# Action of KMno4/H+ and heat in this reaction

The first step uses LAH which is a hydride donor. So H- attacks any one carbonyl carbon and forms an aldehyde and acid attached to the ring.

Then Conc. KMno4 is used. As far as i know, it cleaves c=c and oxidises them into acids. The third step then says heat, so it leads to the formation of the anhydride on both sides.

However, this product has nothing to do with the correct answer which is option (A) . Somehow the product formed consists of 2 seprate 5 membered rings. Any idea on where i went wrong?

edit::

The $$\ce{LiAlH4}$$ will reduce the cyclic anhydride to the diol.
$$\ce{KMnO4}$$ re-oxidises both the alcohol groups (diol formed) to carboxylic acids and also oxidises the double bond to the diacid so the product you have drawn is correct, but it is an intermediate. You then have to consider the effect of heating this (to a pretty high temperature, I would assume).