Wilkinson catalyst hydrogenation mechanism
Trace the Wilkinson catalyst cycle step by step: electron counts, oxidation states, alkene selectivity, two practice questions and a one-week study plan.
This post walks you through the Wilkinson catalyst cycle one step at a time.
You will count electrons, track oxidation states, and solve two practice questions.
What the catalyst is and why it works
Wilkinson's catalyst is chlorido-tris(triphenylphosphine)rhodium(I). Ph stands for the phenyl group, C6H5. It is a red-brown solid that dissolves in solvents such as benzene. It dissolves in the reaction mixture along with the alkene, so it is a homogeneous catalyst. Nickel, palladium and platinum catalysts are usually heterogeneous: they stay as a separate solid.
Start with the structure, because the whole mechanism depends on it:
- Rhodium is +1, a d8 ion.
- The shape is square planar, slightly distorted.
- There are four ligands: one Cl and three PPh3. Each gives two electrons, so the metal has 16 electrons.
The 18-electron rule says a stable organometallic complex tends to have 18 valence electrons. A 16-electron complex is coordinatively unsaturated: it can still bind more ligands, such as an alkene or H2. That is why it works as a catalyst.
The catalytic cycle, step by step
Write the cycle in this order.
- Ligand dissociation. One PPh3 leaves. The 16-electron complex becomes a 14-electron complex with a vacant site. Loss of two ligands, giving a 12-electron species, can also occur.
- Oxidative addition of H2. The H–H bond breaks and two hydrides attach to rhodium. Rh(I) becomes Rh(III), and the electron count rises by two.
- ฯ-complexation of the alkene. The C=C ฯ cloud donates two electrons to rhodium. The alkene takes the vacant site.
- Migratory insertion. One hydride moves from rhodium to a carbon of the alkene, forming a Rh–alkyl bond. This is the rate-limiting step.
- Reductive elimination. The alkyl group and the second hydride join to form the alkane. The alkane leaves, rhodium returns to Rh(I), and the cycle restarts.
Oxidative addition and reductive elimination are opposites. In the first, oxidation state and electron count go up. In the second, both come down. Migratory insertion changes neither.
Electron count and oxidation state at each stage
Exam questions often ask you to fill a table like this one. Practise drawing it from memory.
| Species | Rh oxidation state | Electron count | Step that produces it |
|---|---|---|---|
| RhCl(PPh3)3 | +1 | 16 | Starting complex |
| RhCl(PPh3)2 | +1 | 14 | Ligand dissociation |
| RhH2Cl(PPh3)2 | +3 | 16 | Oxidative addition |
| RhH2Cl(PPh3)2(alkene) | +3 | 18 | Alkene binding |
| RhHCl(alkyl)(PPh3)2 | +3 | 16 | Migratory insertion |
| RhCl(PPh3)2 + alkane | +1 | 14 | Reductive elimination |
The 18-electron species appears only once, with the alkene bound. Every other intermediate has 14 or 16 electrons. Unsaturated intermediates stay reactive, which is what a catalyst needs.
Which alkenes react, and which do not
The insertion step is limited by crowding (steric hindrance) around the metal. So the more substituents on the C=C bond, the slower the reaction.
- Terminal and disubstituted alkenes are good substrates.
- More hindered alkenes react more slowly. Wilkinson's catalyst does not hydrogenate a tetrasubstituted alkene.
- Ethylene gives RhCl(C2H4)(PPh3)2 but is not hydrogenated.
- Alkynes are hard to control. They tend to go to the alkane, passing through the cis-alkene.
The conditions are mild. A terminal alkene with a homogeneous rhodium catalyst is hydrogenated at 1 atm H2 and room temperature. Non-catalytic hydrogenation takes place only at very high temperatures.
Know two contrasts. Crabtree's catalyst is an iridium complex, also square planar and d8. Unlike Wilkinson's catalyst, it hydrogenates tetrasubstituted alkenes. The iridium analogue of Wilkinson's catalyst, IrCl(PPh3)3, undergoes cyclometallation. This gives a saturated Ir(III) complex that is not catalytically active.
How to prepare for this in exams
This topic suits the organometallic sections of CSIR-NET, GATE, IIT-JAM and CUET-PG chemistry. Expect three kinds of questions:
- Counting: find the electron count or oxidation state of a given species.
- Step identification: a change in count or oxidation state is given, and you name the step.
- Selectivity: which substrate reacts fastest, or not at all.
First confirm that your exam or university covers the topic. The chemistry syllabus page on this site lists official syllabus links for Indian universities and institutes. You can search by university, city or state, and tick "Only with syllabus link". The links were checked on 24 October 2026, so confirm the current session on the university's own page. Look for the 18-electron rule, oxidative addition, reductive elimination and homogeneous catalysis. If a heading is missing, read the unit description again before assuming the topic is skipped.
Practice question 1: counting and naming the next step
Question. In the Wilkinson cycle for hydrogenation of a terminal alkene, consider RhH2Cl(PPh3)2, formed after H2 addition. (a) State the oxidation state of rhodium. (b) State the electron count. (c) What happens next?
Solution.
- The ligands are two H−, one Cl− and two neutral PPh3. The complex is neutral, so rhodium is +3, d6.
- Ionic count: Rh3+ gives 6, two H− give 4, Cl− gives 2, two PPh3 give 4. Total 16.
- This 16-electron species has a vacant site. The alkene binds through its ฯ bond and makes an 18-electron species. Migratory insertion follows.
Answer: (a) +3, (b) 16 electrons, (c) the alkene coordinates, then a hydride inserts into the C=C bond.
Check: the starting complex is Rh(I), d8. Rh+ (8) + Cl− (2) + three PPh3 (6) = 16. This matches the table.
Practice question 2: selectivity and a dead end
Question. Three substrates are treated separately with H2 and Wilkinson's catalyst: a terminal alkene, a tetrasubstituted alkene and ethylene. (a) Which is hydrogenated fastest? (b) Which ones are not hydrogenated, and why? (c) Give the electron count of the ethylene complex RhCl(C2H4)(PPh3)2. (d) Which catalyst from this post can hydrogenate the tetrasubstituted alkene?
Solution.
- The rate-limiting insertion step is limited by crowding at the metal. The terminal alkene has the least crowding among the two substituted alkenes, so it reacts fastest.
- The tetrasubstituted alkene is too hindered, so it is not hydrogenated. Ethylene binds to rhodium but is not a substrate for hydrogenation.
- Count the ethylene complex: Rh+ gives 8, Cl− gives 2, C2H4 gives 2, two PPh3 give 4. Total 16. It is Rh(I), square planar, like the parent catalyst.
- Crabtree's iridium catalyst hydrogenates mono-, di-, tri- and tetrasubstituted alkenes.
Answer: (a) the terminal alkene; (b) the tetrasubstituted alkene and ethylene; (c) 16 electrons; (d) Crabtree's catalyst.
A study plan for one week
- Day 1. Learn the 18-electron rule. Count five complexes by the ionic method and again by the neutral method.
- Day 2. Learn oxidative addition and reductive elimination. Write what each changes in oxidation state, electron count and coordination number.
- Day 3. Draw the full cycle from memory and fill the table above without looking.
- Day 4. Learn alkene selectivity and compare Wilkinson's and Crabtree's catalysts.
- Day 5. Redo both practice questions without the solutions. Then change the substrate and solve again.
- Day 6. Check your syllabus link and note the neighbouring topics listed.
- Day 7. Revise the table and cycle. Write down the mistakes you made this week.
Key takeaways
- Wilkinson's catalyst is a 16-electron, square planar Rh(I) d8 complex.
- The cycle: ligand loss, oxidative addition of H2, alkene binding, migratory insertion, reductive elimination.
- Rhodium goes +1 to +3 and back to +1. Only the alkene-bound species has 18 electrons.
- Insertion is rate-limiting, so crowded alkenes react slowly and tetrasubstituted ones do not react.
Frequently asked questions
Why does the catalyst lose a PPh3 first?
The 16-electron complex has no free site for the substrate. Losing one PPh3 gives a 14-electron species with a vacant site, so H2 and the alkene can bind.
Which step is rate-limiting?
Insertion of the alkene into the Rh–H bond. It is limited by crowding around the metal, so the degree of substitution on the alkene controls the rate.
Why is ethylene not hydrogenated?
Ethylene binds and gives RhCl(C2H4)(PPh3)2, but it is not a substrate for hydrogenation. Remember this as a fact for exams.
Is Wilkinson's catalyst heterogeneous?
No. It is homogeneous because it dissolves in the solution together with the alkene.
Can I use it to hydrogenate alkynes?
It is hard to control. Alkynes tend to be reduced to alkanes, with the cis-alkene as an intermediate, so stopping at the alkene is difficult.
