The backbonding, as I know, is proportional to the number of electrons in the d-orbital That's indeed part of it — and that explains why the early d-block metals, Sc and Ti, don't form (neutral) carbonyls at all. However, the backbonding also depends on how willing the metal is to share those d-electrons. $Z_\mathrm{eff}$ increases going across the 3d block,...


As already pointed out in the comments, the second reaction reagent is not exactly an alcohol but an alkoxide. For example, $\ce{EtONa}$ and not $\ce{EtOH}$. This makes the reaction condition for the second reaction basic whereas the first reaction is taking place in acidic conditions.

Only top voted, non community-wiki answers of a minimum length are eligible