The increasing order of acidity of the following compounds based on pKa values:
(A) BrCH$_2$COOH
(B) ClCH$_2$COOH
(C) FCH$_2$COOH
(D) HCOOH
To determine the increasing order of acidity of the compounds based on their pKa values, we need to understand how different substituents affect the acidity of carboxylic acids. Acidity is influenced by the electron-withdrawing or electron-donating nature of substituents:
The presence of electron-withdrawing groups, such as halogens, will stabilize the carboxylate anion formed upon deprotonation, thus increasing acidity. The more electronegative the substituent, the greater the acidity.
Let's analyze each compound's acidity based on the electronegativity of the substituent:
Thus, the increasing order of acidity is as follows: (D) $<$ (A) $<$ (B) $<$ (C)
To determine the increasing order of acidity for the given compounds based on their pKa values, we analyze their structures and the effect of substituents on acidity.
The acidity of carboxylic acids is influenced by the electron-withdrawing or electron-donating nature of substituents attached to the carboxyl group. The more electron-withdrawing the substituent, the more it stabilizes the conjugate base (carboxylate ion), thereby increasing the acidity of the compound.
Let's examine the compounds:
In terms of electron-withdrawing ability, the substituents can be ranked as follows: F > Cl > Br. Fluorine, being the most electronegative, is the strongest electron-withdrawing group among them, followed by chlorine and bromine.
Comparing the acidity:
Therefore, the increasing order of acidity, based on pKa values, is: (D) < (A) < (B) < (C)
In the given reaction sequence, the structure of Y would be: