I'm confused about what is going on here. I gather that step 1 is typically supposed to protect an OH group, but there isn't any in the starting material.

reaction scheme

  • 3
    $\begingroup$ organic-chemistry.org/namedreactions/rubottom-oxidation.shtm $\endgroup$
    – orthocresol
    Oct 28 '17 at 13:01
  • $\begingroup$ What does MCPBA usually do with an electron rich double bond? $\endgroup$
    – Waylander
    Oct 28 '17 at 13:32
  • $\begingroup$ Since step 2 occurs after step 1, epoxidation only occurs after the protection of the OH group. By implication, enolization must have occured before step 2 ( I suspect enolization catalyzed by NEt3). By logical extension the carbonyl oxygen in molecule A is also the oxygen within the -OTMS group. But since the carbonyl oxygen in meta to the methyl group in molecule A, and para to the methyl group in molecule B, I am unsure of what happened. $\endgroup$ Oct 28 '17 at 15:17
  • $\begingroup$ If this was something like TBDMS, I'd expect it to resist hydrolysis, but OTMS normally hydrolyses under acidic conditions as shown. Unless there's stabilisation from the carbonyl? $\endgroup$
    – Beerhunter
    Oct 29 '17 at 8:40
  • $\begingroup$ @Beerhunter- the TMS would go. Possibly a typo in the question, or possibly just left there as a hint as to what's occured $\endgroup$
    – NotEvans.
    Oct 29 '17 at 19:55

In connection with a reasonable base (triethylamine works fine for this), ketones are known to be enolised by silyl chlorides and trapped as silyl enol ethers. This is not only a key process in Mukaiyama aldol reactions but also in your structure.

In general, the combination of $\ce{NEt3}$ and $\ce{TMSCl}$ has three possible ways to enolise your compound:

  1. γ-deprotonation to the methyl group, creating an exocyclic double bond

  2. γ-deprotonation of the ring proton to give a 1-silyloxycyclohexa-1,3-diene system

  3. α-deprotonation to the other side, giving a 2-silyloxycyclohexa-1,3-diene system

By the general rules one is taught to follow, the second option should be identified as the thermodynamically most favoured. However, the third option is typically the kinetically favoured one, so given the correct conditions (low temperatures), we typically expect enolisation by deprotonation the α proton. It helps here that $\ce{TMSCl}$ is a rather small and reactive silyl chloride (Lewis acid), meaning that once it has attached to the keto oxygen deprotonation will be a rather rapid process.

If you left the solution to equilibrise with $\ce{NEt3}$ before quenching you would probably get much more of the γ-deprotonation products.

The next two steps, a Rubottom oxidation, are basically the generation of an epoxide by $\ce{$m$CPBA}$ (as an isolated reaction also known as the Prilezhaev reaction) followed by acid-catalysed epoxide opening which reforms the $\ce{C=O}$ double bond and initiates a Brook migration of the silyl group. The overall mechanism is given in the scheme below.

Mechanism of the Rubottom oxidation
Scheme 1: mechanism of the Rubottom oxidation.[1]


[1]: L. Kürti, B. Csakó, Strategic Applications of Named Reactions in Organic Synthesis, Elsevier Academic Press, 2005, pp. 388–9.

  • $\begingroup$ Brilliant! Though I think there should be an arrow from the O-TMS bond to the O so as to restore the neutral charge, in the final step. $\endgroup$ Oct 30 '17 at 3:48

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