Treatment of 2-vinylbicyclo[2.2.2]oct-5-en-2-ol with potassium hydride

Question

What happens when this bicyclic compound is treated with KH in THF?

Does $\ce{KH}$ simply act as a base here? $\ce{H-}$ from $\ce{KH}$ could potentially act as a nucleophile, but I don't find any electrophilic centre in the compound. What happens?

Answer 1

The reaction in question is one of the early examples of an oxy-Cope rearrangement ^[1] in which a 3-hydroxy-hexadiene undergoes a [3,3]-sigmatropic rearrangement to afford a ketone.

The general mechanism is shown below:

^{Source: Organic Chemistry, Oxford University Press, 2 ed. Clayden, Warren and Greeves}

In a general sense, the reaction is a [3,3]-sigmatropic rearrangement (if you look at the curly arrows, you'll see that we're essentially transposing a sigma bond from one end of the pi system to another).

The presence of the oxygen provides a driving force for the reaction, since the enol(ate) formed during the rearrangement is able to tautomerise, yielding the ketone product.

In the specific case you've given, the presence of a base means that during the sigmatropic rearrangement, the oxygen is deprotonated and therefore anionic. This specific case, using an alkoxide, is a specific kind of reaction known as an anionic oxy-Cope. The anion is thought to help speed up the reaction, to the extent that minimal heat is usually required.

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References:

_{[1] J. Am. Chem. Soc., 1964, 86,5019. DOI: 10.1021/ja01076a067}