Organometallic Chemistry

Complete Study Notes — 47 Exam-Ready Questions

11 Topics 47 Questions 2020–2025
Topic 1: Definition & Classification
1
Define organometallic compound. Give one example of organometallic compound having no metal–carbon bond(s). 2 Marks 2023

Definition: An organometallic compound is one that contains at least one direct metal–carbon bond, where the carbon atom belongs to an organic moiety (e.g., alkyl, aryl, carbene, carbonyl, etc.). Metalloids such as B, Si, Ge, As, Sb are also considered under the purview of organometallic chemistry.

Example with no M–C bond: Wilkinson's Catalyst — [RhCl(PPh3)3]. In this compound, Rh is bonded to Cl and to P atoms of the triphenylphosphine ligands; there is no direct Rh–C bond. Despite this, it is classified as organometallic because the PPh3 ligands are organophosphorus in nature and the compound is central to organometallic chemistry. Another valid example is Ti(OCH2CH3)4 where Ti is bonded to O, not C.

2
Give example of one ionic organometallic compound and one covalent organometallic compound. 2 Marks 2025

Ionic organometallic compound: Cyclopentadienyl sodium — Na+[C5H5]. The bond between Na+ and the cyclopentadienyl anion (Cp) is purely electrostatic (ionic). The large electronegativity difference between Na (0.93) and C (2.55) favours complete electron transfer.

Covalent organometallic compound: Tetramethylsilane — Si(CH3)4 or Dimethylmercury — Hg(CH3)2. The small electronegativity difference between Si/Hg and C results in electron sharing (covalent M–C bonds).

Topic 2: Hapticity
3
Define hapticity with suitable example. 2M(2022) / 3M(2024) 2024, 2022

Hapticity is the number of contiguous atoms of a ligand that are simultaneously bonded to a single metal centre. It is denoted by the Greek letter η (eta) with a superscript, e.g., η5 means five contiguous atoms bind to the metal.

Example 1 — Ferrocene: [Fe(η5-C5H5)2]. All five carbon atoms of each cyclopentadienyl (Cp) ring bond simultaneously to the Fe centre, giving hapticity = 5.

Example 2 — Zeise's salt: K[PtCl32-C2H4)]. Both carbon atoms of ethylene bond simultaneously to Pt, giving hapticity = 2.

4
Give example of naturally occurring M–C σ-bonded species. 1 Mark 2024

Vitamin B12 coenzyme — specifically Methylcobalamin or Adenosylcobalamin.

In methylcobalamin, the cobalt (Co) atom forms a direct σ-covalent bond with the carbon of a methyl group (Co–CH3). In adenosylcobalamin, Co forms a σ-bond with the 5′-carbon of the 5′-deoxyadenosyl group. These are the only known naturally occurring organometallic compounds with direct metal–carbon σ-bonds.

5
Give one example of π-bonded organometallic compound. 1 Mark 2023

Ferrocene[Fe(η5-C5H5)2].

In ferrocene there are no localized Fe–C σ bonds. Instead, the Fe centre interacts with the delocalized π orbitals of both cyclopentadienyl rings. The bonding is entirely of π-character — the metal donates electron density back into the π* orbitals of the ring, making this a classic π-bonded organometallic compound.

Topic 3: 18-Electron & 16-Electron Rules
6
What is the significance of the 18-electron rule? Sketch the MO diagram of Cr(CO)6. 5 Marks 2022

Significance of the 18e rule:

  • Predicts stability: Complexes with 18 valence electrons (filling all bonding and non-bonding MOs) are thermodynamically stable and kinetically inert.
  • Predicts stoichiometry: Helps determine the number and type of ligands a metal can accommodate.
  • Predicts reactivity: Complexes with fewer than 18e tend to be reactive (seek additional ligands); those with more than 18e are rare and unstable.

MO diagram of Cr(CO)6:

Cr(CO)6: Cr is in zero oxidation state, d6 configuration. Each CO donates 2e via σ-donation. Total = 6 (Cr) + 6 × 2 (6 CO) = 18e.

MO picture (Oh symmetry):

  • 6 σ-bonding MOs (from Cr 4s, three 4p, and two 3d eg orbitals): filled by 12 electrons from the 6 CO ligands (each CO donates a lone pair).
  • t2g (dxy, dyz, dxz): these are non-bonding in pure crystal field theory, but are strongly stabilised by π-backbonding — metal d-electrons are donated into the π* orbitals of CO. These 3 orbitals hold the 6 metal d-electrons.
  • eg* (dx²−y², d): strongly antibonding σ* orbitals. Large Δo gap between t2g and eg* ensures the eg* remains empty.

The large Δo (enhanced by π-backbonding) makes it energetically favourable to fill exactly up to t2g (18 electrons) and leave eg* empty — hence the 18e rule is satisfied and the complex is stable.

7
Give an example of a 16-electron compound and draw its MO diagram. 5 Marks 2023

Example: Vaska's Complex — [IrCl(CO)(PPh3)2]

Electron count: Ir(I) d8 = 8e + Cl(2e) + CO(2e) + 2 PPh3(4e) = 16e

MO diagram (Square Planar, D4h):

In square planar geometry, the five d-orbitals split as follows (from lowest to highest energy):

  1. dxz — lowest (non-bonding)
  2. dyz — non-bonding
  3. d — weakly bonding/antibonding
  4. dxy — antibonding
  5. dx²−y² — extremely high energy (strongly antibonding, points directly at the four ligands)

The 8 d-electrons of Ir(I) fill the four lowest d-orbitals completely. The 5th orbital (dx²−y²) is so high in energy that filling it is extremely unfavourable. Thus, the stable configuration has only 16 electrons — the 18e rule is not followed because the 9th orbital is too antibonding to occupy.

8
Key feature of 16-electron organometallic compound. 1 Mark 2025

The key feature of 16e complexes is their high susceptibility to oxidative addition reactions. The empty coordination site (vacant orbital) allows small molecules (H2, CH3I, etc.) to add across the metal centre, increasing both the electron count by 2 (to reach 18e) and the oxidation state by +2.

Topic 4: Applications of Electron Counting Rules
9
Find ‘n’ in: (i) Na2[Fe(CO)n], (ii) [W(η6-C6H6)(CO)n], (iii) [Fe(η5-Cp)(η1-Cp)(CO)n]. 5 Marks 2021

(i) Na2[Fe(CO)n]:

Fe in anionic complex: Fe contributes 8e, charge = 2− adds 2e. Each CO = 2e.

8 (Fe) + 2 (charge) + 2n (CO) = 18 → 10 + 2n = 18 → n = 4

Answer: n = 4, i.e., Na2[Fe(CO)4]

(ii) [W(η6-C6H6)(CO)n]:

W contributes 6e, η6-C6H6 = 6e. Each CO = 2e.

6 (W) + 6 (η6-benzene) + 2n (CO) = 18 → 12 + 2n = 18 → n = 3

Answer: n = 3, i.e., [W(η6-C6H6)(CO)3]

(iii) [Fe(η5-Cp)(η1-Cp)(CO)n]:

Fe = 8e, η5-Cp = 5e, η1-Cp = 1e (only one C bonded). Each CO = 2e.

8 (Fe) + 5 (η5-Cp) + 1 (η1-Cp) + 2n (CO) = 18 → 14 + 2n = 18 → n = 2

Answer: n = 2, i.e., [Fe(η5-Cp)(η1-Cp)(CO)2]

10
Count electrons: (A) 5-C5H5)Fe(CO)2Cl   (B) [PtCl32-C2H4)] 2 Marks 2025

(A) 5-C5H5)Fe(CO)2Cl:

Fe (8) + η5-Cp (5) + 2 CO (4) + Cl (1) = 18e

Answer: 18 electrons

(B) [PtCl32-C2H4)]:

Pt (10) + 3 Cl (3) + η2-C2H4 (2) + charge (1) = 16e

Answer: 16 electrons

11
Find x and y: Fe(η5-Cp)(CO)x(NO)y 2 Marks 2024

Fe = 8e, η5-Cp = 5e. Total so far = 13e. Need 5 more electrons to reach 18e.

Each CO donates 2e, each linear NO donates 3e.

2x + 3y = 5

Testing small integers: x = 1, y = 1 → 2(1) + 3(1) = 5 ✓

Answer: x = 1, y = 1, i.e., Fe(η5-Cp)(CO)(NO)

12
Find a metal nitrosyl isoelectronic with Ni(CO)4. 2 Marks 2023

Ni(CO)4 has 18 electrons: Ni(0) = 10e + 4 × CO(2e) = 8e → total = 18e.

For a metal nitrosyl M(NO)n with 18 electrons:

Each linear NO donates 3 electrons. If n = 4: 4 × 3 = 12e from NO. Need 18 − 12 = 6e from the metal.

Metal with 6 valence electrons = Cr (Group 6, Cr(0) = d6).

Answer: Cr(NO)4 is isoelectronic with Ni(CO)4 (both 18e, both tetrahedral).

13
Ni(CO)4 is stable but [Zn(CO)4]2+ does not exist. Why? 2 Marks 2024

Ni(CO)4: Ni is in the zero oxidation state (d10). Being in a low oxidation state, Ni has expanded 3d orbitals with diffuse electron density that can effectively overlap with the π* orbitals of CO. This extensive π-backbonding strengthens the Ni–C bond and stabilises the complex.

[Zn(CO)4]2+: Although Zn2+ is also d10, the +2 charge creates a very high effective nuclear charge (Zeff). This contracts the 3d orbitals dramatically, making them too small and too tightly held to overlap with the π* orbitals of CO. Without π-backbonding, the Zn–CO bond would be purely σ (weak), which is insufficient to sustain the complex. Hence, [Zn(CO)4]2+ does not exist.

Topic 5: Preparation of Metal Carbonyls
14
Fe + CO at high T, P → ? 2 Marks 2024

Iron reacts with carbon monoxide under high temperature and pressure to form iron pentacarbonyl:

Fe(s) + 5 CO(g) → Fe(CO)5(l)    [200°C, 200 atm]

The product Fe(CO)5 is a volatile, pale yellow liquid. It is a monomeric carbonyl with trigonal bipyramidal geometry.

15
Identify products A and B: 2Fe(CO)5 → A;   Ni(CO)4 + PF3 → B 2 Marks 2024

A — Diiron nonacarbonyl:

2 Fe(CO)5hν/Δ Fe2(CO)9 + CO

Thermal or photolytic dimerisation of Fe(CO)5 produces Fe2(CO)9 with loss of one CO.

B — Substituted carbonyl:

Ni(CO)4 + PF3 → Ni(CO)3(PF3) + CO

PF3 (a π-acceptor ligand similar to CO) substitutes one CO ligand from Ni(CO)4.

16
Structure and magnetic behaviour of Fe2(CO)9. 2 Marks 2022

Structure: Fe2(CO)9 has D3h geometry. It features:

  • One Fe–Fe metal–metal bond
  • 3 bridging CO ligands (μ2-CO) connecting both Fe atoms
  • 6 terminal CO ligands (3 on each Fe)

Magnetic behaviour: Diamagnetic. Each Fe achieves 18e count (Fe–Fe bond contributes 1e to each Fe; 3 terminal CO = 6e; 3 bridging CO = 3e; 1 Fe–Fe bond = 1e; Fe(0) = 8e → total = 18e). With 18e, all electrons are paired.

17
Structures and CO bonding in Mn2(CO)10 and Fe3(CO)12. 2.5 Marks 2021

Mn2(CO)10:

  • Structure: Two square-pyramidal Mn(CO)5 units joined by a direct Mn–Mn bond in a staggered conformation (D4d symmetry).
  • CO bonding: All 10 CO ligands are terminal (η1). There are zero bridging CO ligands.
  • Each Mn achieves 18e: Mn(0) = 7e + 5 CO = 10e + Mn–Mn bond = 1e = 18e.

Fe3(CO)12:

  • Structure: Triangular Fe cluster (C2v symmetry). The three Fe atoms form a triangle with three Fe–Fe bonds.
  • CO bonding: 2 bridging CO (μ2-CO) across one Fe–Fe edge, and 10 terminal CO ligands.
  • Each Fe achieves 18e via metal–metal bonds and terminal/bridging CO.
18
Structure of Co2(CO)8 in solid state. 2 Marks 2024

In the solid state, Co2(CO)8 adopts the bridged isomer (C2v symmetry):

  • One Co–Co metal–metal bond
  • 2 bridging CO ligands (μ2-CO) spanning the two Co atoms
  • 6 terminal CO ligands (3 on each Co)

Each Co achieves 18e: Co(0) = 9e + 3 terminal CO = 6e + 2 bridging CO = 2e + Co–Co bond = 1e = 18e.

Note: In solution, an equilibrium exists between the bridged form and an unbridged form (with all 8 terminal CO and a Co–Co bond).

Topic 7: Synergic Effect & IR Spectroscopy
19
Bonding motifs of CO in metal carbonyls. 2 Marks 2022, 2020

CO bonds to metals in three primary motifs:

  • Terminal (η1, M≡C≡O): CO bonded to a single metal through carbon only. νCO range: 2125–1850 cm−1.
  • Doubly bridging (μ2-CO): CO bridges two metal atoms. νCO range: 1850–1700 cm−1.
  • Triply bridging (μ3-CO): CO bridges three metal atoms in a face-capping manner. νCO range: 1700–1600 cm−1.

As the number of metals bonded to CO increases, more backbonding occurs into π*, weakening the C–O bond and lowering νCO.

20
Effect of metal oxidation state and co-ligand on νCO. 3 Marks 2022, 2020

Effect of metal oxidation state:

Lower oxidation state → more d-electron density on metal → more π-backbonding into CO π* → weaker C–O bond → lower νCO.

[Mn(CO)6]+ (2090 cm−1) > Cr(CO)6 (2000 cm−1) > [V(CO)6] (1860 cm−1)

Mn+ holds d-electrons tightly (minimal backbonding), Cr0 backbonds moderately, V backbonds maximally.

Effect of co-ligand:

  • Strong σ-donor (e.g., PEt3): pumps electron density into the metal → more backbonding to CO → lower νCO.
  • Strong π-acceptor (e.g., PF3): competes with CO for metal d-electrons → less backbonding to CO → higher νCO.
21
What is synergic effect? Give example. 2 Marks 2022

The synergic effect is the mutual reinforcement between two bonding components:

  1. Forward σ-donation: The ligand donates its lone pair into an empty metal orbital (ligand → metal).
  2. Backward π-donation (backbonding): The metal donates d-electrons into the vacant π* orbital of the ligand (metal → ligand).

σ-donation increases electron density on the metal, which in turn strengthens π-backdonation. Conversely, π-backdonation removes electron density from the metal, allowing more σ-donation. Each process reinforces the other — hence "synergic".

Example: In Ni(CO)4, CO donates its lone pair to Ni (σ-donation) and Ni back-donates into the π* orbital of CO (π-backdonation). This synergic bonding is what stabilises the complex.

22
M(PEt3)3(CO)3 shows νCO at 2090/2055 but M(PF3)3(CO)3 at 1937/1847 cm−1. Which is stronger π-bonding? 5 Marks 2021

Analysis:

PEt3 is a strong σ-donor but a poor π-acceptor. It floods the metal with electron density. This excess electron density is offloaded onto the CO ligands via π-backbonding, weakening the C–O bond and raising νCO (the reported 2090/2055 values indicate weak backbonding to CO relative to the other complex).

PF3 is a strong π-acceptor (electronegative F atoms lower the P–F σ* orbitals, making them accessible for metal–→ligand backdonation). PF3 competes with CO for the metal's d-electron density. When PF3 wins this competition, less electron density is available for backbonding into CO π*, so the C–O bond retains more triple-bond character and νCO should be higher.

However, the data shows PF3 complex at lower νCO (1937/1847). This apparent contradiction arises because PF3’s strong π-acceptance also drains electron density from the metal, reducing overall backbonding to CO. The lower νCO in the PF3 complex can be explained if PF3 actually functions as both a good σ-donor and π-acceptor in this system, and the overall electron redistribution favors CO backbonding. The exam interpretation typically concludes:

PF3 is the stronger π-bonding (π-accepting) ligand. The key principle is that PF3 has low-lying acceptor orbitals (like CO) enabling significant π-interaction with the metal, whereas PEt3 is primarily a σ-donor with negligible π-acidity.

23
Comment on νCO values: Cr(CO)6 (1984), [Mn(CO)6]+ (2094), free CO (2143 cm−1). 3 Marks 2024

Free CO (2143 cm−1): No metal is present, so there is no backbonding. The C–O bond retains its full triple-bond character (bond order 3.0), giving the highest stretching frequency.

[Mn(CO)6]+ (2094 cm−1): Mn is in the +1 oxidation state. The positive charge tightly holds the d-electrons (high Zeff), resulting in minimal π-backbonding into CO π*. The C–O bond is only slightly weakened compared to free CO.

Cr(CO)6 (1984 cm−1): Cr is in the zero oxidation state (electron-rich). Massive synergic π-backbonding from Cr(0) into CO π* populates the antibonding orbital of CO, significantly weakening the C–O bond and dropping the frequency by ~160 cm−1 from the cationic complex.

Free CO (2143) > [Mn(CO)6]+ (2094) > Cr(CO)6 (1984)

Trend: lower metal oxidation state → more backbonding → lower νCO.

24
Which has the shortest C–O bond: [V(CO)6] or [Mn(CO)6]+? 3 Marks 2023

[Mn(CO)6]+ has the shortest C–O bond.

Reasoning:

  • [Mn(CO)6]+: Mn is in the +1 oxidation state. The positive charge increases Zeff, tightly holding d-electrons. Very little π-backdonation occurs into CO π*. The C–O bond retains maximum triple-bond character → shortest, strongest C–O bond.
  • [V(CO)6]: V is in the −1 oxidation state (extremely electron-rich). Profuse π-backdonation from V into CO π* populates the antibonding orbital, weakening and lengthening the C–O bond.
Shorter C–O: [Mn(CO)6]+   |   Longer C–O: [V(CO)6]

Shorter C–O bond ⇔ less backbonding ⇔ higher νCO ⇔ higher oxidation state.

Topic 8: Zeise's Salt
25
What is Zeise's salt? Synthesis, structure, nature of M–olefin interaction? 5 Marks 2022, 2020

Zeise's salt: K[PtCl32-C2H4)]·H2O — Potassium trichlorido(η2-ethylene)platinate(II). It is one of the first known organometallic compounds (discovered 1827 by William Zeise).

Synthesis:

K2[PtCl4] + C2H4SnCl2 K[PtCl32-C2H4)] + KCl

SnCl2 acts as a catalyst, facilitating the substitution of one chloride by ethylene.

Structure:

  • Pt(II) is in a square planar geometry (dsp2 hybridisation).
  • Three Cl ligands occupy three corners of the square plane.
  • The 4th coordination site is occupied by the centre of the ethylene C=C double bond.
  • The C–C axis of ethylene is perpendicular to the PtCl3 plane.
  • C=C bond lengthens from 1.34 Å (free ethylene) to ~1.375 Å in the complex.

Nature of M–olefin interaction (Dewar–Chatt–Duncanson model):

  1. σ-Donation: The filled π-orbital of ethylene donates electron density into the vacant dsp2 hybrid orbital of Pt(II) (ethylene → metal).
  2. π-Backdonation: The filled d-orbital of Pt(II) back-donates electron density into the vacant π* orbital of ethylene (metal → ethylene).

This synergic bonding weakens and lengthens the C=C bond while strengthening the Pt–ethylene interaction.

26
Give one synthetic route to Zeise's salt. 2 Marks 2024
K2[PtCl4] + C2H4SnCl2 cat. K[PtCl32-C2H4)] + KCl

Potassium tetrachloroplatinate(II) reacts with ethylene in the presence of SnCl2 catalyst. One Cl is replaced by η2-ethylene, giving the anionic complex [PtCl32-C2H4)] which precipitates as the potassium salt.

Topic 9: Ferrocene
27
How is ferrocene obtained? 2.5M(2021) / 1M(2023) / 2M(2025) 2021, 2023, 2025

Ferrocene is synthesised via a two-step process:

Step 1 — Deprotonation of cyclopentadiene:

C5H6 + KOH → K+[C5H5] + H2O

Cyclopentadiene (CpH) is acidic (pKa ≈ 16) due to aromatic stabilisation of the Cp anion. It is deprotonated by KOH to form cyclopentadienyl potassium.

Step 2 — Reaction with iron(II) chloride:

2 K+[C5H5] + FeCl2 → [Fe(η5-C5H5)2] + 2 KCl

Two equivalents of Cp react with Fe2+ to form ferrocene — an orange crystalline solid.

28
Structure of ferrocene and ruthenocene with conformation. 2.5 Marks 2021

Ferrocene [Fe(η5-C5H5)2]:

  • Sandwich structure: Fe atom centred between two parallel cyclopentadienyl (Cp) rings.
  • Gas phase: Eclipsed conformation (D5h symmetry) — C atoms of one ring align with those of the other.
  • Solid state: Staggered conformation (D5d symmetry) — C atoms of one ring are offset by 36° from those of the other, minimising crystal packing forces.
  • Fe–Cp centroid distance ≈ 1.65 Å; Cp–Cp distance ≈ 3.32 Å.

Ruthenocene [Ru(η5-C5H5)2]:

  • Same sandwich structure but with the larger Ru atom.
  • Eclipsed conformation (D5h) in all phases (gas, solid, solution). The larger Ru atom increases the Cp–Cp distance, eliminating inter-ring H···H repulsions that drive the staggered form in ferrocene. Thus, the eclipsed form (slightly favoured energetically) is always adopted.
29
Compare reactivity of ferrocene vs benzene for Friedel–Crafts acylation and Mannich condensation. 4 Marks 2023

Ferrocene is approximately 106 times more reactive than benzene toward electrophilic aromatic substitution.

Reason: The Fe centre in ferrocene donates electron density into the π-system of the Cp ring (via π-backbonding), making the ring far more electron-rich and hence more nucleophilic toward electrophiles.

Friedel–Crafts Acylation:

  • Benzene: Requires strong Lewis acid catalyst AlCl3. Even then, reaction proceeds slowly.
  • Ferrocene: AlCl3 destroys ferrocene (causes oxidation/ligand exchange). Instead, milder catalysts like H3PO4 or BF3·OEt2 are used. Reaction is fast and high-yielding.

Mannich Condensation:

  • Benzene: Unreactive. The iminium ion electrophile (R2C=NH2+) is too weak to attack benzene.
  • Ferrocene: Reacts smoothly under mild acidic conditions. The enhanced electron density of the Cp ring allows even the weak iminium electrophile to attack successfully.
30
Acylation and Mannich condensation of ferrocene. 3 Marks 2025

Friedel–Crafts Acylation:

[Fe(Cp)2] + (CH3CO)2O →H3PO4 [Fe(Cp)(η5-C5H4-COCH3)] + CH3COOH

Acetyl chloride or acetic anhydride with a mild catalyst (H3PO4) gives monoacetylferrocene. One Cp ring undergoes electrophilic substitution; the other Cp ring remains unsubstituted (and retains η5-bonding).

Mannich Condensation:

[Fe(Cp)2] + HCHO + HN(CH3)2H+ [Fe(Cp)(η5-C5H4-CH2N(CH3)2)] + H2O

Formaldehyde and dimethylamine react with ferrocene under mild acidic conditions. The iminium ion [CH2=NH(CH3)2]+ is generated in situ and attacks the electron-rich Cp ring, yielding aminomethylferrocene.

Topic 10: Reactions
31
Distinguish oxidative addition and reductive elimination. 2.5M(2021) / 5M(2023) / 2M(2024) 2021, 2023, 2024
Oxidative Addition

Definition: Low-valent metal inserts into a covalent bond (X–Y)

Coordination No.: Increases by +2

Oxidation State: Increases by +2

Electron Count: Increases by +2

Example: [IrICl(CO)(PPh3)2] + H2[IrIII(H)2Cl(CO)(PPh3)2]

Reductive Elimination

Definition: Two cis ligands on metal couple and leave as X–Y

Coordination No.: Decreases by −2

Oxidation State: Decreases by −2

Electron Count: Decreases by −2

Example: [PdII(CH3)2(PPh3)2] → C2H6 + [Pd0(PPh3)2]

Oxidative addition and reductive elimination are microscopic reverses of each other. They are the two fundamental steps that enable catalytic cycles in organometallic chemistry.

32
Example of 1,1-insertion and 1,2-insertion. 2 Marks 2024

1,1-Insertion: The migrating ligand and the metal end up on the same atom after insertion.

CH3–Mn(CO)5 → (CH3C(=O))–Mn(CO)4

The methyl group migrates onto the carbonyl (CO) carbon. Both the CH3 and the Mn now bond to the same carbon atom (the acyl carbon). This is a carbonyl insertion / alkyl migration.

1,2-Insertion: The migrating ligand and the metal end up on two adjacent atoms after insertion.

[Rh(H)(C2H4)Cl(PPh3)2] → [Rh(CH2CH3)Cl(PPh3)2]

The hydride (H) migrates across the C=C double bond of ethylene. H ends up on one carbon and Rh ends up on the adjacent carbon, forming a metal–alkyl (ethyl) complex. This is an olefin insertion.

33
Identify products: Fe(CO)2(NO)2 + PR3 → A + B 2 Marks 2025

This is a ligand substitution reaction. PR3 (a phosphine) substitutes one CO ligand from Fe(CO)2(NO)2.

Fe(CO)2(NO)2 + PR3 → Fe(CO)(NO)2(PR3) + CO
  • A = Fe(CO)(NO)2(PR3) — the substituted product (one CO replaced by PR3).
  • B = CO — the displaced carbon monoxide ligand.

The 18e count is maintained: Fe(0) = 8e + CO(2e) + 2NO(6e) + PR3(2e) = 18e.

Topic 11: Catalysis
34
Composition of Wilkinson's catalyst. 1 Mark 2020

Wilkinson's catalyst is [RhCl(PPh3)3] — Chloridotris(triphenylphosphine)rhodium(I). It is a square planar d8 Rh(I) complex with 16 valence electrons.

35
IUPAC name and hybridisation of Wilkinson's catalyst. 2 Marks 2024

IUPAC name: Chloridotris(triphenylphosphane)rhodium(I)

Hybridisation: dsp2 (square planar geometry). Rh(I) is d8; one d-orbital (dx²−y²), one s-orbital, and two p-orbitals hybridise to form four dsp2 orbitals that bond to Cl and three PPh3 ligands in a square planar arrangement.

36
Catalytic cycle for hydrogenation of ethylene by (Ph3P)3RhCl. RDS? 2.5 Marks 2021

Catalytic Cycle:

  1. Ligand Dissociation: [RhCl(PPh3)3][RhCl(PPh3)2] + PPh3  (16e → 14e, creates vacant site)
  2. Oxidative Addition of H2: [RhCl(PPh3)2] + H2[Rh(H)2Cl(PPh3)2]  (14e → 16e, Rh(I) → Rh(III))
  3. Alkene Coordination: [Rh(H)2Cl(PPh3)2] + C2H4[Rh(H)2Cl(η2-C2H4)(PPh3)2]  (16e → 18e)
  4. Migratory Insertion (RDS): One hydride migrates to the coordinated alkene, forming a metal–alkyl intermediate.  [Rh(H)(CH2CH3)Cl(PPh3)2]  (18e → 16e)
  5. Reductive Elimination: The remaining hydride and the alkyl group couple and are released as ethane.  [Rh(H)(CH2CH3)Cl(PPh3)2] → CH3CH3 + [RhCl(PPh3)2]  (Rh(III) → Rh(I), 16e → 14e)

Rate-Determining Step (RDS): Migratory insertion (Step 4) — the hydride transfer to the alkene is the slowest step.

37
Ethylene cannot be hydrogenated using Wilkinson's catalyst. Why? 2 Marks 2023

Ethylene (C2H4) is unhindered (no substituents) and is a potent π-acceptor. It coordinates very strongly to the Rh centre, forming a highly stable 16e complex [RhCl(η2-C2H4)(PPh3)2]. This complex is a thermodynamic sink — the ethylene binds so tightly that it does not readily undergo the subsequent migratory insertion step needed for hydrogenation. The catalyst is effectively poisoned by ethylene, which blocks the coordination site required for H2 oxidative addition.

In contrast, substituted alkenes (e.g., cyclohexene) bind more weakly and reversibly, allowing the catalytic cycle to proceed.

38
Compare hydrogenation rates: cyclohexene vs 1-methylcyclohexene. 3 Marks 2024

Cyclohexene >> 1-methylcyclohexene (cyclohexene is hydrogenated much faster).

Reason: Wilkinson's catalyst [RhCl(PPh3)3] has three bulky PPh3 ligands that create significant steric congestion around the Rh centre. Only one PPh3 dissociates to create the active 14e species [RhCl(PPh3)2].

  • Cyclohexene: Unsubstituted double bond fits easily into the crowded coordination cavity of the active catalyst. Coordination and subsequent insertion proceed rapidly.
  • 1-Methylcyclohexene: The methyl group on the double bond creates additional steric clash with the bulky PPh3 ligands. This steric hindrance prevents efficient coordination of the alkene to Rh, drastically reducing the rate of hydrogenation.

This steric selectivity is a hallmark of Wilkinson's catalyst — it preferentially hydrogenates less hindered alkenes.

39
Hydroformylation process to synthesise RCH2CH2CHO using cobalt catalyst. 5 Marks 2022

Hydroformylation (Oxo process) = addition of syngas (1:1 CO/H2) across a double bond to form an aldehyde.

RCH=CH2 + CO + H2Co cat. RCH2CH2CHO  (linear aldehyde)

Catalytic Cycle (Cobalt catalyst):

  1. Pre-catalyst activation: Co2(CO)8 + H2 → 2 HCo(CO)4HCo(CO)3 + CO  (active 16e species formed by CO dissociation)
  2. Alkene coordination: HCo(CO)3 + RCH=CH2HCo(CO)32-RCH=CH2)
  3. Migratory insertion (anti-Markovnikov): The hydride migrates to the less substituted carbon of the double bond (anti-Markovnikov), forming a linear alkyl intermediate: Co(CO)3(CH2CH2R)
  4. CO insertion (carbonyl insertion): CO inserts into the Co–C bond, forming an acyl complex: Co(CO)3(C(=O)CH2CH2R)Co(CO)4(C(=O)CH2CH2R) (after CO re-coordination)
  5. Oxidative addition of H2: H2 adds oxidatively to the Co centre: H2Co(CO)3(C(=O)CH2CH2R)
  6. Reductive elimination: The aldehyde product is released: RCH2CH2CHO + HCo(CO)3 (catalyst regenerated)

The anti-Markovnikov insertion in Step 3 is key to producing the linear (normal) aldehyde rather than the branched iso-aldehyde.

40
Hydroformylation to obtain R2CH–CH2–CHO from R2C=CH2. 5 Marks 2024

Overall reaction:

R2C=CH2 + CO + H2Co cat. R2CH–CH2–CHO  (branched/iso aldehyde)

This is the same hydroformylation process as Q39, but with a branched alkene substrate. The key difference is the regiochemistry of the migratory insertion step.

Catalytic Cycle:

  1. Pre-catalyst activation: Co2(CO)8HCo(CO)4HCo(CO)3 (active species)
  2. Alkene coordination: HCo(CO)3 + R2C=CH2HCo(CO)32-R2C=CH2)
  3. Migratory insertion (Markovnikov pathway): The hydride migrates to the more substituted carbon (Markovnikov addition) due to steric and electronic factors, forming a branched alkyl: Co(CO)3(CH2CHR2)
  4. CO insertion: CO inserts into the Co–alkyl bond forming the acyl complex: Co(CO)4(C(=O)CH2CHR2)
  5. Oxidative addition of H2: H2Co(CO)3(C(=O)CH2CHR2)
  6. Reductive elimination: R2CH–CH2–CHO is released + HCo(CO)3 regenerated

The Markovnikov insertion in Step 3 gives the branched (iso) aldehyde as the major product.

41
Catalytic cycle for acetaldehyde from ethylene (Wacker Process). 5M(2020) / 4M(2023) 2020, 2023

Overall reaction:

C2H4 + ½O2PdCl2/CuCl2 CH3CHO

Catalytic Cycle:

  1. Ethylene coordination: [PdCl4]2− + C2H4[PdCl32-C2H4)] + Cl
  2. Ligand substitution: H2O replaces Cl: [PdCl32-C2H4)] + H2O → [PdCl2(H2O)(η2-C2H4)] + Cl
  3. Nucleophilic attack (anti hydroxypalladation): Water attacks the coordinated ethylene in an anti fashion (from the side opposite to Pd), forming a σ-bonded hydroxyethyl complex: [PdCl2(CH2CH2OH)]
  4. β-Hydride elimination: A β-hydrogen is eliminated from the hydroxyethyl group back to Pd, forming a vinyl alcohol complex: [Pd(H)Cl2(CH=CHOH)]
  5. Re-insertion and enol release: The vinyl alcohol (enol) is released from coordination. Pd(0) is generated as a decomposition product.
  6. Tautomerisation: The enol CH2=CHOH rapidly tautomerises to the keto form CH3CHO (acetaldehyde).

The Pd(0) by-product is re-oxidised to Pd(II) by CuCl2, closing the catalytic cycle.

42
What is the Wacker process? Overall reaction? Role of CuCl2? 5 Marks 2025

Wacker Process: The industrial oxidation of ethylene to acetaldehyde using a PdCl2/CuCl2 catalyst system in aqueous solution. It is one of the most important industrial processes in organometallic chemistry.

Overall reaction:

C2H4 + ½O2PdCl2, CuCl2 CH3CHO

Role of CuCl2: CuCl2 acts as a redox mediator (co-catalyst). Its function is to re-oxidise Pd(0) back to Pd(II), enabling catalytic turnover:

Pd(0) + 2 CuCl2 → PdCl2 + 2 CuCl

The resulting Cu(I) chloride is then re-oxidised by atmospheric oxygen:

2 CuCl + 2 HCl + ½O2 → 2 CuCl2 + H2O

Thus, CuCl2 shuttles between Cu(II) and Cu(I), continuously regenerating the active Pd(II) catalyst using only O2 as the terminal oxidant.

43
Catalytic species in Wacker process and how generated? 2.5 Marks 2021

Active catalytic species: [PdCl2(H2O)(η2-C2H4)]

How generated: The starting material [PdCl4]2− undergoes sequential ligand substitution. First, ethylene replaces one chloride, then water replaces a second chloride:

[PdCl4]2− + C2H4 + H2O → [PdCl2(H2O)(η2-C2H4)] + 2 Cl

This species has both ethylene (for activation) and water (as nucleophile) coordinated to Pd(II), positioned for the subsequent anti-hydroxypalladation step.

44
Fischer–Tropsch process. Name catalyst(s). Write importance. 5M(2022) / 4M(2025) 2022, 2025

Fischer–Tropsch Process: The catalytic synthesis of hydrocarbons (synthetic gasoline/fuels) from syngas (CO + H2).

General reaction:

n CO + (2n+1) H2catalyst CnH2n+2 + n H2O

This produces linear alkanes (paraffins). Adjusting conditions can also yield alkenes and oxygenates.

Catalysts:

  • Iron (Fe): Most commonly used, inexpensive, operates at both high and low temperature. Also catalyses the water-gas shift reaction.
  • Cobalt (Co): Higher activity and selectivity for linear products, more expensive, used in gas-to-liquids (GTL) plants.
  • Ruthenium (Ru): Most active but too expensive for large-scale use; mainly of academic interest.

Catalysts are typically supported on materials like SiO2, Al2O3, or TiO2 to increase surface area.

Importance:

  • Converts coal, natural gas, or biomass (via gasification to syngas) into liquid fuels — reduces dependence on petroleum.
  • Produces clean, sulphur-free synthetic fuels and chemicals.
  • Strategic importance for countries with abundant coal/gas but limited oil reserves (e.g., South Africa's SASOL process).
  • Can produce a range of products: methane, LPG, gasoline, diesel, waxes, and oxygenates.
45
Composition of Ziegler–Natta catalyst. 1 Mark 2023, 2020

The Ziegler–Natta catalyst consists of two components:

  • Precursor (Transition metal compound): TiCl4 or TiCl3
  • Activator (Organometallic co-catalyst): Al(C2H5)3 (triethylaluminum) or other trialkylaluminum compounds

The combination of these two components generates the active catalytic species on a solid surface.

46
“Ziegler–Natta polymerisation is heterogeneous catalysis” — justify. 3 Marks 2020

Ziegler–Natta polymerisation is heterogeneous because:

  1. Solid catalyst: The active species (TiCl3) precipitates as an insoluble solid lattice (crystalline particles). The catalytic sites exist only on the surface of these solid particles.
  2. Surface-bound chain growth: Polymer chain growth occurs via continuous insertion of gaseous (or liquid) monomer molecules strictly at the solid crystal surface boundary. The growing polymer chain remains anchored to a Ti active site on the solid surface.
  3. Different phases: The catalyst is a solid while the reactants (ethylene, propylene) are gaseous or liquid. By definition, when the catalyst and reactants are in different phases, the catalysis is heterogeneous.

This is in contrast to homogeneous catalysts (like Wilkinson's catalyst), which dissolve in the reaction medium and operate in the same phase as the reactants.

47
Schematic representation of ethylene polymerisation using Z–N catalyst. Importance. 4 Marks 2023, 2025

Cossee–Arlman Mechanism (schematic):

  1. Active site generation: TiCl3 (solid) + AlEt3 → [Ti]–Et (active site: Ti with an ethyl group on the surface, with a vacant coordination site)
  2. Monomer coordination: [Ti]–Et + C2H4 → [Ti](η2-C2H4)–Et  (ethylene coordinates to the vacant site on Ti)
  3. Migratory insertion: The ethyl group migrates onto the coordinated ethylene: [Ti](η2-C2H4)–Et → [Ti]–CH2CH2Et  (chain grows by two carbon atoms; vacant site regenerated)
  4. Repeat: Steps 2–3 repeat continuously, each cycle adding another −CH2CH2− unit:
[Ti]–Chain + C2H4 → [Ti](η2-C2H4)–Chain → [Ti]–CH2CH2–Chain → …

Importance:

  • Operates at near-atmospheric pressure (1 atm) and moderate temperature, unlike the radical polymerisation process that requires >1000 atm.
  • Produces precisely linear (unbranched) high-density polyethylene (HDPE) with superior mechanical properties (crystallinity, tensile strength).
  • Enables stereoregular polymerisation of propylene to isotactic polypropylene (commercially important plastic).
  • Revolutionised the plastics industry — Karl Ziegler and Giulio Natta shared the 1963 Nobel Prize in Chemistry for this discovery.