Advanced MCQs on Wilkinson's Catalyst with Explanations

Wilkinson's Catalyst: Advanced Chemistry MCQs

Must Read Wilkinson's Catalyst: Mechanism, Catalytic Hydrogenation, and Kinetics

1. In the catalytic cycle of hydrogenation using Wilkinson's catalyst, [RhCl(PPh₃)₃], what is the coordination number and oxidation state of the active catalytic species that directly binds to the alkene?

A) 4, +1

B) 5, +3

C) 3, +1

D) 4, +3

View Answer & Explanation

Correct Answer: C) 3, +1

Explanation: The precursor complex is the 16-electron, square planar [RhCl(PPh₃)₃]. However, it is too sterically hindered to act as the true active catalyst. In solution, it undergoes rapid reversible dissociation of a bulky phosphine ligand to form a 14-electron, 3-coordinate active species: RhCl(PPh₃)₃ ⇌ RhCl(PPh₃)₂ + PPh₃. In this active species, Rhodium retains its original +1 oxidation state. This highly unsaturated 3-coordinate intermediate is what undergoes subsequent oxidative addition of H₂.

Exam Reference: Similar to foundational mechanics tested in CSIR NET Chemical Sciences (India).

2. Wilkinson’s Catalyst shows highly selective kinetic rates for alkene hydrogenation. Which of the following alkenes will undergo hydrogenation at the fastest rate?

A) 2-Methylpropene

B) Cyclohexene

C) Z-But-2-ene

D) Ethylene

View Answer & Explanation

Correct Answer: D) Ethylene

Explanation: The rate of hydrogenation by Wilkinson's catalyst is dictated entirely by steric hindrance during the alkene coordination step to the metal center. Unsubstituted, terminal, or sterically unhindered alkenes coordinate much faster than substituted or cyclic ones. Ethylene (ethene) has zero substitution and experiences the lowest steric barrier, allowing it to react exponentially faster. The general rate order is: Ethylene > Terminal alkenes > Cis-internal alkenes > Trans-internal alkenes > Trisubstituted alkenes.

Exam Reference: Adapted from standard organometallic kinetics problems seen in GRE Chemistry Test (USA).

3. During the oxidative addition of H₂ to the active form of Wilkinson's catalyst, what is the stereochemical geometry of the resulting dihydride intermediate?

A) Cis-dihydride, octahedral

B) Trans-dihydride, trigonal bipyramidal

C) Cis-dihydride, trigonal bipyramidal

D) Trans-dihydride, octahedral

View Answer & Explanation

Correct Answer: C) Cis-dihydride, trigonal bipyramidal

Explanation: When H₂ adds via oxidative addition to the 14-electron active catalyst [RhCl(PPh₃)₂], the addition happens in a concerted, cis-fashion. This changes the metal's oxidation state from Rh(I) to Rh(III) and boosts the coordination number from 3 to 5, yielding a 16-electron trigonal bipyramidal (or distorted square pyramidal) complex where the two hydride ligands sit cis to one another.

Exam Reference: Highly representative of advanced stereochemistry questions in GATE Chemistry (India).

4. Which of the following statements regarding the role of oxygen (O₂) in reactions using Wilkinson’s catalyst is correct?

A) Molecular oxygen acts as a co-catalyst by accelerating phosphine dissociation.

B) O₂ oxidizes the PPh₃ ligands to O=PPh₃, irreversibly destroying catalytic activity.

C) O₂ stabilizes the Rh(III) intermediate to prevent reductive elimination.

D) O₂ acts as a sacrificial electron acceptor to keep Rhodium in a +1 state.

View Answer & Explanation

Correct Answer: B) O₂ oxidizes the PPh₃ ligands to O=PPh₃, irreversibly destroying catalytic activity.

Explanation: Wilkinson’s catalyst is exceptionally air-sensitive in solution. Molecular oxygen (O₂) chemically attacks the coordinated or dissociated triphenylphosphine (PPh₃) ligands, converting them into triphenylphosphine oxide (O=PPh₃). Because O=PPh₃ is a very poor ligand for Rh(I), the complex decomposes into inactive clusters, effectively poisoning the catalyst. Thus, these reactions must strictly be performed under an inert atmosphere (N₂ or Ar).

Exam Reference: Oxford University Chemistry Primer / Organic Synthesis Exams (UK).

5. Wilkinson's catalyst can be used to convert an acyl chloride (R-COCl) into an alkyl chloride (R-Cl). What type of reaction is this, and what is the stoichiometric byproduct?

A) Decarboxylation; CO₂

B) Decarbonylation; [RhCl₂(CO)(PPh₃)₂]

C) Decarbonylation; trans-[RhCl(CO)(PPh₃)₂]

D) Hydroformylation; HCHO

View Answer & Explanation

Correct Answer: C) Decarbonylation; trans-[RhCl(CO)(PPh₃)₂]

Explanation: Beyond hydrogenation, Wilkinson's catalyst can perform stoichiometric decarbonylation of aldehydes and acyl chlorides. The acyl chloride undergoes oxidative addition to the Rh(I) center, followed by a migration step. Finally, it undergoes reductive elimination to yield the alkyl chloride (R-Cl). Crucially, the carbonyl group (CO) stays bound to the Rhodium atom, transforming the catalyst into the exceptionally stable, inactive complex trans-[RhCl(CO)(PPh₃)₂] (Vaska's complex analog). Because the CO ligand binds too tightly, the reaction is typically stoichiometric rather than catalytic unless performed at extreme temperatures (>200°C).

Exam Reference: Mimics Advanced Synthetic Organometallic questions in MIT/Caltech Graduate Exams (USA).

6. If the triphenylphosphine (PPh₃) ligands in Wilkinson's catalyst are replaced with a trialkylphosphine ligand like triethylphosphine (PEt₃), what is the observed outcome on the hydrogenation rate of standard olefins?

A) The rate increases due to enhanced electron density on the Rhodium center.

B) The rate decreases dramatically because the trialkylphosphines bind too strongly to dissociate.

C) The rate remains unchanged because the coordination number remains the same.

D) The catalyst becomes a heterogeneous mixture and precipitates out.

View Answer & Explanation

Correct Answer: B) The rate decreases dramatically because the trialkylphosphines bind too strongly to dissociate.

Explanation: Trialkylphosphines (PMe₃, PEt₃) are much stronger σ-donors and are sterically much smaller than the bulky PPh₃. Because they are smaller and more basic, they bind tightly to the Rh center. This heavily suppresses the initial, critical dissociation equilibrium step required to make the 14-electron active intermediate. Without ligand dissociation, the catalyst cannot bind H₂ or the alkene efficiently, rendering it virtually inactive at room temperature.

Exam Reference: Classic coordination chemistry conceptual check found in Joint Admission Test for M.Sc. - IIT JAM (India).

7. What is the correct chronological sequence of fundamental steps occurring in the inner-sphere catalytic cycle of alkene hydrogenation using Wilkinson's Catalyst?

A) Alkene Coordination → Oxidative Addition → Migratory Insertion → Reductive Elimination

B) Oxidative Addition → Alkene Coordination → Migratory Insertion → Reductive Elimination

C) Oxidative Addition → Migratory Insertion → Alkene Coordination → Reductive Elimination

D) Alkene Coordination → Migratory Insertion → Oxidative Addition → Reductive Elimination

View Answer & Explanation

Correct Answer: B) Oxidative Addition → Alkene Coordination → Migratory Insertion → Reductive Elimination

Explanation: The universally accepted mechanism for Wilkinson's hydrogenation follows this sequence: (1) Phosphine Dissociation to make the active 14e catalyst, (2) Oxidative Addition of H₂ to form a 16e dihydride intermediate, (3) Alkene Coordination to step up to an 18e complex, (4) Migratory Insertion of one hydride to create an alkyl group (16e), and (5) Reductive Elimination of the second hydride and alkyl group to yield the final alkane.

Exam Reference: Standard textbook mechanism question universally asked in CSIR-NET (India) and Tokyo University Chemistry Entrance (Japan).

8. When a mixture of a maleate ester (cis-alkene) and a fumarate ester (trans-alkene) is subjected to deuteration (D₂) using Wilkinson's catalyst, the stereochemical outcome of the deuterated product is:

A) Exclusive anti-addition yielding a racemic mixture

B) Exclusive syn-addition yielding stereospecific products

C) Non-stereospecific addition yielding a 50:50 mixture of diastereomers

D) No addition occurs because esters poison the catalyst

View Answer & Explanation

Correct Answer: B) Exclusive syn-addition yielding stereospecific products

Explanation: The migratory insertion and subsequent reductive elimination steps within the inner sphere of the Rhodium complex happen on the same face of the coordinated alkene. Because both Deuterium atoms (D) are transferred sequentially from the metal center to the face of the alkene facing the metal, the reaction proceeds with exclusive syn-addition stereospecificity.

Exam Reference: Advanced Organic Synthesis modules at German Universities (e.g., LMU Munich / Heidelberg).

9. Wilkinson's catalyst can be modified to perform asymmetric (enantioselective) hydrogenation of prochiral alkenes. What modification is required to achieve this?

A) Replacing Cl^- with a chiral counterion like tartrate.

B) Replacing the achiral PPh₃ ligands with chiral phosphine ligands like DIOP or BINAP.

C) Conducting the reaction under a circularly polarized UV light source.

D) Using a chiral protic solvent like (R)-2-butanol.

View Answer & Explanation

Correct Answer: B) Replacing the achiral PPh₃ ligands with chiral phosphine ligands like DIOP or BINAP.

Explanation: By exchanging the achiral triphenylphosphine (PPh₃) ligands with optically active, enantiopure chiral diphosphine ligands (such as DIOP, CHIRAPHOS, or BINAP), the environment around the Rhodium atom becomes asymmetric. This configuration forces the syn-dihydride delivery to occur enantioselectively, yielding products with high enantiomeric excess (ee%). This discovery formed the foundation of the 2001 Nobel Prize in Chemistry.

Exam Reference: International Chemistry Olympiad (IChO) Finals / Nobel-Prize Related University Exam Problems.

10. Consider the complex [RhCl(PPh₃)₃]. What are the total valence electron count (TVEC) of the central metal and its magnetic behavior?

A) 16-electron, Diamagnetic

B) 18-electron, Paramagnetic

C) 16-electron, Paramagnetic

D) 14-electron, Diamagnetic

View Answer & Explanation

Correct Answer: A) 16-electron, Diamagnetic

Explanation: Calculating via the covalent/neutral method: Rhodium (Rh) belongs to Group 9 (9e), Chlorine provides 1e, and three PPh₃ ligands provide 3 × 2 = 6e. Total valence electron count = 9 + 1 + 6 = 16 electrons. For a d⁸ transition metal like Rh(I), a 16-electron count favors a square-planar geometry. In square planar arrangements of d⁸ complexes, the orbital split creates a massive energy gap before the highest d_x²-y² orbital. All 8 d-electrons pair up perfectly in the lower four orbitals, rendering the system diamagnetic.

Exam Reference: Standard coordination chemistry question widely featured in CSIR-NET and GATE examinations.