Homogeneous Alkene Hydrogenation

Introduction

Adding hydrogen to a C=C bond is thermodynamically favourable, releasing roughly 120 kJ mol⁻¹ for a simple alkene, but H₂ does not react with alkenes on its own at a useful rate. Heterogeneous metals such as palladium on carbon have long catalysed the reaction. In 1965 Wilkinson's group reported RhCl(PPh₃)₃, which does the job in solution at room temperature and atmospheric pressure. Homogeneous hydrogenation is now one of the best-understood catalytic cycles and a foundation for asymmetric catalysis.

Core explanation

The catalyst. RhCl(PPh₃)₃ is a red, square-planar Rh(I), d⁸, 16-electron complex. It is a precatalyst: in solution it loses one PPh₃ to give the highly reactive 14-electron RhCl(PPh₃)₂.

The hydride route cycle.

1. Ligand dissociation: RhCl(PPh₃)₃ → RhCl(PPh₃)₂ + PPh₃ (16 → 14 e⁻). 2. Oxidative addition of H₂: RhCl(PPh₃)₂ + H₂ → RhH₂Cl(PPh₃)₂, Rh(III), 16 e⁻. The two hydrides are cis. 3. Alkene coordination: the alkene binds to give RhH₂Cl(PPh₃)₂(alkene), 18 e⁻. 4. Migratory insertion: the alkene inserts into a cis Rh–H bond, giving an alkyl hydride, 16 e⁻. 5. Reductive elimination: the alkyl and remaining hydride combine to release the alkane, regenerating RhCl(PPh₃)₂, Rh(I), 14 e⁻.

Because H₂ adds first, this is called the hydride route . Insertion of the alkene is often turnover-limiting in Wilkinson's system.

Stereochemistry. Both hydrogens come from the metal and are delivered from the face of the alkene bound to rhodium, so addition is syn . For example, hydrogenation of an alkyne-derived cis-alkene labelled with deuterium gives the erythro/threo pattern expected for syn delivery.

Selectivity. The bulky PPh₃ groups make binding of crowded alkenes difficult. Rates fall steeply with substitution: terminal and cis-disubstituted alkenes react fastest, trisubstituted alkenes slowly and tetrasubstituted alkenes hardly at all. This allows selective reduction of the least hindered C=C bond in a polyene. Carbonyl groups, nitro groups and aromatic rings are usually untouched.

Unsaturate route. Cationic rhodium diphosphine catalysts such as [Rh(diphosphine)(solvent)₂]⁺ often bind the alkene first and then add H₂. This order matters in asymmetric hydrogenation, where the alkene–catalyst complex sets the stereochemistry.

Crabtree's catalyst. [Ir(COD)(PCy₃)(py)]⁺ with a weakly coordinating anion is highly unsaturated and electrophilic. It hydrogenates tri- and tetrasubstituted alkenes and can be directed by nearby hydroxyl groups binding to iridium, delivering H₂ to one face with high diastereoselectivity.

Limitations. Wilkinson's catalyst decarbonylates aldehydes, forming inactive RhCl(CO)(PPh₃)₂, and it is poisoned by strongly binding groups such as thiols. It can also isomerise some alkenes via reversible insertion and β-hydride elimination.

Step-by-step reasoning

To follow a Wilkinson hydrogenation:

1. Remove one PPh₃ to reach 14 electrons. 2. Add H₂, raising Rh from +1 to +3. 3. Bind the alkene, reaching 18 electrons. 4. Insert the alkene into Rh–H to give an alkyl hydride. 5. Reductively eliminate the alkane, returning to Rh(I).

Visual explanation

The SIM-CAT-001 cycle shows rhodium moving around the loop, with oxidation-state labels switching between +1 and +3. The two hydrogen atoms are coloured to show that both end up on the same face of the former alkene, making syn addition visible.

Real-world analogy

Imagine a stapler loaded with two staples (the hydrides). A sheet of paper (the alkene) is placed in the jaws, both staples go in from the same side, and the stapler reloads for the next sheet.

Real-world example

Selective homogeneous hydrogenation is used in fine chemical and pharmaceutical synthesis where only one of several double bonds must be reduced, such as reducing a terminal alkene in the presence of a trisubstituted one in terpene chemistry. Large-scale hydrogenations use flammable hydrogen, so industry runs them in pressure-rated, well-ventilated plant with strict ignition controls.

Why?

Why must a phosphine leave before catalysis begins? The 16-electron precatalyst has room for either H₂ or the alkene, but not both. Losing PPh₃ provides the second site, so the metal can hold two hydrides and an alkene at the same time.

Common misconception

"Wilkinson's catalyst reduces any unsaturated group." It is selective for less hindered C=C and C≡C bonds; ketones, esters, nitro groups and arenes are generally left intact, which is one of its main advantages.

Worked example

Question: Give the oxidation state and electron count of rhodium in RhH₂Cl(PPh₃)₂ and in RhH₂Cl(PPh₃)₂(CH₂=CHR).

Reasoning: Two H and one Cl are X ligands, so Rh is +3, d⁶. Neutral count: Rh 9 + 2 H + Cl 1 + 2 PPh₃ 4 = 16. Adding the alkene (2) gives 18.

Answer: Rh(III), 16 electrons; Rh(III), 18 electrons.

Quick check

1. Why is hydrogen added syn in homogeneous hydrogenation by Wilkinson's catalyst? Answer: Both hydrogen atoms are transferred from the rhodium to the same alkene face that is coordinated to the metal.

Exam focus

Draw the hydride-route cycle with oxidation states and electron counts at every stage, explain the role of phosphine loss, and state the substrate selectivity. Contrast the hydride and unsaturate routes and mention Crabtree's catalyst for hindered alkenes.

Advanced insight

Halpern's kinetic studies showed that the most abundant rhodium species in solution are not necessarily the ones on the productive cycle. Dimers such as [RhCl(PPh₃)₂]₂ and the observable dihydride RhH₂Cl(PPh₃)₃ are largely off-cycle, and the productive intermediates are present at tiny concentrations. This insight, that the species you can see may not be the species doing the work, recurs throughout mechanistic catalysis.

Summary

Wilkinson's catalyst, RhCl(PPh₃)₃, loses PPh₃, oxidatively adds H₂ to give a cis-dihydride, binds the alkene, inserts it into Rh–H and reductively eliminates the alkane. Rhodium cycles between +1 and +3. Addition is syn, and the catalyst favours less hindered alkenes. Cationic rhodium systems may follow the unsaturate route, and Crabtree's iridium catalyst reduces hindered alkenes.

Practice questions

1. What is the role of phosphine dissociation in Wilkinson's catalyst? Answer: It creates the extra vacant site needed to hold both the hydrides and the alkene. 2. Rank ethene, cis-2-butene and 2,3-dimethyl-2-butene by rate of hydrogenation with Wilkinson's catalyst. Answer: Ethene > cis-2-butene > 2,3-dimethyl-2-butene, because steric hindrance increases with substitution. 3. Why does Wilkinson's catalyst become inactive in the presence of aldehydes? Answer: It decarbonylates the aldehyde, forming the stable carbonyl complex RhCl(CO)(PPh₃)₂, which is inactive. 4. How does the unsaturate route differ from the hydride route? Answer: In the unsaturate route the alkene binds before H₂ adds; in the hydride route H₂ adds first.