The correct order of decreasing \( pK_a \) values of the following compounds
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B), (C), (D), (A)
A phenol is acidic because its conjugate base, the phenoxide ion, is stabilised by resonance. A lower pKa means a stronger acid, so decreasing pKa means increasing acid strength. An electron-withdrawing group stabilises the phenoxide ion and raises the acidity; an electron-donating group destabilises it and lowers the acidity. The nitro group withdraws by −I from every position, but by the stronger −R effect only from ortho and para.
(B) 4-Methoxyphenol: –\( \ce{OCH3} \) at the para position pushes electrons into the ring by resonance (+R), which destabilises the phenoxide ion. Weaker than phenol: highest pKa (about 10.2).
(C) Phenol: no substituent (pKa 10.0).
(D) 3-Nitrophenol: the meta nitro group acts by −I only (pKa 8.3).
(A) 4-Nitrophenol: the para nitro group withdraws by both −R and −I, delocalising the negative charge onto its oxygen atoms. The strongest acid: lowest pKa (7.1).
In decreasing order of pKa:
\( \displaystyle \text{4-methoxyphenol} \gt \text{phenol} \gt \text{3-nitrophenol} \gt \text{4-nitrophenol} \)
Shortcut: 4-Methoxyphenol has the highest pKa and 4-nitrophenol the lowest. Only Option (1) starts with (B) and ends with (A).