Question:

(i)(ii)(iii)(iv)
The correct order of acidity of above compound is 

Updated On: Apr 11, 2025
  • II>IV>I>III
  • III>IV>II>I
  • IV>II>III>I
  • IV>III>I>II
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The Correct Option is D

Solution and Explanation

Detailed Explanation of the Acidity Order: IV > III > I > II

Acidity of aromatic carboxylic acids (like benzoic acid derivatives) is influenced mainly by two effects:

  • −I effect (inductive electron withdrawal): Electron-withdrawing groups (EWGs) stabilize the conjugate base by dispersing negative charge, increasing acidity.
  • +M effect (mesomeric electron donation): Electron-donating groups (EDGs) through resonance donate electron density to the ring, destabilizing the conjugate base and decreasing acidity.

Let’s evaluate each compound (I to IV):

  1. Compound I: Ortho-substituted benzoic acid
    The substituent is not specified, but if it's an electron-donating group (EDG), ortho effect still makes this more acidic than meta or para substituted EDG groups. 
    Ortho effect: Even EDGs at the ortho position can increase acidity due to steric inhibition of resonance, preventing lone pair donation into the ring.
  2. Compound II: Para-substituted with an electron-donating group (e.g., –OH, –OCH₃)
    The +M effect from the para position increases electron density on the ring, destabilizing the conjugate base. Least acidic.
  3. Compound III: Meta-substituted with an electron-withdrawing group (e.g., –NO₂)
    The −I effect is significant in the meta position (though no resonance with carboxylate anion), which helps stabilize the conjugate base. More acidic than I and II.
  4. Compound IV: Ortho-substituted with a strong electron-withdrawing group (like –NO₂ or –Cl)
    This combines:
    • −I effect from the EWG
    • Ortho effect (steric inhibition of resonance of the carboxyl group)

Final Acidity Order: IV > III > I > II

Correct Answer: (D): IV > III > I > II

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Concepts Used:

Alcohols, Phenols, and Ethers - Functional Groups

In alcohols, the oxygen of the –OH group is attached to carbon by a sigma (σ ) bond formed by the overlap of an sp3 hybridized orbital of carbon with an sp3 hybridized orbital of oxygen.

The bond angle in alcohols is slightly less than the tetrahedral angle (109°-28′). It is due to the repulsion between the unshared electron pairs of oxygen.

In phenols, the –OH group is attached to the sp2 hybridized carbon of an aromatic ring. The carbon-oxygen bond length (136 pm) in phenol is slightly less than that in methanol. This is due to:

  1. The partial double bond character on account of the conjugation of unshared electron pair of oxygen with the aromatic ring.
  2. The sp2 hybridized state of carbon to which oxygen is attached. 

In ethers, the four-electron pairs, i.e., the two bond pairs and two lone pairs of electrons on oxygen are arranged approximately in a tetrahedral arrangement. The bond angle is slightly greater than the tetrahedral angle due to the repulsive interaction between the two bulky (–R) groups. The C–O bond length (141 pm) is almost the same as in alcohols. 
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