Question

Which of the following has the lowest $p{K_a}$ value?
A) 2- Chlorobutanoic acid
B) 3- Chlorobutanoic acid
C) 4- Chlorobutanoic acid
D) Butanoic acid

Answer 1

$p{K_a}$ is known as the acid dissociation constant. Lower the value of $p{K_a}$ the more acidic the substance will be. Higher value of $p{K_a}$ indicates the substance to be weakly acidic. In this question we need to know the effect of substituents on the value of $p{K_a}$ .

Complete Step By Step Answer:
We are given chloro substituted butanoic acids. The position of the chloro substituents varies in the given options. We know that Chloro is an Electron withdrawing group, meaning it pulls the electrons towards itself. The electrons of the C-Cl bond are pulled towards Cl, and the electron density is pulled away from the carboxylate anion. This disperses the charge and creates a stabilizing effect. The more stable the carboxylate ion is the more acidic the acid will be. The chlorine hence acts as an electron withdrawing group.
More the electronegativity of the substituent the $p{K_a}$ decreases and the compound becomes more acidic. The no. of electron withdrawing groups attached to the carbon, the more acidic it becomes.
Since the inductive effects are transmitted through the covalent bonds, the acid strengthening effects fall rapidly with the increase in the no. of sigma bonds between the carboxylic acid and the electron withdrawing group. A distance of 3 sigma bonds almost completely eliminates the inductive effect.
From the given options, 2-chlorobutanoic acid will have the lowest value of $p{K_a}$ and will be the most acidic as it is the closest to the -COOH group.
Hence the correct answer is Option A.

Note:
If electron donating groups are present, then the situation will be reversed. The electron density will increase and the value of $p{K_a}$ will increase, reducing the acidity. The farther the electron donating group, the more acidic it will become.