2
$\begingroup$

This is a question from Org. Chem. by Solomons, Fryhle & Snyder, adapted for JEE by M.S. Chouhan :

enter image description here

  1. In the first step, I expected that the reaction would stop at alkene formation via syn addition of $\ce{H_2}$ since Lindlar catalyst is poisobed and heterogeneous. What are some other other poisoned catalysts that stop at alkene?

  2. In the second step, I expected syn addition of $\ce{D_2}$ since $\ce{Pd-C}$ is a heterogeneous catalyst, forming product (a). But the answer given is (c) — does this mean that some heterogeneous catalyts can give anti addition, or is it just an error on the adapting author's part?

$\endgroup$
7
$\begingroup$
  1. The "Lindlar catalyst" (Pd/CaCO3) is probably the most common way of effecting stereoselective reduction of alkynes to (Z)-alkenes. It is fairly accessible and the reaction is easy to run. But it is not the only way: there are other catalyst systems. One of the more notable ones is probably the Brown catalyst "P2-Ni" (Wikipedia).

    There is a nice review on alkyne semihydrogenation here: Chem. Rev. 2013, 113 (3), 1313–1350, which covers both heterogeneous and homogeneous methods, as well as methods involving other hydrogen sources apart from H2 gas itself. I recommend it to interested readers and especially anybody considering carrying out such a reduction.


  1. The Lindlar hydrogenation of an alkyne forms (Z)-alkenes stereoselectively. Syn hydrogenation of the (Z)-alkene gives product (c), not product (a).

    Reaction scheme

    Note that in the fully reduced alkane, there is free rotation about the central C–C bond. So, having both D atoms pointing "out of the plane" is irrelevant, as you can easily rotate to a conformation in which one D atom is pointing in the plane. However, the configuration of the product cannot be changed, and that is the difference between options (a) and (c).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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