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. 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 :
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To remember: "Cis" starts with C, like "Pd/C" (though usually poisoned). "Trans" involves "Sodium" ($Na$). Symmetrical alkenes always give the same carbonyl products in ozonolysis regardless of geometry.
Step 1: Understanding the Concept:
Alkynes undergo stereoselective partial hydrogenation depending on the reagent used. Lindlar's catalyst (or poisoned Pd/C) leads to syn-addition, while sodium in liquid ammonia leads to anti-addition via a radical mechanism. Step 2: Key Formula or Approach:
1. Alkyne + $H_2$, $Pd/C$ (poisoned) $\rightarrow$ \textit{cis}-alkene.
2. Alkyne + $Na/liq. NH_3$ $\rightarrow$ \textit{trans}-alkene.
3. Physical properties: \textit{Cis}-alkenes are generally more polar and have higher boiling points than \textit{trans}-alkenes. Step 3: Detailed Explanation:
- Product X: Reaction with $Pd/C$ gives \textit{cis}-but-2-ene.
- Product Y: Reaction with $Na/liq. NH_3$ gives \textit{trans}-but-2-ene.
- Statement A: True. \textit{Cis} and \textit{trans} isomers are geometrical isomers (stereoisomers).
- Statement B: Incorrect. \textit{Cis}-but-2-ene (X) is polar ($\mu \neq 0$), whereas \textit{trans}-but-2-ene (Y) has a dipole moment of zero due to symmetry.
- Statement C: True. Due to higher dipole-dipole interactions, the boiling point of \textit{cis} (X) is higher than \textit{trans} (Y).
- Statement D: Incorrect. Both isomers undergo ozonolysis to yield the same product: 2 moles of acetaldehyde ($CH_3CHO$). Step 4: Final Answer:
The incorrect statements are B and D only.
Which of the following points in Figure 2 most accurately represents the nodal surface shown in Figure 1?
