Hydroformylation of Alkenes

Introduction

Hydroformylation, also called the oxo process, adds a hydrogen atom and a formyl group across a carbon–carbon double bond. From propene, carbon monoxide and hydrogen it makes butanal, the precursor to plasticiser alcohols. With production of roughly ten million tonnes of aldehydes and alcohols a year, it is one of the largest homogeneous catalytic processes in the world. It is also a textbook example in which every organometallic step you have met appears in a single cycle.

Core explanation

Net reaction.

RCH=CH₂ + CO + H₂ → RCH₂CH₂CHO (linear) or RCH(CH₃)CHO (branched)

The linear aldehyde is usually wanted, so the linear-to-branched ratio is a key performance figure.

The Heck–Breslow cycle (cobalt). The active catalyst is HCo(CO)₄, an 18-electron Co(I) hydride.

1. CO dissociation gives 16-electron HCo(CO)₃. 2. Alkene coordination gives HCo(CO)₃(alkene), 18 electrons. 3. 1,2-Insertion of the alkene into Co–H gives an alkyl, RCH₂CH₂Co(CO)₃, 16 electrons. Anti-Markovnikov insertion places cobalt on the terminal carbon and leads to the linear product. 4. CO coordination restores 18 electrons. 5. CO migratory insertion forms the acyl RCH₂CH₂C(O)Co(CO)₃, 16 electrons. 6. Oxidative addition of H₂ gives Co(III), 18 electrons. 7. Reductive elimination of the aldehyde regenerates HCo(CO)₃.

Step 6–7 may alternatively occur by reaction with a second HCo(CO)₄.

Cobalt versus rhodium.

Feature Unmodified Co Rh/PPh₃ (low-pressure oxo) --- --- --- Active species HCo(CO)₄ HRh(CO)(PPh₃)₂ and related Pressure about 200–300 bar about 10–20 bar Temperature about 140–180 °C about 80–120 °C Linear:branched about 3–4:1 about 10:1 or better with excess PPh₃ Side reactions more hydrogenation, isomerisation fewer

Rhodium is about a thousand times more active than cobalt, so despite its cost it dominates propene hydroformylation. Cobalt remains in use for longer-chain, internal alkenes, where its isomerising ability helps.

Regioselectivity. Bulky phosphines crowd the metal, favouring insertion that places the metal on the less hindered terminal carbon and so favour the linear aldehyde. Diphosphines with a wide bite angle (around 110–120°, such as BISBI or Xantphos) occupy two equatorial sites of the trigonal-bipyramidal intermediate and give very high linear selectivity.

Water-soluble catalysts. The Ruhrchemie/Rhône-Poulenc process uses a rhodium complex of sulfonated triphenylphosphine (TPPTS) dissolved in water. The aldehyde product forms a separate organic layer, so the expensive rhodium is recovered simply by decanting.

Step-by-step reasoning

To predict a hydroformylation product:

1. Write the alkene and identify its terminal and internal carbons. 2. Place H on one alkene carbon and CHO on the other. 3. Metal on the terminal carbon after insertion gives the linear aldehyde. 4. Metal on the internal carbon gives the branched aldehyde. 5. Use ligand bulk and bite angle to decide which is favoured.

Visual explanation

The SIM-CAT-001 cycle shows the metal hydride at the top. Moving clockwise, the alkene docks, slides into the M–H bond, CO slots in to make an acyl, H₂ adds, and the aldehyde departs. Colour coding follows the hydrogen from H₂ into the CHO group and the one from the hydride into the chain.

Real-world analogy

Picture a production line in which an empty car body (the alkene) receives a wheel (hydrogen) at one station, an engine (CO) at another, and final bolts (the second hydrogen) at the last. The robot arm (the metal) returns to the start after each car, ready for the next.

Real-world example

Butanal from propene hydroformylation is converted by aldol condensation and hydrogenation into 2-ethylhexanol, which is esterified to make plasticisers for PVC. Longer-chain hydroformylation products become detergent alcohols used in household cleaning products.

Why?

Why does a large excess of PPh₃ raise the linear selectivity with rhodium? Extra phosphine keeps two bulky PPh₃ ligands on the metal instead of CO. The crowded metal prefers to bind the less hindered terminal carbon during insertion, giving more linear aldehyde, though the rate drops because phosphine dissociation is suppressed.

Common misconception

"Both new hydrogen atoms in the aldehyde come from the same H₂ molecule." One hydrogen comes from the metal hydride present before the alkene inserts; the other comes from the H₂ that adds later. In a single turnover they originate from different H₂ molecules.

Worked example

Question: Name the linear and branched products from hydroformylation of 1-hexene.

Reasoning: Linear: CHO adds to the terminal carbon, giving a seven-carbon straight-chain aldehyde, heptanal. Branched: CHO adds to C2, giving 2-methylhexanal.

Answer: Heptanal (linear) and 2-methylhexanal (branched).

Quick check

1. Which elementary step in hydroformylation creates the new carbon–carbon bond? Answer: CO migratory insertion, which joins the alkyl group to carbon monoxide to form the acyl ligand.

Exam focus

Draw the full Heck–Breslow cycle with electron counts, name each step and identify where C–H and C–C bonds form. Compare cobalt and rhodium conditions and explain ligand control of linear-to-branched ratio. Mention carbon monoxide's toxicity and flammable hydrogen as reasons why such processes are run in sealed industrial plant.

Advanced insight

Asymmetric hydroformylation of styrene or vinyl acetate with chiral diphosphite or phosphine–phosphite ligands (such as BINAPHOS) gives branched chiral aldehydes with high enantioselectivity, offering short routes to profen anti-inflammatory drugs. Here the branched product is desired, showing how the same cycle can be steered in opposite directions by ligand design.

Summary

Hydroformylation adds H and CHO across an alkene using CO and H₂. The cycle involves CO loss, alkene binding, 1,2-insertion, CO migratory insertion, H₂ oxidative addition and aldehyde reductive elimination. Rhodium phosphine catalysts work under milder conditions and give better linear selectivity than cobalt. Bulky ligands and wide bite angles favour the linear aldehyde.

Practice questions

1. Write the net equation for hydroformylation of propene to the linear product. Answer: CH₃CH=CH₂ + CO + H₂ → CH₃CH₂CH₂CHO (butanal). 2. Give the electron count of HCo(CO)₄ and of the species formed when it loses CO. Answer: HCo(CO)₄ is 18-electron; HCo(CO)₃ is 16-electron. 3. Explain the advantage of the aqueous TPPTS process. Answer: The rhodium catalyst stays in the water layer while the aldehyde forms a separate organic layer, allowing easy separation and recovery of rhodium. 4. Why is rhodium preferred over cobalt despite its much higher price? Answer: Rhodium is far more active and selective, working at much lower pressure and temperature with fewer side reactions, so very little is needed.