When 2-butyne is treate 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 ?

X will have higher dipole moment and higher boiling point than Y.

Y will have higher dipole moment and higher boiling point than X.

X will have lower dipole moment and lower boiling point than Y.

Y will have higher dipole momet and lower boiling point than X.

Hydrocarbons PYQ — JEE Main 2018 · 15 Apr · Shift 2 · Q5 with Solution

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Correct answer: A

2-butyne is: $CH_3-C\equiv C-CH_3$

With $H_2$ /Lindlar's catalyst, partial hydrogenation of an alkyne gives the cis-alkene.

Hence, $X = \text{cis-2-butene}$
With $Na$ /liq. $NH_3$ , dissolving metal reduction of an alkyne gives the trans-alkene.

Hence, $Y = \text{trans-2-butene}$

In cis-2-butene, the two methyl groups are on the same side, so bond dipoles do not cancel completely. Thus, cis isomer has a net dipole moment.
In trans-2-butene, the molecule is more symmetrical, so the bond dipoles cancel to a large extent. Thus, trans isomer has very low (approximately zero) dipole moment.

Therefore, $\mu_X > \mu_Y$

Boiling point depends on intermolecular forces.

Since cis-2-butene has higher dipole moment, it has stronger dipole-dipole attraction.
trans-2-butene is less polar, so intermolecular attraction is weaker.

Hence, $\text{b.p. of } X > \text{b.p. of } Y$

A: $X$ will have higher dipole moment and higher boiling point than $Y$ . ✅
B: Opposite of true. ❌
C: Opposite of true. ❌
D: Says $Y$ has higher dipole moment but lower boiling point. First part is false. ❌

$X = \text{cis-2-butene}, \quad Y = \text{trans-2-butene}$

So the correct statement is Option A.