[Ashutosh]

Ithink LiAlH4 is best reducing agent due to presence of 4 hydrogen atoms in it
how to chose reducing agent for any reaction?
e.g.
(CH3)2CO+2H=(CH3)2CHOH
 this reaction takes place in the presence of Na-Hg.
further reaction is not possible.
But in clemenson reduction,
(CH3)2CO+4H=CH3CH2CH3+H2O
this reaction takes place in the presence of Zn-Hg and conc.HCl.
why the formation of propane in the first reaction is not possible?
is the reason concentration of reagent?
if then can the propane formation propane in the first reaction possible by increasing concentration of Na-Hg to have more no. of H atoms for the reaction

[movies]

I think if you did the reduction under the same conditions both Na-Hg and Zn-Hg could both yield propane.  The difference between these two is that each Na in the Na-Hg can only provide one electron for reduction, while each Zn can provide two electrons for reduction.

LiAlH₄ reacts very differently however, an generally can't reduce alcohols to alkanes.  The difference is in the mechanism.

[Ashutosh]

THANKS MOVIES. MY DIFFICULTY IS CLEARED.