Arrange the following in the increasing order of effective delocalization of negative charge in phenoxide ion:
(A) 2, 4, 6-trinitrophenol
(B) 3, 5-dinitrophenol
(C) 4-methylphenol
(D) 3-nitrophenol
The correct option is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (C) < (D) < (B) < (A)
In a phenoxide ion the negative charge on oxygen is delocalised into the benzene ring. Electron withdrawing groups such as –\( \ce{NO2} \) pull this charge further away and spread it over more atoms. They do so most strongly at the ortho and para positions, where they can take the charge by resonance (−R), and less at the meta position (−I only). Electron donating groups such as –\( \ce{CH3} \) reduce delocalisation.
(C) 4-methylphenoxide: –\( \ce{CH3} \) donates electrons, first (least delocalisation)
(D) 3-nitrophenoxide: one –\( \ce{NO2} \) at meta (−I only): second
(B) 3,5-dinitrophenoxide: two –\( \ce{NO2} \) groups at meta: third
(A) 2,4,6-trinitrophenoxide: three –\( \ce{NO2} \) groups at ortho and para (−R and −I), last (most delocalisation)
The same order is the order of increasing acid strength of the phenols (picric acid is the strongest).
Shortcut: 4-Methylphenol (C) is the lowest and picric acid (A) the highest; only Option (1) starts with C and ends with A.