# Reaction of CpFe(CO)2Cl with phenyllithium

The reaction of $\ce{(C5H5)Fe(CO)2Cl}$ with an equimolar amount of $\ce{Li+[C6H5]-}$ gives a product. Draw its structure and apply the 18 electron rule.

In a question I was given the molecular weight and percentages of each element in the compound. I worked out that there is $\ce{1 Fe}$, $\ce{13 C}$, $\ce{10 H}$ and $\ce{2 O}$.

The product seems to be an iron atom surrounded with ligands, $\ce{2 CO}$ ligands, $\ce{1 C5H5}$ and $\ce{1 [C6H5]-}$ ion.

The 18 electron rule expands to 8 valence electrons from $\ce{Fe}$, 2 electrons from each $\ce{CO}$ ligand, 5 electrons from $\ce{C5H5}$ and 1 electron from the $\ce{[C6H5]-}$ anion: $8+4+5+1 = 18$. But what would the structure look like? Especially how is the $\ce{[C6H5]-}$ anion connected?

Direct displacement gives $\ce{CpFe(CO)2Ph}$, with the phenyl being a simple anion sigma-bonded to the metal. If there are two carbons plus two oxygens in the product, implying the CO ligands are intact, your are done. If some fancy stiff happened with those ligands and the phenyl anion, you must work it out or look it up.