Animated Solution for Chemistry - Organic Chemistry: The increasing order of the $p{K}_a$ values of the following compounds is

Phenol and its Derivatives

Welcome to a classic and highly conceptual problem from Organic Chemistry! We are given four compounds: Phenol ($A$), p-Nitrophenol ($B$), m-Nitrophenol ($C$), and p-Methoxyphenol ($D$).

Our mission is to determine the increasing order of their $p{K}_a$ values. To do this, we must first understand the relationship between $p{K}_a$ and acidic strength, and then analyze the electronic effects of the substituents on the phenol ring.

Before we dive into the molecules, let's establish the ground rules. The acid dissociation constant, $K_a$, is a direct measure of a compound's acidic strength. A higher $K_a$ means a stronger acid.

However, $p{K}_a$ is defined as the negative logarithm of $K_a$:

This mathematical relationship means that $p{K}_a$ is inversely proportional to acidic strength. Therefore, a stronger acid will have a lower $p{K}_a$ value. Our strategy is simple: find the order of acidic strength, and then reverse it to get the order of $p{K}_a$.

Stability is Key

The acidic strength of a phenol derivative depends entirely on the stability of its conjugate base, the phenoxide ion. When phenol loses an $H^{+}$ ion, it forms a phenoxide ion, which is stabilized by resonance.

Any group attached to the benzene ring that helps disperse the negative charge will stabilize the phenoxide ion, thereby increasing the acidity. These are Electron Withdrawing Groups (EWG). Conversely, groups that intensify the negative charge will destabilize the ion, decreasing the acidity. These are Electron Donating Groups (EDG).

Let's analyze compound $D$, p-Methoxyphenol. It has a methoxy ($-OC{H}_3$) group at the para position.

While oxygen is electronegative and exerts a weak $-I$ (inductive) effect, it also possesses lone pairs that it can donate into the benzene ring through resonance. This is a strong $+M$ (mesomeric) effect.

Because the $+M$ effect strongly outweighs the $-I$ effect, the $-OC{H}_3$ group acts as a net Electron Donating Group (EDG). This destabilizes the phenoxide ion, making compound $D$ the weakest acid among the four.

Nitrophenols

Now let's compare compounds $B$ (p-Nitrophenol) and $C$ (m-Nitrophenol). Both contain the nitro ($-N{O}_2$) group, which is a powerful Electron Withdrawing Group (EWG).

In compound $B$, the $-N{O}_2$ group is at the para position. Here, it can stabilize the phenoxide ion through both its strong $-M$ (resonance) effect and its $-I$ (inductive) effect.

In compound $C$, the $-N{O}_2$ group is at the meta position. A crucial rule of organic chemistry is that the mesomeric effect does not operate at the meta position. Therefore, the $-N{O}_2$ group in compound $C$ can only stabilize the ion through its $-I$ effect.

Because compound $B$ benefits from both $-M$ and $-I$ stabilization, it is more acidic than compound $C$. Both $B$ and $C$ are more acidic than unsubstituted phenol ($A$).

Putting it all together, the decreasing order of acidic strength is: $B>C>A>D$

Since $p{K}_a$ is inversely proportional to acidic strength, we simply reverse this order to find the increasing order of $p{K}_a$: $B<C<A<D$

This matches option (b).

As a final thought experiment, consider what would happen if we had ortho-nitrophenol. While it also benefits from $-M$ and $-I$ effects, it forms an intramolecular hydrogen bond between the $-OH$ group and the $-N{O}_2$ group. This "locks" the proton slightly, making it a weaker acid than p-nitrophenol. Always keep an eye out for the ortho effect!