Question

Reaction of $Pd/BaS{O_4}$ with 2-butyne gives predominantly:
A. Cis-2-butene
B. But-1,3-diene
C. 1-Butyne
D. 1-Butene

Answer 1

Try to recall that $Pd/BaS{O_4}$ is known as Lindlar catalyst and is a heterogeneous catalyst which is used for the partial hydrogenation of alkynes. Now by using this you can easily answer the given question.

Complete step by step solution:
It is known to you that $Pd/BaS{O_4}$ is known as Lindlar catalyst and is named after its inventor Herbert Lindlar Wilson.
It is a heterogeneous catalyst which consists of palladium deposited on barium sulphate with traces of lead and quinoline.
Since, palladium is a good absorber of hydrogen and has very high catalytic properties. Therefore, it is poisoned with various forms of lead or sulphur or quinoline in order to reduce its activity of reducing double bonds.
So, $Pd/BaS{O_4}$ is used for the partial hydrogenation of alkynes to alkenes and does not have the ability to reduce double bonds.
Also, the product formed by using Lindlar catalyst i.e. $Pd/BaS{O_4}$ is cis alkene.
Hence, when 2-butyne reacts with $Pd/BaS{O_4}$ it forms cis-2-butene. The reaction is:

In the above hydrogenation reaction, hydrogen atoms get added to the same side(cis) of alkyne and form cis alkenes through syn addition (addition of two substituents on the same side of alkyne or alkene).
Hence, from above we can clearly say that option A is the correct option to the given question.

Note: Students should remember that Hydrogenation of alkynes in presence of $Pd/BaS{O_4}$ is stereoselective and happens through syn addition.
Also, it should be remembered that if $Pd/BaS{O_4}$ is directly used without being poisoned then it will hydrogenate alkynes directly to alkanes. That’s why quinoline is used as a catalytic poison to stop the reaction at alkene.