σ-Bond Metathesis

Introduction

Oxidative addition needs a metal that can give up two electrons. But what about a d⁰ centre such as Zr(IV) or Lu(III), which has no d electrons to give? Such metals still break H–H and C–H bonds, including the famously unreactive C–H bonds of methane. They do so by σ-bond metathesis , a concerted swap of σ-bond partners that leaves the oxidation state untouched. It is the key bond-breaking step in early-metal and lanthanide catalysis.

Core explanation

The reaction. An M–R bond meets an H–R′ bond, and the partners exchange:

L nM–R + H–R′ → L nM–R′ + H–R

The reaction goes through a four-centre, kite-shaped transition state in which M–R and H–R′ weaken while M–R′ and R–H form. Nothing is added to or removed from the metal's set of X ligands, so the oxidation state and electron count are identical before and after.

Why d⁰ metals. Oxidative addition would raise the oxidation state by two. A d⁰ metal such as Sc(III) or Zr(IV) is already in its highest accessible state, so that route is closed. σ-Bond metathesis needs only an empty metal orbital to accept electron density from the incoming σ bond, which d⁰ metals have in abundance.

The transition state. The hydrogen sits at the "tip" of the kite, between the two carbons or between carbon and hydrogen. Hydrogen's spherical 1s orbital can bond to both neighbours at once, which is why the transferring atom is almost always hydrogen. Placing a carbon at the tip would need a directional sp³ orbital to point two ways, which costs far more energy.

Classic examples.

Reaction Metal Result --- --- --- Cp ₂Lu–CH₃ + ¹³CH₄ → Cp ₂Lu–¹³CH₃ + CH₄ Lu(III) methane exchange (Watson) Cp ₂Sc–CH₃ + H₂ → Cp ₂Sc–H + CH₄ Sc(III) hydrogenolysis Cp ₂Sc–CH₃ + C₆H₆ → Cp ₂Sc–C₆H₅ + CH₄ Sc(III) C–H activation Cp₂Zr(H)Cl + R₃Si–H Zr(IV) Si–H/M–H exchange in dehydrocoupling

Rates and selectivity. Rates generally follow H–H > sp² C–H > sp³ C–H, reflecting how easily each bond interacts with the metal. Transition states are tightly ordered, so ΔS‡ is large and negative. Kinetic isotope effects are typically significant, confirming that the X–H bond breaks in the rate-determining step.

Relatives. Late metals can undergo related steps in which the hydrogen is partly transferred to the metal — "σ-complex-assisted metathesis" (σ-CAM) — blurring the line between metathesis and oxidative addition.

Step-by-step reasoning

1. Check the metal's d-electron count. If d⁰, rule out oxidative addition. 2. Identify the M–R bond and the incoming H–R′ bond. 3. Arrange the four atoms M, R, H, R′ in a kite with H at the tip. 4. Swap partners: form M–R′ and R–H. 5. Confirm the oxidation state is unchanged.

Visual explanation

Draw a kite: M at the left corner, R at the top, H at the right tip and R′ at the bottom. Dashed lines connect all four. The M–R and H–R′ bonds are fading; M–R′ and R–H are forming. A curly-arrow diagram shows electrons circulating around the four-membered ring.

Real-world analogy

Two couples at a dance swap partners in one smooth movement without anyone leaving the floor. No one arrives or departs; the pairings simply exchange. σ-Bond metathesis is that simultaneous swap, choreographed around the metal.

Real-world example

In metallocene-catalysed polypropylene production, hydrogen gas is added to control molecular mass. The growing zirconium–polymer chain reacts with H₂ by σ-bond metathesis, releasing a saturated polymer chain and a Zr–H species that starts a new chain.

Why?

Why is the transferring atom nearly always hydrogen? Its 1s orbital is spherical, so it can overlap with two partners at once in the kite. Carbon's directional sp³ orbitals cannot bridge two atoms efficiently, making a carbon-at-the-tip transition state much higher in energy.

Common misconception

"σ-Bond metathesis is just oxidative addition followed by reductive elimination." For d⁰ metals, no M(n+2) intermediate is possible, and none is formed. The exchange is a single concerted step with unchanged oxidation state throughout.

Worked example

Question: Predict the products when Cp ₂Y–CH(SiMe₃)₂ reacts with H₂, and give the yttrium oxidation state throughout.

Reasoning: Y(III) is d⁰, so oxidative addition is impossible. H₂ approaches the Y–C bond; one H goes to Y and the other to the alkyl carbon via a four-centre transition state.

Answer: Cp ₂Y–H and CH₂(SiMe₃)₂; yttrium remains +3 throughout.

Quick check

1. Why can Zr(IV) not undergo oxidative addition of H₂? Answer: Zr(IV) is d⁰ and has no d electrons to donate, so it cannot be oxidised to Zr(VI).

Exam focus

Contrast σ-bond metathesis with oxidative addition: same bonds broken, but no change in oxidation state and no need for d electrons. Draw the four-centre transition state with hydrogen at the tip and link the step to d⁰ early metals and lanthanides.

Advanced insight

Computations show that σ-bond metathesis transition states carry partial hydride character on the central hydrogen, resembling a proton moving between two carbanion-like groups. This explains the strong preference for sp² C–H bonds, whose greater s character stabilises the negative charge. The same model rationalises catalytic hydrosilylation and dehydrocoupling of silanes at d⁰ metals.

Summary

σ-Bond metathesis exchanges partners between an M–R bond and an H–R′ bond through a concerted four-centre transition state with hydrogen at the tip. It leaves oxidation state and electron count unchanged, so it is the route of choice for d⁰ metals and lanthanides that cannot undergo oxidative addition. It enables hydrogenolysis, C–H activation and chain transfer in polymerisation.

Practice questions

1. Write the equation for hydrogenolysis of Cp ₂Lu–CH₃ by H₂. Answer: Cp ₂Lu–CH₃ + H₂ → Cp ₂Lu–H + CH₄. 2. State two ways σ-bond metathesis differs from oxidative addition. Answer: The oxidation state does not change, and the metal does not need d electrons; the step is also a single concerted exchange. 3. Why is ΔS‡ for σ-bond metathesis large and negative? Answer: Two molecules combine into a single, highly ordered four-centre transition state. 4. Which C–H bond, in benzene or in methane, reacts faster with Cp ₂Sc–CH₃, and why? Answer: Benzene's sp² C–H bond, because greater s character stabilises the developing negative charge on carbon and the π system can pre-coordinate.