Sodium borohydride/iodine reduction

Lately I’ve been doing quite a few asymmetric Michael additions using a chiral oxazolidinone, or one of the thio variants, as a chiral auxiliary. Some of the auxiliaries are commercially available but far too expensive for my taste. They can be easily synthesized from amino alcohols which are also commercially available but they too cost more then I’m willing to pay. So, the only option is to make them myself, which is not that bad a choice considering that the corresponding amino acids are not that expensive, especially the L ones. The D ones can be quite pricey. This is how the auxiliaries look like:

There are many ways to reduce amino acids to amino alcohols but my favourite by far is sodium borohydride/iodine combination (.pdf), which is not only non-expensive alternative but also convenient compared to, for example, lithium aluminum hydride (see, I don’t like burning labs). Below is how it looks like chemically written:

So, what is actually happening there? This is how I understand it:

First, iodine is reacting with one sodium borohydride, so you get borane, sodium iodide, and hydrogen iodide. Then hydrogen iodide is reacting with another sodium borohydride, so you get another borane and another sodium iodide, plus some hydrogen. All that borane is actually the stuff that reduces the acid to alcohol. Very nice and convenient. I’ve been doing it in 200 mmol scale and works just great, giving the amino alcohol in good >70% yield after recrystallization from toluene.

That’s it for today. Feel free to stop by again and comment if you like.

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