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Question

Question: Which of the following is the most acidic? A.\({\text{HCOOH}}\) B.\({\text{C}}{{\text{H}}_{\text...

Which of the following is the most acidic?
A.HCOOH{\text{HCOOH}}
B.CH3COOH{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}
C.CH3CH2COOH{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH}}
D.CH3CH2CH2COOH{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH}}

Explanation

Solution

The strength of an acid depends on how easily it can donate a hydrogen ion. This depends on the stability of the conjugate base of the acid. The more stable the conjugate base, the more acidic that compound. The ability to stabilize the conjugate base depends on the substituent groups attached to the acid.

Complete step by step answer:
As we know, acids donate hydrogen ions in a solution. The more easily it can donate hydrogen ions, the more acidic the compound is. The ability to donate hydrogen ions depends upon the stability of the conjugate base. As we know, in a solution the dissociation reaction occurs as:
 - COOH  - COO -  + H + {\text{ - COOH }} \to {\text{ - CO}}{{\text{O}}^{\text{ - }}}{\text{ + }}{{\text{H}}^{\text{ + }}}
The acidity of the compounds depends on the distribution of the negative charge throughout the compound. The better the distribution of charge, the more stable is the conjugate base and more acidic is the compound.
In the question, the acids all have different alkyl groups attached to them. Now, we know that alkyl groups show  + I{\text{ + I}} effect, i.e. they donate electron density. This means that instead of stabilizing the compound, they will destabilize the negative charge on oxygen as they will increase its electron density. This means that higher the number of carbons in the alkyl group, the more  + I{\text{ + I}} effect and less acidic the compound will be.
In option A, the absence of alkyl group means that there will be no  + I{\text{ + I}} effect on the compound, unlike the other compounds and thus, it will be acidic.

\therefore The correct option is option A, i.e. HCOOH{\text{HCOOH}}.

Note:
The presence of  - I{\text{ - I}} effect groups like halogens or even benzene rings help stabilize the conjugate base. In case of benzoic acid, the stability increases in presence of electron withdrawing group and decreases with presence of electron donating group.