a
c
b
a
b
b
Mechanism
Consider the resonance structures of ozone.
Electrophile
capability.
Nucleophile
capability.
O
O
O
O
O
O
O
O
O
O
O
O
These two, charged at
each end, are the useful
ones to think about.
Mechanism - 2
O
O
O
O
O
O
O
O
O
O
O
O
Mechanism - 3
Mechanism - 4
Hydrogenation
No regioselectivity
Syn addition
Heats of Hydrogenation
Consider the
cis vs trans
heats of
hydrogenation
in more
detail…
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
By same reasoning higher degree of substitution provide lower heat of
hydrogenation and are, therefore, more stable.
Reduced heat of Hydrogenation
Increasing substitution
Heats of Hydrogenation
Acid Catalyzed Polymerization
Principle: Reactive pi electrons (Lewis base) can react with Lewis acid. Recall
H
+ H
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 carbocation is an acid!
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.
Examples of Synthetic Planning
OH
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
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…
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)
trans
cis
Br2
Br2
Br
Br
H
H
Br
rotate lower unit
+ enantiomer
Br
H
Br
Br
H
Br
H
Br
meso
This worked! How about
starting with the cis?
racemic mixture
This did not work, gave us
the wrong stereochemistry!
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.
A
C
R
Want meso instead?? Have to
use trans. Two changed!!
Br2
H
Br
+ enantiomer
Br
H
A
T
M
cis
Br2
racemic mixture
Br
H
Br
H
trans
meso