Animated Solution for Chemistry - Organic Chemistry: Arrange the following labelled hydrogens in decreasing order of acidity

Imagine you are a proton ($H^{+}$) trying to break free from a complex organic molecule. Your ability to leave—your acidity—depends entirely on how stable the molecule becomes after you are gone. In this problem, we are presented with a heavily substituted benzene ring featuring four distinct, labelled hydrogens: $a$, $b$, $c$, and $d$.

To determine their decreasing order of acidity, we must first identify the functional groups they belong to. Hydrogen $a$ is part of a terminal alkyne ($-C\equiv C-H$). Hydrogen $b$ belongs to a carboxylic acid group ($-COOH$). Hydrogens $c$ and $d$ are both phenolic protons, meaning they are attached to oxygen atoms directly bonded to the benzene ring.

Before diving into the complex electronic effects of the benzene ring, we can establish our extremes using a fundamental rule of organic chemistry: Carboxylic acids are generally much stronger acids than phenols, and phenols are significantly more acidic than terminal alkynes.

Why is this the case? When a carboxylic acid loses a proton, the resulting negative charge is delocalized equally over two highly electronegative oxygen atoms via resonance. This creates an incredibly stable carboxylate ion. Phenols also stabilize their negative charge through resonance, but the charge is delocalized onto less electronegative carbon atoms within the benzene ring, making it less stable than a carboxylate ion. Finally, terminal alkynes rely solely on the high electronegativity of their $sp$-hybridized carbon (which has 50% s-character) to stabilize the charge, which is the weakest stabilization method of the three.

Carboxylic Acid vs. Alkyne

Applying our golden rule, we can immediately identify the most and least acidic protons. Hydrogen $b$, being part of the carboxylic acid group, is the undisputed heavyweight champion of acidity in this molecule. Conversely, hydrogen $a$, attached to the alkyne, is the least acidic.

So far, we know that the order must start with $b$ and end with $a$. The real challenge lies in differentiating the two phenolic protons in the middle.

Para vs. Meta

Hydrogens $c$ and $d$ are both phenolic, but they are not created equal. Their acidity is heavily influenced by the powerful electron-withdrawing nitro group ($-N{O}_2$) attached to the top of the ring.

Let's look at hydrogen $c$. The $-N{O}_2$ group is located at the para position relative to this hydroxyl group. At the para position, the nitro group exerts two distinct forces: a strong $-R$ (resonance) effect and a $-I$ (inductive) effect. When hydrogen $c$ leaves, the resulting negative charge on the oxygen can delocalize all the way into the oxygen atoms of the nitro group. This extensive delocalization provides massive stabilization to the conjugate base.

Why Meta Misses Out

Now, let's examine hydrogen $d$. The $-N{O}_2$ group is located at the meta position relative to this hydroxyl group. Here is the critical catch: resonance effects do not operate at the meta position. If you draw the resonance structures for the phenoxide ion formed by losing hydrogen $d$, you will see that the negative charge skips the meta carbon entirely.

Therefore, the nitro group can only stabilize the conjugate base of hydrogen $d$ through its $-I$ (inductive) effect. While the inductive effect is helpful, it is significantly weaker than the combined $-R$ and $-I$ effects experienced at the para position.

Because the para position offers much greater stabilization of the negative charge through resonance, hydrogen $c$ is more acidic than hydrogen $d$.

Combining all our findings, the final decreasing order of acidity is: $b>c>d>a$. The carboxylic acid leads the pack, followed by the para-nitrophenol, then the meta-nitrophenol, and finally the terminal alkyne.