Chemistry 212
Spring 2013
Equilibrium Controlled Reactions:
Carbonyl Reactions – 7
Out of Class Applications
Analysis
1. Reading: Read CGWW Chapter 14, pp. 348-354.
2. Try to apply your mechanism(s) for the first two substitution for carbonyl oxygen reactions to these other examples of substitutions for carbonyl
oxygen from Carbonyl-3.
Break C=O,
both - & š -bonds.
1 O
C
2
o.
3
CH3
-> WHAT OVERALL CHANGE OCCURS
(Bonds made and broken)? Illustrate on
the given structures.
The overall reaction requires the breaking 2
N4-H -bonds of the hydroxylamine molecule
and the C2-O1 - & -bonds and making the
C2-N4 - & -bonds and 2 O1-H -bonds.
This reaction has an overall change very
similar to reaction j.
+
NH2OH
4 5
HEE
Break 2 N-H -bond's and
form C=N - & š -bonds.
H2O
H
H
2
C
H
N
6 O H
7
3 CH3
N
4
5 OH
+ H2O
Form
2 O-H -bonds.
1
-> WHERE ARE THE HIGHEST ENERGY ELECTRONS? Circle on original structures.
Potential choices for highest energy e-‘s are:
1. The lone pair e-‘s on the neutral sp2 C=O O1 atom.
2. The lone pair e-‘s on the neutral sp3 O atom of water.
3. The lone pair e-‘s on the neutral sp3 N4 atom of the NH2 group of hydroxylamine.
In this case N4 lone pair e-‘s are HEE. (pKaH ~ 10). The O e-‘s have stronger attractive forces
due to the higher effective nuclear charge of O vs. N.
-> REACTION CONDITIONS
NOTE that this reaction is being run with hydroxylamine in the presence of a small amount of its conjugate acid, hydroxylammonium
ion. Thus both are available to support the reaction.
Carbonyl-7-OCA-Analysis
2
-> HOW CAN THE HIGHEST ENERGY ELECTRONS BE USED IN A SIMPLE REACTION STEP TO BEGIN THE CONVERSION
OF THE REACTANTS TOWARD THE PRODUCTS? Propose a reaction step with original structures using arrows. Write
intermediate product structures below:
If we start with the site of the HEE (N4) and consider what changes
must occur, we find that it must lose 2 H’s and form 2 bonds to the
C2 of the other reactant (ketone). Since HEE’s benefit (lower their
energies) from bond formation, it seems that the most productive
first step should be formation of an N4-C2 bond. We have seen this
initial step in reaction j and the “Uncatalyzed Addition Mechanism”
–(Carbonyl-2 summary).
1 O
C
2
3 + NH OH
2
CH3
4 5
HEE
HEE
H
N O
C
4 5 H
2
H
3 CH3
Step A
H2O
1 O
H
H
N
H 6 O H
7
-> NOW WHERE ARE THE HIGHEST ENERGY ELECTRONS? Circle on your new structures.
As the arrows clearly indicate, this first step moves HEE’s from N4 to O1 a negative O (pKaH~16). This increases energy by ~6 pK units
reasonable for an equilibrium reaction step.
-> HOW CAN THESE HIGHEST ENERGY ELECTRONS BE USED IN A SIMPLE REACTION STEP TO MAKE AND BREAK
BONDS THAT WILL MOVE THE INTERMEDIATE STRUCTURES TOWARD THE PRODUCTS?
4-Centered
To complete the addition reaction we need to transfer a proton
Transition
H
HEE
H

from the positively charged N4 to the negatively charged O1.
State
1 O
1 O
However as we saw with reaction j., this transfer cannot be

N O
N O
accomplished in one step since it would require a 4-centered
C
4 5 H
C
4 5 H
reaction, atoms 1, 2, 4 and the H attached to 4.
2
2
H
H
So we must look for a 2-step route to transfer this proton.
3 CH3
3 CH3
In this case the HEE are on the O1 of the intermediate and, (as
discussed above) O1 must ultimately make 2 bonds to protons
HEE
to form water in the product. So we need to look for a
1
reasonable source of protons, the most acidic proton. Since the
O
H
6
hydroxylammonium ion was in the reaction mixture, its
N
H
positive nitrogen should hold the most acidic proton. This
+
protonation should yield a decrease in energy since the change HO 7 H +
in e- environments is essentially the reverse of that in the step
A.
H
H
C
2
N + 5
OH
4
CH3
3
H
HEE
Step B
H
H
N
HO 7
H
6
+
4
N
+
O
1 C
2 CH
3
3
H
5
OH
3
Carbonyl-7-OCA-Analysis
Once the negative oxygen is protonated (step B), moving the
HEE are again on the phenylhydrazine nitrogen. The most
productive bond this nitrogen can make (HEE make bonds) is HEE
H
a bond to a proton regenerating the catalyst. The proton that
H
N 6
must be removed to move the intermediate toward products is
that on N4 and it is the most acidic proton in the reaction at
HO 7
this point (attached to a positive N). In step C the HEE are
moved from one neutral nitrogen to another, so there should
be little energy change.
We have now completed and addition reaction as we did in
reactions m. & j. of the carbonyl-3 activity and HEE are on N4.
The HEE provided the opportunity to eliminate the original
carbonyl oxygen. As we saw in reaction j. (See Summary of
Class Discussion), since the elimination step is unimolecular, it
is possible for an acid catalyst to participate in the reaction.
This general acid catalysis allows the transfers of the HEE
from one neutral nitrogen to another (See step D). General
acid catalysis is a process in which the proton is added while
the OH is leaving.
All that is left to be done is breaking of the N4-H bond and
regenerating the acid catalyst of the elimination,
phenylhydrazinium ion. Since HEE are on N6, it is set up to
break the N4-H bond and complete the reaction, Step E.
H
H
O
1
+
O1
H
+
HEE 6 H
HO 7
H
C
2 CH
3
3
+
4
N
HEE
Step C
6H
N H
HO 7
+
H
H
5
OH
C
2 CH
3
3
H
N
5
OH
H
O
1
+
4
N
H
5
OH
C
2 CH
3
3
ADDITION INTERMEDIATE
HEE
6H
N H
HO 7
+
H
H
4
N
+
5
4 OH
N +
+
C
2 CH
3
3
H
Step D
HEE 6 H
N
HO 7
Step E
H
O1
H
+
H
6H
N H
HO 7
+
H
+
5
4 OH
N +
+
C
2 CH
3
3
4
N
H
O1
H
5
OH
+
C
2 CH
3
3
H
O1
H
Again we see that Reaction o. follows the Addition-Elimination path. With the initial HEE on one of the reactants and having energy comparable
to O-, the addition is uncatalyzed (the original HEE add directly to the carbonyl carbon –step A-without an initial proton transfer), while the
elimination (step D) is general acid catalyzed.
Carbonyl-7-OCA-Analysis
4
H
1
Step A
O
HEE 1
H2O
C
2
3
CH3
+
NH2OH H
H
4 5
N
HEE
H 6 O H
7
H
O
6H
N H
+
HO 7
+
H
C
2
H
HEE
N + 5
OH
4
H
Step B
H
N
H
6
+
HO 7
CH3
3
4
N
+
H
5
OH
O
1 C
2 CH
3
3
Step C
Addition
Elimination
6H
N H +
HO 7
+
H
4
N
H
5
OH
+
H
C
2 CH
3
3
Step E
O1
H
HEE 6 H
N
HO 7
H
+
5
4 OH
N +
+ H
C
2 CH
3
3
HEE
Step D
O1
H
HEE 6 H
N
HO 7
H +
+
H
General
Acid
Catalysis
H
O1
4
N
C
2 CH
3
3
H
5
OH
5
Carbonyl-7-OCA-Analysis
Breakage of C2=O1,
both - & š -bonds.
n.
O 1
C 2 +
H
Breakage of 2 C4-H -bonds and
formation of C2=C4 - & š -bonds.
O 8
4
C
3
H
5
9 OH
6
7
H2O
O 8
HEE
4

5
2
H
Formation of
2 O1-H -bonds.
6
7 + H2O 1
3
->
WHAT OVERALL CHANGE OCCURS (Bonds made and
broken)? Illustrate on the given structures.
See above. This reaction has an overall change very similar to
reaction m.
->
WHERE ARE THE HIGHEST ENERGY ELECTRONS?
Circle on original structures. sp3 negative O vs. neutral O’s. O9 has
HEE due to increase e—e- repulsion of the negative ion. With HEE
on an O- of a catalyst, we might expect a base catalyzed reaction.
-> HOW CAN THE HIGHEST ENERGY ELECTRONS BE USED IN A SIMPLE REACTION STEP TO BEGIN THE CONVERSION
OF THE REACTANTS TOWARD THE PRODUCTS?
O 8
O 8
Since the HEE will benefit the reaction most and are on an atom (O9)
Step A
that does not become incorporated into the product, its most
H
H
+ H2O 9
6 7
6
productive action would be to remove a proton from a reactant. As
7
5
5
9 OH + H
C4
C4
in Reaction m., the proton that needs to be removed (on C4) is the
HEE
HEE
most acidic proton (pKaH ~ 20 – delocalized conjugate base).
3
3
-> NOW WHERE ARE THE HIGHEST ENERGY ELECTRONS? Circle on your new structures. C4 – see above.
-> HOW CAN THE HIGHEST ENERGY ELECTRONS BE USED IN A SIMPLE REACTION STEP TO BEGIN THE CONVERSION
OF THE REACTANTS TOWARD THE PRODUCTS?
Now the HEE are on C4, which must form 2 bonds
to C2. C2 is a carbonyl carbon atom and can accept
a new bond, so the most productive action for HEE
here (Step B) is to form a C4-C2 bond. This step
moves HEE from delocalized carbanion to the
negative sp3 O1. A favorable change.
O1 must form 2 bonds to protons and break a bond
to C2. Again HEE contribute most by making
bonds, so it most productive possibility is to remove
a proton from water, moving the HEE to O9
hydroxide ion. Step C completes a base catalyzed
addition to the aldehyde carbonyl group. This
mechanism is essentially identical to that of reaction
m. (See Summary of Cass Discussion.)
O 8
H
6
C4 5
HEE
3
HEE
H
+ H2 O 9
7
O 1 O 8
C 4
6
5
2 C
H
3
H
Step B
1O
C
H
2
Step C
7
H
O9
H
HEE
O 1 O 8
C 4
5 6
2 C
H
3
7
H1O H O 8
HEE
4
C
5 6 7 +
2 C
O9
H
H
3
Addition Intermediate
Carbonyl-7-OCA-Analysis
To complete the reaction we need to break the
remaining C4-H and the C2-O1 -bonds and make a
-bond between C2 and C4. The HEE on O9 can
best support these changes by removing the proton
from C4 yielding a delocalized carbanion on C4. As
with step A, this yields an energy increase of about 4
pK units – OK.
The resonance structures to the right illustrate the
delocalization that contributes to lowering the
electron energies of the Step D product.
After Step D the C2-O1 -bonds must be broken
a -bond between C2 and C4 needs to be made.
Once again, the HEE lead us to make the C2-C4
-bond and eliminating the C2-O1 bond. Step C
completes the reaction and regenerates the
hydroxide ion catalyst.
and
6
H O1
O 8
H
C 4
2 C
C
5
6
4
7 +
3 H
H O1
H
C 4
2 C
H O
1
H
C 4
2 C
H O1
C 6
5
3 HEE
8 O
3 HEE
Step D
OH
9
HEE
8 O
C
5
O 8
C 6
5
3 HEE
H O1 H
C 4
2 C
6
H
C 4
2 C
7
Step E
C 4
2 C
C
5
C
5
+ H2O 9
HEE
6
7
3
8 O
H
7
8 O
7
6
+
7
OH 1
HEE
3
Once again the substitution for carbonyl reaction operates by an Addition-Elimination mechanism. In fact this mechanism involves the
same sequence of steps as seen in reaction m. (See Summary of Class Discussion.)
7
Carbonyl-7-OCA-Analysis
O
c. CH3
1
2
C
CH2
3
5
HEE
+
4
CH3
6
H2 O
NH2OH
6
7
CH3
NH3OH
8
1
2
CH2
N
C
3
O 7
H
4
+
H2O 5
CH3
9
This reaction is essentially identical to Reactions j (See Summary of Class Discussion) & o. (See above).
-> WHAT OVERALL CHANGE
OCCURS (Bonds made and broken)?
Illustrate on the given structures.
The overall reaction requires the breaking
2 N6-H -bonds of the hydroxylamine
molecule and the C3-O5 - & -bonds and
making the C3-N6 - & -bonds and 2
O5-H -bonds.
This reaction has an overall change very
similar to reaction j. and o.
->
WHERE ARE THE HIGHEST ENERGY ELECTRONS? Circle on original structures.
Potential choices for highest energy e-‘s are:
1. The lone pair e-‘s on the neutral sp2 C=O O5 atom.
2. The lone pair e-‘s on the neutral sp3 O atom of water.
3. The lone pair e-‘s on the neutral sp3 N6 atom of the NH2 group of hydroxylamine.
In this case N6 lone pair e-‘s are HEE. (pKaH ~ 10). The O e-‘s have stronger attractive forces
due to the higher effective nuclear charge of O vs. N.
-> REACTION CONDITIONS
NOTE that, as with reaction o., this reaction is being run with hydroxylamine in the presence of a small amount of its conjugate acid,
hydroxylammonium ion. Thus both are available to support the reaction.
-> HOW CAN THE HIGHEST ENERGY ELECTRONS BE USED IN A SIMPLE REACTION STEP TO BEGIN THE CONVERSION
OF THE REACTANTS TOWARD THE PRODUCTS? Propose a reaction step with original structures using arrows. Write
intermediate product structures below:
HEE
Reaction conditions favor uncatalyzed addition: If we start with the
Step
A
H
5 O
5 O
site of the HEE (N6) and consider what changes must occur, we find
H2O
N
O
that it must lose 2 H’s and form 2 bonds to the C3 of the other
C
C
6
H
2
reactant (ketone). Since HEE benefit (lower their energies) from
7
+ NH2OH
2
1
1
3
H
H
3 4
H
bond formation, it seems that the most productive first step should
4
6 7
9
N
be formation of an N6-C3 bond. We have seen this initial step in
HEE
H 8 O H
reactions j. and o. and the “Uncatalyzed Addition Mechanism” –
(Carbonyl-2 summary).
-> NOW WHERE ARE THE HIGHEST ENERGY ELECTRONS? Circle on your new structures.
This first step moves HEE’s to O5 a negative O (pKaH~16). This increases energy by ~6 pK units reasonable for an equilibrium reaction step.
Carbonyl-7-OCA-Analysis
8
 HOW CAN THESE HIGHEST ENERGY ELECTRONS BE USED IN A SIMPLE REACTION STEP TO MAKE AND BREAK
BONDS THAT WILL MOVE THE INTERMEDIATE STRUCTURES TOWARD THE PRODUCTS?
The HEE are on the O5 of the intermediate and, (as
discussed above) O5 must ultimately make 2 bonds to
protons to form water in the product. So we need to look
for a reasonable source of protons, the most acidic proton.
Since the hydroxylammonium ion was in the reaction
mixture, its positive nitrogen should hold the most acidic
proton. This protonation should yield a decrease in energy
since the change in e- environments is essentially the
reverse of that in the step A.
Once the negative oxygen is protonated (step B), moving
the HEE are on the phenylhydrazine nitrogen. The most
productive bond this nitrogen can make (HEE make
bonds) is a bond to a proton regenerating the catalyst. The
proton that must be removed to move the intermediate
toward products is that on N6 and it is the most acidic
proton in the reaction at this point. In step C the HEE are
moved from one neutral nitrogen to another, so there
should be little energy change.
We have now completed and addition reaction as we did in
reaction j. & o. before.
The HEE provided the opportunity to eliminate the
original carbonyl oxygen. As we have seen before
(reactions j. & o.), since the elimination step is
unimolecular, it is possible for an acid catalyst to
participate in the reaction. This general acid catalysis
transfers the HEE from one neutral nitrogen to another.
General acid catalysis is a process in which the proton is
added while the OH is leaving.
All that is left to be done is breaking of the N6-H bond and
regenerating the acid catalyst of the elimination,
phenylhydrazinium ion. Since HEE are on N8, it is set up
to break the N6-H bond and complete the reaction, Step E.
HEE
Step B
5
N
C
2
1
H
O
H
5
1
N
C
O
+
6 7 H
4 H
3
H
H
H
N
9
C
2
1
N
6 7 H
4 H
3
8 O H
N
H
9
8 O H
N
C
2
1
9
8 O H
N
H
+
O
6 7 H
4 H
3
H
+
HEE
5 O
H2O
H
HEE
O
Step C
HEE
H
O
2
H
H
O
O
6 7 H
4 H
3
H
5
H2O
H
H
H
9
8 O H
N
ADDITION INTERMEDIATE
H
O
H
HEE
5 O
1
2
N
C
O
1
2
H
+
O
6 7 H
4 H
3
Step D
H
6
4
9
1
6
C
+
H
N
9 + 5 O
O
8 O H
H
H2O
1
2
8 O H
H
Step E
9 + 5 O
N
HEE
H
H
+
H
3
H
H
7
N
2
8 O H
7
H
3
N
H2O
4
N
C
H
6
H
H
+
H
N
H
3
4
H
7
N
C
H
9 + 5 O
8 O H
H
H
9
Carbonyl-7-OCA-Analysis
Again we see that Reaction c. follows the Addition-Elimination path. In this case the addition is uncatalyzed (the original HEE’s add directly to
the carbonyl carbon –step A-without an initial proton transfer), while the elimination (step E) is general acid catalyzed.
Step A
Step B
H
5 O
HEE
H
H
H2O
HEE
5 O
H2 O
5 O
H
C
N
O
N
O
2
NH
OH
+
4
+
2
1
N
9
3
H
H
H
H
6 7 H
2 C
6 7 H H 8 O H
2 C
6 7
1
1
3
3
N
9
N
9
4 H
HEE
4 H
H 8 O H
H 8 O H
Step C
Addition
H2O
Elimination
O
1
2
N
6
C
H
7
H
+
+
Step E
H
9
N
H
3
4
O
8 O H
H2O
1
2
H
7
HEE H
N
C
3
6
H
+
H
N
9
8 O H
H
H
4 +
5 O
H
H
5 O
Step D
H2O
General
Acid 5 O
Catalysis
2 C
1
3
H
HEE
N
O
6 7 H
4 H
+
H
H
H
9
8 O H
N