Question

$H_{3}C-C\equiv C-C{H}_3\underset{Pd-BaS{O}_4}{\overset{H_2}{\to }}\underset{CC{l}_4}{\overset{B{r}_2}{\to }}M$

$H_{3}C-C\equiv C-C{H}_3\underset{liq.N{H}_3}{\overset{Na}{\to }}\underset{CC{l}_4}{\overset{B{r}_2}{\to }}N$

Relationship between $M$ and $N$ is:

• A. Identical
• B. Enantiomers
• C. Diastereomers
• D. Structural isomers

Answer 1

The correct option is C Diastereomers


M and N are diastereomers.
A Lindlar catalyst is a heterogeneous catalyst that consists of palladium deposited on calcium carbonate or barium sulphate which is then poisoned with various forms of lead or sulphur. It is used for the hydrogenation of alkynes to cis alkenes (i.e. without further reduction into alkanes) When the triple bond is (2-butyne) hydrogenated over the Lindlar's catalyst i.e. $Pd/BaS{O}_4$. It gives predominantly cis alkene (2-butene).
The bromines add to opposite faces of the double bond ("anti addition") as cis-alkene gives racemic compound. Sometimes the solvent is mentioned in this reaction - a common solvent is carbon tetrachloride $\left(CC{l}_4\right)$. $CC{l}_4$ actually has no effect on the reaction, it's just to distinguish this from the reaction where the solvent is $H_{2}O$, in which case a bromohydrin is formed.
Sodium will dissolve in liquid ammonia (boiling point $-33{}^{\circ }C$) producing a beautiful deep blue color. When alkynes are present, they will be reduced to the trans (i.e. E) alkene. trans-alkene gives meso product. This makes $Na/N{H}_3$, a useful companion to the Lindlar reduction of alkynes which gives cis-alkenes.