The increasing order of pK$$_a$$ for the following phenols is
(A) 2,4-Dinitrophenol
(B) 4-Nitrophenol
(C) 2,4,5-Trimethylphenol
(D) Phenol
(E) 3-Chlorophenol
Choose the correct answer from the option given below:

We need to arrange the following phenols in increasing order of pK$$_a$$ (from most acidic to least acidic):

(A) 2,4-Dinitrophenol

(B) 4-Nitrophenol

(C) 2,4,5-Trimethylphenol

(D) Phenol

(E) 3-Chlorophenol

Lower pK$$_a$$ = stronger acid (more acidic). Electron-withdrawing groups (EWG) increase acidity (lower pK$$_a$$). Electron-donating groups (EDG) decrease acidity (higher pK$$_a$$).

(A) 2,4-Dinitrophenol: Two strong EWG ($$-\text{NO}_2$$) at ortho and para positions strongly stabilize the phenoxide ion. pK$$_a \approx 4.09$$ — most acidic.

(B) 4-Nitrophenol: One EWG at para position. pK$$_a \approx 7.15$$.

(E) 3-Chlorophenol: Chlorine is weakly electron-withdrawing at meta position. pK$$_a \approx 9.02$$.

(D) Phenol: No substituents. pK$$_a \approx 10.0$$.

(C) 2,4,5-Trimethylphenol: Three methyl groups (EDG) destabilize phenoxide ion. pK$$_a \approx 10.6$$ — least acidic.

$$ \text{(A)} < \text{(B)} < \text{(E)} < \text{(D)} < \text{(C)} $$

$$ (4.09) < (7.15) < (9.02) < (10.0) < (10.6) $$

The correct answer is Option B: (A), (B), (E), (D), (C).