Question

The species that undergoes \(\beta\)-elimination is

• A. \([\mathrm{Rh(C_5H_5)(P(CH_3)_3)(C_2H_5)}]^{+}\)
• B. \([\mathrm{Rh(NH_3)_5(C_2H_5)}]^{2+}\)
• C. \([\mathrm{Pt(P(C_6H_5)_3)_2(C_6H_5)I}]\)
• D. \([\mathrm{Cr(CH_2Si(CH_3)_3)_4}]\)

Hint:

Beta-hydride elimination needs BOTH a hydrogen on the beta carbon AND an open coordination site cis to the alkyl group; check the electron count for a vacant site, and check whether the beta position is even a carbon with an H, phenyl and \(\mathrm{CH_2SiMe_3}\) fail this second test.

Answer 1

The Correct Option is A

Beta-hydride elimination is often treated as if it just needs a beta-hydrogen, but that is only half the story. The metal also needs somewhere for the hydrogen to go, an open coordination site sitting cis to the alkyl group. Go through each complex checking both conditions.

1. $[\mathrm{Rh(C_5H_5)(PMe_3)(C_2H_5)}]^{+}$: add up the electrons this Rh(III) center receives: 6 from $d^6$ Rh(III), 6 from $\eta^5$-cyclopentadienyl, 2 from $\mathrm{PMe_3}$, 2 from ethyl, total 16. A 16-electron complex has a coordination gap, an open site the ethyl group's beta-hydrogen can swing toward. This complex satisfies both conditions and is the one that eliminates.
2. $[\mathrm{Rh(NH_3)_5(C_2H_5)}]^{2+}$: six ligands fully occupy an octahedral Rh(III) center, giving 18 electrons with no gaps anywhere. The ethyl group has beta-hydrogens, but there is no empty site for them to reach, so nothing happens.
3. $[\mathrm{Pt(PPh_3)_2(C_6H_5)I}]$: square-planar Pt(II) is 16-electron and does have an open site, but the group attached to platinum is phenyl, not a simple alkyl. The would-be beta-carbons are locked into an aromatic ring, pulling a hydrogen off one of them would destroy the ring's aromaticity, so this pathway is blocked structurally, regardless of the open site.
4. $[\mathrm{Cr(CH_2Si(CH_3)_3)_4}]$: the metal-bound carbon is $\mathrm{CH_2}$, and one atom further out is silicon, not carbon. Beta-hydride elimination needs a hydrogen on a beta carbon, and there simply is no beta carbon here, only a beta silicon. This ligand is chosen specifically because it cannot eliminate.

Three of the four complexes fail one of the two requirements: (B) has the hydrogen but no open site, (C) has the open site but no usable beta-hydrogen, and (D) has neither a beta carbon nor, therefore, a beta-hydrogen. Only (A) has both an open coordination site and an ethyl group with real beta-hydrogens.

Let's summarize:

• Beta-hydride elimination needs both a beta-C-H bond and an open cis coordination site.
• Saturated 18-electron complexes, like the pentaammine ethyl rhodium, cannot eliminate even with a good beta-hydrogen present.
• Aryl, vinyl, and $\mathrm{CH_2SiMe_3}$-type ligands have no usable beta-hydrogen at all.

The species that undergoes beta-elimination is the 16-electron cyclopentadienyl rhodium ethyl complex, option (A).