Chemistry 652: Organometallic Chemistry - Key
Spring 2013 University of Delaware
Prof. Donald Watson
Problem Set 2 Due Beginning of Class 4/4/13
You may work in groups. No not try to fit answers on these pages, use separate paper.
Do not look up answers.
1)
a) PPh3 is a better sigma donor and poorer pi-acceptor than CO, so the Fe atom is
more electron rich follow substitution. The appearance of the 3rd IR band
indicates that the phosphines are in a cis geometry.
b) Tetracyano ethylene is a very good pi-acceptor, so there is significant pibackbonding from the electron-rich Ir center.
c) The para methyl group does not contribute to an increase in the cone angle as it
is 180 ° from the P-Ar bond axis. The ortho group, on the other hand, is only 30°
and has a significant effect.
d) This indicates that the TBP complex has all of the CO equitoral (and equivalent)
and the phosphines axial.
PPh3
CO
OC Co CO
PPh3
e) Stained alkenes have a high lying homo (and are there for good sigma-donors)
and a low lying lumo (and are there for good pi-acceptors).
f)
Alpha agnostic interaction, small JCH and attenuated C-H IR stretch.
Cl
Me3P
H
PMe3
Ta PMe3
PMe3
g) CO binds stronger to the Fe center (backbonding), does not dissociate and allow
an open coordination site.
H
Fe PPh
3 -PPh3
Ph3P
Me
Me Me
Fe
Ph3P
Me
Me
Fe
+PPh3
H
Ph3P
Me
–
Me
h) Beta-elimination possible with methyl, not with mesityl.
Me
Me
Fe
Ph3P
H
PPh3
Me
M
M
M
H
2.
Me
H
Me
Bridging CO IR stretch ~ 1850 cm-1
O
C
Cp*
Rh
OC
Cp*
Rh
Rh
CO
Cp*
3.
Me
hν
Si H
Ph
Np
R(+)
Mn(CO)3
Me
Mn
OC
CO
hν
Si H
Ph
Np
R(+)
CO
+ CO
OC Mn CO
Me
H
Si
Ph
retention of stereochem
Np
MnI, d6, 18 e–
tetrahedral
MnIII, d4, 18 e–
square pyramidal
Ph
(Ph3P)4Pd
Br
H
R
inversion
Ph
Ph3P
Ph3P
Pd
PPh3
PPh3
Pd(0), d10, 18 e–
square planar
Ph
Br
H
R
H
PPh3
R
Pd
Ph3P
Br
PdII, d8, 16 e–
square planar
+ 2 PPh3
Cl
(Et3P)3Pt
Me
CO2Et
R-(+)-α-ethylchloropropionate
hint: order of reactivity t-BuX > i-PrX > EtX > MeX for these types of reactions
Osborne and Labinger JACS 1974 96 7145
racemization
Et3P
Et3P
Pt
PEt3
PEt3
Pt(0), d10, 18 e–
square planar
CO2Et
H
PEt3
Me
Pt
Et3P
Cl
Cl
Me
CO2Et
+ 2 PEt3
PtII, d8, 16 e–
square planar
R-(+)-α-ethylchloropropionate
O
[(Me3P)4Ir]+Cl–
Ph
Milstein et al. JACS 1982 104 3773
loss of stereochemistry
H
O
Me3P
Me3P
Ir
PMe3
PMe3
IrI, d8, 16 e–
square planar
Cl–
O
Ph
Ph
Me3P Ir
Cl
PMe3
PMe3
IrIII, d6, 18 e–
square planar
4.
Ph
PPh3
Δ
Pt
Ph3P
Ph
Pt(0) s
PPh3
Pt
PPh3
Δ
+ Pt(PPh3)2
Pt
Ph3P
PPh3
Ph3P
PPh3
Pt
PtII, d8, 16 e–
square planar
Pt(0), d10, 14 e–
linear
Ph3P
PPh3
Pt(0), d10, 18 e–
square planar
Ph
1,1-disubstituted
(retention)
This reaction is accelerated in more Lewis basic solvents. Draw a mechanism
explaining why this phenomenon is observed.
Lewis basic solvents accelerate trans- to cis- isomerization.
Ph
PPh3
Pt
Ph3P
PtII, d8, 16 e–
square planar
solv:
PPh3
solv
Pt
PPh3
Ph
solv
RE
PPh3
Pt
Ph3P
solv
Pt(0), d10, 18 e–
square planar
Ph