Question

Correct order of acidic strength: (I) Phenol, (II) p-Cresol, (III) m-Nitrophenol, (IV) p-Nitrophenol

• A. \( \text{IV}>\text{III}>\text{I}>\text{II} \)
• B. \( \text{III}>\text{IV}>\text{I}>\text{II} \)
• C. \( \text{IV}>\text{I}>\text{III}>\text{II} \)
• D. \( \text{III}>\text{I}>\text{IV}>\text{II} \)

Hint:

Acidity depends on how well the phenoxide ion (after losing the O-H proton) can spread out its negative charge. Check whether each substituent can donate that charge into its own atoms through resonance, which only works from the ortho and para positions, never from meta.

Answer 1

The Correct Option is A

Step 1: Electron-withdrawing groups increase acidity by stabilizing the phenoxide ion, while electron-donating groups decrease acidity.
Step 2: The nitro group (\(-\mathrm{NO_2}\)) is a strong electron-withdrawing group. At the para position, it stabilizes the phenoxide ion by both \(-I\) and \(-R\) effects, making p-nitrophenol the most acidic.
Step 3: In m-nitrophenol, the nitro group shows only the \(-I\) effect (no resonance interaction), so it is less acidic than p-nitrophenol but more acidic than phenol.
Step 4: p-Cresol contains a methyl group (\(-\mathrm{CH_3}\)), which is electron-donating and destabilizes the phenoxide ion, making it the least acidic. \[ \text{Acidic strength: } \text{IV}>\text{III}>\text{I}>\text{II} \]

Answer 2

Concept:

• Acidity of a phenol is decided by how well the corresponding phenoxide ion (conjugate base) is stabilized once the O-H proton is removed.
• A substituent stabilizes the phenoxide ion by spreading the negative charge onto itself. This spreading can happen through resonance (only from the ortho and para positions) or through the inductive effect alone (works from any position, including meta).
• More resonance structures that place negative charge on an electronegative atom means greater stability, and therefore greater acidity.

Step 1: Draw the phenoxide ion formed from each compound and track how the negative charge is spread.
For phenol (I), the negative charge on oxygen is delocalized only into the ring carbons. This is the baseline case.

Step 2: Examine p-Nitrophenoxide (from IV).
The nitro group at the para position lets one resonance structure place the negative charge directly on an oxygen atom of the $-NO_2$ group.
Charge sitting on an electronegative oxygen atom is far more stable than charge sitting on carbon, so this is the strongest stabilization of the set.
Hence p-Nitrophenol (IV) is the most acidic.

Step 3: Examine m-Nitrophenoxide (from III).
From the meta position, no resonance structure can place the negative charge on the nitro oxygens, since the meta carbon is never one of the delocalized ring positions reached by phenoxide resonance.
The nitro group can still pull electron density only through the inductive ($-I$) effect, which is weaker than resonance stabilization.
So m-Nitrophenol (III) is more acidic than plain phenol but less acidic than p-Nitrophenol.

Step 4: Examine p-Cresoxide (from II).
The methyl group at the para position pushes electron density into the ring through hyperconjugation, adding extra negative character right where the charge is already sitting.
This destabilizes the phenoxide ion, making p-Cresol (II) the least acidic of the four.

Step 5: Rank all four by stability of their phenoxide ion.
Most stabilized to least stabilized: IV (resonance onto $-NO_2$) $>$ III (inductive only) $>$ I (no substituent effect) $>$ II (electron-donating methyl).

Final Answer: $\text{IV} > \text{III} > \text{I} > \text{II}$