Question

The given zirconocene compound, (η⁵-Cp)₂ZrEt₂, when heated in the presence of an equimolar amount of PMe₃ results in the formation of a compound X which obeys the 18-electron rule. The reaction also resulted in the release of a saturated hydrocarbon.

[Given: Atomic number of Zr = 40]

The structure of compound X is:

• A.
• B.
• C.
• D.

Hint:

For organozirconium complexes, Beta-hydride elimination followed by formation of an alkene complex is a common route. Always check the 18-electron rule to determine the correct structure of the product.

Answer 1

The Correct Option is C

The compound given is a dialkyl zirconocene: (\(η⁵−Cp)₂ZrEt₂. Upon heating in the presence of PMe₃, one of the alkyl groups undergoes beta-hydride elimination, leading to the loss of ethane (C₂H₆), a saturated hydrocarbon.

This results in the formation of a Zr–alkene complex with PMe₃ coordinated to the Zr center. The product should satisfy the 18-electron rule, so we count:

• Zr contributes 4 d-electrons.
• Each Cp ligand contributes 5 electrons: \(2 \times 5 = 10\)
• Alkene donates 2 electrons.
• PMe₃ donates 2 electrons.

Total: \(4 + 10 + 2 + 2 = 18\) electrons. Thus, structure (C) is consistent with the electron count and the product formed from beta-hydride elimination.

Other options:

• (A) Involves formation of a metallacycle, which is inconsistent with beta-hydride elimination and the 18-electron rule.
• (B) and (D) do not provide the correct oxidation state or ligand environment to satisfy 18 electrons.

\[ \boxed{\text{Correct structure of X is (C)}} \]