Transcript Slide 1
Migratory Insertion & Elimination Rxns
A migratory insertion reaction is when a cisoidal anionic and neutral ligand on
a metal complex couple together to generate a new coordinated anionic ligand.
This new anionic ligand is composed of the original neutral and anionic ligands
now bonded to one another.There is NO change in the oxidation state or d
electron-count of the metal center.
migratory insertion
O
O
C
OC
Mn
OC
C
O
OC
CO
Mn(+1)
18eelimination
O
CH3
Mn
CO
+L
OC
CH3
Mn
OC
OC
CH3
ligand addition
acyl
C
O
C
O
Mn(+1)
16e-
Mn(+1)
18e-
CO
L
General Features of Migratory Insertions:
1) No change in formal oxidation state (exception: alkylidenes)
2) The two groups that react must be cisoidal to one another
3) A vacant coordination site is generated by the migratory insertion. Therefore, a
vacant site is required for the back elimination reaction (e.g., b-hydride elimination).
A trapping ligand is often needed to coordinate to the empty site formed from a
migratory insertion in order to stop the back elimination reaction.
4) Migratory insertions are usually favored on more electron-deficient metal centers.
The following are common anionic and neutral ligands that can do migratory
insertion reactions with one another:
Anionic: H-, R- (alkyl), Ar- (aryl), acyl-, O2- (oxo)
Neutral: CO, alkenes, alkynes, carbenes
CO and alkyl migratory insertions (as shown on previous slide) are extremely
important and are often generically referred to as carbonylation reactions.
Hydride and CO migratory insertions to produce formyl groups are not common
due to the thermodynamic instability of the formyl-metal interaction.
Some Electronic effects
R
R
Z
OC
CO
Fe
C
O
+L
CO
OC
O
Fe
CO
L
THF
C
O
best Lewis acid - can coordinate to electron-rich
CO ligands and drain off some e- density
+
Z
+
= Li
> Na
+
+
> (Ph3)2N
strongest coordinating ligand - best trapping ligand
L = PMe3 > PPhMe2 > PPh2Me > CO
most electron-rich alkyl group makes the best nucleophile for
migrating to the electron-deficient CO
R = n-alkyl- > PhCH2-
Z
Migration vs. Insertion
Migration
O
O
C
OC
CO
Mn
OC
CH3
C
O
Mn(+1)
18e-
OC
CH3
Mn
OC
C
O
Mn(+1)
16e-
CO
a MIGRATION rxn involves the
anionic ligand doing a
nucleophillic-like attack on the
neutral ligand. This involves the
anionic ligand moving to the site
where the neutral ligand is
coordinated. An empty
coordination site is left behind.
Insertion
O
C
OC
Mn
OC
C
O
Mn(+1)
18e-
CO
CH3
an INSERTION rxn involves the
neutral ligand moving over to
OC
CO
Mn
O where the anionic ligand is
coordinated and "inserting" into
OC
the anionic ligand-metal bond to
C
CH3 generate the new anionic
O
ligand. An empty coordination
site is left behind from where
Mn(+1)
the neutral ligand originally was
16elocated.
While most systems studied have been shown to do migrations, both are possible.
The following is a system where both are very similar in energy and the solvent
used favors one or the other.
inversion
Et migrates
CH3NO2
Ph3P
Fe*
Ph3P
Ph3P
Et
Fe*
*CO
Fe*
Et
O
C*O
O
Et
C
O
HMPA
CO inserts
Fe*
Ph3P
Et
*CO
O
Fe*
Ph3P
*C
O Et
retention
O
Alkene Migratory Insertions
Zr
CH3
Zr
CH3
Zr
CH3
Zr
CH3
Zr
CH3
Alkene Migratory Insertion – b-Hydride Elimination
migratory insertion
H
H
H
H
M
H
H
H
‡
H
H
H
H
H
H
M
H
H
M
b-hydride elimination
*
*
Nb
H
*
CO
R
Nb
CO
Nb
R
H
*
*
*
H
R
N MR ir r adiat ion of t he Nb-hy dr ide r esonance af f ects the NMR r esonance f or t he alk yl hydr ide,
demonst rating t hat t hey ar e connect ed by t he migr at or y inser tion mechanism
Problem: Why don’t either of the complexes shown below do alkene-hydride
migratory insertions at room temperature?
H
Cl
Ir
Ph3P
PPh3
CO
Et3P
Pt
H
PEt3
Problem: Sketch out and label the two mechanistic steps (in the correct
order) that are occurring for the following reaction.
Ru
H
C
O
Ru
+ PPh3
Ph3P
C
O
CH3
RCN
Rh
Me3P
CH3
PMe3
RCN
Rh
H
Me3P
PMe3
RCN
Rh
H
PMe3
Me3P
CH3
O
O
O
Ph3P
-PPh3
O
Pd
Cl
O
Ph3P
Ph3P
Pd
Pd
PPh3
Me
Ph3P
O
Cl
Cl
O
Me
O
+PPh3
Pd
Cl
PPh3
Agostic C-H to Metal Interactions – “Frozen Migratory Insertion”
*
Co
R3P
*
*
+ H+
Co
Co
H
R3P
H
H
H
R3P
One of the C-H bonds of the methyl group is within bonding distance to the Co center.
This is called an Agostic C-H bond interaction.
Because the C-H bond is sharing some of its s-bond electron density with the
metal, the C-H bond is weakened. This produces some relatively clear-cut
spectroscopic characteristics:
1) nC-H infrared stretching frequency is lowered to the mid-2500 cm-1 region from a normal
value of 2900-3000 cm-1
2) the JC-H coupling constant in the 13C NMR is lowered to around 70-90 Hz from a normal
value of 150 Hz.
3) the 1H chemical shift of the agostic proton is in the -10 to -15 ppm region, much like a
metal-hydride resonance.
Carbene-Alkylidene Migratory Insertions
X
M
CH2 + L
L
X
M
CH2
X = H-, R-, OR-, halideNormally a migratory insertion refers to a neutral ligand reacting with an anionic
ligand to produce a new anionic ligand. But if we electron-count the carbene as a
dianionic ligand (alkylidene), we are reacting a monoanionic ligand (X) with a
dianionic ligand (alkylidene) to make a new monoanionic ligand. This changes the
oxidation state of the metal center and is now formally a reductive coupling
reaction.
In the case of X = H-, the reverse reaction is called an a-hydride abstraction or
elimination.
Ph3C+ = good hydride abstracting reagent
*
*
*
+ Ph3C+
Re
H3C
CH3
Ph3P
Re
-
-H
H3C
Re
CH2
Ph3P
Ph3P
*
Re
H
Ph3P
W
Ph
CH3
Ph3C
W
-H
Ph
CH2
NCMe
W
NCMe
CH2
Ph
Eliminations
H
H
H
b-hydride elimination
H
M
M
H
H
H
M
R
R
O
M
a-hydride elimination
M
R
M
CO
carbonyl elimination
or decarbonylation
R
The key points are:
1) No change in formal oxidation state (exception: alkylidenes)
2) You must have an empty orbital that is cisoidal to the group that you are doing an elimination
reaction on. Alternatively, a cisoidal labile ligand that can easily dissociate to open up an
empty orbital.
migratory insertion
H
H
H
M
H
H
H
CH3
-
‡
H
H
H
M
H
H
H
CH3
CH3
M
But the reverse methyl elimination rxn is very difficult:
H
H
H
methyl elimination
H
H
H
CH3
M
M
rotation of C-C
bond to move CH3
group away from
metal to avoid steric
effects
H
H
CH3
H
H
H
H
‡
H
H
H
H
CH3
M
H
H
CH3
H
M
M
b-hydride elimination
H
CH3
Very
Difficult!
One unusual example of what is believed to be a methyl elimination reaction is
involved in the following transformation (Bergman, JACS, 2002, 124, 4192-4193):
*
*
Rh
Me3P
H3C
N
+ HSiPh3
C
CH3
Rh
Me3P
H3C
C
+ CH4
N
SiPh3
*
*
- CH4
+ HSiPh3
+ NCCH3
Rh
Me3P
N
H3C
*
- NCCH3
Me3P
C
Rh
Me3P
H
H 3C
CH3
Rh
SiPh3
Rh(+5)
Rh
methyl
elimination
H3C
ligand
dissociation
Rh
C
N
Me3P
SiPh3
N
C
H3C
SiPh3
CH3
N
C
C
*
C
Rh
CH3
H3C
Rh
Me3P
SiPh3
N
Note that the isocyanide ligand is
more strongly donating and a better
-backbonding ligand than the starting
acetonitrile. This keeps this from being
a catalytic reaction.
*
N
Ph3Si
*
Rh
Me3P
C
N
Ph3Si
CH3
SiPh3
One reason that the methyl elimination reaction occurs here is that the b-hydride
elimination reaction generates a high energy ketene-imine:
*
*
N SiPh3
Rh
Me3P
N
C
H3C
SiPh3
Rh
Me3P
H
H2C
C
Problem: Identify each step in the following mechanism. Some steps may
have several things occurring.
*
1
*
Co
2
*
Co
H3C
Co
H
H3C
H3C
H 3C
*
+ CH4 +
Co
Me3P
CH2=CH2
PMe3
3
Problem: Sketch out a detailed mechanism and label each step for the
following overall reaction.
Ph3P
+ 2CO
Rh
H3 C
CH3
OC
Rh
PPh3
+
O
H 3C
CH3