Mechanism of S2Cl2 and NaHSO3 [closed]








**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
 
 






---








 **Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)




Closed 3 months ago.








[Improve this question](/posts/174100/edit) 



I learned that NaHSO3 reacts with S2Cl2 to give Na2S4O6 as product. After looking up the structure of Na2S4O6, it confuses me that four sulfur atoms connect together. I couldn't figure out the exact mechanism for this reaction.




Nothing odd happens. The $S-S$ covalent bond is relatively strong and sulfur atoms don't mind at all linking up in chains of four or longer. In that respect they're quite differengt from oxygen atoms, which have smaller, highly localized lone pairs that repel each other strongly.


Mechanistically, the reaction is a simple nucleophilic substitution, where both the nucleophile (in $HSO\_3^-$) and the electrophilic center (in $S\_2Cl\_2$) are sulfur atoms. The bisulfite ion is electron rich due to its charge, and attacks one of the electron poor (due to the electron withdrawing chlorine atom) sulfurs in $S\_2Cl\_2$, expelling the chlorine as $Cl^-$ and forming a $S-S$ bond in addition to the one already present in $S\_2Cl\_2$.
$$\ce{S\_2Cl\_2 + HSO\_3^- -> HS\_3O\_3Cl + Cl^-}$$
This reaction will proceed smoothly as chloride is a weak base and good leaving group, and sulfur has diffuse electron pairs and is a good nucleophile, especially when charged. The resulting intermediate is much more acidic and easily gives off a proton; if there is no other base present sulfurous acid will form, which will partially decompose into $SO\_2$ and $H\_2O$:
$$\ce{HS\_3O\_3Cl + HSO\_3^- <=>> S\_3O\_3Cl^- + H\_2SO\_3}$$
$$\ce{H\_2SO\_3 <=> H\_2O + SO\_2}$$


Now repeat the exact same events on the second sulfur atom in $S\_2Cl\_2$ with a second $HSO\_3^-$, forming the fourth $S-S$ bond, and you get tetrathionate:


$$\ce{S\_3O\_3Cl^- + HSO\_3^- -> S4O6^2- + H+ +Cl-}$$
$$\ce{H+ + HSO\_3^- <=>>H\_2O + SO2}$$


Combining both steps we get an overall stochiometry:
$$\ce{S2Cl2 + 4 HSO3- ->S4O6^2- + 2 Cl- + 2SO2 + 2H2O}$$


 3

The simplest explanation is that the central sulfur atoms in $\ce{S2Cl2}$ are acting as substrates for a nucleophilic substitution reaction. Additional bisulfite ions, or an added base, removes the protic hydrogens. A similar mechanism with alkyl halides as the substrate is one way to introduce sulfonate functions into organic compounds.


[![enter image description here](https://i.stack.imgur.com/CJo2Z.png)](https://i.stack.imgur.com/CJo2Z.png)


 2