What is the mechanism of Salkowski reaction of сholesterols?

I recently came across the Salkowski reaction of cholesterols. Cholesterol in chloroform is treated with concentrated sulfuric acid. A positive test exhibits two distinct layers, the upper chloroform layer gets a red to violet colour, while the lower sulfuric acid layer exhibits a greenish glow.


1. The chemistry behind the reaction is dehydration using $\ce{H2SO4},$ forming cholestadiene.


[![dehydration of сholesterol](https://i.stack.imgur.com/jyI4s.png)](https://i.stack.imgur.com/jyI4s.png)
2. In the second step, the cholestadiene undergoes dimerization.


[![dimerization of cholestadiene](https://i.stack.imgur.com/HOZXc.png)](https://i.stack.imgur.com/HOZXc.png)
3. In the last step, the cholestadiene undergoes sulfonation at the 7,7' position. (I believe I drew the correct structure)


[![sulfonation of cholestadiene](https://i.stack.imgur.com/ojs82.png)](https://i.stack.imgur.com/ojs82.png)


However, I couldn't find a *valid mechanism* for the second and the third steps (dimerization and sulfonation). Can someone explain the mechanism?




The mechanism of dimerization is now clear, Thanks to Mathew Mahindratne.
Ref.[Yoshihisa, *et.al*](https://doi.org/10.1016/0039-128X(78)90110-1)


There are two different isomers of Cholestadiene present, namely 3,5 Cholestadiene **[ I ]**, and 2,4 cholestdiene **[ IV ]**. The plausible mechanism of their formation is shown below. They can undergo dimerization to yield [ II ] & [III].
[![enter image description here](https://i.stack.imgur.com/oxbdF.png)](https://i.stack.imgur.com/oxbdF.png)


[![enter image description here](https://i.stack.imgur.com/0XK40.png)](https://i.stack.imgur.com/0XK40.png)


Dimer [ II ] was formed from two pathways, which were the conversion of the dimer [ III ] and the dimerization of [I] (3,5-cholestadiene).


The dimer [ III ] was formed from [ IV ] (2,4-cholestadiene).


The mechanism of their formation is shown below.
[![enter image description here](https://i.stack.imgur.com/RIKSx.png)](https://i.stack.imgur.com/RIKSx.png)


The formation of the dimer [ II ] and/or the dimer [ III ] varies with the kind of acids employed.
Using a **Bronsted acid** medium results in the formation of **[ II ]**.
For eg: In presence of Trichloroacetic acid and hydrochloric acid (10:1 ratio), dimer [ II ] was obtained.


Using a **Lewis acid** results in the formation of **[ III ]**. For eg: *Zinc chloride* and *Acetyl chloride* (Tschugaeff reaction),  

*Antimony trichloride* and *Acetyl chloride*,  

*Ferric chloride* and *Sulfuric acid* (Zak-Henly reaction),  

*Ferric chloride*, *Perchloric acid*, and *Phosphoric acid* results in the formation of [ III ].


If the solution contains both **Bronsted and Lewis acid**s, then a **mixture** **of products** is formed.


Eg: Both [ II ] and [ III ] are formed from the reactions with


1. Sulfuric acid and Acetic anhydride (Liebermann-Burchard reaction)
2. sulfuric acid (Salkowski reaction)
3. Trichloroacetic acid-SbCl3


Since sulfuric acid, having both Bronsted and Lewis acid character is used in our reaction, a mixture of products will be formed.
Ps: All Bronsted acids are Lewis acids, but not vice versa.


However, the third step (Sulfonation) needs further explanation.


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