When 2-butyne is treated with H$$_2$$/Lindlar's catalyst, compound X is produced as the major product and when treated with Na/liq. NH$$_3$$ it produces Y as the major product. Which of the following statements is correct?

We have the starting alkyne $$CH_{3}-C\equiv C-CH_{3}$$, whose IUPAC name is 2-butyne.

First we consider the reaction with $$H_{2}$$ in the presence of Lindlar’s catalyst. The Lindlar system is a poisoned palladium catalyst that brings about syn addition of one molecule of hydrogen across the triple bond. The syn mode of addition places the two newly added hydrogens on the same side of the former $$\pi$$-bond. As a result the product is the cis (Z) alkene:

$$CH_{3}-C\equiv C-CH_{3} \overset{H_2,\;\text{Lindlar}}{\longrightarrow} \text{CH}_3\text{-CH}=\text{CH-CH}_3 \;(\text{cis})$$

Hence, the compound we call $$X$$ is $$\text{cis-2-butene}$$.

Now we examine the reaction with metallic sodium in liquid ammonia, $$Na/liq.\;NH_{3}$$. This Birch-type reduction donates electrons in anti fashion, giving trans addition of hydrogen atoms across the alkyne. Consequently the product is the trans (E) alkene:

$$CH_{3}-C\equiv C-CH_{3} \overset{Na,\;\text{liq. }NH_3}{\longrightarrow} \text{CH}_3\text{-CH}=\text{CH-CH}_3 \;(\text{trans})$$

Therefore, compound $$Y$$ is $$\text{trans-2-butene}$$.

Next we compare the physical properties of the two alkenes.

For dipole moment we recall that the molecular dipole $$\mu$$ is the vector sum of all individual bond dipoles. In $$\text{cis-2-butene}$$ both $$C-CH_{3}$$ bond dipoles are oriented toward the same side of the double bond, so they reinforce each other and give a net dipole moment different from zero. In $$\text{trans-2-butene}$$ the molecule is planar and the two $$C-CH_{3}$$ bond dipoles lie in opposite directions, so they effectively cancel:

$$\mu_{\text{cis}} \gt 0,\qquad \mu_{\text{trans}}\approx 0$$

Thus, $$X$$ (cis) has the higher dipole moment, while $$Y$$ (trans) has the lower dipole moment.

Boiling point is influenced both by molecular mass and by intermolecular attractions. The masses are identical, so the decisive factor is the strength of the attractive forces. Because $$X$$ possesses a non-zero dipole, it exhibits dipole-dipole interactions in addition to London dispersion forces. These interactions raise its boiling point. In contrast, $$Y$$ has almost no permanent dipole and relies mainly on dispersion forces, giving it a lower boiling point. Experimental data corroborate this reasoning:

$$\text{b.p. of cis-2-butene}\;(X)\approx 3.7^{\circ}\text{C}$$
$$\text{b.p. of trans-2-butene}\;(Y)\approx 1.0^{\circ}\text{C}$$

So we conclude that $$X$$ possesses both the higher dipole moment and the higher boiling point relative to $$Y$$.

Among the given statements, only Option D matches this conclusion.

Hence, the correct answer is Option D.