Question

When 2-butyne is treated with ${{\text{H}}_{\text{2}}}$/ Lindlar’s catalyst, compound X is produced as the major product and when treated with ${\text{Na/liq}}{\text{.N}}{{\text{H}}_{\text{3}}}$ it produces Y as the major product. Which of the following statements is correct?
A.Y will have higher dipole moment and higher boiling point than X
B.Y will have higher dipole moment and lower boiling point than X
C.X will have lower dipole moment and lower boiling point than Y
D.X will have higher dipole moment and higher boiling point than Y

Answer 1

To answer this question you must recall the mechanism of action of ${{\text{H}}_{\text{2}}}$/ Lindlar’s catalyst and ${\text{Na/liq}}{\text{.N}}{{\text{H}}_{\text{3}}}$ on an alkyne. Both the reagents are used to convert alkynes into alkenes.

Complete step by step solution:
The conversion of alkynes to alkenes is achieved by the partial hydrogenation using special catalysts.
The reactions taking place are stereoselective reactions and the product formed may be either cis alkene or a trans alkene depending upon the type of catalyst used.
In hydrogenation using Lindlar’s catalyst, the hydrogen atoms are added on the same side of the molecule. This type of addition is known as syn addition. On treatment of 2-butyne with hydrogen gas and Lindlar’s catalyst, the major product obtained is cis-2-butene.
X is cis-2-butene
In hydrogenation using ${\text{Na/liq}}{\text{.N}}{{\text{H}}_{\text{3}}}$, the hydrogen atoms are added to the opposite sides of the double bond as it proceeds via two single electron transfers from sodium metal to alkene followed by protonation with ammonia. This type of addition is known as anti-addition. The product obtained on treatment of 2-butyne with ${\text{Na/liq}}{\text{.N}}{{\text{H}}_{\text{3}}}$ is trans-2-butene.
Y is trans-2-butene.
In trans- 2-butene, the methyl groups are present on the opposite side of the double bond. This reduces the dipole moment of the molecule. While in cis-2-butene, both the methyl groups are present on the same of the double bond and thus, the molecule has a higher dipole moment.
Thus, we can say that X has a higher dipole moment than Y

The correct answer is D.

Note: Lindlar’s catalyst is a poisoned metal catalyst that is used to perform hydrogenations of alkynes in the presence of hydrogen gas. By poisoned it is meant that the reagent lacks the normal activity of palladium catalysts for reducing double bonds. This is useful when we start with an alkyne and have to go down one step of the oxidation ladder to obtain an alkene. But if we use normal palladium on carbon, we will get full reduction to the alkane.