Why it is not yet possible to produce ethanol selectively in the Fischer Tropsch reaction?
in Fischer-Tropsch synthesis-like reaction there are (roughly) following processes:


$\ce{CO \rightarrow  Surf-CO}$
$\ce{H2 \rightarrow  2 Surf-H}$
$\ce{CO + Surf-(CH2)_{n}-H \rightarrow  Surf-CO-(CH2)_{n}-H}$

3'. $\ce{CO + Surf-H \rightarrow  Surf-CO-H}$
$\ce{Surf-CO-(CH2)_{n}-H + 4 Surf-H \rightarrow
    Surf-(CH2)_{n\,+1}-H + H2O }$
$\ce{Surf-CO-(CH2)_{n}-H + 3 Surf-H \rightarrow 
    HO-(CH2)_{n\,+1}-H }$
$\ce{Surf-CH2-CH2-(CH2)_{n}-H \rightarrow Surf-H  + 
    CH2=CH-(CH2)_{n}-H}$
$\ce{Surf-(CH2)_{n}-H + Surf-H \rightarrow H-(CH2)_{n}-H}$


All processes happen simultaneously, and tuning catalyst and reaction conditions it is possible to hinder some processes. However, as you can see, as long as you have chain growth at all (that is necessary to have $\ce{C2}$ products, like ethanol), you cannot stop $\ce{C2}$ surface particles to participate in further chain growth, producing $\ce{C_{3+}}$ products. So, if chain growth happens at all, a mixture of products with different chain length is always obtained, and there is no way around it. On the other hand, on proper catalyst in proper conditions it is possible to prevent chain growth, but still allow reduction of carbon monoxide, gaining either methane or methanol depending on catalyst. So, in practice it is possible to obtain one of the following


Methane
Mixture of terminal alkenes and alkanes
Methanol
Mixture of linear terminal alcohols (OK, this case is a bit more complicated, as other oxygenates may be produced as well, giving rather complex mixture, but that's beyond the point)


It is still possible to obtain pure ethanol indirectly (using hydration of isolated ethylene), or using other processes. It is also possible to obtain product mixture where some products, say C2-C4 alcohols, will be dominating.