Question

The reduction of 4-octyne with $H_2$ in the presence of $Pd/BaS{O}_4$ and quinoline gives:

• A. trans-4-octene
• B. cis-4-octene
• C. a mixture of cis and trans-4-octene
• D. a completely reduced product $C_8{H}_{18}$

Answer 1

The correct option is B cis-4-octene

The complete reduction of alkynes leads to alkane so while adding $H_2$ to alkyne, if the reduction is to be stopped at the alkene stage, the catalyst needs to be poisoned.
A palladium catalyst $\left(Pd/BaS{O}_4\right)$ is deliberately poisoned with lead/sulphur compounds/Quinoline.
This reagent is known as Lindlar's catalyst. They give only cis product.
Reduction of 4-octyne with $H_2$ and $Pd/BaS{O}_4$ with quinoline gives cis-4-octene. The reaction is shown below.