Question:
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.
\includegraphics[width=0.3\linewidth]{Screenshot 2026-02-04 152309.png
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 :}
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.
\includegraphics[width=0.3\linewidth]{Screenshot 2026-02-04 152309.png
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 :}
\includegraphics[width=0.3\linewidth]{Screenshot 2026-02-04 152309.png
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 :}
Show Hint
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.
Updated On: Feb 4, 2026
- A and B Only
- A and C Only
- B and C Only
- B and D Only
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The Correct Option is D
Solution and Explanation
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.
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.
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