Animated Solution for Chemistry - Organic Chemistry: The correct order of acid strength of the following carboxylic acids is -

When we talk about the strength of an acid in organic chemistry, we are essentially talking about a game of stability. An acid is only as strong as its willingness to give up a proton ($H^{+}$). But why would a molecule want to lose a proton? The answer lies in the stability of what is left behind: the conjugate base.

For carboxylic acids, losing a proton creates a carboxylate anion ($R-CO{O}^{\ominus }$). This anion carries a negative charge. If the group $R$ attached to it can help pull that negative charge away (an electron-withdrawing effect), the anion becomes highly stable, and the acid is strong. Conversely, if the group $R$ pushes more electrons toward the already negative oxygen (an electron-donating effect), the anion becomes unstable, and the acid is weak.

The Power of Hybridization

Let's look at our first two contenders: Compound I ($HC\equiv C-COOH$) and Compound II ($C{H}_2=CH-COOH$). The secret here lies in the hybridization of the carbon atom directly attached to the carboxylate group.

In Compound I, the carbon is part of a triple bond, meaning it is $sp$ hybridized. An $sp$ orbital has $50\%$ s-character. Because s-orbitals are closer to the nucleus, a higher s-character means the nucleus has a stronger grip on the electrons. This makes the $sp$ carbon highly electronegative, exerting a powerful $-I$ (inductive) effect. This strong pull stabilizes the carboxylate anion beautifully, making Compound I the strongest acid in our lineup.

In Compound II, the carbon is part of a double bond, making it $s{p}^2$ hybridized. With $33.3\%$ s-character, it is still electronegative and exerts a $-I$ effect, but it is noticeably weaker than the $sp$ carbon. Thus, Compound II is a strong acid, but not quite as strong as Compound I.

Now, let's jump to Compound III ($p-MeO-C_6{H}_4-COOH$). At first glance, you see a benzene ring, which typically exerts a mild $-I$ effect. But there is a catch—a methoxy ($-OMe$) group at the para position.

The oxygen in the methoxy group has lone pairs that it can donate into the benzene ring through resonance. This is known as the $+R$ effect. This resonance effect is incredibly powerful, pumping electron density through the ring and right onto the carboxylate group. This massive influx of electrons destabilizes the anion, making Compound III a surprisingly weak acid.

Finally, we have Compound IV ($C{H}_3-C{H}_2-COOH$), propanoic acid. Here, the carboxylate is attached to an ethyl group. Alkyl groups are purely $s{p}^3$ hybridized and exert a $+I$ effect. They are electron-donating by nature, which destabilizes the conjugate base.

So, how do we rank Compound III and Compound IV? While Compound III suffers from the strong $+R$ effect of the methoxy group, its carboxylate is still attached to an $s{p}^2$ hybridized carbon of the benzene ring. Compound IV is attached to a purely $s{p}^3$ hybridized carbon. Because $s{p}^2$ carbons are inherently more electronegative than $s{p}^3$ carbons, Compound III retains a slight edge in acidity over Compound IV.

Putting it all together, the order of acid strength is dictated by the stabilizing $-I$ effects and destabilizing $+I/+R$ effects:

This perfectly matches option (D).