Questions: Catalytic Cycles (Wilkinson's Catalyst, Grubbs)

Question 1

In the catalytic cycle of Wilkinson's catalyst for hydrogenation of an alkene, what is the first step after dissociation of one PPh₃ ligand to generate the active 14-electron species?

• A. Migratory insertion of the alkene into a Rh-H bond
• B. Oxidative addition of H₂ to the rhodium center, forming a dihydride complex and increasing the oxidation state from Rh(I) to Rh(III)
• C. Beta-hydride elimination from a coordinated alkane
• D. Reductive elimination of HCl to generate a more reactive Rh(0) species

Answer: (B)

After PPh₃ dissociation, the active species [RhCl(PPh₃)₂] (14e, Rh⁺) undergoes oxidative addition of H₂. The H-H bond breaks, and both hydrogen atoms bond to rhodium, forming [RhH₂Cl(PPh₃)₂] (Rh³⁺, 16e). This is followed by alkene coordination, migratory insertion of the alkene into one Rh-H bond (forming a Rh-alkyl species), and finally reductive elimination to release the alkane product and regenerate the active catalyst. Each step follows the rules of organometallic elementary reactions, and the overall cycle returns the catalyst to its original state.

Question 2

Grubbs' catalyst accomplishes olefin metathesis — the exchange of substituents between two alkenes — through a mechanism involving a metal carbene intermediate (a M=CHR double bond).

• T. True
• F. False

Answer: True

The Chauvin mechanism for olefin metathesis involves a [2+2] cycloaddition between the metal carbene (Ru=CHR in Grubbs' catalyst) and an alkene substrate, forming a metallacyclobutane intermediate. This intermediate then undergoes a retro-[2+2] cycloreversion in the other direction, releasing a new alkene and regenerating a metal carbene with a different substituent. The cycle repeats, redistributing the substituents among alkene partners. Grubbs' ruthenium carbene catalysts are air-tolerant and functional-group tolerant, making them practical for organic synthesis — earning Grubbs the 2005 Nobel Prize.

Question 3

A good homogeneous catalyst must bind substrates strongly enough to activate them but weakly enough to release products. This balance is described as the Sabatier principle.

• T. True
• F. False

Answer: True

The Sabatier principle states that optimal catalytic activity requires intermediate binding strength between catalyst and substrate. If binding is too weak, the substrate cannot be activated (no reaction). If binding is too strong, the product cannot dissociate (catalyst poisoning). Wilkinson's catalyst illustrates this: PPh₃ dissociation creates a coordinatively unsaturated site for substrate binding, but the remaining ligands keep the metal center electron-rich enough to undergo oxidative addition of H₂. After hydrogenation, reductive elimination releases the product because the thermodynamic driving force of alkane formation exceeds the binding energy.

Question 4

Trace the complete catalytic cycle for hydrogenation of ethylene by Wilkinson's catalyst, identifying each elementary step and the electron count at each stage.

Model answer: 1) [RhCl(PPh₃)₃] (16e, Rh⁺) loses one PPh₃ to form [RhCl(PPh₃)₂] (14e, Rh⁺). 2) Oxidative addition of H₂: [RhH₂Cl(PPh₃)₂] (16e, Rh³⁺). 3) Coordination of ethylene: [RhH₂Cl(C₂H₄)(PPh₃)₂] (18e, Rh³⁺). 4) Migratory insertion: ethylene inserts into one Rh-H bond to form [Rh(C₂H₅)HCl(PPh₃)₂] (16e, Rh³⁺). 5) Reductive elimination: C₂H₅ and H couple to release C₂H₆, regenerating [RhCl(PPh₃)₂] (14e, Rh⁺). The PPh₃ re-coordinates to reform the resting state. Each step alternates the electron count between 14-18 electrons, and the oxidation state cycles between Rh(I) and Rh(III).

The elegance of this cycle is that every step is a well-understood elementary reaction. No new chemistry is invented — the catalyst simply orchestrates a sequence of oxidative additions, insertions, and eliminations in the correct order. Understanding the elementary steps enables modification: changing ligands tunes selectivity, changing the metal can change which substrates are activated.