lecture 19 february 16, 2011 transition metals:pd and pt
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Lecture 19 February 16, 2011 Transition metals:Pd and Pt. Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy. William A. Goddard, III, [email protected] 316 Beckman Institute, x3093 - PowerPoint PPT PresentationTRANSCRIPT
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1© copyright 2011 William A. Goddard III, all rights reserved Ch120a-Goddard-L19
Nature of the Chemical Bond with applications to catalysis, materials
science, nanotechnology, surface science, bioinorganic chemistry, and energy
Lecture 19 February 16, 2011
Transition metals:Pd and Pt
William A. Goddard, III, [email protected] Beckman Institute, x3093
Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics,
California Institute of Technology
Teaching Assistants: Wei-Guang Liu <[email protected]>Caitlin Scott <[email protected]>
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Last time
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Compare chemistry of column 10
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Ground state of group 10 column
Pt: (5d)9(6s)1 3D ground statePt: (5d)10(6s)0 1S excited state at 11.0 kcal/molPt: (5d)8(6s)2 3F excited state at 14.7 kcal/mol
Pd: (5d)10(6s)0 1S ground statePd: (5d)9(6s)1 3D excited state at 21.9 kcal/molPd: (5d)8(6s)2 3F excited state at 77.9 kcal/mol
Ni: (5d)8(6s)2 3F ground stateNi: (5d)9(6s)1 3D excited state at 0.7 kcal/molNi: (5d)10(6s)0 1S excited state at 40.0 kcal/mol
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Salient differences between Ni, Pd, Pt
Ni Pd Pt
4s more stable than 3d 5s much less stable than 4d 6s, 5d similar stability
3d much smaller than 4s(No 3d Pauli orthogonality)Huge e-e repulsion for d10
4d similar size to 5s (orthog to 3d,4s
Differential shielding favors n=4 over n=5,
stabilize 4d over 5s d10
2nd row (Pd): 4d much more stable than 5s Pd d10 ground state
3rd row (Pt): 5d and 6s comparable stability Pt d9s1 ground state
Relativistic effects of 1s huge decreased KE contraction stabilize and contract all ns destabilize and expand nd
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Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt
Why is CC coupling so much harder than CH coupling?
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Step 1: examine GVB orbitals for (PH3)2Pt(CH3)
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Analysis of GVB wavefunction
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Alternative models for Pt centers
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energetics
Not agree with experiment
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Possible explanation: kinetics
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Consider reductive elimination of HH, CH and CC from Pd
Conclusion: HH no barrier
CH modest barrierCC large barrier
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Consider oxidative addition of HH, CH, and CC to Pt
Conclusion: HH no barrier
CH modest barrierCC large barrier
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Summary of barriers
But why?
This explains why CC coupling not occur for Pt while CH and HHcoupling is fast
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How estimate the size of barriers (without calculations)
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Examine HH coupling at transition state
Can simultaneously get good overlap of H with Pd sd hybrid and with the other H
Thus get resonance stabilization of TS low barrier
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Examine CC coupling at transition state
Can orient the CH3 to obtain good overlap with Pd sd hybrid OR can orient the CH3 to obtain get good overlap with the other CH3
But CANNOT DO BOTH SIMULTANEOUSLY, thus do NOT get resonance stabilization of TS high barier
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Examine CH coupling at transition state
H can overlap both CH3 and Pd
sd hybrid simultaneously but CH3 cannot
thus get ~ ½ resonance
stabilization of TS
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Now we understand Pt chemistry
But what about Pd?
Why are Pt and Pd so dramatically different
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new
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Pt goes from s1d9 to d10 upon reductive eliminationthus product stability is DECREASED by 12 kcal/mol
Using numbers from QM
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Pd goes from s1d9 to d10 upon reductive eliminationthus product stability is INCREASED by 20 kcal/mol
Using numbers from QM
Pd and Pt would be ~ same
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Thus reductive elimination from Pd is stabilized by an extra 32 kcal/mol than for Pt due to the ATOMIC nature of the states
The dramatic stabilization of the product by 35 kcal/mol reduces the barrier from ~ 41 (Pt) to ~ 10 (Pd)
This converts a forbidden reaction to allowed
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Summary energetics
Conclusion the atomic character of the metal can
control the chemistry
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Examine bonding to all three rows of transition metals
Use MH+ as model because a positive metal is more representative of organometallic and inorganic complexes
M0 usually has two electrons in ns orbitals or else one
M+ generally has one electron in ns orbitals or else zero
M2+ never has electrons in ns orbitals
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Ground states of neutral atoms
Sc (4s)2(3d)
Ti (4s)2(3d)2
V (4s)2(3d)3
Cr (4s)1(3d)5
Mn (4s)2(3d)5
Fe (4s)2(3d)6
Co (4s)2(3d)7
Ni (4s)2(3d)8
Cu (4s)1(3d)10
Sc++ (3d)1
Ti ++ (3d)2
V ++ (3d)3
Cr ++ (3d)4
Mn ++ (3d)5
Fe ++ (3d)6
Co ++ (3d)7
Ni ++ (3d)8
Cu++ (3d)10
Sc+ (4s)1(3d)1
Ti+ (4s)1(3d)2
V+ (4s)0(3d)3
Cr+ (4s)0(3d)5
Mn+ (4s)1(3d)5
Fe+ (4s)1(3d)6
Co+ (4s)0(3d)7
Ni+ (4s)0(3d)8
Cu+ (4s)0(3d)10
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Bond energies MH+
Cr
Mo
Re
Ag
Cu
Au
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Exchange energies:
Get 6*5/2=15 exchange terms5Ksd + 10 KddResponsible for Hund’s rule
Ksd KddMn+ 4.8 19.8 Tc+ 8.3 15.3Re+ 11.9 14.1
kcal/mol
Form bond to H, must lose half the exchange stabilization for the orbital bonded to the H
A[(d1)(d2)(d3)(d4)(d5)(s)]
Mn+: s1d5
For high spin (S=3)
A{(d1)(d2)(d3)(d4)(sdb)[(sdb)H+H(sdb)]()}
sdb is half the time and half the time
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Ground state of M+ metals
Mostly s1dn-1Exceptions:1st row: V, Cr-Cu2nd row: Nb-Mo, Ru-Ag3rd row: La, Pt, Au
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Size of atomic orbitals, M+
Valence s similar for all three rows,5s biggest
Big decrease from La(an 57) to Hf(an 72
Valence d very small for 3d
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Charge transfer in MH+ bonds
electropositive
electronegative
1st row all electropositive
2nd row: Ru,Rh,Pd
electronegative3rd row:
Pt, Au, Hg electronegative
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1st row
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Steigerwald
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2nd row
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3rd row
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Physics behind Woodward-Hoffman Rules
For a reaction to be allowed, the number of bonds must be conserved. Consider H2 + D2
2 bonds TS ? bonds 2 bonds
Bonding2 elect
nonbonding1 elect
antibonding0 elect
Have 3 electrons, 3 MO’s
Have 1 bond. Next consider 4th atom, can we get 2 bonds?
To be allowed must have 2 bonds at TSHow assess number of bonds at the TS. What do the dots mean? Consider first the fragment
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Can we have 2s + 2s reactions for transition metals?
2s + 2s forbidden for organics
X
Cl2Ti Cl2Ti Cl2Ti? ?
2s + 2s forbidden for organometallics?
Cl2Ti Cl2Ti Cl2TiMe
Me
Me
Me
Me
Me
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Physics behind Woodward-Hoffman Rules
Bonding2 elect
nonbonding1 elect
antibonding0 elect
Have 1 bond. Question, when add 4th atom, can we get 2 bonds?
Can it bond to the nonbonding orbital?
Answer: NO. The two orbitals are orthogonal in the TS, thus the reaction is forbidden
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Now consider a TM case: Cl2TiH+ + D2
Orbitals of reactants
GVB orbitals of TiH bond for Cl2TiH+
GVB orbitals of D2
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Is Cl2TiH+ + D2 Cl2TiD+ + HD allowed?
Bonding2 elect
nonbonding1 elect
antibonding0 elect
when add Ti 4th atom, can we get 2 bonds?
Answer: YES. The two orbitals can have high overlap at the TS orthogonal in the TS, thus the reaction is allowed
Now the bonding orbital on Ti is d-like. Thus at TS have
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GVB orbitals at the TS for Cl2TiH+ + D2 Cl2TiD+ + HD
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GVB orbitals for the Cl2TiD+ + HD product
Note get phase change for both orbitals
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Follow the D2 bond as it
evolves into the HD bond
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Follow the TiH bond as it
evolves into the TiD bond
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Barriers small, thus allowed
Increased d character in bond smaller barrier
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Are all MH reactions with D2 allowed? No
Example: ClMn-H + D2
Here the Mn-Cl bond is very polar
Mn(4s-4pz) lobe orbital with Cl:3pz
This leaves the Mn: (3d)5(4s+4pz), S=3 state to bond to the HBut spin pairing to a d orbital would lose
4*Kdd/2+Ksd/2= (40+2.5) = 42.5 kcal/mol
whereas bonding to the (4s+4pz) orbital loses
5*Ksd/2 = 12.5 kcal/mol
As a result the H bonds to (4s+4pz), leaving a high spin d5.
Now the exchange reaction is forbidden
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Show reaction for ClMnH + D2
Show example reactions
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Olefin Metathesis
Diego Benitez, Ekaterina Tkatchouk, Sheng Ding
2+2 metal-carbocycle reactions
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Mechanism: actual catalyst is a metal-alkylidene
R1 R1 R2 R2+
R1 R22
M
R2
R1 R3
M
R2
R1 R3
M
R2
R1 R3
Catalytically make and break double bonds!
OLEFIN METATHESIS
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Ru Olefin Metathesis BasicsRu Olefin Metathesis Basics
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Common Olefin Metathesis CatalystsCommon Olefin Metathesis Catalysts
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Applications of the olefin metathesis reaction
Acc. Chem. Res. 2001, 34, 18-29
http://www.pslc.ws/macrog/pdcpd.htmbulletproof resin
Small scale synthesisto industrial polymers
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History of Olefin Metathesis Catalysts
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Ch120-L20 13/11/02Ch120-L20 13/11/02GODDARD GODDARD 81
Well-defined metathesis catalysts
Ru
PCy3
Ph
Cl
ClNN MesMes
Ru
PCy3
Ph
Cl
ClNN MesMes
R R
R=H, Ph, or -CH2-(CH2)2-CH2-
R R
R=H, Cl
NMo
PhCH3
CH3(F3C)2MeCO
(F3C)2MeCO
iPr iPrRuPCy3
PCy3
Ph
Cl
Cl
1 2 3 4Schrock 1991alkoxy imido molybdenum complexa
Bazan, G. C.; Oskam, J. H.; Cho, H. N.; Park, L. Y.; Schrock, R. R. J. Am. Chem. Soc. 1991, 113, 6899-6907
Grubbs 1991 ruthenium
benzylidene complexb
Grubbs 19991,3-dimesityl-imidazole-2-ylidenes
P(Cy)3 mixed ligand system”c
Scholl, M.; Trnka, T. M.; Morgan, J. P.; Grubbs, R. H. Tetrahedron Lett. 1999, 40, 2247-2250.
Wagener, K. B.; Boncella, J. M.; Nel, J. G. Macromolecules 1991, 24, 2649-2657
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Examples 2nd Generation Grubbs Metathesis Catalysts
General mechanism of Metathesis
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Schrock and Grubbs catalysts make olefin metathesis practical
Schrock catalyst –very active, but destroysmany functional groups
Grubbs catalyst –very stable, high functionalgroup tolerance, but not asreactive as Schrock
Catalysts contain many years of evolutionary improvements