The small steps behind big catalytic cycles

Science · Chemistry · ages 20-22

Name ______________________ Date ____________

With all other ligands unchanged, how do metal oxidation state and electron count change in conventional two-electron oxidative addition?
- Oxidation state rises by 2; count stays unchanged
- Oxidation state falls by 2; count decreases by 2
- Oxidation state rises by 2; count increases by 2
One metal cleaves H-H and binds both fragments as hydride ligands in a two-electron step. Which step is described?
- Oxidative addition of H2
- CO migratory insertion
- Beta-hydride elimination
In CO insertion of a metal alkyl, before any further ligand binds, oxidation state is unchanged and electron count falls by 2.
Circle one: True False
A 14-electron M(I) complex undergoes mononuclear two-electron oxidative addition of an aryl halide. No other ligand changes occur. What is the product?
- A 12-electron M(-I) complex
- A 16-electron M(III) complex
- A 14-electron M(I) complex
What distinguishes CO migratory insertion of a metal alkyl from direct ligand association?
- A free CO binds without joining the metal-bound alkyl group
- The bound alkyl migrates onto CO, forming an acyl ligand
- The CO leaves while the alkyl remains bound to the metal
A saturated 18-electron precursor undergoes oxidative addition after losing a neutral 2-electron ligand. Why can ligand loss help?
- It gives a 16-electron species because ligand loss raises the oxidation state
- It gives a 16-electron species and can make coordination capacity available
- It gives a 20-electron species because ligand loss adds donor electrons
CO insertion changes an 18-electron alkyl complex to a 16-electron acyl. The observed acyl has 18 electrons and no free site. A neutral 2-electron solvent ligand is present. What explains this?
- Solvent binding raised the oxidation state, adding two metal d electrons to the acyl
- CO insertion left an 18-electron acyl before the solvent ligand could bind
- Solvent binding supplied two donor electrons and filled the available position
A metal alkyl transfers a beta hydrogen to the metal and forms a bound alkene. A student calls this reductive elimination. Which correction fits?
- Reductive elimination; the metal couples two groups into a departing molecule
- Oxidative addition; the metal cleaves the bond of an incoming H2 molecule
- Beta-hydride elimination; the metal gains H from its alkyl and forms an alkene

Answer key

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The small steps behind big catalytic cycles W1-mt_VVLGGh8QZ_-s1

Oxidation state rises by 2; count increases by 2 · The metal oxidation state rises by 2. Two new X ligands add 4 donated electrons while the metal d count loses 2, giving a net increase of 2.
Oxidative addition of H2 · The H-H bond is cleaved and both fragments attach to the same metal as X-type ligands. This is oxidative addition.
True · An X-type alkyl and a 2-electron CO become one X-type acyl. The ligand charge is unchanged but the donation decreases by 2.
A 16-electron M(III) complex · Add 2 to both quantities: 14 becomes 16 electrons and oxidation state +1 becomes +3.
The bound alkyl migrates onto CO, forming an acyl ligand · Migratory insertion forms a bond between existing bound ligands. Direct association adds a ligand without that rearrangement.
It gives a 16-electron species and can make coordination capacity available · Loss of a neutral 2-electron ligand lowers the count to 16 without changing oxidation state and can prepare a site for reaction.
Solvent binding supplied two donor electrons and filled the available position · Insertion lowers the count from 18 to 16. Binding the stated solvent donor adds 2, restoring 18 and occupying the available position.
Beta-hydride elimination; the metal gains H from its alkyl and forms an alkene · A beta hydrogen leaves the alkyl to become a metal hydride as an alkene forms. Reductive elimination instead couples bound groups into a leaving molecule.