Question

The major product obtained by the treatment of \((\eta^{5}\text{-}\mathrm{C_5H_5})_2\mathrm{Ni}\) with Na/Hg in ethanol is

• A. \((\eta^{5}\text{-}\mathrm{C_5H_5})(\eta^{3}\text{-}\mathrm{C_5H_5})\mathrm{Ni}\)
• B. \((\eta^{3}\text{-}\mathrm{C_5H_5})_2\mathrm{Ni}\)
• C. \((\eta^{5}\text{-}\mathrm{C_5H_5})(\eta^{3}\text{-}\mathrm{C_5H_7})\mathrm{Ni}\)
• D. \((\eta^{3}\text{-}\mathrm{C_5H_7})_2\mathrm{Ni}\)

Hint:

Reductions of metallocenes often induce \(\eta^{5}\!⇒\!\eta^{3}\) ring slip; protic solvents can trap the reduced ring by protonation. Look for “Na/Hg + ROH” \(⇒\) reduce then protonate: mixed \(\eta^{5}/\eta^{3}\) complexes are common outcomes.

Answer 1

The Correct Option is C

Step 1: Reduction and ring‐slip.
Sodium amalgam reduces nickelocene to an anionic species that undergoes \(\eta^{5}\!\to\!\eta^{3}\) ring slip on one Cp ring.
Step 2: Protonation by ethanol.
The reduced, slipped ring is then protonated by \(\mathrm{EtOH}\), converting \(\mathrm{C_5H_5^-}\) to \(\mathrm{C_5H_7^-}\) while retaining \(\eta^{3}\)-coordination on that ring. The other ring stays \(\eta^{5}\)-bound.
Step 3: Product identity.
The resulting mixed-sandwich complex is \((\eta^{5}\text{-}\mathrm{C_5H_5})(\eta^{3}\text{-}\mathrm{C_5H_7})\mathrm{Ni}\), which matches option (C).
\[ \boxed{(\eta^{5}\text{-}\mathrm{C_5H_5})(\eta^{3}\text{-}\mathrm{C_5H_7})\mathrm{Ni}} \]