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?

Please describe it clearly.It confuses me alot... Please provide side to side difference if possible...

2 Answers
Mar 19, 2018

Both "lithal" and "sodium borohydride" are hydride transfer reagents...

Explanation:

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....

Mar 22, 2018

Here's what I get.

Explanation:

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.

propargylpropargyl

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.

UnsatUnsat

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