Organic reduction

@tapan....

wrt post #7

Ph--CH=CH--CH=O ----- ^LiAlH₄ -------> (fatal for C=C) Ph--CH₂--CH₂--CH₂--OH

BUT

Ph--CH=CH--CH=O ----- ^NaBH₄ -------> (C=C is not affected) Ph--CH=CH--CH₂--OH

oh i wrote ulta.....

Conjugated system

----- ^LiAlH₄ ------>
R--CH=CH--CH=O R--CH=CH--CH₂--OH (C=C is not affected)
----- ^NaBH₄ ------>

BUT

----- ^LiAlH₄ ------>
R--CH=CH--C--CH₃ R-CH₂--CH₂--CH--CH₃ (~100%)
|| ----- ^NaBH₄ ------> |
O OH

r u sure?

see LiAlH4 is an hardnuclophile, so it attacks on more nuclophilic site, so it will attack first on c=o and if taken in excess it will reduce double bond also,

and NaBH4 is a soft nuclophile so it will attack on double bond only and

well, LIAlH4 and NaBH4 never reduce alkenes or alkynes, because they reduce by giving H- ion, which can never approach double or triple bond