Metal Alkyls, Aryls, and Hydrides and Related sBonded Ligands

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Chapter 3

• Metal Alkyls, Aryls, and Hydrides and Related
  s-Bonded Ligands

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Bond dissociation energy (BDE)
LnM-X ? LnM. X.
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Non-transition metals Except B-C, BDE(M-C)
89-280 kJ/mol Transition metals BDE
(M-alkyl) 125 - 260 kJ/mol BDE (M-aryl) 250
- 350 kJ/mol Conclusion transition metal-carbon
bond energies are not so different from those of
main group metals.
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B. Factors influence M-C bond strength

• (1) Metal effect
• For main group metals, BDE (M-CH3), kJ/mol
• B-Me 363 Zn-Me 176
• Al-Me 276 Cd-Me 136
• Ga-Me 247 Hg-Me 122
• BDE(M-C) decreased when going down a group. Why?

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For transition metals, BDE (M-CH3),
kJ/mol, (CO)5Mn-CH3 150 (PhCH2)3Ti-CH2Ph 240 (
CO)5Re-CH3 220 (PhCH2)3Zr-CH2Ph 380 BDE(M-C)
increased when going down a group. Why?
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R effect
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Would you expect M-CH3 bond be stronger or weaker
than
Why?
dp-pp interaction M-C bond distance is
shorter. Orbitals can overlap more effectively.
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Summary

• 1. The M-C bond in transition metal alkyl is not
  as weak as was previously thought. There is no
  inherent instability attached to transition metal
  alkyls.
• 2. The D(M-R) value increases as the atomic
  number increases among the congeners in the same
  group in an opposite trend to non-transition
  metal alkyls.
• 3. The D(M-CH3) value is as strong as the D(M-I)
  values or slightly smaller than that.
• 4. The D(M-R) value decrease in the order M(M-H)
  gtD(M-CF3)gtD(M-Ph)gtD(M-CH3)gtD(M-CH2CH3)gtDM-CH2Ph).

The earlier failure in obtaining TM alkyls is not
due to thermodynamic reasons!
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C. Origin of earlier failure in obtaining TM
alkyls Kinetic consideration.
General questions. Q1. Why many TM alkyl
complexes could not be isolated? There are many
pathways that TM alkyl complexes can be
decomposed.
Q2. What are the common decomposing pathways for
transition metal alkyls?
(The most common route!)
Others, e.g. a,g-eliminations, intramolecular
reactions.
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• Q3. How can we obtain stable M-alkyl complexes?
• Stable M-alkyl complexes could be obtained if the
  above decomposing pathways can be prevented.

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(1) b elimination.

• A. There must be a b-H.
• B. The complex should be able to form
  four-membered co-planar transition state (A).
• C. There is a vacant site cis to the alkyl (e.g
  B. but not C).
• (need or empty orbital to
• D. The olefin formed is stable.
• E. The metal has e- to go to s(C-H).

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Exercises.

• (a) It is difficult for the following compounds
  to undergo b-hydrogen eliminations. Suggest
  reasons.

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• b) Would you expect the following complexes to
  undergo b-H elimination easily?

(c) Which of the following complexes would you
expect to be least stable?
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Further notes on b-elimination.

• (a) b-H elimination on d0 (electron deficient)
  metal complexes may not occur. e.g.

Good s acidic Poor p basic
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(b) Other b-elimination is also possible
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(2) Reductive elimination

• Coordination number decrease by 2.
• Oxidation number increase by 2.

The process is the second most common
decomposition pathway for metal alkyls. e.g.
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(3) Other pathways.

• a-elimination

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g-elimination. e.g.
Summary Common pathways of decomposing
M-alkyls b-H elimination Reductive
elimination others.
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• Exercises. For each pair of complexes listed
  below, which one is less stable?

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2. Preparation of complexes with metal-carbon s
bonds.

• The most common methods to prepare complexes
  containing M-C bonds includes.
• A. Synthesis by alkyl transfer reactions.
• R- M-X ? M-R X-
• B. Synthesis from anionic transition metal
  complexes.
• M- R-X ? M-R X-
• C. Synthesis by oxidative addition reactions.
• M R-X ? X-M-R
• D. Synthesis involving insertion Reactions.
• M-X A ? M-A-X
• E. Synthesis involving elimination reactions.
• M-A-X ? M-X A
• F. Synthesis by attack on coordinated ligands

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A. Synthesis by alkyl transfer reactions.

• R- M-X ? M-R X-
• Typical R- reagents RLi, RMgX, AlR3, AlR2X,
  ZnR2, etc.
• Examples
• Preparation of homoletic alkyl complexes
• R- MXn ? MRn X-

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Examples
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Note different alkylating agents may have
different reactivity, e.g.
ZnR2 is also a weak alkylating reagents. e.g.
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Preparation of alkyl complexes

• R- LnMXy ---------gt LnMRy yX-
• e.g.

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B. Synthesis from anionic transition metal
complexes.

• M- R-X ? M-R X-
• R reagent

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Exercise. Suggest reagents for the following
preparations
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C. Synthesis by oxidative addition reactions.

• Typical oxidative addition reactions

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Examples of preparation of complexes.
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D. Synthesis involving insertion Reactions.

• M-X A ? M-A-X

Example
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E. Synthesis involving elimination reactions.

• M-A-X ? M-X A
• A CO, CO2, SO2, N2, e.g.

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F. Synthesis by attack on coordinated ligands

• We will go to more details late.

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3. Chemical properties of complexes with
metal-carbon s bonds

• (1) b-hydrogen elimination
• e.g.

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(2) Reductive elimination

• e.g.

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(3) a, g-elimination, e.g.
(4) Migratory insertion
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(5) Electrophilic attack on alkyls
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Examples.
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Exercises.

• Predict the products of the following reactions

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(6) Special properties of some M-R complexes

• Small metallocycles show interesting
  rearrangement.

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Related s-bonded ligands complexes

• Group 14 elements
• Comparison between M-CR3 and M-SiR3
• Which bond is longer? (steric)
• Which bond is stronger? (p-interaction)
• Can LnM-SiCH2R undergo b-H elimination?
• In general, M-SiR3 has properties similar to
  M-CR3

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Groups 15-17

• Comparison between M-CR3, M-NR2, M-OR, M-F, M-Cl
• M-CR3
• no dp-pp interaction
• M-X (lone pairs)
• possible dp-pp interaction
• Early transition metal oxophilic/fluorophilic
• p-donor ligands are favored.
• Late transition metal repulsion of filled metal
  orbitals with lone pairs in M-X bond.
• p-acceptor ligands are favored (CO).

dialkylamido
alkoxo
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B. Comparison between M-CR3 and M-OR, M-NR2,
M-F, M-Cl, etc.

• M-CR3 M-X M-X
• no dp-pp interaction possible dp-pp interaction
• Consequences
• M-X can stabilize complexes with less than 18e.
  e.g.
• Why?
• Ans p bonding can make these complexes to have
  formally 18e.

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(2) Similarity between M-CR3 and M-OR

• Both can undergo b-H elimination.

Use of the reaction
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5. Metal hydrides, M-H

• A. Structural types of metal hydrides.
• (1) Terminal hydride or classic hydride, e.g.
• (2) Non-classical hydride or molecular hydrogen
  complexes

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Q1. Would you expect the H2 molecule in the
following complex can rotate freely? why?
Consider the bonding between M and H2.
Rotation will break the p bond between M and H2
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• Q2. LnMH2 can have two forms, depending on M and
  L. Rationalize the structural difference in the
  following pairs.

In terms of bonding
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(3) "Hydrogen-bonding" type interaction H-X s
complexes
Often the interaction can be represented as
For example,
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(4) Bridges hydrides
Examples
For electron counting purpose, we can use the
resonance structures like
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Exercise. 18e or not?
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B. Synthesis of metal hydride complexes.

• Main routes include
• (1) By protonation

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• (2) From hydride donors
• M-X H- donor ? M-H X-
• Typical H- donors NaH, LiAlH4, NaBH4, LiBHEt3 ...

Examples. WCl6 LiBHEt3 PR3 ? WH6(PR3)3
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(3) From H2
Examples
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What are the likely intermediates or transition
states?
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(4) From ligands
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C. Chemical properties of M-H complexes

• LnM-H acidic or hydridic?
• What about a transition metal hydride complexes?
• Can be hydridic and acidic.

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For example,
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• Factors influence hydridic and acidic properties
  of M-H
• In general, increase in electron-richness of
  metal center, will
• ----gt decrease acidic characters
• -----gtincrease hydridic characters

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(a) H-Co(CO)4 vs H-Co(CO)3(PPh3) H-Co(CO)4 is
more H-Co(CO)3(PPh3) is more. Because. (b)
H-Co(CO)3(PMe3) vs H-Co(CO)3(PPh3) H-Co(CO)3(PMe3
) is more. H-Co(CO)3(PPh3) is more. Because. (c
) CpRu(CO)2H vs CpOs(CO)2H CpRu(CO)2H is more
CpOs(CO)2H is more Because.
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(2) Chemical reactions

• a. Deprotonation reactions with nucleophiles
  (bases)
• e.g.
• WH6(PMe3)3 NaH ? NaWH5(PMe3)3 H2
• OsH4(PR3)4 NaH ? NaOsH3(PR3)4 H2

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b. Reactions with electrophiles.

• e.g.
• WH6(PMe3)3 HBF4 3 MeCN ? WH2(MeCN)3(PR3)32
  H2

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c. H atoms transfer reactions

• e.g.

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d. Insertion reactions