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 depends on the stability of the conjugate base formed after losing a proton (i.e., the carboxylate anion). This stability is influenced by the inductive effect of alkyl groups:

- Electron-donating groups (like alkyl groups) destabilize the carboxylate ion by increasing electron density, thus decreasing acidity.
- Electron-withdrawing groups stabilize the anion, increasing acidity.

Now, analyze each acid:
- HCOOH (Formic acid): No alkyl group → Highest acidity.
- CH₃COOH (Acetic acid): One methyl group → lower than formic acid.
- (CH₃)₂CHCOOH: Isopropyl group → stronger +I effect → lower acidity.
- (CH₃)₃CCOOH: Tertiary butyl group → even more +I effect → least acidic.

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