Arrange in **decreasing** order of acidic strength: $\mathrm{CH_3COOH}…

Question

Arrange in **decreasing** order of acidic strength: $\mathrm{CH_3COOH}$, $\mathrm{FCH_2COOH}$, $\mathrm{CF_3COOH}$, $\mathrm{ClCH_2COOH}$.

Accepted Answer

A. $\mathrm{CF_3COOH} > \mathrm{FCH_2COOH} > \mathrm{ClCH_2COOH} > \mathrm{CH_3COOH}$

**Principle.** Electron-withdrawing groups stabilise the carboxylate anion (RCOO⁻), making the acid stronger (lower pKₐ).
**Effects involved.**
1. **Number of halogens.** Three F atoms (in $\mathrm{CF_3COOH}$) pull more electron density than one. So tri-F beats mono-halo by a wide margin.
2. **Electronegativity of the halogen.** F (3.98) is more electronegative than Cl (3.16) — so $\mathrm{FCH_2COOH}$ is more acidic than $\mathrm{ClCH_2COOH}$.
3. **No halogen.** Pure $\mathrm{CH_3COOH}$ has only the +I-donating methyl, which *destabilises* the conjugate base — least acidic of the four.
**Approximate pKₐ values for sanity:** $\mathrm{CF_3COOH}$ 0.23, $\mathrm{FCH_2COOH}$ 2.66, $\mathrm{ClCH_2COOH}$ 2.87, $\mathrm{CH_3COOH}$ 4.76.
Order: $\mathrm{CF_3COOH} > \mathrm{FCH_2COOH} > \mathrm{ClCH_2COOH} > \mathrm{CH_3COOH}$.

Answer

$\mathrm{CH_3COOH} > \mathrm{CF_3COOH} > \mathrm{ClCH_2COOH} > \mathrm{FCH_2COOH}$

Answer

$\mathrm{CF_3COOH} > \mathrm{ClCH_2COOH} > \mathrm{FCH_2COOH} > \mathrm{CH_3COOH}$

Answer

$\mathrm{CH_3COOH} > \mathrm{ClCH_2COOH} > \mathrm{FCH_2COOH} > \mathrm{CF_3COOH}$