Alkene Insertion into Metal–Hydrogen and Metal–Carbon Bonds

Introduction

A coordinated alkene can accept a hydrogen or carbon group from the same metal. Insertion into M–H gives a metal alkyl through hydrometallation; insertion into M–C lengthens an alkyl chain. These steps underlie many hydrogenation, hydrofunctionalisation and olefin-polymerisation mechanisms. Their arrows must preserve atom connectivity: the metal remains attached to one alkene carbon while H or the original carbon group bonds to the other.

Core explanation

An alkene first binds through its π bond to a metal with a suitable vacant or accessible site. In a common 1,2-migratory insertion , an M–H or M–R bond and the coordinated C=C reorganise together. For M–H, one alkene carbon receives H and the other becomes metal-bound, giving M–alkyl. For M–R, R forms a new C–C bond to one alkene carbon and the metal bonds to the other, extending the metal-bound chain by the alkene's carbon count. The formal metal oxidation state generally remains unchanged: one X-type hydride or alkyl becomes one X-type alkyl. The alkene is L-type before insertion and no longer a separate coordinated alkene afterwards.

Geometry matters. The alkene and M–H or M–R group need access to a compatible cis arrangement for the common intramolecular step. A catalyst may first dissociate a ligand or rearrange its coordination sphere to create that geometry. Substituted alkenes can insert in more than one orientation, giving regioisomers; steric and electronic factors, ligand shape and secondary interactions determine the ratio. Alkene stereochemistry can influence the product, but subsequent rotations or elimination may erase a simple stereochemical prediction. Original organometallic teaching notes on migratory insertion discuss hydride and alkyl migration into unsaturated ligands.

The reverse of alkene insertion into M–H is β-hydride elimination from a metal alkyl when a β hydrogen and vacant site are suitably arranged. Insertion and elimination can equilibrate, moving a metal along an alkyl chain (“chain walking”) under some catalysts. For polymerisation, rapid monomer coordination and insertion relative to chain-transfer or β-hydride elimination can make long chains. If elimination or transfer dominates, shorter products result. This competition is more informative than saying only that a catalyst “activates ethene.”

Hydrogenation cycles can combine M–H formation, alkene insertion and product-releasing steps. A single insertion places one H and M across C=C; it does not by itself add H₂ to yield an alkane. Another hydrogen transfer or reductive-elimination event is needed, depending on the catalyst. In chain growth, each ethene insertion adds two carbon atoms to the chain while the chain remains attached to the metal until termination or transfer.

Step-by-step reasoning

1. Draw the alkene π-coordinated to the metal next to M–H or M–R. 2. Label the two alkene carbons before choosing insertion orientation. 3. Form H–C or R–C on one carbon and M–C on the other. 4. Check carbon count, formal oxidation state and open coordination site. 5. Ask whether another alkene inserts, β-hydride elimination reverses or chain transfer ends growth.

Visual explanation

Draw M–H next to CH₂=CH₂ coordinated side-on; an arrow produces M–CH₂–CH₃. Below, draw M–R plus CH₂=CH₂ giving M–CH₂–CH₂–R. Use different colours for the original R and the newly inserted two carbons. A backward arrow from an M–alkyl to alkene plus M–H marks β-hydride elimination where β-H geometry permits.

Real-world analogy

Think of a metal as a clip holding a growing chain. Each arriving two-link unit is inserted between the clip and the existing chain, so the clip remains at the chain end. That captures repeated ethene insertion, though an actual catalyst uses orbital overlap and can choose among several regio- and stereochemical paths.

Real-world example

In coordination polymerisation of ethene, a metal–alkyl site binds ethene and inserts it into its metal–carbon bond. Repetition builds polyethylene. The catalyst's ligand environment affects how fast ethene binds and inserts and how often chain transfer or β-hydride elimination stops growth. This molecular mechanism helps explain why different catalysts yield different molecular-weight distributions.

Why?

Why does metal oxidation state commonly stay constant? A hydride or alkyl is an X-type ligand before insertion; the newly formed metal alkyl is still X-type. The coordinated alkene contributes a neutral L-type interaction that is consumed in bond formation. Organic connectivity changes substantially, but the formal metal charge assignment does not.

Common misconception

“One hydride insertion fully hydrogenates an alkene” is false; it leaves a metal–carbon bond. “Insertion adds a carbon atom to ethene” confuses CO insertion with alkene insertion: ethene contributes two carbons. “Every insertion is irreversible” ignores β-hydride elimination and related reversible pathways.

Worked example

Start with M–CH₃ and one ethene molecule. In a simple orientation, insertion gives M–CH₂–CH₂–CH₃ , a metal-bound propyl chain. The original methyl carbon remains the terminal carbon farthest from M, and the two ethene carbons lie between it and the metal. A second ethene insertion gives a five-carbon M–alkyl, M–(CH₂)₄–CH₃, if no chain transfer or rearrangement occurs. Each insertion adds two carbon atoms; formal metal oxidation state is unchanged in this model.

Quick check

1. What remains after one alkene insertion into M–H: a free alkane or a metal alkyl? Answer: A metal alkyl; a further step is needed to release a fully hydrogenated organic product.

Exam focus

Label atoms and count carbons explicitly. Show the metal attached to one former alkene carbon and the migrating group attached to the other. Distinguish M–H from M–C insertion and name β-hydride elimination as a common reverse or competing step. Use “usually unchanged” for formal oxidation state under standard ligand assignments.

Advanced insight

Chain walking through repeated insertion and β-hydride elimination can produce branched polymers even from ethene. Ligand-controlled insertion selectivity can set stereoregularity when substituted alkenes polymerise. Kinetic isotope effects from M–D versus M–H systems can help locate hydride-transfer involvement, though isotope effects can include pre-equilibria and must be interpreted with a full rate model.

Summary

Coordinated alkenes can insert into metal–hydrogen or metal–carbon bonds. Hydride insertion yields a metal alkyl; alkyl insertion extends a chain, normally without changing formal metal oxidation state. Geometry, regioselectivity and competition with β-hydride elimination or transfer determine products and polymer lengths.

Practice questions

1. How many carbon atoms does each ethene insertion add to a growing M–alkyl chain? Answer: Two carbon atoms. 2. What product class forms directly from ethene insertion into M–H? Answer: A metal ethyl or related metal alkyl species, not a released ethane molecule. 3. What step can reverse an M–H alkene insertion if an appropriate β hydrogen is present? Answer: β-Hydride elimination. 4. Does insertion of a neutral alkene into one M–C bond normally increase formal metal oxidation state by two? Answer: No. One X-type alkyl is replaced by another X-type alkyl, so the formal metal oxidation state ordinarily remains unchanged.