Oxidative Addition/Reductive Elimination 1

Oxidative Addition
An oxidative addition reaction is one in which (usually) a neutral ligand
adds to a metal center and in doing so oxidizes the metal, typically by
2e-. The transferring of the two electrons from the metal to the
incoming ligand breaks a bond in that ligand forming two new anionic
ligands. At least one of these new anionic ligands ends up bonded to the
metal center.
two new anionic
hydride ligands

H H H
OC PPh3 +H2 OC PPh3 OC PPh3
Ir Ir Ir
Ph3P Cl Ph3P Cl Ph3P H
Cl
Ir(+1) Ir(+1) Ir(+3)
16e- 18e- 18e-

There are three main classes of molecules (substrates) that can perform
oxidative additions to metal centers: non-electrophillic, electrophillic,
and “intact.” Collman, Hegedus, Norton & Finke classify these as
Class A, B, and C substrates.
Non-electrophillic (Class A): these molecules do NOT contain
electronegative atoms and/or are not good oxidizing agents. Aside from
H2, they are often considered to be “non-reactive” substrates. These
molecules generally require the presence of an empty orbital on the
metal center in order for them to pre-coordinate prior to being activated
for the oxidative addition rxn.
H2, C-H bonds, Si-H bonds, S-H bonds,
B-H bonds, N-H bonds, S-S bonds, C-C bonds, etc.
H2 is by far the most important for catalytic applications, followed by
Si-H bonds, B-H, N-H, and S-H bonds. C-H bond activation and
functionalization is very important, but still not practical.
 Oxidative Addition/Reductive Elimination 2

Electrophillic (Class B): these molecules do contain electronegative

atoms and are good oxidizing agents. They are often considered to be
“reactive” substrates. These molecules do NOT require the presence of
an empty orbital (18e- is OK) on the metal center in order to perform
the oxidative addition rxn.
X2 (X = Cl, Br, I), R-X, Ar-X, H-X, O2, etc.
The most common substrates used here are R-X (alkyl halides), Ar-X
(aryl halides), and H-X. An example of the oxidative addition of
CH3Br to IrCl(CO)(PPh3)2 is shown below. Note that the starting metal
complex in this case is 16e-:
SN2 nucleophillic attack
δ+ δ− H
Ph3P CO
+CH3Br
OC PPh3 C Br
Ir Ir
Ph3P Cl H
Cl PPh3 H

Ir(+1) Ir(+1)
16e- 16e-
two new anionic
two new anionic ligands
ligands
CH3 CH3
OC PPh3 +Br− OC PPh3
+ Br−
Ir Ir
Ph3P Cl Ph3P Cl
Br
Ir(+3) Ir(+3)
18e- 16e-

Note that the H3C-Br bond is broken on the oxidative addition reaction
generating two new anionic ligands: CH3− and Br−. If the starting
metal complex is 16e- (as shown above) both ligands will usually end up
coordinated to the metal to make an 18e- complex.
 Oxidative Addition/Reductive Elimination 3

In the case of a starting 18e- complex (shown below) only one of the
two anionic ligands (usually the strongest binding) generated from the
oxidative addition will end up coordinated to the metal unless a separate
substitution reaction occurs.
oxidative
CO H H OC CO
OC addition
+ Cl−
OC Re + Cl OC Re
OC CO
OC CO
two new anionic
ligands
Re(-1) -CO Re(+1)
18e- 18e-
OC CO
1) CO ligand dissociation
OC Re 2) η1- to η3-allyl hapticity change

OC

Re(+1)
18e-

In this case the alkyl anion is the best donor ligand and easily “beats
out” the more electronegative and poorly donating Cl− anion. Note that
the alkyl ligand (-CH2CH=CH2) initially coordinated to the Re after the
oxidative addition is an η1-allyl ligand and that it can convert to the
generally more stable η3-allyl on CO ligand dissociation.
“Intact” (Class C): these molecules may or may not contain
electronegative atoms, but they do need to have a double or triple bond
present. Unlike most of the other substrate molecules that break a single
bond and form two separate anionic ligands upon the oxidative addition,
these ligands have double or triple bonds and only one of the π-bonds is
broken leaving the σ-bond intact. The ligand does pick up two electrons
from the metal and becomes a dianionic ligand.
 Oxidative Addition/Reductive Elimination 4

One usually needs a metal center with an empty orbital (16e- or lower
count) in order to pre-coordinate the ligand before the oxidative addition
occurs. The one notable exception to this is O2, which can also act as an
electrophillic (Class B) substrate.
Typical “intact” ligands that can perform an oxidation addition without
fragmenting apart are:
alkenes, alkynes, and O2
One often needs to have electron withdrawing functional groups on the
alkenes or alkynes in order to “soup-up” their electron-withdrawing
ability in order to help promote the transfer of electrons from the metal
to the ligand.
metallocyclopropene
R
R R PMe3
PMe3
Me3P Pt Pt
PMe3
-PMe3 PMe3
R
In this case we have oxidized the Pt center from Pt(0) d10 to Pt(+2) d8
and generated a new dianionic unsaturated alkenyl ligand. Note that we
have broken one of the alkyne π-bonds.
General Features of Oxidative Additions
Because oxidative addition involves oxidation (removal of electrons) of
the metal center, the more electron-rich the metal is the easier the
oxidative addition to the metal center will be. So in comparing two or
more metal complexes to see which will be the most reactive towards a
particular substrate for oxidative addition you would pick the metal
center with the strongest donor ligands, fewest π-acceptor ligands, or
most negative charge. Also remember that the non-electrophillic ligands
(Class A) and “intact” ligands (Class C) usually require that there is an
empty orbital (16e- or lower) on the metal center in order to react.
 Oxidative Addition/Reductive Elimination 5

Kinetic Data for Oxidative Addition Reactions of MX(CO)(PR3)2

Rate Const ∆ H‡ ∆S‡
M X PR3 Reactant (M−1 sec−1) (kcal/mol) (J/mol K)
Ir Cl PPh3 H2 0.67 10.8 −23
Br 10.5 12.0 −14
I > 100
Ir Cl PPh3 O2 3.4 x 10−2 13.1 −21
Br 7.4 x 10−2 11.8 −24
I 30 x 10−2 10.9 −24
Ir Cl PPh3 CH3I 3.5 x 10−3 5.6 −51
Br 1.6 x 10−3 7.6 −46
I 0.9 x 10−3 8.8 −43
Ir Cl P(p-C6H4-OMe)3 CH3I 3.5 x 10−2 8.8 −35
P(p-C6H4-Cl)3 3.7 x 10−5 14.9 −28
Rh Cl PPh3 CH3I 12.7 x 10−4 9.1 −44
P(p-C6H4-OMe)3 51.5 x 10−4 10.2 −43

Data adapted from “Principles and Applications of Organotransition Metal Chemistry”, Coleman,
Hegedus, Norton & Finke, University Press, 1987; refs: Chock & Halpern, JACS, 1966, 88, 3511; Ugo,
Pasini, Fusi, Cenini, JACS, 1972, 94, 7364; Douek & Wilkenson, J. Chem. Soc. (A), 1964, 2604. Rxns
generally run in benzene at 25ºC.

Notice the trends in the table above. The more electron-rich the metal center
(better donating ligands) the faster the oxidative addition reactions. The one
exception is the oxidative addition of
CH3I with the Ir-Cl, Br, I series of
complexes. The slow down is caused by
steric factors caused by the increase in
size of the halide affecting the
nucleophillic attack of the metal dz2
orbital on the CH3I to start the oxidative
addition reaction.
The space-filling figure to the right shows two different
views of the Ir-Cl (left) and Ir-I (right) complexes. The top
view is looking down on the square plane with the halide
oriented to the left. The bottom view is looking down the
halide-Ir axis and illustrates how the larger size of the iodide
causes more steric interactions with the phenyl rings on the
PPh3 restricting their rotation. The PPh3 phenyl rings as
shown in the top view partially block the axial coordination
site. The larger size of the iodide causes pushes the
phenyl rings more towards the metal causing more steric hindrance for incoming substrates.
 Oxidative Addition/Reductive Elimination 6

WARNING: d0 metals can NOT do oxidative additions!! So always

electron count the starting and final metal complexes to check out the
overall electron-count, metal oxidation state and d-electron count!

Oxidative additions are easy to identify IF YOU ELECTRON

COUNT the metal complexes. When an oxidative addition rxn occurs
the metal will be oxidized, usually by 2e-. So, if you start with a metal
in the 0 oxidation state (d8), after the oxidative addition the metal will be
in the +2 oxidation state (d6). Once you get used to looking at
organometallic rxns you will be able to identify common oxidative
additions quite quickly. H2, R-X, and H-SiR3 are three of the most
common substrates that perform oxidative addition reactions in
catalytic cycles.

Problem: H2 will do an oxidative addition most readily to which of

the following complexes. Why?

a) b)
(MeO)3P Cl Me3P Br
Ir Rh
OC P(OMe)3 OC PMe3

O
c) C

Pt CO
Me3P
PMe3
 Oxidative Addition/Reductive Elimination 7

Problem: Cl2 will do an oxidative addition most readily to which of

the following complexes. Why?
a) b) PMe3
CO
OC CO
Re Pd PMe3
OC CO Me3P
CO PMe3

c)
Cl
Ti
Cl

Problem: CH3Br will do an oxidative addition most readily to

which of the following complexes. Why?
a) O b) PMe3

Hf Pt NCMe
NR2
Me3P Me3P
NR2 NCMe

c) CO
(PhO)3P CO
Fe
OC P(OPh)3
 Oxidative Addition/Reductive Elimination 8

Oxidative Coupling
Consider the following reaction:

Ph2
Cl Ph 2
Cl
P P
H 3C N Cr H 3C N Cr
P P
Ph2 Cl Ph2 Cl

Cr(+3) d 3 Cr(+5) d 1
15e- 13e-

The Cr on the right now has two new anionic alkyl ligands forming a
metallocyclopentane ring system. We have done an oxidative addition,
but in forming a new bond between the two ethylene ligand (and losing
the original double bonds) we have coupled the two ligands together.
While this is an oxidative addition, there is a special term for this type of
reaction called oxidative coupling. The metal is being oxidized to
create two new anionic ligands, but the original two neutral ligands also
form a new bond between them, instead of fragmenting apart to make
two new independent anionic ligands.
The driving force for this reaction is the formation of a new C-C σ-bond
(stronger than a π-bond) and the creation of two new strongly donating
anionic ligands that can better donate to the metal even though one has
technically lowered the electron count.
 Oxidative Addition/Reductive Elimination 9

Reductive Elimination

A reductive elimination reaction is the reverse of an oxidative addition.

It is a reaction in which two cisoidal anionic ligands on a metal center
couple together. Each anionic ligand pushes one electron back onto the
metal center (in the case of a monometallic complex) to reduce it by 2e-.
The coupled anionic ligands then usually fall off the metal center as a
neutral molecule.
two cisoidal anionic ligands that
can form a bond between them and eliminate
a neutral
ligand
H H H
OC PPh3 OC PPh3 OC PPh3
Ir Ir Ir + H2
Ph3 P H Ph3 P Cl Ph3 P Cl
Cl
Ir(+1) Ir(+1)
Ir(+3) 18e- 16e-
18e- rarely observed intermediate metal reduced
by 2 e-

Since electron-rich metal complexes favor oxidative addition, the

reverse is true for reductive elimination. Since reductive elimination
involves pushing electrons back onto the metal center from two anionic
ligands that are usually more electronegative than the metal center, it is
best if the metal center is electron deficient. This can be accomplished
by having electron-withdrawing ligands (e.g., CO), cationic charge(s),
and/or coordinative unsaturation (sub-18e- counts).
While reductive elimination can occur from saturated 18e- complexes
(so long as the two ligands that you want to reductively eliminate are
cisoidal to one another), it has been shown that reductive elimination
can be promoted by a ligand dissociation generating an unsaturated and
more electron-deficient metal center.
 Oxidative Addition/Reductive Elimination 10

CH3 CH3
Ph2
P CH3
-I− Ph2
P CH3
Ph2
P CH3
Pt Pt Pt + CH3CH3
P CH3 P CH3 P
Ph2 I ∆ Ph2 Ph2

Pt(+4) Pt(+4) Pt(+2)

18e- 16e- 14e-

+I−
The dissociation of the I- generates Ph2
P CH3
a cationic unsaturated complex. This Pt Goldberg, J. Am. Chem. Soc.
is electron deficient enough to help P I 1995, 117, 6889-6896
promote the reductive elimination of Ph2
ethane (CH3CH3).
Pt(+2)
16e-

In studying the above system, it was also found that one could have
reductive elimination of CH3I from the starting 18e- complex. This
reaction, however, is very reversible due to the high reactivity of CH3I
for doing an oxidative addition back reaction with the electron-rich
neutral Pt(+2) complex to make the Pt(+4) octahedral compound.

reductive elimination The reductive elimination of the CH3I is

kinetically favored. This is because the
Ph2
CH3 Ph2 orbitals around the iodide anion are
P CH3 P CH3
spherically symmetric and this makes
Pt Pt + CH3I
P CH3 P CH3 it much easier to overlap with the alkyl
Ph2 I ∆ Ph2 group orbital to perform the reductive
elimination. The sp3 directed orbitals
Pt(+4) Pt(+2) on the two CH3 groups are more difficult
18e- 16e- to overlap in order to get the reductive
elimination to occur. But the reductive
oxidative addition elimination of the CH3CH3 is thermo-
dynamically considerably more favorable
and the back oxidative addition much
more difficult.
 Oxidative Addition/Reductive Elimination 11

Problem: Which of the following compounds will be most likely to

do a reductive elimination of ethane (CH3-CH3)? Why?

a) CH3 b) NCMe
Me3P CH3 (MeO)3P CH3
V Os
Me3P CO H3C P(OMe)3
CO NCMe

c) PMe3
MeCN CH3
Pt
Ph3P CH3
CH3

Problem: Which of the following mechanisms makes the most sense

for the indicated reductive elimination (direct route or dissociative
route via a cationic intermediate)? Why?
a) CH3
Me3P CH3 Me3P CH3
Pt Pt + H3C CH3
Me3P CH3 Me3P I
I

b) CH3 CH3
Me3P CH3 Me3P CH3
Pt Pt + I
Me3P CH3 Me3P CH3
I

Me3P CH3
Pt + H3C CH3
Me3P I
 Oxidative Addition/Reductive Elimination 12

Binuclear Systems
Complexes with two (or more) adjacent metal atoms can also participate
in oxidative addition and reductive elimination reactions. This often
involves both metal centers. If so, then each metal changes its oxidation
state by only ±1, instead of ±2 as occurs with single metal centers.
This also often involves the making or breaking of a M-M bond,
depending on what is present in the starting complex.
Cl
Cl Cl
Au Au Au
+ Cl2
Me2P PMe2 + Cl2 Me2P PMe2 Me2P PMe2
Au Au Au
Au-Au = 3.0Å Au-Au = 2.6Å Cl Cl Cl Au-Au = 3.1Å

Au(+1) d10 Au(+2) d9 Au(+3) d8

Below is a somewhat usual case of a C-C bond oxidative addition to a

dimetal unit. The reverse reaction could also be considered a reductive
coupling instead of a reductive elimination since the ligand stays
coordinated to the metal center:
O O
C C

ν
hν Ru Vollhardt, JACS,
Ru-Ru = 2.82Å Ru Ru 1983, 105, 1676.
Ru
C
O C C CO ∆ C
O O O C
O
Ru(+1) d7 Ru(+2) d6

The next two examples involve reductive eliminations across a

bimetallic unit:
D3C CD3 D 3C
CH2 H2C CH2 D Me2N NMe2
β-hydride
O O elimination O O O O
Mo Mo Mo Mo Mo Mo
Me2N Me2N - CH2=CD2 Me 2N Me2N
+ D3 C CH2D
bimetallic reductive
O O O O elimination O O

NMe2 NMe2 NMe2 Chisholm

& coworkers
Mo(+3) d3 Mo(+3) d3 Mo(+2) d4
16e- 16e- 14e-
 Oxidative Addition/Reductive Elimination 13

In general bimetallic reductive eliminations occur across two metals

when there is a M-M bond and when one can have good overlap of the
two groups orbitals. Notably, there are very few if any examples of two
alkyl groups performing a bimetallic reductive elimination.
CH3 CH3 M M
or + H3C CH3
M M M M

CH3 H M M
or + H3C H
M M M M

This is due to the very poor overlap of the directed sp3 hybrid orbitals on
the alkyls used to bond to the metal centers. Alkyl and hydride
eliminations have been observed, no doubt due to the spherically
symmetric orbital on the hydride that can overlap with the carbon sp3
hybrid orbital promoting the reductive elimination. The reductive
elimination of two hydrides is well known and quite common.
Problem: Which of the two bimetallic complexes shown below will
be most likely to do a reductive elimination of H2? Why?