# Organometallic Synthesis: Cp2Ti(I)Me from TiCl4

I've been working on this synthesis question for a couple days but I'm a little stuck. The question is:

Using only $$\ce{TiCl_{4}}$$, $$\ce{NaCp}$$, $$\ce{PhLi}$$, $$\ce{MeI}$$, and $$\ce{EtMgBr}$$, synthesize $$\ce{Cp_{2}Ti(I)Me}$$. (should take 4 to 6 steps)

But it seems to me the synthesis can be done in 2 steps like so:

$$\ce{TiCl_{4} + 2NaCp -> Cp_{2}TiCl_{2}}$$

$$\ce{Cp_{2}TiCl_{2} + MeI -> Cp_{2}Ti(I)Me}$$

Which doesn't match the question's "4-6 steps" note, so I'm not sure what I'm missing.

• You need a reductive step to go from Ti(IV) to Ti(III) – Waylander Oct 2 at 8:44
• @waylander Cp2Ti(Me)I is titanium IV. – matt_black Oct 2 at 12:28
• My bad, not enough coffee – Waylander Oct 2 at 12:57