But-2-yne and hydrogen (one mole each) are separately treated with (i) Pd/C and (ii) Na/ liq. NH₃ to give the products X and Y respectively.
But-2-yne + H₂ →(Pd/C) X ; → Y
Identify the incorrect statements.
A. X and Y are stereoisomers.
B. Dipole moment of X is zero.
C. Boiling point of X is higher than Y .
D. X and Y react with O₃/Zn+H₂O to give different products.
Choose the correct answer from the options given below :

Solution

Chemistry ma’am · 3 steps

Pin down X as cis and Y as trans first, then test each statement against that one fact.

1. Identify X. On the Pd/C surface, both H atoms add from the same face (syn addition). With only one mole of H₂, but-2-yne stops at cis-but-2-ene, so X is cis.
2. Identify Y. Na in liquid NH₃ is a dissolving-metal reduction. The H atoms end up on opposite sides (anti addition), so Y is trans-but-2-ene.

   Statement	Check for cis (X) and trans (Y)	Verdict
   A. Stereoisomers	Same connectivity, different arrangement about the C=C	correct
   B. Dipole moment of X is zero	In cis, the two C-CH₃ bond dipoles point to the same side and do not cancel, so μ ≠ 0	incorrect
   C. b.p. of X higher than Y	Polar cis has stronger dipole–dipole forces: about 4 °C against about 1 °C for trans	correct
   D. Different ozonolysis products	Both cleave at C=C to give 2 molecules of CH₃CHO. Ozonolysis destroys the stereochemistry	incorrect

3. So. The incorrect statements are B and D, which is option (a).

The idea to keep. Metal catalyst with H₂ adds syn and gives the cis alkene. Na in liquid NH₃ adds anti and gives the trans alkene. Cis isomers are polar and boil higher, and ozonolysis cannot tell cis from trans.

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