In a Wurtz reaction, is the intermediate formed a carbanion or a free radical?
And if we can assume both mechanisms to be true, how can we decide the reactivity of a tertiary halide?
$\begingroup$
$\endgroup$
1
asked Apr 20, 2015 at 2:27
-
$\begingroup$ I think your second question is fine as is. But, in the context of your first question, I just want to verify for myself that you do mean overall reactivity of a tertiary halide via the Wurtz reaction, irrespective of reaction pathway, and that you are not asking whether the reactivity will be dominated by carbanion-forming v.s. free-radical forming pathways. Wow, ancient question ;) I hope we can lead you to an acceptable answer! $\endgroup$– airhuffJan 13, 2017 at 2:48
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
I think both mechanisms are valid. As far as I know, both are supported by some experimental evidence. Byproducts, like alkanes (other than the expected one) and alkenes, can be explained by the free radical mechanism. On the other hand, the formation of Organometallic compounds can be verified by the ionic mechanism.
The essential thing to note is that the mechanism involves a $\mathrm{S_N2}$ step. And because of that sterically hindered alkyl halides, like tertiary alkyl halides, instead of giving a substitution product yield an elimination product and you get an alkene.
$\begingroup$
$\endgroup$
A mechanism is only a hypothesis which can explain a particular reaction. Wurtz reaction can be explained by both ionic and free radical mechanism.
$\begingroup$
$\endgroup$
Well I personally think it is better explained by free radical mechanism, as using this mechanism, you can easily analyse all the products formed, whether it be symmetrical compound formed from same alkanes, or an unsymmetrical compound.
