Hemiacetals and Acetals, carbonyls and alcohols

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Hemiacetals and Acetals, carbonyls and alcohols
Addition reaction.
(Unstable in Acid Unstable in base)
(Unstable in Acid Stable in base)
Substitution reaction
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Acetals as Protecting Groups
E
Synthetic Problem, do a retrosynthetic analysis
Target molecule
N
Form this bond by reacting a nucleophile with an
electrophile. Choose Nucleophile and Electrophile
centers.
Grignard would react with this carbonyl.
The nucleophile could take the form of an
organolithium or a Grignard reagent. The
electrophile would be a carbonyl.
Do you see the problem with the approach??
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Use Protecting Group for the carbonyl Acetals
are stable (unreactive) in neutral and basic
solutions.
Create acetal as protecting group.
protect
Now create Grignard and then react Grignard with
the aldehyde to create desired bond.
react
Remove protecting group.
deprotect
Same overall steps as when we used silyl ethers
protect, react, deprotect.
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Tetrahydropyranyl ethers (acetals) as protecting
groups for alcohols.
Recall that the key step in forming the acetal
was creating the carbocation as shown
There are other ways to create carbocations
Recall that we can create carbocations in several
ways 1. As shown above by a group
leaving. 2. By addition of H to a CC double
bond as shown next.
This resonance stabilized carbocation then reacts
with an alcohol molecule to yield the acetal.
An acid
This cation can now react with an alcohol to
yield an acetal. The alcohol becomes part of an
acetal and is protected.
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Sample Problem
Provide a mechanism for the following conversion
First examination have acid present and will
probably protonate Forming an acetal. Keep those
mechanistic steps in mind.
Ok, what to protonate? Several oxygens and the
double bond. Protonation of an alcohol can
set-up a better leaving group. Protonation of a
carbonyl can create a better electrophile. We do
not have a carbonyl but can get a similar species
as before.
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Strongly electrophilic center, now can do
addition to the CO
The protonation of the CC
Now do addition, join the molecules
Product
Now must open 5 membered ring here. Need to
set-up leaving group.
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Leaving group leaves.
Followed by new ring closure.
Done. Wow!
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Sulfur Analogs
Consider formation of acetal
Sulfur Analog
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The aldehyde hydrogen has been made acidic
Why acidic?
Sulfur, like phosphorus, has 3d orbitals capable
of accepting electrons violating octet rule.
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Recall early steps from the Wittig reaction
discussed earlier
This hydrogen is acidic.
Why acidic? The P is positive and can accept
charge from the negative carbon into the 3d
orbitals
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Some Synthetic Applications
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Umpolung reversed polarity
What we have done in these synthetic schemes is
to reverse the polarity of the carbonyl group
change it from an electrophile into a nucleophile.
electrophile
nucleophilic
Can you think of two other examples of Umpolung
we have seen?
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Nitrogen Nucleophiles
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Mechanism of Schiff Base formation
Attack of nucleophile on the carbonyl
Followed by transfer of proton from weak acid to
strong base.
Protonation of OH to establish leaving group.
Leaving group departs, double bond forms.
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Hydrazine derivatives
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Note which nitrogen is nucleophilic
Nucleophilic nitrogen
Favored by resonance Less steric hinderance
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Reductive Amination
Pattern R2CO H2N-R ? ? R2CH-NH-R
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Enamines
Recall primary amines react with carbonyl
compounds to give Schiff bases (imines), RNCR2.
Primary amine
But secondary amines react to give enamines
See if you can write the mechanism for the
reaction.
Secondary Amine
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Acidity of a Hydrogens
a hydrogens are weakly acidic
Weaker acid than alcohols but stronger than
terminal alkynes.
Learn this table.
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Keto-Enol Tautomerism
(Note we saw tautomerism before in the hydration
of alkynes.)
Fundamental process
Mechanism in base
Negative carbon, a carbanion, basic, nucleophilic
carbon.
Additional resonance form, stabilizing anion,
reducing basicity and nucleophilicity.
Protonation to yield enol form.
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Details
Base strength
Alkoxides will not cause appreciable ionization
of simple carbonyl compounds to enolate. Strong
bases (KH or NaNH2) will cause complete
ionization to enolate.
Double activation (1,3 dicarbonyl compounds) will
be much more acidic.
For some 1,3 dicarbonyl compounds the enol form
may be more stable than the keto form.
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More details
Nucleophilic carbon
nucleophilicity
Some examples
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Some reactions related to acidity of a hydrogens
Racemization
Exchange
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Oxidation Aldehyde ? Carboxylic
Recall from the discussion of alcohols.
Milder oxidizing reagents can also be used
Tollens Reagent test for aldehydes
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Drastic Oxidation of Ketones
Obtain four different products in this case.
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Reductions two electron
NaBH4
Or LiAlH4
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Reductions Four Electron
Clemmenson
Wolf-Kishner
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Mechanism of Wolf-Kishner, CO ? CH2
Recall reaction of primary amine and carbonyl to
give Schiff base. Here is the formation of the
Schiff base. We expect this to happen.
These hydrogens are weakly acidic, just as the
hydrogens a to a carbonyl are acidic.
Weakly acidic hydrogen removed. Resonance occurs.
Same as keto/enol tautomerism.
Protonation (like forming the enol)
Perform an elimination reaction to form N2.
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Haloform Reaction, overall
The last step which produces the haloform, HCX3
only occurs if there is an a methyl group, a
methyl directly attached to the carbonyl.
a methyl
If done with iodine then the formation of
iodoform, HCI3, a bright yellow precipitate, is a
test for an a methyl group (iodoform test).
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Steps of Haloform Reaction
The first reaction

• All three Hs replaced by
• This must happen
• stepwise, like this

Pause for a sec We have had three mechanistic
discussions of how elemental halogen, X2, reacts
with a hydrocarbon to yield a new C-X bond. Do
you recall them?
Radical Reaction R. X-X ? R-X
X. (initiation required)
Addition to double bond CC X-X ?
Br- (alkene acts as nucleophile, ions)
Nucleophilic enolate anion
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Mechanism of Haloform Reaction-1
Using the last of the three possibilities
One H has been replaced by halogen.
Repeat twice again to yield
Where are we? The halogens have been introduced.
First reaction completed.
But now we need a substitution reaction. We have
to replace the CBr3 group with OH.
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Mechanism of Haloform - 2
This is a substitution step OH- replaces the CX3
and then ionizes to become the carboxylate anion.
Heres how
Attack of hydroxide nucleophile. Formation of
tetrahedral intermediate. Anticipate the attack
Reform the carbonyl double bond. CX3- is ejected.
The halogens stabilize the negative carbon.
Neutralization.
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Cannizaro Reaction
Overall
Restriction no a hydrogens in the aldehydes.
a hydrogens
No a hydrogens
Why the restriction? The a hydrogens are acidic
leading to ionization.
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Mechanism
What can happen? Reactants are the aldehyde and
concentrated hydroxide. Hydroxide ion can act
both as Base, but remember we have no acidic
hydrogens (no a hydrogens). Nucleophile,
attacking carbonyl group.
Attack of nucleophilic HO-
Acid-base
Re-establish CO and eject H- which is
immediately received by second RCHO
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Experimental Evidence
These are the hydrogens introduced by the
reaction. They originate in the aldeyde and do
not come from the aqueous hydroxide solution.
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Kinetic vs Thermodynamic Contol of a Reaction
Examine Addition of HBr to 1,3 butadiene
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Mechanism of reaction.
Allylic resonance
But which is the dominant product?
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Nature of the product mixture depends on the
temperature.
Product mixture at -80 deg 80
20 Product mixture at 40
deg 20 80
Goal of discussion how can temperature control
the product mixture?
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When two or more products may be formed in a
reaction A ? X or A ? B
Thermodynamic Control Most stable product
dominates
Kinetic Control Product formed fastest dominates
Thermodynamic control assumes the establishing of
equilibrium conditions and the most stable
product dominates.
Kinetic Control assumes that equilibrium is not
established. Once product is made it no longer
changes.
Equilibrium is more rapidly established at high
temperature. Thermodynamic control should
prevail at high temperature where equilibrium is
established. Kinetic Control may prevail at low
temperature where reverse reactions are very slow.
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Nature of the product mixture depends on the
temperature.
Product mixture at -80 deg 80
20 Product mixture at 40
deg 20 80
More stable product
Thermodynamic Control
Kinetic Control
Product formed most quickly, lowest Ea
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Formation of the allylic carbocation.
Can react to yield 1,2 product or 1,4 product.
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Most of the carbocation reacts to give the 1,2
product because of the smaller Ea leading to the
1,2 product. This is true at all
temperatures. At low temperatures the reverse
reactions do not occur and the product mixture is
determined by the rates of forward reactions. No
equilibrium.
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Most of the carbocation reacts to give the 1,2
product because of the smaller Ea leading to the
1,2 product. This is true at all
temperatures. At higher temperatures the reverse
reactions occur leading from the 1,2 or 1,4
product to the carbocation. Note that the 1,2
product is more easily converted back to the
carbocation than is the 1,4. Now the 1,4 product
is dominant.
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Diels Alder Reaction/Symmetry Controlled Reactions
Quick Review of formation of chemical bond.
Electron donor
Electron acceptor
Note the overlap of the hybrid (donor) and the s
orbital which allows bond formation.
For this arrangement there is no overlap. No
donation of electrons no bond formation.
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Diels Alder Reaction of butadiene and ethylene to
yield cyclohexene.
We will analyze in terms of the pi electrons of
the two systems interacting. The pi electrons
from the highest occupied pi orbital of one
molecule will donate into an lowest energy pi
empty of the other. Works in both directions A
donates into B, B donates into A.
B HOMO donates into A LUMO
Note the overlap leading to bond formation
LUMO acceptor
LUMO acceptor
A HOMO donates into B LUMO
HOMO donor
HOMO donor
Note the overlap leading to bond formation
B
A
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Try it in another reaction ethylene ethylene
? cyclobutane
LUMO
LUMO
Equal bonding and antibonding interaction, no
overlap, no bond formation, no reaction
HOMO
HOMO
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Reaction Problem
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Synthesis problem
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Mechanism Problem
Give the mechanism for the following reaction.
Show all important resonance structures. Use
curved arrow notation.