Chemistry · Lesson
Oxidative addition, reductive elimination and insertion
Track selected oxidative addition, reductive elimination and migration by conserving atoms and stating the formal electron-count model. Keep a reaction class distinct from proof of its detailed mechanism.
Back to the Chemistry mapBy the end, you can
- Compare formal changes in selected two-electron addition and elimination.
- Distinguish bound-ligand migration from later ligand association.
- Conserve the reacting carbon and hydrogen inventories in alkene insertion.
- Explain the conditional requirements and immediate product of β-hydride elimination.
A transformation name does not specify every pathway
Oxidative addition and reductive elimination organize changes in metal–ligand connectivity and formal oxidation assignment. Oxidative addition can proceed through different concerted, polar or radical pathways. A common single-metal two-electron example changes metal oxidation state by +2, but the reaction name alone does not establish every intermediate, rate law or geometry.
For each selected example here, first state the entity charge, ligand assignments and retained ligands. Then compare atoms, formal metal state and the same valence-count convention. A formal increase in metal oxidation state need not mean that free electrons have escaped the whole reacting system; the electrons are reassigned within a conserved molecular net.
An intact substrate can become two assigned ligands
Choose the formal net IrCl(CO)(PPh₃)₂+H₂→IrClH₂(CO)(PPh₃)₂, retaining the other ligands. With Cl⁻ and two product H⁻ assignments, the metal changes Ir+1, d⁸→Ir+3, d⁶. Ionic ligand donation grows by four from the two hydrides, while the metal d count decreases by two. The valence total therefore changes sixteen→eighteen.
In the stated one-donor-site-per-ligand drawing, coordination sites change four→six, and the whole entity remains neutral. Two H nuclei are conserved. This is the selected single-metal two-electron bookkeeping, not a universal statement that every oxidative addition has that ligand inventory, geometry or mechanism. An intact H₂ ligand before cleavage is a separate connectivity state.
Reductive elimination reverses a selected connectivity change
In the symbolic model PdL₂(CH₃)(H)→PdL₂+CH₄, take L as a neutral monodentate two-electron donor, methyl −1 and hydride −1. The starting neutral entity has Pd+2, d⁸ and sixteen electrons. The returned PdL₂ model has Pd0, d¹⁰ and fourteen. One C and four H atoms appear as methane; both retained L ligands remain in the metal pool.
The two X-type ligand assignments combine into a neutral molecule. Metal d count increases by two while ligand donation decreases by four, giving a net decrease of two in this selected metal valence count. Suitable disposition and actual kinetic viability require evidence. The formal net does not prove that every PdL₂ species is isolable or that methane formation proceeds at a particular rate.
Migration and incoming CO are different steps
For a selected methylmanganese model, CH₃Mn(CO)₅ has Mn+1, d⁶ and eighteen electrons. Migration between its bound methyl and one bound CO gives a C-bound acyl entity Mn(CO)₄[C(O)CH₃]. Mn remains +1; the former neutral CO donor and methyl donor combine into one assigned acyl −1 donor. The immediate formal valence count is sixteen.
The initial and immediate-product inventories are Mn₁C₆O₅H₃. No external carbon has been added in that step. A separate incoming free CO can then associate, restoring an eighteen-electron count and adding one C and one O to the metal entity. Calling the first step CO insertion does not by itself establish that an unbound CO collided directly into the metal–alkyl bond.
A suitable adjacent arrangement of the reacting bound groups is important in the selected migration description, but does not alone prove a rate or a particular intermediate lifetime. Different acyl binding, bridging or spectator-ligand changes require their own ledgers. The diagrams and counts here deliberately retain the C-bound acyl assumption.
Insertion into a metal–H interaction can conserve oxidation state
Choose neutral model RhClL₂H₂(η²-C₂H₄), with two hydrides and a bound neutral alkene. Rh is +3, d⁶ and its count is eighteen. Migration of one hydride to the bound ethene creates an ethyl −1 ligand, giving RhClL₂H(C₂H₅) with Rh still +3 and sixteen electrons. The reacting H⁻ plus C₂H₄ pool becomes C₂H₅⁻; the second hydride remains.
The selected ligand-site count changes six→five, while direct metal–atom contacts change seven→five because the initial alkene has two carbon contacts. These different counts must not be silently equated. A new ligand could subsequently bind, but it is not part of this immediate insertion ledger.
The reverse β-hydride elimination cleaves a suitably accessible C–H interaction at the carbon next to the metal-bound carbon, giving bound hydride and alkene in the selected model. It requires appropriate geometry and accessible adjacent coordination space under the chosen mechanism. It does not necessarily release free alkene immediately. Absence of a β H blocks this simple pathway, while alternative pathways and actual rates need additional evidence.
| Selected change | Formal metal state | Metal valence total | Inventory or subsequent-step boundary |
|---|---|---|---|
| H₂ addition to retained-ligand Ir model | +1→+3 | 16→18 | Two H become two assigned hydrides; entity stays neutral |
| CH₄ elimination from PdL₂(CH₃)(H) | +2→0 | 16→14 | C₁H₄ released; two L retained |
| Bound methyl/CO migration at Mn | +1→+1 | 18→16 | Mn₁C₆O₅H₃ conserved in immediate entity |
| Later free-CO association at Mn | +1→+1 | 16→18 | One additional C and O enter separately |
| One hydride/ethene migration at Rh | +3→+3 | 18→16 | H⁻+C₂H₄→C₂H₅⁻; second hydride retained |
Original chosen mononuclear bookkeeping. L is neutral, monodentate and a two-electron donor. Alkyl, acyl and hydride are assigned −1; alkene and CO are neutral. The site count treats an η²-alkene as one ligand function, with two carbon contacts separately. These classes and immediate products are not complete validated real mechanisms.
Worked examples
Account for both metal and ligand contributions
Use the stated retained-ligand Ir H₂ net.
- Ir formal d count decreases 8→6.
- Two new hydride donors add 4.
- Net valence change −2+4=+2:16→18.
- Whole-entity charge stays zero.
A consistent selected two-electron addition, without a unique mechanistic conclusion.
Conserve the methyl and hydride atoms
Use PdL₂(CH₃)(H)→PdL₂+CH₄.
- CH₃ and H contain C₁H₄ together.
- Pd+2, d⁸ becomes Pd0, d¹⁰.
- Starting count 8+4+2+2=16; returned count 10+4=14.
One neutral methane per formal event, with actual feasibility and rate unspecified.
Keep incoming CO outside the first ledger
Compare CH₃Mn(CO)₅ and Mn(CO)₄[C(O)CH₃].
- Both immediate entities contain Mn₁C₆O₅H₃.
- Mn+1 and the assigned acyl −1 leave the oxidation state unchanged.
- One neutral CO donor function is absorbed into the acyl connectivity, reducing 18→16.
- A later free CO adds its atoms and two-electron donation separately.
Bound-ligand migration followed, if stipulated, by a distinct association step.
Compare sites and atom contacts during ethene insertion
Use the selected Rh hydride/ethene and ethyl entities.
- Starting sites:Cl1, L2, H=2, alkene 1=6; atom contacts 7.
- After migration:Cl1, L2, H=1, ethyl 1=5 contacts and sites.
- Assigned charge sum stays −3, so Rh remains +3.
- Ionic counts 18→16.
Conserved reacting C₂H₆ total including the spectator hydride, with different contact/site conventions explicit.
Try it yourself
In the selected Ir H₂ addition, metal d count falls by two and donation rises by four. What is the net valence-count change?
Show the answer and reasoning
An increase of two electrons.
−2+4=+2 under the same ionic convention; the whole system has not created two electrons.
Before external CO association, how many C and O atoms are in each selected methyl-Mn and immediate acyl-Mn entity?
Show the answer and reasoning
Six C and five O atoms in each.
The methyl supplies one C, and the five original CO units supply five C and five O; migration reorganizes that pool.
Why does the simple β-hydride-elimination model require more than the mere presence of a C–H bond somewhere in the ligand?
Show the answer and reasoning
The H must be an accessible β H with suitable local geometry and coordination space.
A remote C–H bond or an incompatible arrangement does not satisfy this selected elementary-class description.
Does an oxidative-addition name prove a concerted molecular pathway?
Show the answer and reasoning
No.
The reaction class can include different polar, concerted or radical mechanisms; evidence must discriminate them.
What to carry forward
- State the retained ligands and charge before comparing counts.
- Reaction classes do not specify every pathway.
- Keep bound-ligand migration separate from external association.
- Conserve atoms alongside formal electron assignments.
- Distinguish ligand functions, contacts and accessible β-H geometry.
References
Teaching resources and reference material supporting this lesson’s explanations and examples.
- Tutorial on Oxidative AdditionJay A. LabingerSource
- 14.2.2 CO Insertions (Alkyl Migration)Michael Evans text/figure attribution in body; Kathryn Haas page curatorSource
- 10.2 Insertion and Elimination ReactionsM. S. Balakrishna; Prasenjit GhoshSource
- Electron Counting, Formal Oxidation States and the18Electron RuleHandout individual byline unverified; institutional owner/instructor association David Berg verified on UVic profileSource · PDF