Question:
Consider the acidity of the carboxylic acids: I. PhCOOH, II. $o-NO₂C₆H₄COOH$, III. $p-NO₂C₆H₄COOH$, IV. $m-NO₂C₆H₄COOH$. Which of the following order is correct?
Consider the acidity of the carboxylic acids: I. PhCOOH, II. $o-NO₂C₆H₄COOH$, III. $p-NO₂C₆H₄COOH$, IV. $m-NO₂C₆H₄COOH$. Which of the following order is correct?
Show Hint
Consider the acidity of the carboxylic acids: I. PhCOOH, II. $o-NO2C6H4COOH$, III. $p-NO2C6H4COOH$, IV. $m-NO2C6H4COOH$. Which of the following order is correct?
Updated On: Apr 15, 2026
- $I>II>III>IV$
- $II>IV>III>I$
- $II>IV>III$
- $II>III>IV>I$
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Concept
Electron-withdrawing groups (like $-NO_{2}$) increase the acidity of benzoic acid.
Step 2: Analysis
- Ortho-substituted benzoic acids are generally more acidic than benzoic acid due to the "ortho effect" (steric inhibition of resonance). Thus, II is the most acidic.
Step 3: Substituent Position
- The $-I$ and $-M$ effects of $-NO_{2}$ work together at the para position (III). - Only the $-I$ effect works at the meta position (IV).
Step 4: Conclusion
The correct order is: $o-nitro>p-nitro>m-nitro>\text{benzoic acid}$ (II>III>IV>I).
Final Answer: (D)
Electron-withdrawing groups (like $-NO_{2}$) increase the acidity of benzoic acid.
Step 2: Analysis
- Ortho-substituted benzoic acids are generally more acidic than benzoic acid due to the "ortho effect" (steric inhibition of resonance). Thus, II is the most acidic.
Step 3: Substituent Position
- The $-I$ and $-M$ effects of $-NO_{2}$ work together at the para position (III). - Only the $-I$ effect works at the meta position (IV).
Step 4: Conclusion
The correct order is: $o-nitro>p-nitro>m-nitro>\text{benzoic acid}$ (II>III>IV>I).
Final Answer: (D)
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