All I have found on this is that NaHSO3 reduces partially the osmate ester, so breaks off Os and leaves a syn diol. Is it possible to show this step with arrows showing the process of electron movement? I suppose this step is not the main part of the reaction, and mayn't be very important, but I would like to know if it's possible - and if it is, how it looks.


1 Answer 1


After you add the $\ce{OsO_4}$ to make the reaction, you'll need to reduce the remaining $\ce{OsO_4}$ maintaining the $\ce{pH}$ of the solution constant.

In addition:

The hydrolysis of the osmate ester proceeds by simple substitution reactions. The $\ce{NaHSO3}$, a mild reducing agent, facilitates ligand subsitution by reducing the $\ce{Os}$ further.

The Art of Writing Reasonable Organic Reaction Mechanisms, Robert B. Grossman, p 293

And $\ce{NaHSO_3}$ can do both and so you will use it. You also can add $\ce{Na_2SO_3}$.

I found also something interesting on JACS, it's mentioned at least four times in the document that $\ce{NaHCO3}$ is used as buffer. Also there is a lot of information in the references of this article.

While the reaction is normally run under basic conditions ($\ce{K_2CO_3}$, $\ce{pH} = 12.2$, aqueous layer), it is possible to buffer the system with $3$ equiv of $\ce{NaHCO3}$ ($\ce{pH} = 10.3$, aqueous layer).$^{38, 39}$ Buffering of the reaction mixture does not affect the ee, but it can have a beneficial effect on the yield when base-sensitive substrates are used or base-sensitive products are formed. Thus, the AD of allyl or cinnamyl halides (cf. eqs $1$ and $3$) should be performed under buffered conditions to minimize epoxide formation (see eq $1$, the yield with allyl bromide as substrate is between $61$ and $74%$ under buffered conditions; in the absence of $\ce{NaHCO3}$ only $40-50$ % of diol is obtained38). Unfortunately, the AD reaction does not turn over if $\ce{K_2CO_3}$ is replaced entirely by $\ce{NaHCO3}$.

Hartmuth C. Kolb, Michael S. VanNieuwenhze, K. Barry Sharpless Chem. Rev., 1994, 94 (8), p 2494


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.