How This two reducing agent act (LiAlH4 and NaBH4 ) on breaking different type of bond like (c=c,c=o e.t.c) in different compound?

And thus we get....

`RC(=O)H + AlH_4^(-) rarr RCH(-O^(-))H + AlH_3`

...the aluminium reagent has THREE more hydrides to transfer....

Reduction of an olefin, i.e. `C=C`, versus a carbonyl, would require pretty fierce conditions....

On the other hand, sodium borohydride is a bit less potent as a reductant...and a bit less reactive....

Both reagents normally reduce `"C=O"` double bonds to alcohols.

For example,

`"CH"_3"CH=CHCH"_2"CHO" stackrelcolor(blue)("LiAlH"_4color(white)(l) "or NaBH"_4color(white)(mm))(→) "CH"_3"CH=CHCH"_2"CH"_2"OH"`

The only alkene double bonds that either reagent reduces are those in
α,β-unsaturated carbonyl compounds.

`"LiAlH"_4` is the more powerful reducing agent but, surprisingly, `"NaBH"_4` is more likely to reduce a `"C=C"` double bond.

A. Lithium aluminium hydride (LAH)

LAH reduces alkene double bonds only to a minor extent.

However, it reduces propargylic alcohols to alcohol to (`E`)-allylic alcohols.

B. Sodium borohydride

`"NaBH"_4` has a significant tendency to reduce the `"C=C"` bond of α,β-unsaturated ketones.

For example, the reduction of cyclohex-2-enone gives an almost 50 % yield of cyclohexanol.

However, reduction in the presence of `"CeCl"_3` gives only reduction of the carbonyl group.