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Oxidative addition, reductive elimination, migratory insertion and beta-hydride elimination, and how each changes a metal's oxidation state, coordination number and electron count.
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By the end of this lesson you will be able to identify the four elementary steps of organometallic chemistry and follow a metal's oxidation state, coordination number and electron count through each.
From lesson 15 you can count a complex's valence electrons by the ionic method and know that catalysts cycle between sixteen and eighteen electrons, opening a site to bind a reactant. From lesson 2 you can find an oxidation state, and from lesson 25 you know that carbon monoxide and alkenes bind metals by donation and back-donation. This lesson names the handful of moves that make up almost every catalytic cycle.
| Term | What it means |
|---|---|
| Elementary step | A single step of a mechanism, one bond-making or bond-breaking event at the metal. |
| Oxidative addition | A molecule A-B splits and both fragments bind the metal, raising its oxidation state, coordination number and electron count by two. |
| Reductive elimination | Two ligands on the metal join and leave as a molecule, lowering all three by two; the reverse of oxidative addition. |
| Migratory insertion | A ligand moves onto a neighboring unsaturated ligand, such as CO or an alkene, freeing a coordination site. |
| Beta-hydride elimination | A hydrogen on the carbon two atoms from the metal moves to the metal, leaving an alkene. |
| Coordinatively unsaturated | Having fewer than eighteen electrons, and so an open site for a new ligand. |
| Catalytic cycle | A sequence of elementary steps that returns the catalyst to its starting form while turning reactants into products. |
A catalyst works by passing reactants through a sequence of elementary steps, each changing the metal in a predictable way. Four do most of the work.
Oxidative addition. A molecule such as $\mathrm{H_2}$, $\mathrm{CH_3I}$ or an aryl bromide splits its bond across the metal, and both pieces bind it. Each piece counts as an anionic ligand, so the metal's oxidation state rises by two, its coordination number by two, and its electron count by two. Vaska's complex, $\mathrm{IrCl(CO)(PPh_3)_2}$, square-planar iridium(I) with sixteen electrons, adds hydrogen to become octahedral iridium(III) with eighteen. The metal needs an open site and electrons to spare, so electron-rich, sixteen-electron complexes do it best.
Reductive elimination. The reverse: two ligands on the metal join and leave as one molecule, and all three numbers fall by two. It is usually the step that releases the product.
Migratory insertion. A ligand such as a methyl or a hydride moves onto a neighboring carbon monoxide or alkene, making a new bond. The oxidation state does not change, but one site is freed: coordination number down by one, electron count down by two. This is how new carbon-carbon bonds are made at a metal.
Beta-hydride elimination. A metal holding an alkyl group can pull a hydrogen from the carbon two atoms away, the beta carbon, leaving a hydride and a bound alkene. The oxidation state does not change; coordination number up by one, electron count up by two. It needs an open site next to the alkyl group.
Because the changes are fixed, bookkeeping follows a catalyst through its cycle. And it catches impossible proposals: an oxidative addition to an eighteen-electron complex would give twenty, so that complex must first lose a ligand. A metal already in its highest common oxidation state, such as platinum(IV), cannot undergo oxidative addition at all, and one in oxidation state zero cannot undergo reductive elimination.
Another way: picture
Think of the metal as a workbench with a fixed number of spaces. Oxidative addition clamps two new pieces onto the bench, using two spaces; reductive elimination joins two pieces into a finished part and lifts it off, freeing two. Migratory insertion slides one piece onto its neighbor, freeing a space where it had been; beta-hydride elimination snips a small piece off a larger one and puts it in the free space.
Another way: steps
The name of a step follows from which bonds change. If a bond outside the metal breaks and both ends end up on the metal, it is an oxidative addition: $\mathrm{H-H}$, $\mathrm{C-Br}$ or $\mathrm{C-I}$ splitting across a palladium or rhodium center. If two groups on the metal leave as one bonded molecule, it is a reductive elimination: a hydride and a methyl leaving as methane, or an aryl and another carbon group leaving joined.
If a group on the metal ends up bonded to a carbon of a neighboring ligand, and the metal keeps the combined group, it is a migratory insertion: a methyl and a carbon monoxide becoming an acetyl group, $\mathrm{C(O)CH_3}$, or a hydride and an ethylene becoming an ethyl group. And if an alkyl group loses a hydrogen to the metal and becomes an alkene still bound to it, it is a beta-hydride elimination. Drawing the structure before and after, and circling the bonds that changed, settles which it is.
Two further moves appear in almost every cycle without changing the metal much. A ligand can simply leave, lowering the coordination number by one and the electron count by two, which is how an eighteen-electron catalyst opens a site; and a new ligand can bind, the reverse. Neither changes the oxidation state. Many proposed mechanisms fail only because they skip one of these: a step drawn at an eighteen-electron metal that needs an open site is impossible until a ligand has left, so the dissociation has to be written into the cycle as its own step.
Each step has requirements, and many failed catalysts fail because one is not met. Oxidative addition needs an open site, or a ligand that can leave, and an oxidation state two below one the metal can reach; low-valent, electron-rich metals with donor ligands do it fastest. Reductive elimination is favored by an electron-poor metal and by bulky ligands that push the leaving groups together, and the two groups must be next to each other, cis, not across from each other.
Migratory insertion needs the two groups cis as well. Beta-hydride elimination needs a hydrogen on the beta carbon and an empty site cis to the alkyl group; alkyl groups with no beta hydrogen, such as methyl or neopentyl, cannot do it, which is why chemists use them to make stable metal alkyls. And polymerization catalysts must make insertion much faster than beta-hydride elimination, or the growing chain falls off the metal as a short alkene.
A catalytic cycle is a sequence of these steps that returns the metal to where it began. Every oxidative addition in the cycle must be matched by a reductive elimination, so that the oxidation state comes back; every site opened must be filled again. Wilkinson's catalyst for adding hydrogen to alkenes shows the pattern: rhodium(I) loses a phosphine, adds hydrogen by oxidative addition to become rhodium(III), binds the alkene, inserts it into a rhodium-hydrogen bond, and finally eliminates the alkane by reductive elimination, returning to rhodium(I).
The Monsanto process for acetic acid runs the same way: rhodium(I) adds methyl iodide, a methyl migrates onto a carbon monoxide, and the acetyl iodide leaves by reductive elimination, to be hydrolyzed to acetic acid. Following the numbers through each step, as the next lesson does for whole cycles, is how chemists check that a proposed mechanism is possible.
Three checks catch most slips. First, the oxidation state and coordination number move together in additions and eliminations, both by two; if one changes by two and the other by one, the step has been misnamed. Second, the electron count must stay at eighteen or below. An oxidative addition proposed for an eighteen-electron complex is impossible as written; a ligand must dissociate first.
Third, the reachable oxidation states: palladium moves between $0$ and $+2$, rhodium and iridium between $+1$ and $+3$, platinum between $+2$ and $+4$. A step that would take a metal outside its range, such as an oxidative addition to palladium(II) in an ordinary cross-coupling, is not part of the cycle. And migratory insertion and beta-hydride elimination never change the oxidation state: if your answer does, you have treated a neutral rearrangement as an addition or an elimination.
Many drugs, pesticides and the materials in OLED screens are built by palladium-catalyzed cross-coupling, which joins two carbon groups that would never react on their own. Richard Heck, Ei-ichi Negishi and Akira Suzuki shared the 2010 Nobel Prize in Chemistry for these reactions. In a Suzuki coupling, palladium(0) adds an aryl halide by oxidative addition, takes a second carbon group from a boron compound, and releases the two groups joined by reductive elimination.
The whole cycle is oxidative addition, exchange, reductive elimination, with palladium shuttling between $0$ and $+2$. Pharmaceutical chemists choose the phosphine ligands on the palladium to speed whichever step is slowest for their molecules, electron-rich ligands for a stubborn oxidative addition, bulky ones for a slow reductive elimination, reasoning directly from the requirements of each elementary step.
More polyethylene is made than any other plastic, over a hundred million tonnes a year, most of it with metal catalysts based on the work of Karl Ziegler and Giulio Natta, who shared the 1963 Nobel Prize in Chemistry. The chain grows by repeated migratory insertion: an ethylene binds to a titanium or zirconium center next to the growing alkyl chain, and the chain migrates onto it, lengthening by two carbons and freeing the site for the next ethylene.
The chain stops growing when beta-hydride elimination removes it as a long alkene. The ratio of the two rates, insertion against elimination, sets how long the chains become and so the strength of the plastic. Catalyst designers make insertion thousands of times faster than elimination to get the long chains that give milk jugs, pipes and bulletproof fibers their properties.
It is natural to associate oxidation with oxygen or a strong oxidant, and to be surprised that hydrogen, a reducing agent, oxidizes iridium(I) to iridium(III). But the oxidation state is bookkeeping: when H-H splits across the metal, each hydrogen is counted as a hydride, $\mathrm{H^-}$, so the metal must be counted two units more positive. Nothing has been removed from the metal; its electrons have been shared into two new bonds.
A second error is to treat migratory insertion as a change in the metal's oxidation state. The methyl and the acetyl are both single anionic ligands, so the metal's oxidation state is the same before and after; only its coordination number and electron count fall.
Count the starting complex.
$\mathrm{Ir^{I}}, \ \text{CN } 4, \ 8 + 8 = 16 \text{ electrons}$
Iridium(I) is $d^8$; four ligands give eight.
Name the step.
$\mathrm{H{-}H} \text{ splits across Ir}$
Both hydrogens bind: oxidative addition.
Change the oxidation state.
$+1 \to +3$
Two hydride ligands, each counted as $\mathrm{H^-}$.
Change the coordination number.
$4 \to 6$
Square planar becomes octahedral.
Change the electron count.
$16 \to 18$
Iridium(III) is $d^6$; six ligands give twelve: eighteen.
Count the starting complex.
$\mathrm{CH_3Mn(CO)_5}: \ \mathrm{Mn^{I}}, \ \text{CN } 6, \ 18$
Manganese(I) is $d^6$; six ligands give twelve.
Name the step.
$\mathrm{CH_3} \text{ moves onto a CO}$
Migratory insertion, making an acetyl group.
Change the oxidation state.
$+1 \to +1$
The acetyl is still one anionic ligand.
Change the coordination number.
$6 \to 5$
Methyl and carbonyl now make one ligand.
Change the electron count.
$18 \to 16$
An open site appears.
Say what fills the site.
$\text{a new CO binds}: \ 16 \to 18$
Carbon monoxide under pressure traps the acetyl complex.
Count the active catalyst.
$\mathrm{Pd(PPh_3)_2}: \ \mathrm{Pd^{0}}, \ \text{CN } 2, \ 10 + 4 = 14$
Palladium(0) is $d^{10}$; two phosphines give four.
Name the step with bromobenzene.
$\mathrm{C_6H_5{-}Br} \text{ splits across Pd}$
Oxidative addition.
Change the oxidation state.
$0 \to +2$
Phenyl and bromide both count as anions.
Change the coordination number.
$2 \to 4$
Square-planar palladium(II).
Change the electron count.
$14 \to 16$
The stable count for square-planar $d^8$.
Look ahead to the last step.
$\text{reductive elimination}: \ \mathrm{Pd^{II}} \to \mathrm{Pd^{0}}$
Two carbon groups leave joined, and the cycle closes.
Name the step.
$\text{reductive elimination}$
Two ligands leave joined.
Lower the oxidation state and coordination number.
$+3 \to +1, \quad 6 \to 4$
Both fall by two.
Lower the electron count.
Match each elementary step to the changes it makes to the metal's oxidation state, coordination number and electron count.
| $+2, +2, +2$ | $-2, -2, -2$ | $0, -1, -2$ | $0, +1, +2$ | |
|---|---|---|---|---|
| oxidative addition | ||||
| reductive elimination | ||||
| migratory insertion | ||||
| beta-hydride elimination |
Complete the worked solution: $\mathrm{[Rh(CO)_2I_2]^-}$ takes part in an oxidative addition. Find the metal's state afterward.
Raise the oxidation state by two.
$\text{oxidation state after} =$ p
The bond split across the metal gives it two anionic ligands.
Raise the coordination number by two.
$\text{coordination number after} =$ q
Both fragments bind to the metal.
Raise the electron count by two.
$\text{electron count after} =$ r
One more pair of electrons around the metal in the ionic count.
In one elementary step, $\mathrm{CH_3Mn(CO)_5}$ moves CH_3 onto CO. The metal goes from oxidation state $1$ to $1$ and from coordination number $6$ to $5$. Which kind of step is it?
$\mathrm{[Rh(CO)_2I_3(COCH_3)]^-}$ (oxidation state $3$, coordination number $6$, $18$ electrons) loses CH_3COI by reductive elimination; $\mathrm{Pd(PPh_3)_4}$ ($0$, $4$, $18$) loses PPh_3 by ligand loss. For each, in that order, fill in the metal's oxidation state, coordination number and electron count after the step.
| oxidation state after | coordination number after | electron count after | |
|---|---|---|---|
| the first step | |||
| the second step |
$\mathrm{RhCl(PPh_3)_3}$ has $16$ valence electrons. It adds H_2 by oxidative addition. How many valence electrons does the metal have afterward?
Answer: valence electrons after the step
$\mathrm{IrCl(CO)(PPh_3)_2}$, with the metal in oxidation state $1$, coordination number $4$ and $16$ electrons, adds H_2 in one elementary step, oxidative addition. What is the metal's oxidation state after the step?
Answer: oxidation state after the step
A process chemist annotating the steps of several industrial catalytic cycles tracks the metal through each. $\mathrm{Pd(PPh_3)_2}$ ($0$, $14$ electrons) adds C_6H_5Br by oxidative addition; $\mathrm{[Rh(CO)_2I_2]^-}$ ($1$, $16$) adds CH_3I by oxidative addition; $\mathrm{[Rh(CO)_2I_3(COCH_3)]^-}$ ($3$, $18$) loses CH_3COI by reductive elimination. For each, in that order, fill in the oxidation state and electron count after the step.
| oxidation state after | electron count after | |
|---|---|---|
| the first step | ||
| the second step | ||
| the third step |
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$\mathrm{[Rh(CO)_2I_2]^-}$ (oxidation state $1$, coordination number $4$, $16$ electrons) adds CH_3I by oxidative addition; $\mathrm{PtH(CH_3)(PPh_3)_2}$ ($2$, $4$, $16$) loses CH_4 by reductive elimination. For each, in that order, fill in the metal's oxidation state, coordination number and electron count after the step.
| oxidation state after | coordination number after | electron count after | |
|---|---|---|---|
| the first step | |||
| the second step |
You can follow a metal through a step. Explain why hydrogen oxidizes Vaska's complex from iridium(I) to iridium(III), and why an eighteen-electron complex cannot undergo oxidative addition directly.
15. Your turn: $\mathrm{[Rh(CO)_2I_3(COCH_3)]^-}$, rhodium(III), coordination number 6, 18 electrons, eliminates acetyl iodide. What is rhodium's state afterward?, step 3
$18 \to 16$
Back to the square-planar catalyst.