6
$\begingroup$

The reaction in black below was presented in a set of lecture notes as an example of the use of the Tebbe reagent, however it's not immediately obvious (to me, at least) that the product should be the gem-dimethyl compound shown.

enter image description here

Mechanistically, it appears that the Tebbe olefination has taken place once (standard formation of Schrock carbene, [2+2], retro [2+2] to give the alkene and a Ti=O compound as a thermodynamic driving force). The alkene must then however react with more Tebbe reagent (or more Schrock carbene) to form a metallacyclobutane. It's at this stage that I feel it makes less sense: firstly, what is the involvement of the deuterium oxide, but secondly and more importantly, why doesn't the metallacyclobutane just collapse back down to form the thermodynamically favoured alkene product.

I've used Tebbe in the past to turn ketones into terminal methylene groups, but never encountered what appears to me to be an over-reaction (I'd actually always considered Tebbe to be fairly mild and selective).

After some digging around in Comprehensive Organic Synthesis I, there is some mention of this 'over reaction' (1985JOC50). The authors report that treating the ketone 22 with 1 equivalent of Tebbe reagent affords a mixture of the expected methynlenation product along with some of the gem-dimethyl product (i.e. implying that an excess isn't required, but the reaction is just not overly chemoselective for the cyclic ketone over the cyclic alkene).

enter image description here

$\endgroup$
  • $\begingroup$ What are the typical reaction conditions? $\endgroup$ – Zhe Dec 11 '16 at 20:03
  • $\begingroup$ Tebbe reagent and some pyridine to help form the Schrock carbene. Usually use more than 1 eq anyway (i.e. not being super careful to control stoichiometry) but in the JOC paper cited they use a 1:1 ratio and still get some overaddition $\endgroup$ – NotEvans. Dec 11 '16 at 20:08
  • $\begingroup$ I was thinking more along the lines of temperature. Also, I think you already noted this, but you might get an intermediate mixture under thermodynamic control before the quench with water. $\endgroup$ – Zhe Dec 11 '16 at 21:25
  • $\begingroup$ @Zhe Perhaps. It seems like one obtains a mixture anyway (similar to Grignard + ester kind of problem?). In any case, the mechanism of the second step is still puzzling me, esp the breakdown using D2O $\endgroup$ – NotEvans. Dec 12 '16 at 20:13
  • $\begingroup$ Isn't it just acid base chemistry at the end? $\endgroup$ – Zhe Dec 13 '16 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.