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I was reading some enolate chemistry from Carruthers textbook and came across the selective alkylation of unsymmetrical ketones. The reaction involves blocking one alpha-position of the ketone using an enol thioether, as shown below. enter image description here

I cannot understand the mechanism of the final step and while searching, I ended up with this from Sciencedirect (Ref.1)

Removal of the thioenol ether protecting group gives the aldehyde function, and under the acidic conditions the β-hydroxy group eliminates to give the mainly (E)-unsaturated enal. This reaction also works if the ketone is reduced to a secondary alcohol with LiAlH4 (69% yield).

My questions are:

  1. Are enol thioethers the same as thioenol ethers? If not, please draw an example.
  2. If they are the same, does the deprotection take place through an aldehyde intermediate as stated above?
  3. Also, please explain the mechanism of this deprotection.

Ref. 1: Warren J. Ebenezer, Paul Wight, in Comprehensive Organic Functional Group Transformations, 1995
Ref. 2: Modern Methods of Organic Synthesis 4th edition, page no. 10

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2 Answers 2

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Mechanism: 1 -hydroxide anion adds to the =CHSBu carbon in a Michael addition giving the enolate anion of the carbonyl and -CHOH(SBu).

2- The enolate anion deprotonates the newly arrived -OH group to reform the carbonyl group

3- The alkoxide anion displaces BuS- to form the aldehyde

4- retroaldol gives the unsubstituted carbonyl

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  1. What you have here would conventionally be a thioenol ether. I've never heard the term enol thioether or enol sulfide before, but I'd assume it is the same because a thioether which is not in a enol structure would normally be classified as a sulfide.

  2. Yes, it does. As stated it is the beta hydroxy tautomer of the beta aldehyde which is the key intermediate.

  3. I believe the mechanism involes first hydrolysis of the thioenol ether to a thiocarbonyl. Then the thioaldehyde is hydrolysed further to an aldehyde. From here, the key step is a retro claisen condensation to remove the aldehyde group. The presence of base and glycol and water would support this mechanism. Formic acid and some formate esters would be found as by-products.

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  • $\begingroup$ The term enol thioether is written in the book itself, and the given structure is in the enol form. And, what is enol sulfide here? Also, someone please explain the mechanism clearly. $\endgroup$
    – Cyclopropanol
    Commented Oct 11, 2023 at 3:59