Animated Solution for Chemistry - Organic Chemistry: The correct order of acidity for the following compounds is:

Unraveling the Ortho Effect

When comparing the acidic strength of organic molecules, the golden rule is to always look at the stability of the conjugate base. The reaction is an equilibrium:

According to Le Chatelier's principle, any factor that stabilizes the conjugate base will drive the equilibrium to the right, making the parent compound a stronger acid. In this problem, we are comparing four substituted benzoic acids. Let's remove the acidic proton from the carboxylic acid group of each molecule and analyze the resulting carboxylate anions.

Let's start with Compound I (2,6-dihydroxybenzoic acid). When it loses a proton, the resulting carboxylate anion is flanked by two hydroxyl ($-OH$) groups at the ortho positions. These $-OH$ groups are perfectly positioned to form strong intramolecular hydrogen bonds with the negatively charged oxygen atoms of the carboxylate group.

This phenomenon, where a ring-like structure is formed via hydrogen bonding, is called chelation. Because Compound I can form two such hydrogen bonds, its conjugate base is exceptionally stable. This is a classic manifestation of the ortho effect, which generally makes ortho-substituted benzoic acids stronger than their meta and para counterparts.

Moving to Compound II (salicylic acid), we see a similar story. It has one ortho $-OH$ group, which forms a single intramolecular hydrogen bond with the carboxylate group. This stabilizes the anion significantly, making it a strong acid. However, because it only has one hydrogen bond compared to the two in Compound I, its conjugate base is slightly less stable. Therefore, Compound I is more acidic than Compound II.

In Compound III (3-hydroxybenzoic acid), the $-OH$ group is located at the meta position. This is a critical detail. In electrophilic aromatic substitution and resonance stabilization, the resonance ($+R$ or $-R$) effects do not operate effectively from the meta position. The electron density changes induced by resonance bypass the meta carbon.

Therefore, the $-OH$ group in Compound III can only exert its inductive effect ($-I$). Oxygen is more electronegative than carbon, so it pulls electron density away from the ring through the sigma bonds. This electron-withdrawing effect helps to disperse the negative charge on the carboxylate group, stabilizing the conjugate base. While not as powerful as the chelation seen in I and II, this $-I$ effect makes Compound III more acidic than an unsubstituted benzoic acid (and certainly more acidic than Compound IV).

Finally, let's examine Compound IV (4-hydroxybenzoic acid). Here, the $-OH$ group is at the para position. From the para position, the lone pairs on the oxygen atom can fully participate in resonance with the benzene ring.

For the $-OH$ group, the electron-donating resonance effect ($+R$) is significantly stronger than its electron-withdrawing inductive effect ($-I$). This means the $-OH$ group pumps electron density into the ring, which eventually reaches the carbon attached to the carboxylate group. This influx of electron density intensifies the negative charge on the carboxylate anion, severely destabilizing it. Because its conjugate base is the least stable, Compound IV is the weakest acid of the bunch.

By evaluating the stability of the conjugate bases, we can definitively rank the acidic strengths. Compound I is the most stable (double chelation), followed by Compound II (single chelation). Compound III is stabilized by the $-I$ effect, while Compound IV is actively destabilized by the $+R$ effect.

Thus, the correct order of acidity is I > II > III > IV.