Animated Solution for Chemistry - Organic Chemistry: The strongest acid amongst the following compounds is

Unraveling the Inductive Effect

When we talk about the strength of an acid in organic chemistry, we are essentially talking about a microscopic tug-of-war for electrons. An acid's primary job is to donate a proton (${\text{H}}^{+}$). But how willing is it to let that proton go? The secret lies not in the acid itself, but in what is left behind: the conjugate base.

The golden rule to remember is this: Acidic strength is directly proportional to the stability of its conjugate base.

When a carboxylic acid ($\text{R-COOH}$) loses a proton, it forms a carboxylate ion (${\text{R-COO}}^{-}$). This ion carries a negative charge. In the microscopic world, a concentrated negative charge is a source of instability. If the molecule can find a way to spread out or "disperse" this charge, the conjugate base becomes much more stable, and consequently, the original acid is much stronger.

This is where the Inductive Effect steps onto the battlefield.

Groups attached to the carbon chain can either push electrons away or pull electrons towards themselves through the sigma bonds.

1. Electron Donating Groups ($+I$ effect): Alkyl groups like methyl (${\text{CH}}_3-$) are electron donors. They push electron density towards the already negatively charged carboxylate ion. This intensifies the charge, destabilizing the conjugate base. Therefore, acetic acid (${\text{CH}}_3\text{COOH}$) is a relatively weak acid.

2. Electron Withdrawing Groups ($-I$ effect): Highly electronegative atoms like halogens ($\text{F}$, $\text{Cl}$, $\text{Br}$) are electron thieves. They pull electron density away from the carboxylate ion, helping to disperse the negative charge. This stabilizes the conjugate base, making the parent acid much stronger.

If we compare formic acid ($\text{HCOOH}$) and acetic acid (${\text{CH}}_3\text{COOH}$), formic acid is stronger because it lacks the destabilizing $+I$ effect of the methyl group.

Now, let's look at the heavy hitters in our problem: the chlorine-substituted acids. Both ${\text{CH}}_3{\text{CH}}_2{\text{CH(Cl)CO}}_2\text{H}$ and ${\text{ClCH}}_2{\text{CH}}_2{\text{CH}}_2\text{COOH}$ have a chlorine atom exerting a stabilizing $-I$ effect. So, which one wins?

The inductive effect has a critical limitation: it is highly distance-dependent. It operates through the sigma bonds, and its strength drops off rapidly as you move further down the carbon chain.

In ${\text{CH}}_3{\text{CH}}_2{\text{CH(Cl)CO}}_2\text{H}$, the chlorine atom is attached to the $\alpha$-carbon, right next door to the carboxylate group. Its electron-withdrawing pull is incredibly strong here.

In ${\text{ClCH}}_2{\text{CH}}_2{\text{CH}}_2\text{COOH}$, the chlorine atom is banished to the $\gamma$-carbon, three bonds away. Its pull is significantly weakened by the distance.

Because the $-I$ effect is strongest when the electronegative atom is closest to the acidic group, the conjugate base of ${\text{CH}}_3{\text{CH}}_2{\text{CH(Cl)CO}}_2\text{H}$ is the most stable among the choices.

Therefore, ${\text{CH}}_3{\text{CH}}_2{\text{CH(Cl)CO}}_2\text{H}$ is the strongest acid. The complete order of acidic strength is: