enter image description here

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:: enter image description here


1 Answer 1


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).

Product A Heating causes a double dehydration to produce the di-anhydride shown but the right hand pair of acids do not recombine, instead the top acids combine and the bottom acids combine.

Product B Heating dehydrates the right hand diacid to the anhydride, the left hand diacid does a Claisen-ester-type thermal condensation followed by decarboxylation to give a ketone in the left hand ring.

  • $\begingroup$ Got it now! Thanks! Just one doubt. Here a claisen occurs between the 2 acids on the left(product b) . For this a carbanion must be formed right? so should the questions mention a base? I have attached an image for this. Pls tell me if my rxn is correct. $\endgroup$ Apr 5, 2020 at 13:56
  • 1
    $\begingroup$ Yes, that is correct. I think if base was introduced then the salt of the di-acid is less likely to undergo the thermal condensation $\endgroup$
    – Waylander
    Apr 5, 2020 at 14:08

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