1
$\begingroup$

The major product obtained in the following reaction is:

Enter image description here

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?

Improve this question
$\endgroup$
5
  • 4
    $\begingroup$ 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
    Commented Aug 8, 2021 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
    Commented Aug 8, 2021 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
    Commented Aug 8, 2021 at 5:26
  • $\begingroup$ Re "carbonitriles": Just "nitriles"? $\endgroup$
    – Peter Mortensen
    Commented Aug 8, 2021 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 $\endgroup$
    – Nilay Ghosh
    Commented Aug 9, 2021 at 1:42

3 Answers 3

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
0
$\begingroup$

For this question : The Lindlar's catalyst is poisoned by quinoline ,Here since quinoline isn't mentioned the Pd isn't poisoned hence it will act as H2/Pd only. You may be confused because in various textbooks and questions H2/Pd-C is mentioned as poisoned without mentioning quinoline. So Sometimes while solving you "may consider" H2/Pd-C or H2/CaCo3 as poisoned. However here it is as theroy would state it will act as normal H2/Pd which can reduce CN and C=O bonds.

Improve this answer
New contributor
Shreyash Kumar is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
0
$\begingroup$

This isn't Lindlar catalyst. The hydrogenation reaction with Pd/C can reduce C-C, C-O as well as N- O bond. But if we do it selectively then it would reduce the C-C BOND first

Improve this answer
New contributor
Rishabh Kaushal is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
1
  • $\begingroup$ Ps: sometimes jee mains answers keys are really fucked up. But that does not generally happens in Jee Adv. $\endgroup$
    – Rishabh Kaushal
    Commented 56 mins ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.