Which carboxylic acid has the highest $K_a$ (i.e. is the most acidic)?
A$\mathrm{CH_3CH_2COOH}$ (propanoic)
B$\mathrm{CH_3CH(F)COOH}$ ($\alpha$-fluoropropanoic)
C$\mathrm{CH_3CH(Cl)COOH}$ ($\alpha$-chloropropanoic)
D$\mathrm{CH_3CH(Br)COOH}$ ($\alpha$-bromopropanoic)
Answer & Solution
Correct answer: B. $\mathrm{CH_3CH(F)COOH}$ ($\alpha$-fluoropropanoic)
1. The acidity of a substituted carboxylic acid is governed primarily by the INDUCTIVE EFFECT of the substituent on the carboxylate stability.
2. Among the halogens, the ELECTRONEGATIVITY order is F > Cl > Br > I. Higher electronegativity
→ stronger electron-withdrawing inductive effect
→ more stable conjugate base → higher acidity.
3. So among the $\alpha$-halo propanoic acids, the FLUORO derivative is most acidic.
4. The numbers: $K_a$ values (approximate, syllabus style): unsubstituted propanoic $\approx 1.3\times 10^{-5}$.
$\alpha$-F $\approx 2\times 10^{-3}$.
$\alpha$-Cl $\approx 1.4\times 10^{-3}$.
$\alpha$-Br $\approx 1.3\times 10^{-3}$. F-substitute wins.
5. Option A has no substituent at all and is least acidic. Options C and D weaker than B for the reason in step 2.
_Source: syllabus Chemistry Part 2, Ch 8, §8.6 (Carboxylic acids — effect of substituents on $K_a$) + Table 8.4, p. 18–19._
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