Question

The correct order of acidic strength is: \[ CH_3COOH,\quad ClCH_2COOH,\quad Cl_2CHCOOH,\quad Cl_3CCOOH \]

• A. \(CH_3COOH > ClCH_2COOH > Cl_2CHCOOH > Cl_3CCOOH\)
• B. \(ClCH_2COOH > CH_3COOH > Cl_2CHCOOH > Cl_3CCOOH\)
• C. \(Cl_3CCOOH > Cl_2CHCOOH > ClCH_2COOH > CH_3COOH\)
• D. \(Cl_2CHCOOH > Cl_3CCOOH > ClCH_2COOH > CH_3COOH\)

Hint:

Electron-withdrawing groups such as F, Cl, Br, NO\(_2\), and CN increase acidity, whereas electron-donating groups such as alkyl groups decrease acidity.

Answer 1

The Correct Option is C

Concept: Acidic strength of carboxylic acids depends upon the stability of the conjugate base. Electron-withdrawing groups increase acidity through the \((-I)\) effect.

Step 1: Understand the inductive effect. Chlorine is highly electronegative. It withdraws electron density from the carboxylate ion. This stabilizes the negative charge.

Step 2: Compare the number of chlorine atoms. \[ CH_3COOH \] contains no chlorine. \[ ClCH_2COOH \] contains one chlorine. \[ Cl_2CHCOOH \] contains two chlorines. \[ Cl_3CCOOH \] contains three chlorines.

Step 3: Apply inductive effect. Greater number of chlorine atoms means stronger \((-I)\) effect. Therefore acidity increases in the order: \[ CH_3COOH < ClCH_2COOH < Cl_2CHCOOH < Cl_3CCOOH \] Hence the correct decreasing order of acidity is: \[ Cl_3CCOOH > Cl_2CHCOOH > ClCH_2COOH > CH_3COOH \] Therefore option (C) is correct.