The major product obtained in the following reaction is:

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Question source: JEE Main 2019

The molecule shown as product is the official answer given by JEE from the given options.

I have written the possible intermediate molecule formed by reagents. On Google I have obtained information in this PDF document (page 3) that a Lindlar catalyst is poisoned, so that it cannot reduce an aldehyde. Also there is no information of a Lindlar catalyst being used as the reducing agent for nitriles and carbonitriles.

So, is the answer by JEE incorrect or am I missing something?

  • 3
    $\begingroup$ Note: $\ce{Pd/C}$ is not similar to $\ce{Pd/BaSO4}$ or $\ce{Pd/CaCO3}$. So the reagent in your question isn't Lindlar's catalyst in the first place. It behaves like normal $\ce{Pd/H2}$ as far as I've learnt. $\endgroup$
    – TRC
    Aug 8 '21 at 5:10
  • 2
    $\begingroup$ From MasterOrganicChemistry Reagent Guide - Pd/C reduces alkene, alkyne, nitro, cyanide, C=N, azide - and probably most other reducible groups too. $\endgroup$
    – TRC
    Aug 8 '21 at 5:24
  • 2
    $\begingroup$ From Organic Chemistry for JEE Advanced by KS Verma, Cengage publications - It reduces all of the above but they haven't mentioned alkyne and C=N. Most likely missed mentioning those two. $\endgroup$
    – TRC
    Aug 8 '21 at 5:26
  • $\begingroup$ Re "carbonitriles": Just "nitriles"? $\endgroup$ Aug 8 '21 at 14:55
  • $\begingroup$ @PeterMortensen Essentially they are the same, but in some nomenclature rules, carbonitrile is used instead of nitrile: acdlabs.com/iupac/nomenclature/79/r79_566.htm $\endgroup$ Aug 9 '21 at 1:42

There is just a silly mistake. As TRC told, $\ce{Pd/C}$ is not a Lindlar catalyst and it behaves as a normal $\ce{Pd/H2}$. Refer to the link for more information.


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