Identification of Stereochemical Isomers
of [Mo(CO)4(L)2] by Infra-Red
Spectroscopy
AKA Group Theory
Emma Kate Payne, Courtney Arnott, and
Julia Schmitz
Background Information
• As an electron drops from a higher energy state, it creates
both a magnetic and electric vector
• Only those molecules that create changing magnetic
vectors are absorbed by IR
• IR absorption will occur if the vibration between two
atoms has a magnetic oscillating frequency that is the same
frequency as the photon of light emitted
• The more symmetric a molecule the less vibration and less
absorption by the IR
How to Determine Expected IR
Absorption
• Use symmetry operations and a Group
Theory Flow Chart to find point group of
molecule
• Then use a characteristic table to determine
the number of bands expected to show up
on IR
• Try to distinguish between cisMo(CO)4(PPh3)2 and trans-Mo(CO)4(PPh3)2
Group Theory Flow Chart
Characteristic Tables
C2v
FIND THAT cis will have four bands
E
C2
sv(
xz)
sv(
yz)
A1
1
1
1
1
z
x2, y2, z2
A2
1
1
-1
-1
Rz
xy
B1
1
-1
1
-1
x, Ry
xz
B2
1
-1
-1
1
y, Rx
yz
Characteristic Tables (cont.)
D4h
Find that trans has one band
total
2C '
i
2S4
sh
2sv
2sd
1
1
1
1
1
1
-1
-1
1
1
1
-1
-1
1
1
-1
1
-1
1
1
-1
x2 - y2
-1
1
-1
1
1
-1
1
-1
1
xy
2
0
-2
0
0
2
0
-2
0
0
A1u
1
1
1
1
1
-1
-1
-1
-1
-1
A2u
1
1
1
-1
-1
-1
-1
-1
1
1
B1u
1
-1
1
1
-1
-1
1
-1
-1
1
B2u
1
-1
1
-1
1
-1
1
-1
1
-1
Eu
2
0
-2
0
0
-2
0
2
0
0
2
E
2C4
C2
2C2'
'
A1g
1
1
1
1
A2g
1
1
1
B1g
1
-1
B2g
1
Eg
x2 + y2, z2
Rz
(Rx, Ry)
z
(x, y)
(xz, yz)
IR Identification
• In IR the carbon monoxide peak of the cis
molecule will have four peaks
• The carbon monoxide peak of the trans
molecule will have one peak because it is
more symmetrical
Reaction Scheme
Synthesis of Mo(CO)4(pip)2
•
•
•
•
All reactions carried out under N2
1 g MoCO6 suspended in dry toluene
10 ml piperidene added
Reflux 2 hrs  yellow-orange solution,
yellow precipitate
• Filter, dry on vacuum line
• Take weight, IR, melting point
Synthesis of Mo(CO)4(PPh3)2
•
•
•
•
0.5 g MoCO6 suspended in dry CH2Cl2
0.75 g PPh3 added
Reflux 15 min  orange solution, filtered
Filtrate reduced on rotovap, 15 mL MeOH
added, cooled in freezer  yellow crystals
• Filter, dry on vacuum line
• Take weight, IR, melting point
Thermal Isomerization
• 0.5 g Mo(Co)4(PPh3)2 dissolved in dry
•
•
•
•
•
toluene
Reflux 15 min  dark solution
Rotovap to yield an off-white product
Rinse with CH2Cl2
Filter, dry on vacuum line
Take weight, IR, melting point
Experimental Error
• During our second
synthesis, our solvent
boiled off
• No condenser tube
• Thus we thermally
isomerized our
reaction straight to the
trans configuration
What We Obtained
Compound
Melting Point (°C) Yield (g) Percent yield
Mo(CO)4Pip2
249
1.1374
75%
Mo(CO)4(PPH3)2 (cis)
181
0.32
20%
Mo(CO)4(PPH3)2 (trans)
191
0.909
56%
IR of Mo(CO)4Pip2
IR Mo(CO)4(PPh3)2 (Cis)
IR of Mo(CO)4(PPh3)2 (Trans)
Conclusions
• Made both the Cis and Trans Compounds
• Group Theory correctly predicted the
number of bands in both IRs
• Improvements:
• Condenser column on all reactions
• Better N2 conditions
• More careful monitoring of heat/evaporation
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