This is my homework question I tried it hard the question is as follow:


I attempted it as follow:


According to my understanding as acid base reaction are fastest the carbanion will gain a $\ce{H+}$. From the water formed and will the product as I made, but the answer is completely different:


Please guide me where I am wrong.

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    $\begingroup$ You are wrong in thinking that the anion of the chloroalkene will simply pick up a proton. These reactions are run at a high temperature ao there is plenty of energy around for forming unstable species $\endgroup$ – Waylander Jan 23 at 10:04
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    $\begingroup$ You have a leaving group beta to anion. You would have no doubt about what will happen if it were on an alkane. $\endgroup$ – Waylander Jan 23 at 10:36

You are correct in the first step - the decarboxylation gives the anion of the chloroalkene.

It is important to remember that such soda lime reactions are run at high temperature and CaO is a powerful drying agent, so there will be no water present to quench the anion. What happens next would be obvious if you were dealing with an alkane; there is a leaving group beta to an anion so there would be an elimination with loss of Cl-. This is what happens; Cl- leaves to give the triple bond (again remember that this is a high temperature reaction so plenty of energy for forming unstable species). This is clearly unstable within a [6] ring so it dimerises by a 2+2 process to give the product drawn.


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