Question

Consider the given statements regarding major organic products X and Y formed in the following reactions: $\underset{(Major)}{X} \xrightarrow{Na, Liq. NH_3}$ 2- Butyne $\xrightarrow{H_2, Lindlar's catalyst} \underset{(Major)}{Y}$

• A. (I) and (II)
• B. (I) and (III)
• C. (II) and (III)
• D. (III) and (IV)

Answer 1

Answer: (C)

(II) and (III)

Answer 2

1. Identifying the products:

   • Na in liquid NH₃ reduction: This reaction converts 2‐butyne to trans‑2‑butene (X) via anti addition, resulting in a symmetric molecule with negligible net dipole moment.

   • H₂ with Lindlar's catalyst: This method converts 2‑butyne to cis‑2‑butene (Y), which has a dipole moment due to the two methyl groups being on the same side.

2. Analyzing the statements:

   • (I) X will have higher dipole than Y: False, because trans‑2‑butene (X) is symmetric (no dipole) whereas cis‑2‑butene (Y) has a measurable dipole moment.

   • (II) Y will have higher boiling point than X: True, since the dipole in cis‑2‑butene (Y) leads to stronger intermolecular forces (dipole-dipole interactions) resulting in a higher boiling point.

   • (III) X will have higher melting point than Y: True, as the symmetric trans‑2‑butene (X) can pack more efficiently in the solid state, hence a higher melting point.

   • (IV) Y will have higher melting point than X: False, the opposite is true.

Final Answer: The correct statements are (II) and (III).

Summary:

• Explanation:
  • Na/liq NH₃ gives trans‑2‑butene (X): symmetric, low dipole, high melting point.
  • Lindlar’s catalyst gives cis‑2‑butene (Y): polar, higher boiling point, lower melting point.