# What is the role of NaHSO3 in syn diol formation from alkenes with OsO4?

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.

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.

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.
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 $\ce{NaHCO3}$.