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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.

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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).

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  • $\begingroup$ What are the typical reaction conditions? $\endgroup$
    – Zhe
    Commented Dec 11, 2016 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.
    Commented Dec 11, 2016 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
    Commented Dec 11, 2016 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.
    Commented Dec 12, 2016 at 20:13
  • $\begingroup$ Isn't it just acid base chemistry at the end? $\endgroup$
    – Zhe
    Commented Dec 13, 2016 at 1:35

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the Deuterium oxide step seems to be irreversibly, yielding another titanium oxide Tebbe and the side product. Therefore each molecule that reacts is lost, may the reaction be as unfavoured as it is. also the five membered ring constrain may shift equilibrium of the intermediate step towards metallacycle. maybe less side product with cyclohexanone?

if you activate your tebbe reagent with base you may be better on not doing so, since the overreaction demands another coordination to the metal. this may be less favoured when your tebbe reagent is still in its trimethylaluminium stabilized form. it is unknown to me if the Tebbe reagent is then reactive enough for the demanded first reaction.

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