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.

  • $\begingroup$ You need a reductive step to go from Ti(IV) to Ti(III) $\endgroup$ – Waylander Oct 2 '19 at 8:44
  • $\begingroup$ @waylander Cp2Ti(Me)I is titanium IV. $\endgroup$ – matt_black Oct 2 '19 at 12:28
  • $\begingroup$ My bad, not enough coffee $\endgroup$ – Waylander Oct 2 '19 at 12:57

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