See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Arranging Carboxylic Acids by pKa **Key Principle:** Electron-withdrawing groups (like Cl) stabilize the conjugate base ($COO^-$), making the acid stronger (lower pKa). The effect decreases with distance from the carboxyl group. ## Analysis of Each Compound: **(a) $CH_3-CH_2-COOH$** (Propionic acid) - No electron-withdrawing groups - Highest pKa ≈ 4.88 **(b) $CH_3-CH_2-CHCl-COOH$** (Cl at position 2, α-carbon) - Cl is **closest to $COOH$** (adjacent) - Maximum inductive effect → **strongest acid** - Lowest pKa ≈ 2.86 **(c) $Cl-CH_2-CH_2-CH_2-COOH$** (Cl at position 4, far from $COOH$) - Cl is **three bonds away** from carboxyl - Weak inductive effect → pKa ≈ 4.06 **(d) $CH_3-CHCl-CH_2-COOH$** (Cl at position 2, α-carbon) - Cl on α-carbon but one carbon removed from immediate vicinity - Intermediate inductive effect → pKa ≈ 3.98 ## Order of Increasing pKa: $$b < d < c < a$$ **Answer: (A)** ✓ The order reflects that electron-withdrawing Cl strengthens acidity most effectively when closest to the carboxyl group, with diminishing effect at greater distances.