Worked Problem — 4 Step-by-Step Solved Problems — Notes

Q: Rank in order of decreasing acidity: (A) Cl3CCOOH, (B) CH3CH2COOH, (C) HCOOH, (D) ClCH2COOH.

Step 1: Identify inductive effects.

(A) Cl3CCOOH: three Cl atoms, strong cumulative –I effect → best stabilization of COO–
(D) ClCH2COOH: one Cl, moderate –I effect → good stabilization
(C) HCOOH: H has no inductive effect → intermediate
(B) CH3CH2COOH: ethyl group has +I → destabilizes COO– → weakest

Step 2: Rank by conjugate base stability (most stable = strongest acid). (A) > (D) > (C) > (B)

Step 3: Verify with pKa: Cl3CCOOH (0.65) > ClCH2COOH (2.87) > HCOOH (3.75) > CH3CH2COOH (4.87). Correct.

Answer: (A) > (D) > (C) > (B)