# Reduction using sodium borohydride

When we react a ketone with NaBH4, it reduces to alcohol. I coudn't understand the mechanism of this reaction. The first step in my text book states that the hydrogen from NaBH4 attacts the carbonyl carbon. I didnt't understand why this happens. Specifically, I coudn't understand why the Boron didnt attack oxygen instead.

• For starters, the formal charges should provide a good hint as to how the reaction proceeds. – Zhe Nov 5 '18 at 22:50

Sodium borohydride is a complex unit made from the let's call it molecule $$\ce{BH3}$$ and the salt $$\ce{NaH}$$. Group 3 compounds like boron only have what we call an electron sextett instead of an octett. That makes them good Lewis acids, hence they tend to react with Lewis bases to form adducts. A famous example is the $$\ce{H3B-NH3}$$ adduct. In your example the hydride anion $$\ce{H^-}$$ is the Lewis base that forms an adduct with $$\ce{BH3}$$ to give $$\ce{Na[BH4]}$$. This means that at least one $$\ce{H}$$ in $$\ce{[BH4]^-}$$ can be somewhat easily removed again. The others tend to react as well but let's just call the $$\ce{BH3}$$ unit a stable molecule and assume it's a combination of $$\ce{BH3}$$ and $$\ce{NaH}$$. It's just a good way to store this hydride-anion so that it can be easily removed again. So once it sees your carbonyl group that has a positive partial charge on the carbonyl carbon (as the more electronegative oxygen pulls the electrons closer to itself) it can attack there as nucleophile. As you drew it already the double bond will turn to a single bond giving a negative charge on the oxygen.
Next you have to assume that $$\ce{B}$$ and $$\ce{O}$$ form quite stable bonds, hence the formation of a $$\ce{B-O}$$ bond is somewhat favored. And of course once the hydride ion is at the carbon we end up with an electron sextett on the boron again. Therefore it's the same thing, the carbonyl oxygen with an extra negative charge behaves like a strong Lewis-base again forming an adduct with the $$\ce{BH3}$$ unit.