Catalytic Carbonylation

Introduction

Carbon monoxide is a cheap one-carbon building block, available from synthesis gas. Transition metals can stitch it into organic molecules, turning an alkyl group into an acyl group and then into an acid, ester, aldehyde or amide. Catalytic carbonylation combines steps met earlier in this unit — oxidative addition, migratory insertion and reductive elimination — into cycles that run industrially on a vast scale. Ethanoic (acetic) acid manufacture is the classic case, and it is the best-understood cycle in homogeneous catalysis.

Core explanation

The Monsanto process. Methanol and carbon monoxide react to give ethanoic acid, CH₃OH + CO → CH₃COOH, using a rhodium catalyst with an iodide promoter. The rhodium complex does not react directly with methanol. Instead, hydrogen iodide converts methanol into methyl iodide: CH₃OH + HI → CH₃I + H₂O. The metal cycle then processes CH₃I.

The active catalyst is the square-planar anion [Rh(CO)₂I₂]⁻, a 16-electron rhodium(I) d⁸ complex. The cycle has four main steps:

1. Oxidative addition of CH₃I gives the octahedral rhodium(III) anion [CH₃Rh(CO)₂I₃]⁻ (18 electrons). This step is turnover-limiting, so the rate is first order in both rhodium and methyl iodide and roughly zero order in CO pressure. 2. Migratory insertion moves the methyl group onto an adjacent CO, forming the five-coordinate acyl [CH₃C(O)Rh(CO)I₃]⁻ (16 electrons). This is the step that creates the new C–C bond. 3. CO coordination restores an 18-electron six-coordinate acyl complex. 4. Reductive elimination of acetyl iodide, CH₃C(O)I, regenerates [Rh(CO)₂I₂]⁻.

Outside the metal cycle, acetyl iodide is hydrolysed by water to ethanoic acid and HI, which returns to convert more methanol. Thus two linked cycles operate: an organic iodide cycle and a rhodium cycle.

The Cativa process. Iridium analogues with promoters such as ruthenium carbonyl complexes largely replaced rhodium in many plants. For iridium, oxidative addition of CH₃I is much faster, but migratory insertion becomes slower because the Ir–CH₃ bond is stronger. Promoters abstract iodide, opening a site and speeding insertion. The iridium system is stable at lower water content, reducing by-products and the energy spent on drying the acid.

Other carbonylations. Palladium catalysts carbonylate aryl halides: oxidative addition of Ar–X, CO insertion into Pd–Ar and attack of an alcohol or amine on the Pd–acyl give esters or amides. Hydroformylation, covered earlier, is also a carbonylation, adding H and CHO across an alkene. Methanol carbonylation of methyl ethanoate, under anhydrous conditions, gives ethanoic anhydride by a closely related cycle.

Formulae

Net: CH₃OH + CO → CH₃COOH. Electron counts in the rhodium cycle: [Rh(CO)₂I₂]⁻ 16 e (Rh(I), d⁸) → [CH₃Rh(CO)₂I₃]⁻ 18 e (Rh(III), d⁶) → acyl 16 e → CO adduct 18 e → back to 16 e.

Step-by-step reasoning

To analyse any carbonylation cycle:

1. Identify how the organic substrate reaches the metal (often oxidative addition of a halide). 2. Locate the migratory insertion that converts M–R into M–C(O)R. 3. Check each intermediate's oxidation state and electron count. 4. Identify the product-forming step: reductive elimination or nucleophilic attack on the acyl. 5. Confirm that the starting catalyst is regenerated.

Visual explanation

Draw a loop with [Rh(CO)₂I₂]⁻ at the top. CH₃I enters on the right, the methyl slides onto CO at the bottom, CO enters on the left and CH₃COI leaves at the top. Beside it, draw a small second loop showing CH₃OH becoming CH₃I and CH₃COI becoming CH₃COOH.

Real-world analogy

The iodide cycle acts like a shuttle bus at an airport. Methanol cannot board the rhodium "plane" directly, so HI carries it there as methyl iodide, and afterwards collects the acetyl iodide and drops it off as acid before going back for the next passenger.

Real-world example

Most of the world's ethanoic acid — well over ten million tonnes a year — is made by methanol carbonylation. It is used to make vinyl ethanoate for paints and adhesives, cellulose ethanoate for fibres, and purified terephthalic acid, where ethanoic acid is the solvent for oxidation.

Why?

Why is oxidative addition, not insertion, slow for rhodium? Adding CH₃I requires the anionic Rh(I) centre to act as a nucleophile in an SN2-like attack on the methyl carbon. That bimolecular step has a significant barrier, whereas methyl migration to CO in the resulting crowded Rh(III) complex is fast.

Common misconception

"The carbonyl oxygen in ethanoic acid comes from carbon monoxide and the OH comes from methanol." The CO does supply the carbonyl carbon and oxygen, but the OH group comes from water during hydrolysis of acetyl iodide; the methanol oxygen leaves as water when CH₃I is formed.

Worked example

Question: Determine the oxidation state and d-electron count of rhodium in [CH₃C(O)Rh(CO)I₃]⁻.

Reasoning: Acyl and three iodides are X-type ligands (each −1 in ionic counting): total −4. The overall charge is −1, so Rh is +3. Rhodium is in group 9, so d⁹⁻³ = d⁶. Electron count: 6 + 2 (acyl) + 2 (CO) + 6 (three I⁻) = 16.

Answer: Rh(III), d⁶, a 16-electron five-coordinate complex.

Quick check

1. Which step of the Monsanto cycle forms the new carbon–carbon bond in ethanoic acid? Answer: Migratory insertion, in which the methyl group moves onto a coordinated carbon monoxide to form the acetyl ligand.

Exam focus

Draw the full Monsanto cycle with correct oxidation states (Rh(I) and Rh(III)) and electron counts. State the rate-determining step and the resulting rate law, and explain why iodide is essential. Contrast with the Cativa process, where insertion rather than oxidative addition becomes slow.

Advanced insight

In the rhodium process, a competing water–gas shift reaction (CO + H₂O → CO₂ + H₂) consumes CO and forms inactive Rh(III) species such as [Rh(CO)₂I₄]⁻, which can precipitate as RhI₃ where CO is scarce. High water concentrations help keep rhodium in its active form, which explains the high water content of the original process and one of the advantages of iridium.

Summary

Catalytic carbonylation incorporates CO into organic molecules through migratory insertion into a metal–carbon bond. In the Monsanto process, [Rh(CO)₂I₂]⁻ undergoes rate-limiting oxidative addition of CH₃I, methyl migration to CO, CO uptake and reductive elimination of CH₃COI, which water converts into ethanoic acid. The iridium Cativa process shifts the slow step to insertion. Palladium catalysts carbonylate aryl halides to esters and amides.

Practice questions

1. Why is methyl iodide rather than methanol the substrate that reacts with the rhodium complex? Answer: The C–O bond of methanol does not undergo oxidative addition readily, whereas CH₃I adds easily to nucleophilic Rh(I); HI converts methanol into CH₃I. 2. Give the rate law for the Monsanto process and explain it. Answer: Rate = k[Rh][CH₃I]; oxidative addition of CH₃I to the rhodium catalyst is turnover-limiting, so CO pressure has little effect. 3. What product forms when a palladium catalyst carbonylates bromobenzene in the presence of methanol and a base? Answer: Methyl benzoate, formed by oxidative addition, CO insertion to a benzoyl–palladium complex and methanolysis of the acyl. 4. State one advantage of the Cativa process over the Monsanto process. Answer: The iridium catalyst remains stable at low water concentrations, reducing by-products and the energy needed to dry the acid.