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.

propargyl

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.

Unsat

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