Question:
Choose the correct statement.
Choose the correct statement.
Show Hint
Intramolecular H-bonding reduces boiling point; intermolecular H-bonding increases it.
Updated On: Apr 23, 2026
- \(A\) is more volatile than \(B\)
- \(B\) is more volatile than \(A\)
- \(A\) is formed more rapidly at a higher temperature
- \(A\) is formed in higher yield at a low temperature
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Formula / Definition}
\[ \text{Intermolecular H-bonding} \Rightarrow \text{higher boiling point, lower volatility} \]
Step 2: Calculation / Simplification}
\(A\) (ortho-hydroxy benzaldehyde): intramolecular H-bonding \(\Rightarrow\) lower boiling point, more volatile.
\(B\) (para-isomer): intermolecular H-bonding \(\Rightarrow\) higher boiling point, less volatile.
\(\therefore A\) is more volatile than \(B\).
Step 3: Final Answer
\[ A \text{ is more volatile than } B \]
\[ \text{Intermolecular H-bonding} \Rightarrow \text{higher boiling point, lower volatility} \]
Step 2: Calculation / Simplification}
\(A\) (ortho-hydroxy benzaldehyde): intramolecular H-bonding \(\Rightarrow\) lower boiling point, more volatile.
\(B\) (para-isomer): intermolecular H-bonding \(\Rightarrow\) higher boiling point, less volatile.
\(\therefore A\) is more volatile than \(B\).
Step 3: Final Answer
\[ A \text{ is more volatile than } B \]
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