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

Step 1: Recall what beta-hydride elimination needs.
\(\beta\)-hydride elimination is the reverse of alkene insertion, a hydrogen on the carbon beta to the metal moves onto the metal, forming a metal hydride and releasing, or coordinating, an alkene. Two things are needed: (1) a hydrogen present on the carbon beta to the metal, in a geometry that can reach the metal, and (2) an open, cis coordination site on the metal for the developing metal-hydride and alkene to occupy, since the transition state needs the alkyl and the vacant site to be cis, going through an agostic interaction. A coordinatively saturated, 18-electron, complex cannot do this even if the beta-hydrogen is present, because there is nowhere for the mechanism to go.

Step 2: Check option (A), \([\mathrm{Rh(C_5H_5)(PMe_3)(C_2H_5)}]^{+}\).
Counting electrons ionically: \(\mathrm{Rh}\) here is \(\mathrm{Rh(III)}\), \(d^6\) (6 electrons), \(\eta^5\text{-}\mathrm{C_5H_5}^{-}\) donates 6 electrons, \(\mathrm{PMe_3}\) donates 2 electrons, and the ethyl group donates 2 electrons: \(6 + 6 + 2 + 2 = 16\) electrons. This is a 16-electron, coordinatively unsaturated complex, so it has an open coordination site cis to the ethyl group. The ethyl group has beta-hydrogens, on its \(\mathrm{CH_3}\) carbon, and the vacant site lets the molecule reach the four-centered agostic transition state needed for \(\beta\)-hydride elimination. This complex readily undergoes \(\beta\)-hydride elimination.

Step 3: Check option (B), \([\mathrm{Rh(NH_3)_5(C_2H_5)}]^{2+}\).
This is an octahedral, six-coordinate \(\mathrm{Rh(III)}\) complex with all six sites occupied by strong \(\sigma\)-donor ligands (five \(\mathrm{NH_3}\) plus the ethyl group), giving an 18-electron, coordinatively saturated center. Even though the ethyl group has beta-hydrogens, there is no open cis site for the elimination pathway, so this complex is stable toward \(\beta\)-hydride elimination, a classic textbook example of a saturated alkyl complex that does not eliminate.

Step 4: Check option (C), \([\mathrm{Pt(PPh_3)_2(C_6H_5)I}]\).
This square-planar \(\mathrm{Pt(II)}\), 16-electron complex does have an accessible open site, square-planar \(d^8\) complexes are coordinatively unsaturated, but the group here is a phenyl, an aromatic ring bonded to the metal through an \(sp^2\) carbon. The carbons "beta" to platinum are the aromatic ring carbons, and removing one of their hydrogens as a beta-hydride would break the aromaticity of the ring, far too costly energetically. Phenyl, and other aryl or vinyl, groups have no accessible \(\beta\)-hydrogens for this pathway, so this complex cannot undergo \(\beta\)-hydride elimination despite being unsaturated.

Step 5: Check option (D), \([\mathrm{Cr(CH_2Si(CH_3)_3)_4}]\).
The ligand here is \(\mathrm{{-}CH_2Si(CH_3)_3}\), trimethylsilylmethyl. The carbon directly bonded to chromium is \(\mathrm{CH_2}\), and the atom beta to the metal is silicon, not carbon. There is no beta carbon-hydrogen bond at all in this ligand, only a beta \(\mathrm{Si{-}C}\) bond, so the whole \(\beta\)-hydride elimination pathway is structurally unavailable. This is exactly why \(\mathrm{CH_2SiMe_3}\), like neopentyl, is a favorite ligand for making stable homoleptic high-oxidation-state alkyl complexes.

Final Answer:
Only the 16-electron, coordinatively unsaturated ethyl complex in option (A) has both a beta-hydrogen and the open site needed to use it. \[ \boxed{\text{(A)}} \]