Question

When 1 mol of `NaBH_4` is used to reduce a ketone, how many moles of Hydride is used? How do you determine this?

Answer 1

`RC(=O)R' + NaBH_4 rarr RCH(-O^(-)Na^+)R' + BH_3`

Explanation:

For the ketone reduction, while the first hydrogen adds as a hydride, we are usually left with `BH_3`, which is still a reducing agent. `NaBH_4` is usually used 1:1, because work up of `BH_3` is not too vigorous.

When `LiAlH_4` is used, if it is used 1:1, water work up of residual aluminum hydride is VERY VIGOROUS, sometimes so much so that problems can occur. While `LiAlH_4` could deliver 4 equiv hydrides, I used to use the commercial reagent (a gray powder) to deliver 3 hydrides. Work up of the residual salts was fairly innocuous.

Answer 2

Theoretically, 1 mol of `"NaBH"_4` can use 4 mol of `"H"^"-"` to reduce a ketone.

Explanation:

The equation for the first step is:

`"4R"_2"C=O" + "Na"^+"B"color(red)("H")_4^"-" → ("R"_2"C"color(red)("H")"-O")_4"B"^"-""Na"^+`

That makes a ratio of 1 mol of ketone to 0.25 mol of `"NaBH"_4`.

However, an excess of reducing reagent is often used because some of it may react with the solvent.

In the second step, water is added, followed by heating.

This hydrolyzes the boron complex and forms the alcohol:

`("R"_2"CH-O")_4"B"^"-""Na"^+ + 2color(brown)("H")_2"O" stackrelcolor(blue)(Δcolor(white)(m))(→) "4R"_2"CH-O"color(brown)("H") + "Na"^+"BO"_2^"-"`

Thus, one of the `"H"` atoms added to each molecule of the ketone comes from `"NaBH"_4`, and the second `"H"` comes from water.

Any excess reducing agent is also destroyed in this step, since `"NaBH"_4` decomposes in hot water.