Transcript Ethers, Sulfides, Epoxides

Hemiacetals and Acetals, carbonyls and alcohols
Addition
reaction.
(Unstable in
Acid; Unstable
in base)
(Unstable in
Acid; Stable in
base)
Substitution reaction
Acetals as Protecting Groups
Synthetic Problem, do
a retrosynthetic analysis
E
Target molecule
N
Form this bond by reacting a nucleophile with an
electrophile. Choose Nucleophile and Electrophile
centers.
The nucleophile
could take the
form of an
organolithium or
a Grignard
reagent. The
electrophile
would be a
carbonyl.
Grignard would
react with this
carbonyl.
Br-Mg
Do you see the problem with
the approach??
Use Protecting Group for the carbonyl…
Acetals are stable (unreactive) in neutral
and basic solutions.
Create
acetal as
protecting
group.
Now create
Grignard and then
react Grignard with
the aldehyde to
create desired
bond.
protect
react
Remove protecting
group.
Same overall steps as when we used
silyl ethers: protect, react, deprotect.
deprotect
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.
This resonance stabilized
carbocation then reacts with an
alcohol molecule to yield the acetal.
An acid
2. By addition of H+
to a C=C double bond as
shown next.
This cation can now react with an alcohol
to yield an acetal. The alcohol becomes
part of an acetal and is protected.
Sample Problem
Provide a mechanism for the following conversion
O
HCl/H2O
HO
OH
O
O
OH
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.
Strongly electrophilic center, now can do
addition to the C=O
The protonation of the C=C
O
O
H+
O
H
Now do addition, join the molecules
HO
HO
OH
O
Product
O
O
Now must open 5 membered ring here. Need to set-up leaving group.
HO
HO
H+
O
O
H
O
O
Leaving group leaves….
HO
HO
HO
H
H
H
O
O
O
O
O
O
Followed by new ring closure.
Done. Wow!
Sulfur Analogs
Consider formation of acetal
O
OH
OH
O
dry HCl
acetaldehyde
ethanal
Sulfur Analog
O
SH
SH
S
dry HCl
acetaldehyde
ethanal
dithiane
S
O
The aldehyde hydrogen has been made acidic
Bu Li
+ BuH
S
S
S
S
H
Why acidic?
Sulfur, like phosphorus, has 3d orbitals capable of accepting electrons: violating octet
rule.
S
S
S
S
Recall early steps from the Wittig reaction discussed earlier
H
Ph3P:
H
I
Ph3P
+
H
strong base, BuLi
Ph3P
Ph3P
This hydrogen is acidic.
Why acidic? The P is positive and can accept charge from the negative carbon
into the 3d orbitals
PH
Some Synthetic Applications
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.
O
CN
O
CN
electrophile
O
O
OH
O
S
S
nucleophilic
Can you think of two other examples of Umpolung we have seen?
Nitrogen Nucleophiles
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.
Hydrazine derivatives
Note which nitrogen is nucleophilic
O
O
H2N
N
H
NH2
H2N
N
H
Nucleophilic nitrogen
Favored by resonance
Less steric hinderance
NH2
Reductive Amination
Pattern:
R2C=O + H2N-R’
  R2CH-NH-R’
Enamines
Recall primary amines react with carbonyl compounds to give Schiff bases
(imines), RN=CR2.
Primary amine
But secondary amines react to give enamines
See if you can write the mechanism for the reaction.
Secondary Amine
Acidity of a Hydrogens
a hydrogens are weakly acidic
Weaker acid than alcohols but
stronger than terminal
alkynes.
Learn this table….
Keto-Enol Tautomerism
(Note: we saw tautomerism before in the hydration of alkynes.)
Fundamental process
acid or base
catalysis
O
HO
CH2
CH3
keto form
enol form
usually small component
Mechanism in base:
O
:OH-
O
O
HO
H-O-H
CH3
Negative carbon, a
carbanion, basic,
nucleophilic
carbon.
CH2
CH2
Additional resonance
form, stabilizing
anion, reducing
basicity and
nucleophilicity.
CH2
Protonation to
yield enol form.
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.
O
O
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.
H
H
More details…
Nucleophilic carbon
O
nucleophilicity
CH2
Some examples:
O
O
base
O
O
O
R-X
O
R
:OH-
O
O
O
Br-Br
CH3
CH2
CH2Br
Some reactions related to acidity of
a hydrogens
Racemization
Exchange
Oxidation: Aldehyde  Carboxylic
Recall from the discussion of alcohols.
Milder oxidizing reagents can also be used
Ag(NH3)2+
RCHO
RCO2- + Ag
Tollens Reagent test for
aldehydes
“Drastic Oxidation” of Ketones
dichromate, etc
CO2H
HO2C
at high temperature
O
CO2H
HO2C
Obtain four different products in this case.
Reductions: two electron
OH
O
NaBH4
H
Or LiAlH4
OH
O
H2/Pt
H
Reductions: Four Electron
Clemmenson
acid
H
O
H
Zn(Hg), HCl
Wolf-Kishner
base
H
O
H2N-NH2
KOH, heat
H
Mechanism of Wolf-Kishner, C=O  CH2
H
Recall reaction of primary
amine and carbonyl to give
Schiff base. Here is the
formation of the Schiff base.
We expect this to happen.
Weakly acidic
hydrogen removed.
Resonance occurs.
Same as keto/enol
tautomerism.
N
O
N
-
These hydrogens
are weakly acidic,
just as the
hydrogens a to a
carbonyl are acidic.
OH
H
N
H
H
N
C
N
N
H
N
C
O
Here is the resonance for the anion from the keto-enol system
N
H
N
H-O-H
N
Protonation (like
forming the enol)
Perform an
elimination reaction
to form N2.
N
H2N-NH2
H
H
H
C
N
H
C
C
C
O
O
H
-
N
N
OH
H
N
H
N
N
N
H
H-O-H H
H
H
H
Haloform Reaction, overall
O
O
CH3
CX3
CO2-
X2
NaOH
O
CH3
a methyl
NaOH
+ HCX3
The last step which produces the
haloform, HCX3 only occurs if there is
an a methyl group, a methyl directly
attached to the carbonyl.
If done with iodine then the formation of
iodoform, HCI3, a bright yellow
precipitate, is a test for an a methyl
group (iodoform test).
Steps of Haloform Reaction
O
The first reaction:
O
CX3
CH3
X2
NaOH
O
O
CH2X
CH3
All three H’s replaced by
X. This must happen
stepwise, like this:
X2
NaOH
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: C=C + X-X 
+ Br- (alkene acts as nucleophile, ions)
:OH-
O
Nucleophilic enolate anion:
Br
O
O
Br-Br
CH3
CH2
CH2Br
Mechanism of Haloform Reaction-1
Using the last of the three possibilities
:OH-
O
O
O
Br-
Br-Br
C H2
C H3
R
R
CH2Br
One H has been
replaced by halogen.
R
O
Repeat twice again to yield
C Br3
R
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.
Mechanism of Haloform - 2
O
O
CH3
R
CX3
X2
O
NaOH
O-
+ HCX3
R
NaOH
This is a
substitution
step; OHreplaces the CX3
and then ionizes
to become the
carboxylate
anion.
R
Here’s 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.
O
CX3
O
-
CX3
OH
OH
R
R
O
CX3
O
OH
O
O-
OH
R
R
+ -:CX3
R
+ HCX3
Cannizaro Reaction
Overall:
conc. KOH
RCO2- + RCH2OH
2 RCHO
heat
Restriction: no a hydrogens in the aldehydes.
CHO
O
CHO
H
H3C
a hydrogens
No a hydrogens
Why the restriction? The a hydrogens are acidic leading to ionization.
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.
O
O
O
O
R
+
R
H
-
HO :
Attack of
nucleophilic
HO-
+
R
R
H
R
OH
H
O
R
OH
H
R
H
H
O
H
Re-establish C=O and
eject H- which is
immediately received
by second RCHO
O
Acid-base
OH
Experimental Evidence
KOH, H2O
2 RCDO
RD2OH + RCO2-
These are the hydrogens introduced by the
reaction. They originate in the aldeyde and do
not come from the aqueous hydroxide solution.
Kinetic vs Thermodynamic Contol of a
Reaction
Examine Addition of HBr to 1,3 butadiene
H
H
HBr
+
Br
1,2 product
Br
1,4 product
Mechanism of reaction.
Allylic resonance
H
H
H-Br
Br
Br
H
H
Br
Br
1,2 product
But which is the dominant product?
1,4 product
Nature of the product mixture depends on the temperature.
H
H
HBr
+
Br
1,2 product
Product mixture at -80 deg
Product mixture at + 40 deg
80%
20%
Br
1,4 product
20%
80%
Goal of discussion: how can temperature control the product mixture?
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.
Nature of the product mixture depends on the temperature.
H
H
HBr
+
Br
1,2 product
Product mixture at -80 deg
Product mixture at + 40 deg
80%
20%
Br
1,4 product
20%
80%
More stable
product
Thermodynamic Control
Kinetic Control
Product formed most
quickly, lowest Ea
Formation of the allylic carbocation.
Can react to yield 1,2 product or 1,4 product.
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.
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.
Diels Alder Reaction/Symmetry Controlled Reactions
Quick Review of formation of chemical bond.
Electro
n 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.
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
LUMO
acceptor
LUMO
acceptor
HOMO
donor
B
A
Note the
overlap
leading to
bond
formation
A HOMO donates into B LUMO
HOMO
Note the
donor
overlap
leading to
bond
formation
Try it in another reaction: ethylene + ethylene  cyclobutane
LUM
O
HOMO
LUMO
HOMO
Equal bonding and antibonding
interaction, no overlap, no bond
formation, no reaction
Reaction Problem
Br
excess sodium methoxide
Br
Synthesis problem
OEt
HO
using only compounds
having two carbons as
the source of all carbons in
the target molecule
Mechanism Problem
Give the mechanism for the following reaction. Show all
important resonance structures. Use curved arrow notation.
O
OH
aq. acid
+ EtOH
heat
OEt