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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

  2. In the second step, the cholestadiene undergoes dimerization.

    dimerization of cholestadiene

  3. In the last step, the cholestadiene undergoes sulfonation at the 7,7' position. (I believe I drew the correct structure)

    sulfonation of cholestadiene

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

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    $\begingroup$ I'm not sure, but this mechanism looks very "hand-wavey" to me, possibly just to make it a little easier to follow. This paper makes me think it might go through a carbonation intermediate. $\endgroup$ Dec 3, 2022 at 20:21
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    $\begingroup$ @SendersReagent: However, this article before yours has claimed the isolation of final dimer without sulfonic acid groups under Zac-Henry conditions. $\endgroup$ Dec 4, 2022 at 15:28
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    $\begingroup$ I think this article may give you the insight for given chemical formations. $\endgroup$ Dec 4, 2022 at 15:35

1 Answer 1

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The mechanism of dimerization is now clear, Thanks to Mathew Mahindratne. Ref.Yoshihisa, et.al

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].

Mechanism of formation of cholestadiene isomers

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

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 acids, 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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