Why do ketenes participate in [2+2] cycloadditions with their $\ce{C=C}$ bond and not the $\ce{C=O}$ bond? Can the reactivity be reversed somehow?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
Ketenes can participate in [2+2] cycloadditions with their $\ce{C=O}$ double bond. Ketene dimerization is the best known example of such a cycloaddition.
As you add substituents to the ketene $\ce{C=C}$ double bond, you begin to favor cycloadditon across the carbon-carbon double bond at the expense of cycloaddition across the carbonyl double bond.
Some Diels-Alder cycloadditions also involve adding across the ketene carbonyl.
With both the [2+2] and [4+2] cycloadditions often the kinetic product involves cycloaddition across the carbonyl, but on mild heating this cycloadduct can rearrange to the product from cycloaddition across the carbon-carbon double bond, the thermodynamic product.
A very nice review of the synthetic and mechanistic aspects of ketene cycloadditions can be found here.
-
$\begingroup$ "As you add substituents to the ketene C=C double bond..." Do you mean electronegative, electropositive or any kind of substituents and how does that work? $\endgroup$– EJCCommented May 5, 2015 at 19:21
-
$\begingroup$ Any kind, it's not well understood, but the effect is thought to be steric in origin. Look at the pictures of the two transition states at the top of page 10 in the last link in my answer ("here") - see how steric factors (repacing H with any kind of R) can start to impact the TS leading to cycloaddition across the carbonyl more than cycloaddition across the $\ce{C=C}$ double bond. $\endgroup$– ronCommented May 5, 2015 at 20:13