# Simple soda–lime decarboxylation

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.

• 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 – Waylander Jan 23 at 10:04
• You have a leaving group beta to anion. You would have no doubt about what will happen if it were on an alkane. – 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.