But-2-yne is reacted separately with one mole of Hydrogen as shown below :
B ←(Na) CH₃-C≡C-CH₃ →(Pd/C) A
A. A is more soluble than B.
B. The boiling point & melting point of A are higher and lower than B respectively.
C. A is more polar than B because dipole moment of A is zero.
D. Br₂ adds easily to B than A.
Identify the incorrect statements from the options given below :

Solution

Chemistry ma’am · 3 steps

Name A and B first: the metal surface gives cis, sodium in ammonia gives trans. Then test each statement.

1. Identify A. With H₂ on Pd/C, both H atoms add from the catalyst surface on the same side (syn addition). With one mole of H₂, A is cis-but-2-ene.
2. Identify B. Na in liquid NH₃ adds the two H atoms from opposite sides (anti addition). So B is trans-but-2-ene.

   Statement	Fact	Verdict
   A: A more soluble than B	cis has a net dipole and trans has none, so cis is more soluble	correct
   B: A has higher bp and lower mp	cis boils higher (about 4 vs 1 °C) but melts lower, because trans packs better in the solid	correct
   C: A more polar since its dipole moment is zero	the dipole moment of cis is not zero. It is trans whose bond dipoles cancel to zero	incorrect
   D: Br₂ adds more easily to B	cis is less stable because its two methyl groups crowd each other, so cis reacts faster	incorrect

3. So. The incorrect statements are C and D, which is option (d).

The idea to keep. A metal catalyst gives the cis alkene and Na in liquid ammonia gives trans. Cis is more polar, boils higher, melts lower and reacts faster. Trans has zero dipole and packs better.

Answer from the NTA final answer key. Solution by PrepWiser.