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Aldehydes and Ketones
Chapter 16
Structure
Aldehydes
Ketone
Carbonyl
group
O
R
O
R
H
R
sp2
2-pentanone
O
pentanal
O
Examples of Naming
OH
O
O
O
O
pentan-2-one
3-oxopentanal
O
O
2-(1-hydroxyethyl)pentanal
CHO
CH3
H
CHO
H
CO2H
CH3
(E)-3-(but-1-enyl)hexane-2,4-dione
2,3-dimethyl-4-oxobutanoic acid
benzaldehyde
Resonance
O
O

result

O
Extension of resonance
O
O
Boiling points
For compounds of comparable molecular weight…
Alkanes, ethers < aldehydes, ketones < alcohols < carboxylic acids
Dispersion
Forces
Dipole-dipole
Hydrogen Bonding
Water Solubility
Ketones and Aldehydes, like ethers, can function as hydrogen bond
acceptors and smaller compounds have significant water solubility.
Recall Preparation from Alcohols
OH
OH
Can also be
done using
KMnO4 in base
with heat or
bleach in acid
solution
(HOCl).
Na2Cr2O7
acid, 35 deg.
HO
CH2OH
Be sure you
can balance
this kind of
reaction.
O
CO2H
Use PCC to limit oxidation of primary alcohol to the aldehyde.
Secondary are oxidized to ketone.
Primary alcohol
PCC
RCH2OH
RCH=O
Secondary
PCC
R2CHOH
R2C=O
Preparations, con’d
• Reaction of acid chloride and Gilman
O
R
OH
SOCl2
O
Na2Cr2O7
O
RCH2OH
R
Cl
R
R' 2CuLi
1. Li
R'X
2. CuX
But where do we
get these??
R'
Note that we have two possible disconnects available
HOCH2R
R'OH
O
+ R'X
HO
O
R'
R
R
O
+ RX
R'
OH
ROH
HOCH2R'
Example: Prepare 2-butanone from ethyl alcohol
Requirement to start with ethanol suggests a disconnect into two carbon fragments.
CH3CH2OH
O
Done!
CH3CO2H + CH3CH2X
via Gilman and
acid chloride
Aldehydes from carboxylic acids
Reduction
O
R
O
SOCl2
OH
LiAlH(OC(CH3)3)3
R
Cl
O
R
H
And from alcohols, as before:
PCC
RCH2OH
Oxidation
RCHO
A Common Sequence
PCC
RCH2OH
Mg
PX3
R'OH
OH
RCHO
RX
R
R'
R'MgX
Na2Cr2O7
O
Observe these parts at this moment.
R
R'
Reactions
Addition of a nucleophile: Nucleophilic Addition
Good nucleophile, usually basic
O
O
O
+
Nu:
-
Nu
Nu
tetrahedral intermediate
OH
OH
+
Nu
Attack of
nucleophile
occurs on both
sides of carbonyl
group.
Produces both
configurations.
Nu
Overall: H – Nu was added to carbonyl group double bond.
Notice that the CO bond order was reduced from 2 to 1. The addition reduced the bond
order. We will use this idea later.
Reaction can also be done in acid environment.
Nucleophiles not expected to be as strong (why?) but the oxygen may
become protonated making the carbonyl a better electrophile
(why?).
O
OH
acid
OH
OH
+
Nu:
Nu
Nu
Very electronegative,
protonated oxygen. Pulls
the pi electrons into itself
strongly.
Problem: If there is too much acid present the nucleophile may become
protonated, deactivating it
Addition of Grignard (Trumpets Please)
Recall the formation of a Grignard and its addition to an oxirane
O
Mg
R-X
 
R-Mg-X
OMgX
ether
Carbonyls may be added to
in same way…
Mg
R-X
O
R
OMgX
 
R-Mg-X
ether
If a new chiral center is
created both configurations
will be produced.
OMgX
+
R
R
mild acid
OH
OH
R
R
Common Reactions of Grignards
R2R'COH
a tertiary alcohol
R-H
Examine reaction
with ester further.
R'CO2Et
an ester
ROHRXRH(D)
acid,
weak acid
O
R
primary alcohol
O
ROH + R’CH2OHRX + R’CO2HR2C(OH)R’
O
R-Mg-X
H
R'
OH
R''
H
RCH2OH
O
CO2
R'
Both of these
reactions extend
carbon chain &
keep -OH
functionality at
end of chain.
Can extend
further.
primary alcohol
H
RR'R''C-OH
tertiary alcohol
ROH + R’R”CHOH
RX + R’R”CO
RR’C (OH)R”
RCHR'
secondary alcohol
OH
RCO2H
carboxylic acid
ROHRXRCO2H
ROH + R’CH2OHRX + R’CHORCH(OH)R’
Grignard Reacting with an Ester.
Look for two kinds of reactions.
O
OMgX
O
O
R-Mg-X
R-Mg-X
R'
R'
Any alcohol
will do
here.
OEt
OEt
R'
R
R'
R
R
R
Substitution
EtOH
R'COCl
But where does an ester
come from?
OH
Acid
chloride
R'
SOCl2
R'CO2H
R
R
Perhaps this carboxylic acid
comes from the oxidation of a
primary alcohol or reaction of
a Grignard with CO2.
Addition
Synthetic Planning… Use of epoxides and
carbonyls offer different disconnect sites.
Pattern
1. RMgX
HO-C-C-R
HO
O
2. H2O, HCl
epoxide
R
New bond.
Disconnect site.
O
OH
R-Mg-X
carbonyl
R
New bond.
Disconnect
site
Nucleophile
Pattern
HO-C-R
Want this to be
the nucleophile
(Grignard).
Patterns to recognize: carbonyl vs
oxirane
We can create the following fragments of
target molecules by using an
organometallic (carbon nucleophile)
reaction with a carbonyl
O
RMgX
dil. acid
OH
The difference is the
extra CH2 when using
an oxirane.
R
reaction with a oxirane
O
RMgX
dil. acid
OH
R
Synthetic Planning…
Three different disconnects possible
OH
Give synthetic
routes to
O
+ R3-Mg-X
C
R1
R1
R3
R2
R2
O
+ R2-Mg-X
O
R1
R3
+ R1-Mg-X
R2
R3
If none of the Rs are H then these three synthetic
routes may be available.
Example: Synthesize from ethanol
CH3CH2OH
CH3CH2X
O
OH
Done
OH
2
MgX
CH3CO2Et
O
CH3CH2MgX
CH3CO2H
CH3CH2OH
CH3CH2OH
Preliminary Analysis
•Hmmm, even number of carbons, at least that is good; ethanol is a two
carbon molecule.
•Now the problem is to divide it up into smaller fragments.
•Ether linkage is easily constructed. Williamson.
•Two butyl groups attached to the central 2 carbon fragment. Grignard + ester.
Bisulfite Addition
O
OH
O
S
HO
S
HO
O
O
S
O
O
O
O
O
Addition product.
Practical importance: liquid carbonyl compounds can be difficult to purify. The
bisulfite addition products will be crystalline and may be recrystallized.
Addition of Organolithium
Compounds to Carbonyls
Generally the reactions are the same as for Grignards but the lithium
compounds are more reactive (and more difficult to handle).
O
O
Br
OH
Li
Li
mild acid
bromocyclohexane
Decreased reactivity of
electrophile due to steric
hindrance to attack. So we
used the alkyl lithium instead of
a Grignard.
Nucleophiles derived from terminal alkynes
For example, once
formed, the new
alkynyl alcohol can be
hydrated in two ways,
Markovnikov and anti
Markovnikov.
Carefully observe the
structure of the
products, the
relationship of the OH
and the carbonyl.
Can do all the reactions of an
alkyne and an alcohol. But
remember that we have two
acidic groups: the more acidic
OH and the less acidic terminal
alkyne. We discussed this
problem earlier.
Note that the
regioselectivity
used here is only
effective if this
alkyne is terminal.
Otherwise get a
mixture.
Addition of hydrogen cyanide
basic
Think of what the mechanism should be….
Followed by protonation of the alkoxide ion (perhaps by
unionized HCN).
pH issue.
Slightly basic
media so that
HCN has
partially
ionized to
cyanide ion,
the actual
nucleophile.
Follow-up reactions on the
cyanohydrins…
H2O
R
N
RHC
acid catalysis
dehydration
O
NaCN
dil acid
OH
We saw this hydrogenation before.
OH
H2
R
CH2R
R
NH2
N
CH2R
catalyst
R
CH2R
Hydrolysis. Acid, heat
or
base, heat OH
R
CO2H
CH2R
Let’s see what we can do with the mechanism of the
hydrolysis of the nitrile group to a carboxylic acid.
Overall
aq. acid, heat
+ NH3
R
R
CO2H
N
The action is at the nitrile group, CN --> CO2H.
But how does a nitrile group behave? What could be happening?
We are breaking the CN bond; bond order goes from 3 to 0. Probably stepwise.
Chemically speaking: the nitrogen of the nitrile is basic (lone pair) and can
be protonated. This makes it a better electrophile (cf. carbonyl). Multiple
bond can undergo addition (cf. carbonyl) reducing bond order.
Goal: Break the C to N bonding and create C-O bonds.
Considerations: neither the electrophile (RCN) nor the nucleophile (water) is very
reactive. Since we are in acid protonate the CN group to make it a better electrophile.
Then attack it with the water nucleophile to add water. This results in reduction of C-N
bond order and creation of C to O bonds .
Again, we are in acid environment. Let’s protonate something….
Protonate the multiple bonded N atom to make better electrophile and attack with
the nucleophile, water.
Note the bonding pattern here. We have seen it before.
acid
R
R
N
NH
HO
similar to
NH
HO
CH2
which tauermerizes in acid or base, keto-enol
HO
R
R
CH2
O
R
R
+ H+
+ H+
NH
H2O
NH
HO
NH
O
NH2
HOH
What have done so far? Reduced the CN bond order from 3 to 2 and added one
O to the C. Moving in the right direction!
Want to reduce the CN bond order to zero and introduce more O on the C. Keep
going! To induce the water to attack again (adds another O) need to increase
the reactivity of the electrophile. Protonate again!! On the O.
R
HOH
acid
R
NH2
O
O
R
NH2
OH2
H
Initial equilibrium with acid
NH2
HO
Now want to get rid of the NH2. We have all the O’s we need.
We know what we have to do. Have to get the N protonated to
make it a good leaving group.
OH2
H
R
HO
NH2
reposition the H+
+ NH3
O
R
R
HO
NH3
HO
Done.
O
Wittig Reaction
O
H
I
PPh3
Bu-Li
R
R
or NaH
Substitution Elimination
Example, synthesize
R
R
CHCH3
or combine them the other way…
Wittig Reaction Mechanism
Ph3P:
Acidic
hydrogen
H
H
I
Ph3P
+
Nucleophilic substitution
Nucleophilic
center
Phosphonium ylide
H
strong base, BuLi
Ph3P
Ph3P
R
O
O
R
O
R
R
+
Ph3P
R
Ph3P
Ph3P
R
oxaphosphetane
betaine
R
R
O
R
Ph3P
oxaphosphetane
Ph3PO
R
Friedel Crafts Acylation
R
O
O
R
Cl
AlCl3
And then all the
reactions of ketones…
Formation of Hydrates, carbonyls
and water.
Carbonyl side of equilibrium is usually favored.
Hemiacetals and Acetals, carbonyls and alcohols
Addition
reaction.
(Unstable in
Acid; Unstable
in base)
(Unstable in
Acid; Stable in
base)
Substitution reaction
Formation of Hemiacetals, catalyzed by either
acid or base. Let’s do it in Base first.
But first let’s take stock.
We have an addition reaction.
Just mixing a carbonyl and an
alcohol do not cause a
Use Base to
reaction.
set-up good
necleophile.
One of them must be made a
better reactant.
Carbonyl can be made
into a
Poor nucleophile
better electrophile by
protonating in acid.
Good nucleophile
Alcohol can become a better
nucleophile in base by
ionization.
An addition of the
alcohol to the carbonyl
has taken place. Same
mechanism as
discussed earlier.
hemiacetal
Alternatively, hemiacetal formation
in Acid
Protonation of
carbonyl (making the
oxygen more
electronegative)
Attack of the (poor)
nucleophile on (good)
electrophile.
Deprotonation
Overall, we have added
the alcohol to the
carbonyl.
Hemiacetal to Acetal, Acid Only
Protonate the hemiacetal,
setting up leaving group.
Departure of leaving
group.
Attack of
nucleophile
Deprotonation
Substitution
reaction, cf
S 1.
Equilibria
Generally, the hemiacetals and acetals are only a minor component of an
equilibrium mixture. In order to favor formation of acetals the carbonyl
compound and alcohol is reacted with acid in the absence of water. Dry
HCl) The acetals or hemiacetals maybe converted back to the carbonyl
compound by treatment with water and acid.
An exception is when a cyclic hemiacetal can be formed (5 or 6 membered
rings).
Hemiacetal of D-Glucose
The alcohol
The carbonyl
Try following
the
stereochemistry
here for
yourself
The hemiacetal can form with two different configurations at the carbon of the
carbonyl group. The carbon is called the anomeric carbon and the two
configurations are called the two anomers. The two anomers are
interconverted via the open chain form.
Stabilities of the Anomers…
Here note the alternating up-down relationships.
More stable b form, with the
OH of the anomeric carbon is
equatorial
Less stable a form.
Here see the cis relationship of these two OH groups, one
must be axial.
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