Question

Arrange in the order of decreasing $p{{k}_{b}}$
(A)$F-CH2CH2COOH$
(B)$Cl2-CH-CH2COOH$
(C)$F-CH2COOH$
(D)$Br-CH2CH2COOH$

Answer 1

Hint ${{k}_{a}}$ represents acid dissociation constant and $p{{k}_{b}}$ is negative logarithm of ${{k}_{a}}$ ${{k}_{b}}$ represents base dissociation constant and $p{{k}_{b}}$ is negative logarithm of ${{k}_{b}}$ and measures the strength of base. The more stable the conjugate base, the stronger the acid. The stronger the base, the weaker is its conjugate acid and vice-versa.

Complete step by step solution:
According to question, given compounds are acid and their strength is determined by stability of the conjugate base i.e. $R-COO-$ in the following reaction
$R-COOH\to R-COO-+H+$
Stability of acid depends upon the stability of carboxylate ions. In other words, the more stable the carboxylate ion, the more stable the acid will be.
Now, option (C) is most acidic because fluorine is the most electronegative element that will stabilize the negative charge on the oxygen atom resulting in the lower energy and greater stability. So, the conjugate base will be stabilized. Hence, it will be most acidic.
In option (B), two chlorine are present which also have affinity for electrons being electronegative in nature and stabilise the negative charge on oxygen atom resulting in stable conjugate base. So this is the second most acidic compound after option (A).
Now, option (A) and option (D), the only difference is the halogen atoms. Bromine is less electronegative than fluorine will stabilize the conjugate base less than fluorine. Therefore, option (A) will be more acidic. Order of acidic strength is given as:
(C)> (B)> (A)> (D)
Stronger the acidic strength, lower is the $p{{k}_{a}}$ value. Therefore, decreasing order of $p{{k}_{a}}$ value of these acids will be:
(D) > (A) > (B) > (C)
We know that $p{{k}_{a}}$ and $p{{k}_{b}}$ are related as
$p{{k}_{a}}+p{{k}_{b}}=14$
Higher the $p{{k}_{a}}$ value, lower is $p{{k}_{b}}$ value. Decreasing order of $p{{k}_{b}}$ value is: (C)> (B)> (A)> (D)

Note: Effect on the acidity of the acid decreases gradually if the distance between the substituent group and carboxylic acid increases. In other words, substituent groups are more effective if these are present at $\alpha$-position.