Question

The correct order of decreasing acidity of the following aliphatic acids is:

• A. $CH_{3}COOH$ $>$ (CH_{3})_{2}CHCOOH $>$$ (CH_{3})_{3}CCOOH$ $>$ HCOOH
• B. $CH_3COOH > (CH_3)_2CHCOOH > (CH_3)_3CCOOH$
• C. $HCOOH > (CH_3)_3CCOOH > (CH_3)_2CHCOOH > CH_3COOH$
• D. $(CH_3)_3CCOOH > (CH_3)_2CHCOOH > CH_3COOH > HCOOH$

Hint:

More alkyl groups mean stronger electron-donating (+I) effect, which reduces acid strength by destabilizing the conjugate base.

Answer 1

The Correct Option is B

The acidity of carboxylic acids is determined by the stability of their conjugate bases (carboxylate anions) formed after proton loss. This stability is modulated by the inductive effect of substituents:

- Electron-donating groups (e.g., alkyl groups) destabilize the carboxylate anion by increasing electron density, thereby reducing acidity.
- Electron-withdrawing groups stabilize the anion, thus enhancing acidity.

Analysis of specific acids:
- HCOOH (Formic acid): Lacks an alkyl group, exhibiting the highest acidity.
- CH₃COOH (Acetic acid): Possesses one methyl group, resulting in lower acidity than formic acid.
- (CH₃)₂CHCOOH: Features an isopropyl group, which exerts a stronger positive inductive effect (+I), leading to reduced acidity.
- (CH₃)₃CCOOH: Contains a tertiary butyl group, demonstrating an even greater +I effect and consequently, the least acidity.

Therefore, the order of decreasing acidity is as follows:
$$ \text{HCOOH}>\text{CH}_3\text{COOH}>(\text{CH}_3)_2\text{CHCOOH}>(\text{CH}_3)_3\text{CCOOH} $$