Chemistry 211
2012
Equilibrium Controlled Reactions:
Carbonyl Reactions – 3
Summary of Class Discussion
C. Exploration: Mechanisms of Addition Reactions
1. Most complex reactions occur through a sequence of simple reaction steps as illustrated in Carbonyl-2. Use the proposed analytical process
outlined below to help you write a reasonable reaction mechanism (sequence of simple reaction steps illustrated with curved arrows) for addition
reaction d. reproduced below. These reactions are equilibrium reactions (fast in both the forward and reverse directions). So, it is important that
energies of the highest energy electrons in all intermediate products in your mechanisms be kept as low as possible (no more than ~5 pK units
above the original HEE). One very high energy intermediate will essentially stop the progress of the reaction along that path. One way to
keep the energy low in creating a proposed mechanism is to use the highest energy electrons in each reaction step to form a bond to assist
the reactants in moving to the next intermediate. Recall that forming a bond to the HEE will cause a relatively large decrease in energy.
This energy decrease will help offset any energy increases when other bonds are broken.
O
O
d.
CH3
CH2
+
C
CH2
H
CH2
CH3
H
OH
C
CH2
H
H2O
O
C
CH2
CH3
O
H
CH2
C
CH
H
CH3 CH2
Proposed Analytical Process for Finding a Relatively Low Energy Pathway from Reactants to Products
a. Find Initial HEE: Identify the highest energy e-'s at the beginning of the reaction. (everything inside the blue box above) Don't forget to
consider the solvent and catalyst. (structures over and under the reaction arrows)
Indicate reactant HEE’s on structures above and provide a warrant below indicating how your group chose them.
The reaction conditions are strongly basic with the highest energy e- pair on HO-. Hydroxide ion has its lone pair e-'s on a negative sp3
oxygen. The only other lone pairs of e-'s are on the solvent, H2O, with the lone pair e-'s on a neutral sp3 oxygen and the carbonyl
oxygen, which is a neutral sp2 oxygen. Both the charge and the hybridization of the carbonyl oxygen indicate that its lone pair e-'s will
have lower energy than those on the hydroxide oxygen because of both stronger e--nuclear attraction due to higher s-character
(hybridization effect) and less e--e- repulsion due to less excess e- density (a less negative environment). At the same time the lone pair
e-'s on the neutral oxygen atoms of water will have lower energy than those of the hydroxide oxygen because of less e--e- repulsion due
to less excess e- density (less negative environment). So the order of lone pair e- energies should be HO- highest, then the H2O and the
carbonyl oxygen should be lowest. Therefore, hydroxide ion, holding the highest energy e- pair, should be the most likely species to
react in the first step since its energy will be lowered more than the other lower energy e-'s. It also, sets parameters for HEE through
out the mechanism. pKaH 16 + ~5 = 21 pK units as the upper limit for HEE on any intermediate.
2
Carbonyl Reactions – 3
b. Determine bonds made and broken: Identify the bonds that need to be made and broken to convert the reactants to the products. (Identified
previously for rxn d.)
d.
C-O -bond broken
CH3
CH2
H
C-C -bond made
C
C
O
O
+
CH3
H
CH2
H
H
-
C
C
H
H
OH
H2O
H O
O-H -bond made
CH3
CH2
C-H -bond broken
H
C
O
CH
C
CH2
CH3
CH2
H
The overall reaction requires the breakage of a C-H bond on the -carbon of one butanal (butyraldehyde) molecule, formation of a
-bond between a hydrogen and the carbonyl oxygen of a second butanal molecule, breakage of the carbonyl π-bond and the formation
of a -bond between the -carbon atom of the butanal molecule that lost a hydrogen and the carbonyl carbon of the other butanal
molecule.
c. Use HEE’s to make some of the needed bond changes & draw resulting structures: Start from the HEE and consider how they can be
used to make or break one or two of the bonds (a simple reaction step) that move the reactants toward the products without raising the
energies of the new HEE by more than ~5 pK units above the original HEE. Draw the arrows that indicate the bond changes and then
draw the intermediate structures that result from your arrows for the reaction step. If there seem to be more than one possibility, draw
arrows and structures for all possibilities.
[NOTE: A reaction step may include no more than two different reactant structures. The probability of properly aligning more than two
separate structures is so small that such "ter-reactant" processes are unlikely.]
We should start by assuming that the HEE will have the best effect on the overall reaction if they make a bond in the first step because
they will decrease in energy more than any other electrons in the reaction. With that assumption, there was only one suggestion for the
first step in this reaction: HEE can act as a base by breaking the C-H bond of the -carbon atom of butanal.
Thus, hydroxide can lower the energy of its electrons by forming a bond to
(1a) A carbonyl carbon atom (Act as a nucleophile).
Or
(1b) A proton (Act as a base).
Let's consider the two results!
(1a) Addition to the Carbonyl Carbon of Butanal:
As we discussed in class, this reaction step transfers HEE from O- to O- and should be a reasonable reaction because there is
very little change in the energy of HEE. -hydrogen is the most acidic proton available for the base because the conjugate
base is delocalized onto the adjacent carbonyl oxygen as indicated by the resonance structures below. Removal of any other
H+ would produce a lone pair of electrons isolated on a carbon atom (pKaH ~40).
..O..
C
CH3 CH2 CH2
H
+
-...
...OH
.. ..
..
.. O ..
O
C
CH3 CH2 CH2
H
H
Carbonyl Reactions - 3
3
(1b) Deprotonation of the of the -Carbon of Butanal:
As we discussed in class, the -hydrogen is the most acidic proton available for the base because the conjugate base HEE are
delocalized onto the adjacent carbonyl oxygen as indicated by the resonance structures below (pKaH~ 20 – see acid-Base 8 activity).
Removal of any other H+ would produce a lone pair of electrons isolated on a carbon atom (pKaH ~40).
..O..
C
H
H
CH3 CH2 CH
.. O ..-
..
-
C
..
-...
...OH
H
+ CH3 CH2
..O..
C
H
+
C
CH3 CH2 CH
..
.
H2O.
H
Accomplishments of the steps:
(1a) Breaks the C=O -bond required in the reaction and creates a new C-O -bond.
(1b) Breaks the C-H -bond required in the reaction, creates an O-H -bond and transfers the lone pair of e-’s to the -carbon of
the butanal creating a delocalized anion
Potential problems:
(1a) New C-O bond formed is not present in the products.
(1b) New O-H bond formed is not present in the products.
Potential opportunities:
(1a) We didn’t see any initially.
(1b) The new HEE are on a carbon atom that will need to be connected to the other aldehyde. So we pursued this option.
d. Find new HEE & Compare Energies with Initial HEE: Identify the HEE in your intermediate structures (products of the reaction step). If
you are evaluating more than one possibility, identify the HEE in each set of intermediate structures. Compare the energies of each group of
the HEE to that of the HEE at the beginning of the reaction. Any set of intermediates is acceptable if the HEE are no more than ~5 pK
units higher in energy than those at the beginning of the reaction. If there is more than one acceptable intermediate, choose the one with the
lowest energy HEE first. Provide a warrant for your choice of the first reaction step.
ANALYSIS OF THE REACTION STEP (1b)
Overall Change in Structure:
-In reaction step (1) an H-O -bond is formed while a C-H -bond is broken and the lone pair of e-’s is moved from a negative sp3
oxygen to a delocalized carbanion.
Change in Energy Due to Structural Changes:
-The net change in energy of the highest energy pair of e-’s is unfavorable by ~ 4 pK units (-OH pKaH ~16 to -carbanion
pKaH ~20) within the limits set by the original HEE.
4
Carbonyl Reactions – 3
e. Use HEE of your first intermediate to make some additional needed bond changes & draw resulting structures for the second reaction
step: Identify the bond changes that remain to be made to get to the products. Start from the HEE in the first intermediate and consider
how they can be used to make or break one or two of the bonds that move the intermediate toward the products. Draw the arrows that
indicate the bond changes and then draw the next intermediate structures that result from your arrows below. As before, if there seem to be
more than one possibility, draw arrows and structures for all possibilities. (Same process as in c.) Provide a warrant for your choice for the
second reaction step.
Most likely Course of the Reaction:
The HEE are on the -carbon of the “enolate ion” of butanal, which was formed in step (1b) Thus, this ion is likely to be involved in
the next step.
Bonds yet to be made or broken:
C-C -bond made
d.
C-O -bond broken
CH2
O
+
C
CH3
C
H
-
O
H
CH2
OH
O-H -bond made
C
CH3
C
H
H
H
H2O
CH3
C
H
CH
CH2
C-H -bond broken
O
H
C
CH2
H
H
O
CH3
CH2
It appeared that the most valuable action new HEE could take would be to make the C-C bond between it and the carbonyl
carbon of the other butanal molecule.
Step (2) Addition of -anion formed in step (1b) to the carbonyl carbon on the second aldehyde molecule.
C
C
C
H + CH3 CH2 CH2
H
CH3 CH2 CH2
+
O
H
H
CH3
H
CH
..
..
..
CH3 CH2 CH
. . .. ..O
..O..
..O..
..O..
C
H
H
..
.
O.
+
H
CH2
Overall Change in Structure for step (2):
-- Accomplishes the breakage of the C=O π-bond required in the reaction.
-- Creates the new C-C -bond required in the reaction and transfers the lone pair of e-’s to a negative sp3 oxygen.
f. Find the new HEE in intermediate 2 & compare their energy with that of the Initial HEE: This is the same process as in d. Be sure to
always compare new HEE to the original HEE identified in a.
Change in Energy Due to Structural Changes in Step (2):
-- This step is a favorable process. The HEE are moved from a higher energy delocalized position to a lower energy negative sp3
oxygen position. (-carbanion pKaH ~20 to -O pKaH ~16 essentially the reverse of the energy change is step (1b))
-- The lowering of the energy of the lone pair e-’s and changes in bonding are both favorable so step (2) is a very reasonable
energy change.
Carbonyl Reactions - 3
5
g. Continue repeating steps e. & f. until you reach the product structures.
Use New HEE to make some additional needed bond changes & draw resulting structures: (Same process as in c & e.)
After step (2), the HEE are on the negative sp3 oxygen created by the addition to the carbonyl carbon. Thus, this oxygen is likely to be
involved in the next reaction step.
After step (2) there is only one bond change left to complete: protonating the negative oxygen formed in step (2) creating the needed
O-H bond and reforming the hydroxide ion catalyst consumed in step (1b). So, the formation of products and regeneration the HO
catalyst is accomplished by the proton transfer indicated in step (3.).
Step (3)
Protonation of the Addition Intermediate formed in step (2) to Produce the Final Product and Regenerate the Base
Catalyst.
. . ..
- ..O
C
CH3
CH2
H
CH2
+
CH2
H
..
..O
H
H
C
H
C
CH
CH3
..O..
..
..O
CH3
CH2
H
CH2
C
CH
CH3
..O..
H
-..
+ ..OH
..
CH2
Overall Change in Structure:
-- Creates the new O-H -bond required in the reaction and transfers the highest energy pair of e-’s to another negative sp3
oxygen regenerating the HO
catalyst consumed in step (1.)
Change in Energy Due to Structural Changes:
-- This reaction produces little or no change in free energy since the highest energy pair of e-’s is moved from one negative sp3 oxygen
to another negative sp3 oxygen and there is no net change in the types of bonds.
6
Carbonyl Reactions – 3
Final Mechanism
Steps (1b), (2) & (3)
..O..
C
C
CH3 CH2 CH2
HEE
C
C
H
H
O HEE
..
+ CH3
+
H
..
..
-...
. ..OH
H
CH2
CH3 CH2 CH2
H
..O..
+
H
CH3 CH2 CH
H
..O..
C
H
Initial HEE pKaH 16
New HEE pKaH 20
All HEE within < 5 pK unit of the initial pKaH.
..
..
..O..
.. . HEE ..O .
C
CH3 CH2 CH2
CH3
O
H
CH
..O..H
C
+
H
..
..O
C
CH3 CH2 CH2
H
CH2
Next HEE pKaH 16
H
CH3
H
CH
..O..
C
H
CH2 + -....
OH
HEE ..
HEE return to catalyst
This is a well-known reaction called the ALDOL CONDENSATION because of the aldehyde
(ALD) and alcohol (OL) functional groups in the product.
(See CGW p. 615)