Lecture 19 February 16, 2011 Transition metals:Pd and Pt · Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt Why is CC coupling

1© copyright 2011 William A. Goddard III, all rights reservedCh120a-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]>

mailto:[email protected]

mailto:[email protected]


Last time


Compare chemistry of column 10


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


Salient differences between Ni, Pd, Pt

Ni Pd Pt

4s more stable than 3d 5s much less stable than 4d 6s, 5d similar stability3d 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


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?


Step 1: examine GVB orbitals for (PH3)2Pt(CH3)


Analysis of GVB wavefunction


Alternative models for Pt centers



energetics

Not agree with experiment


Possible explanation: kinetics


Consider reductive elimination of HH, CH and CC from Pd

Conclusion: HH no barrier

CH modest barrierCC large barrier


Consider oxidative addition of HH, CH, and CC to Pt

Conclusion: HH no barrier

CH modest barrierCC large barrier


Summary of barriers

But why?

This explains why CC coupling not occur for Pt while CH and HHcoupling is fast


How estimate the size of barriers (without calculations)


Examine HH coupling at transition state

Can simultaneously get good overlap of H with Pd sdhybrid and with the other H

Thus get resonance stabilization of TS low barrier


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 CH3But CANNOT DO BOTH SIMULTANEOUSLY, thus do NOT get

resonance stabilization of TS high barier


Examine CH coupling at transition state

H can overlap both CH3 and Pd

sd hybrid simultaneously but CH3 cannot

thus get ~ ½resonance

stabilization of TS


Now we understand Pt chemistry

But what about Pd?

Why are Pt and Pd so dramatically different


new


Pt goes from s1d9 to d10 upon reductive eliminationthus product stability is DECREASED by 12 kcal/mol

Using numbers from QM


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


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


Summary energetics

Conclusion the atomic character of the metal can

control the chemistry


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


Ground states of neutral atoms

Sc (4s)2(3d)Ti (4s)2(3d)2V (4s)2(3d)3Cr (4s)1(3d)5Mn (4s)2(3d)5Fe (4s)2(3d)6Co (4s)2(3d)7Ni (4s)2(3d)8Cu (4s)1(3d)10

Sc++ (3d)1Ti ++ (3d)2V ++ (3d)3 Cr ++ (3d)4Mn ++ (3d)5Fe ++ (3d)6 Co ++ (3d)7 Ni ++ (3d)8 Cu++ (3d)10

Sc+ (4s)1(3d)1Ti+ (4s)1(3d)2V+ (4s)0(3d)3Cr+ (4s)0(3d)5Mn+ (4s)1(3d)5Fe+ (4s)1(3d)6Co+ (4s)0(3d)7Ni+ (4s)0(3d)8Cu+ (4s)0(3d)10


Bond energies MH+

Cr

Mo

Re

Ag

Cu

Au


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[(d1α)(d2α)(d3α)(d4α)(d5α)(sα)]

Mn+: s1d5

For high spin (S=3)

A{(d1α)(d2α)(d3α)(d4α)(sdbα)[(sdb)H+H(sdb)](αβ−βα)}sdb is α half the time and β half the time


Ground state of M+ metalsMostly s1dn-1Exceptions:1st row: V, Cr-Cu2nd row: Nb-Mo, Ru-Ag3rd row: La, Pt, Au


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


Charge transfer in MH+ bondselectropositive

electronegative

1st row all electropositive

2nd row: Ru,Rh,Pd

electronegative3rd row:

Pt, Au, Hg electronegative






1st row


39© copyright 2011 William A. Goddard III, all rights reservedCh120a-Goddard-L19Schilling


Steigerwald




2nd row








3rd row














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


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


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


Now consider a TM case: Cl2TiH+ + D2

Orbitals of reactants

GVB orbitals of TiH bond for Cl2TiH+

GVB orbitals of D2


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


GVB orbitals at the TS for Cl2TiH+ + D2 Cl2TiD+ + HD


GVB orbitals for the Cl2TiD+ + HD productNote get phase change for both orbitals


Follow the D2 bond as it

evolves into the HD bond


Follow the TiHbond as it

evolves into the TiD bond


Barriers small, thus allowed

Increased d character in bond smaller barrier


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


Show reaction for ClMnH + D2

Show example reactions


Olefin Metathesis

Diego Benitez, Ekaterina Tkatchouk, Sheng Ding

2+2 metal-carbocycle reactions


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


R� O�� M�� B��R� O�� M�� B��


C�� O�� M�� C��C�� O�� M�� C��


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


History of Olefin Metathesis Catalysts

Ch120Ch120--L20 13/11/02L20 13/11/02GODDARD GODDARD 81

Well-defined metathesis catalysts

RuPCy3

Ph

Cl

ClNN MesMes

R R

R=H, Cl

NMo

PhCH3

CH3(F3C)2MeCO(F3C)2MeCO

iPr iPrRuPCy3

PCy3

Ph

Cl

Cl

1 2 3Schrock 1991alkoxy imido molybdenum complexa

Grubbs 1991 ruthenium

benzylidenecomplexb

Grubbs 19991,3-dimesityl-imidazole-2-ylidenes

P(Cy)3 mixed ligand system”c

Bazan, G. C.; Oskam, J. H.; Cho, H. N.; Park, L. Y.; Schrock, R. R. J. Am. Chem. Soc. 1991, 113, 6899-6907

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. Macromolecules1991, 24, 2649-2657


Ru

NNMes MesCl

Cl PhPCy3

Ru

NNMes MesCl

Cl PhPy

Ru

NNMes MesCl

ClO

i-Pr

slow initiating catalyst ultra-fast-initiating catalystfast-initiating catalyst

Ru

Cl

IMesCl

RLPh

IMes

Ru

LCl

ClInitiation

Ru

Cl

IMesCl

Cl

IMesClR

R2

RuR3

Ru

Cl

IMesCl

R1

PropagationR2R3

R3

R2

R1

+

Examples 2nd Generation Grubbs Metathesis Catalysts

General mechanism of Metathesis


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

Documents

Lecture 19 February 16, 2011 Transition metals:Pd and Pt · Mysteries from experiments on oxidative addition and reductive elimination of CH and CC bonds on Pd and Pt Why is CC coupling