Why does lialH4 and NaBH4 reduce C=O but not C=C ? what is the reason

Generally, neither LiAlH4 nor NaBH4 is able to reduce an isolated C=C But——

Cinnamaldehyde, C6H5-CH=CH-CHO when added slowly to an excess of the LiAlH4 dissolved in ether (normal addition), gives C6H5CH2CH2-OH, whereas in “inverse addition method” where LiAlH4/ether is added slowly to cinnamaldehyde, so that LiAlH4 is never in excess, the product is cinnamyl alcohol, C6H5-CH=CH-CH2OH.

The exact process and products of reduction of conjugated carbonyl compounds are dependent on the nature of reductant.

In case of conjugated aldehyde or ketone, the nucleophilic attack takes place at either the β-carbon (conjugate addition) or at the carbonyl carbon (direct addition).

According to the HSAB Principle, the vinylic β-carbon is a "soft" electrophilic center and prefer to react with "soft" nucleophiles whereas the carbonyl carbon is a "hard" electrophilic center and hence prefers to react with "hard" nucleophiles.

NaBH4 is a rather soft nucleophile and thus it prefers in some cases to react with 4-position or reduces conjugated >C=C

The extent of reduction to saturated alcohol is usually greater with NaBH4 than with LiAlH4.