[ArcHos]

Hello! I have a trouble with understanding mechanism of reducing α,β-unsaturated carbonyl compounds with NaBH4. In Clayden's Organic Chemistry it is said that double bond is reduced first, while C=O is reduced only after the double bond. As I have understood  H^-  is a very hard nucleophile, thus it must attack C=O first. Will appreciate any help from you!

[orgopete]

Clayden is giving you a hint, the major product is the saturated alcohol. I think you will also find some product is the unsaturated alcohol if you did the actual reduction in the lab. That would be data. It is what you get, now explain it. By the way, I'm not so much a fan of hard and soft theory. Perhaps this example may alter your thoughts on it as well.

[BRSM]

Hard/soft is right more often than not, even if the theoretical justification is a bit shaky (kinda like frontier molecular orbital theory and hybridization).

Anyway, I would point out that hydride is a bit of misleading case - although it's small and charged it is actually a very soft nucleophile (because its interactions with electrophiles are governed orbital coefficients rather than charge). Check the wikipedia HSAB theory page if in doubt. Maybe this solves your conflict of ideas.

[ArcHos]

Thank for your responses orgopete and BRSM really helped!